Artin approximation for polynomial equations and its desingularization proof

Written by GPT-6.1 Sol (OpenAI), Codex, Ultra. Self-checked by the writing AI. CC0 1.0.

A formal solution satisfies polynomial equations in a completion. Approximation asks for an exact solution on an étale neighbourhood with the same prescribed finite-order values. To keep both requirements, we encode the equations and the congruences in a single finite presentation.

The proof proceeds from smooth factorization to a pointed étale neighbourhood. Sections 1–3 give that passage and the finite coefficient system for an embedded torus. Section 4 proves the desingularization theorem that supplies the smooth factorization; Section 5 contains the algebra used in its proof.

1. The exact approximation statement

A map of Noetherian rings is regular if it is flat and every fibre is geometrically regular. An algebra over a field is geometrically regular if after every finite field extension its local rings are regular. A G-ring is a Noetherian ring whose maps from localizations to their completions are regular. The relevant implication for a local G-ring is therefore immediate from its definition: \(R\to\widehat R\) is regular. The broader local criterion and permanence statements, including the property for local rings essentially of finite type over \(\mathbb Z\), are supplied among the foundation arguments below.

Theorem 1.1 (Artin approximation in a pointed étale neighbourhood). Let \((R,\mathfrak m)\) be a Noetherian local G-ring. Let

For every positive integer \(N\), there exist a finitely presented étale \(R\)-algebra \(R'\), a maximal ideal \(\mathfrak m'\subset R'\) over \(\mathfrak m\), and a solution \(b\in(R')^n\), such that

Here \(\rho:R'\to\widehat R\) is the canonical lift of the specified residue-field point. This is a homomorphism; no assertion that the whole étale algebra embeds into the completion is required. The statement allows arbitrary characteristic, imperfect residue fields, nilpotent base elements, and nonregular local rings.

Lemma 1.2 (the canonical comparison map). A finitely presented étale \(R\)-algebra and a specified \(R\)-map \(R'\to R/\mathfrak m\) have a unique compatible map \(R'\to\widehat R\).

Proof. Starting with the specified map to \(R/\mathfrak m\), lift successively to \(R/\mathfrak m^r\). The kernel of \(R/\mathfrak m^{r+1}\to R/\mathfrak m^r\) is \(\mathfrak m^r/\mathfrak m^{r+1}\). Its square is zero because \(2r\geq r+1\). Formal etaleness gives a unique lift of the already chosen map at each step. Denote these compatible \(R\)-maps by \(\phi_r:R'\to R/\mathfrak m^r\). By definition \(\widehat R=\varprojlim_r R/\mathfrak m^r\): it is the subring of the product formed by compatible tuples, with addition and multiplication componentwise. Therefore \(x\mapsto(\phi_r(x))_r\) is an \(R\)-algebra map \(R'\to\widehat R\), inducing the specified residue map. Conversely any such map, composed with the coordinate projections \(\widehat R\to R/\mathfrak m^r\), gives these same \(\phi_r\) by induction and formal uniqueness. Its value on every \(x\) is consequently the same tuple. This proves existence and uniqueness using the inverse-limit definition alone, with no Noetherian or completion-quotient theorem needed for this lemma. Formal etaleness is part of the defining lifting property of an etale algebra in AG-CA-17; its nilpotent lifting criterion is Theorem 1.2 of that earlier lesson. ∎

Lemma 1.3 (a smooth residue-field point has an étale lift). If \(B\) is a smooth finitely presented \(R\)-algebra and \(B\to\kappa=R/\mathfrak m\) is an \(R\)-map, there are a finitely presented étale \(R\)-algebra \(R'\), a point \(R'\to\kappa\), and an \(R\)-map \(B\to R'\) inducing the original point.

Proof. Use the local standard form proved in AG-CA, Formally smooth, unramified and étale ring maps, Theorem 5.1. At the kernel of the given point, localize once to write

This presentation includes any localization variable and its equation. Choose \(r_{c+1},\ldots,r_s\in R\) lifting the point's coordinates in the remaining variables. Substitute these elements in the equations, keeping the first \(c\) variables, and invert the substituted \(\Delta\) and the original denominators. Call the resulting finitely presented algebra \(R'\). Here is its entire lifting test. Given a map \(R'\to D/J\) with \(J^2=0\), lift its variable values to a vector \(x\in D^c\). The inverted polynomials stay units: if \(uv=1+j\) with \(j\in J\), then \(u\) has inverse \(v(1-j)\). The equation errors form a vector \(h(x)\in J^c\), and the Jacobian \(H(x)\) is invertible by its unit determinant and the cofactor identity. Taylor expansion of each monomial, dropping every product of two elements of \(J\), gives \(h(x+z)=h(x)+H(x)z\) for \(z\in J^c\). Thus \(z=-H(x)^{-1}h(x)\) gives a solution. If \(y\) is any second lifted solution, the same identity gives \(H(x)(y-x)=0\), so \(y=x\). The inverse variables are then forced by the unit equations. For \(J^N=0\), repeat this unique lifting through \(D/J^{r+1}\to D/J^r\), whose kernel is square-zero for \(r\geq1\). This proves formal etaleness, including \(c=0\); finite presentation proves etaleness by the earlier programme definition. This is the square-Jacobian specialization of AG-CA-17, Theorem 4.1, with uniqueness proved here as well. The original residue coordinates define \(R'\to\kappa\), because all inverted elements have nonzero residues. Its kernel is maximal with residue field \(\kappa\), since \(R\to\kappa\) is surjective. Substitution gives \(B_g\to R'\), hence \(B\to R'\), inducing exactly the original point. ∎

Proof of Theorem 1.1. The equality $R/\mathfrak m^N\simeq\widehat R/\mathfrak m^N\widehat R$ is proved in the earlier AG-CA lesson Completion, Proposition 2.2. Choose \(c_i\in R\) representing \(\widehat a_i\) modulo \(\mathfrak m^N\widehat R\), and choose finitely many generators \(d_1,\ldots,d_M\) of \(\mathfrak m^N\). Write

Put

The tuple \(\widehat u\) gives an \(R\)-map \(A\to\widehat R\). Since \(R\to\widehat R\) is regular, the desingularization theorem proved by the sequence ending at Theorem D.38, and the elementary factorization criterion proved at the corresponding foundation lemma, give a factorization

with \(B\) smooth and finitely presented over \(R\). Compose with the residue map \(\widehat R\to\kappa\), then apply Lemma 1.3. The images \(u_{il}\in R'\) solve the \(g_j\) exactly. Set \(b_i=c_i+\sum_l d_lu_{il}\). Substitution in (1.2) gives \(f_j(b)=0\) exactly. Under the canonical map of Lemma 1.2,

This proves the prescribed congruence without demanding that the auxiliary \(u_{il}\) themselves be close to \(\widehat u_{il}\). The exact written foundations and the earlier programme proofs used in this argument are registered in §6. ∎

Corollary 1.4 (henselian formulation). If \(R\) is henselian, the solution can be chosen in \(R\). In general it gives a solution in the henselization \(R^h\), descending to the pointed étale neighbourhood just constructed.

Proof. The section criterion proved in AG-CA, Henselian local rings and henselization, Theorem 1.2 gives \(R'\to R\) with the specified residue map. Its composition with \(R\to\widehat R\) is the map \(\rho\) by Lemma 1.2, so (1.1) is preserved. For \(R^h\), use the pointed étale colimit and its universal property proved in that lesson, Lemma 4.1 and Theorem 4.2. Compatibility with the map into the complete henselian ring follows from the same uniqueness. Here the principal monic chart consumed by that section-criterion proof is supplied by the full earlier structural étale proof, Lemma D2.2, whose integral closure input is the affine conductor theorem, Theorem C4.3. Its polynomial normality input is proved in the preliminary coefficient argument, Corollary B1.2. To spell out the actual section step, localize the selected étale algebra to $(R[T]/(F))_h$, with $F$ monic and $F'$ nonzero at the given residue root $\bar a$. The henselian simple-root property gives $a\in R$ with $F(a)=0$ and $a\bmod\mathfrak m=\bar a$. Every denominator has nonzero residue at $\bar a$, hence is a unit at $a$, and evaluation defines the section. The resulting map $R'\to R$ induces the prescribed residue map; formal étale uniqueness along the quotients $R/\mathfrak m^n$ proves its composition with completion is $\rho$. Thus this corollary uses these written earlier proofs rather than an external conductor proof tag. ∎

2. Returning from a local model to the original base

Lemma 2.1 (spreading finite presentation data). Let \(R_0\) be Noetherian and \(\mathfrak p\subset R_0\). A pointed étale algebra, a finite tuple, and finitely many equations over \((R_0)_{\mathfrak p}\), as in Theorem 1.1, spread to \((R_0)_s\) for some \(s\notin\mathfrak p\). Their solution therefore gives a pointed étale neighbourhood over \(R_0\).

Proof. In the construction of Lemma 1.3, every polynomial coefficient, every fixed nonpivot coordinate, and every chosen inverse has a denominator outside \(\mathfrak p\). Invert their product in \(R_0\). The finitely many equations for the tuple and the finitely many witnesses verifying them likewise hold after multiplying by finitely many additional denominators; invert their product too. The standard étale presentation spreads with its invertible Jacobian determinant, so its spread algebra is étale by the same direct Newton calculation. Regard it as an \(R_0\)-algebra through the étale localization \(R_0\to(R_0)_s\); composing two formally étale, finitely presented maps remains étale because a square-zero lifting problem is solved uniquely in two successive steps. The chosen residue map survives and gives a point over \(\mathfrak p\) with unchanged residue field. The defining equations remain exact, and localization at the selected point gives the original local comparison. ∎

This proof uses the specific standard étale algebra produced above and therefore does not invoke an external spreading theorem for an arbitrary presentation. Pulling the resulting neighbourhood back to the original scheme preserves every defining equation and the specified fibre point.

3. The finite system for an embedded torus

The approximation step in the torus lesson begins after the formal deformation and effectivity argument has produced an affine torus \(\widehat T\) over the completed local model and a closed homomorphism \(\widehat T\hookrightarrow G_{\widehat R}\). The present lesson proves the approximation input; it does not claim to supply that preceding effectivity argument.

Lemma 3.A (finite splitting over a complete local ring). Let \((A,\mathfrak m,\kappa)\) be a local ring with \[ A\simeq\varprojlim_{n\geq1}A/\mathfrak m^n. \] No Noetherian, reducedness, normality or perfection hypothesis is imposed. Suppose a torus \(T/A\) has an affine etale splitting chart \(U\to\operatorname{Spec}A\) and a point of its closed fibre with finite separable residue field \(K/\kappa\). Then a finite free, faithfully flat, etale algebra \(E/A\) splits \(T\). We may choose \(E\) local and complete, with residue field a finite Galois extension \(L/\kappa\) containing \(K\), and with an action of \(\Gamma=\operatorname{Gal}(L/\kappa)\) for which \[ E\otimes_AE\xrightarrow{\sim}\prod_{\sigma\in\Gamma}E, \qquad x\otimes y\longmapsto\bigl(x\sigma(y)\bigr)_\sigma. \tag{3.2} \] Here a torus is etale locally a product of copies of \(\mathbf G_m\). Starting instead with the fpqc multiplicative-type definition gives this same condition by AG-GS-05, Theorem 7.7; its finite torsion-free character-group step is proved in AG-RG-S04, Lemma P0.10. Every torus therefore has such a chart and residue point: choose an affine finite-presentation part of an etale splitting cover meeting the closed fibre, and a closed point of that fibre. Its residue field is finite over \(\kappa\) by the finite-type field-algebra argument in AG-RG-S04, Lemma P0.9. The quotient sequence for differentials gives \(\Omega_{K/\kappa}=0\), because the chart is formally unramified. The finite-field differential calculation in AG-CA-17, Lemma 7.1 then proves separability.

Furthermore, a torus \(T_0/\kappa\), with character lattice split by \(L\), has a torus lift over \(A\) split by this same \(E\). Its character lattice and descent action reduce compatibly modulo every \(\mathfrak m^n\).

Proof. A finite separable extension embeds in a finite normal separable extension: adjoin all the roots of the minimal polynomials of a finite generating list. The finite splitting-field construction and separable-tower calculations of AG-RG-S02, Lemma 4.A, prove finiteness and separability. Every embedding permutes the root sets, so the resulting extension \(L/\kappa\) is normal. The same embedding count gives \(|\Gamma|=[L:\kappa]\). We give the whole primitive-element step here. Write \(n=[L:\kappa]\), choose a finite generating list \(b_1,\ldots,b_s\), and list the \(n\) embeddings \(\sigma_i\) into a finite splitting field. If \(\kappa\) is infinite, every pair of distinct embeddings differs on some generator, so \(\sum_j(\sigma_i(b_j)-\sigma_{i'}(b_j))X_j\) is a nonzero polynomial. Their finite product is nonzero, because a polynomial ring over a field is a domain, as follows by multiplying leading coefficients successively in each variable. It cannot vanish on every tuple in \(\kappa^s\): induction on the variables first makes a nonzero coefficient polynomial nonzero and then avoids the finitely many roots in the last variable, using the root bound of Lemma 4.A. Choose such a tuple \((c_j)\). Then \(\alpha=\sum_j c_jb_j\) has \(n\) different embedding images. Its minimal polynomial has at least \(n\) roots, hence degree at least \(n\); the tower-degree formula bounds that degree by \(n\), so \(L=\kappa(\alpha)\). If \(\kappa\) is finite, a finite basis makes \(L\) a finite set. The complete cyclic-subgroup calculation of Lemma 4.A makes \(L^\times\) cyclic. A generator \(\alpha\) generates \(L\) as a field because every nonzero element is its power. Degree one is included by choosing \(\alpha=1\). Let \(\bar f\in\kappa[X]\) be the monic minimal polynomial of this \(\alpha\), and lift its coefficients to a monic \(f\in A[X]\). Put \[ E=A[t]=A[X]/(f),\qquad d=\deg(f). \] Successively subtracting leading multiples of the monic polynomial \(f\) reduces every polynomial to degree less than \(d\). A nonzero multiple \(qf\) has degree \(\deg(q)+d\), since its leading coefficient is the nonzero leading coefficient of \(q\). Thus the remainder is unique, and \(1,t,\ldots,t^{d-1}\) is a basis making \(E\) free of positive finite rank over \(A\); it is faithfully flat because tensoring a module with this free module gives a direct sum of \(d\) copies. Its quotient by \(\mathfrak mE\) is \(L\). If \(e\in E\) has nonzero residue, multiplication by \(e\) on this basis is invertible modulo \(\mathfrak m\). Its determinant is a unit of \(A\), and the cofactor calculation proved in Lemma 3.B gives an inverse to multiplication by \(e\). Thus \(e\) is a unit, \(E\) is local with maximal ideal \(\mathfrak mE\), and coordinatewise completeness gives \[ E\simeq\varprojlim_n E/\mathfrak m^nE. \] Separability makes \(f'(t)\) a unit by the same argument. A root in a quotient by a square-zero ideal lifts uniquely by replacing any lift \(a\) with \(a-f'(a)^{-1}f(a)\). Polynomial expansion proves existence, and \(f(b)-f(a)=(b-a)(f'(a)+(b-a)H)\) proves uniqueness. Iteration gives the corresponding assertion for every nilpotent ideal. This proves formal etaleness of \(E/A\); its finite presentation makes it etale.

The residue point of \(U\) gives an algebra map \(\mathcal O(U)\to K\to L\). Formal etaleness of \(U/A\) lifts this map uniquely through \(E/\mathfrak m^{n+1}E\to E/\mathfrak m^nE\), whose kernel is square-zero for \(n\geq1\). Taking the compatible limit gives a map \(\mathcal O(U)\to E\), hence a section of \(U_E\to\operatorname{Spec}E\). Pulling back the splitting on \(U\) therefore splits \(T_E\).

We now construct (3.2), including its automorphisms. For every \(\sigma\in\Gamma\), the residue \(\sigma(\alpha)\in L\) is a simple root of \(\bar f\). In the complete local ring \(E\), Newton's recursion \[ a_{j+1}=a_j-f'(a_j)^{-1}f(a_j) \] starting from any lift of this residue gives \(f(a_j)\in(\mathfrak mE)^{2^j}\) and \(a_{j+1}-a_j\in(\mathfrak mE)^{2^j}\). Completeness and separatedness give a root \(a_\sigma\), and the difference formula above gives its uniqueness with that residue. Distinct \(a_\sigma\) have unit differences. If a polynomial \(P\) vanishes at all these roots, monic division by \(X-a_1\) gives \(P=(X-a_1)Q\), since the remainder is \(P(a_1)=0\). At each other root \(a_\sigma\), the factor \(a_\sigma-a_1\) is a unit, so \(Q(a_\sigma)=0\). Repeating the division shows that \(\prod_\sigma(X-a_\sigma)\) divides \(P\). If \(\deg(P)<d\), monicity forces \(P=0\). The difference between \(f\) and this product has degree less than \(d\) and vanishes at every root, so \[ f(X)=\prod_{\sigma\in\Gamma}(X-a_\sigma). \] The map \(t\mapsto a_\sigma\) defines an \(A\)-algebra endomorphism \(g_\sigma\) of \(E\). The two roots \(g_\sigma g_\tau(t)\) and \(a_{\sigma\tau}\) have the same residue, so uniqueness gives \(g_\sigma g_\tau=g_{\sigma\tau}\). Also \(a_1=t\); hence these maps are automorphisms and lift the Galois action on \(L\).

The isomorphism \(E\otimes_AE=E[X]/(f)\) and evaluation at the \(a_\sigma\) now give (3.2). Explicitly, the polynomials \[ \ell_\sigma(X)=\prod_{\tau\ne\sigma} \frac{X-a_\tau}{a_\sigma-a_\tau} \] give the inverse to evaluation: a tuple \((b_\sigma)_\sigma\) is represented by \(\sum_\sigma b_\sigma\ell_\sigma(X)\). Thus the isomorphism uses no division by \(|\Gamma|\). Its compatibility on triple products is exactly \(g_\sigma g_\tau=g_{\sigma\tau}\). This is the required finite etale Galois torsor.

For the final assertion use the specified finite Galois field \(L\) splitting \(T_0\), so that the algebra \(E\) just constructed is the same one as in the statement. Its character group \(M\) is finite free by AG-RG-S04, Lemma P0.10, and the split character calculation of AG-GS-05, Theorem 2.2. The descent of \(T_0\) along \(L/\kappa\) gives its \(\Gamma\)-action on \(M\): pull back a character along each descent isomorphism; the cocycle gives the group law. Theorem 2.2 identifies these maps with the corresponding lattice maps. On \(E[M]\) put \[ \sigma(ae^m)=g_\sigma(a)e^{\sigma m}. \tag{3.3} \] The group law and (3.2) make this a descent datum. Multiplication and every Hopf operation commute with it, since addition in \(M\) commutes with its action. The actual module and algebra effectivity proof in AG-RG-S04, Lemmas A1.1--A1.2, gives the invariant Hopf algebra \[ C=E[M]^\Gamma,\qquad E\otimes_AC\simeq E[M]. \tag{3.4} \] Tensor products and Hopf identities descend by this isomorphism and faithful flatness. Finite presentation descends here by a finite-generator argument. Express the finitely many Laurent algebra generators over \(E\) as finite \(E\)-linear combinations of elements of \(C\). Those elements generate \(C\), since the cokernel of their generated subalgebra becomes zero over \(E\). For a resulting finite polynomial surjection \(A[Z]\to C\), its kernel becomes a finite ideal after tensoring with \(E\), because \(E[M]\) is finitely presented. To see that finiteness for this particular surjection, compare it with a finite presentation of \(E[M]\): express each presentation's generators in the other, and substitute these expressions in the finitely many relations and composite-generator identities. They generate the new kernel. Write a finite list of its generators as finite sums of elements of the original kernel with coefficients in \(E\). Their finitely many kernel components generate the original kernel by faithful flatness. Thus \(C\) is finitely presented. It defines a torus \(T_A\), because (3.4) splits it on the finite etale faithfully flat cover.

Put \(A_n=A/\mathfrak m^n\) and \(E_n=E\otimes_AA_n\). Tensoring (3.2)--(3.4) with \(A_n\) gives the same torsor and the same character action over \(E_n\): \[ E_n\otimes_{A_n}(C\otimes_AA_n)\simeq E_n[M]. \tag{3.5} \] The algebra on the left descends the datum (3.3) modulo \(\mathfrak m^n\); uniqueness in Lemma A1.2 proves that it is its effective descent. This proves compatibility for every \(n\), even though \(A\to A_n\) need not be flat. It does not assume that taking invariants commutes with an arbitrary quotient. At \(n=1\), the descent datum is precisely the given one on \(L[M]\), so its descended torus is \(T_0\).

More generally, any compatible character data split by these \(E_n\) and identified with the same lattice have no additional nilpotent transition parameters. Comparing coefficients in \(\Delta(\sum c_me^m)=(\sum c_me^m)\otimes(\sum c_me^m)\) gives \(c_m^2=c_m\), \(c_mc_{m'}=0\) for \(m\ne m'\), and \(\sum c_m=1\). Each \(E_n\) is local, so exactly one coefficient is \(1\). A character is therefore one monomial; a torus automorphism is one integral lattice automorphism. A compatible automorphism reducing to a fixed one is that same one. Hence a compatible lattice descent action is the fixed action in (3.3), and (3.5) gives its actual compatible descent. This completes both assertions. \(\square\)

Choose a finite presentation of the Hopf algebra of \(\widehat T\). Lemma 3.A supplies a finite free, etale, faithfully flat splitting algebra \(E/\widehat R\); choose the resulting splitting isomorphism

Lemma 3.B (perfect trace, over an arbitrary base). Let $A$ be a commutative ring and $C$ a commutative $A$-algebra free of finite rank $d$. If the determinant of the pairing $(c,z)\mapsto\operatorname{Tr}_{C/A}(cz)$ is a unit, then $C/A$ is finitely presented and formally etale, hence etale. If $d>0$, it is faithfully flat. These assertions require no Noetherian, reducedness or characteristic hypothesis.

Proof. First, the elementary matrix identities used here hold over every commutative ring. Define a determinant by the signed permutation sum and a cofactor by the corresponding smaller signed determinant. Expanding along a row or column gives the determinant. A matrix with two equal rows or columns has determinant zero: pair each permutation with the permutation interchanging those positions; their contributions have opposite signs, including in characteristic two. Consequently, if $M$ is square, multiplying it on either side by its transposed cofactor matrix gives $(\det M)I$: a diagonal entry is the row or column expansion, and an off-diagonal entry is the expansion of a determinant with two equal rows or columns. A unit determinant therefore gives the inverse $(\det M)^{-1}\operatorname{adj}(M)$. For size zero the determinant is one. To verify the converse implication as well, expand $\det(UV)$ by multilinearity in its columns. The expansion chooses, for each column of $UV$, one column of $U$ with its coefficient from $V$. A tuple repeating a column of $U$ contributes zero by the cancellation just proved. Every remaining tuple is a permutation of all the columns of $U$, whose determinant is its permutation sign times $\det(U)$. Summing its coefficients with these signs is precisely $\det(V)$. Thus $\det(UV)=\det(U)\det(V)$, including for size zero. If $UV=I$, this equation gives $\det(U)\det(V)=1$, so an invertible matrix has a unit determinant. For square matrices $U,V$, the finite double sum also gives $\operatorname{tr}(UV)=\operatorname{tr}(VU)$. Thus conjugating a multiplication matrix by an invertible change of basis preserves its trace. Multiplication matrices, their traces and their determinants commute with every scalar extension, entry by entry.

Suppose first that $d>0$, and choose a basis $u_1,\ldots,u_d$. Write $u_i u_j=\sum_k m_{ijk}u_k$ and $1=\sum_i a_i u_i$. There is an explicit finite algebra presentation $$ C=A[X_1,\ldots,X_d]/\bigl(X_iX_j-\sum_k m_{ijk}X_k,\;1-\sum_i a_iX_i\bigr). $$ Indeed, repeatedly replace a product of two variables by the displayed linear combination, reducing each monomial to degree at most one. Replace its constant term by that constant times $\sum_i a_iX_i$. Every quotient element is therefore represented by $\sum_i b_iX_i$. The map to $C$ is surjective, and if this representative maps to zero, linear independence of the $u_i$ makes every $b_i$ zero. This proves the presentation without an assumption on ideals of $A$.

The inverse pairing matrix supplies a trace-dual basis $v_1,\ldots,v_d$, characterized by $\operatorname{Tr}(u_jv_i)=\delta_{ij}$. In particular $x=\sum_i u_i\operatorname{Tr}(v_ix)$ and $x=\sum_i\operatorname{Tr}(u_ix)v_i$. To check the second equality, pair both sides with each $u_j$ and use invertibility of the pairing; the first is the ordinary basis-coordinate equality. Put $$ e=\sum_i u_i\otimes v_i\in C\otimes_A C. $$ The $A$-linear map $$ \Phi:C\otimes_A C\longrightarrow\operatorname{End}_A(C),\qquad \Phi(a\otimes b)(x)=a\operatorname{Tr}(bx) $$ is an isomorphism, with inverse $F\mapsto\sum_i F(u_i)\otimes v_i$. The first coordinate equality proves $\Phi\Phi^{-1}(F)=F$; the second proves $\Phi^{-1}\Phi(a\otimes b)=a\otimes b$, and pure tensors generate. Since $\Phi(e)$ is the identity, both $(c\otimes1)e$ and $(1\otimes c)e$ have image multiplication by $c$. Thus $$ (c\otimes1)e=(1\otimes c)e\qquad(c\in C). \tag{3.6} $$ Let $s=\sum_i u_iv_i$. The diagonal coefficient of multiplication by $c$ in position $i$ is $\operatorname{Tr}(v_i c u_i)$. Summing those coefficients gives $\operatorname{Tr}(c)=\operatorname{Tr}(cs)$ for every $c$. Invertibility of the pairing now gives $$ \sum_i u_iv_i=1. \tag{3.7} $$ No division by $d$, or assertion that $d$ is a unit, occurs.

We prove the lifting property directly. Let $J\subset D$ be a square-zero ideal of a commutative $A$-algebra, and let $\bar f:C\to D/J$ be an $A$-algebra map. The free-module lifting calculation in AG-RG-S01, Lemma 1.0 gives an $A$-linear lift $l:C\to D$. Its value $l(1)$ is $1+j$ with $j\in J$, and has inverse $1-j$. Multiplying the whole lift by that inverse makes $l(1)=1$ while preserving $A$-linearity and its reduction. Give $J$ its $C$-module action via $\bar f$: multiply by any lift of $\bar f(c)$, which is independent of that lift because $J^2=0$.

The defect $b(c,z)=l(cz)-l(c)l(z)$ lies in $J$, is $A$-bilinear and has $b(c,1)=b(1,c)=0$. Associativity in $C$ and $D$ gives, by expanding the four terms, $$ c\,b(z,t)-b(cz,t)+b(c,zt)-b(c,z)\,t=0. \tag{3.8} $$ Define the $A$-linear correction $h(c)=\sum_i u_i b(v_i,c)$ in $J$. Apply the $A$-linear map $a\otimes z\mapsto a b(z,t)$ to (3.6); it gives $c h(t)=\sum_i u_i b(cv_i,t)$. Using (3.8) with $(v_i,c,t)$ and then (3.7), we obtain $$ c h(t)-h(ct)+h(c)t =\sum_i u_i\bigl(b(v_ic,t)-b(v_i,ct)+b(v_i,c)t\bigr) =\sum_i u_iv_i b(c,t)=b(c,t). $$ Consequently $f=l+h$ is multiplicative: the defect of this new map is $b(c,t)+h(ct)-c h(t)-h(c)t=0$, since $h(c)h(t)=0$. Also $h(1)=0$. Thus $f$ is a unital $A$-algebra lift of $\bar f$.

For uniqueness, the difference $\delta$ of two such lifts takes values in $J$ and satisfies $\delta(ct)=c\delta(t)+\delta(c)t$; expand their products and use $J^2=0$. Apply $a\otimes z\mapsto\delta(a)z$ to (3.6). The difference between its two sides is $\delta(c)\sum_i u_iv_i=\delta(c)$, by that identity and (3.7). It is zero, so every lift is unique. If $J^N=0$, lift successively through $D/J^{r+1}\to D/J^r$, for $1\le r<N$. Its kernel is square-zero because $2r\ge r+1$. Existence and uniqueness at each step prove formal etaleness for every nilpotent ideal. Together with the explicit finite presentation, this is etaleness by the programme definition.

For $d=0$, $C$ is the zero algebra with presentation $A/(1)$. If a unital map $C\to D/J$ exists and $J$ is nilpotent, then $D/J=0$, so $1\in J$ and nilpotence forces $D=0$. There is exactly one lift. Finally, for positive $d$, tensoring an $A$-module with $C$ gives $d$ copies of that module; this is exact and detects whether it is zero. This proves faithful flatness. $\square$

This is the finite-free case of the discriminant criterion in the freely accessible Stacks Project, Lemma 49.3.1, Tag 0BJF. The complete lifting proof above supplies the result used here.

Return to the splitting algebra $E/\widehat R$ from Lemma 3.A. Its trace determinant is a unit, as we now check rather than adding an unsupported inverse variable. By (3.2), after scalar extension from $\widehat R$ to $E$ the algebra becomes $\prod_{\sigma\in\Gamma}E$. Multiplication in this product is diagonal in the coordinate idempotent basis. Its trace pairing matrix is the identity, hence invertible. The preceding trace and basis-change calculations imply that the original pairing matrix becomes invertible over $E$, so its determinant $\Delta$ becomes a unit there. Then $(\widehat R/\Delta\widehat R)\otimes_{\widehat R}E=0$. Faithful flatness from Lemma 3.A detects zero modules, giving $\widehat R/\Delta\widehat R=0$, which means that $\Delta$ is already a unit of $\widehat R$.

Fix the finite free basis of $E$ supplied by Lemma 3.A. Its multiplication table, unit and multiplication matrices define finitely many coefficient variables; associativity, commutativity and the unit laws are finite polynomial equations. Add a variable $w$ and the equation $w\Delta=1$ for the trace determinant. The original completed coefficient tuple satisfies this equation by the preceding argument. After approximation it still satisfies the same equation exactly. Lemma 3.B makes the resulting finite free algebra etale and, since its unchanged rank is positive, faithfully flat.

Add coefficients for the finite presentations, Hopf maps, the isomorphism (3.1), its inverse, and the homomorphism to \(G\). Laurent expressions can be written polynomially by adding inverse variables and relations \(Z_iW_i=1\). The maps need to be specified on finite generating sets. Relations between their images need finitely many witnesses in the finitely generated relation ideals. The two composites of the splitting isomorphism and its inverse are tested on finite generators; equality of these composites with the identity is recorded by their witnesses. The same method records comultiplication, counit, inverse, and the homomorphism identities on a finite list of generators. Coassociativity and every needed descent identity are likewise checked on finite generators, using the finite multiplication tables on the finite overlap algebras. Each expression and witness has finite support.

Retain the closed immersion in this same system. The pullback \(\mathcal O(G_{\widehat R})\to\mathcal O(\widehat T)\) is surjective. Choose preimages of a finite list of algebra generators of \(\mathcal O(\widehat T)\), and include the equations asserting that their images are those generators. After approximation these equations still prove surjectivity; consequently the resulting homomorphism is a closed immersion.

We have obtained one finite polynomial system over the local model, solved by the original completed coefficient tuple. Apply Theorem 1.1 at order one. The defining equations hold exactly after approximation, while all the chosen coefficients retain their original residues. The preserved splitting maps identify the source Hopf algebra with a Laurent polynomial Hopf algebra on the finite étale faithfully flat cover. Thus it is a torus, with the specified special fibre and closed embedding. If a prescribed higher jet is needed, use that order instead. Lemma 2.1 spreads the data and the neighbourhood back to an open part of the finite-type model; base change returns them to the original base.

The local model is essentially of finite type over \(\mathbb Z\). The G-ring permanence argument in §5 is therefore a consumed part of this application, including its coefficient-ring inputs. It must be proved before the approximation step is declared complete. An excellence label or an external Stacks link does not replace this work.

Scope of the supporting algebra. The approximation theorem and the desingularization theorem concern Noetherian rings. Every base ring of the finite-type algebra diagrams in their proof is Noetherian. In the foundational complete-intersection arguments below we state and prove the exact Noetherian forms consumed there; arguments explicitly stated over arbitrary rings retain that wider scope. Polynomial presentations with arbitrarily many variables occur only in the two-term cotangent comparison, whose elementary presentation and filtered-colimit proofs are written without Noetherian assumptions. This distinction preserves arbitrary characteristic, imperfect residue fields, and the full Noetherian local G-ring statement.

4. Desingularization proof sequence

We prove approximation by passing through presentation changes, infinitesimal lifts and improvements of singular ideals. Local resolutions are constructed separately for separable and inseparable residue fields, then assembled into the smooth-colimit theorem. Smoothness throughout means the conormal condition F.6. Each displayed list of arrows specifies the maps in the relevant commuting diagram.

Singular ideals

The singular ideal records the complement of the smooth locus. Its radical formulation lets us compare smooth neighbourhoods across different finite presentations.

Definition D.1: singular ideal

Reading source: Stacks, Tag 07C5.

For any ring map $R\to A$, let $U_{A/R}$ be its smooth locus in $\operatorname{Spec}A$, with smoothness at a prime understood as in F.7. Its singular ideal $H_{A/R}$ is the radical ideal characterized by $$V(H_{A/R})=\operatorname{Spec}A\setminus U_{A/R}.$$ No finite-presentation hypothesis on $A/R$, or quasi-compactness hypothesis on $U_{A/R}$, is part of this definition.

Verification. The definition of smoothness at a prime writes $$U_{A/R}=\bigcup_{\substack{g\in A\\ R\to A_g\ \mathrm{smooth}}}D(g).$$ Indeed a witnessing $g$ gives smoothness at every prime avoiding it, and every smooth prime has such a witness. Thus $U_{A/R}$ is open. Write its closed complement as $V(J)$ and set $H_{A/R}=\sqrt J$. A prime contains $J$ exactly when it contains $\sqrt J$, since a power belonging to a prime forces its base element into that prime. This proves existence.

Here is the radical-ideal uniqueness argument used below. For any ideal $J$ and $x\notin\sqrt J$, the ideals containing $J$ and avoiding $\{1,x,x^2,\ldots\}$ have a maximal member $L$, by the chain-union argument and Zorn's lemma. If $uv\in L$ while $u,v\notin L$, maximality provides a power of $x$ in each of $L+(u)$ and $L+(v)$. Their product is again a power of $x$ and belongs to $L$, a contradiction. Hence $L$ is a prime containing $J$ and avoiding $x$. It follows that $$\sqrt J=\bigcap_{\mathfrak q\in V(J)}\mathfrak q,$$ where the empty intersection is $A$. Two radical ideals with the same closed set are therefore equal. This also covers $A=0$ and an empty singular locus. The smooth open need not have a finite subcover, as the fully proved example D.6 shows. ∎

Lemma D.2: find strictly standard

Reading source: AI Integrated Stacks, pinned smoothing, lines 141–181.

Fix a finite presentation $$P=R[x_1,\ldots,x_n],\qquad I=(f_1,\ldots,f_m),\qquad A=P/I.$$ At a prime $\mathfrak q$ where $A/R$ is smooth, one can select equally many equations and variables, indexed by sets $V$ and $U$ of size $c\leq\min(n,m)$, and find $a,a'\in A$ such that $$a\notin\mathfrak q,\qquad a=a'\det(\partial f_j/\partial x_i)_{j\in V,\ i\in U},\qquad \widetilde a f_\ell\in(f_j:j\in V)+I^2\quad(1\leq\ell\leq m).$$ Here $\widetilde a$ is any polynomial lift of $a$. The last condition is equivalently the relation $a[f_\ell]\in\langle[f_j]:j\in V\rangle$ in $I/I^2$.

Proof. The task is to choose a nonvanishing minor and then kill a finite conormal quotient. Let $M=I/I^2$ and $K=\kappa(\mathfrak q)$. Smoothness at the prime supplies $s\notin\mathfrak q$ with $A_s/R$ smooth, by F.7. The full comparison F.11, localization F.65, and definition F.6 give $$0\longrightarrow M_s\xrightarrow{\mathrm d}A_s^n \longrightarrow\Omega_{A_s/R}\longrightarrow0.$$ This sequence splits because its quotient is projective. Consequently the differential on $M\otimes_A K$ is injective. The finite list $[f_j]$ spans that vector space; extract a basis from this list and call its index set $V$. If its length is $c$, injectivity implies $c\leq n$, while its construction implies $c\leq m$.

To justify a choice of $U$, use the following row calculation over $K$. When $c>0$, a nonzero entry in the first column supplies a pivot row. Subtract its multiples from the other rows. The lower block of the remaining $c-1$ columns is independent: a dependence there could be completed by a multiple of the first column to give a dependence of all $c$ columns. Induction selects $c-1$ lower rows with nonzero determinant. The determinant on these rows and the pivot row is the pivot times this nonzero determinant. Subtracting multiples of the pivot row did not change it, so it is a minor of the original columns. For $c=0$ use the determinant $1$. Thus the corresponding Jacobian minor $\delta$ belongs to $A\setminus\mathfrak q$.

Now form the finite $A$-module $$Q=M/\langle[f_j]:j\in V\rangle.$$ Its residue at $\mathfrak q$ vanishes, and F.10 gives $Q_{\mathfrak q}=0$. For each generator of $Q$ choose a localization witness outside $\mathfrak q$ annihilating it. Their finite product $t$ annihilates $Q$ and still lies outside $\mathfrak q$; use $t=1$ when $Q=0$. Put $a=t\delta$ and $a'=t$. Then $aQ=0$ gives every stated conormal relation. Lifting its coefficients to $P$ gives the displayed polynomial-ideal relation. A different lift of $a$ changes its product with $f_\ell$ by an element of $I^2$, so the relation does not depend on that lift. No cancellation of $a$, $\delta$ or another possible zero divisor occurs. The empty equation and variable cases are included by the same construction. ∎

The next definition distinguishes witnesses using a single Jacobian minor from witnesses using a sum of minors. D.15 compares these two conditions with smoothness after localization.

Definition D.3: strictly standard

Reading source: Stacks, Tag 07C7.

Let $R\to A$ be finitely presented. A witness for either standard-element condition consists of $$P=R[x_1,\ldots,x_n],\qquad I=(f_1,\ldots,f_m),\qquad A=P/I,\qquad 0\leq c\leq\min(n,m).$$ All Jacobian entries below are reduced to $A$. For $a\in A$, choose any lift $\widetilde a\in P$ when writing a polynomial-ideal condition.

The element $a$ is elementary standard if some such witness and $a'\in A$ satisfy both

and

The element $a$ is strictly standard if some witness and coefficients $a_U\in A$, indexed by the $c$-element subsets of $\{1,\ldots,n\}$, satisfy both

and

Use increasing order on each selected variable set in the determinants. Changing the lift $\widetilde a$ adds an element of $I$; multiplying this difference by $f_{c+j}\in I$ puts it in $I^2$. Thus both relation conditions are independent of the lift. Equivalently they assert $a[f_{c+j}]\in\langle[f_1],\ldots,[f_c]\rangle$ in the $A$-module $I/I^2$. The minor condition and this relation condition are separate requirements.

For $c=0$ there is one minor, the empty determinant $1$, and the first equation imposes no restriction on $a$. The second equation then reads $\widetilde a I\subset I^2$. For $m=c$ the second equation has an empty list of tests. These conventions include empty polynomial presentations and the zero algebra. ∎

Lemma D.4: parse equation strictly standard one

Reading source: Stacks, Tag 07ET.

In the presentation of D.3, the strict minor condition for $a$ gives an $A$-linear map $\psi:A^n\to A^c$ for which the composite $$A^c\xrightarrow{e_j\mapsto[f_j]}I/I^2\xrightarrow{\mathrm d}A^n\xrightarrow{\psi}A^c$$ is $a1_c$. Conversely the existence of this map gives the strict minor condition for $a^c$. These assertions concern the minor condition alone; the remaining-equation condition still has to be imposed independently.

Proof. Identify $A^n$ with the free differential module on $dx_1,\ldots,dx_n$. Its conormal map is well-defined: for $u,v\in I$, the product rule gives $d(uv)=u\,dv+v\,du$, which vanishes after reduction to $A$. The composite of its first two maps has the $n\times c$ matrix $$M_{ij}=\frac{\partial f_j}{\partial x_i}\bmod I.$$ For a selected variable set $U$, let $E_U:A^n\to A^c$ be coordinate selection and put $M_U=E_UM$. The determinant in D.3s1 is $\det M_U$: transposition preserves the determinant by reversing the permutation indexing in its signed expansion.

The arbitrary-ring adjugate and determinant identities are proved in F.70. If $a=\sum_U a_U\det M_U$, they give the explicit map with matrix $$B=\sum_U a_U\operatorname{adj}(M_U)E_U,\qquad BM=a1_c.$$ This is the required $\psi$. Conversely write $B$ for any proposed $\psi$. The same proved multilinear expansion gives $$a^c=\det(BM)=\sum_{|U|=c}\det(B^U)\det(M_U),$$ where $B^U$ selects the columns indexed by $U$. This is the required minor expression for $a^c$, without cancellation or an invertibility assumption. If $c=0$, both modules are zero, the map is unique, and $a^0=1$ has its expression using the empty minor. Zero rings obey the same formulas. ∎

Lemma D.5: elkik

Reading source: Stacks, Tag 07CA.

For a finitely presented $R$-algebra $A$, let $J_e$ and $J_s$ be the ideals generated by all its elementary standard and strictly standard elements, respectively. Then $$H_{A/R}=\sqrt{J_e}=\sqrt{J_s}.$$ In particular $A_a$ is smooth over $R$ whenever $a$ is strictly standard.

Proof. First fix a strictly standard witness from D.3 and put $B=A_a$ and $N=(I/I^2)\otimes_A B$. Its relation condition says that the first $c$ equation classes generate $N$: divide each relation by the unit $a$. Write $q:B^c\to N$ for this surjection and $d:N\to B^n$ for the conormal differential. D.4 gives $\psi:B^n\to B^c$ with $$\psi d q=a1_c.$$ If $q(v)=0$ this identity gives $av=0$, hence $v=0$. Consequently $q$ is an isomorphism. The map $\chi=q\,a^{-1}\psi:B^n\to N$ satisfies $\chi d=1_N$. Thus $$B^n=d(N)\oplus\ker\chi.$$ Its quotient by $d(N)$ is a finite projective module, being a direct summand of $B^n$, and $d$ has zero kernel.

The actual localization still has a finite presentation over $R$: if $h\in P$ lifts $a$, it is $P[z]/(f_1,\ldots,f_m,hz-1)$. The quotient isomorphism follows by adjoining the specified inverse of $h$. The full presentation comparison F.11 and localization calculation F.65 identify its first cotangent homology and differential module with the kernel and cokernel just calculated. Therefore the smoothness definition F.6 proves that $R\to B$ is smooth. If $B=0$, these modules are zero and the same finite presentation and criterion apply.

Each singular prime contains every such $a$, since a prime avoiding $a$ has the smooth neighbourhood $A_a$. Hence $V(H_{A/R})\subset V(J_s)$. Setting every other minor coefficient to zero makes an elementary witness a strict witness, so $J_e\subset J_s$. If a prime is smooth, D.2 supplies an elementary witness outside it; relabel its selected variables and equations to place them first, absorbing any ordering sign in the coefficient. Such a prime cannot contain $J_e$. These statements give $$V(J_e)\subset V(H_{A/R})\subset V(J_s)\subset V(J_e).$$ The radical/prime equality proved in D.1 now proves both asserted equalities of ideals. Neither this argument nor the conormal splitting assumes that the base is Noetherian or that the smooth locus is quasi-compact. ∎

Example D.6: not quasi compact

Reading source: AI Integrated Stacks, pinned smoothing, lines 285–292.

A smooth locus can fail to be quasi-compact even for a finitely presented map. For any field $k$, take $$R=k[x,y_1,y_2,\ldots]/(xy_i\mid i\geq1),\qquad A=R/(x).$$ In $\operatorname{Spec}A$, its smooth locus is precisely $$U=\bigcup_{i\geq1}D(y_i) =\operatorname{Spec}A\setminus\{\mathfrak m\},\qquad \mathfrak m=(y_1,y_2,\ldots).$$ No finite family of these distinguished opens covers $U$. In particular $U$ is not quasi-compact.

Proof. The quotient identifies $A$ with $k[y_1,y_2,\ldots]$. The map $R\to A$ is of finite presentation: it is the quotient by the single generator $x$, with no algebra variables required. The base ring $R$ is not assumed Noetherian. Evaluation at all $y_i=0$ gives $A/\mathfrak m=k$, so $\mathfrak m$ is maximal and is the only prime containing every $y_i$.

Here is the ideal calculation. Setting all $y_i$ to zero gives a retraction $R\to k[x]$ of the natural inclusion $k[x]\to R$, so that inclusion is injective. For $I=(x)\subset R$, the relations $xy_i=0$ give $xr=xr(x,0,0,\ldots)$ for every $r\in R$. Therefore $$ I=xk[x],\qquad I^2=x^2k[x],\qquad I/I^2=k\,\bar x\simeq A/\mathfrak m. $$ In the last identification every $y_i$ annihilates $\bar x$. For the zero-variable polynomial presentation $R\twoheadrightarrow A$, the conormal complex is $[I/I^2\to0]$. The presentation comparison in Lemma F.11 identifies it with the naive cotangent complex used in Definition F.6.

First suppose $y_i\notin\mathfrak q$. Inverting $y_i$ in $R$ makes $x=0$, hence $A_{y_i}=R_{y_i}$. As an $R$-algebra this is $$ R[t]/(h),\qquad h=y_it-1. $$ The element $h$ is a nonzerodivisor over this arbitrary base ring. Indeed, if $h\sum_{n=0}^Nb_nt^n=0$, its constant coefficient gives $b_0=0$, and its successive coefficients $y_ib_{n-1}-b_n=0$ give every $b_n=0$. Consequently $(h)/(h^2)$ is free on $\bar h$ over $R[t]/(h)$: multiplication by $h$ gives a surjection onto it, and cancellation of the nonzerodivisor $h$ shows that its kernel is exactly $(h)$. An $R$-derivation of $R[t]$ is determined by its value on $t$; the product rule gives $d(t^n)=nt^{n-1}dt$. Thus $\Omega_{R[t]/R}=R[t]dt$, and this presentation has conormal complex $$ A_{y_i}\,\bar h\xrightarrow{\ y_i\ }A_{y_i}\,dt. $$ Its differential is an isomorphism, since $y_i$ is a unit. Both first homology and the differential module are zero, and the zero module is finite projective. The displayed presentation is finite, so Definition F.6 proves $R\to A_{y_i}$ smooth. Definition F.7 proves smoothness at every point of $D(y_i)$.

We must also exclude the exceptional point $\mathfrak m$ from every smooth neighbourhood. Let $g\in A\setminus\mathfrak m$. Its constant coefficient $c=g(0,0,\ldots)$ is nonzero. Regard its polynomial expression as a lift $\widetilde g\in R$. A presentation of $A_g$ over $R$ is $$ P=R[t]\twoheadrightarrow A_g,\qquad J=\ker(P\to A_g)=(x,\widetilde g t-1). $$ The class of $x$ in $J/J^2$ has relative differential zero, because $x$ belongs to the base ring $R$. This class is nonzero. To see this, put $C=(R/(x^2))_{\widetilde g}$. The $R$-algebra map $P\to C$ sending $t$ to $\widetilde g^{-1}$ maps $J$ into the square-zero ideal $(x)C$, and maps $J^2$ to zero. It therefore maps the class of $x$ in $J/J^2$ to $x\in C$. The assignments $x\mapsto\epsilon$ and $y_i\mapsto0$ define $$ R/(x^2)\longrightarrow k[\epsilon]/(\epsilon^2). $$ They respect all the relations, and send $\widetilde g$ to the unit $c$. The map extends to $C$ and sends $x$ to the nonzero element $\epsilon$; its nonzeroness follows directly from the degree bound for a multiple of $\epsilon^2$. Hence $x\ne0$ in $C$, and the class of $x$ is a nonzero kernel element in the conormal complex $[J/J^2\to A_gdt]$. By the same presentation comparison, $H_1(\operatorname{NL}_{A_g/R})\ne0$. Definition F.6 shows that $R\to A_g$ is not smooth. This holds for every $g\notin\mathfrak m$, so Definition F.7 shows that $R\to A$ is not smooth at $\mathfrak m$.

Every other prime omits some $y_i$, proving that the smooth locus is exactly the displayed open $U$. In the notation of Definition D.1, this also gives $H_{A/R}=\mathfrak m$.

Finally the open cover $U=\bigcup_{i\geq1}D(y_i)$ has no finite subcover. Given a finite set $F$ of indices, choose $j\notin F$. Evaluation $y_j\mapsto1$ and $y_i\mapsto0$ for $i\ne j$ defines a surjection $A\to k$ with prime kernel $\mathfrak q_j$. This point lies in $D(y_j)\subset U$ but in none of the $D(y_i)$ with $i\in F$. Thus no finite subfamily covers $U$, and $U$ is not quasi-compact. $\square$

Lemma D.7: strictly standard base change

Reading source: Stacks, Tag 07CC.

Let $R\to A$ be finitely presented and $R\to R'$ any ring map. An elementary standard element $a$ remains elementary standard as $a\otimes1$ in $A\otimes_R R'$. The same assertion holds for strictly standard elements. No flatness assumption on $R'$ is needed.

Proof. Take the finite presentation and witness from D.3 and set $P'=R'[x_1,\ldots,x_n]$. The polynomial base-change map $P\otimes_R R'\to P'$ sends $x^E\otimes r'$ to $r'x^E$. Its inverse sends the coefficient of every monomial to the same tensor; the polynomial monomials are the free $R$-basis of $P$, so these maps are inverse. Tensor right exactness identifies $$A\otimes_R R'=P'/I',\qquad I'=(f'_1,\ldots,f'_m),$$ where $f'_j$ is obtained by changing coefficients along $R\to R'$.

For a monomial $r x^E$, differentiation multiplies its coefficient by the integer $E_i$ and reduces the $i$th exponent by one. Changing coefficients commutes with this operation, including when that integer becomes zero in $R'$. Consequently every selected Jacobian entry and its determinant is the image of the old one. The elementary single-minor identity, or the strict sum-of-minors identity, maps to exactly the required new identity with coefficients $a'\otimes1$ or $a_U\otimes1$.

For the relation condition, choose a polynomial lift $\widetilde a$ and write each old relation with its finitely many coefficients in $(f_1,\ldots,f_c)+I^2$. Its image expresses $\widetilde a' f'_{c+j}$ in $(f'_1,\ldots,f'_c)+(I')^2$, because products of two elements of $I$ map to products of two elements of $I'$. These are exactly D.3e2 or D.3s2 for the new witness. No injectivity of a tensor map was used. Empty minors, empty relation lists and a zero base-changed algebra satisfy the same identities. ∎

Lemma D.8: final solve

Reading source: Stacks, Tag 07EU.

Suppose $R\to A\xrightarrow{\rho}\Lambda$, with $A$ finitely presented over $R$ and $H_{A/R}\Lambda=\Lambda$. Then $\rho$ factors through an $R$-smooth algebra: $$A\longrightarrow B\longrightarrow\Lambda.$$

Proof. If $\Lambda=0$, choose $B=0$. Its presentation $R/(1)$ is finite and its cotangent homology and differential module are zero, so F.6 makes it smooth. Assume henceforth $\Lambda\ne0$.

Let $J_s$ be the ideal generated by strictly standard elements. D.5 gives $H_{A/R}=\sqrt{J_s}$. We first check that extending the smaller ideal also gives the unit ideal. Choose $h_1,\ldots,h_t\in H_{A/R}$ and $\lambda_i\in\Lambda$ with $\sum_i\rho(h_i)\lambda_i=1$. For each $i$ some $n_i\geq1$ has $h_i^{n_i}\in J_s$. Raise that identity to $$N=1+\sum_i(n_i-1).$$ Every term in its distributive expansion uses some $h_i$ at least $n_i$ times: otherwise its total degree would be at most $N-1$. Each term therefore belongs to $J_s\Lambda$. Thus $1\in J_s\Lambda$.

Expanding the finite expressions in this ideal gives strictly standard elements $a_1,\ldots,a_r$ and $\mu_i\in\Lambda$ with $\sum_i\rho(a_i)\mu_i=1$. Here $r>0$, since the target is nonzero. Define $$B=A[T_1,\ldots,T_r]/\left(\sum_i a_iT_i-1\right).$$ The map sending $T_i$ to $\mu_i$ gives $B\to\Lambda$ and extends the prescribed $\rho$. Appending these finite variables and this one relation to a finite presentation of $A$ proves that $B/R$ is finitely presented.

In $B$, the elements $a_i$ generate one. Their principal opens therefore cover its spectrum: a prime containing all of them would contain the displayed unit. For each $i$ there is the explicit isomorphism $$B_{a_i}\cong A_{a_i}[T_j:j\ne i].$$ The map from right to left includes the remaining variables; its inverse substitutes $T_i=a_i^{-1}(1-\sum_{j\ne i}a_jT_j)$. Substitution kills the relation and both composites fix their generators. D.5 proves $A_{a_i}/R$ smooth. Adjoining the finitely many remaining polynomial variables is smooth directly: its empty-relation presentation has zero first cotangent homology and a finite free differential module on those variables, by F.11 and AG-CA-15, Theorem 5.1. F.6 therefore gives smoothness over $A_{a_i}$. The entire composition proof F.29 makes $B_{a_i}/R$ smooth. The proved finite principal-cover criterion F.67 now makes $B/R$ smooth. Zero localizations create empty charts and are already included in those criteria. This proves the required factorization over an arbitrary base ring. ∎

Presentations of algebras

The presentation change in D.9 uses a symmetric algebra over $A$. Over a Noetherian base its finite generators also give finite presentation. These constructions are among the results originating with Elkik.

Lemma D.9: improve presentation

Reading source: Stacks, Tag 07CE.

For a finitely presented algebra $A$ over an arbitrary ring $R$, there is an $R$-algebra map $A\to C$ of finite type with a retraction $C\to A$ such that:

1. Whenever $R\to A_a$ is a local complete intersection, $A_a\to C_a$ is smooth and $C_a$ has a finite polynomial presentation over $R$ with free conormal module.
2. Whenever $A_a/R$ is smooth, $\Omega_{C_a/R}$ is free.

Proof. Fix $P=R[X_1,\ldots,X_n]\twoheadrightarrow A$ with $I=(f_1,\ldots,f_m)$ and let $$M=I/I^2,\qquad K=\ker(A^m\longrightarrow M),\qquad C=\operatorname{Sym}_A(M).$$ The map $A^m\to M$ sends $e_j$ to $[f_j]$. The symmetric algebra and its degree-zero augmentation are constructed in F.128; their composite on $A$ is identity. Its universal property identifies $C$ with the quotient of $P[Y_1,\ldots,Y_m]$ by an ideal $J$ generated by $I$ and the linear forms whose coefficient classes belong to $K$. In fact maps out of this quotient are exactly maps $A\to D$ together with linear maps $A^m\to D$ killing $K$, hence linear maps $M\to D$. The resulting maps are inverse because they fix $A$ and every $Y_j$. This proves the kernel description and finite type even when $K$ is not finite.

Fix $a\in A$ and write $B=A_a$. Suppose first $R\to B$ is a local complete intersection. The arbitrary-ring presentation comparison F.94.A supplies local Koszul-regular equation lists for the inverse-variable polynomial presentation of $B$. Such a list gives a conormal basis: correct a relation modulo the square of its ideal by coefficients in the ideal; the resulting exact relation is a first Koszul cycle, hence a boundary, whose coefficients belong to the ideal. This proves independence as well as generation of its equation classes. F.72 identifies this locally free conormal module with $M_a\oplus B$. Its summand $M_a$ is locally finite flat and finitely presented, since both summands are generated by projections of a finite basis. The local finite-flat proof S01, Lemma 1.2 therefore makes $M_a$ free at each local ring; its finite presentation permits extending that basis to a principal neighbourhood. Thus $M_a$ is locally finite free.

Here are the finiteness and global splitting steps. Choose finitely many principal charts $D(b_i)$ covering $\operatorname{Spec}B$ on which $M_a$ is free. A finite subcover exists because the covering elements generate the unit ideal, witnessed by a finite expression for $1$. On each chart the surjection $B_{b_i}^m\to(M_a)_{b_i}$ splits and its kernel $(K_a)_{b_i}$ is a finite direct summand. Lift its finite generating list to fractions of elements of $K_a$ and collect their numerators in a finite submodule $K_0\subset K_a$. Then $(K_a/K_0)_{b_i}=0$ for every $i$. Any module element vanishing on these charts is zero: a power of each $b_i$ kills it, and these powers generate $1$, since a prime containing them would contain all the $b_i$. Hence $K_0=K_a$, so $M_a$ is finitely presented.

For each local splitting $s_i:M_{ab_i}\to B_{b_i}^m$, finite presentation identifies $$\operatorname{Hom}_B(M_a,B^m)_{b_i}=\operatorname{Hom}_{B_{b_i}}(M_{ab_i},B_{b_i}^m).$$ To verify this identity, present $M_a$ by finite free modules; both Hom modules are the kernels of the same finite matrix map after exact localization. Clear denominators of $s_i$, then multiply by a further power of $b_i$ to kill the error on a finite generating list. This gives $t_i:M_a\to B^m$ with $\pi t_i=b_i^{e_i}1_{M_a}$. Choose $r_i$ satisfying $\sum_i r_i b_i^{e_i}=1$. The map $\sum_i r_it_i$ splits $\pi$. Consequently $$M_a\oplus K_a\cong B^m,$$ and both summands are finite projective. This argument also handles $B=0$ directly by its zero modules.

The identities in F.128 give $C_a=\operatorname{Sym}_B(M_a)$, which is smooth and finitely presented over $B$. The inverse-variable presentation makes $B/R$ finitely presented, and finite substitution F.28 therefore makes $C_a/R$ finitely presented as well. The conormal calculation F.102 identifies the kernel conormal of $B[Y]\to C_a$ with $K_a\otimes_B C_a$. On principal charts where the split modules are free, this kernel is generated by a subset of polynomial coordinates. Those coordinates form a regular sequence: multiplication by each new variable is injective by polynomial coefficient comparison, even over an arbitrary ring. The regular-list cone calculation in F.118 gives their Koszul exactness in positive degrees; that direction requires no Noetherian hypothesis. Thus F.130 supplies the actual left injection in the conormal transitivity sequence.

If $h\in P$ lifts $a$, introduce an inverse variable $X_0$ and use the equation ideals $I'=(I,hX_0-1)$ and $J'=(J,hX_0-1)$. The explicit localization calculation F.72 and the conormal sequence just justified give $$0\longrightarrow C_a\oplus(M_a\otimes_B C_a)\longrightarrow J'/(J')^2\longrightarrow K_a\otimes_B C_a\longrightarrow0.$$ The last module is projective because a split finite free presentation stays split after tensoring. Hence $$J'/(J')^2\cong C_a\oplus((M_a\oplus K_a)\otimes_B C_a)\cong C_a^{m+1}.$$ Together with the explicit polynomial quotient for $C_a$ this proves (1). This conormal calculation uses only finite projectivity of $M_a$ and $K_a$, so it also applies when those properties are supplied by the smooth branch below.

For (2), suppose $B/R$ is smooth, without first assuming (1). F.6, the full presentation comparison F.11 and localization F.65 give the split exact conormal sequence for the original $n$ variables: $$0\longrightarrow M_a\longrightarrow B^n\longrightarrow\Omega_{B/R}\longrightarrow0.$$ Thus $M_a$ is finite projective and $M_a\oplus\Omega_{B/R}\cong B^n$. F.128 gives smoothness of $C_a/B$; its derivation universal property also identifies $\Omega_{C_a/B}=M_a\otimes_B C_a$, since a derivation from the symmetric algebra is exactly a linear map on its degree-one generators. The full cotangent transitivity calculation F.42 now gives $$0\longrightarrow\Omega_{B/R}\otimes_B C_a\longrightarrow\Omega_{C_a/R}\longrightarrow M_a\otimes_B C_a\longrightarrow0.$$ The right term is projective, so the sequence splits and its middle term is $C_a^n$. This proves (2). Zero algebras and empty variable or equation lists obey the same identities. ∎

Proposition D.10: lift smooth

Reading source: Stacks, Tag 07M8.

Let $R\twoheadrightarrow R_0=R/I$ be a surjection of Noetherian rings. A syntomic $R_0$-algebra $A_0$ has a syntomic lift $A$ over $R$, with a specified isomorphism $A/IA\cong A_0$. If $A_0/R_0$ is smooth, the lift can be chosen smooth. This is the auxiliary Noetherian lifting assertion used in desingularization.

Proof. If $A_0=0$, take $A=0$: its presentation $R/(1)$ is finite, it is flat, its cotangent modules are zero, and its fibres are empty. Assume $A_0\ne0$ and syntomic. Choose a finite polynomial presentation $A_0=R_0[X]/J_0$ and put $P_0=J_0/J_0^2$. F.88 makes $P_0$ finite projective. By F.129 the map is a local complete intersection, so D.9 constructs $$C_0=\operatorname{Sym}_{A_0}(P_0)$$ with a free-conormal polynomial presentation and a smooth map $A_0\to C_0$. Its finite presentation follows from F.128 and finite substitution F.28. The smooth/syntomic and composition proofs F.83 and F.32 make $C_0/R_0$ syntomic.

The complete equation-basis argument F.55 yields $$C_0=R_0[Y_1,\ldots,Y_v]/(\bar f_1,\ldots,\bar f_c),$$ with these classes a conormal basis. Flatness identifies that conormal after residue-field base change with the fibre conormal, by right exact tensor and the injection of the equation ideal into the polynomial ring. The fibres are locally complete intersections by syntomicness. F.57 therefore makes their point-dimensions $v-c$; applying this at their finitely many minimal primes gives dimension $v-c$ for every nonempty fibre.

Lift each $\bar f_j$ to the full polynomial ring $R[Y_1,\ldots,Y_v]$ and set $C^{\mathrm{pre}}=R[Y]/(f_1,\ldots,f_c)$. Its reduction is exactly $C_0$. F.66(1) gives $g\equiv1\pmod{IC^{\mathrm{pre}}}$ such that $C=(C^{\mathrm{pre}})_g$ is a relative global complete intersection. Thus $C/IC=C_0$, with no localization of the special fibre, and F.76 makes $C/R$ syntomic.

Choose a complementary finite projective module $Q_0$ with $P_0\oplus Q_0=A_0^N$. Apply the complete lifting proof F.117 to the finite projective $C_0$-module $Q_0\otimes_{A_0}C_0$. It gives a finitely presented étale $C\to C'$ inducing $C'/IC'\cong C_0$ and a finite projective $C'$-module $Q$ reducing to that specified module. Put $D=\operatorname{Sym}_{C'}(Q)$. It is smooth over $C'$ by F.128, and the proved composition statements make $D/R$ syntomic: étale maps are smooth by their zero cotangent complex and finite presentation, smooth maps are syntomic, and the base rings here are Noetherian finite-type algebras. The actual symmetric-algebra identities from F.128 give $$\begin{aligned} D/ID&\cong\operatorname{Sym}_{C_0}(Q_0\otimes_{A_0}C_0)\\ &\cong\operatorname{Sym}_{A_0}(Q_0)\otimes_{A_0}\operatorname{Sym}_{A_0}(P_0)\\ &\cong\operatorname{Sym}_{A_0}(Q_0\oplus P_0)\cong A_0[Z_1,\ldots,Z_N]. \end{aligned}$$ Choose $z_i\in D$ lifting these variables and set $E=D/(z_1,\ldots,z_N)$. Reduction of this quotient gives the specified $E/IE=A_0$.

We verify the claimed syntomic neighbourhood of every $\mathfrak q\in V(IE)$ rather than assuming it. Let $\mathfrak t$ be its preimage in $D$ and $\mathfrak p=\mathfrak q\cap R$. Then $I\subset\mathfrak p$. By F.87 choose $u\notin\mathfrak t$ and a relative global complete-intersection presentation $$D_u=R[W_1,\ldots,W_m]/(a_1,\ldots,a_s).$$ In the fibre over $k=\kappa(\mathfrak p)$ this is the localization of $(A_0\otimes_{R_0}k)[Z_1,\ldots,Z_N]$ at the image of $u$. Its ordered $Z_i$ are a regular sequence at the prime in question: multiplication by a polynomial variable is injective over any coefficient ring, successive quotients remove one variable, and exact localization preserves each injection. Their ideal stays proper at this prime because it comes from $\mathfrak q$ of the quotient.

Lift each $z_i$ to a polynomial $b_i$ in $R[W]$. By F.77, the fibre $\bar a_j$ form a regular sequence in the corresponding polynomial local ring. The just proved regularity of $Z_i$ in its quotient makes $(\bar a_1,\ldots,\bar a_s,\bar b_1,\ldots,\bar b_N)$ a regular sequence there. F.57 gives fibre point-dimension $m-s-N$ for $$E_{\bar u}=R[W]/(a_1,\ldots,a_s,b_1,\ldots,b_N).$$ Apply F.66(3) to this actual $s+N$-equation presentation. It supplies a relative global complete-intersection neighbourhood of $\mathfrak q$, hence a syntomic one by F.76. A further localization of $E_{\bar u}$ is a principal localization of $E$: write its inverted element as $e/\bar u^r$ and invert $\bar u e$. This proves the claim at every point of $V(IE)$.

Let $W$ be the union of these syntomic neighbourhoods and write its closed complement as $V(L)$ in $\operatorname{Spec}E$. Since $V(L)\cap V(IE)=\varnothing$, the ideal $L+IE$ is all of $E$; otherwise a containing maximal ideal would give a point of that intersection. Choose $h\in L$ with $h\equiv1\pmod{IE}$. Then $D(h)\subset W$, and $E_h/IE_h=A_0$ exactly. The finite principal-cover criterion F.62 makes $A=E_h$ syntomic over $R$: its spectrum has a finite subcover by the already constructed syntomic charts, using the finite expression for $1$ in their covering ideal. This proves the first assertion, including $N=0$.

Now assume $A_0/R_0$ smooth. For every prime $\mathfrak q\in V(IA)$ the fibre of $A/R$ is the same as the fibre of $A_0/R_0$ over its contracted base prime. Those fibres are smooth by the following calculation, which does not assume that the residue-field map is flat. F.83 covers $A_0$ by standard smooth principal charts; choose finitely many by their unit-ideal cover. Tensor right exactness changes each polynomial quotient to the quotient by the coefficient-changed equations. Monomial differentiation commutes with changing coefficients, so its chosen Jacobian determinant is the image of the old unit and is still a unit. F.85 makes every fibre chart smooth. The finite expression for $1$ in the covering ideal remains such an expression after base change, so these charts cover the fibre; F.67 makes the whole fibre smooth. Since $A/R$ is finitely presented and flat, F.45 gives a smooth neighbourhood of each such $\mathfrak q$. Write the complement of their union as $V(L')$. Exactly as above choose $h'\in L'$ with $h'\equiv1\pmod{IA}$. Then $A_{h'}/IA_{h'}=A_0$, while F.67 makes $A_{h'}/R$ smooth. No hypothesis that $I$ is nilpotent or lies in the Jacobson radical is needed for either shrinking step. ∎

For Noetherian $R$, F.87 gives relative global complete-intersection neighbourhoods of a syntomic map. The next lemma proves the required arbitrary-base extension and constructs a symmetric-algebra bundle over $\operatorname{Spec}(A)$ with a retraction whose total space is a relative global complete intersection over $R$.

Lemma D.11: syntomic complete intersection

Reading source: Stacks, Tag 07CG.

For a syntomic map $R\to A$ over any ring $R$, there is a smooth $A$-algebra $C$ and a retraction $C\to A$ such that $$C=R[x_1,\ldots,x_n]/(f_1,\ldots,f_c)$$ is flat over $R$ and every nonempty fibre has dimension $n-c$.

Proof. We first prove the arbitrary-base local complete-intersection assertion needed to apply D.9. Fix a finite presentation $A=R[X_1,\ldots,X_N]/(u_1,\ldots,u_s)$ and a prime $\mathfrak q$ of $A$ over $\mathfrak p$ of $R$. In the polynomial local ring of its $k=\kappa(\mathfrak p)$ fibre, syntomicness and F.58 give a complete-intersection equation ideal. Its conormal is free by F.57. Select from the $u_i$ a list whose conormal residue classes are a basis. Nakayama and the unit determinant of its coordinate matrix make its local conormal classes a basis; the final assertion of F.57 makes that selected list generate the fibre ideal regularly.

Model the local map $R_{\mathfrak p}\to A_{\mathfrak q}$ using F.59. Explicitly, for finite-type $\mathbf Z$-subalgebras $R_0\subset R$ containing all equation coefficients, put $T_0=R_0[X]/(u_i)$ and contract the chosen primes to $\mathfrak p_0,\mathfrak q_0$. The stages are $R_{0,\mathfrak p_0}\to(T_0)_{\mathfrak q_0}$. Adding numerators and denominators of any finite set of fractions proves cofinality in the stated local-model system. The models are Noetherian, their transitions are exactly localized base changes, and their colimits are the given local rings. Since $A_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ and is a finite module over itself, F.21 gives a stage where $(T_0)_{\mathfrak q_0}$ is flat over $R_{0,\mathfrak p_0}$.

We verify its closed-fibre complete-intersection property. Let $k_0=\kappa(\mathfrak p_0)$. There is an embedding $k_0\subset k$ by contraction of primes and the fraction-field construction. The map from the polynomial fibre local ring $V_0$ at the contracted prime to the final polynomial fibre local ring $V$ is local and flat, being field extension followed by localization. The selected equations generate the final fibre ideal. The finite $V_0$-module consisting of the model ideal modulo their span becomes zero after tensoring with $V$, so it is zero: a finite module with zero tensor along a local flat map has zero residue-field tensor, and Nakayama kills it. Here the target residue field extends the source residue field, so that residue-field tensor is faithful on vector spaces.

The same argument applies to each positive homology module of the selected finite free Koszul complex. These modules are finite because $V_0$ is Noetherian. Flat tensor commutes with their kernels and images, so their tensors are zero, since the final list is regular. Nakayama again gives zero homology in the model. The list lies in the maximal ideal, and F.118 makes it regular in $V_0$. Thus the model has precisely the closed fibre required by the Noetherian-base theorem F.87.

Apply that theorem to obtain $g_0\notin\mathfrak q_0$ with $B_0=(T_0)_{g_0}$ a relative global complete intersection over $R_0$. Its scalar extension is exactly $A_g$, with $g\notin\mathfrak q$. Write $B_0=P_0/(b_1,\ldots,b_t)$ using its finite polynomial complete-intersection presentation. By F.126 its augmented Koszul complex is exact, and by F.76 its last quotient $B_0$ is $R_0$-flat. Every complex term is also $R_0$-flat, being finite free over the polynomial ring.

All successive kernels in this augmented complex are $R_0$-flat. For $0\to Z\to F\to Q\to0$ with $F,Q$ flat, quotient flatness makes tensor preserve the left injection. For any inclusion of modules, the resulting map after tensoring with $Z$ is injective by embedding its two tensor products into those for $F$. Thus $Z$ is flat. Induct upwards from the flat augmentation quotient. Tensoring each short exact sequence with $R$ now remains exact. The actual base-changed equation list for $A_g$ is therefore Koszul-regular. This is the needed arbitrary-base assertion; it does not apply F.87 directly to a non-Noetherian base. The comparison F.94.A transports these presentations to the fixed finite polynomial presentation of $A$ at every prime, proving that $R\to A$ is a local complete intersection.

Let $I=(u_i)$ and $M=I/I^2$. D.9 with $a=1$ constructs $C=\operatorname{Sym}_A(M)$ with its degree-zero augmentation $C\to A$, proves $M$ finite projective and $A\to C$ smooth, and supplies a polynomial presentation over $R$ with free conormal. The finite-presentation proof F.128 and finite substitution F.28 make $C/R$ finitely presented. The complete equation-basis construction F.55 gives a presentation $C=R[x_1,\ldots,x_n]/J$ with $J=(f_1,\ldots,f_c)$ and those classes a basis of $J/J^2$.

On a finite principal cover of $\operatorname{Spec}A$ trivializing $M$, the symmetric-algebra identities in F.128 make $C$ a polynomial algebra over each chart. Those algebras are $A$-flat. Tensoring an injection with $C$ and detecting its kernel on the cover proves $A$-flatness of $C$ globally. The given flatness of $A/R$ consequently makes $C/R$ flat.

For a base residue field $k$, the same identities give $C_k=\operatorname{Sym}_{A_k}(M\otimes_A A_k)$. The original syntomic hypothesis makes $A_k$ locally a complete intersection. Trivializing the module and restricting to complete-intersection charts makes $C_k$ a polynomial algebra over such a chart. Its regular equations remain regular after polynomial extension and localization, because both are flat; the ambient polynomial local rings are regular. F.58 therefore proves that $C_k$ is locally a complete intersection.

Flatness of $C/R$ makes tensoring $0\to J\to R[x]\to C\to0$ injective on the left. Hence $J\otimes_R k$ is the fibre equation ideal $\bar J$. Right exactness applied to $J^2\to J\to J/J^2\to0$ identifies its conormal: the image of $J^2\otimes_R k$ consists exactly of products of the fibre ideal generators. Thus $$\bar J/\bar J^2\cong(J/J^2)\otimes_R k\cong C_k^c.$$ At a minimal prime of a nonzero $C_k$, F.57 identifies this rank with the height of its polynomial preimage, since the quotient local ring has dimension zero. Each component therefore has dimension $n-c$ by AG-CA-09, Theorem 4.3. The field algebra has finitely many components, so its dimension is their maximum, also $n-c$. The augmentation fixes $A$, and is the required retraction.

If $A=0$, take $C=0$ with presentation $R/(1)$. It is flat, the identity $A\to C$ is smooth, and all fibres are empty. Empty variable or regular lists use the same constructions. No condition has been added to the arbitrary base ring. ∎

Lemma D.12: smooth standard smooth

Reading source: Stacks, Tag 07CH.

Every smooth algebra $A$ over an arbitrary ring $R$ is a retract of an $R$-algebra $B$ that is standard smooth over $R$, with the map $A\to B$ itself smooth. Thus $B$ has a finite polynomial presentation with an invertible square Jacobian minor on its first equation variables.

Proof. If $A=0$, set $B=0$. The presentation $R[X]/(1)$ has Jacobian entry $0$, which is a unit in the zero ring, and the identity is its smooth retraction. Suppose $A\ne0$.

The smooth branch of D.9 constructs $C=\operatorname{Sym}_A(I/I^2)$, with $A\to C$ smooth and an augmentation $C\to A$. In this branch the conormal sequence is split by F.6, so the same finite-projective conormal calculation in D.9 applies, independently of any syntomic implication over the base. Appending the inverse variable for $1$ gives a finite polynomial presentation of $C$ with free conormal. F.55 replaces it by $$C=R[X_1,\ldots,X_n]/(f_1,\ldots,f_c)$$ whose equation classes are a conormal basis. F.29 makes $C/R$ smooth. It is nonzero, because it retracts onto $A$.

By F.6 and the full comparison F.11, the conormal differential in this presentation is a split injection. Write its matrix with equations as rows as $J=(\partial f_j/\partial X_i)$, of size $c\times n$. A left inverse of the transposed conormal map gives an $n\times c$ matrix $T$ over $C$ such that $JT=1_c$. Reduction at any maximal ideal of $C$ gives $c\le n$ by vector-space dimension. Lift its entries to polynomials $\widetilde T_{i\ell}$.

Set $E=C[Y_1,\ldots,Y_c]$ and introduce redundant coordinates $Z_i$ by the equations $$h_i=X_i-\sum_{\ell=1}^c\widetilde T_{i\ell}(X)Y_\ell-Z_i.$$ Eliminating $Z_i$ gives inverse generator maps for $$E=R[X,Y,Z]/(f_1,\ldots,f_c,h_1,\ldots,h_n).$$ Let $G$ be the determinant of its square Jacobian on the selected variables $(X_1,\ldots,X_n,Y_1,\ldots,Y_c)$, using the displayed equation order. Modulo $Y$ in $C$ its matrix is $$\begin{pmatrix}J&0\\1_n&-T\end{pmatrix}.$$ Interchange the two row blocks, which contributes $(-1)^{nc}$, then subtract $J$ times the first block row from the second. The resulting matrix is block upper triangular with diagonal $1_n,JT=1_c$. Its determinant is $1$. Thus the image $g$ of $G$ in $E$ satisfies $g\bmod(Y)=(-1)^{nc}$.

Put $B=E_g$. Evaluation at $Y=0$ sends $g$ to that unit and therefore extends the augmentation to $B\to C$. Its composite with $C\to B$ is identity; composing with $C\to A$ gives the required retraction of $A\to B$.

We verify both remaining smooth maps directly. The polynomial extension $C\to E=C[Y_1,\ldots,Y_c]$ has an empty-relation conormal complex: its degree-one term is zero, and its differential module is finite free on $dY_1,\ldots,dY_c$, by AG-CA-15, Theorem 5.1. Presentation comparison F.11 and the definition F.6 therefore give smoothness. The localization is the finite presentation $B=E[U]/(gU-1)$, with inverse maps adjoining $g^{-1}$ or evaluating $U$ there. Its conormal is generated by the single equation class, whose differential is $g\,dU$. This coefficient is a unit in $B$. Hence the surjection $B\to (gU-1)/(gU-1)^2$, followed by the conormal differential to $B\,dU$, is an isomorphism. Both maps are consequently isomorphisms. Its kernel and differential cokernel are zero, so F.11 and F.6 prove smoothness of $E\to B$. Applying the entire composition proof F.29 to $A\to C\to E\to B$ proves that $A\to B$ is smooth. These calculations use no Noetherian or fibre-dimension hypothesis.

Finally adjoining the inverse of $G$ writes $$B=R[X,Y,U,Z]/(f_1,\ldots,f_c,h_1,\ldots,h_n,UG-1).$$ Order the selected variables as $(X,Y,U)$, before all the $Z_i$. There are $n+c+1$ equations and selected variables. The Jacobian has the earlier square block in its upper left, zeros above the last column, and $G$ in its last diagonal position. Its determinant is therefore $G^2$, a unit since $UG=1$. Renaming the variables in this order gives the standard smooth presentation asserted. Empty blocks have determinant $1$; the argument includes $c=0$ and $n=0$ over any base ring. ∎

Lemma D.13: colimit standard smooth

Reading source: Stacks, Tag 07CI.

If an $R$-algebra $\Lambda$ is a filtered colimit of smooth $R$-algebras, then it is a filtered colimit of standard smooth $R$-algebras, over an arbitrary base ring $R$.

Proof. First consider a map $A\to\Lambda$ with $A$ finitely presented over $R$. Choose representatives for its finite generator list at one common smooth stage. Each of its finitely many defining relations is zero at a later stage, by the equality rule for filtered colimits. A common successor therefore gives a factorization $A\to B\to\Lambda$ with $B/R$ smooth. By D.12, $B$ is a retract of a standard smooth $R$-algebra $C$. The composite $A\to B\to C$ followed by $C\to B\to\Lambda$ is the original map. Thus every such finite-presentation map factors through a standard smooth algebra.

Take as objects finite polynomial presentations of standard smooth $R$-algebras together with maps to $\Lambda$, and as arrows algebra maps respecting those maps. This is a small category: the finite coefficient lists form a set, and each map to $\Lambda$ is determined by a finite tuple of elements. The presentation of $R$ with empty variable and equation lists is an object. Two objects have a common successor by applying the factorization just proved to the map from their tensor product to $\Lambda$; that tensor product has the finite presentation with their disjoint variables and both relation lists. To equalize parallel arrows $u,v\colon B\rightrightarrows C$, choose generators $b_1,\ldots,b_s$ of $B$ and append $u(b_i)-v(b_i)$ to the relations of $C$. The resulting finitely presented quotient maps to $\Lambda$; its factorization through a standard smooth object supplies an arrow out of $C$ equalizing $u,v$, since equality on generators implies equality on all polynomials in them. These constructions prove that the category is filtered.

Let $L$ be its colimit. The canonical map $L\to\Lambda$ is surjective: factor $R[T]\to\Lambda$ with $T\mapsto\lambda$ to represent any chosen $\lambda$ at one of the objects. For injectivity, if $b\in B$ has zero image, the finitely presented algebra $B/(b)$ maps to $\Lambda$ and factors through an object. The resulting successor kills $b$, so its colimit class is zero. For two representatives with equal images, first move them to a common successor and use the same argument on their difference. Hence $L\to\Lambda$ is an isomorphism and every stage is standard smooth. Empty lists and zero rings obey the same constructions. ∎

Lemma D.14: standard smooth include generators

Reading source: Stacks, Tag 07EY.

Suppose $A/R$ is standard smooth and $E=\{a_1,\ldots,a_n\}\subset A$ is finite. There is a standard smooth presentation $$A=R[x_1,\ldots,x_{n+m}]/(f_1,\ldots,f_c),\qquad c\ge n,$$ whose first $n$ coordinate classes are the specified $a_i$ in any chosen order, and whose Jacobian minor on its first $c$ variables is a unit.

Proof. Start with $A=R[y_1,\ldots,y_m]/(g_1,\ldots,g_d)$ and rename its variables so that the $d\times d$ minor on $y_1,\ldots,y_d$ is a unit. Choose polynomials $h_i(y)$ mapping to $a_i$. Adjoin variables $x_i$ with equations $x_i-h_i(y)$. Substitution defines an isomorphism $$R[x_1,\ldots,x_n,y_1,\ldots,y_m]/(x_i-h_i,g_j)\longrightarrow A.$$ Its inverse sends the old $y_j$ classes to themselves. The two composites fix all $y_j$, and fix $x_i$ because its new relation identifies it with $h_i(y)$; they are therefore identity maps.

Order the equations by the $n$ new graph equations followed by the $d$ old equations. With equations as rows, the minor on $(x_1,\ldots,x_n,y_1,\ldots,y_d)$ is $$\begin{pmatrix}1_n&-(\partial h_i/\partial y_j)\\0&(\partial g_i/\partial y_j)\end{pmatrix}.$$ In the signed permutation formula, each of its first $n$ columns has only its diagonal entry nonzero. The surviving terms therefore give exactly the old $d\times d$ determinant, which is a unit. Transposing yields the determinant convention with variables as rows and equations as columns. Now set $c=n+d$ and rename $y_j=x_{n+j}$, placing the unselected $y$ variables last. Since $d\le m$, this is the required minor on the first $c$ variables. The first $n$ classes are the prescribed elements. Empty blocks use determinant $1$, so the proof includes an empty $E$ or an empty old equation list. ∎

Lemma D.15: compare standard

Reading source: AI Integrated Stacks, pinned smoothing, lines 718–850.

For a finitely presented $R$-algebra $A$ and $a\in A$, put $B=A_a$. Number the following six properties:

1. $B/R$ is smooth.
2. $B/R$ is smooth and $\Omega_{B/R}$ is stably free.
3. $B/R$ is smooth and $\Omega_{B/R}$ is free.
4. $B/R$ admits a standard smooth presentation.
5. $a$ satisfies the strictly standard condition of D.3.
6. $a$ satisfies the elementary standard condition of D.3.

Their comparisons are:

• (a) $(4)\Rightarrow(3)\Rightarrow(2)\Rightarrow(1)$.
• (b) $(6)\Rightarrow(5)$.
• (c) $(6)\Rightarrow(4)$.
• (d) $(5)\Rightarrow(2)$.
• (e) Property (2) gives an $e_0$ such that $a^e$ is strictly standard for every $e\geq e_0$.
• (f) Every sufficiently large positive power $a^e$ is elementary standard exactly when $B/R$ is smooth and $\Omega_{B/R}$ has a finite basis $\mathrm db_1,\ldots,\mathrm db_d$ represented by elements $b_i\in A$. Having even one elementary positive power implies both these conditions.

In (f) the representatives must come from $A$, rather than merely from $B$. Property (4) by itself does not give elementary powers over an arbitrary base. The proof ends with a complete example establishing that distinction.

Proof of (a)--(e). By F.85, a standard smooth algebra is smooth and its differentials are free. A free module is stably free by adding the zero free module; forgetting this condition gives (a). For (b), take the coefficients of the other minors to be zero in D.3s1; transposing the one square minor does not change its determinant. D.3e2 and D.3s2 are the same relation condition.

For (c), use an elementary presentation $A=P/I$, where $P=R[x_1,\ldots,x_n]$, $I=(f_1,\ldots,f_m)$ and the first $c$ equations have minor $\Delta$. Lift $a$ to $h\in P$. Write $$B=A_a=P[x_0]/J,\qquad J=(x_0h-1,f_1,\ldots,f_m),$$ and $K=(x_0h-1,f_1,\ldots,f_c)$. The localization conormal calculation F.72 and D.3e2 give $J=K+J^2$: after inverting $a$, the classes of the remaining $f_j$ are combinations of the first $c$ classes. The module $J/K$ is finite, since $J$ has the displayed finite generating list, and satisfies $J/K=J(J/K)$. Choose generators $v_i$ and write $v_i=\sum_j b_{ij}v_j$ with $b_{ij}\in J$. Multiplication by the adjugate of $(\delta_{ij}-b_{ij})$, whose identity is proved in F.70, shows that $$g=\det(\delta_{ij}-b_{ij})\in1+J,\qquad gJ\subset K.$$ For the zero module choose $g=1$. Thus $J_g=K_g$. Since $g$ maps to $1$ in $B$, adjoining an inverse for it does not change $B$, and $$B=P[x_0,z]/(x_0h-1,f_1,\ldots,f_c,gz-1).$$ For these ordered equations select $(x_0,x_1,\ldots,x_c,z)$. The first $c+1$ rows have zero last column. Their first $c+1$ columns form the block matrix with upper left entry $h$, lower left zero, and lower right Jacobian minor $\Delta$. The last diagonal entry of the whole matrix is $g$. Its determinant is therefore $gh\Delta$. In $B$, $g=1$, $h=a$ is a unit, and D.3e1 gives $a=a'\Delta$, so $\Delta^{-1}=a'/a$. This proves standard smoothness, including the empty minor case $c=0$.

For (d), D.3s2 makes the first $c$ equation classes generate $M=(I/I^2)_a$. D.4 and D.3s1 give a map back from $B^n$ whose composite with the map $B^c\to M\to B^n$ is multiplication by the unit $a$. Thus $B^c\to M$ is both surjective and injective, and the conormal differential is a split injection. D.5 proves that $B=A_a$ is smooth. The localized conormal sequence, by F.65 and F.72, now gives $$M\oplus\Omega_{B/R}\cong B^n,\qquad M\cong B^c.$$ Consequently $\Omega_{B/R}\oplus B^c\cong B^n$, exactly the required stable freeness.

For (e), start with a finite presentation $A=R[X_1,\ldots,X_n]/I$ and $B=A_a$. If $B=0$, the equality $1/1=0$ means $a^L=0$ for some $L$. For every $e\ge L$, take $c=0$ in any finite presentation: the empty minor is $1$ and all multiplied relation classes are zero. Such powers are elementary, hence strictly standard. Assume $B\ne0$. Smoothness, the conormal description of the naive cotangent complex, and F.65/F.72 give the split exact sequence $$0\to M=(I/I^2)_a\to B^n\to\Omega_{B/R}\to0.$$ If $\Omega_{B/R}\oplus B^s\cong B^t$, then $M\oplus B^t\cong B^{n+s}$. Add $t$ variables $Y_i$ mapping to zero in $A$. The kernel of this enlarged presentation is $J=(I,Y_1,\ldots,Y_t)$, and $$J/J^2\cong I/I^2\oplus A^t.$$ Indeed, evaluation at $Y=0$ gives the first component, and the coefficients of the terms linear in $Y$ modulo $I$ give the second; their kernel consists precisely of products of two elements of $J$. The differential sends the new generators to the independent $\mathrm dY_i$. Thus $(J/J^2)_a$ is finite free and its differential is a split injection into $B^{n+t}$.

Choose representatives $f_1,\ldots,f_c\in J$ of a basis after localization: a basis element is a fraction of an element of $J/J^2$, and multiplying each basis element by its own denominator, a unit in $B$, still gives a basis. Extend them to a finite generating list $f_1,\ldots,f_m$ of $J$. Finiteness follows from F.44, or directly from the old finite list and the new variables. Each remaining class is zero in the localization of $$Q=(J/J^2)/\langle f_1,\ldots,f_c\rangle.$$ For each of its finitely many generators some power of $a$ kills that class. Choose a common exponent $L$. For a polynomial lift $h$ of $a$ this says $$h^L f_j\in(f_1,\ldots,f_c)+J^2\qquad(j>c).$$ Multiplying by further powers of $h$ gives D.3s2 for every $a^e$, $e\ge L$.

Let $F$ be the $(n+t)\times c$ matrix of the differentials of the first $c$ equations over $A$. The split injection after localization has a matrix left inverse $P$ over $B$. Choose $u$ and a matrix $P_0$ over $A$ with $(P_0)_a=a^uP$. Every entry of $P_0F-a^u1_c$ becomes zero in $B$, so a common further power $a^v$ kills all those finitely many entries. Hence $$P_1F=a^{u+v}1_c,\qquad P_1=a^vP_0,$$ as an exact equality over $A$. Increase $u+v$ to a positive integer $k$ by multiplying $P_1$ by an additional power of $a$ if necessary. No power of $a$ has been cancelled. F.70 puts $a^{ck}$ in the ideal of the $c\times c$ minors of $F$. Reduction at a prime of the nonzero ring $B$ shows $c\le n+t$. For every $e\ge\max(L,ck,1)$, multiply the minor coefficients by $a^{e-ck}$ to obtain D.3s1, together with the already established D.3s2. When $c=0$, the minor ideal is $A$ and the same conclusion uses the empty determinant $1$. This proves (e) over arbitrary rings, including zero divisors. ∎

Proof of (f). Put $B=A_a$. If $B=0$, some power $a^L$ is zero. The zero differential module has the empty basis, and, as in (e), every $a^e$ with $e\ge\max(L,1)$ is elementary using $c=0$ and the empty minor $1$. Assume $B\ne0$.

First suppose $B$ is smooth and $\mathrm db_1,\ldots,\mathrm db_d$ is a basis with $b_i\in A$. Choose finite algebra generators $z_1,\ldots,z_n$ of $A$ over $R$ and the surjection $$P=R[X_1,\ldots,X_n,Y_1,\ldots,Y_d]\longrightarrow A, \qquad X_i\longmapsto z_i,\quad Y_j\longmapsto b_j.$$ Its kernel $J$ is finitely generated by F.44. The smoothness definition F.6, the presentation description of the naive cotangent complex F.11, and localization F.65 give the exact sequence $$0\longrightarrow M=(J/J^2)_a\xrightarrow{\mathrm d} B^n\oplus B^d\longrightarrow\Omega_{B/R}\longrightarrow0.$$ Here the two summands have coordinates $\mathrm dX_i$ and $\mathrm dY_j$. The map from the second summand to $\Omega_{B/R}$ is an isomorphism, because its images are the chosen basis. Consequently projection onto the first summand restricts to an isomorphism $M\to B^n$: for each $v\in B^n$, there is a unique $w\in B^d$ whose differential image cancels that of $v$, so $(v,w)$ is in the image of $M$; a vector of $M$ with first component zero has second component zero as well. Injectivity of $\mathrm d$ proves uniqueness in $M$.

Let $\eta_1,\ldots,\eta_n$ be the resulting basis of $M$ whose projected differentials are the standard basis of $B^n$. Write each $\eta_i$ as a fraction of a class represented by $f_i\in J$, with denominator $a^{k_i}$. Thus the classes of $f_i$ are $a^{k_i}\eta_i$, still a basis over $B$. Their square Jacobian on $X_1,\ldots,X_n$ is diagonal after passage to $B$, with entries $a^{k_i}$. In particular the image of $$\delta=\det(\partial f_i/\partial X_j)_{i,j=1}^n\in A$$ is a unit in $B$. Extend $f_1,\ldots,f_n$ to a finite generating list $f_1,\ldots,f_m$ of $J$. The finite module $$Q=(J/J^2)/\langle f_1,\ldots,f_n\rangle$$ has zero localization at $a$. A common power $a^L$ kills its finite generating list. For a polynomial lift $h\in P$ of $a$, this means $$h^L f_j\in(f_1,\ldots,f_n)+J^2\qquad(j>n).$$ All higher powers satisfy the same relation condition.

Since $\delta$ is a unit in $B$, choose $u\in A$ and $r\ge0$ with $\delta(u/a^r)=1$ in $B$. The equality in this localization supplies $v\ge0$ such that $$a^v(\delta u-a^r)=0\quad\hbox{in }A, \qquad a^{r+v}=\delta(a^v u).$$ Increase the exponent to a positive $K$ by multiplying this equality by an additional power of $a$ if necessary. This gives $a^K=\delta u_0$ in $A$, without cancelling a power of a possible zero divisor. Every $e\ge\max(K,L,1)$ then satisfies both D.3e1 and D.3e2 for the presentation $A=P/J$, with $c=n$ and the selected variables $X_i$. This proves the forward direction, including $n=0$, when $\delta$ is the empty determinant $1$.

Conversely, suppose $a^r$ is elementary for some $r\ge1$. Inverting $a^r$ is the same as inverting $a$, since $a^{-1}=a^{r-1}(a^r)^{-1}$. Part (c), applied to $a^r$, proves that $B$ is standard smooth, hence smooth. Take its elementary presentation $A=R[x_1,\ldots,x_N]/I$ with $I=(g_1,\ldots,g_m)$ and selected minor on $x_1,\ldots,x_c$. In $B$, D.3e2 makes every remaining conormal class a combination of the first $c$. Therefore $$\Omega_{B/R}=B^N/\langle\mathrm dg_1,\ldots,\mathrm dg_c\rangle.$$ The selected $c\times c$ Jacobian block is invertible. Subtracting a unique combination of these $c$ differential rows eliminates the first $c$ coordinates of any vector in $B^N$; a combination with those coordinates zero has all its coefficients zero. Hence the images of $\mathrm dx_{c+1},\ldots,\mathrm dx_N$ form a basis of this quotient. Their representatives are elements of $A$. The forward direction now proves that every sufficiently large power is elementary. This proves (f). ∎

Why standard smoothness alone is insufficient. Take $$R=\mathbf Z,\qquad A=\mathbf Z[t^2,t^3]\subset\mathbf Z[t],\qquad a=t^2.$$ This algebra has the finite presentation $$A\cong\mathbf Z[X,Y]/(Y^2-X^3),\qquad X\longmapsto t^2,\quad Y\longmapsto t^3.$$ Indeed division by the monic polynomial $Y^2-X^3$ writes every class uniquely as $p(X)+Yq(X)$. After substitution, the first summand has even exponents and the second has odd exponents at least $3$; their disjoint monomials prove injectivity. The image is precisely the indicated subalgebra.

Inverting $a$ gives $$B=A_a=\mathbf Z[t,t^{-1}],\qquad t=Y/X,\quad t^{-1}=Y/X^2.$$ These identities follow from $Y^2=X^3$ and give both inclusions with the Laurent ring. The presentation $B=\mathbf Z[T,U]/(TU-1)$ is standard smooth: select $U$ for its one equation, whose derivative is the unit $T$. Every derivation of $B$ is uniquely determined by its value on $t$, with $\mathrm d(t^{-1})=-t^{-2}\mathrm dt$. Thus $\Omega_{B/\mathbf Z}=B\,\mathrm dt$.

The units of $\mathbf Z[t,t^{-1}]$ are exactly $\pm t^k$. To check this, the lowest and highest nonzero exponents of a product add, since their coefficients are nonzero products in the domain $\mathbf Z$. For a product equal to $1$, each factor must therefore have width zero and be a monomial; its integer coefficient must be $1$ or $-1$.

Every $b\in A$ has the form $b_0+\sum_{k\ge2}b_k t^k$, with integer coefficients and a finite sum. Consequently $$\mathrm db=\left(\sum_{k\ge2}k b_k t^{k-1}\right)\mathrm dt.$$ This coefficient cannot be a unit: if it were $\pm t^j$, necessarily $j\ge1$, and its sole coefficient would require $(j+1)b_{j+1}=\pm1$, impossible in $\mathbf Z$. A basis of the free rank-one module $B\,\mathrm dt$ must contain one generator; tensoring with $\mathbf Q(t)$ verifies that its length is one. Hence it has no basis consisting of differentials of elements of $A$. By the converse in (f), no positive power of $a$ is elementary standard, despite standard smoothness of $B$.

There is no conflict with (e). In the original cusp presentation the two one-by-one minors are $-3X^2$ and $2Y$, and in $A$ $$X^3=(-X)(-3X^2)+(-Y)(2Y).$$ The relation condition is empty because there is only one equation. Thus $a^3$, and every $a^e$ with $e\ge3$ after multiplication by $a^{e-3}$, is strictly standard. This exact example distinguishes the sum-of-minors condition from the single-minor condition over the same arbitrary base.

The lifting problem

D.18 lifts a smooth filtered presentation across a nilpotent ideal when the target remains flat. Its two preliminary steps use the explicit smooth presentations constructed in the preceding section.

Lemma D.16: lift once

Reading source: Stacks, Tag 07CK.

Let $R\to\Lambda$ be a ring map and $I\subset R$ an ideal with $I^2=0$. Suppose that $\Lambda/I\Lambda$ is a filtered colimit of smooth $R/I$-algebras. Every map $\varphi:A\to\Lambda$ from a finitely presented $R$-algebra factors as $$A\longrightarrow B/J\longrightarrow\Lambda,$$ where $B/R$ is smooth and $J\subset IB$ is a finitely generated ideal. Neither $R$ nor $I$ is required to be Noetherian or finite.

Proof. Write $\bar R=R/I$ and $\bar A=A/IA$. By D.13, the induced map $\bar A\to\Lambda/I\Lambda$ factors through a standard smooth $\bar R$-algebra $C$. This factorization occurs at a finite stage: take representatives for the finite generators of $\bar A$ and then a common successor where all its finitely many equations vanish.

There is a finite presentation $C=\bar A[t_1,\ldots,t_r]/(\bar g_1,\ldots,\bar g_s)$ relative to this specified map. To see the finiteness directly, choose $C=\bar R[t]/(k_1,\ldots,k_v)$ and generators $a_1,\ldots,a_u$ of $\bar A$. Represent their images by polynomials $v_i(t)$. Over $\bar A[t]$, the finite equations $k_j(t)$ and $a_i-v_i(t)$ give exactly $C$: substituting for the $a_i$ recovers its displayed $\bar R$-presentation. Include these equations in the list $\bar g$.

Lift each $\bar g_i$ to $g_i\in A[t]$ and the images of the $t_j$ in $\Lambda/I\Lambda$ to $\lambda_j\in\Lambda$. Evaluation gives a finite expression $$\varphi(g_i)(\lambda)=\sum_{j=1}^{q_i}\epsilon_{ij}\mu_{ij},\qquad \epsilon_{ij}\in I,\quad\mu_{ij}\in\Lambda.$$ Introduce one variable $z_{ij}$ for each of these terms and set $$T=A[t,z]/\bigl(g_i-\sum_j\epsilon_{ij}z_{ij}\bigr).$$ This is finitely presented over $R$, by combining the finite presentation of $A$ with the displayed lists. The assignments $a\mapsto\varphi(a)$, $t_j\mapsto\lambda_j$, $z_{ij}\mapsto\mu_{ij}$ define $T\to\Lambda$, and reduction gives $$T/IT\cong C[z].$$ Adjoining free variables to a standard smooth presentation preserves its invertible Jacobian minor: retain the equations and their pivot variables. Thus choose $$T/IT=(R/I)[X_1,\ldots,X_n]/(\bar f_1,\ldots,\bar f_c)$$ with $\bar\Delta=\det(\partial\bar f_i/\partial X_j)_{1\leq i,j\leq c}$ invertible. Lift the $f_i$ to $R[X]$, put $\Delta=\det(\partial f_i/\partial X_j)_{1\leq i,j\leq c}$, and form $$B=R[X_1,\ldots,X_n,W]/(f_1,\ldots,f_c,W\Delta-1).$$ On pivots $X_1,\ldots,X_c,W$, its Jacobian has upper right block zero and lower right entry $\Delta$; its determinant is $\Delta^2$, a unit in $B$. It is therefore standard smooth, hence smooth by F.85. Its reduction is precisely $T/IT$, since $W$ becomes the inverse of $\bar\Delta$.

We construct a map $\psi:B\to T$ inducing that isomorphism. Choose lifts $y_i\in T$ of the displayed coordinate classes in $T/IT$, and let $L=IT$, so $L^2=0$. The element $\Delta(y)$ is a unit: a lift $d$ of its modular inverse satisfies $\Delta(y)d=1+\ell$ with $\ell\in L$, and $(1+\ell)^{-1}=1-\ell$. Let $H=(\partial f_i/\partial X_j(y))_{1\leq i,j\leq c}$. Its determinant is $\Delta(y)$, so the cofactor identity proved in F.70 gives an inverse matrix. Every $f_i(y)$ lies in $L$. Choose $$ (u_1,\ldots,u_c)^{\mathsf t}=-H^{-1}(f_1(y),\ldots,f_c(y))^{\mathsf t}\in L^c$$ and set $u_j=0$ for $j>c$. Expanding each monomial, all terms with two correction factors vanish in $L^2$, and hence $$ f_i(y+u)=f_i(y)+\sum_{j=1}^c\frac{\partial f_i}{\partial X_j}(y)u_j=0.$$ The new determinant is still a unit because its reduction is unchanged. Set $\psi(X_i)=y_i+u_i$ and $\psi(W)=\Delta(y+u)^{-1}$. These assignments satisfy every equation of $B$ and lift the required reduction map.

The map $\psi$ is onto without a finiteness assumption on $I$. For $t\in T$, lift its reduction to $b\in B$ and write $t-\psi(b)=\sum_\nu\eta_\nu t_\nu$ with $\eta_\nu\in I$. Lift each reduction of $t_\nu$ to $b_\nu\in B$. Since $I(IT)=0$, one obtains $t=\psi(b+\sum_\nu\eta_\nu b_\nu)$. Its kernel $J$ lies in $IB$, because $B/IB\to T/IT$ is injective. To prove that $J$ is finite, compose the finite polynomial presentation $R[X,W]\twoheadrightarrow B$ with $\psi$. The kernel of the resulting surjection to the finitely presented algebra $T$ is finite by F.44. Its images in $B$ generate $J$. Thus $T\cong B/J$, and the maps $A\to T\to\Lambda$ give the claimed factorization. Empty equation lists use $\Delta=1$ and require no correction coordinates. ∎

Lemma D.17: lift twice

Reading source: Stacks, Tag 07CL.

Let $R\to\Lambda$ be flat, let $I^2=0$, and suppose that $\Lambda/I\Lambda$ is a filtered colimit of smooth $R/I$-algebras. If $B/R$ is smooth, $\varphi:B\to\Lambda$ is an $R$-map, and $J\subset IB$ is a finitely generated ideal with $\varphi(J)=0$, there are maps $$B\xrightarrow{\alpha}B'\xrightarrow{\beta}\Lambda$$ with $B'/R$ smooth, $\alpha(J)=0$, and the exact equality $\beta\alpha=\varphi$.

Proof. We first give the square-zero lifting calculation needed below. For any smooth $R$-algebra $B$, any $R$-algebra $E$, any ideal $L\subset E$ with $L^2=0$, and any $R$-map $\gamma:B\to E/L$, there is a lift $B\to E$. Choose a finite polynomial presentation $P=R[Z_1,\ldots,Z_n]\twoheadrightarrow B$, with kernel $K$, and lifts $e_i\in E$ of $\gamma(Z_i)$. Evaluation of a relation $f\in K$ lies in $L$ and vanishes on $K^2$. It gives a $B$-linear map $$\delta:K/K^2\longrightarrow L,$$ where $L$ is a $B$-module through $\gamma$; its $E/L$-module structure is well defined because $L^2=0$. Indeed evaluation of $pf$ is evaluation of $p$ times evaluation of $f$, and that multiplication on $L$ depends only on $p$ modulo $K$ and $E$ modulo $L$.

By the definition F.6 and presentation comparison F.11, the sequence $$0\longrightarrow K/K^2\xrightarrow{d}B^n\longrightarrow\Omega_{B/R}\longrightarrow0$$ is exact and its last module is finite projective. This sequence splits: realize that projective module as a direct summand of a finite free module, lift the free basis through the surjection, and restrict the lift to the summand. The resulting section defines a retraction $r:B^n\to K/K^2$ of $d$. Put $\ell=\delta r:B^n\to L$ and $c_i=\ell(dZ_i)$. Evaluate polynomials instead at $e_i-c_i$. The monomial expansion, using $L^2=0$, gives for each $f\in K$ $$ f(e-c)=f(e)-\sum_i\frac{\partial f}{\partial Z_i}(e)c_i =\delta([f])-\ell(d[f])=0.$$ In the second equality the derivative coefficients act on $L$ through their images under $\gamma$, so no choice of their lifts changes the product. Evaluation therefore descends to $B\to E$ and lifts $\gamma$. This proves the needed lifting statement in full arbitrary-ring scope.

Now suppose first that $J=(h)$. Write $h=\sum_{i=1}^q\epsilon_i b_i$ with $\epsilon_i\in I$ and $b_i\in B$. Consider the row map $v:R^q\to R$, $v(a)=\sum_i\epsilon_i a_i$, and let $K_v=\ker v$, $H_v=\operatorname{im}v$. Flatness preserves the two exact sequences $0\to K_v\to R^q\to H_v\to0$ and $0\to H_v\to R$. Consequently $$\ker(v\otimes_R\Lambda)=\operatorname{im}(K_v\otimes_R\Lambda\longrightarrow\Lambda^q).$$ The vector $(\varphi(b_i))_i$ belongs to this kernel, since $\varphi(h)=0$. An element of a tensor product is a finite sum of pure tensors. Thus there are finitely many $\lambda_j\in\Lambda$ and columns $(a_{ij})_i\in K_v$ with $$\varphi(b_i)=\sum_{j=1}^m a_{ij}\lambda_j,\qquad \sum_i\epsilon_i a_{ij}=0\quad\text{for each }j.$$ This derives the finite relation directly from flatness, including the case of a zero vector and an empty tensor sum.

Set $C=B[T_1,\ldots,T_m]/(b_i-\sum_j a_{ij}T_j)$. Its finite presentation follows by appending these variables and equations to a finite presentation of $B$. The displayed relations define a map $C\to\Lambda$ extending $\varphi$, with $T_j\mapsto\lambda_j$. By D.16 this map factors through $E/J'$, where $E/R$ is smooth and $J'\subset IE$ is finite. In particular $(J')^2=0$. Apply the lifting calculation above to $B\to C\to E/J'$ and obtain $\alpha:B\to E$. Let $\beta:E\to E/J'\to\Lambda$. Since the reduction of $\alpha$ is the original map through $C$, one has $\beta\alpha=\varphi$ exactly.

Choose lifts $\xi_j\in E$ of the images of $T_j$. The equations in $C$ give $\alpha(b_i)=\sum_j a_{ij}\xi_j+\theta_i$ for some $\theta_i\in J'$. Therefore $$\alpha(h)=\sum_j\Bigl(\sum_i\epsilon_i a_{ij}\Bigr)\xi_j +\sum_i\epsilon_i\theta_i=0.$$ The first sum vanishes by the column relations, the second because $IJ'\subset I^2E=0$. This proves the one-generator case, with $B'=E$.

For $J=(h_1,\ldots,h_s)$, apply that construction successively to the image of $h_i$ in the smooth algebra obtained at the preceding step. Each such image lies in $I$ times that algebra, and its map to $\Lambda$ is zero because the composite to $\Lambda$ remains $\varphi$. Each new map preserves the previously killed generators. After finitely many steps their composite kills every generator, hence the whole ideal $J$, and still factors $\varphi$ exactly. If $J=0$, take $B'=B$, $\alpha=1$, and $\beta=\varphi$. ∎

Proposition D.18: lift

Reading source: Stacks, Tag 07CM.

Suppose that $R\to\Lambda$ is flat, $I\subset R$ is nilpotent, and $\Lambda/I\Lambda$ is a filtered colimit of smooth $R/I$-algebras. Then $\Lambda$ is a filtered colimit of smooth $R$-algebras.

Proof. Begin with $I^2=0$. Given any $R$-map $A\to\Lambda$ from a finitely presented algebra, D.16 gives maps $A\to B/J\to\Lambda$ with $B/R$ smooth, $J\subset IB$ finite. Let $\varphi:B\to\Lambda$ be the induced map; it kills $J$. The flat map and the same square-zero ideal meet every hypothesis of D.17. Obtain $B\to B'\to\Lambda$ with $B'/R$ smooth and $J$ killed in $B'$. The first arrow descends to $B/J\to B'$, and its composite to $\Lambda$ is the given quotient map because D.17 gives equality of the maps from $B$. Thus $A\to B'\to\Lambda$ is an exact factorization of the original map. Criterion F.90, whose filtered-category construction and colimit isomorphism are proved there, gives the assertion for square-zero ideals.

Induct on a positive integer $n$ with $I^n=0$. For $n=1$ the ideal is zero and the assertion is the hypothesis. For $n\geq2$, set $J=I^{n-1}$, so $J^2=0$ because $2n-2\geq n$. Over $\bar R=R/J$ put $\bar\Lambda=\Lambda/J\Lambda$ and $\bar I=I/J$. Then $\bar I^{n-1}=0$ and $\bar\Lambda/\bar I\bar\Lambda=\Lambda/I\Lambda$. The map $\bar R\to\bar\Lambda$ is flat: for every $\bar R$-module $M$ the maps on pure tensors give $$M\otimes_{\bar R}(\Lambda/J\Lambda)\cong M\otimes_R\Lambda.$$ Here tensors involving $J\Lambda$ vanish because $JM=0$; this verifies the inverse maps as well. An exact sequence of $\bar R$-modules is exact over $R$, so the original flatness makes the displayed tensor functor exact. The induction hypothesis therefore expresses $\bar\Lambda$ as a filtered colimit of smooth $\bar R$-algebras. Apply the square-zero case already proved to $J\subset R$ and the original flat map $R\to\Lambda$. Its quotient is exactly this smooth colimit, hence it expresses $\Lambda$ as a filtered colimit of smooth $R$-algebras. This completes the induction without assuming that the nilpotent ideal is finitely generated. ∎

The lifting lemma

D.19 constructs a finitely presented algebra whose smooth locus lifts the prescribed quotient data.

Lemma D.19: lifting

Reading source: Stacks, Tag 07CP.

Let \(R\) be Noetherian, let \(\pi\in R\), let \(\Lambda\) be an \(R\)-algebra, and assume \[ \operatorname{Ann}_R(\pi)=\operatorname{Ann}_R(\pi^2), \qquad \operatorname{Ann}_\Lambda(\pi)=\operatorname{Ann}_\Lambda(\pi^2). \] Given a finitely presented \(R/\pi^2R\)-algebra \(\bar C\) and a map \(\bar C\to\Lambda/\pi^2\Lambda\), there is a finitely presented \(R\)-algebra \(D\), a map \(D\to\Lambda\), and commuting maps \[ \begin{array}{ccccc} R/\pi^2R&\longrightarrow&\bar C&\longrightarrow&\Lambda/\pi^2\Lambda\\ \downarrow&&\downarrow&&\downarrow\\ R/\pi R&\longrightarrow&D/\pi D&\longrightarrow&\Lambda/\pi\Lambda. \end{array} \] The map \(R\to D\) is smooth at every prime avoiding \(\pi\). It is also smooth at each prime containing \(\pi\) whose inverse image in \(\bar C\) is smooth over \(R/\pi^2R\). At these latter primes, \(\bar C/\pi\bar C\to D/\pi D\) is smooth as well.

Proof. Put \(P=R[x_1,\ldots,x_n]\) and choose lifts \(f_1,\ldots,f_m\in P\) of a finite relation list for \[ \bar C=(R/\pi^2R)[x_1,\ldots,x_n]/(\bar f_1,\ldots,\bar f_m). \] Write \(\bar P=(R/\pi^2R)[x]\) and \(\bar J=(\bar f_1,\ldots,\bar f_m)\). Thus the kernel of \(P\to\bar C\) is \((f_1,\ldots,f_m)+\pi^2P\); \(\pi^2\) need not be in the chosen list.

At a smooth prime, D.2 supplies a principal neighbourhood, selected equation indices \(E\), and equally many variable indices \(U\). On that neighbourhood the selected Jacobian minor \(\delta\) is a unit and the selected classes generate \(\bar J/\bar J^2\). They are a basis: their composite with the conormal differential projected onto the \(U\)-coordinates is invertible, so a relation among them is zero. Shrink the polynomial neighbourhood to make \(\bar J\) itself generated by these selected equations. Here are the details of that shrinking. The finite module \[ M=(\bar J/(\bar f_j:j\in E))_a \] satisfies \(M=\bar J_aM\). The determinant form of F.10 gives \(b'\in\bar J_a\) with \((1+b')M=0\). Write \(b'=b/a^N\), with \(b\in\bar J\). Replacing the polynomial \(a\) by \(a(a^N+b)\) inverts \(1+b'\). Its image in \(\bar C\) is the old \(a^{N+1}\), so this replacement preserves the principal open in \(\operatorname{Spec}\bar C\).

These principal opens cover the smooth locus. They have a finite subcover for an algebraic reason: \(\bar C\) is Noetherian by S01, Lemma 1.3, so the ideal generated by all their defining elements is generated by finitely many of those elements. To obtain such a list, express a finite ideal generating list as finite sums of the original elements and collect those that occur. Their principal opens have the same union. Index the resulting charts by \(k=1,\ldots,r\). An empty smooth locus uses \(r=0\).

Clear the finitely many denominators in the equality of localized relation ideals on each chart and replace its \(a_k\) by a sufficiently large positive power. There are then exact identities in \(P\) \[ a_kf_\ell=\sum_{j\in E_k}h_{k\ell}^jf_j+\pi^2g_{k\ell}. \tag{D.19a} \] For \(\ell\in E_k\), take \(h_{k\ell}^j=a_k\delta_{\ell j}\) and \(g_{k\ell}=0\). The selected conormal classes remain a basis on \(\bar C_{a_k}\), and the selected Jacobian minor \(\delta_k\) remains invertible there.

In \(Q=P[z_1,\ldots,z_m]\), put \[ F_\ell=f_\ell-\pi z_\ell,\qquad p_{k\ell}=a_kz_\ell-\sum_{j\in E_k}h_{k\ell}^jz_j-\pi g_{k\ell}, \] and define \(D=Q/(F_\ell,p_{k\ell})_{\ell,k}\). All lists are finite. Choose lifts \(\lambda_i\in\Lambda\) of the prescribed images of \(x_i\), and write \(f_\ell(\lambda)=\pi^2\mu_\ell\). Send \(x_i\) to \(\lambda_i\) and \(z_\ell\) to \(\pi\mu_\ell\). The \(F_\ell\) vanish. Evaluation of (D.19a) shows that \[ \pi^2\left(a_k(\lambda)\mu_\ell- \sum_{j\in E_k}h_{k\ell}^j(\lambda)\mu_j-g_{k\ell}(\lambda)\right)=0. \] The annihilator hypothesis in \(\Lambda\) makes its product with \(\pi\) zero; that product is exactly the image of \(p_{k\ell}\). This defines \(D\to\Lambda\). Reduction of \(F_\ell=0\) modulo \(\pi\) defines \(\bar C\to D/\pi D\), and the variable assignments verify every square in the displayed diagram.

The polynomial identity \[ \pi p_{k\ell}=-a_kF_\ell+\sum_{j\in E_k}h_{k\ell}^jF_j \tag{D.19b} \] follows by substituting (D.19a). After inverting \(\pi\), all \(p_{k\ell}\) follow from the \(F_\ell\), and \(F_\ell=0\) eliminates \(z_\ell=f_\ell/\pi\). Thus \(D_\pi=R_\pi[x]\), a polynomial localization smooth over \(R\) by the empty-equation and inverse-variable cases of F.85.

Fix \(k\), and retain only its selected equations and its own compatibility equations: \[ D_k=Q/(F_j:j\in E_k,\ p_{k\ell}:\ell\notin E_k). \] There is a surjection \(D_k\to D\). Let \(\mathfrak q\subset D\) contain \(\pi\) and lie over a smooth prime of \(\bar C\); choose \(k\) with \(a_k\notin\mathfrak q\), and let \(\mathfrak q_k\) be its inverse image in \(D_k\). In \(D_k/\pi D_k\), after inverting \(a_k\), each \(z_\ell\) for \(\ell\notin E_k\) is uniquely \[ z_\ell=a_k^{-1}\sum_{j\in E_k}h_{k\ell}^jz_j. \] Substitution and its reverse on the remaining variables give the actual isomorphism \[ (D_k/\pi D_k)_{a_k} \cong(\bar C/\pi\bar C)_{a_k}[z_j:j\in E_k]. \tag{D.19c} \] Here the equality of the relation ideals modulo \(\pi^2\) also gives their equality modulo \(\pi\).

There are \(m\) equations in this presentation of \(D_k\). Choose as pivot variables \(x_i\), \(i\in U_k\), followed by \(z_\ell\), \(\ell\notin E_k\). Their Jacobian modulo \(\pi\), with selected equations first, has blocks \[ \begin{pmatrix} (\partial f_j/\partial x_i)_{j\in E_k,i\in U_k}&0\\ *&a_k1_{m-|E_k|} \end{pmatrix}. \] Its determinant is \(\delta_k a_k^{m-|E_k|}\), invertible at \(\mathfrak q_k\). If \(\Delta_k\) denotes the determinant before reduction, \((D_k)_{a_k\Delta_k}\) is standard smooth: adjoining an inverse of \(a_k\Delta_k\) adds one equation and one pivot, whose block determinant is \((a_k\Delta_k)\Delta_k\). F.85 proves smoothness. F.83, with the Noetherian base \(R\), proves flatness of this smooth chart. Localization is exact by S01, Lemma 1.1, so \(T=(D_k)_{\mathfrak q_k}\) is flat over \(R\).

We now show that every additional relation of \(D_k\to D\) vanishes in \(T\). First (D.19b), with the \(k\)-relations zero and \(a_k\) invertible, gives \(F_\ell=0\) for all \(\ell\). For another chart \(k'\), set \[ b_{kk'\ell}^j=a_{k'}h_{k\ell}^j- \sum_{j'\in E_{k'}}h_{k'\ell}^{j'}h_{kj'}^j \quad(j\in E_k). \] Reduce (D.19a) modulo \(\pi^2\) and express \(a_ka_{k'}[\bar f_\ell]\) in the conormal basis indexed by \(E_k\) in the two possible orders. The difference is \(\sum_{j\in E_k}b_{kk'\ell}^j[\bar f_j]=0\). Basis independence therefore makes every coefficient zero in \(\bar C_{a_k}\). Equivalently, \[ b_{kk'\ell}^j\in ((f_j:j\in E_k)+\pi^2P)_{a_k}. \] Its image in \((D_k)_{a_k}\) belongs to \(\pi(D_k)_{a_k}\), since every selected \(f_j\) equals \(\pi z_j\). Directly expanding the definitions, modulo \(\pi\), gives \[ a_kp_{k'\ell}-a_{k'}p_{k\ell} +\sum_{j'\in E_{k'}}h_{k'\ell}^{j'}p_{kj'} =\sum_{j\in E_k}b_{kk'\ell}^jz_j. \] Consequently \(p_{k'\ell}\in\pi T\). Identity (D.19b) for \(k'\) also gives \(\pi p_{k'\ell}=0\) in \(T\).

Flat tensor applied to the multiplication-kernel sequences identifies \[ \operatorname{Ann}_T(\pi^e) =\operatorname{im}(\operatorname{Ann}_R(\pi^e)\otimes_RT\to T) \qquad(e=1,2). \] Indeed, factor multiplication through its image and tensor its short exact kernel sequence and that image's injection into \(R\); flatness preserves both. The assumed equality over \(R\) thus gives equality of these annihilators in \(T\). For \(p=p_{k'\ell}=\pi u\), the equality \(\pi p=0\) means \(\pi^2u=0\); it follows that \(p=\pi u=0\). This kills all additional generators.

Finally this local calculation supplies genuine neighbourhoods. The kernel of \(D_k\to D\) is generated by the finite lists of additional \(F_\ell\) and \(p_{k'\ell}\). For each generator, vanishing at \(\mathfrak q_k\) gives an element outside \(\mathfrak q_k\) killing it. Their product \(s\) kills the entire kernel. The surjection therefore becomes an isomorphism after inverting \(a_k\Delta_ks\), which still avoids \(\mathfrak q_k\). This identifies a principal neighbourhood of \(\mathfrak q\) with a localization of the standard smooth chart, proving the required smoothness of \(R\to D\) by F.85 and F.7. Modulo \(\pi\), (D.19c) identifies the same neighbourhood with a localization of a polynomial algebra over \(\bar C/\pi\bar C\): denominators in the \(a_k\)-localized polynomial algebra can be cleared by a power of \(a_k\). F.85 proves its smoothness over that base. This gives the second claimed smoothness assertion as well. Empty equation or chart lists use determinant and denominator product \(1\); the same constructions cover zero algebras. ∎

The desingularization lemma

D.20 combines a lifting witness with extra equations to enlarge the singular ideal in the target.

Lemma D.20: desingularize

Reading source: Stacks, Tag 07CR.

Let \(R\) be Noetherian, let \(\pi\in R\), and let \(\Lambda\) be an \(R\)-algebra with \(\operatorname{Ann}_\Lambda(\pi)=\operatorname{Ann}_\Lambda(\pi^2)\). Suppose \(A\) is finitely presented over \(R\), a map \(A\to\Lambda\) is given, the image of \(\pi\) is strictly standard in \(A/R\), and a section \[ \rho:A/\pi^4A\longrightarrow R/\pi^4R \] agrees with the prescribed map to \(\Lambda/\pi^4\Lambda\). Then there is a factorization \(A\to B\to\Lambda\), with \(B/R\) finitely presented, such that \[ \mathfrak aB\subset H_{B/R},\qquad \mathfrak a=\operatorname{Ann}_R \bigl(\operatorname{Ann}_R(\pi^2)/\operatorname{Ann}_R(\pi)\bigr). \] No annihilator equality in \(R\) is assumed.

Proof. Choose the strict witness from D.3: \[ A=R[x_1,\ldots,x_n]/(f_1,\ldots,f_m),\quad 0\le c\le\min(n,m),\quad I=(f_1,\ldots,f_m). \] It gives a sum-of-minors identity for \(\pi\) and the relations \(\pi f_\ell\in(f_1,\ldots,f_c)+I^2\) for \(\ell>c\). Subtract lifts in \(R\) of the finitely many values \(\rho(x_i)\) from the variables. This translation is a polynomial automorphism, preserves both witness identities, and makes \(\rho(x_i)=0\). Hence \(f_j(0)\in\pi^4R\), and the actual image of \(x_i\) in \(\Lambda\) has the form \(\pi^4\lambda_i\).

Write the linear terms of \(f_j\) as \(\sum_i r_{ji}x_i\), and let \(M=(r_{ji})_{j\le c,i\le n}\). Evaluate the sum-of-minors identity at \(\rho\). There are coefficients \(d_U\in R\) and \(u\in1+\pi^3R\) such that \[ u\pi=\sum_{|U|=c}d_U\det(M_U). \] Apply F.70 to \(M^{\mathsf t}\) and transpose its resulting identity. This supplies an \(n\times c\) matrix \(S=(s_{ij})\) with \(MS=u\pi1_c\).

Introduce \(v_1,\ldots,v_c,w_1,\ldots,w_n\), set \(L_i=\sum_j s_{ij}v_j+\pi w_i\), and substitute \(x_i=\pi^2L_i\). For \(j\le c\), constant terms are divisible by \(\pi^4\), linear terms become \[ \pi^2\sum_i r_{ji}L_i =\pi^3u v_j+\pi^3\sum_i r_{ji}w_i, \] and every monomial of degree \(d\ge2\) becomes divisible by \(\pi^{2d}\). Dividing these specified powers in the coefficient expressions, rather than cancelling a ring element, constructs a polynomial \(G_j\) and an exact identity \[ f_j(\pi^2L)=\pi^3g_j,\qquad g_j=u v_j+\sum_i r_{ji}w_i+\pi G_j. \tag{D.20a} \] For example, a constant term \(\pi^4a_j\) contributes \(a_j\) to \(G_j\), and a degree-\(d\) term with coefficient \(b_\alpha\) contributes \(b_\alpha\pi^{2d-4}L^\alpha\). Thus this construction is valid even when \(\pi\) is a zero divisor.

Let \(h_i=x_i-\pi^2L_i\), and define \[ B=R[x,v,w]/(h_i,\ g_j:j\le c,\ f_\ell:1\le\ell\le m). \] Eliminating \(x_i\) using the \(h_i\) is an actual inverse substitution isomorphism. The two required maps send the old \(x_i\) to their classes in \(B\), and send \[ x_i\longmapsto\pi^4\lambda_i,\qquad v_j\longmapsto0,\qquad w_i\longmapsto\pi\lambda_i \] in \(\Lambda\). The \(h_i\) and \(f_\ell\) vanish there. Each \(g_j\) has image \(t\in\pi\Lambda\) by (D.20a) modulo \(\pi\), and satisfies \(\pi^3t=0\). The annihilator equality implies \(\operatorname{Ann}_\Lambda(\pi^e)=\operatorname{Ann}_\Lambda(\pi)\) for every \(e\ge1\): if \(\pi^e b=0\) for \(e\ge2\), apply the equality to \(\pi^{e-2}b\) and reduce the exponent successively. Writing \(t=\pi b\) now gives \(\pi^4b=0\), hence \(t=0\). All relations therefore vanish, and the factorization is exact.

After inverting \(\pi\), the \(h_i\) uniquely eliminate \(w_i\), and (D.20a) makes the \(g_j\) consequences of the \(f_j\). Consequently \[ B_\pi\cong A_\pi[v_1,\ldots,v_c]. \] D.5 makes \(A_\pi\) smooth over \(R\); a polynomial extension is smooth by F.85, and F.29 proves smoothness of \(B_\pi/R\).

For primes containing \(\pi\), first use the algebra \[ B'=R[v,w]/(g_1,\ldots,g_c). \] The Jacobian on the pivot variables \(v_j\) is \(u1_c+\pi(\partial G_j/\partial v_k)\), so its determinant \(\Delta\) is \(1\) modulo \(\pi\). Thus \(B'_\Delta\) is standard smooth by the inverse-variable Jacobian calculation in F.85. It is flat over the Noetherian base \(R\) by F.83. Notice also the explicit reduction \(B'/\pi B'\cong(R/\pi R)[w]\), obtained by eliminating \(v_j=-\sum_i r_{ji}w_i\).

The natural surjection \(B'\to B\) has kernel generated by \[ F_\ell=f_\ell(\pi^2L)\quad(\ell>c); \] the first \(c\) equations have already vanished by (D.20a). Let \(\mathfrak q\subset B\) contain \(\pi\) and avoid some \(r\in\mathfrak a\), and let \(\mathfrak q'\) be its inverse image in \(B'\). Then \(\Delta\notin\mathfrak q'\), and \(T=B'_{\mathfrak q'}\) is flat over \(R\), since it is a localization of that flat chart. Every \(F_\ell\) belongs to \(\pi^2T\): its substituted variables have this divisibility and its constant term belongs to \(\pi^4R\).

Substitute in the strict relation condition and use the first \(c\) equations already killed in \(B'\). It gives \[ \pi F_\ell=\sum_{a,b>c}b_{\ell ab}F_aF_b. \] For the finite \(T\)-module \(N=(\pi F_\ell:\ell>c)\), divisibility of \(F_a\) by \(\pi^2\) gives \(N\subset\pi N\); the reverse inclusion holds for every ideal. Since \(\pi\) lies in the maximal ideal of \(T\), F.10 gives \(N=0\). Thus \(\pi F_\ell=0\).

Flatness identifies each multiplication kernel and preserves their quotient sequence, giving the canonical isomorphism \[ \bigl(\operatorname{Ann}_R(\pi^2)/\operatorname{Ann}_R(\pi)\bigr) \otimes_RT \cong \operatorname{Ann}_T(\pi^2)/\operatorname{Ann}_T(\pi). \] For the kernel identification, tensor the short exact sequence through the image of multiplication by \(\pi^e\), and its injection into \(R\), for \(e=1,2\). Tensor their annihilator inclusion and quotient sequence next. The element \(r\) kills the left-hand module and is a unit in \(T\), so the quotient on the right is zero. Write \(F_\ell=\pi b_\ell\). From \(\pi F_\ell=0\) and the resulting annihilator equality in \(T\), we obtain \(F_\ell=0\).

To obtain a smooth neighbourhood rather than only an isomorphism of local rings, choose for each of the finitely many \(F_\ell\) an element of \(B'\setminus\mathfrak q'\) killing it; these exist by the localization equality criterion. Their product \(s\) kills the whole kernel. Hence \(B'_{\Delta s}\to B_{\Delta s}\) is an isomorphism, and \(\Delta s\) avoids \(\mathfrak q\). This is a localization of a standard smooth algebra, so F.85 and F.7 prove smoothness of \(R\to B\) at \(\mathfrak q\).

Every prime avoiding \(\pi\) is smooth by the earlier description of \(B_\pi\). Every other prime avoiding \(\mathfrak aB\) has some such \(r\) and is smooth by the chart just constructed. Thus every nonsmooth prime contains \(\mathfrak aB\). The radical/prime intersection formula proved in D.1, applied to \(H_{B/R}\), gives \(\mathfrak aB\subset H_{B/R}\). Empty lists use determinant and product \(1\), so the argument includes \(c=0\), \(c=m\), and zero rings. ∎

Lemma D.21: desingularize strictly standard

Reading source: Stacks, Tag 07CT.

Let \(R\) be Noetherian, let \(\pi\in R\), and let \(A\to\Lambda\), \(D\to\Lambda\) be \(R\)-algebra maps, with \(A/R\) and \(D/R\) finitely presented. Assume \[ \operatorname{Ann}_R(\pi)=\operatorname{Ann}_R(\pi^2),\qquad \operatorname{Ann}_\Lambda(\pi)=\operatorname{Ann}_\Lambda(\pi^2), \] that \(\pi\) is strictly standard in \(A/R\), and that a map \(A/\pi^4A\to D/\pi^4D\) agrees with both prescribed maps to \(\Lambda/\pi^4\Lambda\). There are compatible maps \(A\to B\), \(D\to B\), \(B\to\Lambda\), with \(B/R\) finitely presented, such that \[ H_{D/R}B\subset H_{B/D}, \qquad H_{D/R}B\subset H_{B/R}. \]

Proof. Let \(C=A\otimes_RD\), with map to \(\Lambda\) sending \(a\otimes d\) to the product of their prescribed images. A finite presentation \(A=R[x]/(f)\) gives \(C=D[x]/(f)\), so \(C/D\) is finitely presented. The strict witness changes coefficients to \(D\) by D.7. The given map \(\phi:A/\pi^4A\to D/\pi^4D\) gives the section \[ C/\pi^4C\longrightarrow D/\pi^4D,\qquad a\otimes d\longmapsto\phi(\bar a)\bar d. \] The tensor relations make this well-defined; on the \(D/\pi^4D\) factor it is the identity. Its composition to \(\Lambda/\pi^4\Lambda\) is the original product map by the assumed compatibility.

The ring \(D\) is Noetherian by S01, Lemma 1.3. Apply D.20 with base \(D\) and algebra \(C\). Its hypotheses require the annihilator equality in \(\Lambda\), already assumed, and require no such equality in \(D\). We obtain \(C\to B\to\Lambda\), with \(B/D\) finitely presented, and \[ \mathfrak aB\subset H_{B/D},\qquad M=\operatorname{Ann}_D(\pi^2)/\operatorname{Ann}_D(\pi),\quad \mathfrak a=\operatorname{Ann}_D(M). \] The two tensor-factor maps give \(A\to B\) and \(D\to B\), with the desired exact composites to \(\Lambda\). If \(D=R[u]/(g)\) and \(B=D[v]/(h)\), choose polynomial lifts \(\widetilde h\) in \(R[u,v]\). Substitution in both directions gives \(B=R[u,v]/(g,\widetilde h)\), proving finite presentation over \(R\).

Fix a prime \(\mathfrak q\subset D\) where \(R\to D\) is smooth. A smooth principal neighbourhood is flat over the Noetherian ring \(R\) by F.83; exact localization S01, Lemma 1.1 makes \(D_{\mathfrak q}\) flat as well. Tensor the multiplication-kernel sequences for \(\pi,\pi^2\) as in D.20. The equality of the original annihilators over \(R\) gives \[ \operatorname{Ann}_{D_{\mathfrak q}}(\pi) =\operatorname{Ann}_{D_{\mathfrak q}}(\pi^2),\qquad M_{\mathfrak q}=0. \] The module \(M\) is finite: its two annihilator ideals are finite over the Noetherian ring \(D\), and \(M\) is their quotient. Choose generators \(m_i\). For each, vanishing at \(\mathfrak q\) gives \(s_i\notin\mathfrak q\) with \(s_im_i=0\). Their product \(s\) kills every generator; hence \(s\in\mathfrak a\setminus\mathfrak q\). For \(M=0\), use \(s=1\).

At a prime \(\mathfrak r\subset B\) above \(\mathfrak q\), this element remains outside \(\mathfrak r\). Since \(s\in\mathfrak aB\subset H_{B/D}\), D.1 shows that \(D\to B\) is smooth at \(\mathfrak r\). To verify smooth composition at the level of actual neighbourhoods, choose \(t\in D\setminus\mathfrak q\) with \(D_t/R\) smooth. Inside a smooth \(D\)-neighbourhood of \(\mathfrak r\), F.83 supplies a principal neighbourhood \(B_g\) that is standard smooth over \(D\). The denominator conversion in its proof makes \(g\) an element of \(B\setminus\mathfrak r\). Base change to \(D_t\) preserves the standard smooth presentation and its unit determinant by F.85(5), so \(B_{tg}/D_t\) is smooth. F.29 now makes \(B_{tg}/R\) smooth. Thus \(R\to B\) is smooth at every such \(\mathfrak r\).

An element of \(H_{D/R}\) avoided by a prime of \(B\) is avoided by its contraction in \(D\), so that contraction is smooth. We have just proved that the prime of \(B\) is smooth for both \(B/D\) and \(B/R\). Consequently every nonsmooth prime for either map contains \(H_{D/R}B\). Intersecting these primes and using the radical/prime formula in D.1 proves both claimed ideal inclusions. Empty singular loci and zero algebras use the same convention that the intersection of no primes is the unit ideal. ∎

Lemma D.22: desingularize lifting apply

Reading source: Stacks, Tag 07F0.

Let \(R\) be Noetherian, let \(\pi\in R\), and let \(A\to\Lambda\) be an \(R\)-algebra map with \(A/R\) finitely presented. Assume \[ \operatorname{Ann}_R(\pi)=\operatorname{Ann}_R(\pi^2),\qquad \operatorname{Ann}_\Lambda(\pi)=\operatorname{Ann}_\Lambda(\pi^2), \] and that \(\pi\) is strictly standard in \(A/R\). Given a factorization \[ A/\pi^8A\longrightarrow\bar C\longrightarrow\Lambda/\pi^8\Lambda \] with \(\bar C\) finitely presented over \(R/\pi^8R\), there is a factorization \(A\to B\to\Lambda\), with \(B/R\) finitely presented, such that \(B_\pi/R_\pi\) is smooth and \[ H_{\bar C/(R/\pi^8R)}(\Lambda/\pi^8\Lambda) \subset \bigl(\sqrt{H_{B/R}\Lambda}+\pi^8\Lambda\bigr)/\pi^8\Lambda. \]

Proof. If multiplication by \(u\) on a module has \(\ker u=\ker u^2\), its kernels for all positive powers agree. Indeed \(u^{r+1}v=0\) puts \(u^{r-1}v\) in \(\ker u^2=\ker u\), hence \(u^rv=0\); repeating lowers the exponent to one. Thus the annihilators of \(\pi^4\) and \(\pi^8\) agree in both \(R\) and \(\Lambda\).

Use D.19 with parameter \(\pi^4\) and algebra \(\bar C\). It constructs \(D/R\) finitely presented, a map \(D\to\Lambda\), and commuting maps \[ \bar C/\pi^4\bar C\longrightarrow D/\pi^4D \longrightarrow\Lambda/\pi^4\Lambda. \] It proves smoothness of \(D/R\) away from \(\pi\), and also at a prime over \(\pi\) if its contraction to \(\bar C\) is smooth over \(R/\pi^8R\). Compose the reduced map from \(A\) with this diagram. The resulting \(A/\pi^4A\to D/\pi^4D\) agrees with the prescribed maps to \(\Lambda/\pi^4\Lambda\). Therefore D.21 constructs compatible maps \(A\to B\), \(D\to B\), \(B\to\Lambda\), with \(B/R\) finitely presented and \[ H_{D/R}B\subset H_{B/R}. \tag{D.22a} \] If a prime \(\mathfrak b\subset B\) avoids \(\pi\), its contraction to \(D\) is smooth and hence avoids \(H_{D/R}\). Inclusion (D.22a) makes \(\mathfrak b\) avoid \(H_{B/R}\) as well, so \(B/R\) is smooth at that prime by D.1. Its smooth principal neighbourhood can be localized further at \(\pi\) by F.67. Viewed over \(R_\pi\), it is smooth: a polynomial presentation over \(R\) in which \(\pi\) is invertible localizes to the same presentation over \(R_\pi\). Exact localization S01, Lemma 1.1 identifies its kernel ideal and conormal quotient with their localizations. Since \(\pi\) is already invertible on the presented algebra, its two conormal terms and their differential are unchanged. Definition F.6 and comparison F.11 give the assertion. Every prime of \(B_\pi\) has such a neighbourhood, and F.67 gives global smoothness over \(R_\pi\).

Set \(J=H_{B/R}\Lambda\). For each prime \(\mathfrak l\subset\Lambda\) containing \(J\), let \(\mathfrak b,\mathfrak d\) be its contractions to \(B,D\). They contain \(H_{B/R},H_{D/R}\), respectively, so both contracted points are nonsmooth. The preceding conclusion forces \(\pi\in\mathfrak l\). The prime \(\mathfrak d/\pi^4D\) now contracts along the quotient diagram to a prime \(\mathfrak c\) of \(\bar C\). If this were smooth, D.19 would make \(\mathfrak d\) smooth, a contradiction. Consequently \(H_{\bar C/(R/\pi^8R)}\subset\mathfrak c\).

The contraction \(\mathfrak c\) is also the inverse image of \(\mathfrak l/\pi^8\Lambda\) under \(\bar C\to\Lambda/\pi^8\Lambda\). To check this equality, reduce that map modulo \(\pi^4\) and use the commuting diagram through \(D/\pi^4D\); both quotient primes are the contractions of the same \(\mathfrak l\), which contains \(\pi\). Thus the ideal generated by the images of \(H_{\bar C/(R/\pi^8R)}\) lies in every such \(\mathfrak l/\pi^8\Lambda\).

For an explicit radical argument, let \(K\subset\Lambda\) be the inverse image of that generated ideal. The preceding paragraph and \(\pi\in\mathfrak l\) give \(K\subset\mathfrak l\) for every prime containing \(J\). The prime-intersection proof in D.1 yields \(K\subset\sqrt J\), as well as \(\pi\in\sqrt J\). In particular \(\pi^8\Lambda\subset\sqrt J\); taking the image of \(K\subset\sqrt J\) in \(\Lambda/\pi^8\Lambda\) is precisely the claimed inclusion. If no prime contains \(J\), D.1 gives \(\sqrt J=\Lambda\), and the same argument holds. This proves the result for an arbitrary target \(\Lambda\), including zero quotients. ∎

Warmup: reduction to a base field

The preceding lifting constructions allow D.26 to reduce the smooth-colimit theorem to a field base. We fix the global assumptions before proving that reduction.

Situation D.23: global

Reading source: AI Integrated Stacks, pinned smoothing, lines 1915–1918.

Throughout the reduction to a field, $R$ and $\Lambda$ are Noetherian and the map $R\to\Lambda$ is regular. For this fixed map, the abbreviation PT holds means that there is a filtered system of smooth $R$-algebras with colimit $\Lambda$ as an $R$-algebra.

Lemma D.24: product

Reading source: Stacks, Tag 07F3.

For \(i=1,2\), let \(R_i\to\Lambda_i\) be a regular map of Noetherian rings as in D.23. If each \(\Lambda_i\) is a filtered colimit of smooth \(R_i\)-algebras, then \(\Lambda_1\times\Lambda_2\) is a filtered colimit of smooth \(R_1\times R_2\)-algebras.

Proof. Put \(R=R_1\times R_2\), with orthogonal idempotents \(e_1=(1,0)\), \(e_2=(0,1)\). For any \(R\)-module \(M\), the maps \[ M\longrightarrow e_1M\times e_2M,\quad m\longmapsto(e_1m,e_2m), \qquad (m_1,m_2)\longmapsto m_1+m_2 \] are inverse, since \(e_1+e_2=1\) and \(e_1e_2=0\). Tensor products split by the same rule: \[ M\otimes_R(\Lambda_1\times\Lambda_2) \cong(e_1M\otimes_{R_1}\Lambda_1)\times (e_2M\otimes_{R_2}\Lambda_2). \] The maps send a pure tensor to its two component tensors; summing the inclusions gives the inverse, and cross tensors vanish because the idempotent can be moved to either factor. This verifies the formula by S01, Lemma 1.0. Exact sequences split into their two component sequences, so component flatness proves flatness of the product map.

Every ideal in a finite product splits in this fashion into two ideals. Finite generator lists for those component ideals give a finite generator list for the original ideal. Hence \(R\) and \(\Lambda_1\times\Lambda_2\) are Noetherian. A prime contains exactly one of \(e_1,e_2\): it contains at least one since their product is zero, and cannot contain both since their sum is one. It therefore comes from exactly one component, with that component's residue field. Tensoring with this residue field kills the other component and identifies the fibre with the corresponding fibre of \(R_i\to\Lambda_i\). Geometric regularity of these fibres proves regularity of the product map, so it is again in Situation D.23.

We prove smoothness of each product algebra used below. Let \(C_i/R_i\) be smooth and let \(C=C_1\times C_2\). Then \(C_{e_i}=C_i\), with inverse maps given by projection and inclusion of the selected component; the other component vanishes when \(e_i\) is made invertible. Also \(R_{e_i}=R_i\). Indeed inverting an idempotent forces it to equal one, so this localization is the quotient \(R/(1-e_i)\). The inverse-variable presentation \(R[T]/(e_iT-1)\), whose derivative \(e_i\) is a unit, proves \(R_i/R\) smooth by F.85. Thus F.29 makes \(C_i/R\) smooth. The two elements \(e_i\) generate one in \(C\), so F.35 gives finite presentation of \(C/R\) from that of the \(C_{e_i}/R\). At every prime one \(e_i\) is avoided; the corresponding localization is smooth over \(R\). The full gluing result F.67 therefore proves \(C/R\) smooth.

Choose the assumed presentations \[ \Lambda_i=\operatorname*{colim}_{a\in I_i}C_{i,a}. \] The category \(I_1\times I_2\) is filtered: it is nonempty, successors of two objects are chosen in each coordinate, and parallel arrows are equalized in each coordinate. The preceding argument makes every \(C_{1,a}\times C_{2,b}\) smooth over \(R\). Component maps induce \[ \operatorname*{colim}_{(a,b)\in I_1\times I_2} (C_{1,a}\times C_{2,b})\longrightarrow\Lambda_1\times\Lambda_2. \] Any pair on the right has representatives at a pair of stages, proving surjectivity. If a represented pair maps to zero, each component becomes zero at a later stage by the filtered equality criterion. The pair of these stages kills the represented pair, proving injectivity. The same criterion applied to differences proves equality detection, and the maps preserve sums, products and units componentwise. The displayed map is consequently an \(R\)-algebra isomorphism. Iterating gives the same conclusion for any finite nonempty product; zero components obey these identical idempotent calculations. ∎

Lemma D.25: delocalize base

Reading source: Stacks, Tag 07F4.

Let \(R\to A\to\Lambda\) be ring maps with \(A/R\) finitely presented, and let \(S\subset R\) be multiplicative. Suppose \(S^{-1}A\to B'\to S^{-1}\Lambda\) is a factorization with \(B'/S^{-1}R\) smooth. Then the original map factors as \(A\to B\to\Lambda\), with \(B/R\) finitely presented and the image of some \(s\in S\) elementary standard in \(B/R\) in the sense of D.3.

Proof. If \(0\in S\), take \(B=A\) and \(s=0\). In its finite presentation choose \(c=0\): the empty minor is one with coefficient zero, and multiplying every extra equation by zero gives zero. This verifies both elementary standard conditions. Hence assume \(0\notin S\).

The retraction D.12 replaces \(B'\) by a standard smooth algebra: compose the original map into its retract inclusion, and compose its retraction with the original target map. These composites preserve the original factorization. Write \(A=R[x_1,\ldots,x_n]/(g_1,\ldots,g_t)\), with coordinate images \(\lambda_i\in\Lambda\). The graph-equation construction D.14, applied to this ordered list of images, gives \[ B'=S^{-1}R[x_1,\ldots,x_{n+m}]/(f_1,\ldots,f_c), \qquad n\le c\le n+m, \] with the first \(n\) variables representing the given images and with the minor on its first \(c\) variables invertible. Repeated images are permitted: each graph equation has its own variable and the same identity Jacobian block.

For \(i>n\), write the image of the old coordinate \(x_i\) in \(S^{-1}\Lambda\) as \(\lambda_i/d_i\), with \(d_i\in S\). Introduce the new coordinate \(y_i=d_ix_i\), leaving \(y_i=x_i\) for \(i\le n\). Express each old equation in the new coordinates by substituting \(x_i=d_i^{-1}y_i\). The reverse substitution is \(y_i=d_ix_i\), so these changes give inverse polynomial-algebra maps over \(S^{-1}R\). The new coordinates map to \(\lambda_i/1\). Differentiating the substituted equations multiplies the column for an added selected variable by \(d_i^{-1}\); the determinant remains a unit. Clear the finite coefficient denominators in each equation by multiplying it by \(u_j\in S\). This does not change the localized relation ideal and multiplies the determinant by the unit \(\prod_j u_j\). Rename the new coordinates \(x_i\) and the scaled equations \(f_j\). Now \[ P=R[x_1,\ldots,x_{n+m}],\qquad f_j\in P,\qquad \Delta=\det(\partial f_j/\partial x_i)_{1\le j,i\le c} \] has invertible image in \(B'\).

A polynomial representative of its inverse gives \(b_0,\ldots,b_c\in S^{-1}P\) with \(1=b_0\Delta+\sum_j b_jf_j\). Choose \(d\in S\) and \(c_j\in P\) representing all \(db_j\). The polynomial \(d-c_0\Delta-\sum_jc_jf_j\) is zero after localization; some \(e\in S\) therefore annihilates it in \(P\). With \(s_0=ed\) and \(a_j=ec_j\), we have the exact identity \[ s_0=a_0\Delta+\sum_{j=1}^c a_jf_j\quad\text{in }P. \tag{D.25a} \] The annihilating multiplier is required even when \(P\) has \(S\)-torsion; no element of \(S\) has been cancelled in \(P\).

Each original relation \(g_k\) is zero in \(S^{-1}(P/(f_1,\ldots,f_c))\), so choose \(s_k\in S\) and \(v_{kj}\in P\) with \(s_kg_k=\sum_jv_{kj}f_j\). Each \(f_j(\lambda)\) is zero in \(S^{-1}\Lambda\), so choose \(t_j\in S\) with \(t_jf_j(\lambda)=0\) in \(\Lambda\). Set \[ h_j=t_jf_j,\quad B=P/(h_1,\ldots,h_c,g_1,\ldots,g_t),\quad s=s_0\prod_{k=1}^t s_k\prod_{j=1}^c t_j. \] These finite relations give finite presentation. The original coordinates define \(A\to B\); evaluation of all coordinates at \(\lambda_i\) defines \(B\to\Lambda\), since it kills both relation lists. Their composite is the given map.

Here is the full witness for elementary standardness. For each \(k\), multiply its relation identity by \(s_0\prod_{\ell\ne k}s_\ell\prod_jt_j\). Replacing \(t_jf_j\) by \(h_j\) gives \[ sg_k=\sum_{j=1}^c s_0\Bigl(\prod_{\ell\ne k}s_\ell\Bigr) \Bigl(\prod_{q\ne j}t_q\Bigr)v_{kj}h_j \quad\text{in }P. \] Thus \(sg_k\in(h_1,\ldots,h_c)\), which implies D.3's extra-relation condition even without its allowed \(I^2\) term. Multiplication of (D.25a) by \(\prod_jt_j\) gives \[ s_0\prod_jt_j =a_0\det(\partial h_j/\partial x_i)_{1\le j,i\le c} +\sum_{j=1}^c a_j\Bigl(\prod_{q\ne j}t_q\Bigr)h_j. \] The determinant equality follows by factoring \(t_j\), a base scalar, from equation row \(j\). In \(B\), the second sum is zero. Multiplying by \(\prod_ks_k\) expresses the image of \(s\) as this selected minor times the image of \(a_0\prod_ks_k\), exactly D.3's other condition. Every product with an empty index set is one. This proves the assertion also for empty equation lists and zero target algebras. ∎

Lemma D.26: reduce to field

Reading source: Stacks, Tag 07F5.

If PT holds for every regular map \(K\to\Lambda\) of Noetherian rings with \(K\) a field, then PT holds for every regular map \(R\to\Lambda\) of Noetherian rings, as in D.23.

Proof. Suppose a map fails PT. Each quotient \(R/I\to\Lambda/I\Lambda\) is regular by F.123; its rings are Noetherian because ideals in a quotient lift to ideals in the original ring. The set of failing quotient ideals is nonempty and has a maximal member by the ascending-chain condition. Replace the map by that quotient. Hence every further nonzero ideal gives PT. The new \(R\) is nonzero: a unital map from the zero ring has zero target, which is already a smooth algebra over itself and a constant filtered colimit.

The nilradical \(N\) is finitely generated because \(R\) is Noetherian. If its generators satisfy \(a_i^{n_i}=0\), every monomial of degree \(1+\sum_i(n_i-1)\) contains some \(a_i^{n_i}\), so that power of \(N\) is zero. If \(N\ne0\), maximality gives PT after quotienting by \(N\), and flatness of the original regular map allows D.18 to lift PT back to \(R\). This is a contradiction. Thus \(R\) is reduced.

We supply the finite-prime decomposition needed for its total fraction ring. Every radical ideal of a Noetherian ring is a finite intersection of prime ideals. Otherwise take a maximal counterexample \(I\) by the ascending-chain condition. It is proper and not prime, so there are \(a,b\notin I\) with \(ab\in I\). Both \(\sqrt{I+(a)}\) and \(\sqrt{I+(b)}\) are strictly larger and hence finite intersections of primes (the unit ideal is the empty intersection). Their intersection is \(I\): if \(x\) lies in both, then \(x^m\in I+(a)\) and \(x^n\in I+(b)\) for some positive \(m,n\), whence \(x^{m+n}\in I+(ab)=I\); radicalness gives \(x\in I\). Combining the two finite lists contradicts the choice of \(I\).

Apply this to the radical zero ideal of the nonzero reduced ring, and remove repeated primes and any prime containing another in the list. We obtain \[ 0=\bigcap_{i=1}^r\mathfrak p_i \] with a nonempty, pairwise incomparable list. These are exactly the minimal primes. Indeed the product of the listed ideals is contained in their intersection, so any prime over zero contains at least one listed prime: otherwise choose an element from each \(\mathfrak p_i\) outside it, whose product contradicts primality. If a prime lies strictly inside \(\mathfrak p_i\), the same observation puts a listed \(\mathfrak p_j\) inside it, contradicting incomparability. This proves finiteness, minimality and the zero intersection directly.

For each \(i\), choose \[ y_i\in\bigcap_{j\ne i}\mathfrak p_j\setminus\mathfrak p_i \] by multiplying one element of \(\mathfrak p_j\setminus\mathfrak p_i\) for each \(j\ne i\); for \(r=1\) take \(y_1=1\). An element outside all \(\mathfrak p_i\) is a nonzerodivisor: test an annihilated product in the domains \(R/\mathfrak p_i\), then use their zero intersection. Conversely \(a\in\mathfrak p_i\) satisfies \(ay_i=0\), while \(y_i\ne0\), and is a zerodivisor. Hence the multiplicative set of nonzerodivisors is \(S=R\setminus\bigcup_i\mathfrak p_i\).

Let \(d=\sum_i y_i\in S\), \(Q=S^{-1}R\), and \(e_i=y_i/d\). Fractions define an injection \[ Q\longrightarrow\prod_i\operatorname{Frac}(R/\mathfrak p_i): \] the denominators are nonzero in every component, and a fraction in its kernel has numerator in their zero intersection. In component \(i\), \(e_i\) equals one; in all other components it equals zero. Therefore \(e_i^2=e_i\), \(e_ie_j=0\) for \(i\ne j\), and \(\sum_i e_i=1\). Projection by these idempotents and summation give inverse maps \(Q\cong\prod_i e_iQ\).

Each \(K_i=e_iQ\) is a field. A nonzero \(e_i a/s\) has \(a\notin\mathfrak p_i\). The element \[ u=e_i a+(1-e_i) =\frac{y_i a+\sum_{j\ne i}y_j}{d} \] is a unit of \(Q\): its numerator avoids \(\mathfrak p_i\) because \(y_i a\) does, and avoids every other \(\mathfrak p_j\) because only \(y_j\) survives there. Its numerator is thus in \(S\), giving the fraction inverse. The inverse of \(e_i a/s\) inside \(e_iQ\) is \(e_i s u^{-1}\). This proves the claimed field decomposition with actual inverse ring maps.

The localization \(Q\to\Lambda_Q=S^{-1}\Lambda\) is a regular map of Noetherian rings. To verify Noetherianness of either localization, extend the contraction of an ideal: each fraction in the ideal has its numerator in that contraction, and finite generators of the contraction generate the localized ideal. For a \(Q\)-module \(M\), the tensor maps of S01, Lemma 1.0 give \[ M\otimes_Q\Lambda_Q\cong M\otimes_R\Lambda. \] Here elements of \(S\) are already invertible on \(M\), so the maps on pure tensors and their inverse respect the fraction relations. Exactness of the right side gives flatness. For a prime \(\mathfrak q\subset Q\), its contraction \(\mathfrak p\subset R\) avoids \(S\), and \(\mathfrak q=S^{-1}\mathfrak p\): a fraction belongs to \(\mathfrak q\) exactly when its numerator does, because its denominator is a unit. Conversely a prime disjoint from \(S\) extends to a prime, since \(S^{-1}(R/\mathfrak p)\) is a domain. The quotient \(Q/\mathfrak q\) is this localized domain and its fraction field is \(\operatorname{Frac}(R/\mathfrak p)\), so its residue field equals \(\kappa(\mathfrak p)\). Tensor associativity from S01, Lemma 1.0 identifies its fibre with \(\Lambda\otimes_R\kappa(\mathfrak p)\), geometrically regular by hypothesis. The idempotents split \(\Lambda_Q=\prod_i e_i\Lambda_Q\). Each component \(K_i\to e_i\Lambda_Q\) is flat by the same component tensor calculation in D.24, and has the corresponding regular fibre. Its rings are Noetherian, since it is a quotient of the respective localization. The assumed field case and the finite-product conclusion of D.24 give PT for \(Q\to\Lambda_Q\).

Take any finitely presented \(R\)-algebra \(A\to\Lambda\). Localizing its finite generator and relation lists gives a finitely presented \(Q\)-algebra. Criterion F.90 factors it through a smooth \(Q\)-algebra. By D.25 there is \(A\to A_1\to\Lambda\), with \(A_1/R\) finitely presented and \(\pi\in S\) elementary standard in \(A_1/R\). It is strictly standard by D.3: put its elementary coefficient at the selected variable set and put all other minor coefficients equal to zero.

Since \(R\ne0\) and multiplication by \(\pi\) is injective, \(\pi^8\ne0\). Maximality of the failing quotient therefore gives PT for \(R/\pi^8R\to\Lambda/\pi^8\Lambda\), including when this quotient is zero. Apply F.90 to obtain \[ A_1/\pi^8A_1\longrightarrow\bar C\longrightarrow\Lambda/\pi^8\Lambda \] with \(\bar C/(R/\pi^8R)\) smooth. All powers of \(\pi\) act injectively on \(R\), and flatness makes them injective on \(\Lambda\) by tensoring \(0\to R\xrightarrow{\pi^n}R\), using S01, Lemma 1.0. Thus both annihilator equalities hold for parameters \(\pi\) and \(\pi^4\).

Apply D.19 with parameter \(\pi^4\). It supplies \(D/R\) finitely presented, \(D\to\Lambda\), and the compatible map \[ A_1/\pi^4A_1\longrightarrow\bar C/\pi^4\bar C \longrightarrow D/\pi^4D\longrightarrow\Lambda/\pi^4\Lambda. \] Every prime of \(D\) avoiding \(\pi\) is smooth by that lemma. Every prime containing \(\pi\) contracts to a smooth point of \(\bar C\), since the whole map \(\bar C/(R/\pi^8R)\) is smooth, and is therefore smooth as well. The actual neighbourhood criterion F.67 makes \(D/R\) globally smooth.

Finally D.21, applied to \(A_1,D,\pi\) and the displayed quotient map, constructs \(A_1\to B\to\Lambda\), with \(B/R\) finitely presented and \(H_{D/R}B\subset H_{B/R}\). Definition D.1 gives \(H_{D/R}=D\), so \(H_{B/R}=B\). Thus every prime of \(B\) is smooth, and F.67 makes \(B/R\) smooth. Composing with \(A\to A_1\) gives the required smooth factorization of every finitely presented input. F.90 now gives PT for the supposedly failing map, a contradiction. Consequently no failing quotient exists and PT holds for the original map. ∎

Local tricks

Situation D.27: local

Reading source: Stacks, Tag 07F7.

Fix a Noetherian ring \(R\), an \(R\)-algebra \(A\) of finite presentation, an \(R\)-algebra map \(A\to\Lambda\), and a prime \(\mathfrak q\) of \(\Lambda\). The target need not be Noetherian. For any intermediate \(R\)-algebra \(T\) with a specified map to \(\Lambda\), write \[ \mathfrak h_T=\sqrt{H_{T/R}\Lambda}, \] where \(H_{T/R}\) is the singular ideal of D.1 and its extension is generated by its images in \(\Lambda\).

A resolution at \(\mathfrak q\) is a factorization \(A\to B\to\Lambda\) through an \(R\)-algebra \(B\) of finite presentation for which \[ \mathfrak h_A\subset\mathfrak h_B, \qquad \mathfrak h_B\not\subset\mathfrak q. \] We say that the specified local problem is resolvable when such a factorization exists. Both ideals in this condition belong to the same target ring \(\Lambda\); finite presentation of \(B\) is over the fixed base \(R\). The second condition means that some element of \(\mathfrak h_B\) lies outside \(\mathfrak q\), while the first requires preservation of the entire original singular radical.

Lemma D.28: lift solution

Reading source: Stacks, Tag 07F8.

Work in D.27. Let \(r\ge1\), and choose \(\pi_1,\ldots,\pi_r\in R\) whose images lie in \(\mathfrak q\). For each \(i\), suppose that \(\pi_i\) is strictly standard in \(A/R\), and that multiplication by \(\pi_i\) and by \(\pi_i^2\) has the same kernel on each of \[ R/(\pi_1^8,\ldots,\pi_{i-1}^8)R, \qquad \Lambda/(\pi_1^8,\ldots,\pi_{i-1}^8)\Lambda. \] If the problem obtained by quotienting \(R,A,\Lambda\) by all \(\pi_i^8\), at the image of \(\mathfrak q\), is resolvable, then the original problem is resolvable.

Proof. First establish a base-change inclusion used in the single-element step. If \(R\to R'\) is any map and \(A'=A\otimes_R R'\), then \[ H_{A/R}A'\subset H_{A'/R'}. \tag{D.28a} \] Indeed, a prime of \(A'\) whose contraction is smooth has a standard smooth principal neighbourhood after taking a smooth neighbourhood of that contraction and applying F.83. The coefficient base-change construction F.85(5) makes its inverse image standard smooth over \(R'\). Thus every nonsmooth prime of \(A'\) contracts to a nonsmooth prime of \(A\), and contains the images of \(H_{A/R}\). The prime-intersection verification in D.1 gives (D.28a), including an empty nonsmooth locus. No flatness was required.

Consider \(r=1\), and abbreviate \(\pi=\pi_1\). Put \(\bar R=R/\pi^8R\), \(\bar A=A/\pi^8A\), \(\bar\Lambda=\Lambda/\pi^8\Lambda\), and \(\bar{\mathfrak q}=\mathfrak q/\pi^8\Lambda\). The assumed resolution provides \(\bar A\to\bar C\to\bar\Lambda\), with \(\bar C/\bar R\) finitely presented, such that \[ \sqrt{H_{\bar A/\bar R}\bar\Lambda} \subset\sqrt{H_{\bar C/\bar R}\bar\Lambda} \not\subset\bar{\mathfrak q}. \tag{D.28b} \] The two kernel equalities and strict standardness are exactly the hypotheses of D.22. Let \(A\to B\to\Lambda\) be its factorization, and set \(\mathfrak h_B=\sqrt{H_{B/R}\Lambda}\). The proof of D.22 shows that every prime containing \(H_{B/R}\Lambda\) contains \(\pi\); D.1 consequently gives \(\pi\in\mathfrak h_B\). Its other conclusion is \[ H_{\bar C/\bar R}\bar\Lambda \subset \mathfrak h_B/\pi^8\Lambda. \tag{D.28c} \] The ideal on the right is radical. To check this directly, lift a class whose power belongs to it to \(x\in\Lambda\). Then \(x^a\in\mathfrak h_B+\pi^8\Lambda=\mathfrak h_B\), so \(x\in\mathfrak h_B\). Taking radicals in (D.28c) is therefore legitimate.

Write \(\rho:\Lambda\to\bar\Lambda\) for the quotient. Inclusion (D.28a) gives \[ \rho(\mathfrak h_A) \subset\sqrt{H_{\bar A/\bar R}\bar\Lambda} \subset\mathfrak h_B/\pi^8\Lambda: \] an element of \(\mathfrak h_A\) has a power in \(H_{A/R}\Lambda\), whose image belongs to \(H_{\bar A/\bar R}\bar\Lambda\). Since \(\ker\rho=\pi^8\Lambda\subset\mathfrak h_B\), this proves \(\mathfrak h_A\subset\mathfrak h_B\). If \(\mathfrak h_B\subset\mathfrak q\), then (D.28c), after taking radicals, would put \(\sqrt{H_{\bar C/\bar R}\bar\Lambda}\) inside \(\bar{\mathfrak q}\), contradicting (D.28b). Thus \(\mathfrak h_B\not\subset\mathfrak q\). D.22 supplies the required finite presentation of \(B/R\), so this is a resolution with both radical conditions verified.

For \(r>1\), quotient first by \(\pi_1^8\). The base \(R'=R/\pi_1^8R\) is Noetherian: any ascending chain of its ideals lifts to one in \(R\). The finite polynomial presentation of \(A\) changes coefficients to give the finite presentation of \(A'=A/\pi_1^8A\) over \(R'\). The image \(\mathfrak q'\) is prime because \(\pi_1^8\Lambda\subset\mathfrak q\). For the remaining elements \(\pi_2,\ldots,\pi_r\), each successive quotient is precisely the original quotient by \(\pi_1^8,\ldots,\pi_{i-1}^8\); hence its two kernel hypotheses are unchanged. Strict standardness survives by D.7. The induction hypothesis gives a resolution of this first quotient problem. Apply the single-element step just proved to \(\pi_1\) to lift it to the original problem. This completes the induction, with no Noetherian assumption on \(\Lambda\). ∎

Lemma D.29: delocalize weak

Reading source: AI Integrated Stacks, pinned smoothing, lines 2147–2227.

Use the local setting D.27, and set $\mathfrak p=R\cap\mathfrak q$. Suppose $\mathfrak q$ is minimal over $\mathfrak h_A$ and the problem over $R_{\mathfrak p}$ with input $A_{\mathfrak p}$ and target $\Lambda_{\mathfrak q}$ is resolvable. Then the original map from $A$ factors through a finitely presented $R$-algebra $C$ whose singular ideal satisfies $$H_{C/R}\Lambda\not\subset\mathfrak q.$$

Proof. We construct the required algebra by turning a localized smooth presentation into a finite system over $R$. A resolution over the local target has singular radical outside its maximal ideal. Its extended singular ideal must therefore be the unit ideal: a proper ideal in that local ring lies in the maximal ideal, and so does its radical. Apply D.8 to replace that intermediate algebra by a smooth one, then D.12 to obtain a standard smooth factorization of the prescribed map from $A_{\mathfrak p}$.

Write $A=R[x_1,\ldots,x_n]/(g_1,\ldots,g_t)$ and denote the target images of $x_i$ by $\lambda_i$. By D.14, the smooth factorization has a presentation $$D=R_{\mathfrak p}[x_1,\ldots,x_n,u_1,\ldots,u_m]/(f_1,\ldots,f_c),\qquad c\geq n,$$ containing that specific generator list, with a chosen Jacobian determinant invertible. D.14 uses a separate identity equation for each prescribed generator, so repeated images are allowed.

Multiplying each $f_j$ by a coefficient denominator outside $\mathfrak p$ puts every equation in $P=R[x,u]$. These multipliers are units over $R_{\mathfrak p}$, so the chosen determinant $\delta$ remains invertible there. Lift its inverse and the identity proving it is an inverse to polynomial expressions over $R_{\mathfrak p}$. Clear their finite denominator lists and annihilate the remaining localization error by an element outside $\mathfrak p$. This yields $s_0\notin\mathfrak p$ for which $\delta$ is invertible in $P_{s_0}/(f_j)$. Likewise, since each $g_i$ vanishes in $D$, there is $s_i\notin\mathfrak p$ such that $s_i g_i\in(f_j)$ in $P$. For $$s=s_0\prod_{i=1}^t s_i,\qquad D_0=P_s/(f_j),$$ we have a standard smooth $R_s$-algebra receiving $A$, and $D_0/R$ is smooth by principal localization and smooth composition.

The target images of the finitely many $u_\ell$ can be written $\nu_\ell/h$ with one $h\in\Lambda\setminus\mathfrak q$. If there are no such variables, choose $h=1$. For integers $d_j$ large enough to remove negative powers, form $$F_j(x,U,v)=v^{d_j}f_j(x,U/v)\in R[x,U,v].$$ Each value $F_j(\lambda,\nu,h)$ becomes zero in $\Lambda_{\mathfrak q}$. Choose a localization annihilator for each value and multiply them to obtain $\lambda_0\notin\mathfrak q$ with $$\lambda_0 F_j(\lambda,\nu,h)=0\quad\text{in }\Lambda\quad\text{for every }j.$$ Define the finite presentation $$C=R[w,x_1,\ldots,x_n,U_1,\ldots,U_m,v]/ \bigl(g_i(x),\,wF_j(x,U,v)\bigr).$$ The assignments $w\mapsto\lambda_0$, $x_i\mapsto\lambda_i$, $U_\ell\mapsto\nu_\ell$, $v\mapsto h$ respect all equations and give $A\to C\to\Lambda$.

Let $b=wvs$ in $C$. All three factors of its image avoid $\mathfrak q$, so its image does too. In $C_b$ substitute $u_\ell=U_\ell/v$. Inverting $w,v,s$ changes each equation $wF_j=0$ into $f_j=0$; the $g_i$ are already consequences of these equations over $R_s$. The inverse substitution $U_\ell=vu_\ell$ proves, on generators and relations, the exact isomorphism $$C_b\cong D_0[w,w^{-1},v,v^{-1}].$$ The right side is smooth over $R$ by polynomial extension, localization and F.29. Thus $b\in H_{C/R}$ by D.1 and F.67. Its image outside $\mathfrak q$ proves the claimed noncontainment. Every annihilator used belongs to a finite list; the argument requires no Noetherian property of $\Lambda$. ∎

Lemma D.30: delocalize height zero

Reading source: AI Integrated Stacks, pinned smoothing, lines 2229–2274.

In the setting D.27, put $\mathfrak p=R\cap\mathfrak q$. Assume $\mathfrak q$ is minimal over $\mathfrak h_A$, that the localized problem $R_{\mathfrak p}\to A_{\mathfrak p}\to\Lambda_{\mathfrak q}$ is resolvable, and that $\dim\Lambda_{\mathfrak q}=0$. Then a resolution exists over the original base $R$ and target $\Lambda$.

Proof. We need an algebra that improves the singular ideal at $\mathfrak q$ while retaining the original singular radical. The Noetherian ring here is $R$. Its finite-type algebra $A$ is Noetherian by AG-CA-03, Theorem 2.1, so write $H_{A/R}=(a_1,\ldots,a_r)$.

A zero-dimensional local ring has just one prime, its maximal ideal: another prime would give a strict chain of length one. The radical/prime identity proved in D.1 therefore identifies the nilradical of $\Lambda_{\mathfrak q}$ with $\mathfrak q\Lambda_{\mathfrak q}$. Since each $a_i$ maps into this prime, it has nilpotent image in the localization. Choose one positive exponent $N$ annihilating their finite list there. Clearing the localization zeros gives $\lambda\in\Lambda\setminus\mathfrak q$ such that $$\lambda a_i^N=0\quad(1\leq i\leq r)$$ in $\Lambda$. If the list is empty, take $N=1,\lambda=1$. This step uses nilpotence of individual elements, without asserting that the whole maximal ideal is nilpotent or that the local target is Artinian.

By D.29 there is a finite-presentation factorization $A\to C\to\Lambda$ with $\mathfrak h_C\not\subset\mathfrak q$. Choose $c\in H_{C/R}$ with image outside $\mathfrak q$; otherwise that ideal and its radical would be contained in the prime. We can present $C$ as $$C=A[x_1,\ldots,x_n]/(f_1,\ldots,f_m).$$ Indeed a finite $R$-generating list also generates over $A$, and the resulting polynomial ring over Noetherian $A$ has a finitely generated kernel.

Introduce $r$ variables $y_i$, one variable $z$, and $rm$ variables $t_{ij}$. Set $$B=A[x_1,\ldots,x_n,y_1,\ldots,y_r,z,t_{ij}]/ \left(f_j-\sum_{i=1}^r y_it_{ij},\,zy_i\right).$$ Send $x_i$ to its given image, $y_i$ to $a_i^N$, $z$ to $\lambda$, and every $t_{ij}$ to zero. The equations vanish by the chosen annihilators and the map on $C$, so this defines $A\to B\to\Lambda$. The finite lists make $B$ finitely presented over $R$.

There are two kinds of smooth principal neighbourhoods. On the first, invert $z$; the equations force every $y_i=0$, giving $$B_z=C[z,z^{-1},t_{ij}].$$ Since $c$ lies in the singular ideal, $C_c/R$ is smooth by F.67. Polynomial extension and localization then make $B_{zc}/R$ smooth. For the second kind, invert $y_\ell$. We must have $z=0$, and each equation eliminates exactly one variable: $$t_{\ell j}=y_\ell^{-1}\left(f_j-\sum_{i\ne\ell}y_it_{ij}\right).$$ The assignments in this formula and their inverse inclusions identify $$B_{y_\ell}=A[x_j,y_i,y_\ell^{-1},t_{ij}\ (i\ne\ell)].$$ After also inverting $a_\ell$, this is smooth over $R$ because $A_{a_\ell}/R$ is smooth. Consequently $zc$ and every $a_\ell y_\ell$ belong to $H_{B/R}$.

Their target images are $\lambda c$ and $a_\ell^{N+1}$ respectively. The first avoids $\mathfrak q$, proving $\mathfrak h_B\not\subset\mathfrak q$. The others put all $a_\ell$ in the radical $\mathfrak h_B$, hence give $\mathfrak h_A\subset\mathfrak h_B$. These are precisely both resolution requirements. ∎

For separable residue extensions, a suitable sequence in the base permits the local lifting construction to produce a resolution.

Lemma D.31: ogoma

Reading source: Stacks, Tag 07FC.

Let \(A\) be Noetherian, \(M\) a finite \(A\)-module, \(S\subset A\) multiplicative, and \(\pi\in A\). Suppose that multiplication by \(\pi\) and \(\pi^2\) has equal kernels on \(S^{-1}M\). There is one \(s\in S\) such that, for every positive integer \(n\), \[ \ker(s^n\pi:M\to M)=\ker((s^n\pi)^2:M\to M). \]

Proof. Let \[ N=\{m\in M:\pi^2m/1=0\text{ in }S^{-1}M\}. \] It is a submodule of the finite module \(M\), so S01, Lemma 1.3 gives a finite generating list \(u_1,\ldots,u_t\). By the assumed equality of localized kernels, \(\pi u_j/1=0\) for every \(j\). The fraction-zero calculation of S01, Lemma 1.1 gives \(s_j\in S\) with \(s_j\pi u_j=0\) in \(M\). Put \(s=\prod_j s_j\), using \(s=1\) for an empty list. Multiplication by \(s\pi\) kills every generator and hence all of \(N\).

Fix \(n>0\) and suppose \((s^n\pi)^2m=0\). Since \(s\) is a unit after localization, this implies \(\pi^2m/1=0\), so \(m\in N\). Therefore \(s\pi m=0\), and multiplication by \(s^{n-1}\) gives \(s^n\pi m=0\). The reverse inclusion follows by applying \(s^n\pi\) a second time to an element of its kernel. The same \(s\) works for all \(n\). The argument also covers \(0\in S\) and \(M=0\), without cancelling \(s\) inside \(M\). ∎

Lemma D.32: find sequence

Reading source: Stacks, Tag 07FD.

Let \(\Lambda\) be Noetherian, let \(I\subset\mathfrak q\) with \(\mathfrak q\) prime, and fix positive integers \(n,e\). Suppose that \(L=\Lambda_{\mathfrak q}\) is regular local of dimension \(d\) and that \(\mathfrak q^nL\subset IL\). There are \(\pi_1,\ldots,\pi_d\in\Lambda\) satisfying \[ (\pi_1,\ldots,\pi_d)L=\mathfrak qL, \qquad \pi_i^n\in I, \] and, for every \(1\le i\le d\), \[ \operatorname{Ann}_{\Lambda/(\pi_1^e,\ldots,\pi_{i-1}^e)\Lambda}(\pi_i) =\operatorname{Ann}_{\Lambda/(\pi_1^e,\ldots,\pi_{i-1}^e)\Lambda}(\pi_i^2). \] The prescribed exponents \(n,e\) are retained.

Proof. Set \(S=\Lambda\setminus\mathfrak q\). Regularity says that \(\mathfrak qL/(\mathfrak qL)^2\) has dimension \(d\). Lift a basis to \(\mathfrak qL\); finite-module Nakayama, S01, Lemma 1.2, makes these \(d\) lifts generate \(\mathfrak qL\). Each lift is a fraction with numerator in \(\mathfrak q\) and denominator in \(S\). Multiplying each by its denominator replaces it by \(a_i\in\mathfrak q\) and preserves the minimal generating property, since these denominators are units in \(L\).

The containment \(\mathfrak q^nL\subset IL\) gives \(a_i^n/1=0\) in \(S^{-1}(\Lambda/I)\). Exact localization and the fraction-zero criterion S01, Lemma 1.1 provide \(v_i\in S\) with \(v_i a_i^n\in I\). Replace \(a_i\) by \(v_i a_i\). Its \(n\)th power is \(v_i^{n-1}(v_i a_i^n)\in I\), and the list still minimally generates the localized maximal ideal. No change of \(n\) was made. Later multiplication of an entry by an element of \(S\) also preserves these two properties.

Now choose the entries in order. Suppose \(\pi_1,\ldots,\pi_{i-1}\) have been fixed, with their required annihilator equalities, and keep the remaining entries as their current \(a_j\). The whole list is a minimal generating list of \(\mathfrak qL\), so it is regular by the full parameter proof AG-CA-12, Theorem 6.1. The power theorem AG-CA-12, Proposition 1.3, applied with exponents \(e\) on the first \(i-1\) entries and exponent one on the rest, therefore makes \(a_i\) injective on \[ L/(\pi_1^e,\ldots,\pi_{i-1}^e)L. \] Its square is injective too, so their kernels are both zero.

Take the finite \(\Lambda\)-module \(M_i=\Lambda/(\pi_1^e,\ldots,\pi_{i-1}^e)\Lambda\). S01, Lemma 1.1 identifies its localization with the displayed quotient of \(L\). Apply D.31 to \(M_i,S,a_i\). It supplies \(w_i\in S\) for which multiplication by \(w_i a_i\) and its square has the same kernel on \(M_i\); choose the positive power one in that lemma and set \(\pi_i=w_i a_i\). This is exactly the required \(i\)th annihilator equality.

The earlier equalities involve only their own entry and the entries preceding it, none of which has changed. The \(n\)th-power condition remains true because \(I\) is an ideal. In \(L\), the unit \(w_i\) preserves the generating list and its successive regularity: the preceding power ideals stay the same, and multiplication by a unit preserves injectivity on every quotient. Thus the next step has the same hypotheses. After \(d\) steps the list satisfies all assertions. If \(d=0\), Nakayama gives \(\mathfrak qL=0\); the empty list already satisfies them, with both given exponents unchanged. ∎

Lemma D.33: resolve special

Reading source: Stacks, Tag 07FE.

In D.27, let the base be a field \(k\), let \(\Lambda\) be Noetherian, and let \(\mathfrak q\) be minimal over \(\mathfrak h_A\). Assume that \(L=\Lambda_{\mathfrak q}\) is regular local and that its residue field \(K=\kappa(\mathfrak q)\) is separable over \(k\). Then the problem has a resolution.

Proof. Write \(\mathfrak m=\mathfrak qL\) and \(d=\dim L\). If \(d=0\), regularity says that \(\mathfrak m/\mathfrak m^2=0\). The finite-module Nakayama lemma F.10 makes \(\mathfrak m=0\), so \(L=K\). By F.23, this field is a filtered colimit of smooth \(k\)-algebras. Finite presentation and F.90 factor \(A\to L\) through one such algebra. The resulting localized resolution returns to \(\Lambda\) by D.30. Henceforth \(d>0\).

Set \(I=H_{A/k}\Lambda\). Minimality gives \(\sqrt{IL}=\mathfrak m\). Choose finite generators of \(\mathfrak m\), and, for each generator, a positive power belonging to \(IL\). A monomial whose degree exceeds the sum of these powers minus the number of generators contains one of those powers. Consequently there is \(n>0\) with \(\mathfrak m^n\subset IL\). The Hilbert theorem AG-CA-03, Theorem 2.1 makes \(A\) Noetherian; choose \(H_{A/k}=(a_1,\ldots,a_r)\).

Introduce the polynomial base \(P=k[x_1,\ldots,x_d]\) and the finitely presented algebra \[ B=A[x_1,\ldots,x_d,z_{ij}]/(x_i^n-\sum_j a_jz_{ij}). \] On the chart where \(a_j\) is invertible, each equation eliminates \(z_{ij}\). This chart is a polynomial algebra over \(A_{a_j}[x_1,\ldots,x_d]\), and is smooth over \(P\). On inverting \(x_i\), the equation gives \(1=\sum_j a_jx_i^{-n}z_{ij}\); thus these \(a_j\)-charts cover \(B_{x_i}\). Smooth gluing F.67 proves that \(B_{x_i}/P\) is smooth.

Apply D.9 over \(P\). It supplies \(B\to C\) and a retraction \(C\to B\), with each \(C_{x_i}/P\) smooth and its differential module finite free. Its symmetric-algebra construction is smooth also on every \(a_j\)-chart. The algebra \(C\) is finitely presented, since it is finite type over the Noetherian ring \(P\). By D.15(e), choose a common \(c>0\) such that every \(x_i^c\) is strictly standard in \(C/P\), and put \(e=8c\).

Use D.32 with the fixed \(n,e,I\). Obtain elements \(\pi_i\in\Lambda\) generating \(\mathfrak m\) in \(L\), with \[ \pi_i^n=\sum_j a_j\lambda_{ij} \quad\hbox{in }\Lambda, \] and equal kernels for \(\pi_i\) and \(\pi_i^2\) on each quotient by the earlier \(\pi_j^e\). Sending \(x_i\) to \(\pi_i\) and \(z_{ij}\) to \(\lambda_{ij}\) defines \(B\to\Lambda\); the retraction defines \(C\to\Lambda\).

It is enough to resolve this new problem over \(P\). In fact the smooth \(a_j\)-charts put every \(a_j\) in \(H_{C/P}\). A \(P\)-resolution through \(T\) therefore preserves \(\mathfrak h_A\). A smooth \(P\)-neighbourhood is smooth over \(k\), so \(H_{T/P}\subset H_{T/k}\). Finite presentation over \(P\) is finite presentation over \(k\), by adjoining the finitely many \(x_i\) to a presentation. These observations give both conditions of an original \(k\)-resolution.

For the new problem apply D.28 to \(x_1^c,\ldots,x_d^c\). On the source quotients multiplication by a remaining polynomial variable is injective, as equality of polynomial coefficients shows. On a target quotient, equality \(\ker u=\ker u^2\) implies \(\ker u^s=\ker u\) for all \(s\ge1\): if \(u^{s+1}v=0\), applying the equality to \(u^{s-1}v\) gives \(u^sv=0\), and induction finishes the argument. Thus D.32 supplies the equal kernels for \(\pi_i^c\) and \(\pi_i^{2c}\). Since \((x_i^c)^8=x_i^e\), D.28 reduces the task to the quotient by these \(e\)-th powers.

Localize that quotient at its target prime. The base and target are \[ S=P_{(x_1,\ldots,x_d)}/(x_1^e,\ldots,x_d^e), \qquad L_0=L/(\pi_1^e,\ldots,\pi_d^e). \] The contracted prime is \((x_1,\ldots,x_d)\), because reduction of a polynomial in the \(\pi_i\) is its constant in \(k\subset K\). The polynomial local base is regular by iterating AG-CA-14, Proposition 3.3 from the field. The regular local ring \(L\) has dimension \(d\), and its maximal ideal is generated by the \(d\) elements \(\pi_i\). The minimal-parameter theorem AG-CA-12, Theorem 6.1 makes this list regular. The criterion F.46 gives flatness of \(P_{(x)}\to L\); quotient base change gives flatness of \(S\to L_0\).

The maximal ideals of \(S\) and \(L_0\) are nilpotent: every monomial of degree \(d(e-1)+1\) contains an \(e\)-th power. The closed fibre of \(S\to L_0\) is \(K/k\). F.23 expresses it as a smooth colimit, and the nilpotent lifting theorem D.18 does the same for \(L_0/S\). By F.90, the finitely presented algebra \(C\otimes_P S\) factors through a smooth \(S\)-algebra.

To return to the unlocalized quotient, let its target prime be \(\overline{\mathfrak q}\). Its target localization has dimension zero by the preceding nilpotence calculation. If the quotient input's singular radical escapes \(\overline{\mathfrak q}\), its identity factorization already resolves the problem. Otherwise \(\overline{\mathfrak q}\) is minimal over that radical, and D.30 delocalizes the smooth factorization. D.28 lifts the quotient resolution to \(P\to C\to\Lambda\). The change-of-base argument above then gives the required resolution over \(k\). ∎

Inseparable residue fields

For inseparable residue extensions, we first enlarge the coefficient field and control the necessary denominators. The resulting finite system then permits a local resolution.

Lemma D.34: helper

Reading source: Stacks, Tag 07FG.

Let \(k\) be a field of characteristic \(p>0\), and let \((\Lambda,\mathfrak m,K)\) be an Artinian local \(k\)-algebra. If \(H_1(L_{K/k})\) is finite-dimensional over \(K\), then \(\Lambda\) is a directed union of Artinian local \(k\)-subalgebras \(A\), essentially of finite type over \(k\), such that \(A\to\Lambda\) is faithfully flat and \(\mathfrak m_A\Lambda=\mathfrak m\).

Proof. Keep the specified map \(i:k\to\Lambda\) throughout. Its reduction identifies \(k\) with a subfield of \(K\). By AG-CA-03, Theorem 4.2, choose generators \(u_1,\ldots,u_d\) of \(\mathfrak m\) and a least \(N\ge1\) with \(\mathfrak m^N=0\). We use the following internal prime-field lifting calculation. For a field \(E\) of characteristic \(p\), F.119 proves the p-basis presentation \[ E=E^p[T_b:b\in\mathcal B]/(T_b^p-b^p:b\in\mathcal B). \] In a square-zero test \(f:E\to C/J\), define a map on \(E^p\) by sending \(a^p\) to the pth power of any lift of \(f(a)\). The pth root in \(E\) is unique, and the value is independent of the lift because \(j^p=0\) for \(j\in J\). The Frobenius identities prove addition and multiplication for this map. Choose any lifts of the images of the basis elements \(b\). Their pth powers are exactly the prescribed values of \(b^p\), so the displayed presentation gives an \(\mathbb F_p\)-map \(E\to C\) lifting \(f\). This proves formal smoothness for every such field, including an infinite p-basis.

It lifts maps across a nilpotent ideal by its successive powers: the kernel from the quotient by \(J^{a+1}\) to the quotient by \(J^a\) is square zero for \(a\ge1\). In particular a residue section \(s:K\to\Lambda\) exists. Define \(\psi_s:K[X_1,\ldots,X_d]\to\Lambda\) by this section and \(X_j\mapsto u_j\).

We first prove that the section can be chosen so that, for a field \(k\subset F_0\subset K\) finitely generated over \(k\), the subring \(\psi_s(F_0[X])\) contains \(i(k)\). This is automatic for \(N=1\), with \(s\) the identity and \(F_0=k\). Suppose \(N>1\) and put \(I=\mathfrak m^{N-1}\). Then \(I^2=0\) and \(\mathfrak m I=0\), so scalar multiplication on \(I\) factors through \(K\). Applying the induction assertion to \(\Lambda/I\) gives a section \(s_0\) and a finitely generated field \(F\supset k\) containing the prescribed image in \(s_0(F)[\bar u]\). Lift \(s_0\) to \(s\). The ring \[ T=F[X_1,\ldots,X_d]/(X_1,\ldots,X_d)^N \] maps onto \(s_0(F)[\bar u]\). The kernel is contained in the ideal of positive-degree terms, because a nonzero constant from \(F\) has nonzero residue. It is therefore nilpotent. Formal smoothness of \(k/\mathbb F_p\) lifts its specified map to a map \(v:k\to T\). The difference \(e=i-\psi_s v:k\to I\) is an \(\mathbb F_p\)-derivation: expand a product, use agreement modulo \(I\), and discard the product of the two errors in \(I^2\).

Choose the absolute differential basis \((dz_j)_{j\in J}\) supplied by F.119. Formal smoothness and F.17 give \(H_1(L_{K/\mathbb F_p})=0\). Consequently F.131 identifies the kernel in \[ \Omega_{k/\mathbb F_p}\otimes_kK\longrightarrow\Omega_{K/\mathbb F_p} \] with \(H_1(L_{K/k})\). A finite basis of this kernel uses only finitely many coordinates \(dz_j\); let \(J_0\) contain their supports. The images of the other basis vectors are linearly independent, since a relation supported outside \(J_0\) would also have to be supported inside it. Define a \(K\)-linear map on their span by \(dz_j\mapsto e(z_j)\). Extend a basis of this span to a basis of \(\Omega_{K/\mathbb F_p}\), assign zero on the extra vectors, and use the universal differential map to obtain a derivation \(\delta:K\to I\).

The formula \(s_1=s+\delta\) gives another ring section: the derivation identity accounts for the two cross terms, and \(I^2=0\) discards the last term. In evaluating \(v(a)\), the constant coefficient is \(a\in k\subset F\). Changes to positive-degree coefficients are annihilated by their monomials, using \(\mathfrak m I=0\). Thus \(\psi_{s_1}v(a)-\psi_s v(a)=\delta(a)\). The remaining error has the form \[ e(a)-\delta(a)=\sum_{j\in J_0}a_j w_j, \qquad da=\sum_j a_j dz_j,\quad a_j\in k, \quad w_j=e(z_j)-\delta(z_j)\in I. \] Each \(w_j\) is a linear combination of the finitely many degree-\((N-1)\) monomials \(u^E\), with coefficients in \(K\). Indeed these monomials generate \(I\), and a coefficient may be replaced by its residue because \(\mathfrak m I=0\). Adjoin all these coefficients to \(F\) to obtain \(F_0\). When multiplying a \(w_j\), the scalar \(i(a_j)\) can equally be represented by \(s_1(a_j)\), since their difference belongs to \(\mathfrak m\). The displayed error and \(\psi_{s_1}v(a)\) therefore both lie in \(\psi_{s_1}(F_0[X])\). This proves the assertion for the actual structure map. Fix this section henceforth and call it \(s\).

The map \(\psi_s\) is surjective. Subtract a residue representative, write the remainder using the \(u_j\), and do the same for its coefficients. After \(N\) steps the remainder lies in \(\mathfrak m^N=0\). This produces a polynomial expression for each element. By the Hilbert basis theorem AG-CA-03, Theorem 2.1, its kernel has finitely many generators \(g_1,\ldots,g_b\). Enlarge \(F_0\) by their finitely many coefficients to a field \(F_*\). For each finitely generated intermediate field \(F_*\subset F\subset K\), set \[ A_F=F[X_1,\ldots,X_d]/(g_1,\ldots,g_b). \] The presentation gives \(K\otimes_F A_F=\Lambda\). Choose a basis of \(K/F\) containing \(1\). As an \(A_F\)-module this tensor product is a direct sum of copies of \(A_F\); its \(1\)-coordinate makes the natural map injective, exact, and faithful on every module. Thus \(A_F\subset\Lambda\) and this inclusion is faithfully flat. It contains \(i(k)\) by the preceding assertion.

The ideal generated by the \(X_j\) has \(N\)th power zero, as may be checked in the injective image. Its quotient is \(F\), since each \(g_j\) has zero constant coefficient. The monomials of degree below \(N\) span \(A_F\) over \(F\). Finite dimension makes it Artinian; the nilpotent ideal is its unique maximal ideal, since an element with nonzero constant term is invertible by the finite geometric series. That ideal extends to \(\mathfrak m\). The residue field \(F/k\) is finitely generated, hence is the fraction field of the finite-type domain generated by a field-generator list. F.40 now makes \(A_F\) essentially of finite type for its actual, possibly different, \(k\)-structure map.

Two finite field-generator lists have a common successor, their joint generated field. Every polynomial expression in \(\Lambda\) uses finitely many coefficients of \(K\); adjoining them to \(F_*\) places that element in some \(A_F\). These actual subrings therefore form a directed union equal to \(\Lambda\), with all the required properties. ∎

Lemma D.34A: Artinian denominator enlargement

Reading sources: the unit enlargement in Stacks, Tag 07FI and the finite étale construction in Tag 04GK. The auxiliary statement and the calculations below are written here.

Let \(k\) have characteristic \(p>0\), and let \(A\to B\) be a flat local map of Noetherian Artinian local \(k\)-algebras with \(\mathfrak m_AB=\mathfrak m_B\) and \(\mathfrak m_B^N=0\). Denote their residue fields by \(F\subset K\). Given finitely many units \(b_j\in B\), there is a local factorization \(A\to A'\to B\) such that \(A\to A'\) is essentially smooth, \(A'\to B\) is faithfully flat, \(\mathfrak m_{A'}B=\mathfrak m_B\), and a \(p\)-power of each \(b_j\) belongs to the subring \(A'\subset B\). The residue extension is a finite tower of simple transcendental and finite separable extensions. At a transcendental step the unit itself is adjoined.

Proof. The local faithful-flatness theorem AG-CA-08, Theorem 2.1 makes the given map faithfully flat, hence injective. For one unit \(b\), write \(\beta\in K^\times\) for its residue.

If \(\beta\) is transcendental over \(F\), take \(A'=A[T]_{\mathfrak m_AA[T]}\) and send \(T\) to \(b\). Every denominator has a nonzero polynomial residue at \(\beta\), hence maps to a unit of \(B\). This defines a local, essentially smooth map with maximal ideal \(\mathfrak m_AA'\) and residue \(F(\beta)\). The ring is Noetherian by the Hilbert basis theorem and localization. Its maximal ideal is nilpotent. Each successive ideal-power quotient is a finite-dimensional vector space over its residue field, since each power is a finite module. These quotients give finite length, so \(A'\) is Artinian.

In the remaining algebraic case, take \(A'=A\) if \(\beta\in F\). Otherwise let \(f\) be its monic irreducible polynomial. If \(f'=0\), every positive exponent occurring in \(f\) is divisible by \(p\), so \(f(T)=f_1(T^p)\). The polynomial \(f_1\) is irreducible, since a factorization would give one of \(f\) by substitution. Repeat this operation while the derivative vanishes. The degrees decrease, so it stops at an irreducible polynomial with nonzero derivative. Its roots include \(\beta^{q_0}\) for a \(p\)-power \(q_0\). Bézout with the derivative proves this root separable. Thus \(F_1=F(\beta^{q_0})\) is finite separable over \(F\).

Lift its monic minimal polynomial to \(h\in A[T]\), and put \(A'=A[T]/(h)\). Monic division makes \(A'\) finite free over \(A\). Its reduction is the field \(F_1\); the ideal \(\mathfrak m_AA'\) is nilpotent and is its unique maximal ideal. It is Artinian by the same finite-length argument. Bézout for \(\bar h,\bar h'\) makes \(h'\) invertible in the reduction, and a finite geometric series lifts its inverse across the nilpotent ideal. The one-equation Jacobian proof AG-CA-17, Proposition 6.1 makes \(A'/A\) étale.

Here is the actual map to \(B\). Choose \(r_0\in B\) reducing to \(\beta^{q_0}\). Its derivative value is a unit, and \(h(r_0)\in\mathfrak m_B\). Define \[ r_{a+1}=r_a-h'(r_a)^{-1}h(r_a). \] Polynomial expansion shows that \(h(r_{a+1})\) belongs to the square of the ideal containing \(h(r_a)\): the constant and linear terms cancel, and all remaining terms contain at least two copies of the correction. Consequently \(h(r_a)\in\mathfrak m_B^{2^a}\). The residues stay fixed and the derivatives stay units. Once \(2^a\ge N\), this gives an exact root and the required homomorphism \(A'\to B\). Two roots with that residue are equal: their difference times the divided-difference polynomial is zero, and that polynomial has the nonzero derivative as residue and is a unit. This is the simple-root construction without an external lifting step.

In all three cases \(A'\) is \(A\)-flat and \(A'/\mathfrak m_AA'=F'\), its residue field. Tensor a free \(A\)-resolution of \(F\) with \(A'\); flatness makes it a free \(A'\)-resolution of \(F'\). Tensoring further with \(B\) gives \[ \operatorname{Tor}^{A'}_1(F',B)=\operatorname{Tor}^{A}_1(F,B)=0. \] Apply the Noetherian local criterion AG-CA-08, Theorem 4.2 with local source \(A'\), local target \(B\), and the finite \(B\)-module \(B\). It proves \(A'\to B\) flat. Locality makes it faithfully flat and injective, so \(A'\) is the asserted subring. Its maximal ideal still generates \(\mathfrak m_B\).

In the two algebraic cases choose a unit \(x\in A'\) with residue \(\beta^{-q_0}\), taking \(q_0=1\) if \(\beta\in F\). Then \(xb^{q_0}-1\in\mathfrak m_B\). For a \(p\)-power \(q_1\ge N\), the characteristic-\(p\) binomial identity yields \[ (xb^{q_0})^{q_1}=1, \qquad b^{q_0q_1}=x^{-q_1}\in A'. \] In the transcendental case \(b\) itself already belongs to \(A'\). Repeat for the finite unit list. Each earlier power remains in every later subring, and the maps remain essentially smooth by composition.

For use in D.35, retain the natural differential information. A simple transcendental residue step has \[ \Omega_{F(\beta)/k} =(\Omega_{F/k}\otimes_F F(\beta))\oplus F(\beta)\,d\beta, \] by polynomial differentials and the inverse rule in localization. In a finite separable step, differentiating the monic equation solves its new differential uniquely, because its derivative is invertible; it leaves precisely the old differential space after scalar extension. Both residue extensions are formally smooth by F.49, so their relative first homology vanishes by F.17. The exact field sequence F.131 therefore gives the natural isomorphism \[ H_1(L_{F'/k})=H_1(L_{F/k})\otimes_F F'. \] Composition gives these formulas for the whole finite tower. In particular its first-homology image in \(H_1(L_{K/k})\) is unchanged. ∎

Lemma D.35: solution modulo

Reading source: Stacks, Tag 07FH. The global representatives and their denominators are constructed explicitly below.

Let \(k\) have characteristic \(p>0\), let \(\Lambda\) be a Noetherian geometrically regular \(k\)-algebra, and choose a prime \(\mathfrak q\), an integer \(n\ge1\), and a finite subset \(E\subset\Lambda_{\mathfrak q}/\mathfrak q^n\Lambda_{\mathfrak q}\). There is a map \(\varphi:P=k[y_1,\ldots,y_m]\to\Lambda\), for some \(m\ge0\), such that, with \(\mathfrak p=\varphi^{-1}(\mathfrak q)\):

1. \(P_{\mathfrak p}\to\Lambda_{\mathfrak q}\) is flat and \(\mathfrak p\Lambda_{\mathfrak q}=\mathfrak q\Lambda_{\mathfrak q}\).
2. Its quotient map has a local Artinian factorization \[ P_{\mathfrak p}/\mathfrak p^nP_{\mathfrak p} \longrightarrow D\longrightarrow \Lambda_{\mathfrak q}/\mathfrak q^n\Lambda_{\mathfrak q}, \] with the first arrow essentially smooth and the second faithfully flat.
3. The image of \(D\) contains \(E\).

Proof. Write \(L=\Lambda_{\mathfrak q}\), \(\mathfrak m=\mathfrak qL\), \(K=L/\mathfrak m\), and \(B=L/\mathfrak m^n\). The injection in F.133 embeds \(H_1(L_{K/k})\) in the finite-dimensional space \(\mathfrak m/\mathfrak m^2\). By D.34, choose a local Artinian subring \(A\subset B\) containing \(E\), essentially of finite type over \(k\), faithfully flat to \(B\), with \(\mathfrak m_AB=\mathfrak m_B\). Let \(F\subset K\) be its residue field. This field is finitely generated over \(k\): present \(A\) as a localization of a finite-type algebra, take its residue quotient, and express the resulting field as fractions of the image of the finite algebra-generator list.

The differentials of such a field-generator list span \(\Omega_{F/k}\), since the product and inverse rules differentiate every rational expression. Select a basis \(d\theta_1,\ldots,d\theta_t\) from this spanning list and lift its elements to \(A\). Each lift has a representative \(a_j/b_j\in L\), with \(a_j\in\Lambda\) and \(b_j\notin\mathfrak q\).

By the left injection in F.131, \(H_1(L_{F/k})\otimes_FK\) embeds in \(H_1(L_{K/k})\). Denote its image in \(\mathfrak m/\mathfrak m^2\) by \(V\). If \(n\ge2\), the elements of \(\mathfrak m_A\) span this cotangent space over \(K\): its identification with the cotangent space of \(B\) and the equality of extended maximal ideals prove that assertion. Choose elements whose images are a basis of a complement to \(V\). Lift them to fractions \(c_i/d_i\), with \(c_i\in\mathfrak q\) and \(d_i\notin\mathfrak q\). For \(n=1\), select instead representatives \(c_i\in\mathfrak q\) of a complement in \(\mathfrak m/\mathfrak m^2\) and put \(d_i=1\). Their images in \(B=K\) are zero, so are already in \(A\).

Apply D.34A to the finitely many units represented by \(b_j,d_i\) in \(B\). Obtain \(A\subset A'\subset B\), with \(A\to A'\) essentially smooth, \(A'\to B\) faithfully flat, and unchanged extended maximal ideal. For appropriate \(p\)-powers \(q_j,r_i\), all four kinds of global elements \[ a_jb_j^{q_j-1},\quad b_j^{q_j},\quad c_id_i^{r_i-1},\quad d_i^{r_i} \] have images in \(A'\). The first and third are the fractional lifts multiplied by the indicated powers; the second and fourth are supplied by the enlargement. Record in addition every original denominator which was adjoined at a transcendental residue step. It itself has its image in \(A'\).

Let \(F'\) be the residue field of \(A'\). The residues of the first two kinds of elements, together with these recorded transcendental denominators, have differentials spanning \(\Omega_{F'/k}\). Indeed their quotient is \(\theta_j\), so the inverse rule expresses each old \(d\theta_j\) in their span. D.34A adds just one new differential at each transcendental step and none at a finite separable step. Select a differential basis from this finite list, and call its actual global representatives \(\lambda_1,\ldots,\lambda_s\in\Lambda\). This argument does not set \(d(b_j^{q_j})\) to zero in \(F'\): the root \(b_j\) need not belong to that field. Every selected representative has its class in \(A'\).

Define \(P'=k[y_1,\ldots,y_s]\to\Lambda\) by these elements, let \(\mathfrak p'\) be the inverse image of \(\mathfrak q\), and set \(K_0=\kappa(\mathfrak p')\subset F'\). Since all the basis differentials come from \(K_0\), \(\Omega_{F'/K_0}=0\). The extension is finitely generated, because \(F'/k\) is finitely generated. The equality proved in F.98 gives \[ -\dim_{F'}H_1(L_{F'/K_0}) =\operatorname{trdeg}_{K_0}F'. \] Its two nonnegative dimensions must both be zero. Finitely many algebraic field generators give a finite extension by multiplying their degrees; the zero first homology gives separability by F.91. Thus \(F'/K_0\) is finite separable. The field sequence F.131, whose relative first homology and differential module here vanish, identifies \[ H_1(L_{K_0/k})\otimes_{K_0}K =H_1(L_{F'/k})\otimes_{F'}K =H_1(L_{F/k})\otimes_FK. \] The last identity is the natural one retained in D.34A, so its image in \(\mathfrak m/\mathfrak m^2\) is still \(V\).

Apply F.109 using the selected differential basis. It gives a flat map \(P'_{\mathfrak p'}\to L\) and a regular quotient \(L/\mathfrak p'L\). Its written conormal square identifies the image of the source cotangent space with the preceding first-homology image \(V\). Append the global elements \(\lambda_{s+i}=c_id_i^{r_i-1}\) to form \(P=k[y_1,\ldots,y_m]\to\Lambda\). Their cotangent classes are the originally chosen complementary classes multiplied by the nonzero scalars \(\bar d_i^{r_i}\), and therefore still form a complementary basis. Their residues are zero. Hence \(P_{\mathfrak p}\) is the localization of \(P'_{\mathfrak p'}[y_{s+1},\ldots,y_m]\) at the prime generated by \(\mathfrak p'\) and these variables.

Polynomial local regularity is proved in AG-CA-14, Proposition 3.3. The old regular parameters and the appended variables are a regular parameter list for this source: its maximal ideal is generated by them and its cotangent space is the old cotangent space plus one independent class for each new variable. Their images are a regular sequence in \(L\). The old list remains regular by the flat map from \(P'_{\mathfrak p'}\), applying flat base change to each successive multiplication injection. The new list is a minimal parameter list on the regular local quotient \(L/\mathfrak p'L\), hence is regular by AG-CA-12, Theorem 6.1. Concatenating these lists proves the assertion. F.46 now makes \(P_{\mathfrak p}\to L\) flat. Their cotangent classes span \(\mathfrak m/\mathfrak m^2\); Nakayama F.10 gives \(\mathfrak pL=\mathfrak m\). This proves (1).

All polynomial generators have their classes in \(A'\), so the map to \(B\) factors through \(A'\). An element outside \(\mathfrak p\) has nonzero residue and maps to a unit of this local ring. Moreover \(\mathfrak p^n\) maps to zero in \(B\), hence in \(A'\subset B\). We obtain \[ R=P_{\mathfrak p}/\mathfrak p^nP_{\mathfrak p} \longrightarrow A'\longrightarrow B. \] The ring \(R\) is Artinian: it is Noetherian local with nilpotent maximal ideal, and its finite ideal-power quotients have finite dimension over its residue field. Its composite map to \(B\) is the flat base change of (1). Since \(A'\to B\) is faithfully flat, F.48 makes \(R\to A'\) flat. The ideals \(\mathfrak pA'\) and \(\mathfrak m_{A'}\) become equal in \(B\); faithful flatness detects equality by tensoring their quotient, so they are equal. The residue extension is the finite separable extension \(F'/K_0\) just proved. F.40 makes \(A'\) a finite \(R\)-module. Since \(R\) is Noetherian, it is a finitely presented algebra, by taking finite module generators and a finite kernel in their polynomial presentation. Its only fibre is the finite separable residue field. The proved étale criterion F.16 therefore makes \(R\to A'\) étale, in particular essentially smooth. Take \(D=A'\). It retains \(E\subset A\), proving (2) and (3). ∎

Lemma D.36: enlarge solution modulo

Reading source: Stacks, Tag 07FI.

Use the notation and a local Artinian factorization of D.35: \[ R=k[y_1,\ldots,y_m]_{\mathfrak p}/\mathfrak p^n \longrightarrow D\longrightarrow B=\Lambda_{\mathfrak q}/\mathfrak q^n\Lambda_{\mathfrak q}, \] where \(R\to D\) is essentially smooth and \(D\to B\) is flat. For every \(\lambda\in\Lambda\setminus\mathfrak q\), there are a positive \(p\)-power \(q\) and a local Artinian factorization \(R\to D\to D'\to B\) with \(D\to D'\) essentially smooth, \(D'\to B\) faithfully flat, and \(\lambda^q\) in the image of \(D'\).

Proof. The polynomial-source maximal ideal extends to that of \(B\), by D.35(1). We verify that the maximal ideal of \(D\) does also, even for a factorization other than the particular étale one constructed there. The closed fibre \(D/\mathfrak m_RD\) is Artinian local and essentially smooth over the field \(R/\mathfrak m_R\). The regularity theorem AG-CA-18, Theorem 2.1, followed by localization, makes it regular. An Artinian local ring has dimension zero. For a regular local ring of dimension zero its cotangent dimension is zero; its finite maximal ideal equals its square, and Nakayama makes that ideal zero. Thus this fibre is a field and \(\mathfrak m_D=\mathfrak m_RD\). Extending to \(B\) gives \(\mathfrak m_DB=\mathfrak m_B\).

All rings required by D.34A are Noetherian. The source \(R\) is Noetherian, and the essentially smooth algebra \(D\) is a localization of a finitely presented \(R\)-algebra, so the Hilbert basis theorem and localization make it Noetherian. The ring \(B\) is the displayed Artinian quotient of the Noetherian local ring \(\Lambda_{\mathfrak q}\), and \(\mathfrak m_B^n=0\). The flat local map \(D\to B\) is faithfully flat by the local criterion. The image of \(\lambda\) is a unit because its residue is nonzero. Apply the one-unit case of D.34A. It supplies \(D\subset D'\subset B\), an essentially smooth first map, a faithfully flat last map, and the exact power containment \(\lambda^q\in D'\). Composition with \(R\to D\) gives the required factorization. ∎

Lemma D.37: resolve general

Reading source: Stacks, Tag 07FJ.

Let \(k\) have characteristic \(p>0\). In D.27, suppose that \(\Lambda\) is Noetherian and geometrically regular over \(k\), and that \(\mathfrak q\) is minimal over \(\mathfrak h_A\). Then the problem has a resolution.

Proof. Put \(L=\Lambda_{\mathfrak q}\), \(\mathfrak m=\mathfrak qL\), \(d=\dim L\), and \(H=H_{A/k}=(a_1,\ldots,a_r)\). These generators exist by the Hilbert theorem AG-CA-03, Theorem 2.1. The ring \(L\) is regular by geometric regularity. As in D.33, minimality and a finite maximal-ideal generating list give \(N>0\) with \(\mathfrak m^N\subset HL\).

Over \(P_0=k[x_1,\ldots,x_d]\) form \[ B=A[x_1,\ldots,x_d,z_{ij}]/(x_i^{2N}-\sum_j a_jz_{ij}). \] Eliminating \(z_{ij}\) on an \(a_j\)-chart proves smoothness there; the equation makes those charts cover \(B_{x_i}\). By F.67, \(B_{x_i}/P_0\) is smooth. Apply D.9 to obtain \(B\to C\) with retraction \(C\to B\). The smooth charts of \(C\) have finite free differentials. Since \(C/P_0\) is finitely presented by the Hilbert theorem, D.15(e) gives one \(c>0\) for which all \(x_i^c\) are strictly standard. If \(d=0\), take \(c=1\). Fix, before making any Artinian approximation, \[ e=8c,\qquad n=\max\{N+dc,\ d(e-1)+1\}. \tag{D.37a} \]

Apply D.35 to the images of a finite \(k\)-algebra generating list of \(A\) in \(\overline L=L/\mathfrak m^n\). It gives a polynomial algebra \(P=k[y_1,\ldots,y_m]\) mapping to \(\Lambda\), its contracted prime \(\mathfrak p\), and local maps \[ S_0=P_{\mathfrak p}/\mathfrak p^nP_{\mathfrak p} \longrightarrow D\longrightarrow\overline L. \] The first map is essentially smooth, the second is flat, and the generator images belong to \(D\). Also \(P_{\mathfrak p}\to L\) is flat and \(\mathfrak pL=\mathfrak m\). Its closed fibre is a field, so the dimension formula AG-CA-11, Theorem 5.1 gives \(\dim P_{\mathfrak p}=d\).

Choose \(\pi_1,\ldots,\pi_d\in\mathfrak p\) representing a regular parameter list of \(P_{\mathfrak p}\). Polynomial local rings are regular by iteration of AG-CA-14, Proposition 3.3, starting with the field. The minimal-parameter theorem AG-CA-12, Theorem 6.1 makes their minimal generating list regular. Their images generate \(\mathfrak m\) and form a regular parameter list of \(L\), by the same theorem and \(\dim L=d\). Let \[ R=P[t_1,\ldots,t_d],\qquad \gamma_i=\pi_it_i. \] Throughout, localization of this polynomial ring at the coefficients means \[ R_S=(P\setminus\mathfrak p)^{-1}R=P_{\mathfrak p}[t_1,\ldots,t_d]. \] It is distinct from localization at a prime of \(R\).

First arrange the source kernel conditions. By F.75, the \(\gamma_i\) form a regular sequence in \(R_S\). Powers of earlier entries preserve regularity by AG-CA-12, Proposition 1.3. Inductively, on the finite \(R\)-module \(R/(\gamma_1^e,\ldots,\gamma_{i-1}^e)\), the localized kernels for \(\gamma_i\) and \(\gamma_i^2\) are zero. D.31 supplies \(s_i\in P\setminus\mathfrak p\) such that replacing \(\pi_i\) by \(s_i\pi_i\) makes these kernels equal globally. Earlier equalities are unchanged, because their entries and quotient ideals have not changed. Such scaling is by local units, so both local parameter lists remain regular. Continue through all \(i\), using the modified \(\pi_i\) from now on.

Next construct units \(\delta_i\in\Lambda\setminus\mathfrak q\), coefficients \(\lambda_{ij}\in\Lambda\), and an essentially smooth Artinian enlargement \(D\to D'\) with flat local map \(D'\to\overline L\), such that \[ (\delta_i\pi_i)^{2N}=\sum_j a_j\lambda_{ij} \quad\hbox{in }\Lambda, \tag{D.37b} \] and both \(\delta_i\) and \(\lambda_{ij}\) have images in \(D'\). We additionally require equal kernels for \(\delta_i\pi_i\) and its square modulo the earlier \((\delta_h\pi_h)^e\). A flat local map into the nonzero local ring \(\overline L\) is faithfully flat, and hence injective, by AG-CA-08, Theorem 2.2. Thus every ring in this construction may be regarded as a subring of \(\overline L\). In particular all relations among the chosen \(A\)-generator images already hold in \(D\), so there is a map \(A\to D\).

Suppose the construction has reached \(i-1\). The element \(\pi_i^N\) lies in \(H\overline L\); faithful flatness contracts this ideal to \(HD'\). Write \(\pi_i^N=\sum_j a_jd_j\) in \(D'\), and lift the \(d_j\) to \(u_j\in L\). The error lies in \(\mathfrak m^n\subset H\mathfrak m^{n-N}\), since \(n\ge N\). Thus choose \(v_j\in\mathfrak m^{n-N}\) with \[ \pi_i^N=\sum_j a_j(u_j+v_j)\quad\hbox{in }L. \] Set \(w_j=\pi_i^N(u_j+v_j)\). Then \(\pi_i^{2N}=\sum_j a_jw_j\), and the residue of \(w_j\) in \(\overline L\) is \(\pi_i^Nd_j\), because \(\pi_i^Nv_j\in\mathfrak m^n\).

Here is the global denominator calculation. Choose \(s_0\notin\mathfrak q\) so that every \(s_0^{2N}w_j\) is represented by \(\widetilde\mu_j\in\Lambda\). The error \[ r_i=(s_0\pi_i)^{2N}-\sum_j a_j\widetilde\mu_j \] vanishes in \(L\). Hence some \(v\notin\mathfrak q\) annihilates \(r_i\), by the fraction-zero criterion of S01, Lemma 1.1. Set \(s=s_0v\) and \(\mu_j=v^{2N}\widetilde\mu_j\). Now \((s\pi_i)^{2N}=\sum_j a_j\mu_j\) in \(\Lambda\), and \(\mu_j/s^{2N}=w_j\) in \(L\).

Use D.36 to put \(s^h\) in a further \(D'\), for some \(h>0\). Replace \(s\) by \(s^h\) and \(\mu_j\) by \(s_{\rm old}^{2N(h-1)}\mu_j\). The exact equation survives, and the new coefficient residue is \(s^{2N}\pi_i^Nd_j\in D'\). In \(L/((\delta_1\pi_1)^e,\ldots,(\delta_{i-1}\pi_{i-1})^e)\), multiplication by \(s\pi_i\) is injective, because the preceding entries are powers of a regular parameter list scaled by units. Apply D.31 on the corresponding finite \(\Lambda\)-module to obtain \(s'\notin\mathfrak q\) for which every \((s')^q s\pi_i\), \(q>0\), has equal global kernels with its square. D.36 puts one such \((s')^q\) in an enlargement of \(D'\). Take \[ \delta_i=(s')^qs,\qquad \lambda_{ij}=(s')^{2Nq}\mu_j. \] This proves (D.37b), the residue assertions, and the required target kernel equality. Later enlargements retain all earlier assertions. The finite induction constructs the full list; when \(d=0\) this construction is empty and \(D'=D\).

Map \(R\to\Lambda\) by the given \(P\)-map and \(t_i\mapsto\delta_i\). Give \(R\) its \(P_0\)-structure by \(x_i\mapsto\gamma_i\). Equation (D.37b) defines \(B\to\Lambda\) by \(x_i\mapsto\delta_i\pi_i\) and \(z_{ij}\mapsto\lambda_{ij}\). The retraction defines \(C\to\Lambda\), and these maps induce \[ C_R=C\otimes_{P_0}R\longrightarrow\Lambda. \] Every \(a_j\)-chart of \(C_R/R\) is smooth, so the singular radical of this new input contains \(\mathfrak h_A\). A resolution over \(R\) returns to one over \(k\): finite presentation composes with the polynomial base, and smooth composition gives \(H_{T/R}\subset H_{T/k}\).

Let \(I=(\gamma_1^e,\ldots,\gamma_d^e)\subset R\). D.7 transports the strictly standard elements \(x_i^c\) to \(\gamma_i^c\). The arranged source and target kernel equalities hold also for their \(c\)-th powers, by the kernel induction in D.33. D.28 therefore lifts a resolution modulo \(I\) to an \(R\)-resolution. For \(d=0\), \(I=0\) and this lifting step is unnecessary.

We construct the quotient resolution explicitly. Put \(J=(\pi_1^e,\ldots,\pi_d^e)\subset P\), and let \(\mathfrak r\) be the inverse image of \(\mathfrak q\) in \(R\). Each \(t_i\notin\mathfrak r\), since \(\delta_i\notin\mathfrak q\). Consequently \[ IR_{\mathfrak r}=JR_{\mathfrak r}, \qquad IL=JL. \tag{D.37c} \] The parameters generate \(\mathfrak pP_{\mathfrak p}\) and \(\mathfrak m\). The bound \(n\ge d(e-1)+1\) in (D.37a) gives, by expansion into monomials, \[ \mathfrak p^nP_{\mathfrak p}\subset JP_{\mathfrak p}, \qquad\mathfrak m^n\subset JL. \tag{D.37d} \] For \(d=0\), both maximal ideals are zero, so these containments hold as well.

Set \(S_1=S_0[t_1,\ldots,t_d]\). Bars below denote elements in \(D'\subset\overline L\). Each \(\overline\delta_i\) is a unit of \(D'\): its image is a unit and the map is local. Define \[ B\longrightarrow D'[t_1,\ldots,t_d], \qquad x_i\longmapsto\overline\pi_it_i,\quad z_{ij}\longmapsto \overline\lambda_{ij}\overline\delta_i^{-2N}t_i^{2N}. \tag{D.37e} \] Injectivity of \(D'\to\overline L\) transports (D.37b) to \(D'\). The image of the \(i\)-th defining equation in (D.37e) is therefore \[ t_i^{2N}\left(\overline\pi_i^{2N} -\overline\delta_i^{-2N}\sum_j\overline a_j\overline\lambda_{ij}\right)=0. \] Its restriction to \(P_0\) is exactly the prescribed \(x_i\mapsto\pi_it_i\) in \(S_1\), so the map extends to \(B\otimes_{P_0}S_1\). Evaluation \(t_i\mapsto\overline\delta_i\) recovers the prescribed images \(\delta_i\pi_i,\lambda_{ij}\) in \(\overline L\).

The map \(S_0\to D'\) is essentially smooth by construction and composition. Express it as a localization of a smooth finitely presented \(S_0\)-algebra. Localization is the filtered colimit of its principal localizations at finite products of the inverted elements. These stages are smooth, and adjoining the \(t_i\) gives smooth \(S_1\)-stages with colimit \(D'[t_i]\). Finite presentation and F.90 factor (D.37e) through one smooth \(S_1\)-stage \(T\), followed by the specified map to \(\overline L\). Composing with \(C\to B\) gives the same smooth factorization for \(C\otimes_{P_0}S_1\).

By (D.37d) there are quotient maps \(S_1\to P_{\mathfrak p}[t_i]/J\) and \(\overline L\to L/JL\). Base changing \(T\) gives a smooth factorization over \(P_{\mathfrak p}[t_i]/J\). Localize further at \(\mathfrak r\). Every denominator then has a unit image in the local target, so its map extends. By (D.37c) this is a smooth factorization of the localized problem modulo \(I\).

The target \(L/JL\) has nilpotent maximal ideal: degree \(d(e-1)+1\) monomials vanish. Thus its dimension is zero. If the singular radical of the unlocalized quotient input escapes the image of \(\mathfrak q\), the identity factorization resolves that quotient. Otherwise this prime is minimal over the radical, and D.30 delocalizes the smooth factorization just obtained. D.28 then lifts it before quotienting by \(I\). Finally the polynomial change of base returns the resolution to \(k\). All constructions above include \(d=0\), empty parameter lists, and the specified target maps. ∎

The main theorem

The final argument selects a maximal unsuccessful singular radical and uses the local resolutions to enlarge it, giving a contradiction.

Theorem D.38: popescu

Reading source: Stacks, Tag 07GC.

If \(R\to\Lambda\) is a regular map between Noetherian rings, then \(\Lambda\), as an \(R\)-algebra, is a filtered colimit of smooth \(R\)-algebras.

Proof. The reduction D.26 reduces the assertion to a field \(k\) and a Noetherian geometrically regular \(k\)-algebra \(\Lambda\). By D.8, it is enough that every finite type factorization \(k\to A\to\Lambda\) can be followed by \(A\to B\to\Lambda\) with \(B\) finite type and \(\mathfrak h_B=\Lambda\). All such algebras are finitely presented by the Hilbert theorem.

Assume some factorization has no such successor. Among the singular radicals of all these unsuccessful factorizations choose a maximal one, say \(\mathfrak h_A\). A nonempty collection of ideals in a Noetherian ring has a maximal member: otherwise successive strict enlargements would contradict the ascending chain condition. The ideal \(\mathfrak h_A\) is proper, since if it were \(\Lambda\), \(A\) itself would be a successor.

There is a prime \(\mathfrak q\) minimal over \(\mathfrak h_A\). Here are the existence details. Zorn's lemma first gives an ideal maximal among the proper ideals containing \(\mathfrak h_A\): a chain union is still proper, since it cannot contain \(1\). Such an ideal is prime, because if \(a\) is outside it, the ideal and \(a\) generate \(1\), and multiplication of that relation by \(b\) shows that \(ab\) in the ideal forces \(b\) into it. Now order all primes containing \(\mathfrak h_A\) by reverse inclusion. A chain intersection is a proper ideal containing \(\mathfrak h_A\), and is prime: if \(ab\) belongs to the intersection while \(a,b\) do not, two members excluding them have a smaller member excluding both, contradicting its primality. Zorn's lemma therefore supplies the desired minimal prime.

Geometric regularity makes \(\Lambda_{\mathfrak q}\) regular local. In positive characteristic D.37 resolves this problem. In characteristic zero every field extension is separable in the sense of F.5: each finitely generated subextension has a finite transcendence basis followed by a finite algebraic extension, whose irreducible polynomials have nonzero derivatives and hence distinct roots. Thus D.33 resolves it in that case.

In either case obtain \(A\to C\to\Lambda\) with \[ \mathfrak h_A\subset\mathfrak h_C,\qquad \mathfrak h_C\not\subset\mathfrak q. \] Because \(\mathfrak h_A\subset\mathfrak q\), the first inclusion is strict. Maximality of the unsuccessful ideal \(\mathfrak h_A\) means that \(C\) has a successor with unit singular radical. Composing its factorization with \(A\to C\) supplies the forbidden successor of \(A\), a contradiction. Hence every finite type input has the required successor. D.8 supplies the smooth filtered presentation, and D.26 returns the assertion to the original regular map. ∎

5. Foundation statements and their free-source proofs

The flatness arguments use projective resolutions, the two ways of computing Tor, and the connecting maps of a short exact sequence. These constructions are proved for arbitrary modules in Algebra and sheaf cohomology before reductive groups, Lemma 1.A (Reductive group schemes).

We begin with the two-term conormal complex and the field and dimension conventions used below. The numbered arguments then establish the presentation, flatness, regularity and completion results needed in Section 4. Section 6 collects the earlier algebra lessons used along the way.

The two-term conormal complex. A polynomial presentation $P=R[X_s]\twoheadrightarrow S$ with kernel $I$ gives the complex $[I/I^2\to\bigoplus_s S\,dX_s]$, in degrees $-1$ and $0$; the map sends the class of $f$ to $df$. We denote it by $\operatorname{NL}(P\to S)$. One presentation is always available: take a variable $X_s$ for each $s\in S$ and send it to $s$. For maps $R\to R'$ and $S\to S'$ forming a commuting square, a map of presentations is a compatible map of their polynomial algebras lifting $S\to S'$. The comparison and homotopy formulas below prove that the resulting two-term object does not depend on the presentation. The localization argument concerns this complex itself.

The inseparable exponent calculation. Let $\alpha$ be algebraic over a field $F$ of characteristic $p$. Its irreducible polynomial is $f(T)=g(T^{p^s})$ for a maximal $s\geq0$. Such $s$ exists because $f$ has finite positive degree. Maximality gives $g'\ne0$. Any factorization of $g$ would give a factorization of $f$, so $g$ is irreducible, and its relatively prime derivative makes it separable. The element $\alpha^{p^s}$ is a root of $g$, hence is separable over $F$. This proves the field-theoretic exponent choice used in the enlargement argument.

Dimension conventions. The dimension of a ring is the supremum of the lengths of its finite chains of primes. For a prime $\mathfrak p$, its height is $\dim R_{\mathfrak p}$. In a Noetherian local ring of dimension $d$, a system of parameters consists of $d$ elements whose ideal has radical the maximal ideal. Existence is proved in AG-CA-11, Theorem 2.1. Minimal primes and irreducible components are treated in AG-CA-01, Theorem 4.1, Lemma 4.2 and Theorem 4.3. Polynomial-ring dimension and the finite-type domain height formula are AG-CA-11, Theorem 6.1 and AG-CA-09, Theorem 4.3, respectively.

Definition F.1: lci

Reading sources: Stacks, Tag 00SL; AI Integrated Stacks, pinned Algebra, lines 37874–37881.

For an arbitrary ring map $R\to S$, the word syntomic means that the following three requirements hold together: $S$ is a finitely presented $R$-algebra, it is flat as an $R$-module, and, for every $\mathfrak p\in\operatorname{Spec}R$, the fibre algebra $$S(\mathfrak p)=S\otimes_R\kappa(\mathfrak p)$$ is a local complete intersection over $\kappa(\mathfrak p)$ in the sense of F.2. The phrase flat local complete intersection over $R$ has this same meaning.

Definition F.2: lci field

Reading sources: Stacks, Tag 00S9; AI Integrated Stacks, pinned Algebra, lines 37249–37264.

Let $S$ be a finite-type algebra over a field $k$. To call $S$ a global complete intersection over $k$ is to require a finite presentation $$S=k[x_1,\ldots,x_n]/(f_1,\ldots,f_c)$$ for which its Krull dimension is exactly $n-c$. A local complete intersection over $k$ means that one can cover $\operatorname{Spec}S$ by principal opens $D(g_i)$, with each algebra $S_{g_i}$ satisfying this global condition. The zero algebra is included among the global complete intersections by convention; it therefore causes no exception when an open or a fibre is empty.

Definition F.3: lci local ring

Reading sources: Stacks, Tag 00SD; AI Integrated Stacks, pinned Algebra, lines 37446–37458.

Suppose $S$ is a local $k$-algebra essentially of finite type, where $k$ is a field. The condition complete intersection over $k$ asks for a regular local $k$-algebra $R$, itself essentially of finite type, together with a regular sequence $$f_1,\ldots,f_c\in\mathfrak m_R$$ and a $k$-algebra isomorphism $R/(f_1,\ldots,f_c)\cong S$. Thus the presenting ring is local and regular, and the equations lie in its maximal ideal. All these requirements belong to the definition.

Definition F.4: relative global complete intersection

Reading sources: Stacks, Tag 00SP; AI Integrated Stacks, pinned Algebra, lines 37943–37951.

For a ring map $R\to S$, a relative global complete-intersection presentation consists of $$S=R[x_1,\ldots,x_n]/(f_1,\ldots,f_c)$$ such that, for each $\mathfrak p\in\operatorname{Spec}R$ with nonempty fibre, the ring $S\otimes_R\kappa(\mathfrak p)$ has dimension $n-c$. The map is a relative global complete intersection if it has at least one such presentation. When a particular displayed presentation is called a relative global complete intersection, the fibre condition refers to its displayed values of $n$ and $c$. Empty fibres impose no dimension condition.

Definition F.5: separable field extension

Reading sources: Stacks, Tag 030O; AI Integrated Stacks, pinned Algebra, lines 10135–10146.

Consider a field extension $K/k$. It is separably generated when there is a transcendence basis $(x_i)_{i\in I}$ such that $$K/k(x_i\mid i\in I)$$ is algebraic and separable. The index set may be infinite. The condition separable over $k$ used here is tested on every intermediate field $K'$ finitely generated over $k$: each extension $K'/k$ must be separably generated in the preceding sense. These are the two field-extension conditions used in the separable and inseparable arguments.

Definition F.6: smooth

Reading sources: Stacks, Tag 00T2; AI Integrated Stacks, pinned Algebra, lines 38526–38533.

Smoothness of $R\to S$ requires finite presentation together with the following condition on the naive cotangent complex: $$H_1(\operatorname{NL}_{S/R})=0,\qquad \Omega_{S/R}\text{ is a finite projective }S\text{-module}.$$ Equivalently, that two-term complex is quasi-isomorphic to a finite projective module concentrated in degree zero. To verify this equivalence directly, map the two-term complex to its degree-zero cokernel. This map induces an isomorphism on degree-zero homology; it is a quasi-isomorphism exactly when the degree-minus-one kernel, denoted $H_1$, vanishes. Its cokernel is $\Omega_{S/R}$. Presentation independence is proved in F.11.

Definition F.7: smooth at prime

Reading sources: Stacks, Tag 00TB; AI Integrated Stacks, pinned Algebra, lines 38953–38960.

Given $R\to S$ and $\mathfrak q\in\operatorname{Spec}S$, smoothness at $\mathfrak q$ means that some principal neighbourhood of this point has a smooth coordinate algebra over $R$. Explicitly, there must be $$g\in S\setminus\mathfrak q\quad\text{with}\quad R\to S_g\text{ smooth}.$$ This definition applies to any ring map; finite presentation is required of the witnessing localization through F.6.

Definition F.8: unramified

Reading sources: Stacks, Tag 00UT; AI Integrated Stacks, pinned Algebra, lines 42897–42912.

For $R\to S$, both unramified conditions require $\Omega_{S/R}=0$. Their finiteness requirements differ: unramified requires finite type, whereas G-unramified requires finite presentation.

At a prime $\mathfrak q$ of $S$, each condition means that it holds on some principal neighbourhood $D(g)$ of $\mathfrak q$. Thus unramified at $\mathfrak q$ requires $g\notin\mathfrak q$ with $S_g/R$ of finite type and $\Omega_{S_g/R}=0$; G-unramified at $\mathfrak q$ requires such a $g$ with $S_g/R$ finitely presented and the same differential-module vanishing.

Lemma F.9: CM over regular flat

Reading source: AI Integrated Stacks, pinned Algebra, lines 34036–34074.

Let $R\to S$ be local, with both rings Noetherian and local. Suppose $R$ is regular, $S$ is Cohen–Macaulay, and $$\dim S=\dim R+\dim(S/\mathfrak m_RS).$$ Then $S$ is flat as an $R$-module.

Proof. We reduce the dimension of the regular source by one while preserving the dimension equality. Write $d=\dim R$. If $d=0$, regularity says $\mathfrak m_R/\mathfrak m_R^2=0$. Nakayama, F.10, applies to the finite ideal $\mathfrak m_R$ and gives $\mathfrak m_R=0$. Thus $R$ is a field, and a basis makes every $R$-module free, hence flat.

For $d>0$, consider the finitely many associated primes of $S$. The Cohen–Macaulay component theorem AG-CA-12, Theorem 4.1, says that they are its minimal primes $\mathfrak q_i$ and that $\dim(S/\mathfrak q_i)=\dim S$. None contains $\mathfrak m_RS$. Indeed that containment would give a surjection $S/\mathfrak m_RS\to S/\mathfrak q_i$, forcing the dimension of the latter to be at most that of the former, whereas the hypothesis gives $$\dim(S/\mathfrak m_RS)=\dim S-d<\dim S.$$ Consequently every contraction $\mathfrak p_i=R\cap\mathfrak q_i$ is strictly smaller than $\mathfrak m_R$.

The prime-avoidance calculation F.80 now chooses $$x\in\mathfrak m_R\setminus\bigl(\mathfrak m_R^2\cup\textstyle\bigcup_i\mathfrak p_i\bigr).$$ To check its hypotheses, $\mathfrak m_R$ is not contained in any $\mathfrak p_i$, and is not contained in $\mathfrak m_R^2$, since otherwise Nakayama would again give $d=0$. There is just one nonprime ideal in this avoidance list.

The element $x$ avoids all associated primes of $S$, so it is a nonzerodivisor on $S$. The complete regular-sequence and Cohen–Macaulay arguments of AG-CA-12, Sections 2 and 4, make $S/xS$ Cohen–Macaulay with dimension $\dim S-1$. AG-CA-14, Theorem 1.1 and Corollary 1.2, make $x$ a nonzerodivisor of $R$ and $R/xR$ regular of dimension $d-1$. The quotient map is still local, and its closed fibre is $S/\mathfrak m_RS$. Hence it satisfies $$\dim(S/xS)=\dim(R/xR)+\dim(S/\mathfrak m_RS).$$ Induction gives flatness of $S/xS$ over $R/xR$.

The two-term free resolution of $R/xR$ has differential multiplication by $x$. Its tensor with $S$ therefore has zero first homology, because $x$ is a nonzerodivisor on $S$. Thus $\operatorname{Tor}_1^R(R/xR,S)=0$. Apply the entire proper-ideal local criterion AG-CA-08, Corollary 4.4, with ideal $(x)$ and module $S$, which is finite over the stipulated Noetherian local overring $S$. The quotient is flat and its required Tor vanishes. This proves flatness over $R$ and completes the induction. ∎

Lemma F.10: NAK

Reading sources: Stacks, Tag 00DV; AI Integrated Stacks, pinned Algebra, lines 3732–3813.

For a ring $R$, an ideal $I$, and an $R$-module $M$, the following twelve forms of Nakayama's lemma hold. Write $\operatorname{rad}(R)$ for the Jacobson radical.

1. If $M$ is finite and $M=IM$, some $f\in1+I$ annihilates $M$.
2. Under the conditions of (1), $I\subset\operatorname{rad}(R)$ implies $M=0$.
3. If $N,N'\subset M$, $M=N+IN'$ and $N'$ is finite, some $f\in1+I$ satisfies $fM\subset N$ and $M_f=N_f$.
4. Under the conditions of (3), $I\subset\operatorname{rad}(R)$ implies $M=N$.
5. For a module map $N\to M$ with $M$ finite, surjectivity of $N/IN\to M/IM$ implies surjectivity of $N_f\to M_f$ for some $f\in1+I$.
6. Under the conditions of (5), $I\subset\operatorname{rad}(R)$ makes the original map $N\to M$ surjective.
7. If $M$ is finite and the images of $x_1,\ldots,x_n$ generate $M/IM$, these same elements generate $M_f$ over $R_f$ for some $f\in1+I$.
8. Under the conditions of (7), $I\subset\operatorname{rad}(R)$ implies that $x_1,\ldots,x_n$ generate $M$.
9. If $I$ is nilpotent, $M=IM$ implies $M=0$, with no finiteness requirement on $M$.
10. If $I$ is nilpotent and $M=N+IN'$ for submodules $N,N'$, then $M=N$, with no finiteness requirement on $N'$.
11. If $I$ is nilpotent and $N/IN\to M/IM$ is surjective, then $N\to M$ is surjective, without module finiteness assumptions.
12. If $I$ is nilpotent and a possibly infinite family $(x_\alpha)$ generates $M/IM$, it generates $M$.

Proof. The determinant assertion F.10 / nakayama, namely (1), supplies all finite-module cases. Choose a generating column $y=(y_1,\ldots,y_r)^{\mathsf t}$ of $M$. The equality $M=IM$ expresses each $y_i$ as an $I$-linear combination of the $y_j$, so for a matrix $Z$ with entries in $I$ we have $(1_r-Z)y=0$. By the proved adjugate identity F.70, $$\operatorname{adj}(1_r-Z)(1_r-Z)=\det(1_r-Z)1_r.$$ Consequently $f=\det(1_r-Z)$ kills every generator. Reduction modulo $I$ gives $f\equiv1$, as required. An empty generating list uses $f=1$.

If $I\subset\operatorname{rad}(R)$, F.34 proves that every element of $1+I$ is a unit. Multiplying $fM=0$ by its inverse proves (2).

For (3) and (4), put $Q=M/N$. Since the image of $N'$ generates $Q$ and $N'$ is finite, $Q$ is finite. The hypothesis $M=N+IN'$ gives $Q=IQ$. Apply (1) to obtain $fQ=0$, equivalently $fM\subset N$, and apply (2) for the equality $Q=0$ when the radical condition holds. Once $fM\subset N$, any fraction $m/f^a$ in $M_f$ equals $(fm)/f^{a+1}$ and comes from $N_f$. The inclusion $N_f\subset M_f$ follows from exact localization. Thus the stated equality of localized modules is proved as well.

For a map $u:N\to M$, let $Q=\operatorname{coker}u$. Right exactness identifies its quotient $Q/IQ$ with the cokernel of $N/IN\to M/IM$. The assumed surjectivity therefore gives $Q=IQ$. If $M$ is finite, so is $Q$. Assertion (1) makes $Q_f=0$, and exact localization makes $u_f$ surjective, proving (5). Assertion (2) gives $Q=0$ in the radical case, proving (6). Apply these conclusions to $R^n\to M$, $(r_i)\mapsto\sum r_ix_i$, to get (7) and (8).

Finally suppose $I^a=0$. Repeated substitution into $M=IM$ yields $M=I^aM=0$, proving (9) for an arbitrary module. In (10), $Q=M/N$ satisfies $Q=IQ$, whether or not the image of $N'$ is finite, so (9) gives $Q=0$. For (11), take the same cokernel $Q$ of the given map and use its zero quotient modulo $I$. For (12), quotient by the submodule generated by all $x_\alpha$; its quotient modulo $I$ is zero, so (9) kills it. This proves all twelve assertions, including empty generating lists and arbitrary nilpotent ideals. ∎

Lemma F.11: NL homotopy

Reading sources: Stacks, Tag 00S1; AI Integrated Stacks, pinned Algebra, lines 36700–36815.

Let $u:R\to R'$ and $\phi:S\to S'$ be compatible with the structure maps $R\to S$ and $R'\to S'$. Choose polynomial presentations $\alpha:P\twoheadrightarrow S$ and $\alpha':P'\twoheadrightarrow S'$, allowing arbitrary sets of variables. Then:

1. There is a map $\psi:P\to P'$ whose restriction to $R$ is $u$ and for which $\alpha'\psi=\phi\alpha$.
2. Any two such maps induce chain-homotopic maps $\operatorname{NL}(\alpha)\to\operatorname{NL}(\alpha')$.
3. Composing presentation maps composes their induced complex maps.
4. If $u$ and $\phi$ are isomorphisms, every induced map is a homotopy equivalence and a quasi-isomorphism.

In particular the comparison with the canonical presentation identifies $\operatorname{NL}(\alpha)$ with $\operatorname{NL}_{S/R}$ up to homotopy, and these identifications respect the stated functorialities up to homotopy.

Proof. Write $P=R[x_a\mid a\in E]$, $I=\ker\alpha$ and $I'=\ker\alpha'$. For each $a$, choose a polynomial in $P'$ lifting $\phi(\alpha(x_a))$. Sending the variables to these choices and the coefficients through $u$ gives $\psi$ by the polynomial universal property. Each polynomial uses only finitely many variables, so this construction also works when $E$ is infinite. It proves (1).

The two terms of the complex are $$C_1=I/I^2,\qquad C_0=\bigoplus_{a\in E}S\,dx_a,\qquad d:C_1\to C_0,\quad [f]\mapsto df.$$ The map is well-defined because $d(fg)=f\,dg+g\,df$ vanishes modulo $I$ when $f,g\in I$. Its target counterpart is $C'_1=I'/(I')^2\to C'_0=\Omega_{P'/R'}\otimes_{P'}S'$. Regard these target modules as $S$-modules through $\phi$. The induced maps are $$\psi_1([f])=[\psi(f)],\qquad \psi_0(dx_a)=d\psi(x_a).$$ The map on $I/I^2$ is well-defined since $\psi(I)\subset I'$. On a monomial the product rule gives $\psi_0(df)=d\psi(f)$, since differentiation over $R'$ kills the transformed coefficient from $R$. Addition gives this identity for every polynomial. Thus these are complex maps.

For another choice $\chi$, both evaluations of any $p\in P$ have the same residue in $S'$. Set $$D(p)=\psi(p)-\chi(p)\pmod{(I')^2}\in I'/(I')^2.$$ Both choices give the same action of $P$ on this module: their difference lies in $I'$, which kills $I'/(I')^2$. That action factors through $S$, because the image of $I$ is in $I'$. The exact product identity $$\psi(pq)-\chi(pq) =\psi(p)\bigl(\psi(q)-\chi(q)\bigr) +\bigl(\psi(p)-\chi(p)\bigr)\chi(q)$$ therefore becomes the derivation rule for $D$. It is additive and vanishes on $R$. Define the $S$-linear map $$h:C_0\longrightarrow C'_1,\qquad h(dx_a)=D(x_a).$$ Expanding a monomial by the derivation rule shows $h(dp)=D(p)$ for every polynomial $p$, including constant and empty-variable cases.

Both homotopy identities can now be checked without a diagram omission. For $f\in I$, $$h(d[f])=D(f)=\psi_1([f])-\chi_1([f]).$$ For each free differential generator, $$d'h(dx_a)=d\bigl(\psi(x_a)-\chi(x_a)\bigr) =\psi_0(dx_a)-\chi_0(dx_a).$$ The first identity holds on all relation classes and the second on a free basis, so $$\psi_1-\chi_1=hd,\qquad \psi_0-\chi_0=d'h.$$ These are exactly a chain homotopy, proving (2).

For a subsequent presentation map $\eta:P'\to P''$, the relation-class map sends $[f]$ successively to $[\psi(f)]$ and $[\eta\psi(f)]$. The differential map sends $dx_a$ to $d\psi(x_a)$ and then to $d(\eta\psi(x_a))$, by the same monomial product-rule calculation. Thus the composite equals the map of the composite presentation, proving (3).

If the two structural square maps are isomorphisms, apply (1) to their inverse square to get a presentation map in the reverse direction. Both composites lie over identity squares. By (2), their complex maps are homotopic to the respective identity maps. This is the required homotopy equivalence. A homotopy changes a degree-one map by $hd$ and a degree-zero map by $d'h$; the former is zero on the kernel and the latter has zero class in the cokernel. Thus homotopic maps act equally on both homology modules. The two reverse maps consequently induce inverse isomorphisms on kernel and cokernel, proving the quasi-isomorphism assertion. Taking identity structural maps and the canonical polynomial presentation proves the last assertion, with composition compatibility already established. ∎

Lemma F.12: Noetherian field extension

Reading source: AI Integrated Stacks, pinned Algebra, lines 6331–6347.

If $A$ is a Noetherian algebra over a field $k$, then extending scalars from $k$ to a finitely generated field extension $K$ leaves it Noetherian.

Proof. Select a finite field-generating list $\xi_1,\ldots,\xi_r$ for $K/k$, and put $B=k[\xi_1,\ldots,\xi_r]\subset K$. The field of fractions of $B$ is $K$, since it contains both $k$ and all chosen field generators. Therefore $$K=(B\setminus\{0\})^{-1}B.$$ A polynomial surjection $k[T_1,\ldots,T_r]\to B$ gives a surjection $A[T_1,\ldots,T_r]\to A\otimes_kB$. The Hilbert-basis and quotient arguments in AG-CA-03, Theorem 2.1, make the latter algebra Noetherian.

Localizing it at the images of the nonzero elements of $B$ gives $A\otimes_kK$: the maps in both directions send a pure tensor with a fraction to that fraction of the corresponding pure tensor. They are inverse by the tensor and localization universal properties. AG-CA-03, Theorem 2.1, also proves that a localization of a Noetherian ring is Noetherian, giving the result. For a finite extension the same conclusion follows from its finite $k$-basis, which makes the tensor algebra a finite free $A$-module; its ideals are then finite $A$-submodules. The argument requires finite field generation, rather than finite degree. ∎

Lemma F.13: another variant local criterion flatness

Reading source: AI Integrated Stacks, pinned Algebra, lines 24942–24990.

Consider a commutative square of local homomorphisms between Noetherian local rings $$ \begin{array}{ccc} R&\longrightarrow&R'\\ \downarrow&&\downarrow\\ S&\longrightarrow&S'. \end{array} $$ Let $I\subsetneq R$ and let $M$ be a finite $S$-module. Put $I'=IR'$ and $M'=M\otimes_SS'$. Assume $S'$ is a localization of $B=S\otimes_RR'$, the module $M/IM$ is flat over $R/I$, and the natural map $$\operatorname{Tor}_1^R(M,R/I) \longrightarrow\operatorname{Tor}_1^{R'}(M',R'/I')$$ is zero. Then $M'$ is flat over $R'$.

Proof. We prove the needed Tor surjection and then apply the proper-ideal criterion. All free-resolution comparisons, naturality and balance used here are the complete arbitrary-module calculus of AG-RG-S01, Lemma 1.A.

Choose an $R$-free presentation $0\to K\to F\to M\to0$, with no finiteness requirement on $F$. Set $T=\operatorname{Tor}_1^R(M,R/I)$, and let $U$ be the image of $K/IK$ in $F/IF$. The free-presentation kernel formula gives exact sequences $$0\to T\to K/IK\to U\to0,\qquad 0\to U\to F/IF\to M/IM\to0.$$ Since the last quotient is flat, the proved flat-syzygy criterion AG-CA-07, Theorem 6.1, makes $U$ flat over $R/I$. Tensoring the first sequence with an $R/I$-module $N$ is consequently left exact. Tensoring the second is left exact because its last quotient is flat. Their kernel comparison gives the natural identity $$\operatorname{Tor}_1^R(M,N)=T\otimes_{R/I}N.$$ This uses only the kernel formula for a free presentation; it does not assume that $K$ is flat.

Take $N=R'/I'$. In $F_0=F\otimes_RR'$ let $K_0$ be the image of $K\otimes_RR'$. Then $F_0$ is $R'$-free and $F_0/K_0=M_0=M\otimes_RR'$. The surjection $K\otimes_RR'\to K_0$ remains surjective after tensoring with $R'/I'$. A cycle in $$\ker\bigl(K_0\otimes_{R'}R'/I'\to F_0\otimes_{R'}R'/I'\bigr)$$ therefore has a lift in $K\otimes_RR'/I'$. That lift maps to zero in the same free-module target, so is already a source cycle. Thus $$\operatorname{Tor}_1^R(M,R'/I') \twoheadrightarrow\operatorname{Tor}_1^{R'}(M_0,R'/I').$$ The preceding natural identity identifies its source with $T\otimes_RR'$.

Write $S'=V^{-1}B$, so $M'=V^{-1}M_0$. We justify localization of the last Tor group with its natural $B$-action. Choose an $R'$-free resolution of $R'/I'$. Balance identifies Tor with the homology after tensoring this resolution with $M_0$; multiplication by $B$ acts on each tensor term. Exact localization of $B$-modules commutes with kernels and images, and tensoring the same resolution with $V^{-1}M_0$ is the localization of each term. Hence $$\operatorname{Tor}_1^{R'}(M',R'/I') =V^{-1}\operatorname{Tor}_1^{R'}(M_0,R'/I').$$ The preceding surjection is $B$-linear: multiplication by $S$ acts by the functorial module maps on $M$, while $R'$ acts by scalar extension; the comparison calculus in Lemma 1.A makes these actions compatible. Localizing that surjection consequently gives $$V^{-1}(T\otimes_RR')\twoheadrightarrow \operatorname{Tor}_1^{R'}(M',R'/I').$$ The hypothesis says the original natural map from $T$ to this target is zero. Its $R'$-linear extension is zero, and its extension across $V$ is still zero, because the target is a module over $V^{-1}B$. The displayed surjection is therefore the zero map, forcing its target to vanish.

Finally $M'/I'M'$ is obtained from the flat $R/I$-module $M/IM$ by scalar extension to $R'/I'$ and localization at $V$. Scalar extension preserves flatness. For the localization, an injection of $R'/I'$-modules remains injective after tensoring with the unlocalized flat module, and remains injective after exact localization of the resulting $B$-modules. This proves flatness of the localized quotient. The ideal $I'$ is proper because $R\to R'$ is local and $I\subset\mathfrak m_R$. The module $M'$ is finite over the stipulated Noetherian local ring $S'$. AG-CA-08, Corollary 4.4, applies with this quotient flatness and the Tor vanishing just proved, and gives $R'$-flatness. No flatness assumption on $R\to R'$ was added. ∎

Lemma F.14: base change relative global complete intersection

Reading sources: Stacks, Tag 00SS; AI Integrated Stacks, pinned Algebra, lines 38076–38106.

For a relative global complete-intersection presentation $$S=R[x_1,\ldots,x_n]/(f_1,\ldots,f_c),$$ all three of the following presentations have the same property:

1. For any $R\to R'$, the coefficientwise base change $S\otimes_RR'=R'[x_1,\ldots,x_n]/(f_1,\ldots,f_c)$.
2. If $g\in S$ is represented by $h\in R[x_1,\ldots,x_n]$, the principal localization $$S_g=R[x_1,\ldots,x_n,Z]/(f_1,\ldots,f_c,hZ-1).$$
3. If $f\in R$ already has a unit image in $S$, its presentation over $R_f$ obtained by viewing the same equations in $R_f[x_1,\ldots,x_n]$.

Proof. The fibre condition of F.4 is a dimension condition. We first prove the field-extension calculation it requires. Let $A$ be a finite-type $k$-algebra and $K/k$ any field extension. Its nilradical is nilpotent and has finitely many minimal primes, by AG-CA-03, Proposition 2.2. Nilpotence persists after tensoring with $K$ and does not affect dimension, since every prime contains every nilpotent element.

For the reduced quotient, let its minimal primes be $\mathfrak q_1,\ldots,\mathfrak q_r$. Their product is zero, being contained in their zero intersection. After field extension the product of their extended ideals is still zero. Every prime contains one of these extended ideals: otherwise choose one element outside the prime from each ideal and multiply, contradicting primality. Thus the extended spectrum is the finite union of the closed spectra of $$(A/\mathfrak q_i)\otimes_kK.$$ Every prime chain belongs to one of these closed sets, by choosing an ideal contained in its smallest prime. Dimension is therefore the maximum of their dimensions.

For each domain $A/\mathfrak q_i$, Noether normalization AG-CA-09, Corollary 3.2, supplies an injective finite polynomial subalgebra $k[t_1,\ldots,t_{d_i}]$. Exact field tensor preserves its injection and finite module generators. The corresponding $K$-algebra is thus integral and injective over $K[t_1,\ldots,t_{d_i}]$. The entire integral-dimension theorem AG-CA-05, Theorem 5.1, gives dimension $d_i$. Taking the maximum proves $\dim(A\otimes_kK)=\dim A$. The zero algebra has empty spectrum and causes no nonempty-fibre case.

For (1), a fibre at a prime of $R'$ is the scalar extension of the original fibre at its contraction, by associativity of tensor products and the induced residue-field inclusion. A nonempty new fibre therefore has the original dimension $n-c$ by the calculation above. The displayed coefficientwise presentation proves (1).

For (2), the field case of F.77 proves that each original nonempty fibre is equidimensional of dimension $n-c$. Indeed its minimal-prime height calculation applies to its polynomial presentation over the fibre field, without any Noetherian assumption on $R$. A principal open in a surviving irreducible component has the same dimension as that component: its coordinate domain has unchanged fraction field, and AG-CA-09, Theorem 4.2, computes dimension by its transcendence degree. The localized nonempty fibre therefore still has dimension $n-c$. Adjoining $Z$ and $hZ-1$ adds one variable and one equation, preserving the difference $n-c$. Evaluation at $Z=g^{-1}$ and adjoining that inverse give both directions of the claimed algebra isomorphism.

For (3), multiplication gives $S\otimes_RR_f\cong S$, because the image of $f$ already has an inverse. Its inverse sends $s$ to $s\otimes1$; the forced inverse of $f$ verifies the other composite on fractions. Apply (1) to this base localization. All three arguments apply over arbitrary base rings and to empty fibres. ∎

Lemma F.15: change base NL

Reading sources: Stacks, Tag 00S4; AI Integrated Stacks, pinned Algebra, lines 36963–36981.

Let $\alpha:P\twoheadrightarrow S$ be a polynomial presentation over $R$, with an arbitrary variable set, and let $R\to R'$ be flat. Put $P'=P\otimes_RR'$, $S'=S\otimes_RR'$, and let $\alpha':P'\twoheadrightarrow S'$ be the resulting presentation. There is an isomorphism of two-term complexes $$\operatorname{NL}(\alpha)\otimes_RR' =\operatorname{NL}(\alpha)\otimes_SS' \cong\operatorname{NL}(\alpha').$$ Consequently the canonical comparison $$\operatorname{NL}_{S/R}\otimes_SS'\longrightarrow \operatorname{NL}_{S'/R'}$$ is a homotopy equivalence.

Proof. Write $I=\ker\alpha$. Tensoring $0\to I\to P\to S\to0$ with the flat $R'$ identifies $I'=\ker\alpha'$ with $I\otimes_RR'$. The square $(I')^2$ is the image of $I^2\otimes_RR'$: a product of two elementary tensors from $I'$ is the image of the product of their $I$-entries, and every product generator of $I^2$ is obtained this way. Tensoring the quotient therefore gives $$I'/(I')^2\cong(I/I^2)\otimes_RR'.$$ For any $S$-module $E$, the pure-tensor maps $$E\otimes_SS'\longrightarrow E\otimes_RR', \quad e\otimes(s\otimes r')\longmapsto se\otimes r', \qquad e\otimes r'\longmapsto e\otimes(1\otimes r')$$ are inverse. They give the other tensor description of the conormal term.

The polynomial differential term is free on the same $dx_a$ over $S$ and over $S'$. Differentiating a polynomial commutes with changing its coefficients, since every monomial derivative multiplies by the same integer exponent and lowers the same variable degree. Thus both differentials agree on every relation class, proving the complex isomorphism. Each polynomial has finite support, so no finite-variable assumption enters.

Apply F.11 to compare each presentation with the canonical one. Tensoring its explicit homotopy identities with $S'$ preserves those identities. The complex isomorphism just proved, between these comparison homotopy equivalences, therefore identifies the displayed canonical map up to homotopy and supplies its inverse up to homotopy. This proves the final assertion without assuming that $S$ is flat over $R$. ∎

Lemma F.16: characterize etale

Reading source: AI Integrated Stacks, pinned Algebra, lines 40942–40975.

Let $\mathfrak q\in\operatorname{Spec}S$ lie over $\mathfrak p\in\operatorname{Spec}R$. The map $R\to S$ is étale at $\mathfrak q$ if the following four hypotheses hold: it is finitely presented; $R_{\mathfrak p}\to S_{\mathfrak q}$ is flat; the ideal $\mathfrak pS_{\mathfrak q}$ is the maximal ideal of $S_{\mathfrak q}$; and $\kappa(\mathfrak q)/\kappa(\mathfrak p)$ is finite separable.

Proof. Write $k=\kappa(\mathfrak p)$ and $L=\kappa(\mathfrak q)$. The stipulated maximal-ideal equality identifies the local fibre ring with the field $L$. Since $L/k$ is finite, the complete isolated-fibre argument F.56 supplies a principal neighbourhood of $\mathfrak q$ having no other point in this fibre. Its finite presentation is preserved by F.35. On this neighbourhood the fibre algebra $B=S\otimes_Rk$ has exactly one prime. It is local: every nonunit lies in a maximal ideal, and there is only one such ideal. Therefore its localization at that prime is $B$ itself. This localization is the given local fibre $L$, so $B=L$.

We prove directly the smoothness and differential vanishing of this finite separable field fibre. A finite $k$-basis generates $L$ as a field; adjoining its entries gives a finite tower $k=E_0\subset E_1\subset\cdots\subset E_t=L$, with $E_i=E_{i-1}(\alpha_i)$. Let $p_i(T)$ be the monic minimal polynomial of $\alpha_i$ over $E_{i-1}$. The complete monic-division and Bezout arguments AG-RG-S02, Lemma 4.A give $E_i=E_{i-1}[T]/(p_i)$. Each $p_i$ divides the separable minimal polynomial of $\alpha_i$ over $k$, now regarded over $E_{i-1}$, so $p_i'(\alpha_i)\ne0$.

Choose polynomial representatives in the preceding variables for all coefficients of $p_i$, obtaining $P_i(T_1,\ldots,T_i)$. Successive monic division gives $$L=k[T_1,\ldots,T_t]/(P_1,\ldots,P_t).$$ The Jacobian over $L$ is triangular with diagonal $p_i'(\alpha_i)$, so is invertible. The equation classes give a surjection $L^t\to I/I^2$, where $I=(P_1,\ldots,P_t)$, and its composite with the conormal differential to $L^t$ is this invertible Jacobian. The surjection must also be injective, and the conormal differential is an isomorphism. Hence its kernel and cokernel vanish. Definition F.6 and presentation comparison give smoothness of $L/k$ and $\Omega_{L/k}=0$. For $L=k$, use the empty tower and the two zero modules.

The local flatness hypothesis and the finite presentation now meet every condition of F.45, giving a smooth principal neighbourhood of $\mathfrak q$. The complete arbitrary-base chart result F.83 gives a further neighbourhood with a standard smooth presentation $$S=R[X_1,\ldots,X_n]/(f_1,\ldots,f_c).$$ Every inverted element has nonzero residue in $L$, so the fibre remains exactly $L$. By F.85(2), the differential module is free on $dX_{c+1},\ldots,dX_n$. The full differential scalar-extension calculation AG-CA-15, Theorem 2.1, identifies its fibre with $\Omega_{L/k}=0$. Thus $L^{n-c}=0$, forcing $n=c$.

We check the lifting definition of étaleness on this final chart. Given a map to $D/J$ with $J^2=0$, lift its variable tuple to $a\in D^n$. The error vector $f(a)$ lies in $J^n$. The square Jacobian $A(a)$ is invertible: its determinant is a unit modulo $J$, a lift of an inverse gives a product $1+j$ with inverse $1-j$, and the adjugate identity F.70 gives the matrix inverse. Take $$b=a-A(a)^{-1}f(a).$$ Expanding each monomial and dropping terms with two entries from $J$ gives $f(b)=f(a)+A(a)(b-a)=0$. Thus $b$ defines a lift. A second lifting root has the form $b+\epsilon$ for $\epsilon\in J^n$. The same expansion gives $A(b)\epsilon=0$, and $A(b)$ is invertible by its unchanged reduction modulo $J$, so $\epsilon=0$. Every square-zero test therefore has exactly one lift. With finite presentation this is étaleness by the programme definition. The chart contains $\mathfrak q$, proving the claim for the original map. Empty variable lists use the unique empty tuple and determinant $1$; no Noetherian or perfect-field hypothesis was introduced. ∎

Lemma F.17: characterize formally smooth field extension

Reading source: AI Integrated Stacks, pinned Algebra, lines 45959–45969.

For any field extension $K/k$, the vanishing of $H_1(\operatorname{NL}_{K/k})$ is equivalent to formal smoothness of $K$ as a $k$-algebra.

Proof. Choose a polynomial surjection $P=k[X_a\mid a\in E]\to K$, allowing an infinite set $E$, with kernel $I$. By F.11, the homology in question is the kernel of $$d:I/I^2\longrightarrow\Omega_{P/k}\otimes_PK.$$ Its cokernel is $\Omega_{K/k}$, by the conormal calculation. When this kernel is zero, the sequence with these three modules is short exact. A vector-space basis of $\Omega_{K/k}$ can be lifted to the middle module. Extending those lifts $K$-linearly splits the surjection, and hence the sequence.

The entire arbitrary-variable criterion AG-CA-17, Theorem 3.1, proves that this split conormal sequence is equivalent to formal smoothness. Its hypotheses apply to the chosen polynomial presentation over $k$ with target $K$ and impose no finite generation. Conversely, formal smoothness gives a split exact conormal sequence by that same complete theorem, so its left map has zero kernel. F.11 identifies that kernel with the presentation-independent first cotangent homology, also denoted $H_1(L_{K/k})$ in the reading source. This proves both directions for arbitrary field extensions. ∎

Lemma F.18: characterize smooth kbar

Reading sources: Stacks, Tag 00TS; AI Integrated Stacks, pinned Algebra, lines 39962–40064.

Let $S$ be a finite-type algebra over an algebraically closed field $k$, and let $\mathfrak m$ be a maximal ideal. Write $$d=\dim S_{\mathfrak m},\qquad e=\dim_{\kappa(\mathfrak m)} \bigl(\Omega_{S/k}\otimes_S\kappa(\mathfrak m)\bigr).$$ The following four conditions are equivalent: $S_{\mathfrak m}$ is regular local; $e\leq d$; $e=d$; and some $g\in S\setminus\mathfrak m$ has $S_g/k$ smooth.

Proof. The complete rational-point result AG-CA-18, Proposition 3.1, gives $\kappa(\mathfrak m)=k$ and $$\Omega_{S/k}\otimes_Sk\cong\mathfrak m/\mathfrak m^2.$$ The embedding-dimension theorem of AG-CA-14, Section 1, says $d\leq\dim_k\mathfrak m/\mathfrak m^2$, with equality exactly for regularity. Thus the first three conditions are equivalent. The same complete AG-CA-18 proposition proves that smoothness at this closed rational point implies regularity, giving the implication from the fourth to the first.

For the converse we construct a smooth chart and identify its entire relation ideal. Suppose $S_{\mathfrak m}$ is regular. Choose a finite polynomial presentation $S=k[x_1,\ldots,x_n]/I$, shifting each generator by its residue so that the preimage of $\mathfrak m$ is $\mathfrak n=(x_1,\ldots,x_n)$. The exact conormal sequence at this point is $$I/\mathfrak nI\longrightarrow\mathfrak n/\mathfrak n^2 \longrightarrow\mathfrak m/\mathfrak m^2\longrightarrow0.$$ Since the middle dimension is $n$ and the last is $d$, select $c=n-d$ elements $f_1,\ldots,f_c\in I$ whose linear classes form a basis of the kernel. Put $J=(f_1,\ldots,f_c)$ and $T=k[x]/J$, with maximal ideal $\mathfrak m'=\mathfrak n/J$. The corresponding conormal sequence shows that the cotangent space of $T_{\mathfrak m'}$ has dimension $d$. This local ring surjects onto $S_{\mathfrak m}$, so its dimension is at least $d$; its embedding dimension bounds it above by $d$. It is therefore regular of dimension $d$.

By AG-CA-14, Theorem 1.1, $T_{\mathfrak m'}$ is a domain. The surjection to $S_{\mathfrak m}$ has zero kernel. To prove this, a nonzero kernel would put every lifted quotient prime strictly above $(0)$. A prime chain of length $d$ in the quotient exists because its dimension is the finite integer $d$. Prepending $(0)$ to the lifted chain would give length $d+1$ in $T_{\mathfrak m'}$, a contradiction. Thus the local surjection is an isomorphism.

The independent linear coefficient rows of the $f_i$ have a nonzero $c$-variable Jacobian minor $\Delta$ at this point. Existence of such a minor follows from the explicit pivot induction in D.2; for $c=0$ take $\Delta=1$. Reorder the pivot variables first. The presentation $$T_\Delta=k[x_1,\ldots,x_n,u]/(f_1,\ldots,f_c,u\Delta-1)$$ has a selected Jacobian determinant $\Delta^2$: the last column is zero in the first $c$ rows, its last entry is $\Delta$, and the first square block has determinant $\Delta$. This determinant is a unit in the quotient. The complete principal-localization standard-smooth calculation F.85 therefore proves $T_\Delta/k$ smooth.

It remains to pass from $T$ to $S$. The kernel of $T\to S$ is finite since $T$ is Noetherian, and becomes zero at $\mathfrak m'$ by the local isomorphism. For each of its finite generators take a localization annihilator outside $\mathfrak m'$. Their product $a$ still avoids $\mathfrak m'$ and kills the entire kernel. Exact localization of the surjection then gives $T_a\cong S_{\bar a}$, hence $$T_{a\Delta}\cong S_{\overline{a\Delta}}.$$ The left side is a principal localization of the smooth algebra $T_\Delta$, and F.85(4) supplies its smooth presentation. On the right $g=\overline{a\Delta}$ avoids $\mathfrak m$. This gives the fourth condition, with the exact kernel eliminated on the whole neighbourhood. Empty equation lists obey the same construction. The algebraically closed ground-field assumption is retained throughout. ∎

Lemma F.19: charpoly

Reading source: Stacks, Tag 00DX.

For a square matrix \(M\) of size \(n\) over any commutative ring \(R\), let \(P(T)=\det(T1_n-M)\). Then \(P(M)=0\) as an endomorphism of \(R^n\).

Proof. Define determinants by the signed permutation formula. Grouping its terms according to the entry in a fixed row gives the cofactor expansion. If that row is replaced by another row, the same expansion is zero: permutations pair with the transposition of the two equal rows, and the paired terms have opposite signs. The column calculation is identical. Thus, for every square matrix \(C\), the transpose of its signed cofactor matrix, denoted \(\operatorname{adj}(C)\), satisfies \[ C\operatorname{adj}(C)=\operatorname{adj}(C)C=(\det C)1_n. \] This argument uses addition and multiplication only, so it applies over \(R[T]\) as well.

Assume \(n>0\), write \[ \operatorname{adj}(T1_n-M)=\sum_{j=0}^{n-1}C_jT^j, \qquad P(T)=\sum_{j=0}^n p_jT^j,\quad p_n=1. \] The degree bound follows because each cofactor is a determinant of size \(n-1\). Comparing coefficients in the first adjugate identity gives \[ -MC_0=p_0 1_n,\qquad C_{j-1}-MC_j=p_j1_n\ (1\leq j<n),\qquad C_{n-1}=1_n. \] Multiply the equation indexed by \(j\) on the left by \(M^j\), the constant equation by \(1_n\), and the last equation by \(M^n\). Adding them cancels every \(M^{j+1}C_j\) against its next occurrence. The result is \(\sum_{j=0}^n p_jM^j=0\). No substitution into a polynomial with noncommuting matrix coefficients, and no division, is involved. If \(n=0\), the empty determinant gives \(P=1\); its value is the identity of the zero module, which is the zero endomorphism. ∎

Lemma F.20: codimension

Reading source: AI Integrated Stacks, pinned Algebra, lines 29578–29596.

Let $S'\twoheadrightarrow S$ be a surjection of finite-type algebras over a field $k$. For a prime $\mathfrak p\subset S$, let $\mathfrak p'\subset S'$ be its inverse image, and let $x,x'$ denote the corresponding points. Then $$\dim_{x'}\operatorname{Spec}S'-\dim_x\operatorname{Spec}S =\operatorname{ht}_{S'}\mathfrak p'-\operatorname{ht}_S\mathfrak p.$$

Proof. The induced map $S'/\mathfrak p'\to S/\mathfrak p$ is an isomorphism: surjectivity descends, and the definition of $\mathfrak p'$ makes its kernel zero. Its fraction-field map therefore identifies $\kappa(\mathfrak p')$ with $\kappa(\mathfrak p)$ as extensions of $k$. Call their common transcendence degree $e$.

The complete point-dimension argument AG-CA-09, Theorem 6.1, applies separately to the two finite-type algebras and gives $$\dim_{x'}\operatorname{Spec}S'=\dim S'_{\mathfrak p'}+e,\qquad \dim_x\operatorname{Spec}S=\dim S_{\mathfrak p}+e.$$ Here point dimension is the minimum of the dimensions of open neighbourhoods. Localization identifies chains of primes in $S'_{\mathfrak p'}$ with chains in $S'$ ending below $\mathfrak p'$; taking their supremum gives $\dim S'_{\mathfrak p'}=\operatorname{ht}_{S'}\mathfrak p'$. The same chain correspondence gives $\dim S_{\mathfrak p}=\operatorname{ht}_S\mathfrak p$. All dimensions are finite by that earlier finite-type theorem. Subtract the two displayed equalities; the identical terms $e$ cancel and give the assertion. No equidimensionality assumption is needed. ∎

Lemma F.21: colimit eventually flat

Reading source: AI Integrated Stacks, pinned Algebra, lines 34105–34145.

Consider a directed system of local maps $R_\lambda\to S_\lambda$ between Noetherian local rings, with local transition maps, and let $R=\operatorname*{colim}_\lambda R_\lambda$ and $S=\operatorname*{colim}_\lambda S_\lambda$. Suppose that for $\mu\geq\lambda$, $S_\mu$ is the specified localization of $S_\lambda\otimes_{R_\lambda}R_\mu$. Let $M_\lambda$ be finite $S_\lambda$-modules with $M_\mu=M_\lambda\otimes_{S_\lambda}S_\mu$, and put $M=\operatorname*{colim}_\lambda M_\lambda$. These are exactly the kinds of models supplied by F.59 and F.60. If $M$ is $R$-flat, then $M_\mu$ is $R_\mu$-flat at some stage.

Proof. Fix $\lambda$, write $\mathfrak m=\mathfrak m_\lambda$, and form $$T=\operatorname{Tor}_1^{R_\lambda}(M_\lambda,R_\lambda/\mathfrak m) =\ker\bigl(\mathfrak m\otimes_{R_\lambda}M_\lambda\longrightarrow M_\lambda\bigr).$$ The free-resolution and ideal-kernel calculation is AG-RG-S01, Lemma 1.A. The ideal $\mathfrak m$ is finite over $R_\lambda$, so its tensor with the finite $S_\lambda$-module $M_\lambda$ is finite over $S_\lambda$. Since $S_\lambda$ is Noetherian, its kernel $T$ is finite as well. Choose generators $\xi_1,\ldots,\xi_a$ over $S_\lambda$.

For $\nu\geq\lambda$, put $I_\nu=\mathfrak mR_\nu$. The transition of each $\xi_i$ lies in $$\ker(I_\nu\otimes_{R_\nu}M_\nu\longrightarrow M_\nu) =\operatorname{Tor}_1^{R_\nu}(M_\nu,R_\nu/I_\nu).$$ The colimit of the ideals $I_\nu$ is $\mathfrak mR$: a finite sum of images of generators has a representative at a common stage, and equality of two representatives holds at a later stage. The same finite sums and the finitely many tensor relations show $$\operatorname*{colim}_{\nu\geq\lambda} (I_\nu\otimes_{R_\nu}M_\nu) \cong(\mathfrak mR)\otimes_R M.$$ Multiplication from the last module into $M$ is injective, because tensoring the inclusion $\mathfrak mR\subset R$ with the flat module $M$ preserves injectivity. Each $\xi_i$ consequently has zero image in this colimit. The definition of a directed colimit says that its representative is already zero at some later stage. Choose a common successor $\mu$ for these finitely many witnesses. The entire natural map $$T\longrightarrow\operatorname{Tor}_1^{R_\mu}(M_\mu,R_\mu/\mathfrak mR_\mu)$$ is now zero, since it is compatible with the scalar map $S_\lambda\to S_\mu$ and kills every chosen generator.

Apply the full local-square criterion F.13 to $(R_\lambda,S_\lambda,R_\mu,S_\mu)$ and $I=\mathfrak m$. This ideal is proper and its extension is proper because $R_\lambda\to R_\mu$ is local. The module $M_\lambda/\mathfrak mM_\lambda$ is flat over the field $R_\lambda/\mathfrak m$: any vector-space basis expresses it as a free module, without a finite-dimensional assumption. The localized tensor description and finite $S_\lambda$-module hypothesis are exactly those stipulated above. The zero Tor map just proved supplies the last hypothesis of F.13, giving $R_\mu$-flatness of $M_\mu$. ∎

Lemma F.22: colimit formally etale

Reading source: AI Integrated Stacks, pinned Algebra, lines 42732–42747.

Let $(S_i)$ be a filtered system of formally étale $R$-algebras. Then $\operatorname*{colim}_iS_i$ is formally étale over $R$.

Proof. Set $S=\operatorname*{colim}_iS_i$. Test the formal lifting property on an $R$-algebra $A$, an ideal $J\subset A$ with $J^2=0$, and an $R$-map $u:S\to A/J$. The restriction $u_i:S_i\to A/J$ has exactly one lifting $v_i:S_i\to A$, by the assumed formal étaleness.

If $\phi_{ij}:S_i\to S_j$ is a transition map, then $v_j\phi_{ij}$ and $v_i$ both lift $u_i$. Uniqueness on $S_i$ forces $v_j\phi_{ij}=v_i$. The colimit universal property therefore gives a map $v:S\to A$ whose restrictions are all $v_i$. Its reduction agrees with $u$ on every stage and hence on $S$. If $v'$ is another lifting, its restriction to each $S_i$ equals $v_i$, so the same universal property gives $v'=v$.

For a general nilpotent ideal $J$ with $J^N=0$, apply the square-zero test successively to $A/J^{n+1}\to A/J^n$ for $1\leq n<N$. Its kernel $J^n/J^{n+1}$ has square zero because $2n\geq n+1$. Each lifting step exists uniquely; induction reaches $A$ and also proves uniqueness there. Thus either formulation of the formal étale test holds. No finite-presentation condition on any stage or on the colimit enters the proof. ∎

Lemma F.23: colimit syntomic

Reading source: AI Integrated Stacks, pinned Algebra, lines 46090–46123.

Every extension of fields $K/k$ is a filtered union of $k$-subalgebras that are global complete intersections. If $K/k$ is separable, it is a filtered union of smooth $k$-subalgebras.

Proof. We first place an arbitrary finite set $E\subset K$ in one of the asserted complete intersections. The subfield $L=k(E)$ has a finite transcendence basis $t_1,\ldots,t_d$ and is finite over $F=k(t_1,\ldots,t_d)$. Choose a finite tower $$F=L_0\subset L_1\subset\cdots\subset L_r=L,\qquad L_i=L_{i-1}(y_i).$$ Write $e_i$ for the degree of the monic minimal polynomial of $y_i$ over $L_{i-1}$. Successive monic division, proved in AG-RG-S02, Lemma 4.A, makes the products $\prod_{j<i}y_j^{a_j}$ with $0\leq a_j<e_j$ an $F$-basis of $L_{i-1}$. Expand each coefficient of the minimal polynomial in this basis. Its basis coefficients belong to $F$; finitely many rational-function denominators occur.

Choose one nonzero $h\in k[t_1,\ldots,t_d]$ clearing these denominators, and put $B=k[t_1,\ldots,t_d,1/h]$. The expansions give monic polynomials $$P_i\in B[Y_1,\ldots,Y_i],\qquad C=B[Y_1,\ldots,Y_r]/(P_1,\ldots,P_r).$$ At each successive quotient monic division gives a free module on $1,Y_i,\ldots,Y_i^{e_i-1}$. Hence $C$ is finite free over $B$ on the corresponding product monomials. Multiplication by a monic polynomial in a polynomial ring is injective even over a ring with zero divisors: in a nonzero product, the leading coefficient of the other factor survives unchanged as the leading coefficient. Applying this fact at each stage proves that the displayed equation list is a regular sequence.

After tensoring with $F$, these same quotients are the chosen minimal-polynomial tower, so $C\otimes_BF=L$. Freeness over the domain $B$ makes the map $C\to C\otimes_BF$ injective: a nonzero element of $B$ cannot kill a nonzero coordinate in a free basis. We may thus regard $C$ as a subalgebra of $L$. Expand the finitely many elements of $E$ in its product basis over $F$, and enlarge $h$ to clear their coefficients as well. The construction with this enlarged $B$ then contains $E$.

The coefficient ring $B$ has dimension $d$ by the finite-type-domain dimension theorem AG-CA-09, Theorem 4.2; its fraction field is $F$. The finite injective integral map $B\to C$ preserves dimension by AG-CA-05, Theorem 5.1. Encoding $1/h$ by a variable $Z$ gives $$C\cong k[t_1,\ldots,t_d,Y_1,\ldots,Y_r,Z]/ (P_1,\ldots,P_r,hZ-1),$$ where the coefficients of the $P_i$ are polynomial representatives modulo $hZ-1$. This presentation has $d+r+1$ variables, $r+1$ equations, and dimension $d$, so it satisfies the global complete-intersection definition F.4 over $k$.

If the extension is separable, F.5 makes the finitely generated subfield $L$ separably generated. The entire forward construction in F.64 supplies a smooth finite-type domain $D\subset L$ with fraction field $L$. For every element of $E$ choose a numerator and a nonzero denominator in $D$. Localizing at their finite product gives a smooth algebra containing $E$, by the principal-localization calculation F.85.

Finally, in either case two constructed subalgebras have finite sets of algebra generators over $k$. Apply the corresponding construction to the union of those generator sets. The resulting subalgebra contains both, proving directedness under inclusion. Applying the construction to a singleton includes each element of $K$. The union, and thus its filtered colimit, is exactly $K$. Empty towers and transcendence bases use the same argument with their empty products equal to one. ∎

Lemma F.24: colimits NL

Reading source: AI Integrated Stacks, pinned Algebra, lines 36983–36999.

For a filtered system of ring maps $R_\lambda\to S_\lambda$, put $R=\operatorname*{colim}R_\lambda$ and $S=\operatorname*{colim}S_\lambda$. The canonical two-term conormal complexes satisfy $$\operatorname{NL}_{S/R}\cong \operatorname*{colim}_\lambda\operatorname{NL}_{S_\lambda/R_\lambda}.$$

Proof. Use the canonical presentation $P_\lambda=R_\lambda[X_s\mid s\in S_\lambda]\twoheadrightarrow S_\lambda$, with kernel $I_\lambda$. A transition sends coefficients along the base map and sends $X_s$ to the variable named by the image of $s$. Thus distinct variables may become equal; injective transition maps are not assumed. Let $P=R[X_s\mid s\in S]$ and $I=\ker(P\to S)$.

Every element of $P$ uses finitely many coefficients, variables and monomials. They all have representatives at one common stage, giving surjectivity of $\operatorname*{colim}P_\lambda\to P$. If a representative polynomial is zero in $P$, the finitely many identifications among its variable labels and the finitely many coefficient equalities hold at a common later stage. The representative is zero there. This proves injectivity and identifies the two polynomial rings.

A representative polynomial belongs to $I$ precisely when its evaluation becomes zero in $S$. That one finite equality holds at a later stage, so the polynomial can then be represented in $I_\mu$. If an element of $I_\lambda$ becomes zero in $I\subset P$, it is zero at a later polynomial stage, and hence at a later ideal stage. Therefore $I=\operatorname*{colim}I_\lambda$. An element of $I^2$ is a finite sum of products of two elements of $I$. Choose a common stage for all those factors and their kernel equalities; their products lie in $I_\mu^2$. Vanishing is again detected in a later polynomial ring. This proves $I^2=\operatorname*{colim}I_\lambda^2$.

Filtered colimits of the relevant modules preserve short exact sequences. Indeed, if a left representative maps to zero in the middle colimit, it is zero in the middle at a later stage, where injectivity makes it zero on the left. If a middle representative maps to zero on the right, at a later stage exactness expresses it as an image from the left. A right representative has a middle preimage at its own stage. These arguments also work for modules over the varying rings: the actions of the colimit rings are defined using common representatives. Applying them to $I_\lambda^2\subset I_\lambda$ gives $$I/I^2\cong\operatorname*{colim}_\lambda I_\lambda/I_\lambda^2.$$

The degree-zero term is $\bigoplus_{s\in S}S\,dX_s$. Its vectors have finite support; a common-stage argument for their coefficients and labels, with later-stage detection of zero, identifies it with the colimit of $\bigoplus_{s\in S_\lambda}S_\lambda\,dX_s$. Formal differentiation of a polynomial is a finite coefficient calculation and respects every transition, including variable identifications. Consequently the two module identifications intertwine the maps $[f]\mapsto df$. They give the asserted isomorphism of entire two-term complexes. ∎

Lemma F.25: complete local Noetherian domain finite over regular

Reading source: AI Integrated Stacks, pinned Algebra, lines 46654–46712.

Every complete Noetherian local domain $A$ contains a regular complete local subring $A_0$ such that $A$ is finite over $A_0$ and the residue fields agree. The subring can be chosen as a power-series ring over a field or over a Cohen ring.

Proof. Write $\mathfrak m$ for the maximal ideal and $D=\dim A$. The complete coefficient-ring construction is AG-CA-19S, Theorem 5.1. Its arbitrary-residue-field step uses F.119, including the full degree-$p$ argument: $T^p-a^p=(T-a)^p$, and the coefficient $-da$ of a proper monic factor, with $1\leq d<p$, forces the claimed degree. Thus this construction applies without a perfect-residue-field assumption.

In equal characteristic it embeds a coefficient field $k\subset A$. Choose a system of $D$ parameters $x_1,\ldots,x_D$ by AG-CA-11, Theorem 2.1. In mixed characteristic the coefficient ring is a Cohen DVR $C$ with uniformizer $p$ and residue field $k$. It embeds in $A$: a nonzero kernel ideal in this DVR would contain a power of $p$, whereas the domain $A$ has characteristic zero. The element $p$ is consequently a nonzerodivisor in $A$, and AG-CA-11, Theorem 3.2, gives $\dim A/pA=D-1$. Choose parameters $x_1,\ldots,x_{D-1}$ in that quotient and lift them to $A$.

In the respective cases put $$B=k[[X_1,\ldots,X_D]] \quad\hbox{or}\quad B=C[[X_1,\ldots,X_{D-1}]],$$ and let $\mathfrak n$ be its maximal ideal. The ideal $I$ generated by the proposed images of $\mathfrak n$ is $(x_1,\ldots,x_D)$ or $(p,x_1,\ldots,x_{D-1})$. The parameter property gives $\sqrt I=\mathfrak m$. AG-CA-03, Proposition 2.2, supplies an integer $b$ with $\mathfrak m^b\subset I$, while $I\subset\mathfrak m$. Hence $$\mathfrak m^{bn}\subset I^n\subset\mathfrak m^n\qquad(n\geq1).$$ The two filtrations are cofinal. Completeness and separatedness of $A$ for its maximal-ideal filtration therefore imply the same properties for its $I$-adic filtration. Evaluation $X_i\mapsto x_i$ defines a ring map $B\to A$: the tail of a power series of degree at least $n$ evaluates into $I^n$, and finite truncation followed by a separated limit respects addition and multiplication. This is the evaluation calculation of AG-CA-19S, Theorem 6.1.

The local quotient $A/I$ has nilpotent maximal ideal and residue field $k$. Each maximal-ideal layer is finite-dimensional over $k$, since $\mathfrak m$ is finite and its powers are generated by finitely many products. The finite-layer proof in AG-CA-03, Theorem 4.2, therefore makes $A/I$ finite-dimensional. Choose elements $a_1,\ldots,a_s\in A$ lifting a $k$-basis of this quotient.

These elements generate $A$ over $B$. To verify this directly, approximate any $a\in A$ modulo $I$ by $\sum_jc_{j1}a_j$ with coefficients lifted from $k$ to $B$. If coefficients have been chosen through order $n$, the error lies in $I^n$. Express it as a finite sum of degree-$n$ products of the generators of $I$ times elements of $A$. Replace each latter element modulo $I$ by its residue-basis expansion. Lifting those coefficients to $B$ gives corrections in $\mathfrak n^n$ and leaves an error in $I^{n+1}$. Induction constructs compatible coefficients $c_{jn}$ modulo $\mathfrak n^n$. The complete ring $B$ supplies limits $c_j$; continuity of evaluation and $I$-adic separatedness give $a=\sum_jc_ja_j$. Thus $A$ is a finite $B$-module.

The ring $B$ is the maximal-ideal completion of the polynomial local ring over $k$ or $C$: their quotients by each maximal-ideal power have the same finite polynomial truncations, including the coefficient reductions modulo powers of $p$ in the second case. AG-CA-14, Proposition 3.3, proves regularity of that polynomial local ring, and AG-CA-11, Theorem 5.1, gives its dimension $D$. Completion preserves this dimension and regularity by AG-CA-19, Theorem 4.1. Hence $B$ is a regular local domain of dimension $D$.

Let $J$ be the evaluation kernel. If it contains a nonzero element $f$, the one-equation theorem AG-CA-11, Theorem 3.2, gives $\dim B/(f)=D-1$, and the further quotient $B/J$ has dimension at most $D-1$. But $B/J$ embeds in the finite integral extension $A$, so AG-CA-05, Theorem 5.1, gives $\dim B/J=\dim A=D$, a contradiction. Therefore $J=0$. Take $A_0$ to be the resulting embedded copy of $B$. Its residue map is the coefficient identification with $k$, and the finiteness has already been proved. If $D=0$, the Noetherian local domain is its own residue field, so $A_0=A$ gives the empty-variable case directly. ∎

Lemma F.26: completion at quasi finite prime

Reading source: AI Integrated Stacks, pinned Algebra, lines 32412–32464.

Let $R$ be Noetherian, let $R\to S$ be finite type, and let $\mathfrak q\subset S$ be quasi-finite over $\mathfrak p=\mathfrak q\cap R$. There is an algebra $D$ with $$\widehat{R_{\mathfrak p}}\otimes_RS \cong\widehat{S_{\mathfrak q}}\times D,$$ whose first projection is the natural map to the completed local ring.

Proof. The full affine conductor theorem Zariski Main, Theorem C4.3 produces an integral subalgebra $C\subset S$ and $g\in C\setminus\mathfrak q$ such that $C_g=S_g$. Choose a finite set of $R$-algebra generators of $S_g$. Each is a fraction with numerator in $C$ and denominator a power of $g$. Let $S'\subset C$ be generated over $R$ by those numerators and by $g$. All these generators are integral. Repeated monic reduction, as in AG-CA-05, Section 1, shows that their bounded product powers generate $S'$ as an $R$-module. Thus $S'$ is finite over $R$, and the chosen generators give $S'_g=S_g$.

Put $\mathfrak q'=\mathfrak q\cap S'$. Localizing the last equality at the corresponding primes gives $S'_{\mathfrak q'}=S_{\mathfrak q}$, including their maximal ideals, so their maximal-adic completions are naturally identified. By F.27, $$T'=\widehat{R_{\mathfrak p}}\otimes_RS' \cong A\times D',\qquad A=\widehat{S_{\mathfrak q}}.$$ Let $e\in T'$ be the idempotent $(1,0)$. Its image in $T=\widehat{R_{\mathfrak p}}\otimes_RS$ gives an explicit product map $$T\longrightarrow eT\times(1-e)T,\qquad t\longmapsto(et,(1-e)t),$$ with inverse addition. The structural map $A=eT'\to eT$ is unital. The image of $g$ in $A$ is a unit because $g\notin\mathfrak q'$, and remains a unit in $eT$.

Localizing $T'\to T$ at $g$ gives an isomorphism, by $S'_g=S_g$ and the tensor/localization universal properties. Its $e$-factor is $$A=A_g\xrightarrow{\ \sim\ }(eT)_g=eT.$$ This identifies the first factor with the required completion, and the second is $D=(1-e)T$.

It remains to determine the projection on all of $T$. On $\widehat{R_{\mathfrak p}}$ and $S'$ it agrees with the completed-local structural maps by the construction in F.27. Given $s\in S$, the equality $S'_g=S_g$ writes $s/1=b/g^n$ with $b\in S'$. Equality of these fractions means that some $g^a$ kills $g^ns-b$ in $S$. Therefore $g^{n+a}s=g^ab\in S'$. Both the factor projection and the natural map $S\to\widehat{S_{\mathfrak q}}$ have the same value on this element and on the unit $g$. Multiplying by the inverse of $g^{n+a}$ makes their values on $s$ equal. The values on all pure tensors are now equal, proving the assertion about the first projection. ∎

Lemma F.27: completion finite extension

Reading source: AI Integrated Stacks, pinned Algebra, lines 24144–24184.

For a finite map of Noetherian rings $R\to S$ and a prime $\mathfrak p\subset R$, let $\mathfrak q_1,\ldots,\mathfrak q_r$ be all primes of $S$ above $\mathfrak p$. Then $$\widehat{R_{\mathfrak p}}\otimes_RS \cong\prod_{i=1}^r\widehat{S_{\mathfrak q_i}},$$ where each local ring is completed at its own maximal ideal.

Proof. Replace the base by $R_{\mathfrak p}$ and the finite algebra by $S_{\mathfrak p}$; tensor associativity gives the same left side, and localizing at each prime gives the same right side. Write $\mathfrak m$ for the resulting base maximal ideal. The integral lying-over and maximal-ideal results AG-CA-05, Theorems 3.3–3.4, show that the maximal ideals of this finite algebra are precisely the finitely many $\mathfrak q_i$, and all contract to $\mathfrak m$.

For $n\geq1$, the ring $R/\mathfrak m^n$ has a finite filtration whose successive quotients are finite-dimensional over the residue field: each $\mathfrak m^j/\mathfrak m^{j+1}$ is generated by finitely many products of generators of $\mathfrak m$. It is therefore Artinian. The finite module $S/\mathfrak m^nS$ has finite length over this ring, by the finite-module argument AG-CA-03, Theorem 3.3, and consequently is itself Artinian. Its primes are the images of the $\mathfrak q_i$: every prime over $\mathfrak m^n$ contracts to $\mathfrak m$ and is maximal by integrality.

The entire Artinian product theorem AG-CA-03, Theorem 4.2, now gives the canonical localization decomposition $$S/\mathfrak m^nS \cong\prod_{i=1}^r S_{\mathfrak q_i}/\mathfrak m^nS_{\mathfrak q_i}.$$ These maps commute with reduction from order $n+1$ to order $n$, because every projection is the corresponding localization map. Inverse limits of a finite product are computed componentwise: a compatible tuple sequence is exactly a tuple of compatible sequences. Hence the $\mathfrak mS$-adic completion of $S$ is $$\varprojlim_n S/\mathfrak m^nS \cong\prod_i\varprojlim_n S_{\mathfrak q_i}/\mathfrak m^nS_{\mathfrak q_i}.$$

In the $i$th local ring the radical of $\mathfrak mS_{\mathfrak q_i}$ is its maximal ideal $\mathfrak q_iS_{\mathfrak q_i}$: a prime containing the former corresponds to a prime above $\mathfrak m$, and localization leaves only $\mathfrak q_i$. AG-CA-03, Proposition 2.2, supplies $b_i$ such that $$(\mathfrak q_iS_{\mathfrak q_i})^{b_in} \subset\mathfrak m^nS_{\mathfrak q_i} \subset(\mathfrak q_iS_{\mathfrak q_i})^n.$$ These cofinal filtrations have canonically identical completions. The right-hand limits are thus $\widehat{S_{\mathfrak q_i}}$. Finally AG-CA-19, Theorem 3.1, identifies the left-hand limit with $\widehat R\otimes_RS$, because $S$ is a finite $R$-module. Every identification respects the maps from $R$ and $S$, giving the asserted natural product. If there are no primes above $\mathfrak p$, then $S_{\mathfrak p}=0$, and both sides are the zero ring, with the empty product interpreted as that ring. ∎

Lemma F.28: compose finite type

Reading source: AI Integrated Stacks, pinned Algebra, lines 547–571.

Finite-type maps compose, and finitely presented maps compose. Given $R\to S'\to S$, if $S/R$ is finite type then $S/S'$ is finite type. If $S/R$ is finitely presented and $S'/R$ is finite type, then $S/S'$ is finitely presented.

Proof. Choose generators $y_1,\ldots,y_a$ for $S'$ over $R$ and $x_1,\ldots,x_b$ for $S$ over $S'$. Each coefficient of a polynomial expression in the $x_j$ is itself a polynomial in the $y_i$ with coefficients in $R$. Substitution shows that the combined finite list generates $S$ over $R$, proving the first claim.

For the second claim take finite presentations $$S'=R[Y_1,\ldots,Y_a]/(u_1,\ldots,u_c),\qquad S=S'[X_1,\ldots,X_b]/(v_1,\ldots,v_d).$$ Lift the finitely many coefficients of the $v_j$ through $R[Y]\twoheadrightarrow S'$ to obtain $\widetilde v_j\in R[Y,X]$. Then $$R[Y,X]/(u_1,\ldots,u_c,\widetilde v_1,\ldots,\widetilde v_d) \cong S.$$ Indeed quotienting first by the $u_i$ gives $S'[X]$, where the remaining equations are exactly the $v_j$. This is a finite presentation of the composite.

If $S$ is generated by a finite list over $R$, every element is already an $R$-coefficient polynomial in that list, hence also an $S'$-coefficient polynomial. This proves the third claim without a finiteness assumption on $S'$.

For the fourth claim write $$S=R[X_1,\ldots,X_b]/(f_1,\ldots,f_d),\qquad S'=R[Y_1,\ldots,Y_a]/J.$$ The ideal $J$ need not be finite. For each $i$ choose $h_i(X)$ representing the image of $\bar Y_i$ in $S$. Form the finitely presented $S'$-algebra $$Q=S'[X_1,\ldots,X_b]/ (f_1,\ldots,f_d,\bar Y_1-h_1(X),\ldots,\bar Y_a-h_a(X)).$$ Evaluation at the specified elements of $S$ defines $Q\to S$. Conversely the images of the $X_j$ in $Q$ satisfy every $f_j$, giving an $R$-map $S\to Q$. Their composite on $S$ fixes its $R$-algebra generators. The other composite fixes each $X_j$, and it fixes every coefficient from $S'$ because it fixes $\bar Y_i=h_i(X)$ and $S'$ is generated by those elements over $R$. Both composites are identities, so $Q\cong S$ as $S'$-algebras. This proves finite presentation over $S'$ without making any extra assumption on $J$. ∎

Lemma F.29: compose smooth

Reading source: AI Integrated Stacks, pinned Algebra, lines 39039–39053.

If $R\to S$ and $S\to T$ are smooth ring maps, then $R\to T$ is smooth.

Proof. First translate the cotangent condition into the earlier lifting criterion. For a polynomial presentation of $S/R$, F.11 identifies its conormal complex with the canonical one. F.6 gives zero kernel and a projective differential cokernel, so the conormal sequence is split exact. AG-CA-17, Theorem 3.1 then proves that $S$ is formally smooth over $R$. The same argument proves that $T$ is formally smooth over $S$.

In a lifting problem $T\to C/J$ over $R$, where $J$ is nilpotent, first lift its restriction $S\to C/J$ to an $R$-map $S\to C$. This supplies an $S$-algebra structure on $C$ whose reduction is exactly the given one. Formal smoothness of $T/S$ now lifts the original map to an $S$-map $T\to C$. In particular it is an $R$-map. Thus $T/R$ is formally smooth, by an explicit composition of the two lifts.

Finite presentation is equally explicit. Write $S=R[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$ and $T=S[y_1,\ldots,y_m]/(g_1,\ldots,g_s)$, and lift each coefficient in the $g_j$ to a polynomial in the $x_i$. The resulting polynomials $\widetilde g_j$ give $T=R[x_i,y_l]/(f_i,\widetilde g_j)$: the two maps fixing the generators are inverse by the quotient universal property. Applying the same entire conormal criterion AG-CA-17, Theorem 3.1 to this finite presentation gives zero first homology and finite projective differentials. F.11 transports them to the canonical complex, so F.6 proves that $R\to T$ is smooth. The lifting and presentation arguments include zero algebras and empty lists. ∎

Lemma F.30: compose standard smooth

Reading source: AI Integrated Stacks, pinned Algebra, lines 38815–38878.

A composite of standard smooth ring maps is standard smooth.

Proof. Take standard smooth presentations $$S=R[X_1,\ldots,X_n]/(f_1,\ldots,f_c),\qquad T=S[Y_1,\ldots,Y_m]/(g_1,\ldots,g_d).$$ Reorder the variables in each list so that the first $c$ and first $d$ variables respectively give the chosen square Jacobian minors. Denote their determinants by $\Delta_f$ and $\Delta_g$; they are units in $S$ and $T$. Choose polynomial representatives $\widetilde g_j\in R[X,Y]$ for the coefficients of every $g_j$. Quotienting first by $(f)$ and then by the remaining equations gives $$T\cong R[X_1,\ldots,X_n,Y_1,\ldots,Y_m]/ (f_1,\ldots,f_c,\widetilde g_1,\ldots,\widetilde g_d).$$

Use the pivot variables $X_1,\ldots,X_c,Y_1,\ldots,Y_d$ and write Jacobian rows by equations. In $T$ its selected matrix has the form $$\begin{pmatrix}A&0\\ B&C\end{pmatrix},\qquad A=(\partial f_i/\partial X_j),\quad C=(\partial g_i/\partial Y_j).$$ The upper-right entries are zero because the $f_i$ contain no $Y$ variables. Reduction of the derivatives of $\widetilde g_i$ with respect to $Y_j$ equals the derivatives of $g_i$: changing polynomial coefficients commutes with the monomial differentiation rule.

In the permutation expansion of this determinant, a nonzero term must assign the first $c$ rows to the first $c$ columns; otherwise one of those rows uses a zero upper-right entry. The remaining rows then use the remaining columns. Factoring the two independent permutation sums gives $\det A\det C=\Delta_f\Delta_g$. The image of the first unit and the second given unit are units in $T$, so their product is a unit. The displayed finite presentation is therefore standard smooth.

If a presentation includes a principal localization at $h$, replace it by one more variable $Z$ and equation $hZ-1$. With $Z$ added last to the pivot list, the old equation rows have zero $Z$ entry and the new last diagonal entry is $h$. The selected determinant is the old determinant times $h$, which is invertible in the localized quotient. This calculation encodes all the stated localized presentations before applying the preceding composition argument. Empty equation lists use determinant one and the same permutation calculation. ∎

Lemma F.31: composition essentially of finite type

Reading source: AI Integrated Stacks, pinned Algebra, lines 13129–13138.

Composites of essentially finite-type ring maps are essentially finite type. Composites of essentially finitely presented ring maps are essentially finitely presented.

Proof. Let $R\to S\to T$ be the maps. Write $S=U^{-1}A$, where $A$ is a finite-type $R$-algebra, and write $T=V^{-1}(S[Y_1,\ldots,Y_m]/J)$. Set $P=A[Y_1,\ldots,Y_m]$ and let $K$ be the kernel of $P\to S[Y]/J$. Its quotient $B=P/K$ is finite type over $R$, even if $J$ and $K$ have infinitely many generators.

The map $U^{-1}P\to S[Y]/J$ has kernel exactly $U^{-1}K$. Indeed every element of $U^{-1}P$ is $f/u$; its image is zero precisely when the image of $f$ is zero, because the image of $u$ is a unit. This identifies $U^{-1}B$ with $S[Y]/J$. For each $v\in V$, choose a fraction $b_v/u_v$ representing it in $U^{-1}B$. Let $W\subset B$ be generated multiplicatively by the images of $U$ and all the $b_v$. Since the $u_v$ are already inverted, inverting $b_v/u_v$ is equivalent to inverting $b_v$. The localization universal property in both directions gives $T\cong W^{-1}B$, proving the finite-type assertion. No single denominator for all relations has been assumed.

For finite presentation, choose $A$ finitely presented over $R$ and $J=(j_1,\ldots,j_d)$ in $U^{-1}P$. Each $j_i$ has the form $f_i/u_i$ with $f_i\in P$ and $u_i\in U$. The images of $f_1,\ldots,f_d$ generate the same ideal after localizing at $U$, since all their denominators are units there. Therefore $$S[Y]/J\cong U^{-1}\bigl(A[Y]/(f_1,\ldots,f_d)\bigr).$$ The algebra in parentheses is finitely presented over $R$: combine the finite presentation of $A$ with these finitely many polynomial equations, as in F.28. As before, incorporate the numerator representatives of $V$ into the final multiplicative set. Thus $T$ is a localization of a finitely presented $R$-algebra. The two arguments also cover empty variable lists and the zero algebra. ∎

Lemma F.32: composition syntomic

Reading source: AI Integrated Stacks, pinned Algebra, lines 38404–38453.

For arbitrary ring maps $R\to S\to T$:

1. If both maps are syntomic, their composite is syntomic.
2. If both maps admit relative global complete-intersection presentations, their composite admits such a presentation.

Proof. Begin with (2). Choose presentations $$S=R[X_1,\ldots,X_n]/(f_1,\ldots,f_c),\qquad T=S[Y_1,\ldots,Y_m]/(h_1,\ldots,h_d).$$ Lifting each coefficient of $h_j$ to $R[X]$ gives polynomials $\widetilde h_j$ and the exact presentation $$T\cong R[X,Y]/(f_1,\ldots,f_c,\widetilde h_1,\ldots,\widetilde h_d).$$ To check F.4, fix $\mathfrak p\subset R$ and let $k=\kappa(\mathfrak p)$, $A=S\otimes_Rk$ and $B=T\otimes_Rk$. Empty $B$ imposes no dimension condition. If $B\ne0$, then $A\ne0$, and $\dim A=n-c$. Every nonempty fibre of $A\to B$ has dimension $m-d$: it is a field extension of a corresponding fibre of $S\to T$, so the coefficient-change calculation F.14 applies.

We prove both dimension inequalities for $B$. Given a minimal prime $\eta$ of $B$, let $\mathfrak a=\eta\cap A$, $F=\kappa(\mathfrak a)$ and $L=\kappa(\eta)$. The finite-type domain $A/\mathfrak a$ has dimension $\operatorname{trdeg}_kF\leq\dim A=n-c$, by AG-CA-09, Theorem 4.2. The prime induced by $\eta$ in $B\otimes_AF$ has residue field $L$. That fibre has dimension $m-d$, so applying the same domain dimension theorem to its prime quotient gives $\operatorname{trdeg}_FL\leq m-d$. Transcendence degrees add here. To check this, choose a finite transcendence basis $U$ of $F/k$ and one $V$ of $L/F$. The elements of $V$ are independent over $F$, hence over $k(U)$; together with the independence of $U$ this proves independence of $U\cup V$ over $k$. Algebraicity of $F/k(U)$ and $L/F(V)$, and transitivity of algebraicity, show that $L/k(U,V)$ is algebraic. Thus $$\dim(B/\eta)=\operatorname{trdeg}_kL =\operatorname{trdeg}_kF+\operatorname{trdeg}_FL \leq n-c+m-d.$$ The Noetherian finite-type algebra $B$ has finitely many minimal primes, and its dimension is the maximum of their quotient dimensions. This proves the upper bound.

For the lower bound, the polynomial presentation of $B$ has $n+m$ variables and $c+d$ equations. A prime minimal over its equation ideal has height at most $c+d$, by the full height theorem AG-CA-11, Theorem 3.1. The polynomial domain dimension formula AG-CA-09, Theorem 4.3, makes its quotient dimension at least $n+m-c-d$. Such a prime exists because $B\ne0$. Consequently $\dim B\geq n+m-c-d$. Equality follows, proving (2) over arbitrary $R$.

For (1), first suppose that $R$ is Noetherian. Finite presentation of both syntomic maps makes $S$ and $T$ Noetherian by the Hilbert basis theorem. At $\mathfrak q\subset T$, with contraction $\mathfrak r\subset S$, the entire neighbourhood criterion F.87 supplies $g\in S\setminus\mathfrak r$ and $h\in T\setminus\mathfrak q$ such that $R\to S_g$ and $S\to T_h$ have relative global complete-intersection presentations. By F.14, $S_g\to T_{gh}$ has one too. Assertion (2) gives such a presentation for $R\to T_{gh}$. These opens cover $\operatorname{Spec}T$. Its quasi-compactness AG-CA-01, Proposition 2.3 gives a finite subcover. Each chart is syntomic by F.87, and the complete gluing proof F.62 gives syntomicity of $T/R$.

Now let $R$ be arbitrary. Finite presentation follows from F.28, and flatness follows by composing the two exact tensor functors, as in AG-CA-07, Proposition 3.2. For $\mathfrak p\subset R$ use the same $k,A,B$ as above. The map $k\to A$ is syntomic: finite presentation and flatness survive base change, and its only field fibre is the given local complete intersection. The map $A\to B$ is also syntomic. Its finite presentation and flatness are the base changes of those of $S\to T$. At a prime $\mathfrak a\subset A$ over $\mathfrak r\subset S$, its fibre is $$B\otimes_A\kappa(\mathfrak a) \cong\bigl(T\otimes_S\kappa(\mathfrak r)\bigr) \otimes_{\kappa(\mathfrak r)}\kappa(\mathfrak a).$$ Choose the principal global complete-intersection cover of the original field fibre, as in F.2. Scalar extension carries that cover to a cover, and F.14 carries each of its presentations to a global complete-intersection presentation. Thus the displayed fibre is a local complete intersection. Since the base $k$ is Noetherian, the case already proved applies to $k\to A\to B$. It makes $B/k$ syntomic, in particular a local complete intersection. Every field fibre of $T/R$ has now been checked. Together with flatness and finite presentation, the syntomic definition F.1 proves (1) for arbitrary bases. ∎

Lemma F.33: conormal module localize

Reading source: AI Integrated Stacks, pinned Algebra, lines 37203–37225.

Suppose that $R[X_1,\ldots,X_n]\twoheadrightarrow S$ and $R[Y_1,\ldots,Y_m]\twoheadrightarrow S_g$ have kernels $I$ and $J$. There is an isomorphism of $S_g$-modules $$(I/I^2)_g\oplus S_g^{\oplus m} \cong J/J^2\oplus S_g^{\oplus n}.$$

Proof. Put $C=S_g$, $P=R[X]$ and $Q=R[Y]$. Choose $f\in P$ mapping to $g$. Exact localization gives $P_f\twoheadrightarrow C$ with kernel $I_f$. Lift the image of each $Y_j$ to $u_j\in P_f$, and lift the image of each $X_i$ to $v_i\in Q$.

The simultaneous evaluation $W=P[Y]=Q[X]\to C$ has kernel $L=(J,X_i-v_i(Y))$: quotienting by the graph equations eliminates $X$ and leaves exactly $Q/J$. Translation by $v_i(Y)$ makes $W=Q[Z_1,\ldots,Z_n]$ and $L=(J,Z_1,\ldots,Z_n)$. In this polynomial ring $$L^2=(J^2,\ JZ_1,\ldots,JZ_n,\ Z_iZ_j\mid1\leq i,j\leq n).$$ Therefore reading the constant term modulo $J^2$ and each linear coefficient modulo $J$ gives an isomorphism $L/L^2\to J/J^2\oplus C^n$. Its inverse sends a relation class to its constant representative and each coordinate to the class of the corresponding $Z_i$. Changes of representatives lie in the displayed square, so both maps are well defined and inverse.

The image of $f$ is a unit in $C$, and $L/L^2$ is a $C$-module. Multiplication by $f$ on it is consequently invertible. Localizing thus gives the same module, namely $$L/L^2\cong L_f/L_f^2.$$ On $W_f=P_f[Y]$, the same evaluation has kernel $(I_f,Y_j-u_j)$: eliminating $Y$ leaves $P_f/I_f=C$. Translate $Y_j$ by $u_j$. The identical constant-and-linear calculation now gives $$L_f/L_f^2\cong I_f/I_f^2\oplus C^m.$$ The ideal-product fraction formulas identify $I_f/I_f^2$ with $(I/I^2)_g$. Combining these isomorphisms gives the assertion. No free summand was cancelled, and no flatness of $C$ over $R$ was used. Empty variable lists and $C=0$ satisfy the same formulas. ∎

Lemma F.34: contained in radical

Reading source: Stacks, Tag 0AME.

For an ideal \(I\) of a commutative ring \(R\), the inclusion \(I\subset\operatorname{Jac}(R)\) is equivalent to invertibility of every \(1+i\), \(i\in I\). Under either condition, every element invertible modulo \(I\) is already invertible in \(R\). Here the Jacobson radical is the intersection of all maximal ideals.

Proof. We first justify the maximal-ideal test for a unit. A nonunit \(a\) generates a proper ideal. The partially ordered set of proper ideals containing \((a)\) has an upper bound for every chain, its union: that union cannot contain \(1\), since then a member would contain \(1\). Zorn's lemma supplies a maximal proper ideal containing \(a\). Conversely no unit belongs to a proper ideal. Therefore an element is a unit exactly when it belongs to no maximal ideal.

Suppose \(I\) is contained in every maximal ideal. If a maximal ideal contained \(1+i\), it would contain its difference with \(i\), namely \(1\). This is impossible, so the unit test makes every \(1+i\) invertible. Conversely, if a maximal ideal \(\mathfrak m\) fails to contain \(I\), maximality gives \(I+\mathfrak m=R\). Write \(1=i+b\) with \(i\in I\) and \(b\in\mathfrak m\). The element \(b=1-i\) has the stipulated form, yet is not a unit. This contradiction proves the inclusion.

If \(a\bmod I\) is a unit, lift one inverse to \(b\in R\). Then \(ab=1+i\) for some \(i\in I\). The element \(b(ab)^{-1}\) is an actual inverse of \(a\), proving the last assertion. In the zero ring the empty intersection is the whole ring and its unique element is its identity and a unit; the conclusions hold there too. ∎

Lemma F.35: cover upstairs

Reading source: AI Integrated Stacks, pinned Algebra, lines 4346–4405.

Let $g_1,\ldots,g_r\in S$ generate the unit ideal. The $R$-algebra $S$ is finite type, respectively finitely presented, if and only if every $S_{g_i}$ has that property.

Proof. We will use a finite-cover calculation for arbitrary modules. In a ring $A$, suppose $\sum_i c_it_i=1$ and $N_{t_i}=0$ for all $i$. For $n\in N$, choose positive $e_i$ with $t_i^{e_i}n=0$. In the expansion of $(\sum_i c_it_i)^b$ with $b=1+\sum_i(e_i-1)$, every monomial has an exponent at least $e_i$ in one of the $t_i$. Multiplication by $n$ kills every monomial. Since the expanded element is one, $n=0$. Thus $N=0$ without a finiteness hypothesis.

Choose $h_i$ with $\sum_i h_ig_i=1$. If every $S_{g_i}$ is finite type, choose a finite $R$-algebra generator list for each, written as fractions $a_{ij}/g_i^{b_{ij}}$. Let $S_0\subset S$ be generated over $R$ by all $a_{ij},g_i,h_i$. The map $(S_0)_{g_i}\to S_{g_i}$ is surjective by the chosen generators. It is injective: a numerator from $S_0$ whose fraction vanishes in $S_{g_i}$ is killed by a power of $g_i$ in $S$, hence also in the subring $S_0$. Therefore $(S/S_0)_{g_i}=0$ as a module over $S_0$. The $g_i$ generate one in $S_0$, so the cover calculation gives $S=S_0$. This proves finite type.

If every chart is finitely presented, the finite-type case supplies $S=P/J$ for a finite polynomial ring $P=R[X_1,\ldots,X_n]$. Choose lifts $u_i,v_i\in P$ of $g_i,h_i$. The kernel of $$P[Z]\longrightarrow S_{g_i},\qquad Z\longmapsto g_i^{-1},$$ is $JP[Z]+(u_iZ-1)$, since its quotient is exactly $(P/J)[Z]/(g_iZ-1)$. The entire presentation-independence proof F.44 makes this kernel finitely generated. Express each of its finite generators as a finite sum of multiples of elements of $J$ and of $u_iZ-1$. Collect the finitely many coefficients $a_{ij}\in J$ used in those sums. They, together with $u_iZ-1$, generate the same kernel: they belong to it and generate each chosen generator.

Set $$J_0=\bigl(\sum_i v_iu_i-1,\ a_{ij}\text{ for all }i,j\bigr)\subset J,\qquad A=P/J_0.$$ The images of $u_i$ generate one in $A$. The preceding kernel equalities give isomorphisms $A_{u_i}\cong S_{g_i}$. Indeed adjoining $Z$ with $u_iZ-1$ identifies both quotients with the same localization, and every element of $J_0$ also belongs to the original kernel. The kernel $J/J_0$ of $A\twoheadrightarrow S$ therefore vanishes on this cover. The arbitrary-module calculation gives $J=J_0$, so the original finite polynomial presentation has a finite relation ideal.

Conversely, localization adds only $g_i^{-1}$ to a finite generating list. For a finite presentation $S=P/(f_1,\ldots,f_d)$ and a lift $u_i$ of $g_i$ it gives the exact finite presentation $S_{g_i}=P[Z]/(f_1,\ldots,f_d,u_iZ-1)$. If the cover is empty, its unit-ideal condition forces $S=0$; the presentation $R/(1)$ handles both assertions directly. Empty generator lists cause no exception. ∎

Lemma F.36: criterion flatness fibre

Reading source: AI Integrated Stacks, pinned Algebra, lines 34374–34496.

Let $R\to S\to S'$ be local maps of local rings and write $\mathfrak m=\mathfrak m_R$. Suppose that both $S/R$ and $S'/R$ are essentially finitely presented, that $M\ne0$ is a finitely presented $S'$-module, that $M$ is $R$-flat, and that $M/\mathfrak mM$ is flat over $S/\mathfrak mS$. Then $S$ is $R$-flat and $M$ is $S$-flat. The original rings need not be Noetherian.

Proof. We construct compatible Noetherian local models, then apply the Noetherian fibre criterion twice through the two flatness hypotheses. The full finite-expression construction F.59 writes $R=\operatorname*{colim}R_\lambda$, where the $R_\lambda$ are local rings essentially of finite type over $\mathbf Z$, with maximal ideals $\mathfrak p_\lambda$ and local transitions. Their maximal ideals have colimit $\mathfrak m$.

An essentially finitely presented algebra that is local is a localization at one prime of a finitely presented algebra: all elements outside the inverse image of its maximal ideal are units, so the two localization universal properties identify the rings. Choose such finite presentations for $S$ and $S'$. To model their intervening map, choose images in $S'$ of the finite generators of $S$, represented by polynomials with finitely many denominators. Put those denominators among the inverted elements, and impose the finite graph equations equating the chosen source generators with their images. Eliminating the graph variables recovers the presentation of $S'$ over $R$ after localization. The source relations vanish there; any vanishing of their finitely many representatives is witnessed by finitely many additional denominators. Include these witnesses too. All the coefficients and finite equality witnesses can be placed at one stage of F.59.

For later stages we consequently have presentations $$S_\lambda=(R_\lambda[X]/(f_\lambda))_{\mathfrak q_\lambda},\qquad S'_\lambda=(R_\lambda[X,Y]/(f_\lambda,g_\lambda))_{\mathfrak q'_\lambda},$$ where the primes are contracted from the actual local target rings. The two maps are local. The coefficient lists are fixed and finite, so the Hilbert basis theorem makes all these model rings Noetherian. For $\mu\geq\lambda$, tensoring the displayed polynomials and relations gives exactly $$S_\lambda\otimes_{R_\lambda}R_\mu\longrightarrow S_\mu,\qquad S'_\lambda\otimes_{S_\lambda}S_\mu\longrightarrow S'_\mu$$ as localizations. The same description holds with the final rings $R,S,S'$.

Write a finite module presentation $(S')^a\xrightarrow{A}(S')^b\to M\to0$. By F.60, its finitely many matrix entries occur at a common stage. Define $M_\lambda=\operatorname{coker}((S'_\lambda)^a\xrightarrow{A_\lambda}(S'_\lambda)^b)$. Right exactness of tensor gives $$M_\mu\cong M_\lambda\otimes_{S'_\lambda}S'_\mu,\qquad M\cong M_\lambda\otimes_{S'_\lambda}S'.$$ In particular every $M_\lambda$ is nonzero, because a zero one would make the final tensor zero. It is finite over the stipulated Noetherian local overring $S'_\lambda$.

The $R$-flatness of $M$ and the localized tensor descriptions allow F.21 to be applied to $R_\lambda\to S'_\lambda$ and $M_\lambda$. It gives a stage where $M_\lambda$ is $R_\lambda$-flat. At every later stage this persists: scalar extension preserves the exact tensor functor, and localization of an algebra-module preserves its injection tests. The final module and the later ones have precisely these two forms.

Next put $$A_\lambda=S_\lambda/\mathfrak p_\lambda S_\lambda,\qquad B_\lambda=S'_\lambda/\mathfrak p_\lambda S'_\lambda,\qquad N_\lambda=M_\lambda/\mathfrak p_\lambda M_\lambda.$$ These are local Noetherian algebra models with finite target modules. The local transitions and right exactness give their colimits $S/\mathfrak mS$, $S'/\mathfrak mS'$ and $M/\mathfrak mM$. Reduction of the same polynomial presentations along the residue-field maps $R_\lambda/\mathfrak p_\lambda\to R_\mu/\mathfrak p_\mu$ shows that $$B_\lambda\otimes_{A_\lambda}A_\mu\longrightarrow B_\mu$$ is a localization, and that $N_\mu=N_\lambda\otimes_{B_\lambda}B_\mu$. Thus this quotient system has every model property required by F.21. The assumed flatness of its limit over $S/\mathfrak mS$ gives a later stage where $N_\lambda$ is $A_\lambda$-flat. Passing to a common successor retains the first flatness assertion as well.

At that stage the local maps $R_\lambda\to S_\lambda\to S'_\lambda$ and the nonzero finite module $M_\lambda$ satisfy the full criterion F.37. It gives $R_\lambda$-flatness of $S_\lambda$ and $S_\lambda$-flatness of $M_\lambda$. Finally $S$ is a localization of $S_\lambda\otimes_{R_\lambda}R$, so it is $R$-flat. The module $M$ is a localization of $M_\lambda\otimes_{S_\lambda}S$, so it is $S$-flat. These conclusions use the exact scalar-extension and localization functors and impose no new hypothesis on the original rings. ∎

Lemma F.37: criterion flatness fibre Noetherian

Reading source: AI Integrated Stacks, pinned Algebra, lines 24999–25045.

Let $R\to S\to T$ be local maps of Noetherian local rings, let $\mathfrak m=\mathfrak m_R$, and let $M\ne0$ be finite over $T$. If $M$ is $R$-flat and $M/\mathfrak mM$ is flat over $S/\mathfrak mS$, then $M$ is faithfully flat over $S$ and $S$ is flat over $R$.

Proof. Set $I=\mathfrak mS$. Locality gives $I\subset\mathfrak m_S$, so it is proper. The map $$u:\mathfrak m\otimes_RM\longrightarrow I\otimes_SM,\qquad x\otimes z\longmapsto x\otimes z$$ is surjective: write an element of $I$ as a finite sum of products $sx$ and move each $s$ to the module factor. Its composite with multiplication $I\otimes_SM\to M$ is injective by $R$-flatness. Therefore $u$ is injective as well as surjective, and multiplication from $I\otimes_SM$ is injective. The ideal-kernel Tor calculation of AG-RG-S01, Lemma 1.A, gives $\operatorname{Tor}_1^S(S/I,M)=0$.

Apply the full proper-ideal criterion AG-CA-08, Corollary 4.4, to the source $S$, local overring $T$, finite $T$-module $M$ and ideal $I$. Its flat quotient condition is exactly the assumed flatness of $M/IM$ over $S/I$, and the Tor condition has just been proved. Thus $M$ is $S$-flat. Since $M$ is finite and nonzero over the local ring $T$, Nakayama F.10 gives $M/\mathfrak m_TM\ne0$. The quotient $M/\mathfrak m_SM$ surjects onto it, by locality of $S\to T$, so this closed fibre is nonzero too. The complete flat-module criterion AG-CA-08, Theorem 1.1, gives faithful flatness over $S$.

Let $L=\operatorname{Tor}_1^R(R/\mathfrak m,S)$. The ideal sequence over $R$ and its Tor-kernel calculation give the exact sequence of $S$-modules $$0\longrightarrow L\longrightarrow\mathfrak m\otimes_RS \longrightarrow I\longrightarrow0.$$ Tensor with the flat $S$-module $M$. Tensor associativity identifies its middle term with $\mathfrak m\otimes_RM$, and its final term is $I\otimes_SM$. Their map is the isomorphism $u$ already proved, so $L\otimes_SM=0$. Faithfulness forces $L=0$. AG-CA-08, Theorem 4.2, applied to the local map $R\to S$ and the module $S$, finite over its own Noetherian local overring, gives $R$-flatness of $S$. Both conclusions are established with the stated target-module finiteness, without assuming $M$ finite over $R$ or $S$. ∎

Lemma F.38: descent regular

Reading source: AI Integrated Stacks, pinned Algebra, lines 48353–48371.

If $R\to S$ is faithfully flat and $S$ is a regular ring, then $R$ is regular. In particular the source is Noetherian even if this is not separately assumed.

Proof. First prove that $R$ is Noetherian. For an ideal $I\subset R$, its extension $IS$ is finite because $S$ is Noetherian. Express a finite generating list for $IS$ using finitely many elements $a_1,\ldots,a_r$ of $I$, and let $I_0=(a_1,\ldots,a_r)$ in $R$. Then $I_0S=IS$. Flatness identifies $I\otimes_RS$ with $IS$ and $I_0\otimes_RS$ with $I_0S$, by tensoring their inclusions into $R$. Right exactness consequently gives $(I/I_0)\otimes_RS=0$. Faithfulness implies $I=I_0$. Every ideal is finite, proving Noetherianity.

Fix $\mathfrak p\subset R$. Faithful flatness supplies a prime $\mathfrak q\subset S$ above it, by AG-CA-08, Theorem 2.1. Put $A=R_{\mathfrak p}$, $B=S_{\mathfrak q}$ and $k=\kappa(\mathfrak p)$. The induced map $A\to B$ is flat and local, hence faithfully flat by that same theorem. The ring $B$ is regular local by the hypothesis.

Take a free resolution $F_\bullet\to k$ over the Noetherian local ring $A$. Tensoring with the flat algebra $B$ gives a free resolution of $k\otimes_AB$ over $B$. The complex computing its self-Tor is $$(F_\bullet\otimes_AB)\otimes_B(k\otimes_AB) \cong(F_\bullet\otimes_Ak)\otimes_AB.$$ Exact tensor with $B$ commutes with kernels and images, so the entire arbitrary-module resolution calculus AG-RG-S01, Lemma 1.A, gives $$\operatorname{Tor}_i^A(k,k)\otimes_AB \cong\operatorname{Tor}_i^B(k\otimes_AB,k\otimes_AB).$$ The regular-local global-dimension theorem AG-CA-14, Theorem 2.2, makes the right-hand groups zero for sufficiently large $i$. Faithfulness makes the corresponding source groups zero.

The full minimal-resolution construction AG-CA-13, Theorem 2.2, gives a resolution of $k$ by finite free $A$-modules whose differentials have entries in $\mathfrak m_A$. Tensoring it with $k$ makes every differential zero. Thus its degree-$i$ free term modulo $\mathfrak m_A$ is exactly $\operatorname{Tor}_i^A(k,k)$. Eventual vanishing makes those finite free terms zero, and the resolution truncates. Therefore $k$ has finite projective dimension. The reverse implication in AG-CA-14, Theorem 2.2, proves that $A$ is regular local. This holds at every source prime, proving regularity of $R$. ∎

Lemma F.39: dimension fibres bounded open upstairs

Reading source: AI Integrated Stacks, pinned Algebra, lines 32674–32690.

For a finite-type ring map $R\to S$ and $\mathfrak q\subset S$ over $\mathfrak p\subset R$, put $$n=\dim_{\bar{\mathfrak q}}\operatorname{Spec} (S\otimes_R\kappa(\mathfrak p)),$$ the dimension in a neighbourhood of its point in the fibre. Some principal open neighbourhood of $\mathfrak q$ has fibre point-dimension at most $n$ at every point. The base ring is arbitrary.

Proof. The fibre is a finite-type algebra over $k=\kappa(\mathfrak p)$ and hence Noetherian. By the definition of point dimension and the complete field point-dimension theorem AG-CA-09, Theorem 6.1, choose a principal neighbourhood of $\bar{\mathfrak q}$ in that fibre whose algebra has dimension exactly $n$. Its defining element is a fraction $b/a$ with $b\in S$ and $a\in R\setminus\mathfrak p$. Since $a$ is a fibre unit, the same fibre open is defined by $b$; also $b\notin\mathfrak q$. Replace $S$ temporarily by $S_b$.

Apply the full Noether-normalization proof AG-CA-09, Lemma 2.1 and Corollary 3.2, to this fibre. It supplies $n$ algebraically independent parameters and a finite map from their polynomial algebra to the fibre. These parameters lift to elements $t_1,\ldots,t_n\in S_b$. Indeed that proof repeatedly replaces a generator by it minus an integer power of the last generator, so its chosen parameters are polynomials with integer coefficients in the original fibre generators. Dividing a relation by its nonzero leading coefficient proves integrality and does not change those parameter expressions.

The finite-type map $R[T_1,\ldots,T_n]\to S_b$, $T_i\mapsto t_i$, is quasi-finite at the chosen point. Its fibre there is a further fibre of the displayed finite $k[T]$-algebra, and is therefore finite-dimensional over the contracted polynomial prime's residue field. The entire open-locus proof Zariski Main, Theorem C5.1, using its algebraic Theorem C4.3 over arbitrary bases, gives a principal open around that point on which this polynomial map is quasi-finite everywhere. A defining fraction in $S_b$ has numerator $c\in S$; the resulting neighbourhood in the original spectrum is $D(bc)$.

For any $\mathfrak p'\subset R$, let $U=S_{bc}\otimes_R\kappa(\mathfrak p')$. The map $\kappa(\mathfrak p')[T]\to U$ is quasi-finite. This can be checked directly on its fibres: the polynomial base ring is a quotient and localization of $R[T]$ along $\mathfrak p'$, so a point and its residue field are those of the corresponding prime of $R[T]$, and its fibre is the fibre of the quasi-finite polynomial map just constructed.

At a minimal prime $\eta$ of a nonzero $U$, its residue field is finite over the residue field of its contracted polynomial prime, by the full isolated-fibre criterion F.56. The latter field has transcendence degree at most $n$ over $\kappa(\mathfrak p')$, since it is generated as a field by the images of $n$ polynomial variables. Finite algebraic extension preserves this transcendence degree. AG-CA-09, Theorem 4.2, therefore gives $\dim(U/\eta)\leq n$. Taking the maximum over its finitely many minimal primes gives $\dim U\leq n$. Every neighbourhood dimension at a point of this fibre is at most its total dimension, proving the required bound on $D(bc)$. Empty fibres have no point to check. The earlier conductor proof includes the polynomial-normality argument Zariski Main over an arbitrary base, Corollary B1.2; no Noetherian assumption on $R$ or $S$ has entered the argument. ∎

Lemma F.40: essentially of finite type into artinian local

Reading source: Stacks, Tag 07DT.

Let \(R\to S\) be any ring map, with \(S\) Artinian local, maximal ideal \(\mathfrak m\), and residue field \(K=S/\mathfrak m\). Each of the following properties of \(S/R\) is equivalent to the same property of \(K/R\): finite, of finite type, and essentially of finite type.

Proof. The Artinian length theorem AG-CA-03, Theorem 4.2 gives a finite composition series of the \(S\)-module \(S\). Every simple factor is \(K\): a nonzero element of a simple module generates it, and its annihilator is the unique maximal ideal. If \(K\) is finite over \(R\), each factor is therefore a finite \(R\)-module. Finite generation passes through an extension directly: lift finitely many generators of the quotient and adjoin finitely many generators of the submodule. Induction on the composition series makes \(S\) finite over \(R\). The converse follows by taking the images of module generators in \(K\). This proves the finite assertion for an arbitrary source ring.

For finite type, a quotient keeps the images of any finite list of algebra generators. Conversely, lift finite \(R\)-algebra generators of \(K\) to \(s_1,\ldots,s_a\in S\). The polynomial map \(P=R[X_1,\ldots,X_a]\to S\), \(X_i\mapsto s_i\), is surjective on residues. Thus \(K\) is cyclic as a \(P\)-module. The finite assertion just proved makes \(S\) a finite \(P\)-module. Choose its module generators \(u_1,\ldots,u_b\). Every element is a sum of polynomials in the \(s_i\) times the \(u_j\), so \(s_i,u_j\) together generate \(S\) as an \(R\)-algebra.

Finally, a quotient of a localization of a finite-type algebra is the localization of its quotient, by the fraction formula. Hence essential finite type passes from \(S\) to \(K\). For the converse write \(K=T^{-1}C\), with \(C\) a finite-type \(R\)-algebra, and let \(C_0\) be the image of \(C\) in \(K\). The domain \(C_0\) has fraction field \(K\): every element of the displayed localization is a fraction of its elements, and the field \(K\) contains the inverse of every nonzero element of \(C_0\).

Lift a finite list of \(R\)-algebra generators of \(C_0\) to \(S\), and let \(A\subset S\) be the \(R\)-subalgebra they generate. Put \(\mathfrak p=A\cap\mathfrak m\). Then \(A/\mathfrak p=C_0\), so \(\kappa(\mathfrak p)=K\). Every element of \(A\setminus\mathfrak p\) maps to a unit in \(S\); thus the inclusion extends to \(A_{\mathfrak p}\to S\). Applying the finite assertion to this map makes \(S\) a finite \(A_{\mathfrak p}\)-module. Choose generators \(v_1,\ldots,v_c\) and set \(B=A[v_1,\ldots,v_c]\subset S\). This is a finite-type \(R\)-algebra. Localizing \(B\) at \(A\setminus\mathfrak p\) maps injectively into \(S\), because those denominators are units there, and surjectively because the \(v_j\) generate over \(A_{\mathfrak p}\). Therefore \(S\) is that localization of \(B\). This proves the last equivalence without a finiteness assumption on \(R\). ∎

Lemma F.41: etale standard smooth

Reading source: AI Integrated Stacks, pinned Algebra, lines 40737–40760.

Every étale ring map $R\to S$ has a standard smooth presentation with equally many variables and equations: $$S\cong R[X_1,\ldots,X_n]/(f_1,\ldots,f_n),\qquad \det(\partial f_i/\partial X_j)\in S^\times.$$

Proof. If $S=0$, take $S=R[X]/(1)$. Its Jacobian determinant is zero, which is the unit element in the zero ring, so it has the required property.

Suppose $S\ne0$ and choose a finite polynomial presentation $P=R[X_1,\ldots,X_n]\twoheadrightarrow S$ with kernel $I$. The complete formal/conormal comparison AG-CA-17, Theorem 3.1 and Section 4, says that the conormal differential of an étale presentation is an isomorphism: $$I/I^2\xrightarrow{\ \sim\ }\bigoplus_{i=1}^nS\,dX_i,\qquad [f]\longmapsto\sum_i(\partial f/\partial X_i)dX_i.$$ Specifically, formal smoothness gives a split injection and formal unramifiedness gives zero cokernel; étaleness supplies both. Thus $I/I^2$ is finite free.

The entire global construction F.55 replaces this presentation by $S=R[Y_1,\ldots,Y_m]/J$ with $J=(g_1,\ldots,g_c)$ and the classes of the $g_j$ a basis of $J/J^2$. Apply the same étale conormal comparison to this new presentation. In its actual equation and differential bases it is the Jacobian map $S^c\xrightarrow{H}S^m$. It is an isomorphism. Choose any maximal ideal of the nonzero ring $S$ and reduce this inverse pair modulo it. The resulting vector-space isomorphism forces $c=m$.

The matrix $H$ is now square and has an inverse over $S$. The determinant product identity gives $\det(H)\det(H^{-1})=1$, so its determinant is a unit. Relabeling these $m$ variables and $m$ equations produces the asserted presentation of the whole algebra. It proves global standard smoothness, with no Noetherian-base assumption or restriction to a neighbourhood. ∎

Lemma F.42: exact sequence NL

Reading source: AI Integrated Stacks, pinned Algebra, lines 36838–36913.

Let $A\to B\to C$ be ring maps. Choose surjections $P=A[x_s\mid s\in\mathcal S]\to B$ and $Q=B[y_t\mid t\in\mathcal T]\to C$, with kernels $I,J$. Put $E=A[x_s,y_t]$ and $K=\ker(E\to C)$. Variable sets may be infinite. Define $$N_A=(I/I^2)\otimes_BC,\quad N_B=K/K^2,\quad N_C=J/J^2,$$ $$E_A=\bigoplus_sC\,dx_s,\quad E_B=E_A\oplus\bigoplus_tC\,dy_t,\quad E_C=\bigoplus_tC\,dy_t.$$ Their conormal differentials give the commutative diagram with exact rows $$\begin{array}{ccccccccc} 0&\longrightarrow&E_A&\xrightarrow{i}&E_B&\xrightarrow{p}&E_C&\longrightarrow&0\\ &&\uparrow d_A&&\uparrow d_B&&\uparrow d_C\\ &&N_A&\xrightarrow{a}&N_B&\xrightarrow{b}&N_C&\longrightarrow&0. \end{array}$$ No injectivity of $a$ is asserted. With the first cotangent homology $H_1(L_{-/ -})$ computed by these presentation complexes, the resulting canonical exact sequence is $$\begin{gathered} H_1(\operatorname{NL}_{B/A}\otimes_BC)\longrightarrow H_1(L_{C/A})\longrightarrow H_1(L_{C/B})\\ \xrightarrow{\delta} C\otimes_B\Omega_{B/A} \longrightarrow\Omega_{C/A}\longrightarrow\Omega_{C/B}\longrightarrow0. \end{gathered}$$ It extends the differential sequence AG-CA-15, Theorem 3.1. If both $\operatorname{Tor}_1^B(\Omega_{B/A},C)$ and $\operatorname{Tor}_2^B(\Omega_{B/A},C)$ vanish, there is moreover a natural identification $$H_1(\operatorname{NL}_{B/A}\otimes_BC) \cong H_1(L_{B/A})\otimes_BC.$$

Proof. The map $i$ includes the $dx_s$ coordinates, and $p$ kills those coordinates and retains the $dy_t$. Their row is split exact. Define $$a([f]\otimes c)=[\widetilde c f]\in K/K^2,\qquad b([h])=[h\bmod IE]\in J/J^2,$$ where $\widetilde c\in E$ lifts $c$. Changing this lift adds an element of $K$ times $f\in I\subset K$, hence an element of $K^2$. Changing $f$ by $I^2$, or using a different polynomial lift of a coefficient in $B$, also changes the result by $K^2$. Multiplication of coefficient representatives proves the tensor-balancing relation, so $a$ is well defined. The map $E\to Q$ has kernel $IE$, maps $K$ onto $J$, and maps $K^2$ onto $J^2$. Therefore the preimage of $J^2$ in $K$ is $K^2+IE$: subtract a finite sum of products of lifts from $K$ to leave an element of $IE$. The image of $a$ is exactly $(IE+K^2)/K^2$. This proves exactness at $N_B$ and $N_C$.

Differentiation gives $d_B(\widetilde c f)=c\,df$, since $f$ evaluates to zero in $C$. Hence $d_Ba=id_A$. Projecting a polynomial derivative to its $dy_t$ part gives precisely its derivative in $Q$ relative to $B$, so $pd_B=d_Cb$. These identities prove commutativity of the entire diagram.

We now give the connecting map and check every exactness assertion. For $z\in\ker d_C$, choose $y\in N_B$ with $b(y)=z$. Then $p(d_By)=d_Cz=0$, so the split exact upper row gives a unique $v\in E_A$ with $i(v)=d_By$. Set $\delta(z)=[v]\in E_A/\operatorname{im}d_A$. If the lift $y$ changes, its difference is $a(x)$ for some $x\in N_A$; the corresponding $v$ changes by $d_Ax$. Thus this class is well defined.

A cycle $y\in\ker d_B$ mapping to zero under $b$ is $a(x)$, and $0=d_Ba(x)=i(d_Ax)$ forces $d_Ax=0$, since $i$ is injective. This proves exactness at $H_1(L_{C/A})$. At $\ker d_C$, the class $\delta(z)$ is zero precisely when the above $v$ is $d_Ax$ for some $x$; then $y-a(x)$ is a cycle lifting $z$. Conversely a cycle lift has $v=0$. This proves exactness at $H_1(L_{C/B})$.

At $E_A/\operatorname{im}d_A$, a class $[v]$ maps to zero in $E_B/\operatorname{im}d_B$ precisely when $i(v)=d_By$ for some $y$. Its image $b(y)$ is then a cycle in $N_C$, and the construction gives $\delta(b(y))=[v]$. At the next term, if $p(w)=d_Cz$, choose $y$ with $b(y)=z$. Then $p(w-d_By)=0$, so $w-d_By=i(v)$ for a $v\in E_A$. Thus $[w]$ comes from $[v]$. Conversely every image of $i$ is killed by $p$. Finally $p$ is surjective, so its map on the cokernels is surjective as well. These calculations prove the complete displayed sequence.

The entire presentation comparison F.11 identifies the kernels and cokernels with the stated cotangent homology and differential modules. The sequence is canonical under those identifications. Indeed comparison maps are obtained by lifting each source polynomial variable to the new presentation. Make the lifts for $B$ first, then lift the target variables for $C$ and their coefficients; this gives comparison maps on $P,Q,E$ respecting $I,J,K$. Their relation-class and derivative maps commute with $a,b,i,p$. Transporting a chosen lift $y$ transports its derivative and the unique $v$, so the connecting classes commute too. F.11 proves that alternative choices are homotopic and that the reverse comparisons induce inverse homology maps. Hence the induced sequence and its maps are independent of the chosen presentations.

For the final assertion write $\operatorname{NL}_{B/A}=[F_1\xrightarrow{d}F_0]$, with $F_0$ free, put $N=\operatorname{im}d$, $H=\ker d$ and $\Omega=\operatorname{coker}d$. The short exact sequence $0\to N\to F_0\to\Omega\to0$ and the first Tor hypothesis give $$0\longrightarrow N\otimes_BC\longrightarrow F_0\otimes_BC \longrightarrow\Omega\otimes_BC\longrightarrow0.$$ The long Tor sequence and freeness of $F_0$ identify $\operatorname{Tor}_1^B(N,C)$ with $\operatorname{Tor}_2^B(\Omega,C)$, using the full resolution calculus of AG-RG-S01, Lemma 1.A. The second hypothesis therefore makes tensoring $0\to H\to F_1\to N\to0$ give $$0\longrightarrow H\otimes_BC\longrightarrow F_1\otimes_BC \longrightarrow N\otimes_BC\longrightarrow0.$$ Since $N\otimes_BC$ injects into $F_0\otimes_BC$, these two sequences identify $H\otimes_BC$ with the kernel of the tensor differential. This is the claimed natural first-homology identification. No flatness of $F_1$ and no finite-variable assumption has been added. ∎

Lemma F.43: extension

Reading source: Stacks, Tag 0519.

Let \(R\) be any ring and let \(0\to M_1\to M_2\to M_3\to0\) be exact. If the two end modules are finite, so is the middle module; a quotient of a finite module is finite. If both end modules are finitely presented, so is the middle module. A quotient of a finitely presented module by a finite submodule is finitely presented. Finally, if \(M_2\) is finite and \(M_3\) is finitely presented, then \(M_1\) is finite.

Proof. Identify \(M_1\) with its image in \(M_2\). To generate the middle module, take generators of \(M_1\) and lifts of generators of \(M_3\). Subtracting a combination of the latter from an arbitrary element leaves an element of \(M_1\). Images of any generating list generate a quotient. These observations prove the two assertions about finite modules.

We first establish the presentation calculation needed in the other assertions. Suppose \(q : G\twoheadrightarrow N\) is a finite free presentation with finitely generated kernel \(L\), and \(p : F\twoheadrightarrow N\) is any surjection from a finite free module. Choose maps \(u : F\to G\) and \(v : G\to F\) by lifting their finite bases, so that \(qu=p\) and \(pv=q\). For every \(z\in\ker p\), \[ z=v(uz)+(1-vu)z, \qquad uz\in L. \] Both \(v(L)\) and \((1-vu)F\) lie in \(\ker p\), since \(pv=q\) and \(pvu=qu=p\). Thus that kernel is generated by the images under \(v\) of a finite generating list of \(L\), together with the finitely many columns of \(1-vu\). In particular the kernel of every finite free surjection onto a finitely presented module is finite. The choices of lifts here use only the defining property of a free basis.

If \(M_2\) is finite, choose a finite free surjection \(F\to M_2\). When \(M_3\) is finitely presented, the kernel of the composite \(F\to M_3\) is finite by this calculation. Its image in \(M_2\) is exactly \(M_1\): an element mapping to zero in \(M_3\) has any chosen lift in that kernel. Hence \(M_1\) is finite.

Next write a finitely presented \(M_2\) as \(F/K\), with \(F\) finite free and \(K\) finite. If \(M_1\) is generated by \(a_1,\ldots,a_h\), choose lifts \(\widetilde a_i\in F\). The kernel of \(F\to M_3\) is \[ K+R\widetilde a_1+\cdots+R\widetilde a_h. \] Indeed an element in that kernel maps to a combination of the \(a_i\) in \(M_2\); subtracting the corresponding lifted combination puts it in \(K\). This is a finite kernel and gives the asserted finite presentation of the quotient.

Finally choose finite free presentations \(p_i : F_i\twoheadrightarrow M_i\), with finite kernels \(K_i\), for \(i=1,3\). Lift the basis of \(F_3\) to obtain \(\ell : F_3\to M_2\), and use \[ F_1\oplus F_3\longrightarrow M_2, \qquad (a,b)\longmapsto p_1(a)+\ell(b). \] It is onto by the first generator argument. For each generator \(b_j\) of \(K_3\), the element \(\ell(b_j)\) lies in \(M_1\); choose \(a_j\in F_1\) with \(p_1(a_j)=-\ell(b_j)\). Its kernel is generated by \((K_1,0)\) and the finitely many pairs \((a_j,b_j)\). To see this, the second coordinate of any kernel element lies in \(K_3\). Subtract a combination of these pairs to make that coordinate zero; the first coordinate then lies in \(K_1\). We have constructed a finite generating list of relations, so \(M_2\) is finitely presented. Empty generating lists and zero modules satisfy the same formulas. No Noetherian assumption is used. ∎

Lemma F.44: finite presentation independent

Reading source: Stacks, Tag 00R2.

If an \(R\)-algebra \(S\) is finitely presented, the kernel of every surjection \(\alpha : R[X_1,\ldots,X_n]\twoheadrightarrow S\) is a finitely generated ideal.

Proof. Fix one presentation \[ S=R[Y_1,\ldots,Y_m]/(f_1,\ldots,f_h). \] Choose \(g_i(Y)\) representing \(\alpha(X_i)\), and, using the surjectivity of \(\alpha\), choose \(h_j(X)\) representing the class of \(Y_j\). In the polynomial ring \(P=R[X_1,\ldots,X_n]\), form the finite ideal \[ J=\bigl(f_1(h(X)),\ldots,f_h(h(X)),\; X_i-g_i(h(X))\ (1\leq i\leq n)\bigr). \] Each generator maps to zero under \(\alpha\), by the choices of representatives, so \(J\subset\ker\alpha\).

For the reverse inclusion, take \(F(X)\in\ker\alpha\). The polynomial \(F(g(Y))\) represents zero in the fixed presentation; hence there are polynomials \(c_l(Y)\) with \[ F(g(Y))=\sum_{l=1}^h c_l(Y)f_l(Y). \] Substitute \(Y=h(X)\). This puts \(F(g(h(X)))\) in \(J\). Also \[ F(X)-F(g(h(X)))\in \bigl(X_i-g_i(h(X)):1\leq i\leq n\bigr). \] Here is the polynomial identity behind this last inclusion, valid over an arbitrary ring. Replace the variables one at a time and telescope the differences. At a single variable, every monomial difference is divisible by its coordinate difference, because \(U^a-V^a=(U-V)\sum_{b=0}^{a-1}U^{a-1-b}V^b\) for \(a>0\); the difference is zero for \(a=0\). Adding the monomials proves the inclusion without division by any coefficient. Thus \(F\in J\), proving \(\ker\alpha=J\). The argument includes either empty variable list, an empty relation list and the zero algebra. ∎

Lemma F.45: flat fibre smooth

Reading source: AI Integrated Stacks, pinned Algebra, lines 39115–39149.

Let $R\to S$ be a ring map and let $\mathfrak q\subset S$ lie over $\mathfrak p\subset R$. Suppose that some $s\in S\setminus\mathfrak q$ makes $S_s$ finitely presented over $R$, that $R_{\mathfrak p}\to S_{\mathfrak q}$ is flat, and that $S\otimes_R\kappa(\mathfrak p)$ is smooth at the point of this fibre induced by $\mathfrak q$. Then $R\to S$ is smooth at $\mathfrak q$, over an arbitrary base ring.

Proof. Replace $S$ by $S_s$ and keep the same notation for the prime. Choose a finite presentation $S=P/I$, where $P=R[X_1,\ldots,X_n]$ and $I=(a_1,\ldots,a_m)$. Let $\mathfrak r$ be the inverse image of $\mathfrak q$ in $P$, and put $$A=R_{\mathfrak p},\quad k=\kappa(\mathfrak p),\quad Q=P_{\mathfrak r},\quad B=S_{\mathfrak q}.$$ Thus $0\to I_{\mathfrak r}\to Q\to B\to0$ is exact over $A$, and $B$ is $A$-flat.

The full arbitrary-module Tor construction AG-RG-S01, Lemmas 1.0 and 1.A gives exactness after tensoring this sequence with $k$. Indeed use a free resolution of $k$ to compute $\operatorname{Tor}_1^A(k,B)$; tensoring with flat $B$ preserves the kernels and images of that exact resolution, so its positive homology is zero. The natural long exact Tor sequence therefore gives $$0\longrightarrow I_{\mathfrak r}/\mathfrak pI_{\mathfrak r} \longrightarrow Q/\mathfrak pQ\longrightarrow B/\mathfrak pB\longrightarrow0.$$ Set $\bar Q=Q/\mathfrak pQ$, $\bar B=B/\mathfrak pB$ and $J=\ker(\bar Q\to\bar B)$. The injection identifies $J$ with $I_{\mathfrak r}/\mathfrak pI_{\mathfrak r}$. Quotient and localization commute by their fraction maps, giving $\bar Q=k[X]_{\bar{\mathfrak r}}$ and $\bar B=\bar Q/J$. These are local rings, $J$ lies in the maximal ideal of $\bar Q$, and $J$ is generated by the finitely many images of the $a_j$.

The smooth fibre has a smooth principal neighbourhood. Definition F.6, presentation comparison F.11, and the full conormal criterion AG-CA-17, Theorem 3.1 make that neighbourhood formally smooth over $k$: its injective conormal map has projective cokernel, so the sequence splits. The localization lifting proof in AG-CA-17, Theorem 1.2, gives formal smoothness of its local ring $\bar B$. This argument does not assert that the local ring is finitely presented.

Apply formal smoothness to the square-zero surjection $\bar Q/J^2\to\bar B$, lifting the identity of $\bar B$ to a $k$-algebra section $\sigma$. Then $$D(q)=q\bmod J^2-\sigma(q\bmod J)\in J/J^2$$ is a $k$-derivation. Expanding $q=\sigma(\bar q)+D(q)$ and $q'=\sigma(\bar q')+D(q')$, the product of the two correction terms is zero, giving the Leibniz rule. For $j\in J$, $D(j)=[j]$. The universal differential construction thus supplies a left inverse of $$J/J^2\longrightarrow\Omega_{\bar Q/k}\otimes_{\bar Q}\bar B=\bar B^n.$$ The differential basis and its localization are the complete AG-CA-15, Theorems 5.1 and 2.2. In coordinates this conormal map sends $[j]$ to its partial derivatives. Consequently $J/J^2$ is a finite projective direct summand of $\bar B^n$.

Choose from $a_1,\ldots,a_m$ elements $f_1,\ldots,f_c$ whose reduced conormal classes form a basis over the residue field $\kappa(\mathfrak q)$. The finite Nakayama argument F.10 makes the map $\bar B^c\to J/J^2$ surjective. Projectivity splits it. Its kernel is a finite direct summand of $\bar B^c$, and its residue quotient is zero by the chosen residue basis; a second application of F.10 makes this kernel zero. Thus the chosen classes are a free basis.

Let $H$ be their $c$-by-$n$ equation-by-variable Jacobian over $\bar B$. The conormal map has matrix $H^{\mathsf T}$ and a left inverse, so its reduction modulo the maximal ideal is injective; hence $c\leq n$. The full maximal-minor calculation F.70, applied to this $n$-by-$c$ matrix and its left inverse with scalar $1$, makes its maximal minors generate one in $\bar B$. Some minor is therefore a unit in this local ring. Transposing the selected minor preserves its determinant. Let $\Delta\in P$ be the corresponding $c$-variable Jacobian determinant. Its residue in $\kappa(\mathfrak q)$ is nonzero, so $\Delta\notin\mathfrak r$. If $c=0$, use the empty basis and $\Delta=1$.

The chosen $f_i$ generate the whole ideal after localizing at $\mathfrak r$. First $J=(\bar f_1,\ldots,\bar f_c)+J^2$. Its finite quotient by the displayed generators is equal to $J$ times that quotient. Since $J$ lies in the local maximal ideal, F.10 gives $J=(\bar f_1,\ldots,\bar f_c)$. Now the finite $Q$-module $$M=I_{\mathfrak r}/(f_1,\ldots,f_c)Q$$ has quotient $M/\mathfrak pM=0$, by the exact tensor identification of $J$. The ideal $\mathfrak pQ$ lies in the maximal ideal of $Q$, so F.10 gives $M=0$. This is where the flat local quotient is used; neither ring has needed a chain condition.

For each original generator $a_j$ of $I$, write its image in $Q$ as a finite combination of the $f_i$. Clear all coefficient denominators, and also the multipliers outside $\mathfrak r$ witnessing the resulting equalities in the localization. Their finite product $v\notin\mathfrak r$ gives $$I_v=(f_1,\ldots,f_c)P_v,\qquad S_{\bar v\bar\Delta}\cong R[X]_{v\Delta}/(f_1,\ldots,f_c).$$ This uses the fraction equality criterion, including its annihilator factor, so no possible zero divisor has been cancelled.

Adjoin $Z$ with equation $v\Delta Z-1$. The selected Jacobian block for the $c$ pivot variables and $Z$ has zero $Z$ entries in the first $c$ rows; its determinant is $\Delta(v\Delta)=v\Delta^2$, a unit in the quotient. The entire arbitrary-base standard-smooth proof F.85 applies to this finite presentation and proves its smoothness. In terms of lifting, it inverts this square block, corrects the lifted pivot coordinates by the negative error vector, and the monomial expansion kills all terms containing two corrections in a square-zero ideal. Unit lifts then extend evaluation through the inverse-variable relation, exactly as proved in F.85.

Finally express the elements inverted in this neighbourhood of $S_s$ as fractions with powers of $s$ as denominators. Inverting $s$ and their finitely many numerators gives one principal neighbourhood in the original $\operatorname{Spec}S$ containing $\mathfrak q$. Its smoothness proves the conclusion, including empty pivot lists. ∎

Lemma F.46: flat over regular

Reading source: Stacks, Tag 07DY.

Let \(R\to S\) be a homomorphism of Noetherian local rings. Suppose \(R\) is regular and one regular system of parameters \(x_1,\ldots,x_d\) maps to an \(S\)-regular sequence. Then \(S\) is flat over \(R\). A regular sequence here has proper generated ideal.

Proof. The \(x_i\) generate the maximal ideal of \(R\). Their images belong to the maximal ideal of \(S\), because the ideal they generate is proper in that local ring. Thus the given map is local, even though this was not a separate assumption.

For \(0\leq i\leq d\), put \[ R_i=R/(x_1,\ldots,x_i),\qquad S_i=S/(x_1,\ldots,x_i)S. \] These are nonzero Noetherian local rings and their induced maps are local. The written regular-parameter theorem AG-CA-12, Theorem 6.1 makes the chosen list regular in \(R\). Therefore, for every \(i<d\), multiplication by \(x_{i+1}\) is injective on \(R_i\); the assumed regularity in \(S\) gives the same injection on \(S_i\).

At \(i=d\), the source \(R_d\) is the residue field of \(R\), so \(S_d\) is flat over it. Indeed a vector-space basis identifies tensoring with that space with a direct sum of copies of a module, which preserves injections.

We descend from \(i+1\) to \(i\). Since \(x_{i+1}\) is a nonzerodivisor of \(R_i\), \[ 0\longrightarrow R_i\xrightarrow{x_{i+1}}R_i \longrightarrow R_{i+1}\longrightarrow0 \] is a free resolution. Tensoring it with \(S_i\) computes \[ \operatorname{Tor}_1^{R_i}(R_{i+1},S_i) =\ker(x_{i+1} : S_i\to S_i)=0. \] The quotient \(S_i/x_{i+1}S_i=S_{i+1}\) is flat over \(R_{i+1}\) by the descending induction. Apply the complete ideal local criterion AG-CA-08, Corollary 4.4 and equation (6) to \(R_i\to S_i\), the proper ideal \((x_{i+1})\subset R_i\), and the module \(S_i\). This module is finite over the target ring \(S_i\), being generated by \(1\); it need not be finite over the source. The Tor vanishing and quotient flatness prove that \(S_i\) is \(R_i\)-flat. The induction reaches \(S_0=S\) over \(R_0=R\). If \(d=0\), the initial field argument already proves the conclusion. ∎

Lemma F.47: flat over regular with regular fibre

Reading source: Stacks, Tag 031E.

Let \((R,\mathfrak m)\to(S,\mathfrak n)\) be a flat local map of Noetherian local rings. If \(R\) and the closed fibre \(F=S/\mathfrak mS\) are regular, then \(S\) is regular.

Proof. Put \(d=\dim R\) and \(e=\dim F\). Regularity supplies generators \(x_1,\ldots,x_d\) of \(\mathfrak m\) and generators \(\overline y_1,\ldots,\overline y_e\) of the maximal ideal \(\mathfrak n/\mathfrak mS\) of \(F\). Choose lifts \(y_j\in\mathfrak n\). Every \(z\in\mathfrak n\) reduces to a combination of the \(\overline y_j\); lifting the coefficients and subtracting that combination leaves an element of \(\mathfrak mS=(x_1,\ldots,x_d)S\). Consequently \[ \mathfrak n=(x_1,\ldots,x_d,y_1,\ldots,y_e)S. \] Its cotangent space \(\mathfrak n/\mathfrak n^2\) is therefore spanned by at most \(d+e\) vectors over the residue field of \(S\).

The flat local dimension formula, with precisely the stated Noetherian and flatness hypotheses, is proved in AG-CA-11, Theorem 5.1; it gives \(\dim S=d+e\). Its Proposition 4.2 proves that the dimension of a Noetherian local ring is at most the dimension of its cotangent space: lifts of a cotangent basis generate the maximal ideal by finite Nakayama, and the parameter/height theorem bounds dimension by that generator count. Combining these inequalities gives \[ d+e=\dim S\leq \dim_{\kappa(\mathfrak n)}\mathfrak n/\mathfrak n^2 \leq d+e. \] Equality of dimension and embedding dimension is the definition of regularity, so \(S\) is regular. The empty source or fibre parameter lists when \(d=0\) or \(e=0\) obey the same calculation. No finite-type assumption on the ring map is needed. ∎

Lemma F.48: flatness descends more general

Reading source: Stacks, Tag 0584.

Let \(S\to S'\) be a flat map of \(R\)-algebras, let \(M\) be any \(S\)-module, and put \(M'=S'\otimes_S M\). If \(M\) is flat over \(R\), so is \(M'\). If \(S\to S'\) is faithfully flat, the converse holds as well.

Proof. Test an arbitrary injection \(f:N\hookrightarrow N_1\) of \(R\)-modules. Write \[ Z=\ker\bigl(N\otimes_R M\xrightarrow{f\otimes1}N_1\otimes_R M\bigr). \] Both tensor products, their map and \(Z\) are \(S\)-modules, using the action on \(M\). Flatness of \(S'\) over \(S\) preserves this kernel. The balancing relations give inverse isomorphisms \[ S'\otimes_S(N\otimes_R M)\ \longleftrightarrow\ N\otimes_R M', \qquad s'\otimes(n\otimes m)\ \longleftrightarrow\ n\otimes(s'\otimes m), \] and the same formula for \(N_1\). These maps commute with \(f\), so \[ S'\otimes_S Z\cong \ker\bigl(N\otimes_R M'\longrightarrow N_1\otimes_R M'\bigr). \] If \(M\) is \(R\)-flat, \(Z=0\), and the displayed kernel vanishes for every \(f\). If \(M'\) is \(R\)-flat, the right side is zero; faithful flatness then detects \(Z=0\). In either direction the injection characterization of flatness, proved for arbitrary modules in AG-CA-07, Theorem 2.1, gives the claimed flatness. No module in this calculation is required to be finite. ∎

Lemma F.49: formally smooth extensions easy

Reading source: AI Integrated Stacks, pinned Algebra, lines 45971–46012.

For a field extension $K/k$, each of the following conditions implies formal smoothness of $K$ over $k$: it is purely transcendental; it is separable algebraic; or it is separable in the sense of F.5. Infinite generator sets and infinite algebraic extensions are allowed.

Proof. In the first case write $K=k(x_j\mid j\in\mathcal J)$. Given a $k$-algebra $D$, a square-zero ideal $I$ and a $k$-map $K\to D/I$, choose lifts $a_j\in D$ of the variable images. Polynomial evaluation at these lifts is well defined because each polynomial involves finitely many variables. Every nonzero polynomial in the $x_j$ is a unit in $K$, so its evaluated image is a unit modulo $I$. Units lift: if $bc=1+r$ with $r\in I$, then $bc(1-r)=1$ since $r^2=0$. Thus all the nonzero polynomials have unit evaluations in $D$. The polynomial map extends uniquely, for these chosen variable lifts, through the fraction-field localization to a $k$-map $K\to D$. Its reduction is the given map, proving the lifting property.

For a separable algebraic extension, a finite subset of $K$ generates a finite separable subextension: successive degrees of its finitely many algebraic generators are finite and multiply along the tower. These subfields are directed under inclusion, since the union of two finite generator lists is finite, and their union is $K$. The full separable-generator tower proof AG-CA-17, Proposition 6.3 gives a unique square-zero lift on every such finite field. The derivatives of its separable minimal polynomials are units, so its successive one-equation lifting steps apply. Consequently every finite subextension is formally étale. The entire compatible-lifts proof F.22 makes their colimit formally étale, and hence formally smooth.

In the third case use the directed family of finitely generated subfields $K_i/k$. By F.5 each has a separating transcendence basis and is a finite separable extension of the resulting rational-function field. The first two cases give formal smoothness of both maps in that tower; the full composition lifting proof AG-CA-17, Theorem 1.2 gives formal smoothness of $K_i/k$. The arbitrary-field criterion F.17 therefore gives $H_1(L_{K_i/k})=0$.

Now F.24 identifies the conormal complex of $K/k$ with the filtered colimit of these complexes. Its representative proof of exactness of filtered module colimits also identifies their homology: a cycle is represented at a stage after its finite differential equality holds, and a boundary equality is witnessed at a common later stage. Thus $H_1(L_{K/k})=\operatorname*{colim}_iH_1(L_{K_i/k})=0$. Applying F.17 again proves formal smoothness of $K/k$. This uses the field criterion, rather than assuming a general permanence theorem for filtered colimits of formally smooth algebras. ∎

Lemma F.50: geometrically regular

Reading source: AI Integrated Stacks, pinned Algebra, lines 48753–48798.

Let $A$ be a Noetherian $k$-algebra, where $k$ is a field. Regularity of $A\otimes_kK$ for every finitely generated field extension $K/k$ is equivalent to regularity of $A\otimes_kk'$ for every finite purely inseparable field extension $k'/k$.

Proof. The forward implication includes the finite purely inseparable extensions among the finitely generated ones. For the converse, fix a finitely generated $K/k$. The full separabilization construction F.69 gives finite purely inseparable extensions $k'/k$ and $K'/K$, inside a commuting field square, with $K'/k'$ separably generated. This last extension is finitely generated: a finite field-generating list for $K/k$ and finitely many generators for $K'/K$ still generate $K'$ after enlarging the coefficient field to $k'$.

The entire forward smooth-domain construction F.64 supplies a smooth finite-type domain $D\subset K'$ over $k'$ with fraction field $K'$. Its explicit separating-generator equations, denominator clearing and derivative localization have been proved there. They also give the smooth field model recognized by AG-CA-18, Theorem 3.2.

By assumption $A'=A\otimes_kk'$ is regular. It is Noetherian by F.12. Form $C=A'\otimes_{k'}D$. The coefficient-change polynomial presentation makes it finitely presented over $A'$, and formal smoothness survives base change by AG-CA-17, Theorem 1.2. Hence $C/A'$ is smooth. It is also Noetherian, being finite type over the Noetherian ring $A'$.

The complete smooth-flatness and regular-fibre results AG-CA-18, Theorems 6.1 and 2.1, make this map flat with regular residue-field fibres. At a prime of $C$, the localized source is regular, and its closed fibre is the local ring of one of those regular fibres. The full ascent calculation F.47 therefore makes the target local ring regular. This proves regularity of $C$.

Localizing at all nonzero elements of $D$ gives $$C\otimes_DK'=A'\otimes_{k'}K'\cong A\otimes_kK'.$$ Every local ring of this localization is a local ring of the original regular ring, so it is regular. Finally choose a basis of the finite field extension $K'/K$. As a module over $B=A\otimes_kK$, the algebra $A\otimes_kK'$ is the direct sum of $[K':K]\geq1$ copies of $B$, by tensoring this basis. Tensoring an arbitrary $B$-module with that algebra therefore takes this positive finite number of copies; it preserves exactness and detects zero modules. Thus the map $A\otimes_kK\to A\otimes_kK'$ is faithfully flat. The full arbitrary-source regularity descent proof F.38 makes $A\otimes_kK$ regular. This holds for every chosen $K$, proving the equivalence without a perfect-field assumption. ∎

Lemma F.51: geometrically regular descent

Reading source: AI Integrated Stacks, pinned Algebra, lines 48815–48834.

Let $A\to B$ be a faithfully flat map of $k$-algebras. If $B$ is geometrically regular over the field $k$, then $A$ is geometrically regular over $k$. No separate Noetherianity hypothesis on $A$ is needed.

Proof. The trivial extension $k/k$ is among the field tests for geometric regularity, so $B$ is regular. The full regularity descent proof F.38 makes $A$ regular as well, deriving its Noetherianity by its finite-ideal argument.

For every finite purely inseparable extension $k'/k$, set $A'=A\otimes_kk'$ and $B'=B\otimes_kk'$. Both are Noetherian by F.12. Tensor associativity identifies $B'$ with $A'\otimes_AB$, and for an arbitrary $A'$-module $M$ it gives the inverse pure-tensor maps $$M\otimes_{A'}(A'\otimes_AB)\cong M\otimes_AB,\qquad m\otimes(a'\otimes b)\longmapsto a'm\otimes b,\quad m\otimes b\longmapsto m\otimes(1\otimes b).$$ The balancing relations verify these maps and their inverse composites. Restriction from $A'$-modules to $A$-modules preserves exact sequences and detects the zero module. Since $B$ is faithfully flat over $A$, the displayed tensor identification proves that $B'$ is faithfully flat over $A'$.

Geometric regularity makes $B'$ regular. Applying F.38 gives regularity of $A'$ for every such $k'$. The complete finite purely inseparable criterion F.50 now applies to the Noetherian algebra $A$ and proves its geometric regularity over $k$. ∎

Lemma F.52: geometrically regular over subfields

Reading source: AI Integrated Stacks, pinned Algebra, lines 48851–48866.

Let $A$ be a Noetherian algebra geometrically regular over the field $k$.

1. For any field extension $E/k$ such that $C=A\otimes_kE$ is Noetherian, $C$ is regular.
2. Under this same Noetherian tensor hypothesis, $C$ is geometrically regular over $E$.
3. If $k$ is a directed union of subfields $k_i$ and $A$ is geometrically regular over every $k_i$, then $A$ is geometrically regular over $k$.

Proof. For (1), write $E$ as the union of its intermediate fields $E_i$ generated by finite subsets over $k$. They are directed, since the union of two finite generator sets is finite. Put $C_i=A\otimes_kE_i$. An $E_i$-basis of $E$ containing $1$ gives $$C=C_i\otimes_{E_i}E\cong\bigoplus_{\beta\in\mathcal B_i}C_i.$$ Here the structural inclusion $C_i\to C$ is the coordinate belonging to $1$. To obtain such a basis, take a maximal independent set containing $1$; if it did not span, a vector outside its span could be adjoined, contradicting maximality. The tensor/direct-sum calculation is AG-RG-S01, Lemma 1.0. It proves both injectivity of this inclusion and flatness of $C$ over $C_i$, even when the basis is infinite. Every element of $C$ uses finitely many field coefficients and so lies in one $C_i$.

Suppose that $C$ is Noetherian and fix $\mathfrak q\subset C$. In the Noetherian local ring $T=C_{\mathfrak q}$ choose a finite generating list $x_1,\ldots,x_r$ for its maximal ideal $\mathfrak m$. Write $x_j=a_j/s_j$, with $a_j\in\mathfrak q$ and $s_j\notin\mathfrak q$. Put all their finitely many field coefficients into a common $E_i$, so that every $a_j,s_j$ lies in $C_i$. Let $\mathfrak q_i=\mathfrak q\cap C_i$ and $B_i=(C_i)_{\mathfrak q_i}$. The fractions $x_j$ now belong to $B_i$, and their numerators lie in its maximal ideal.

The map $B_i\to T$ is local and flat: it is obtained by localizing the free scalar-extension map above, and exact localization preserves its injection tests. Its extended maximal ideal is contained in $\mathfrak m$ and contains every $x_j$. Hence $$\mathfrak q_iT=\mathfrak m,\qquad T\otimes_{B_i}\kappa(\mathfrak q_i)=T/\mathfrak m.$$ The source $B_i$ is Noetherian by F.12, and is regular because $E_i/k$ is finitely generated and $A$ is geometrically regular. The displayed closed fibre is a field, thus regular. The entire flat-local ascent proof F.47 makes $T$ regular. An empty generating list gives the same argument when $\mathfrak m=0$. This proves regularity at every prime of $C$; if $C=0$, there are no primes to test.

For (2), let $F/E$ be a finite purely inseparable extension. F.12 makes $C\otimes_EF$ Noetherian. Tensor associativity identifies it with $A\otimes_kF$, so (1), applied to the field extension $F/k$, makes it regular. The complete finite purely inseparable criterion F.50 proves geometric regularity of $C$ over $E$.

For (3), we construct an actual finite purely inseparable field model. Let $k'/k$ be finite purely inseparable. If it is trivial, use $k_i'=k_i$ at any index. Otherwise the characteristic is $p>0$. Choose a $k$-basis $b_1=1,b_2,\ldots,b_d$ of $k'$. A common exponent $e$ satisfies $b_j^{p^e}\in k$ for every $j$, since the extension is purely inseparable and the list is finite. Write its finitely many multiplication constants $$b_jb_\ell=\sum_h c_{j\ell h}b_h,\qquad c_{j\ell h}\in k.$$ Choose $k_i$ containing all $c_{j\ell h}$ and all $b_j^{p^e}$. This is possible because the union is directed and the list is finite.

The subspace $V=\bigoplus_j k_i b_j\subset k'$ contains $1$ and is closed under multiplication by those identities. It is a finite-dimensional domain over $k_i$. Multiplication by any nonzero $v\in V$ is an injective endomorphism of that finite-dimensional vector space, hence surjective; a preimage of $1$ gives its inverse. Therefore $V$ is a field, denoted $k_i'$. Moreover every $v=\sum_j a_jb_j$ in it has $$v^{p^e}=\sum_j a_j^{p^e}b_j^{p^e}\in k_i.$$ Thus $k_i'/k_i$ is finite purely inseparable. This checks the exponent condition for every element of the field, beyond just its chosen generators.

The multiplication map $k\otimes_{k_i}k_i'\to k'$, $a\otimes v\mapsto av$, is an algebra isomorphism: its basis $1\otimes b_j$ maps to the given $k$-basis $b_j$, proving both bijectivity and compatibility with the prescribed products. Consequently $$A\otimes_kk'\cong A\otimes_{k_i}k_i'.$$ The right side is regular by the hypothesis over $k_i$. F.50 proves geometric regularity over $k$. In characteristic zero all finite purely inseparable extensions are trivial, so the same conclusion follows from the trivial case. All three assertions are established. ∎

Lemma F.53: grothendieck regular sequence

Reading source: AI Integrated Stacks, pinned Algebra, lines 24513–24526.

Let $R\to P$ be a flat local map of Noetherian local rings, and let $\mathfrak m=\mathfrak m_R$. If the reductions of $f_1,\ldots,f_c\in P$ form a regular sequence in $P/\mathfrak mP$, then the list is regular in $P$, and every $P/(f_1,\ldots,f_i)$ is flat over $R$.

Proof. Put $P_i=P/(f_1,\ldots,f_i)$, including $P_0=P$. The regular-sequence hypothesis includes a nonzero final fibre quotient. Consequently each $\bar f_i$ belongs to the maximal ideal of the local closed fibre: a unit there would make the final quotient zero. Its lift $f_i$ belongs to the maximal ideal of $P$ as well. Every $P_i$ is thus a nonzero Noetherian local quotient, and $R\to P_i$ remains local. Its closed fibre is $$P_i/\mathfrak mP_i \cong(P/\mathfrak mP)/(\bar f_1,\ldots,\bar f_i).$$

Induct on $i$, keeping $R$-flatness of $P_i$. The initial assertion is the assumed flatness of $P$. On $P_i$ consider multiplication by $f_{i+1}$. Both its source and target are finite over this actual Noetherian local ring, and its target is $R$-flat by induction. The displayed fibre description and the original regular-list hypothesis make its reduction injective. The full lifting-injectivity proof AG-CA-08, Lemma 5.1 gives injectivity on $P_i$ and $R$-flatness of its cokernel $P_{i+1}$. This proves the induction.

Every multiplication in the list has now been proved injective on its preceding quotient, and the final quotient is nonzero, so the list is regular. The same induction has proved flatness of each successive quotient. If $c=0$, the list is empty and the only quotient is the original flat ring $P$, giving both assertions directly. ∎

Lemma F.54: henselian cat finite etale

Reading source: Stacks, Tag 04GK.

For a henselian local ring \((R,\mathfrak m,k)\), reduction \(S\mapsto S/\mathfrak mS\) is an equivalence from finite étale \(R\)-algebras to finite products of finite separable extensions of \(k\). Ring maps are unital, and the zero algebra is the empty product.

Proof. The written idempotent and finite-algebra theorem AG-CA-20, Theorem 2.2 decomposes a finite \(R\)-algebra into finitely many local factors. A factor of an étale algebra is étale, since it is localization at its component idempotent. For each such factor \(S_i\), reduction is an étale \(k\)-algebra and is local: every maximal ideal of a finite algebra contracts to \(\mathfrak m\), and \(S_i\) has only one. The field classification AG-CA-17, Theorem 7.2 makes this reduction a finite separable field. In particular \(\mathfrak mS_i\) is its maximal ideal. Every \(S_i\) is henselian by AG-CA-20, Proposition 3.1.

Construct a lift of a finite separable field extension \(L/k\) as follows. Choose finitely many field generators and use their successive fields \[ k=L_0\subset L_1\subset\cdots\subset L_a=L, \qquad L_i=L_{i-1}(\alpha_i). \] Each minimal polynomial is monic separable. Starting with \(T_0=R\), lift its coefficients to a monic \(h_i\in T_{i-1}[X]\) and define \(T_i=T_{i-1}[X]/(h_i)\). Its bounded powers of \(X\) are a free \(T_{i-1}\)-basis, by monic division. Reduction modulo \(\mathfrak m\) is \(L_i\). Every maximal ideal contracts to \(\mathfrak m\) by integrality, so the field reduction makes \(T_i\) local with maximal ideal \(\mathfrak mT_i\). The derivative \(h_i'\) has nonzero residue in that field and is consequently a unit. The one-equation Jacobian proof AG-CA-17, Theorem 4.1 and Section 6 makes \(T_i/T_{i-1}\) étale. Composition makes \(T_a/R\) finite étale, with free rank \([L:k]\) and reduction \(L\). Finite products of these lifts provide every desired residue algebra.

We next show explicitly that every local finite étale \(R\)-algebra \(S\), with residue \(L\), is isomorphic to this constructed \(T_a\). Given the residue identification, lift a map \(T_{i-1}\to S\) through the next step by choosing the root of \(h_i\) reducing to \(\alpha_i\). The simple-root Hensel statement, AG-CA-20, Theorem 1.2, gives exactly one such root in the henselian local ring \(S\), and hence exactly one map \(T_i\to S\) at each step. This constructs \(\phi:T_a\to S\) inducing the chosen isomorphism on residues.

By the complete arbitrary-base proof in S04, Lemma E4.2, a finite étale algebra is finite locally free. Its proof first obtains a finite module presentation from monic algebra equations, then applies Lemma A1.3 to a finitely presented flat module. Since \(R\) is local, \(S\) is finite free, of rank \(\dim_k(S/\mathfrak mS)=[L:k]\). Both modules in \(\phi\) therefore have this same finite rank. In chosen bases its matrix reduces to an invertible matrix over \(k\), so its determinant lies outside \(\mathfrak m\) and is a unit of \(R\). The adjugate matrix divided by this unit is its inverse. Thus \(\phi\) is an isomorphism of modules, and its bijective algebra map is an algebra isomorphism.

Now let \(S_1,S_2\) be local finite étale algebras and let \(L_1\to L_2\) be a \(k\)-field map between their residues. Replace \(S_1\) by its isomorphic tower \(T_a\). Starting with \(R\to S_2\), lift each successive generator to the unique simple root whose residue is its prescribed image in \(L_2\). The same Hensel calculation supplies a unique \(R\)-algebra map \(S_1\to S_2\) inducing the field map. Conversely, any algebra map has those generator values, so it is this unique lift. The map on residue fields is defined because \(\mathfrak mS_i\) is the maximal ideal.

For products, a map to a local target sends each source component idempotent to \(0\) or \(1\); orthogonality and their sum \(1\) force exactly one component to be chosen. The residue map chooses precisely that component, and the local result lifts its field map uniquely. Doing this independently for each target factor proves a bijection on all Hom sets. The empty-product cases follow from unital maps directly: there is one map to the zero ring and none from it to a nonzero ring. We have proved full faithfulness and essential surjectivity, hence the asserted equivalence, over every henselian local source. ∎

Lemma F.55: huber

Reading source: Stacks, Tag 07CF.

Suppose a finitely presented $R$-algebra $S$ has a finite polynomial presentation $S=P/I$, where $P=R[X_1,\ldots,X_n]$, and $I/I^2$ is free over $S$. Then $S$ has another finite polynomial presentation whose defining equations themselves form a basis of its conormal module.

Proof. F.44 makes $I$ finitely generated. Thus $I/I^2$ has a finite basis: when $S\ne0$, the supports of a finite generating list in any free basis have finite union, which must contain the entire basis; when $S=0$, choose its empty conormal basis directly. Lift this basis to $f_1,\ldots,f_c\in I$ and put $K=(f_1,\ldots,f_c)$. The basis property gives $I=K+I^2$. The finite $P$-module $U=I/K$ consequently satisfies $U=IU$.

Choose generators $u_1,\ldots,u_t$ of $U$ and coefficients $b_{ij}\in I$ with $u_i=\sum_j b_{ij}u_j$. The adjugate identity proved in F.70 applied to $1_t-(b_{ij})$ shows that $$g=\det(1_t-(b_{ij})),\qquad gU=0,\qquad g\in1+I.$$ For $U=0$ take $g=1$. Hence $gI\subset K$ and $I_g=K_g$. The image of $g$ in $S$ is exactly $1$, so $P_g/I_g=S$. It follows, with both inverse maps given by adjoining or evaluating $g^{-1}$, that $$S=P[T]/(f_1,\ldots,f_c,gT-1).$$

Let $L$ be the displayed equation ideal. After localization at $g$, the classes of $f_i$ form the given basis of $I_g/I_g^2$. The explicit inverse-variable conormal calculation F.72, applied to $P_g/I_g$ and this presentation, identifies $$L/L^2\cong S^c\oplus S.$$ The first basis vectors are $[f_i]$ and the last is $[gT-1]$: the last equation has derivative $g$, a unit, in its new variable, and the remaining classes are read by evaluation at $T=g^{-1}$. These are exactly the claimed equation classes. No Noetherian hypothesis or cancellation of a free summand is used. Empty bases use the same determinant convention; if $S=0$, all the displayed modules and bases are interpreted over the zero ring. ∎

Lemma F.56: isolated point fibre

Reading source: Stacks, Tag 00PK.

Let $R\to S$ be of finite type and let $\mathfrak q\in\operatorname{Spec}S$ lie over $\mathfrak p$. Put $k=\kappa(\mathfrak p)$, $T=S\otimes_R k$, and let $x\in\operatorname{Spec}T$ be the corresponding point. The following conditions are equivalent: $x$ is isolated; $T_x=S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite-dimensional over $k$; some $g\in S\setminus\mathfrak q$ has $x$ as the only point of its fibre neighbourhood; the fibre point-dimension at $x$ is zero; $x$ is closed and $\dim T_x=0$; and $\kappa(x)/k$ is finite and $\dim T_x=0$.

Proof. All the geometry can be checked in the finite-type field algebra $T$. Such an algebra and its principal localizations are Noetherian by AG-CA-03, Theorem 2.1. We will use the following consequence of its Theorem 4.2. A zero-dimensional finite-type $k$-algebra $B$ is Artinian, with finitely many local factors and nilpotent radical $J$. Every residue field is finite over $k$ by AG-CA-06, Theorem 2.1. Each $J^i/J^{i+1}$ is finite over the product $B/J$ of those fields: $B$ is Noetherian, so the ideals involved are finite. Its coordinate summands are consequently finite-dimensional over the residue fields, hence over $k$. The finite filtration by powers of $J$ proves that $B$ is finite-dimensional over $k$. The factor idempotents give a principal singleton neighbourhood of each of its points.

If $x$ is isolated, choose $h\in T$ with $D(h)=\{x\}$. The algebra $T_h$ has just one prime, so has dimension zero and is finite-dimensional by the preceding calculation. Every element outside its unique maximal ideal is a unit. Its localization at that ideal is therefore itself, and $T_x=T_h$ is finite-dimensional.

Conversely, a finite-dimensional $T_x$ is Artinian: chains of its ideals are chains of subspaces of a finite-dimensional $k$-space. Each of its prime quotients is a finite-dimensional domain, hence a field, because multiplication by a nonzero element is an injective, and thus surjective, linear map. Thus $\dim T_x=0$, and its residue field is finite over $k$. For a prime of a finite-type $k$-algebra, finiteness of the residue extension implies closedness: $T/x$ embeds into the finite-dimensional field $\kappa(x)$, so is a finite-dimensional domain and the same multiplication argument makes it a field. The converse follows from the cited weak Nullstellensatz.

Suppose now that $x$ is closed and $\dim T_x=0$. The prime corresponding to $x$ is both maximal and minimal. There are finitely many other minimal primes, by AG-CA-03, Proposition 2.2. From each choose an element outside $x$ and take their product $h$; use $h=1$ if there are none. A prime in $D(h)$ contains none of those other minimal primes. Since every prime contains a minimal prime, it must contain $x$ and therefore equal $x$. Hence $D(h)=\{x\}$ and $x$ is isolated. This proves the equivalence of isolation, finite local fibre algebra, and the two residue-field/local-dimension conditions.

An isolated point has point-dimension zero, using its singleton open neighbourhood. Conversely, zero point-dimension supplies an open neighbourhood of dimension zero: the infimum defining that dimension is an attained nonnegative integer. Take a principal neighbourhood inside it. Its algebra has dimension zero, so the Artinian factor calculation above supplies a smaller principal singleton neighbourhood of $x$. Thus zero point-dimension implies isolation as well.

Finally, every $h\in T=S_{R\setminus\mathfrak p}/\mathfrak pS_{R\setminus\mathfrak p}$ is represented by $g/s$ with $g\in S$ and $s\in R\setminus\mathfrak p$. The image of $s$ is a fibre unit, so $D(h)=D(\bar g)$, and $h\notin x$ means $g\notin\mathfrak q$. Conversely the fibre of $D(g)$ is exactly $D(\bar g)$. This proves the neighbourhood condition over the original ring, with no Noetherian assumption on $R$. ∎

Lemma F.57: lci

Reading source: AI Integrated Stacks, pinned Algebra, lines 37341–37441.

Let $k$ be a field, $S$ a finite-type $k$-algebra, and $\mathfrak q$ a prime of $S$. Fix any polynomial presentation $S=P/I$, with $P=k[X_1,\ldots,X_n]$, and let $\mathfrak Q\subset P$ lie over $\mathfrak q$. Put $$c=\operatorname{ht}\mathfrak Q-\operatorname{ht}\mathfrak q =n-\dim_{\mathfrak q}\operatorname{Spec}S.$$ The last equality is the point-codimension formula F.20. The following are equivalent.

1. Some $S_g$, with $g\notin\mathfrak q$, is a global complete intersection over $k$.
2. $I_{\mathfrak Q}$ has a generating list of length $c$.
3. $(I/I^2)_{\mathfrak q}$ has a generating list of length $c$.
4. $(I/I^2)_{\mathfrak q}$ is free of rank $c$ over $S_{\mathfrak q}$.
5. $I_{\mathfrak Q}$ is generated by a regular sequence in $P_{\mathfrak Q}$.

Whenever these hold, every list of $c$ elements spanning $I_{\mathfrak Q}/\mathfrak QI_{\mathfrak Q}$ is a regular sequence in $P_{\mathfrak Q}$.

Proof. Write $V=P_{\mathfrak Q}$ and $J=I_{\mathfrak Q}$. The whole polynomial-local regularity proof AG-CA-14, Proposition 3.3, starting with $k$, makes $V$ regular. The whole AG-CA-12, Theorem 6.1 makes it Cohen–Macaulay. Its dimension is $\operatorname{ht}\mathfrak Q$, whereas $V/J=S_{\mathfrak q}$ has dimension $\operatorname{ht}\mathfrak q$. In particular $J$ is proper and finite, and the dimension drop is $c$.

Generators of $J$ give generators of $J/J^2=(I/I^2)_{\mathfrak q}$. Conversely lift $c$ generators of this quotient to $f_1,\ldots,f_c\in J$. Their quotient module $N=J/(f_1,\ldots,f_c)V$ satisfies $N=JN$. It is finite, and $J$ lies in the maximal ideal of $V$, so the determinant proof F.10 gives $N=0$. Thus (2) and (3) are equivalent. The entire expected-dimension criterion AG-CA-12, Theorem 4.2, now says that any such generating list is regular: it lies in the maximal ideal and lowers the dimension of the Cohen–Macaulay ring by exactly its length. A list spanning $J/\mathfrak QJ$ generates $J$ by the same Nakayama argument, proving the final assertion too.

We spell out the conormal basis for a regular generating list. The regular-list cone proof F.118 makes its first Koszul homology zero. If $\sum_i a_if_i\in J^2$, write this sum as $\sum_i b_if_i$ with every $b_i\in J$. The vector $(a_i-b_i)$ is a first Koszul cycle, hence a boundary. Each coefficient of a degree-two Koszul boundary belongs to $J$, because its differential has coefficients among the $f_i$. Thus every $a_i\in J$. This proves independence, as well as generation, of the classes of the $f_i$ in $J/J^2$. Consequently (3) implies (4); (4) plainly implies (3). The same criterion proves (2) implies (5). If (5) uses a regular list of length $e$, successive regular quotients of a Cohen–Macaulay local ring lower dimension by one, by the complete depth and dimension proofs in AG-CA-12, Sections 2 and 4. Hence $$\operatorname{ht}\mathfrak q =\dim(V/J)=\dim V-e=\operatorname{ht}\mathfrak Q-e,$$ so $e=c$, proving (5) implies (2).

Suppose (2) holds. Choose generators $f_i=a_i/s_i$ with $a_i\in I$ and $s_i\notin\mathfrak Q$. Multiplication by these units shows that the $a_i$ also generate $J$ and form a regular list there. Since $I$ is finite over the Noetherian polynomial ring, each generator of $I/(a_1,\ldots,a_c)$ is killed by some element outside $\mathfrak Q$. Their product $h'\notin\mathfrak Q$ gives $I_{h'}=(a_1,\ldots,a_c)_{h'}$.

Choose a principal neighbourhood $D(h'')$ of $\mathfrak q$ on which the dimension is its point dimension. Here is the finite-component construction from F.20: for every minimal prime $\mathfrak p$ of $S$ not contained in $\mathfrak q$, choose an element in $\mathfrak p\setminus\mathfrak q$ and multiply these finitely many elements. The product kills exactly the components not passing through the point; any surviving component has its original dimension, since a nonempty principal open in a finite-type domain over a field has the same transcendence degree and dimension. Thus the maximum surviving component dimension is the point dimension, by the whole AG-CA-09, Theorem 6.1, calculation. Lift this product to $h''\in P\setminus\mathfrak Q$ and put $h=h'h''$. Its image $g$ is outside $\mathfrak q$, and $$S_g\cong k[X_1,\ldots,X_n,Z]/ (a_1,\ldots,a_c,hZ-1),\qquad \dim S_g=n-c.$$ This is the required global complete-intersection presentation: there are $n+1$ variables, $c+1$ equations and the prescribed dimension difference. This proves (1).

Conversely suppose (1), and choose a global complete-intersection presentation $S_g=k[Y_1,\ldots,Y_m]/H$, with $H=(b_1,\ldots,b_t)$ and $\dim S_g=m-t$. Every minimal prime $\mathfrak a$ over $H$ has height at most $t$ by the entire AG-CA-11, Theorem 3.1. Its component has dimension $m-\operatorname{ht}\mathfrak a$ by the polynomial height formula in AG-CA-09, Theorem 4.3. Hence each component has dimension at least $m-t$, while their maximum is the given dimension $m-t$. All components therefore have dimension exactly $m-t$. The point formula of AG-CA-09, Theorem 6.1, gives this same dimension at $\mathfrak qS_g$.

At the polynomial prime $\mathfrak Q''$ over that point, F.20 consequently gives $$t=\operatorname{ht}\mathfrak Q''-\operatorname{ht}\mathfrak q.$$ The actual ambient polynomial local ring is regular and Cohen–Macaulay as above. The expected-dimension criterion makes its given $b_1,\ldots,b_t$ regular. The conormal calculation already proved makes $(H/H^2)_{\mathfrak q}$ free of rank $t$. The entire graph-presentation calculation F.33, localized at this point, supplies $$S_{\mathfrak q}^{\,t+n} \cong (I/I^2)_{\mathfrak q}\oplus S_{\mathfrak q}^{\,m}.$$ Thus the conormal module is a finite direct summand of a finite free module. Tensor injections restricted to this summand prove flatness; the finite complementary summand gives finite presentation. The full local finite-flat proof AG-RG-S01, Lemma 1.2 makes it free. Reducing the displayed isomorphism to the residue field computes its rank as $$t+n-m=n-\dim_{\mathfrak q}\operatorname{Spec}S=c.$$ This proves (4). Empty equation lists are included in the conormal and dimension calculations, so every implication and the last assertion also hold when $c=0$. ∎

Lemma F.58: lci at prime

Reading source: Stacks, Tag 00SG.

For a finite-type $k$-algebra $S$ and $\mathfrak q\in\operatorname{Spec}S$, the following are equivalent: $S_{\mathfrak q}$ is a complete intersection in the sense of F.3; it has a global complete-intersection principal neighbourhood; and the kernel of any finite polynomial presentation of $S$ is generated by a regular sequence in the polynomial local ring above $\mathfrak q$. Equivalently, there is a principal neighbourhood that is locally a complete intersection over $k$.

Proof. Write $C=S_{\mathfrak q}$ and $P=k[X_1,\ldots,X_n]_{\mathfrak q'}\twoheadrightarrow C$. The field $k$ is a regular ring: its only prime localization is itself, with dimension and maximal-ideal generating number both zero. Iterating the entire polynomial-regularity proof AG-CA-14, Proposition 3.3 makes $k[X_1,\ldots,X_n]$ a regular ring. Its localization at the displayed prime is therefore regular local, by the definition of a regular ring. A regular generating list for its kernel therefore gives the ambient ring required by F.3. Conversely suppose F.3 gives $C=Q/(a_1,\ldots,a_e)$, where $Q$ is regular local and essentially of finite type over the same field $k$, and this list is regular in its maximal ideal. We compare this local ambient with $P$ explicitly.

Present $Q$ as a quotient of a polynomial local ring $V=k[Y_1,\ldots,Y_m]_{\mathfrak n}$. Both $V$ and $Q$ are regular local. AG-CA-14, Proposition 1.4, makes the kernel of $V\to Q$ generated by part of a regular parameter system. Append lifts of the $a_i$. The resulting list generates $\ker(V\to C)$ and is regular: its first successive quotients are those by the parameter list, and its remaining multiplication injections are precisely those of the given regular list in $Q$.

Let $U=k[X,Y]_{\mathfrak r}$, where $\mathfrak r$ is the preimage of the maximal ideal of $C$ under the map using both sets of coordinates, and put $L=\ker(U\to C)$. Lift the images of $X_i$ to $V$ and denote them by $u_i(Y)$; these are fractions with denominators outside $\mathfrak n$. Those denominators are units in $U$. The quotient by $X_i-u_i(Y)$ is $V$, including its indicated localization: evaluating an element outside $\mathfrak r$ gives an element outside $\mathfrak n$, hence a unit, and every fraction defining $V$ already occurs in $U$. Translation of polynomial variables shows that the graph equations form a regular sequence. They remain regular after localization by exactness, and their quotient is nonzero. Appending the preceding regular list in $V$ consequently gives a regular generating list for $L$.

In the other direction lift the images of $Y_j$ to fractions $v_j(X)$ in $P$. The identical graph calculation gives $$U/(Y_j-v_j(X))\cong P,\qquad L=(\ker(P\to C),Y_j-v_j(X))U.$$ The constant and linear coefficient calculation of F.33 works in this localized polynomial ring as well: first invert the finitely many denominators of the $v_j$, translate $Y_j$ by $v_j$, and then localize at $\mathfrak r$. Modulo $L^2$ its constant terms are read modulo $\ker(P\to C)^2$, its graph-linear coefficients modulo $\ker(P\to C)$, and terms of graph-degree at least two vanish. The resulting mutually inverse maps give $$L/L^2\cong \ker(P\to C)/\ker(P\to C)^2\oplus C^m.$$

The regular list for $L$ makes its conormal finite free by F.74. Its rank is $\dim U-\dim C$ by AG-CA-12, Corollary 6.2. Hence the first summand in the display is finite projective, and is finite free over the local ring $C$ by S01, Lemma 1.2. Reduction to its residue field computes its rank. The polynomial height formula AG-CA-09, Theorem 4.3 gives $\dim U=\dim P+m$, since the two polynomial primes have the same residue field $\kappa(\mathfrak q)$. The rank is thus $\dim P-\dim C$. F.57 now supplies a regular generating list for $\ker(P\to C)$ and a global complete-intersection principal neighbourhood of $\mathfrak q$. This proves the assertion for every chosen polynomial presentation, including the local ambients in F.3.

Conversely a global complete-intersection neighbourhood has the regular local equation lists proved in F.57, so gives F.3 at $\mathfrak q$. It is also locally a complete intersection: apply F.57 at each of its points. A locally complete-intersection principal neighbourhood contains a further global complete-intersection principal neighbourhood of the specified point by the defining local condition. If the further inverted element is $b/g^r$ in $S_g$, that neighbourhood is $S_{gb}$, with $gb\notin\mathfrak q$. This proves the last equivalence. All graph and rank calculations allow empty coordinate or equation lists. ∎

Lemma F.59: limit essentially finite presentation

Reading source: AI Integrated Stacks, pinned Algebra, lines 33722–33770.

Let $R\to S$ be a local map of local rings, with $S$ essentially finitely presented over $R$. There are a directed set and local maps $R_\lambda\to S_\lambda$ such that their colimit is $R\to S$, each $R_\lambda$ is essentially finite type over $\mathbf Z$, each $S_\lambda$ is essentially finite type over $R_\lambda$, and, for $\lambda\leq\mu$, $$S_\lambda\otimes_{R_\lambda}R_\mu\longrightarrow S_\mu$$ is localization at a prime. All model rings are Noetherian, although $R$ and $S$ need not be.

Proof. Write $S=(R[X_1,\ldots,X_n]/(f_1,\ldots,f_c))_{\mathfrak q}$. Such a prime localization represents the given local essentially finitely presented algebra: in any localization presentation, every element outside the inverse image of the local maximal ideal is a unit, so the two localization universal properties give the prime localization. Locality of $R\to S$ makes $\mathfrak q\cap R=\mathfrak m_R$.

Let $\Lambda$ be the finite subsets of $R$ containing all coefficients of the $f_j$, ordered by inclusion. It is directed under finite union. Let $A_\lambda\subset R$ be the subring generated by such a subset over the image of $\mathbf Z$, put $\mathfrak p_\lambda=A_\lambda\cap\mathfrak m_R$, and set $R_\lambda=(A_\lambda)_{\mathfrak p_\lambda}$. Elements outside $\mathfrak p_\lambda$ are units in $R$, so this localization embeds in $R$: a fraction becoming zero has zero numerator, since its denominator is a unit there and $A_\lambda$ is a subring. The maps $R_\lambda\to R_\mu$ are therefore inclusions and are local, with maximal ideals contracted from $\mathfrak m_R$. Every element of $R$ occurs in some finite subset, proving $\operatorname*{colim}R_\lambda=R$.

Define $$C_\lambda=R_\lambda[X_1,\ldots,X_n]/(f_1,\ldots,f_c),$$ where the fixed coefficient list lies in $R_\lambda$. Let $\mathfrak q_\lambda$ be the inverse image of the maximal ideal of $S$ under $C_\lambda\to S$, and set $S_\lambda=(C_\lambda)_{\mathfrak q_\lambda}$. These are local rings, and $R_\lambda\to S_\lambda$ is local because the maximal-ideal contractions agree. We verify the integer base and the permanence steps explicitly. A nonzero ideal of $\mathbf Z$ has a least positive member $d$. Repeated subtraction of $d$ writes any nonnegative member as $qd+r$ with $0\le r<d$; the remainder is in the ideal and minimality forces $r=0$. Negative members are handled by changing their sign. Thus every integer ideal is principal, including the zero ideal, so $\mathbf Z$ is Noetherian.

Each $A_\lambda$ is a quotient of a finite-variable polynomial ring over $\mathbf Z$, and the Hilbert basis theorem AG-CA-03, Theorem 2.1, makes that polynomial ring Noetherian. An ideal of a quotient is the image of its inverse-image ideal, so is finitely generated. An ideal of a localization equals the localization of its contraction: if $a/s$ is in the ideal, then $a/1$ is too, and the converse follows by multiplying by $1/s$. Its contracted ideal has finite generators, whose localized images generate the original ideal. These calculations prove quotient and localization permanence and make $R_\lambda,C_\lambda,S_\lambda$ Noetherian. Their finite polynomial lists and prime localizations give the two essential finite-type assertions; composition of such maps was fully proved in F.31.

Polynomial scalar extension gives $C_\lambda\otimes_{R_\lambda}R_\mu\cong C_\mu$. The tensor with $S_\lambda$ initially inverts the images of $C_\lambda\setminus\mathfrak q_\lambda$. Those images avoid $\mathfrak q_\mu$, by their definition as contractions from the same target. Thus $\mathfrak q_\mu$ defines a prime of that tensor localization. Localizing at it inverts the remaining elements outside $\mathfrak q_\mu$ and gives exactly $S_\mu$. The fraction universal property proves both inverse maps, establishing the stipulated transition formula. The identical argument with the final base $R$ gives $S$ as a prime localization of $S_\lambda\otimes_{R_\lambda}R$.

For the colimit of the target rings, every fraction in $S$ uses finitely many coefficients in its numerator and denominator, so it has a representative in some $S_\lambda$. If two representatives become equal in $S$, their cross difference is killed in $R[X]/(f)$ by a denominator outside $\mathfrak q$. Lift that denominator to a polynomial and express the killed difference as a finite sum $\sum_jh_jf_j$ in $R[X]$. These polynomials and their equality use finitely many coefficients. Put them into one common later $R_\mu$. The denominator still avoids $\mathfrak q_\mu$, and the same ideal-membership equality holds there, so the representatives are already equal in $S_\mu$. This proves injectivity as well as surjectivity of the colimit map. The structural maps from the base agree throughout, giving the entire claimed colimit of local ring maps. ∎

Lemma F.60: limit module essentially finite presentation

Reading source: AI Integrated Stacks, pinned Algebra, lines 33804–33859.

For a local essentially finitely presented map $R\to S$ and a finitely presented $S$-module $M$, the models in F.59 can be taken at sufficiently large stages with finitely presented $S_\lambda$-modules $M_\lambda$ such that $$M_\lambda\otimes_{S_\lambda}S_\mu\cong M_\mu \quad(\lambda\leq\mu),\qquad \operatorname*{colim}_\lambda M_\lambda\cong M.$$ In particular each module is finite over its model algebra, and all base/prime-localization assertions of F.59 remain valid.

Proof. Choose a finite presentation $$S^a\xrightarrow{H}S^b\longrightarrow M\longrightarrow0.$$ The finitely many entries of $H$ are elements of the colimit $S$. The representative proof in F.59 puts them in a common $S_{\lambda_0}$. Restrict to stages $\lambda\geq\lambda_0$ and let $H_\lambda$ be their images there. Define $$M_\lambda=\operatorname{coker} \bigl(S_\lambda^a\xrightarrow{H_\lambda}S_\lambda^b\bigr).$$ This is a finite presentation over $S_\lambda$. Right exactness of tensor and the fixed free bases identify the tensor of this presentation with the presentation given by $H_\mu$, proving the asserted transition isomorphism. The same calculation gives $M_\lambda\otimes_{S_\lambda}S\cong M$.

We verify the colimit explicitly. Any element of $M$ is represented by a vector in $S^b$, whose finitely many coordinates occur at a common model stage, giving surjectivity from the module colimit. If $x_\lambda\in S_\lambda^b$ represents a class that is zero in $M$, its image is $Hy$ for some finite vector $y\in S^a$. Choose a later stage representing every coordinate of $y$. The vector equality $x=Hy$ is a finite list of ring equalities, so F.59 detects all of them at one common successor. There $x_\mu=H_\mu y_\mu$, proving that the original class is zero in $M_\mu$. This proves injectivity of the colimit map. The representative and transition calculations respect addition and the ring actions, so the isomorphism is an isomorphism of modules over $\operatorname*{colim}S_\lambda=S$.

No ring models have been changed except for removing an initial segment of the directed set, which leaves all colimits unchanged: every earlier representative has an image at a common successor with $\lambda_0$. Hence their local maps, essential finite-type properties and prime tensor-localizations are exactly those proved in F.59. The arguments include $a=0$, $b=0$ and the zero module. ∎

Lemma F.61: local dimension zero henselian

Reading source: AI Integrated Stacks, pinned Algebra, lines 44097–44121.

Every local ring $(R,\mathfrak m)$ of dimension zero is henselian. No Noetherian or nilpotent-maximal-ideal hypothesis is required.

Proof. The only prime of $R$ is $\mathfrak m$: any other prime is properly below the maximal ideal and would give a chain of length one. Every $u\in\mathfrak m$ is nilpotent. Indeed, if no power of $u$ were zero, $R_u$ would be nonzero. A maximal proper ideal of this ring exists by the chain-union and Zorn argument, and its inverse image would be a prime of $R$ avoiding $u$. It cannot be the unique prime $\mathfrak m$, a contradiction.

Use the simple-root definition of henselianity in AG-CA-20, Section 1. Let $f\in R[T]$ be monic, and let $\bar a$ be a root of its reduction with $\bar f'(\bar a)\ne0$. Choose any lift $a_0\in R$. The element $f'(a_0)$ is a unit because its residue is nonzero, and $\epsilon_0=f(a_0)\in\mathfrak m$ has some zero power $\epsilon_0^N=0$.

Inductively set $$a_{j+1}=a_j-f'(a_j)^{-1}f(a_j).$$ Every $a_j$ has residue $\bar a$, so each inverse in this formula exists. Expanding the polynomial by the binomial formula gives, for all $a,h$, $$f(a+h)=f(a)+f'(a)h+h^2Q(a,h)$$ for a polynomial $Q$ over $R$. Substitution of the correction therefore gives $$f(a_{j+1})\in(f(a_j)^2) \subset(\epsilon_0^{\,2^{j+1}}).$$ The correction remains in $\mathfrak m$, as needed for the residue induction. Choose $j$ with $2^j\geq N$; then $f(a_j)=0$. This supplies the required lifted root in finitely many steps, using only the nilpotence of the initial error.

For completeness it is unique with this residue. If $a,b$ are two such roots, polynomial subtraction factors as $$0=f(b)-f(a)=(b-a)\bigl(f'(a)+(b-a)H(a,b)\bigr).$$ The second factor has the nonzero residue $\bar f'(\bar a)$ and is a unit, so $b=a$. All simple residue roots have thus been lifted, proving henselianity. ∎

Lemma F.62: local syntomic

Reading source: AI Integrated Stacks, pinned Algebra, lines 37927–37941.

Let $R\to S$ be any ring map. If finitely many principal opens $D(g_i)$ cover $\operatorname{Spec}S$ and every $R\to S_{g_i}$ is syntomic, then $R\to S$ is syntomic.

Proof. The assumed cover means the $g_i$ generate the unit ideal. The whole finite-equation gluing proof F.35 gives finite presentation of $S/R$ from the chart finite presentations.

To prove flatness, take an arbitrary injection $M'\to M$ of $R$-modules and let $K$ be the kernel of $M'\otimes_RS\to M\otimes_RS$. Exactness of localization identifies $K_{g_i}$ with the kernel after tensoring with $S_{g_i}$; it is zero by chart flatness. If $x\in K$, some power $g_i^{e_i}$ kills $x$ for each $i$. These powers still generate the unit ideal: otherwise a maximal ideal containing all of them would contain all the $g_i$. A linear combination equal to one then kills $x$. Thus $K=0$, proving flatness without a finiteness condition on these modules.