Artin's axioms

Course exposition by GPT-6.1 Sol (OpenAI), Codex, Ultra; desingularization and supporting algebraic proofs by GPT-6 Astra (OpenAI), Codex, Ultra, October 2026. This lesson's independently written expression is dedicated to CC0 1.0. The mathematical work of the Stacks Project authors and the other cited authors is credited throughout. The component record identifies the expression, sources and writing credits.

An algebraic stack has smooth coordinates, whereas a moduli problem usually arrives as a rule assigning a groupoid to each scheme. Artin's criterion constructs coordinates from that rule. The construction has three stages: obtain a versal deformation over a complete local ring, approximate it by a family of finite type, and enlarge its versal locus to an open set. A second argument explains why a flat presentation, even one with inseparable fibres, can be replaced by a smooth presentation.

Read Algebraic stacks first. We use the quotient and bootstrap theorems of The bootstrap theorem, Schlessinger's existence theorem for versal formal objects, the deformation theory of square-zero extensions and the naive cotangent complex, and the commutative algebra of completion and henselization. Precise imported statements appear at the end. All fibre products of groupoids below are 2-fibre products.

1. Infinitesimal objects, arrows and patching

Except in Section 6, let \(S\) be locally Noetherian and let \(\mathcal X\) be a category fibred in groupoids on schemes over \(S\). The subscript “fppf” on this category specifies its underlying site; it does not by itself assert descent.

A finite type field over \(S\) means a field \(k\) with a morphism \(\operatorname{Spec}k\to S\) of finite type. This is more restrictive than being a finitely generated extension of the residue field at an arbitrary point of \(S\). A point of a scheme is a finite type point if its inclusion as the spectrum of its residue field is of finite type. Such a point is closed in some open neighbourhood.

Fix \(x_0\in\mathcal X(k)\). Its predeformation groupoid \(\mathcal F_{k,x_0}(A)\), for an Artinian local \(S\)-algebra \(A\) with specified residue field \(k\), consists of pairs

An arrow respects \(\iota\). Pullback along a ring map gives a covariant functor on Artinian rings. The fibre at \(k\) is equivalent to the terminal groupoid: the marking removes the automorphisms of the chosen special object. Automorphisms of its deformations can remain.

1.1. The Rim–Schlessinger condition

For a diagram of Artinian local \(S\)-algebras

whose spectra are of finite type over \(S\), condition (RS) requires the restriction functor

to be an equivalence. The right side remembers the identification on \(A\). Applying (1.2) to marked objects proves that \(\mathcal F_{k,x_0}\) is a deformation category whenever \(\mathcal X\) has (RS).

Condition (RS\(^*\)) requires the same equivalence for arbitrary \(S\)-algebras, with \(A_2\to A\) surjective and its kernel square zero. Filtering a nilpotent ideal into square-zero quotients shows that it implies (RS). The quantifier “arbitrary” is essential in Section 5, where products of modules need not be finite. These definitions are [Stacks, Tags 06L9 and 0CXN].

Lemma 1.1 (the groupoid operations preserve patching). If \(\mathcal X,\mathcal Y,\mathcal Z\) satisfy (RS), then \(\mathcal X\times_{\mathcal Y}\mathcal Z\) does too. The assertion also holds for (RS\(^*\)).

Proof. A patched object consists of an \(\mathcal X\)-object and a \(\mathcal Z\)-object on each side, and an arrow between their images in \(\mathcal Y\). Patch the two objects by essential surjectivity for \(\mathcal X\) and \(\mathcal Z\). Full faithfulness for \(\mathcal Y\) supplies the unique arrow between their images which restricts to the specified arrows. For morphisms, patch the two component arrows by full faithfulness and test their compatibility after restriction; faithfulness for \(\mathcal Y\) proves the compatibility upstairs. This proves both essential surjectivity and full faithfulness. \(\square\)

Lemma 1.2 (algebraic stacks have (RS)). Every algebraic stack satisfies (RS).

Proof. First, a diagram (1.1) is a pushout of affine schemes, and maps from its pushout to an algebraic space patch uniquely. One can check the latter assertion on étale scheme coordinates: after a finite étale cover of the Artinian local pushout, lift the common residue-field point to an étale chart; formal étaleness lifts that choice through the nilpotent thickenings. The two maps then land in a common affine chart, where the assertion is exactly the fibre product identity for rings. Unique lifts make the resulting maps compatible on overlaps, and the algebraic-space sheaf condition descends them. This also proves uniqueness.

For objects of an algebraic stack, use a smooth scheme atlas. After a finite étale cover of \(\operatorname{Spec}P\), the object over \(A_1\) lifts to the atlas: choose a finite separable residue-field point of its smooth atlas fibre and lift through the Artinian thickening. Its restriction to \(A\), together with the given identification, lifts across \(A_2\to A\) by smoothness. The two atlas maps agree on \(A\), and their closed points lie in a common affine atlas chart. The affine pushout patches them to a map from \(\operatorname{Spec}P\).

This constructs the desired object étale locally. The Isom spaces of two such objects are algebraic spaces, so the first paragraph proves that (1.2) is fully faithful. Consequently the local objects have unique compatible descent identifications, including the cocycle equality, and descend to the required object. \(\square\)

The stronger assertion for arbitrary affine pushouts along a thickening is [Stacks, Tag 07WM]. Its general flat-space patching input is not needed for the proof just given.

1.2. Two linear invariants

Write \(k[\epsilon]=k[\epsilon]/(\epsilon^2)\). For a deformation category, define

Patching the split extensions \(k\oplus M\) makes these \(k\)-vector spaces. Addition uses the map \(k[M\oplus N]\to k[M]\) induced by addition of vectors, and scalar multiplication uses multiplication on the square-zero ideal. For infinitesimal automorphisms, addition agrees with composition; terms involving two elements of that ideal vanish.

More generally, under (RS\(^*\)) there are \(A\)-linear functors \(T_x(M)\) and \(\operatorname{Inf}_x(M)\) for \(x\in\mathcal X(A)\). The first is the set of marked lifts to \(A[M]=A\oplus M\), and the second is the automorphism group of the trivial marked lift. If \(A'\twoheadrightarrow A\) has square-zero kernel \(I\), then, when lifts exist, their isomorphism classes form a torsor under \(T_x(I)\). The automorphisms of every marked lift are canonically \(\operatorname{Inf}_x(I)\).

To see the torsor assertion, use

Applying (RS\(^*\)) produces a difference between two lifts. On three lifts, the identity \((a_3-a_1)=(a_2-a_1)+(a_3-a_2)\) proves the addition law. Fixing one lift converts differences into the claimed simply transitive action. All constructions commute with maps of square-zero extensions. This is the mechanism behind [Stacks, Tag 07Y6].

Lemma 1.3 (field change and fibre products). Under (RS), a finite extension \(l/k\) gives

For \(w=(x,z,\alpha)\) in \(\mathcal W=\mathcal X\times_{\mathcal Y}\mathcal Z\) there is an exact sequence

Proof. For an Artinian \(l\)-algebra \(B\), apply (RS) to \(B\times_l k\). It identifies marked \(k\)-deformations over that ring with marked \(l\)-deformations over \(B\). The field-change result for linear deformation functors then gives (1.5), including inseparable finite extensions.

In (1.6), the second map is the difference between the two induced infinitesimal automorphisms in \(\mathcal Y\). The boundary twists the trivial pair of deformations by an infinitesimal automorphism of its identifying arrow. Its kernel consists precisely of twists obtainable from component automorphisms. A pair of component deformations comes from \(\mathcal W\) exactly when their \(\mathcal Y\)-images are isomorphic as marked deformations; choices of such an isomorphism differ by the boundary. These observations verify exactness at each term. The constructions by split extensions make all the maps linear. The same proof gives (1.6) with \(T_x(M)\) and \(\operatorname{Inf}_x(M)\) under (RS\(^*\)). \(\square\)

References: [Stacks, Tags 07WW, 07WY and 07YT]. For a naive obstruction complex \(E_x\) as in Section 5.3, the analogous field-change maps on \(H^i(E_x\otimes^{\mathbf L}k)\) are isomorphisms for \(i=0,1\): they are dual to the identifications of \(T\) and \(\operatorname{Inf}\). The indices here are \(0,1\), not \(-1,0\).

Lemma 1.4 (finiteness for an algebraic stack). If an algebraic stack has a smooth atlas locally of finite type over \(S\), the two spaces (1.3) are finite dimensional.

Proof. After a finite separable field extension, lift the object to the atlas. Its tangent space is a quotient of the atlas tangent space, since smoothness lifts deformations. On an affine atlas chart \(\operatorname{Spec}B\) over \(\operatorname{Spec}\Lambda\), that space is \(\operatorname{Hom}_B(\Omega_{B/\Lambda},k)\), which is finite dimensional because \(B\) is of finite type. The infinitesimal automorphisms inject into the tangent space at the identity point of the relation space; an étale scheme chart of that space is again locally of finite type. Apply the same argument and descend finiteness by (1.5). \(\square\)

This proof uses the finite module of differentials, so it also covers inseparable residue-field extensions. The reference is [Stacks, Tag 07X1].

2. Limits, complete local rings and approximation

2.1. Limit preservation includes arrows

The category \(\mathcal X\) is limit preserving if, for every directed inverse system of affine \(S\)-schemes with limit \(T=\lim T_i\),

is an equivalence. An object must descend to some stage. An arrow between descended objects must descend after increasing that stage, and two arrows equal on \(T\) must become equal at a later stage. Requiring only descent of objects is weaker. For a morphism \(\mathcal X\to\mathcal Y\), “limit preserving on objects” refers to descent of a lift of an already descended \(\mathcal Y\)-object, including its specified identifying arrow. See [Stacks, Tags 07XK and 06CT].

Finite presentation of algebras, modules, étale charts and their finite diagrams gives the following useful test. An algebraic space \(Z\to B\) is locally of finite presentation exactly when its functor on affine \(B\)-schemes preserves directed limits. One checks this on étale affine charts; finite presentation descends the chart and its equality data, and the étale sheaf condition glues the descended morphisms. The converse on an affine chart is the ring criterion for finite presentation, followed by étale descent.

Lemma 2.1 (testing on finite presentation bases). Suppose \(\mathcal Y\) is limit preserving on objects and both categories are Zariski stacks. To establish representability of \(f:\mathcal X\to\mathcal Y\) by algebraic spaces, it suffices to establish it over affine tests \(V\) locally of finite presentation over \(S\), with a uniform allowable size bound. Once representability is known, any property stable under base change and fppf local on the target can be tested on those same \(V\).

Proof. An affine \(V\) mapping into \(\operatorname{Spec}\Lambda\subset S\) is a limit of affine schemes of finite presentation over \(\Lambda\). The object testing \(f\) descends to a stage \(V_i\); its entire fibre category is therefore represented by the base change of the representing space over \(V_i\). For general \(V\), do this on affine opens. On overlaps the representing spaces have canonical compatible isomorphisms, since they represent the same category. Zariski descent glues them. The size bound permits the coproduct of their charts in the chosen site.

For the property assertion, the same calculation identifies the general fibre morphism with a base change of a finite presentation test. Base change and locality prove the claim. \(\square\)

References: [Stacks, Tags 07WI and 06CT]. Fibre products preserve limit preservation: descend both component objects, then their identifying arrow, and finally the finitely many equalities required for a morphism.

2.2. Formal effectiveness is a groupoid assertion

A formal object over a Noetherian complete local \(S\)-algebra \(R\), with residue field of finite type over \(S\), is a compatible system \(\xi_n\in\mathcal X(R/\mathfrak m^n)\), including its transition isomorphisms. It is effective if it is the completion of an object of \(\mathcal X(R)\). The stronger assertion used in axiom [4] is the equivalence

The limit on the right is a groupoid of compatible systems. The equivalence must include all compatible arrows, not only the existence of algebraizations. These equivalences also respect the local base morphisms in the category of formal objects. See [Stacks, Tag 07X3].

Lemma 2.2 (formal objects of algebraic stacks). Every algebraic stack satisfies (2.2).

Proof. A compatible system of maps to a scheme has its closed point in one affine open. Ring maps into the complete ring \(R=\lim R/\mathfrak m^n\) then give the unique algebraized map. For an algebraic space, choose an étale chart at the formal closed point. After a finite étale extension \(R'/R\), formal étaleness gives compatible lifts to that chart. The scheme assertion algebraizes them. Their identifying arrows algebraize by the same argument on the chart relation, and faithfulness verifies the cocycle. Étale descent gives the unique map over \(R\).

For a stack, two objects have an algebraic-space Isom sheaf. The preceding paragraph proves full faithfulness in (2.2). To algebraize a formal object, choose a smooth atlas and a finite separable residue-field point of its first fibre. The corresponding finite étale extension \(R'\) of the complete local ring lifts this point. Smoothness successively lifts it at every order. The resulting maps to the scheme atlas algebraize. Over \(R'\otimes_RR'\), a finite product of complete local rings, full faithfulness algebraizes the formal descent arrows. It also proves their cocycle on the triple product and identifies the descended object's completion with the specified formal object. Effective étale descent finishes the proof. \(\square\)

Equivalences commute with 2-fibre products. Thus (2.2), just like (RS), passes to a fibre product of three categories satisfying it.

2.3. The approximation input and its use

A Noetherian ring is a G-ring when the completion map of each of its local rings has geometrically regular fibres. The approximation results used here are proved in Appendix B, with both separable and inseparable residue-field arguments included.

Artin approximation. For a regular homomorphism between Noetherian rings, Popescu's theorem expresses the target as a filtered colimit of smooth algebras over the source. Consequently, let $B$ be a henselian Noetherian local G-ring and let a finite polynomial system with coefficients in $B$ have a solution in $\widehat B$. For each $N\geq1$, there is an exact solution in $B$ with the same reduction modulo $\mathfrak m_B^N$. If $B$ is local without the henselian assumption, one obtains the solution in a pointed étale neighbourhood with unchanged residue field. See the proofs of Popescu's theorem, henselian approximation, and pointed approximation; the source locators are Stacks Tags 07GC, 07QY and 07QZ.

For families, it is useful to preserve the tangent cone together with the prescribed finite-order object.

Lemma 2.3 (approximating a family). Suppose $\mathcal X$ is limit preserving on objects. Let $x_R\in\mathcal X(R)$, where $R$ satisfies the complete local and finite-type residue-field hypotheses in (2.2), and let $s\in S$ be the image of its closed point. Assume that $\mathcal O_{S,s}$ is a G-ring. Given $N\geq1$, there are an $S$-algebra $A$ of finite type, a maximal ideal $\mathfrak n$ of $A$, and an object $x_A\in\mathcal X(A)$ with compatible identifications

The first map is an $S$-algebra isomorphism, the second is an isomorphism over that map, and the last is an isomorphism of graded algebras over the common residue field.

Proof. Apply Theorem 6.1 in B.6.1, including its complete proof, to $x=x_R$ and the specified $N$. Its hypotheses are exactly those stated here: it uses limit preservation on objects, the G-ring condition at $s$, and the finite-type residue field, without an additional requirement on characteristic or on descent of arrows.

The proof there approximates a finite presentation together with its syzygy matrix at an order $M\geq\max(N,c+1)$, where $c$ is a common Artin–Rees constant. The exact perturbed syzygies and Lemma A.2 identify the initial ideals, giving the graded algebra isomorphism. Its final finite-type descent preserves the residue field, every maximal-ideal quotient, and the selected marking on the object. Restriction from order $M$ to order $N$ therefore yields all three identifications in (2.3). ∎

The source result is Stacks Tag 07XB. The complete family construction is kept at B.6.1, and its finite-complex perturbation argument is proved in Appendix A.

3. From versality to a smooth chart

A formal object \(\xi\) over \(R\) is versal if its functor of marked pullbacks to Artinian local rings is smooth: a pullback over \(B\) can be lifted across every small extension \(B'\twoheadrightarrow B\), together with the specified identification with a deformation over \(B'\). A small extension has one-dimensional residue-field kernel.

For \(x\in\mathcal X(U)\), with \(U\) locally of finite type over \(S\), say that \(x\) is versal at \(u\) when

is smooth. Maps from Artinian local rings with that closed point factor uniquely through the completion of \(\mathcal O_{U,u}\); thus (3.1) is exactly versality of the completed object.

The Schlessinger theorem used as a prerequisite says that (RS) and finite-dimensional tangent space give a versal formal object with Noetherian complete local parameter ring [Stacks, Tag 06IW]. It concerns groupoids and retains their markings; it does not require absence of automorphisms.

Lemma 3.1 (algebraizing a versal chart). If \(\xi\) is effective and versal, \(\mathcal X\) is limit preserving on objects, and \(\mathcal O_{S,s}\) is a G-ring at its image point, then \(\xi\) is isomorphic to the completion of a finite type family \(x_A\) at a point with the same residue field.

Proof. Apply Lemma 2.3 with \(N=2\). Versality successively lifts the identifying map at order two to a compatible map from \(\xi\) to the completion of \(x_A\). Its ring map \(R\to\widehat{A_{\mathfrak n}}\) is surjective: it is surjective on the residue field and cotangent space, and complete Nakayama gives surjectivity at each order, then on the inverse limit. The two rings have equal dimensions of all graded pieces by (2.3). The surjections on their Artinian quotients therefore have equal finite lengths and are isomorphisms. Their inverse limit is an isomorphism, including the family and its transition identifications. \(\square\)

Reference: [Stacks, Tag 07XH]. Approximation alone would not preserve versality; the graded-ring comparison supplies the missing step.

Lemma 3.2 (versal everywhere implies smooth). Suppose \(\mathcal X\) has representable diagonal, satisfies (RS), and is limit preserving. If \(U\) is locally of finite type over \(S\) and \(x\in\mathcal X(U)\) is versal at every finite type point, then \(U\to\mathcal X\) is smooth.

Proof. Its base change by an affine \(V\to\mathcal X\) locally of finite presentation over \(S\) is an algebraic space \(Z=U\times_{\mathcal X}V\). The finite presentation limit test and the fibre product argument in Section 2 show that \(Z\) is locally of finite presentation over \(S\). Choose an étale scheme chart \(W\to Z\).

At a finite type point \(w\), the induced field-valued point of \(U\) has finite residue extension of the image point. Lemma 1.3 and smoothness under field change for deformation categories therefore preserve versality. The deformation category of \(Z\) is the fibre product of those of \(U\) and \(V\) over that of \(\mathcal X\). Smoothness of (3.1) passes to this fibre product and then through \(W\to Z\). The Artinian lifting test for a morphism of schemes locally of finite type over a locally Noetherian scheme [Stacks, Tag 02HX] makes \(W\to V\) smooth at \(w\).

The smooth locus is open. Every nonempty closed subset of a scheme contains a point closed in an affine open, hence a finite type point. Thus this locus is all of \(W\). Smoothness descends through the étale chart to \(Z\to V\). Lemma 2.1 extends the conclusion to arbitrary tests \(V\). \(\square\)

The same argument at a single finite type point proves the version used later: for finite type \(U,V\), versality at \(u\) makes \(U\times_{\mathcal X}V\to V\) smooth at each finite type point above \(u\). The representable diagonal is locally of finite type because its Isom functors preserve limits. These statements explain [Stacks, Tags 07XD and 07XP].

Openness of versality asks that a versal finite type point of every finite type family have an open neighbourhood on whose finite type points the family is versal. For \(f:\mathcal X\to\mathcal Y\), it means openness for every base change by a finite type scheme over \(\mathcal Y\). In particular, openness for a diagonal is a separate condition.

4. The axioms and the two Artin criteria

Here is the numbering of [Stacks, Tag 07XJ].

Fix small models of the field groupoids when recording the bound in [−1]; the construction chooses representatives of their isomorphism classes. For a functor in sets, [0] says étale sheaf, [3] concerns only tangents, [4] is bijectivity of completion, and [5] asks openness for the functor itself. This is [Stacks, Tag 07XZ].

Theorem 4.1 (Artin with representable diagonal). Suppose \(\mathcal O_{S,s}\) is a G-ring for every finite type point \(s\) of \(S\). Suppose \(\mathcal X\) has representable diagonal, satisfies [−1]–[3], every formal object is effective, and openness of versality holds for \(\mathcal X\). Then \(\mathcal X\) is an algebraic stack. If it is a functor in sets, it is an algebraic space.

Proof. For every finite type field object \(x_0\), Schlessinger's theorem gives a versal formal object. Effectiveness and Lemma 3.1 give a finite type scheme family through \(x_0\), at a point with exactly its residue field. Shrink that scheme by openness of versality. Lemma 3.2 makes the resulting morphism to \(\mathcal X\) smooth. Axiom [−1] allows their disjoint union \(U\).

For an affine finite presentation test \(V\to\mathcal X\), the image of \(U\times_{\mathcal X}V\to V\) is open. It contains every finite type point \(v\): the chart chosen for the object over \(\kappa(v)\), and its specified special point, give a point of this fibre product above \(v\). The complement must consequently be empty. Lemma 2.1 proves surjectivity on every test.

The relation \(R=U\times_{\mathcal X}U\) is an algebraic space, with smooth projections. The quotient stack of this smooth groupoid is algebraic by Lesson 5. It agrees with \(\mathcal X\) as an étale stack: smooth surjections have local sections for the étale topology, so every \(\mathcal X\)-object has local coordinates in \(U\); its arrows and their composition are exactly \(R\). Étale descent identifies the two stacks, proving also the fppf stack condition.

For a functor in sets the quotient has no automorphisms. The smooth relation is an equivalence relation. The flat bootstrap of Lesson 2 identifies its sheaf quotient as an algebraic space, and the same étale local-coordinate argument identifies that space with the original functor. \(\square\)

This proves [Stacks, Tags 07Y1 and 07Y4], including the functor criterion's assumption that its diagonal is representable by algebraic spaces.

4.1. Recovering the diagonal

The second diagonal

is the identity section of the inertia category. Its representability says that the condition \(\alpha=\operatorname{id}\), for a family of automorphisms, is represented by an algebraic space over its parameter scheme. For two isomorphisms \(\alpha,\beta:x\to y\), equality is the condition \(\alpha^{-1}\beta=\operatorname{id}_x\). Consequently (4.1) is representable exactly when every Isom functor has representable diagonal. This is the direct equality test in [Stacks, Tag 07WG].

Theorem 4.2 (Artin's stack criterion). Let \(S\) be locally Noetherian, with \(\mathcal O_{S,s}\) a G-ring at every finite type point. Let \(\mathcal X\) satisfy [−1]–[5]. If its second diagonal (4.1) is representable by algebraic spaces, then \(\mathcal X\) is an algebraic stack.

Proof. Take an affine \(V\) locally of finite presentation over \(S\), and two objects \(x_1,x_2\in\mathcal X(V)\). Their Isom category is a setoid: an object is an isomorphism between the two fixed objects, and it has no nontrivial automorphisms after the two identifications are fixed. Thus its isomorphism-class functor \(I\) is an étale sheaf, by [0]. Its diagonal is representable by the equality test above.

We verify the remaining functor axioms for \(I\). The bound [−1] follows from the bound on arrows in \(\mathcal X\). Limits and (RS) follow by forming the fibre product of \(\mathcal X\to\mathcal X\times_S\mathcal X\) with \(V\), using Lemma 1.1 and the full arrow version of limit preservation. The six-term sequence (1.6) shows that its tangent space is an extension of subquotients of the tangents and infinitesimal automorphisms of \(\mathcal X\) and \(V\); all are finite dimensional. Lemma 2.2 gives effectiveness for the scheme \(V\), and axiom [4] for \(\mathcal X\), together with full faithfulness, algebraizes the identifying arrows. Completion therefore commutes with this fibre product. Finally [5] for \(\Delta_{\mathcal X}\) is exactly openness of versality for \(I\).

The base \(V\) is locally Noetherian. Its local rings at finite type points are essentially of finite type over the corresponding local rings of \(S\), so they are G-rings [Stacks, Tag 07PV]. The functor case of Theorem 4.1 now makes \(I\) an algebraic space over \(V\). Lemma 2.1 extends this assertion from finite presentation tests to all tests; its size bound is the field-arrow bound just used, together with the size of the finite type charts constructed in Theorem 4.1. Thus \(\Delta_{\mathcal X}\) is representable.

Apply the stack case of Theorem 4.1. Its hypotheses are now established, and its smooth atlas proves the conclusion. \(\square\)

Reference: [Stacks, Tag 07Y5]. Neither representability of the first diagonal nor algebraicity is assumed in this argument. The G-ring condition is used at the approximation stage, before constructing the atlas.

5. Two ways to prove openness

5.1. A failed lift can be placed in a family

We first isolate the argument which lets infinitesimal information detect an open locus. Suppose \(\mathcal X\) has (RS\(^*\)), representable diagonal, and limit preservation. Work with \(x\in\mathcal X(A)\), where \(U=\operatorname{Spec}A\) is of finite type over \(\operatorname{Spec}\Lambda\subset S\).

Lemma 5.1 (a witness to nonversality). If \(x\) is not versal at a finite type point \(u\), there is a square-zero extension \(C\twoheadrightarrow A\) with kernel \(\kappa(u)\), and a marked lift \(y\in\mathcal X(C)\), which admits no retraction to the family \(x\) on any open neighbourhood of \(u\).

Here a retraction means a morphism \(r:\operatorname{Spec}C\to U\) restricting to the identity on \(U\), and an isomorphism \(y\simeq r^*x\) restricting to the marking \(y|_U\simeq x\).

Proof. Failure of versality is witnessed by a small extension \(B'\twoheadrightarrow B\), a map \(A\to B\) with closed point \(u\), and a lift of the resulting object over \(B'\), for which the specified lifting problem has no solution. Take \(C=A\times_B B'\). Its kernel is the one-dimensional \(\kappa(u)\)-module of the small extension. Condition (RS\(^*\)) patches \(x\) and the object over \(B'\) to \(y\). A retraction near \(u\) pulls back to \(\operatorname{Spec}B'\), whose sole point lies over \(u\). It would supply exactly the forbidden lift, with its required identification. \(\square\)

Lemma 5.2 (retraction near a versal point). Let \(u_0\) be a versal finite type point. Every marked lift of \(x\) over a square-zero extension \(D\twoheadrightarrow A\) which is of finite type over \(\Lambda\) admits a retraction after shrinking around \(u_0\). The same assertion holds for a possibly infinite extension whenever its marked lift descends to such a finite type extension.

Proof. For the descended family \(x_D\), form \(Z=U\times_{\mathcal X}\operatorname{Spec}D\). The marking gives a section \(i:U\to Z\) over the thickening \(U\subset\operatorname{Spec}D\). The single-point version of Lemma 3.2 shows that \(Z\to\operatorname{Spec}D\) is smooth at \(i(u_0)\). Shrink \(U\) so it is smooth along the whole section. The infinitesimal lifting property of this smooth algebraic space lifts \(i\) across the affine square-zero thickening. The lifted map to \(Z\) supplies both a retraction to \(U\) and the desired isomorphism of objects. Pullback gives the final assertion. \(\square\)

These lemmas also show that versality is preserved under generalization among finite type points. Indeed, descend the witness of Lemma 5.1 to a finite type \(\Lambda\)-subalgebra of \(C\) containing lifts of generators of \(A\). Its map to \(A\) is surjective and its kernel square zero. If \(u\) specialized to a versal \(u_0\), Lemma 5.2 would retract it on a neighbourhood of \(u_0\), hence on a neighbourhood of \(u\), a contradiction.

The following modest Noetherian topological facts will be used. Every subset has the same closure as its points maximal under generalization, and an infinite dense subset of a Noetherian scheme has a countable dense subset [Stacks, Tags 0G2R and 0G2F]. Consequently, if the nonversal finite type points accumulate at a versal \(u_0\), one can choose a countable sequence \(u_i\) of nonversal points, with no specializations between them, whose closure contains \(u_0\). No individual \(u_i\) specializes to \(u_0\).

This gives the strong-effectiveness criterion of [Stacks, Tag 0CXU]. If every compatible system over surjective ring towers \(R_n\), with square-zero \(\ker(R_m\to R_n)\) for \(m\geq n\), is effective over \(\lim R_n\), then openness of versality holds. To prove it, take the witnesses \(C_i\to A\) just constructed and their finite fibre products

Their kernels over \(A\) are direct sums of \(\kappa(u_i)\), with square-zero multiplication. Condition (RS\(^*\)) patches their marked objects compatibly. Strong effectiveness gives an object over \(\lim R_n\). Limit preservation descends that object to a finite type subalgebra surjecting onto \(A\). Lemma 5.2 gives a retraction near \(u_0\). Pulling it to any \(C_i\) whose point lies in that neighbourhood contradicts Lemma 5.1. Thus the accumulating sequence cannot exist. This proves the criterion, with its stronger tower hypothesis explicitly distinguished from axiom [4].

5.2. Product-compatible obstruction modules

An obstruction theory consists of \(A\)-linear functors \(\mathcal O_x:\operatorname{Mod}_A\to\operatorname{Mod}_A\), functorial in the object, base ring and module, and elements

for each square-zero extension. They must commute with maps of deformation situations, and must satisfy

Functoriality includes the identity and composition laws for maps \((A,x,M)\to(B,x|_B,N)\), with an \(A\)-linear map \(M\to N\). An obstruction module without the equivalence (5.3) is insufficient for the theorem below. See [Stacks, Tag 07YG].

Theorem 5.3 (the product criterion). Suppose \(\mathcal X\) has representable diagonal, (RS\(^*\)), and limit preservation. Suppose it has an obstruction theory such that for every object \(x\) and every countable collection of modules \(M_i\),

Then \(\mathcal X\) satisfies openness of versality.

Proof. If openness fails at \(u_0\), choose the sequence and marked witnesses \(y_i/C_i\) of Section 5.1. Form the ring

It surjects onto \(A\) with square-zero kernel \(M=\prod_i\kappa(u_i)\). The obstruction \(o_x(D)\) maps to \(o_x(C_i)=0\) for every \(i\). Injectivity in (5.4) makes it zero, so choose a marked lift \(y/D\).

For each \(i\), its restriction and \(y_i\) differ by \(t_i\in T_x(\kappa(u_i))\), using the torsor of Section 1.2. The first condition of (5.4) supplies \(t\in T_x(M)\) with all these components. Replace \(y\) by \(t\cdot y\). Its restrictions are now isomorphic to the specified \(y_i\) as marked lifts. In particular the isomorphisms respect \(x\), which is necessary for the next step.

Limit preservation descends \(y\) to a finite type \(\Lambda\)-subalgebra \(D_0\subset D\). Increase \(D_0\) to include lifts of generators of \(A\). Then \(D_0\twoheadrightarrow A\) is a finite type square-zero extension, and its family restricts to \(x\) with the inherited marking. Lemma 5.2 retracts it near \(u_0\). Pull this retraction through \(D_0\to D\to C_i\). Since \(u_0\) lies in the closure of the \(u_i\), some \(u_i\) lies in that neighbourhood. The retraction of the corresponding marked \(y_i\) contradicts its defining property.

This proves the theorem directly; no effectiveness assertion for objects over infinite towers is assumed. \(\square\)

This is [Stacks, Tag 0CYF]. The proof only needs compatibility of obstruction classes for maps with fixed quotient \(A\), but (5.2) has been stated with the usual full functoriality.

5.3. Naive cotangent obstruction theories

Use cohomological indexing. For \(A=\Lambda[z_1,\ldots,z_r]/J\), the naive cotangent complex is

A square-zero surjection \(A'\to A\) with kernel \(I\) gives \(\mathrm{NL}_{A/A'}\simeq I[1]\). The successive maps of naive cotangent complexes have zero composition, but in general do not form a distinguished transitivity triangle.

A naive obstruction theory assigns \(E_x\in D^-(A)\) and \(\xi_x:E_x\to\mathrm{NL}_{A/\Lambda}\), together with

Its required compatibilities are as follows. Base change \(A\to B\) gives maps \(E_x\to E_{x|_B}\) in \(D(A)\), compatible with \(\xi\), identities and compositions. The identifications (5.6) respect the module and base-change maps of Section 1.2. The class of \(E_x\to\mathrm{NL}_{A/\Lambda}\to\Omega_{A/\Lambda}\) is the canonical deformation obtained by pulling \(x\) along \(a\mapsto(a,da)\) into \(A[\Omega_{A/\Lambda}]\). Finally, \(x\) lifts across \(A'\to A\) exactly when

is zero in \(D(A)\). All these requirements, not just a complex of finite modules, enter the definition [Stacks, Tag 07YP].

Lemma 5.4 (the fibre criterion). Assume (RS\(^*\)) and the tangent and obstruction properties just specified for \(x/A\). At a finite type point \(u\) with residue field \(k\), consider

These conditions imply versality. If \(u\) is closed in \(U\), versality implies them.

Proof. For a bounded-above complex \(E\), \[ \operatorname{Ext}^i_A(E,k) \simeq\operatorname{Hom}_k(H^{-i}(E\otimes_A^{\mathbf L}k),k). \tag{5.9} \] Resolve by free modules and split the resulting complex of vector spaces to obtain this identity.

The second condition of (5.8), by (5.9) and the canonical-element compatibility, says that \(\operatorname{Der}_\Lambda(A,k)\to T_x(k)\) is surjective. For a small extension \(B'\to B\) with residue field \(k\), the obstruction to lifting \(A\to B\) is in \(\operatorname{Ext}^1_A(\mathrm{NL}_{A/\Lambda},k)\). The first condition makes its map to \(\operatorname{Ext}^1_A(E_x,k)\) injective. If the pulled-back object lifts to \(B'\), (RS\(^*\)) gives a lift of \(x\) to \(A\times_BB'\), so (5.7) kills that image. The ring obstruction is therefore zero. Choose a lift of the ring map. Its object may differ from the prescribed marked object by a tangent class, but the surjectivity on tangents adjusts the lift by a derivation to remove the difference. This solves the lifting problem, including its identifying arrow, and proves sufficiency.

For necessity, tangent surjectivity follows immediately from versality. Suppose the first map in (5.8) were not onto. Choose a functional on \(H^{-1}(\mathrm{NL}\otimes^{\mathbf L}k)\) which kills the image and is nonzero. Extend it to \(J/J^2\otimes_Ak\), and denote the resulting functional by \(\lambda\). With \(P=\Lambda[z_1,\ldots,z_r]\), set

Since \(u\) is closed, the nonzero image is all of \(k\); the kernel \(I\) of \(A'\to A\) is \(k\) and has square zero. The differential of the discarded relation lies in the kernel of \(J/J^2\otimes k\to k^{\oplus r}\). Equivalently, the image under this differential of \(\ker\lambda\) equals the image of all \(J/J^2\otimes k\): subtract a multiple of an element in the differential kernel on which \(\lambda=1\). Thus \(\Omega_{A'/\Lambda}\otimes k\to\Omega_{A/\Lambda}\otimes k\) is an isomorphism. The extension is essential: a section would add the nonzero derivation to \(I=k\), contrary to that isomorphism of differential fibres.

By (5.9), the choice of \(\lambda\) makes \(E_x\to I[1]\) zero. Therefore \(x\) lifts to \(A'\). Artin–Rees gives \(n\) with \((\mathfrak m')^n\cap I=0\). The extension \[ B'=A'/(\mathfrak m')^n\longrightarrow B=A/\mathfrak m^n \] still has kernel \(k\), and remains essential. Versality lifts the quotient map from the completed local ring of \(A\) to \(B\) to a map into \(B'\), with the lifted object. Because the maximal ideal of \(B'\) has \(n\)-th power zero, that map factors through \(B\) and is a section, a contradiction. The first map of (5.8) must be surjective. \(\square\)

This supplies the essential-extension argument behind [Stacks, Tags 07YJ and 07YN], without assuming a transitivity triangle for (5.5).

Theorem 5.5 (naive obstruction theory gives openness). Suppose \(\mathcal X\) has (RS\(^*\)) and a naive obstruction theory. If every \(E_x\) over a finite type base has finitely generated cohomology modules, then \(\mathcal X\) satisfies openness of versality.

Proof. Shrink a finite type family so its versal point \(u_0\) is closed and the base is affine. Set \(C=\operatorname{Cone}(\xi_x)\). It is bounded above with finite cohomology, since \(A\) is Noetherian and the naive cotangent complex has finite modules. By Lemma 5.4 and the cohomology exact sequence of the cone, \(H^{-1}(C\otimes^{\mathbf L}k(u_0))=0\).

Represent \(C\) by a bounded-above complex of finite free modules locally near \(u_0\). Over the residue field it is exact in degree \(-1\). Choose bases for the image of the incoming differential and a complement mapping isomorphically onto the image of the outgoing differential. Invert the finitely many corresponding matrix minors. Cancelling these invertible blocks removes the whole degree \(-1\) term. The remaining complex consequently has \(H^{-1}(C\otimes^{\mathbf L}M)=0\) for every module on that open neighbourhood.

In particular the fibre cone has zero \(H^{-1}\) at every point there. Its cohomology sequence gives both conditions (5.8), and their sufficient direction proves versality at every finite type point in the neighbourhood. \(\square\)

Reference: [Stacks, Tag 07YU]. This criterion needs the functoriality and canonical tangent class in (5.6)–(5.7). It does not replace them by the bare assertion that obstructions happen to lie in a finite module.

6. Flat groupoids over an arbitrary base

In this section \(S\) is any scheme. No Noetherian or G-ring hypothesis is imposed. The flat bootstrap for sheaves from Lesson 2 will be used, while the stack version will be proved.

6.1. Finite sources and restriction of scalars

Lemma 6.1 (maps from a finite locally free space). If \(Z\to B\) is finite locally free and \(X\to Z\) is an algebraic space, the functor \[ \operatorname{Res}_{Z/B}(X)(T)= \operatorname{Mor}_{Z}(Z_T,X) \tag{6.1} \] is an algebraic space. Restriction of scalars carries a surjective étale morphism to a surjective étale morphism.

Proof. We explain the étale assertion first. For \(W\to Z\) étale, its section functor over \(B\) is étale and representable. This can be checked over an affine \(B\). If \(W\to Z\) is separated, a section is an open-and-closed subscheme of \(W_T\), finite locally free over \(T\), mapping isomorphically to \(Z_T\).

The finite-part construction of Lesson 2 applies to the separated, flat, locally finitely presented, locally quasi-finite map \(W\to B\), without the extra requirement of containing an identity section. Over the affine base, separated locally quasi-finite recognition first makes \(W\) a scheme. The construction's proof is unchanged: over a henselian local base, isolate a selected finite union of fibre components; scheme Zariski main and idempotent lifting extend it to a finite open-and-closed part; finite presentation descends it to an étale neighbourhood. Equality of two such parts is an open-and-closed condition, computed by the ranks of their two finite locally free differences. The étale quotient of these charts represents the finite-part functor.

Within that functor, the condition of mapping isomorphically to \(Z_T\) is open. The map from a selected finite part to \(Z_T\) is finite étale. Its rank is one exactly on the isomorphism locus; the complement of that locus is closed in \(Z_T\), and its finite image in \(T\) is closed. Thus the section functor is an open subspace of the finite-part space and is étale over \(B\).

For a general \(W\), cover it by a disjoint union \(W'\) of affine schemes over the affine base. Its map to \(Z\), and its map to \(W\), are separated. Sections of \(W'\to Z\) therefore form an algebraic space by the preceding argument. They cover the section functor of \(W\to Z\) étale locally: over a strictly henselian local base, the finite scheme \(Z\) is a product of henselian local factors with separably closed residue fields. A surjective étale cover has a section on each factor. Their finite union of choices descends to an étale neighbourhood of the base. The same separated calculation describes every fibre of the map of section functors. The étale bootstrap of Lesson 2 now proves the general assertion.

For an étale map \(X'\to X\), the fibre of its restriction of scalars at a map \(Z_T\to X\) is precisely the section functor of \(Z_T\times_XX'\to Z_T\). This proves representability and étaleness, and the preceding local section argument proves surjectivity when the original map is surjective.

Now trivialize the finite locally free algebra of \(Z\) on affine base opens, and take an étale cover \(X'\to X\) with \(X'\) a disjoint union of affine schemes. For a finite union of these affines, maps from \(Z_T\) are described by algebra maps into the finite free algebra of \(Z_T\). Choose its basis; coordinates for the images of generators and polynomial relations describe an affine scheme. The requirement that a map be over \(Z\) imposes the corresponding equations. Infinite presentations cause no difficulty for representability. Every map from a quasi-compact \(Z_T\) uses only finitely many of the affine components. The finite-component functors are open subfunctors, since \(Z_T\to T\) is finite and the excluded image has closed image in \(T\). They glue to represent restriction of scalars for \(X'\).

The étale assertion already proved gives a representable surjective étale map from that space to (6.1). The sheaf condition follows from descent of morphisms. The étale bootstrap proves (6.1). \(\square\)

This proves the restriction-of-scalars input [Stacks, Tag 05YF], including the étale assertion [Stacks, Tag 05YD]. The finite-part argument also explains why separatedness was needed only in an intermediate chart. No separatedness of \(X\) is assumed.

6.2. The finite Hilbert stack used here

Let \(\mathcal H_d\), for \(d\geq1\), classify finite locally free schemes \(Z\to T\) of degree \(d\). Let \(\mathcal H_d(X)\) also remember a map \(Z\to X\). These objects are finite maps, not necessarily closed subschemes.

Lemma 6.2. Both \(\mathcal H_d\) and \(\mathcal H_d(X)\), for an algebraic space \(X\), are algebraic stacks.

Proof. Put an arbitrary commutative unital algebra structure on the free module of rank \(d\). Its multiplication coefficients \(c_{ij}^k\) and unit coefficients satisfy finitely many equations expressing commutativity, associativity and the two unit identities. They define an affine scheme \(\operatorname{Alg}_d\) over \(\mathbf Z\). Change of basis gives an action of the smooth group \(\operatorname{GL}_d\).

A finite locally free algebra has a Zariski locally chosen basis. The basis changes and all their compatibility equations identify its stack with \([\operatorname{Alg}_d/\operatorname{GL}_d]\), after base change to \(S\). This quotient is algebraic by the smooth quotient theorem of Lesson 5. Passing between the algebra groupoid and the finite-scheme groupoid uses \(\varphi\mapsto\operatorname{Spec}(\varphi^{-1})\) on isomorphisms; this inversion makes the functor covariant.

For a fixed finite scheme \(Z/T\), the fibre of \(\mathcal H_d(X)\to\mathcal H_d\) is the space of maps \(Z\to X\). Lemma 6.1 represents it. Pull back a smooth atlas of \(\mathcal H_d\); its fibre is an algebraic space, and its smooth scheme atlas supplies a smooth atlas of \(\mathcal H_d(X)\). The diagonal is represented by the same finite-source equality construction. \(\square\)

References: [Stacks, Tags 05YQ and 05YS]. Finite locally free algebras and their isomorphisms descend because their modules, multiplication and unit descend.

Suppose \(q:U\to\mathcal Y\) is representable, surjective, flat and locally of finite presentation, with \(U\) an algebraic space and \(\mathcal Y\) an fppf stack. Define \(\mathcal H_d(U/\mathcal Y)\) to have objects

Lemma 6.3. The diagonal of \(\mathcal Y\) is representable, and every \(\mathcal H_d(U/\mathcal Y)\) is an algebraic stack.

Proof. For two objects over a scheme \(T\), choose an fppf scheme cover \(T'\to T\) lifting both to \(U\). Their Isom sheaf over \(T'\) is the pullback of the algebraic space \(U\times_{\mathcal Y}U\) along their two coordinate maps. Its map to the original Isom sheaf is representable, flat, locally of finite presentation and surjective, being the base change of \(T'\to T\). The flat sheaf bootstrap makes the original Isom sheaf algebraic. Thus the diagonal is representable.

The map \[ \mathcal H_d(U/\mathcal Y)\longrightarrow \mathcal H_d(U)\times\mathcal Y \tag{6.3} \] is representable: over a fixed \((Z/T,f,y)\), it is \(\operatorname{Res}_{Z/T}\operatorname{Isom}_{\mathcal Y}(y|_Z,q(f))\), an algebraic space by Lemma 6.1.

Choose a smooth scheme atlas \(P\to\mathcal H_d(U)\), and set \(W=P\times_{\mathcal H_d(U)}\mathcal H_d(U/\mathcal Y)\). This is a stack in setoids. An automorphism over the fixed \(Z\) and \(f\) restricts to the identity on \(y|_Z\); the finite locally free map \(Z\to T\) is an fppf cover because \(d>0\), so descent of arrows makes it the identity on \(y\).

By (6.3), \(V=W\times_{\mathcal Y}U\) is an algebraic space over \(P\times_SU\). Its map to the sheaf associated to \(W\) is representable, flat, locally of finite presentation and surjective. The flat sheaf bootstrap makes \(W\) an algebraic space. Finally \(W\to\mathcal H_d(U/\mathcal Y)\) is representable smooth and surjective, as the base change of \(P\). The smooth stack recognition of Lesson 5 proves algebraicity. \(\square\)

This proves the inputs [Stacks, Tags 05XW, 05YH and 06CI] without presupposing the flat stack theorem.

6.3. Why complete intersections give smooth coordinates

In (6.2), the pair \((f,\alpha)\) is a map \(Z\to U_y=U\times_{\mathcal Y}T\). Let \(\mathcal H_{d,\mathrm{lci}}(U/\mathcal Y)\) be the subcategory where this map is unramified and a local complete intersection.

This is an open substack. To check it over a family, use étale scheme charts. Unramifiedness is the vanishing condition for the finite module of relative differentials, and the complete-intersection locus in a flat finitely presented family is open and commutes with base change. The scheme versions follow from finite presentations and the regular-sequence criterion; their algebraic-space formulations are [Stacks, Tags 05X8 and 06CE]. Since \(Z\to T\) is finite, the image of the bad locus is closed in \(T\). Its complement represents exactly the indicated subfunctor. Thus the inclusion is a representable open immersion, including arbitrary base changes.

Lemma 6.4 (lifting the finite slice). Let \(T\subset T'\) be an affine square-zero thickening. Let \(X'\to T'\) be flat and locally of finite presentation, put \(X=X'\times_{T'}T\), and let \(Z\to X\) be unramified lci with \(Z\to T\) finite locally free of degree \(d\). Then it extends to \(Z'\to X'\), with \(Z'\to T'\) finite locally free of the same degree and \(Z=Z'\times_{T'}T\).

Proof. If \(I\) is the ideal of \(T\) in \(T'\), flatness and the regular-immersion conormal calculation give

The last module is finite locally free because \(Z\to X\) is unramified lci. The scheme conormal statements, transported by étale charts, are [Stacks, Tags 06CB and 06CC]. Since \(Z\) is affine, (6.4) splits.

Construct the universal first-order neighbourhood of \(Z\) over \(X'\). Étale locally an unramified map is a closed immersion into an étale neighbourhood, and this neighbourhood is cut out by the square of its ideal. Uniqueness of infinitesimal lifts between the étale neighbourhoods glues these constructions. Its square-zero ideal on \(Z\) is \(\mathcal C_{Z/X'}\). Quotient this ideal by the chosen summand \(\mathcal C_{Z/X}\). The resulting neighbourhood \(Z'\) has ideal \(I\otimes\mathcal O_Z\) and a map to \(X'\).

The natural multiplication map identifies its ideal with the image of \(I\), so the square-zero flatness criterion gives flatness of \(Z'\to T'\) and the stated base-change identity. A nilpotent thickening of an affine scheme is affine. Lift finitely many module generators for the finite algebra of \(Z/T\); they generate the algebra of \(Z'/T'\), since the cokernel equals \(I\) times itself and \(I^2=0\). Thus \(Z'\to T'\) is finite. Locally lift a basis of its reduction of rank \(d\). The resulting map from the free module of rank \(d\) is surjective by the same argument. Flatness makes its kernel remain exact after reduction modulo \(I\), so that kernel is equal to \(I\) times itself and is zero. The algebra is therefore finite locally free of rank \(d\). The open lci condition above ensures that its map to \(X'\) still belongs to the indicated substack. \(\square\)

This is the independently established lifting argument of [Stacks, Tag 06D8]. It also proves that \(\mathcal H_{d,\mathrm{lci}}(U/\mathcal Y)\to\mathcal Y\) is formally smooth on objects: apply the lemma to \(U_{y'}\to T'\) for a lift \(y'\) of the base object.

These morphisms are limit preserving on objects. A finite locally free algebra, its map into the fixed locally finitely presented algebraic space \(U_y\), and its finitely many compatibility data descend along affine limits. The finite source uses only finitely many charts; the coefficient construction in Lemma 6.1 then involves finite presentations. The open unramified lci condition descends after increasing the stage. This is the finite presentation argument of [Stacks, Tag 06CH].

Their union over \(d\geq1\) is surjective on field objects. Indeed \(U_y\) is a nonempty algebraic space locally of finite presentation over a field \(k\). Take an affine étale chart and a closed point in its Cohen–Macaulay locus, which is nonempty on a finite type scheme. A system of parameters in the Cohen–Macaulay local ring is a regular sequence. After shrinking the chart, its zero scheme is supported only at that point and is finite over \(k\). Call it \(Z\). The regular immersion followed by the étale map is unramified lci; \(Z\) is a nonzero finite \(k\)-algebra, so it is finite locally free of some degree \(d\). It gives an object (6.2). The scheme inputs are [Stacks, Tags 045U and 0570]; they impose no perfection assumption on \(k\).

Theorem 6.5 (flat stack bootstrap). If \(q:U\to\mathcal Y\) is a representable surjective flat locally finitely presented morphism from an algebraic space to an fppf stack, then \(\mathcal Y\) is an algebraic stack.

Proof. Lemma 6.3 gives its representable diagonal and the algebraic stacks \(\mathcal H_d(U/\mathcal Y)\). Their lci opens are algebraic too. Choose smooth scheme atlases for these opens, and take their union \(P\).

The map \(P\to\mathcal Y\) is representable because \(\mathcal Y\) has representable diagonal. It is limit preserving on objects and formally smooth on objects: both properties compose, using the lifted identifying arrows in their definitions, and the atlases are smooth. For a representable morphism these two properties mean locally finite presentation and formal smoothness respectively. Test the definitions on \(P\times_{\mathcal Y}T\); its objects are actual morphisms to that algebraic space, so the limit and infinitesimal criteria apply. The infinitesimal criterion for a finitely presented morphism makes \(P\to\mathcal Y\) smooth.

The finite slices above cover every field object, allowing field extensions to lift to \(P\). This is surjectivity for a representable morphism. Thus \(P\to\mathcal Y\) is a smooth surjective scheme atlas. Lesson 5's recognition theorem proves the conclusion. \(\square\)

This proves [Stacks, Tag 06DC]. The uses of limits, formal smoothness and field-surjectivity are the precise notions of [Stacks, Tags 06CT, 06CZ and 06D4].

Theorem 6.6 (flat groupoids are algebraic). For any groupoid \(R\rightrightarrows U\) in algebraic spaces over \(S\), if both projections are flat and locally of finite presentation, then \([U/R]\) is an algebraic stack.

Proof. Lesson 5's diagonal calculation represents the Isom sheaves of the quotient: they are locally the pullbacks of the endpoint map \(R\to U\times_SU\), and the flat sheaf bootstrap descends them. In particular \(U\to[U/R]\) is representable.

On an fppf local chart \(a:T\to U\) of a quotient object, its pullback is \(R\times_{s,U,a}T\to T\). This is flat and locally of finite presentation by base change, and is surjective because the identity gives a section. These properties descend fppf locally. Theorem 6.5 therefore applies to \(U\to[U/R]\). \(\square\)

Reference: [Stacks, Tags 06FG and 06FI]. This completes the flat-groupoid theorem used as a stated forward reference in Quotient stacks and Deligne–Mumford stacks, including examples such as \(B\mu_p\).

For orientation, call a map of stacks algebraic when all its scheme-base fibres are algebraic stacks [Stacks, Tag 05XX]. If the target is algebraic, an algebraic map has algebraic source: pull back an atlas and then take an atlas of that fibre. A map from an algebraic stack to a stack with representable diagonal is algebraic for the same reason, using a source atlas. Finally an algebraic map with target having representable diagonal gives representable source diagonal: over each target Isom space, the source Isom fibre is the Isom space of the corresponding algebraic fibre stack. These atlas and Isom calculations prove the three permanence statements without changing the meaning of “representable”.

7. A complete example and a failure of effectiveness

7.1. Line bundles on a proper flat curve

Let \(C\to S\) be proper, flat and of finite presentation, with fibres of dimension at most one. Let \(S\) be locally Noetherian with the G-ring condition of Theorem 4.2. Define \(\mathcal{Pic}_{C/S}(T)\) to be the groupoid of invertible sheaves on \(C_T\), with all their isomorphisms. This is the Picard stack; we do not divide out line bundles pulled back from \(T\).

We will use two precise scheme results. For a proper flat finitely presented scheme and a finitely presented sheaf flat over the base, its derived direct image is perfect and commutes with arbitrary base change [Stacks, Tag 0B91]. For a proper scheme over a complete Noetherian ring, completion gives an equivalence between coherent sheaves and compatible systems of coherent sheaves on its infinitesimal neighbourhoods [Stacks, Tag 088C; EGA III, Theorem 5.1.5]. The latter includes morphisms.

For a line bundle \(N\) on \(C_A\), the first result and the dimension bound give

with \(P^i\) finite projective \(A\)-modules. To justify the two-term range, its derived fibres have cohomology only in degrees \(0,1\), by coherent cohomology on proper schemes of dimension at most one. A bounded finite-projective representative can then be shortened: cancel the split surjections at its highest excessive degrees and the split injections at its lowest excessive degrees. Fibrewise exactness and Nakayama make these cancellations valid on the base. On an affine base they give (7.1); the construction also shows the asserted Tor amplitude. Thus (7.1) computes base change to every \(A\)-algebra and tensoring with every \(A\)-module.

[−1]. A finite affine cover of \(C_k\), finite presentations of line bundles on that cover, and matrices describing their transitions give a uniform cardinal bound for their isomorphism classes and arrows as \(k\) ranges over finite type fields over \(S\). The cardinal can be enlarged once to contain the affine base presentations and the field models. This supplies the site-size condition.

[0]. Invertible sheaves and their isomorphisms have effective fpqc descent, by descent of quasi-coherent modules followed by the local rank-one condition. In particular this is an étale stack.

[1]. A line bundle on \(C_A\), with \(A=\operatorname{colim}A_i\), is of finite presentation. Finite affine covers, its local presentations and the finitely many transition maps descend to a stage. Their inverse identities and the rank-one condition hold after increasing the stage. The same finite-presentation argument descends isomorphisms and tests equality of isomorphisms. Thus the full groupoids preserve limits.

[2]. Flatness of \(C/S\) makes the structure rings of its affine charts preserve the fibre product (1.1) under base change. Finite projective modules patch over that fibre product, as proved in Exercise 8.2 below. Apply this to the locally free rank-one modules on the charts. Full faithfulness patches their transition maps and checks the cocycle, so the patched sheaf is invertible. This proves (RS), and the same argument with arbitrary square-zero extensions proves (RS\(^*\)).

[3]. The transition-function calculation gives

Indeed, write a first-order change of transition units as \(g_{ij}(1+\epsilon a_{ij})\). The cocycle equation is the additive Čech cocycle equation for \(a_{ij}\); changing frames adds a coboundary. An automorphism reducing to the identity is \(1+\epsilon a\) with a global section \(a\). The spaces are finite dimensional by proper coherent cohomology.

[4]. For a complete Noetherian local ring \(R\), apply Grothendieck existence to \(C_R\) and a compatible system \(L_n\). It gives a coherent \(L\), including all compatible morphisms. It is invertible near the closed fibre: completing the local stalk along the base ideal gives a free module of rank one, and the local adic criterion, or faithful flatness of completion after localization at the point, gives local freeness there. The non-invertible locus is closed. Its image in \(\operatorname{Spec}R\) is closed by properness and misses the closed point. A nonempty closed subset of a local spectrum contains that point, so the locus is empty. Thus \(L\) is invertible everywhere. Full faithfulness algebraizes isomorphisms and their inverses, proving (2.2).

[5] for the stack. For \(L\in\mathcal{Pic}_{C/S}(A)\) and an \(A\)-module \(M\), the same Čech calculation gives \[ T_L(M)=H^1(C_A,p^*M),\qquad \operatorname{Inf}_L(M)=H^0(C_A,p^*M). \tag{7.3} \] The obstruction to lifting \(L\) across \(A'\to A\) lies in \(H^2(C_A,p^*I)\): lift local frames and transition units; their failure to satisfy the cocycle is a Čech 2-cocycle, and changing the transitions changes it by a coboundary. Its vanishing is precisely the ability to glue a lift. Formula (7.1) for \(\mathcal O_C\) shows that this \(H^2\) is zero for every \(I\). Thus zero obstruction modules give a functorial obstruction theory.

Furthermore \(H^1(C_A,p^*M)=\operatorname{coker}(P^0\otimes_AM\to P^1\otimes_AM)\). Finite projective modules commute with products under tensoring, and products of module sequences are exact. Hence \(T_L\) commutes with countable products, and the zero obstruction functor has the required injectivity. Theorem 5.3 proves openness.

[5] for the diagonal, and the second diagonal. For two line bundles \(L_1,L_2\), put \(N=L_1^\vee\otimes L_2\). Their Hom functor is the affine scheme of vectors in \(P^0\) annihilated by the differential in (7.1). The universal vector gives a section of \(N\). It is an isomorphism exactly where its zero locus in the proper curve has empty image in the parameter scheme. That condition is open, so \(\operatorname{Isom}(L_1,L_2)\) is a scheme locally of finite presentation over the base.

Its diagonal is representable, hence the stack's second diagonal is representable. Openness of versality for these Isom schemes follows from the ordinary Artinian smoothness criterion and the open smooth locus, using étale scheme charts if necessary. This checks the other part of [5].

All the axioms, including their arrow conditions, have now been verified. Theorem 4.2 proves that \(\mathcal{Pic}_{C/S}\) is an algebraic stack. The proof did not replace this stack by its sheaf of isomorphism classes.

7.2. The affine line loses a formal automorphism

Let \(R\) be a complete discrete valuation ring with uniformizer \(t\). On \(\mathbf A^1_R\), the trivial line bundle has automorphism group \(R[x]^\times=R^\times\): \(R\) is a domain, so a unit polynomial has degree zero.

Over \(R/t^n\), however, \(1+tx\) is a unit, with inverse

The units and their inverses form compatible systems. They give an automorphism of the formal trivial line bundle in \(\lim_n\mathcal{Pic}_{\mathbf A^1/R}(R/t^n)\). The corresponding element \(1+tx\) in \(\lim_n(R/t^n)[x]\) is not the image of a unit of \(R[x]\), since that image would be constant. Thus completion is not full on automorphisms, even though the formal object under discussion is already the completion of a trivial line bundle. Axiom [4] fails. This is exactly the groupoid distinction needed in (2.2).

8. Exercises and complete solutions

Exercise 8.1 (medium: limits of algebraic stacks). Show that an algebraic stack locally of finite presentation over \(S\) is limit preserving. Do not assume quasi-compactness or quasi-separatedness.

Solution. Choose a smooth atlas \(U\to\mathcal X\) with \(U\) locally of finite presentation over \(S\). Then \(R=U\times_{\mathcal X}U\) is smooth over \(U\), hence locally of finite presentation over \(S\). The morphism \(R\to U\times_SU\) is locally of finite presentation by the finite-presentation permanence rule on charts. Descent along the atlas therefore shows that the diagonal of \(\mathcal X\) is locally of finite presentation.

For descended objects over \(T_i\), their Isom space is locally of finite presentation over \(T_i\); its affine limit criterion gives descent and eventual equality of all arrows. This proves full faithfulness in (2.1).

For essential surjectivity, pull the atlas back to an object over an affine \(T\). Choose an étale scheme chart of that smooth algebraic-space cover and finitely many affine opens whose images cover \(T\). Their disjoint union \(V\to T\) is a smooth surjection of finite presentation, with a map to \(U\) and descent arrows in \(R\). The affine schemes \(V\), \(V\times_TV\) and its triple product are of finite presentation over \(T\). Their finite equations descend to a sufficiently late \(T_i\), as do their maps to the locally finitely presented spaces \(U\) and \(R\). The finitely many identity and cocycle equalities descend after a further increase. Smoothness and surjectivity of the covering descend too. Effective fppf descent in the already algebraic stack produces the object over that \(T_i\). Its pullback is the original object. Only a finite portion of the atlas was used for this affine test, so no global compactness assumption was introduced.

Exercise 8.2 (medium: module patching). Verify (RS) for finite locally free modules, including the arrows.

Solution. For (1.1), the ring \(P\) is local, with residue field that of \(A_1\). A patched pair of finite free modules must have equal rank, since their restrictions to \(A\) are isomorphic. Choose bases. The gluing isomorphism is an invertible matrix \(g\) over \(A\). Lift its entries to \(A_2\); its determinant lifts a unit because \(\ker(A_2\to A)\) is nilpotent, so the lifted matrix is invertible. Change the second basis by that matrix. The gluing is now the identity, and the pair is the restriction of \(P^r\).

An arrow between two such pairs is a pair of matrices over \(A_1,A_2\) which agree over \(A\). Entrywise it is a unique matrix over \(P\). If the arrow is an isomorphism, its compatible inverse matrices patch as well. This proves full faithfulness and essential surjectivity. Markings at the residue field are preserved by doing the same construction with their specified identifications.

The general finite-projective version, used for (RS\(^*\)) in Section 7, is obtained by adding complements. Write the module over \(A_1\) as the image of an idempotent in \(A_1^n\). Lift its restriction to an idempotent in \(A_2^n\). Its image is isomorphic to the given second module with its specified reduction: maps between finite projectives lift across the nilpotent ideal, and maps inverse modulo that ideal are inverse after multiplying by an invertible correction. Add the two complementary summands. The enlarged pair is free on both sides, and its gluing matrix lifts as above. Its two projection idempotents, now equal over \(A\), patch to an idempotent over \(P\). Its image is the desired finite projective module. Equivalently that image is the fibre product of the original two modules. Maps patch entrywise in the free ambient modules and respect the idempotents. This proves the finite-projective Milnor patching lemma for arbitrary \(A_1\) and a square-zero surjection \(A_2\to A\), including all arrows.

Exercise 8.3 (medium: vector-bundle tangents). For a vector bundle \(E\) on a proper curve \(C/k\), prove \[ T_E\mathcal{Bun}_C=\operatorname{Ext}^1_C(E,E) =H^1(C,\mathcal E nd(E)). \tag{8.1} \]

Solution. On a finite affine trivializing cover, write its matrices as \(g_{ij}\). A first-order change has the form \(g_{ij}+\epsilon h_{ij}\). Linearizing the cocycle and the frame changes identifies its class with an additive Čech 1-cocycle in \(\mathcal E nd(E)\), modulo coboundaries. Since \(C\) is separated, affine intersections compute the cohomology of this quasi-coherent sheaf. This gives the last term of (8.1), and addition is exactly the split-extension addition of Section 1.

There is also an intrinsic description. A first-order bundle \(E'\) gives the extension \[ 0\to E\xrightarrow{\epsilon}E'\to E\to0 \] as an \(\mathcal O_C\)-module; flatness over \(k[\epsilon]\) identifies the kernel with \(E\). Conversely, an extension gives the middle term an \(\epsilon\)-action by the composite of its quotient to \(E\) and its inclusion of \(E\). Locally the extension splits because \(E\) is locally free, and the resulting module is a free deformation over \(\mathcal O_C[\epsilon]\). Thus extensions and marked first-order bundles are equivalent on isomorphism classes. Their Baer addition matches the cocycle addition. Finally \(\mathcal H om(E,-)\) is exact locally, so global derived Hom from \(E\) is cohomology of \(E^\vee\otimes-\). This identifies \(\operatorname{Ext}^1(E,E)\) with \(H^1(\mathcal E nd(E))\). Infinitesimal automorphisms, for comparison, are \(H^0(\mathcal E nd(E))\).

Exercise 8.4 (hard: the formal Picard groupoid). For a smooth proper curve \(C/k\), verify [3] and [4] for its line-bundle stack.

Solution. Formula (7.2) gives the tangent and infinitesimal automorphism spaces at any line bundle after every finite field extension. Proper coherent cohomology makes \(H^0\) and \(H^1\) finite dimensional, proving [3]. No assumption that automorphisms vanish is made.

Given a complete Noetherian local \(k\)-algebra \(R\), Grothendieck existence on the proper scheme \(C_R\) algebraizes every compatible system of line bundles to a coherent sheaf. The completed local modules along the closed fibre are free of rank one, so the sheaf is invertible near that fibre. Its closed non-invertible locus has proper closed image in \(\operatorname{Spec}R\); such an image cannot be nonempty while missing the closed point. The sheaf is therefore a line bundle on all of \(C_R\). Full faithfulness of existence algebraizes every compatible isomorphism. Applying it also to the inverse and using faithfulness proves that the algebraized map is an isomorphism. Consequently completion is an equivalence of groupoids, not merely a surjection on objects. This proves [4], and explains precisely where properness excludes the phenomenon in (7.4).

Appendix A. Artin–Rees perturbation and graded quotients

The approximation in Lemma 2.3 preserves an entire finite relation complex. We prove the two perturbation facts that make its associated graded conclusion valid, with their exact congruence order.

A.1. The perturbation theorem

Let \(A\) be Noetherian and let \(I\subset\operatorname{Jac}(A)\). For a map \(q\colon M\to N\) of finite modules, say that \(c\geq0\) is an Artin–Rees constant for \(q\) when \[ q(M)\cap I^nN\subset q(I^{n-c}M)\qquad(n\geq c). \] Artin–Rees applied to the finite submodule \(q(M)\subset N\) gives such a constant. We may enlarge a constant, so two given maps admit one common constant.

Lemma A.1. Suppose \[ L\xrightarrow f M\xrightarrow g N, \qquad L\xrightarrow{f'}M\xrightarrow{g'}N \] are complexes of finite \(A\)-modules, the first is exact at \(M\), and \(c\) is an Artin–Rees constant for both \(f\) and \(g\). Assume \[ (f-f')(L)\subset I^{c+1}M, \qquad(g-g')(M)\subset I^{c+1}N. \tag{AR.1} \] Then the second complex is exact at \(M\), and the same \(c\) is an Artin–Rees constant for \(g'\).

Proof. Fix \(n\geq c\) and \(a\in M\) such that \(g'(a)\in I^nN\). We show that we can subtract elements of \(f'(L)\) until the remaining element lies in \(I^{n-c}M\). Such subtractions do not change \(g'(a)\), because the perturbed sequence is a complex.

Suppose at an intermediate step that \(a\in I^rM\), with \(0\leq r<n-c\). Equation (AR.1) gives \[ g(a)=g'(a)+(g-g')(a) \in I^nN+I^{r+c+1}N=I^{r+c+1}N. \] Artin–Rees for \(g\) supplies \(a_1\in I^{r+1}M\) with \(g(a_1)=g(a)\). Exactness of the first sequence supplies \(b\in L\) with \(a-a_1=f(b)\). If \(r\geq c\), Artin–Rees for \(f\) permits choosing \(b\in I^{r-c}L\), since \(f(b)\in I^rM\). If \(r<c\), keep any such \(b\).

In either case \[ a=f'(b)+a_2, \qquad a_2=a_1+(f-f')(b)\in I^{r+1}M. \] Indeed, for \(r\geq c\) the error is in \(I^{c+1}I^{r-c}M=I^{r+1}M\); for \(r<c\) it is in \(I^{c+1}M\subset I^{r+1}M\). Replace \(a\) by \(a_2\). The order rises by one while \(g'(a)\) remains fixed. After finitely many steps we obtain \[ (g')^{-1}(I^nN)\subset f'(L)+I^{n-c}M. \tag{AR.2} \] Applying \(g'\) proves \(g'(M)\cap I^nN\subset g'(I^{n-c}M)\), including the case \(n=c\), when no adjustment was needed.

If \(g'(a)=0\), (AR.2) holds for every \(n\geq c\). Its class in the finite module \(M/f'(L)\) lies in every power of \(I\). The Krull-intersection theorem, with \(I\subset\operatorname{Jac}(A)\), makes that class zero. Hence \(\ker g'\subset\operatorname{im}f'\). The converse follows from \(g'f'=0\). No completeness of \(A\), freeness of the modules, or exactness at an endpoint has been used. \(\square\)

A.2. Equality of the initial submodules

Lemma A.2. Under the preceding hypotheses put \(Q=N/g(M)\) and \(Q'=N/g'(M)\). The identity on \(\operatorname{gr}_I N\) induces a canonical isomorphism \[ \operatorname{gr}_I Q\simeq\operatorname{gr}_I Q' \] of graded \(\operatorname{gr}_I A\)-modules. If \(N=A\) and the two images are ideals, this is an isomorphism of graded algebras.

Proof. In degree \(n\geq0\) the quotient filtration gives \[ (\operatorname{gr}_I Q)_n =I^nN/\bigl(I^{n+1}N+(g(M)\cap I^nN)\bigr). \tag{AR.3} \] We compare the two denominator submodules inside \(I^nN\). If \(n\leq c\), (AR.1) says that \(g(a)\) and \(g'(a)\) differ by an element of \(I^{n+1}N\). For \(g(a)\in I^nN\), its perturbed image is therefore also in \(I^nN\). This gives one containment of denominators.

If \(n>c\), write an element of \(g(M)\cap I^nN\) as \(g(a)\) with \(a\in I^{n-c}M\), using the constant for \(g\). Then \[ g(a)-g'(a)\in I^{c+1}I^{n-c}N=I^{n+1}N. \] Again \(g'(a)\in I^nN\), giving the containment. The perturbation lemma supplies the same constant for \(g'\), so exchanging \(g\) and \(g'\) proves the reverse containment in every degree. The quotients in (AR.3) are thus identical subquotients of \(\operatorname{gr}_I N\). Their identifications respect its graded module action. For ideals in \(A\), they respect multiplication as well, proving the algebra assertion. \(\square\)

A.3. Flat base change of the constant

Lemma A.3. Let \(A\to B\) be flat, with both rings Noetherian. An Artin–Rees constant \(c\) for \(q\colon M\to N\) remains a constant for \(q_B\) relative to \(IB\).

Proof. For \(n\geq c\), the constant condition is equivalent to \[ q^{-1}(I^nN)\subset\ker q+I^{n-c}M. \] The left side is the kernel of \(M\to N/I^nN\). Flat tensor product preserves that kernel and the kernel of \(q\), and identifies the quotients and powers with their base changes. Tensor the displayed inclusion with \(B\), then apply \(q_B\); the desired containment follows. \(\square\)

A.4. Use in family approximation

In Lemma 2.3 the relation presentation is a complex of finite free modules over the complete local presentation ring \(P\): \[ P^{\oplus t}\xrightarrow f P^{\oplus r} \xrightarrow{(b_1,\ldots,b_r)}P\longrightarrow R\longrightarrow0. \] The columns of \(f\) generate the syzygies of the ideal generators. Choose a common constant \(c\) for its two displayed maps, relative to the maximal ideal of \(P\). Polynomial approximation must preserve the equations saying that the new generators and new syzygies still form a complex, and must approximate all these coefficients to order at least \(c+1\). Its order must also be at least the requested \(N\).

The perturbation lemma gives exactness at \(P^{\oplus r}\). The graded quotient lemma identifies the two associated graded quotient algebras, rather than merely their Hilbert functions. The quotient filtration is the maximal-ideal filtration of the local quotient, so this is precisely the graded comparison asserted in (2.3). Completion of the approximating Noetherian local ring preserves its residue quotients and associated graded algebra; thus the comparison returns to that local ring. For a finite type ring and its localization at a maximal ideal, every quotient by a power of that maximal ideal is unchanged by localization, since each element outside the ideal is already a unit in the quotient. The comparison consequently survives the finite-stage descent of the approximating object.

This argument explains why approximating only the ideal generators is insufficient: the exact perturbed syzygy equations supply (AR.2), and (AR.2) supplies the Artin–Rees constant for the perturbed map needed for the reverse graded containment.

The mathematical source is the Stacks Project authors, More on Algebra, labels lemma-approximate-complex and lemma-approximate-complex-graded (Tags 07VE–07VF), and lemma-works-flat-extension, at AI Integrated Stacks Project revision 565b10e987aba5969b21145a0833f42d69f96790. That section credits Conrad–de Jong for part of the material.

Prerequisites and proof scope

The arguments above establish the space and stack forms of Artin's criterion, openness from naive obstruction theories, and algebraicity of flat groupoid quotients. They also establish the product-compatible obstruction criterion and the criterion using stronger effectiveness. The following inputs retain their stated prerequisite scope.

• The formal-deformation input is Schlessinger's existence of a versal formal object under (S1), (S2) and finite-dimensionality of the tangent space [Stacks, Tag 06IW], together with the field-change, linearity and small-extension tests. The deformation inputs are the cotangent obstruction for ring maps, conormal transitivity for regular immersions and square-zero flatness. Their algebraic-space forms are obtained on étale charts [Stacks, Tags 06CB, 06CC and 06BH].
• Appendix B proves general Néron desingularization, both polynomial-approximation forms and permanence of G-rings [Stacks, Tags 07GC, 07QY, 07QZ and 07PV]. The full marked-family construction appears in B.6.1 and is used in Section 2.3. Appendix A supplies the finite-complex Artin–Rees argument [Stacks, Tags 07VE and 07VF]. The comparison lemma's sixth assertion is false over arbitrary bases, as shown by the integral cusp; its exact differential-basis criterion is proved at that lemma. The desingularization chain uses the valid fifth assertion.
• The ordinary-scheme Artinian lifting test at a point [Stacks, Tag 02HX] is developed in the supporting chain below. Earlier lessons supply étale lifting over henselian local rings, descent of finite presentations along affine limits, scheme Zariski main and idempotent lifting. The unramified and lci open-locus results for flat finitely presented families, and the Cohen–Macaulay slicing results, are used with the precise formulations [Stacks, Tags 05X8, 06CE, 045U and 0570].
• Section 5.1 uses the Noetherian topological assertions [Stacks, Tags 0G2F and 0G2R]. The Picard example uses coherent cohomology finiteness and its dimension bound, perfect direct image with arbitrary base change [Stacks, Tag 0B91], and Grothendieck existence for proper schemes [Stacks, Tag 088C; the proper-support extension is Tag 088E]. Existence includes morphisms. Section 7 explains the passage from coherent sheaves to line bundles.

The pushout statement [Stacks, Tag 07WM] for arbitrary affine thickenings is stronger than what this lesson needs. Lemma 1.2 proves the required Artinian case. The general theorem's flat-space patching input remains a prerequisite rather than a proof supplied here.

References

• [Stacks] The Stacks Project authors, Artin's Axioms, Tags 07T0, 07T2, 07WM, 06L9, 07WT, 07WW, 07WY, 07X3, 07XA, 07XK, 07XD, 07XP, 07XJ, 07XZ, 07Y0, 07Y1, 07Y3, 07Y4, 07Y5, 0CXN, 0CXR, 07Y6, 07YF, 07YG, 0CYF, 07YJ, 07YP, 07YT and 07YU. See the AI Integrated Stacks reader.
• [Stacks] Criteria for Representability, Tags 05XF, 05XH, 06CT, 06CZ, 06D4, 05XX, 06DB, 06DC, 06FG and 06FI, including the finite Hilbert stack and restriction-of-scalars results used in Section 6. See the chapter reader.
• [Stacks] Smoothing Ring Maps, Tags 07GC, 07QY and 07QZ; More on Algebra, Tags 07PV, 07VE and 07VF; Formal Deformation Theory, Tag 06IW; and the scheme-theory inputs specified above. The smoothing chapter keeps the upstream tags. The AI Integrated Stacks edition's additions and corrections have not been reviewed by the Stacks Project maintainers.
• [EGA] A. Grothendieck, with J. Dieudonné, Éléments de géométrie algébrique III, Theorem 5.1.5. Its proper-scheme existence theorem is used through [Stacks, Tag 088C].

Appendix B. Artin approximation and general Néron desingularization

Numbers in this appendix refer to its own sections and statements. References to another course or proof provider include that work's title.

We prove that a regular map of arbitrary Noetherian rings is a filtered colimit of smooth algebras, including the positive-characteristic case with inseparable residue fields. Smooth factorization then gives solutions to finite polynomial systems at any prescribed approximation order. A henselian local G-ring supplies the solution in the original ring; a local G-ring without henselianity supplies it in a pointed étale neighbourhood with the same residue field.

The constructions have two jobs. Singularity ideals record where a finite presentation still fails to be smooth, and the lifting lemmas enlarge its smooth locus. Once a smooth factorization exists, a change of polynomial variables encodes the desired finite-order congruence and smooth-section lifting produces an exact solution.

B.1. Mathematical sources and authorship

The mathematical references are the Stacks Project authors' Smoothing Ring Maps, the marked-family approximation in Artin's Axioms, and the specified results in More on Algebra and Commutative Algebra. The AI Integrated Stacks edition is pinned to revision 565b10e987aba5969b21145a0833f42d69f96790. Popescu proved the principal theorem. The Stacks treatment follows Richard Swan, whose account credits André's notes and Ogoma's arguments. References to those papers here follow Swan’s exposition.

The mathematical exposition, marked-family approximation and Artin–Rees arguments are by GPT-6.1 Sol (OpenAI), Codex, Ultra. The desingularization and supporting algebraic proofs are by GPT-6 Astra (OpenAI), Codex, Ultra. The component record specifies their contributions and the cited mathematical sources. This course expression is dedicated to CC0 1.0.

The source authors are the Stacks Project authors (Copyright (C) 2005–2025 Johan de Jong), with separately credited contributions in the AI Integrated Stacks edition.

B.2. The finite factorization criterion

Rings are commutative with identity, and homomorphisms preserve it. A smooth algebra is finitely presented and formally smooth. The supporting proofs below establish its Jacobian and flat-fibre characterizations. A regular homomorphism is flat with geometrically regular Noetherian fibres. For a Noetherian algebra over a field, geometric regularity can be tested by finite purely inseparable field extensions. This condition does not require every residue field to be separable over the base.

Lemma 2.1. For $R\to\Lambda$, the following statements are equivalent: $\Lambda$ is a filtered colimit of smooth $R$-algebras; every map $A\to\Lambda$ with $A$ finitely presented over $R$ admits a factorization $A\to B\to\Lambda$ with $B/R$ smooth.

Proof. Given a filtered presentation of $\Lambda$, a map from $A$ is specified by finitely many elements satisfying finitely many equations. A common stage contains these elements, and a later stage makes all the equations hold. This yields the factorization.

Conversely, choose a set of representatives for smooth $R$-algebras mapping to $\Lambda$, retaining the maps to $\Lambda$ as part of the objects. The object $R\to\Lambda$ makes this category nonempty. For two objects, form their tensor product over $R$ and apply the assumed factorization to its map to $\Lambda$; the resulting smooth algebra receives both objects. For parallel arrows $u,v\colon B\rightrightarrows C$, choose finitely many algebra generators $b_i$ of $B$. The quotient of $C$ by the elements $u(b_i)-v(b_i)$ is finitely presented. Factoring its map to $\Lambda$ through a smooth algebra equalizes the two arrows. These are exactly the filteredness conditions.

The colimit of this category maps onto $\Lambda$: apply the hypothesis to $R[T]\to\Lambda$ for each desired element. To test injectivity, represent two elements at one common stage. If their images in $\Lambda$ agree, quotient that stage by their difference and factor once more through a smooth algebra. The two elements agree at the resulting stage, hence in the colimit. The canonical map is therefore a bijective ring homomorphism. ∎

This finite-data argument supplies the categorical content of the filtered-colimit criterion without using Popescu's theorem.

B.3. Popescu's theorem and the proof mechanism

Theorem 3.1 (Popescu). If $R\to\Lambda$ is regular and both rings are Noetherian, then $\Lambda$ is a filtered colimit of smooth algebras over $R$.

The complete constructions in B.9 prove the theorem. We describe their dependencies here to explain how the factorization criterion applies.

For a finite presentation $A/R$ mapping to $\Lambda$, let $H_{A/R}$ define the complement of its smooth locus, and put $\mathfrak h_A=\sqrt{H_{A/R}\Lambda}$. A resolution at a prime $\mathfrak q$ containing $\mathfrak h_A$ gives a finite presentation $A\to B\to\Lambda$ for which $\mathfrak h_A\subseteq\mathfrak h_B$ but $\mathfrak h_B\nsubseteq\mathfrak q$. Thus the radical ideal grows strictly. No finite-presentation assumption is made on $\Lambda/R$.

The singularity-ideal lemmas express the relevant elements by Jacobian minors together with conditions on the remaining relations. The improved-presentation construction uses the symmetric algebra of the conormal module and retains an algebra retraction. Over the original smooth locus its differentials become free. The proved fifth assertion of the comparison lemma then produces strictly standard powers. The unrestricted conormal identification needs the correction recorded in B.7; the projective case required here is valid.

Two further constructions do the main algebraic work. A flat nilpotent deformation of an ind-smooth algebra is ind-smooth: lift a presentation in the square-zero case and use the equational flatness criterion to kill finitely many relation errors, then iterate through square-zero quotients. The lifting and desingularization lemmas, under their explicit annihilator hypotheses, convert a smooth factorization modulo $\pi^8$ into an enlargement of the smooth locus. Their calculations use $\pi^8$, $\pi^4$ and $\pi^2$ at different stages; these exponents cannot be replaced by an appeal to simple-root Hensel lifting.

To reduce to a field, suppose the finite-factorization property fails. Choose a maximal failing quotient ideal $I\subset R$, using Noetherianity, and replace the map by $R/I\to\Lambda/I\Lambda$. Every nonzero further quotient now has the property. If this new source had a nonzero nilradical, that nilradical would be nilpotent; flat nilpotent lifting would contradict failure. The source is therefore reduced. Its total fraction ring is a finite product of fields. The field case gives a smooth factorization there, and finitely many denominator clearings give a strictly standard nonzerodivisor $\pi$ over the reduced source. Apply the quotient property modulo $\pi^8$, and then lift and desingularize. Flatness preserves the nonzerodivisor in $\Lambda$, giving the annihilator equalities needed to return to $R$.

For the field case, choose a prime minimal over $\mathfrak h_A$. In characteristic zero the residue extension of the regular local target is separable, and the regular-parameter construction provides a resolution. In positive characteristic, finite polynomial subalgebras and local Artinian approximations capture the needed coefficients even when the residue extension is inseparable or infinitely generated. Geometric regularity provides the injection $H_1(L_{K/k})\to\mathfrak m/\mathfrak m^2$ used for the finite-dimensional bound. Ogoma's annihilator argument prepares parameters before localization; the auxiliary variables $t_i$ and their evaluation maps then return the construction to the original finite presentation. B.9 gives both residue-field proofs and all these return maps.

Repeated resolutions terminate because the radical ideals $\mathfrak h_A$ cannot increase indefinitely in the Noetherian target. At the terminal stage $\mathfrak h_B=\Lambda$, choose $f_i\in H_{B/R}$ and $\lambda_i\in\Lambda$ such that $\sum_i f_i\lambda_i=1$, and set

Evaluation at the $\lambda_i$ gives $C\to\Lambda$. The opens $D(f_i)$ cover $\operatorname{Spec}C$, and on each one its defining equation eliminates $Z_i$. Thus $C_{f_i}$ is a polynomial algebra over the smooth algebra $B_{f_i}$; smoothness on this cover proves $C/R$ smooth. Lemma 2.1 finishes the proof. A zero target is handled by the smooth zero algebra. The local dimension-zero cases are treated explicitly in B.7 and in the corresponding resolution proofs. ∎

B.4. Approximating a finite polynomial system

Theorem 4.1. Let $(R,\mathfrak m,k)$ be a henselian Noetherian local G-ring, with completion $\widehat R$. Suppose $f_1,\ldots,f_s\in R[X_1,\ldots,X_n]$ vanish at $a\in\widehat R^n$. For any $N\geq1$ there is $b\in R^n$ at which all the $f_j$ vanish and for which $a_i-b_i\in\mathfrak m^N\widehat R$ for every $i$.

Proof. The maximal-ideal G-ring condition, together with Noetherian completion and its flatness, says $R\to\widehat R$ is regular. Choose $c_i\in R$ representing $a_i$ modulo $\mathfrak m^N$, and generators $d_1,\ldots,d_M$ of $\mathfrak m^N$. Write $a_i=c_i+\sum_\ell d_\ell a_{i\ell}$. Substitute

The tuple $(a_{i\ell})$ solves this new finite system in $\widehat R$. It is enough to obtain any solution to the new system in $R$, since the substitution then forces the required congruences.

For existence without a congruence condition, take the $R$-subalgebra of $\widehat R$ generated by the finitely many solution coordinates. It is finitely presented because $R$ is Noetherian. Theorem 3.1 and Lemma 2.1 factor its inclusion through a smooth $R$-algebra $B$. Reducing $B\to\widehat R$ gives $B\to k$. The smooth-section lifting proof produces an étale $R$-algebra $R'$ with $R'/\mathfrak mR'=k$ and a map $B\to R'$ lifting this residue evaluation. The construction uses a finite projective conormal complement and slicing, so it does not assume a single simple-root presentation for $B$.

The henselian section criterion gives an $R$-algebra map $R'\to R$. Composing the maps sends the original coordinate algebra into $R$ and solves its equations. Apply this existence argument to the $g_j$, and use $b_i=c_i+\sum_\ell d_\ell y_{i\ell}$. ∎

There is no upper bound on the allowed order $N$. Order zero, if included by convention, follows from order one.

B.5. The pointed étale-neighborhood version

Theorem 5.1. Retain a Noetherian local G-ring $(R,\mathfrak m,k)$, the equations $f_j$ and the formal solution $a$, but drop henselianity. For each $N\geq1$, there are an étale $R$-algebra $R'$, a maximal ideal $\mathfrak m'$ over $\mathfrak m$, an identification $\kappa(\mathfrak m')=k$, and a solution $b\in(R')^n$. Under the pointed local map $R'_{\mathfrak m'}\to\widehat R$, the differences $a_i-b_i$ lie in $\mathfrak m^N\widehat R$. Equivalently, the canonical identification $\widehat{R'_{\mathfrak m'}}\simeq\widehat R$ identifies these differences with elements of $(\mathfrak m')^N\widehat{R'_{\mathfrak m'}}$.

Proof. Use the same substitution as in Theorem 4.1, then factor the algebra generated by the new formal coordinates through a smooth algebra $B$. Smooth-section lifting gives $B\to R'$ with $R'/\mathfrak mR'=k$. Images of the new coordinates in $R'$ solve the $g_j$, so the substituted tuple $b$ solves the $f_j$.

Take $\mathfrak m'=\mathfrak mR'$. The displayed residue quotient makes it maximal and fixes its residue-field identification. The complete local ring $\widehat R$ is henselian; apply its section criterion to the chosen point of $R'\otimes_R\widehat R$. This gives the local map $R'_{\mathfrak m'}\to\widehat R$. After applying it, each difference $a_i-b_i$ is a sum of multiples of the $d_\ell$, so lies in the required power of $\mathfrak m$.

To identify the completions, formal étaleness lifts the selected residue map uniquely through every quotient $R/\mathfrak m^q$. Over this Artinian base the selected local branch of the étale algebra is finite étale: its field fibre is finite and the base maximal ideal is nilpotent. Its finite flat module is free of rank one, because the residue algebra is $k$. The unit generates it by Nakayama, and the structural map from $R/\mathfrak m^q$ is an isomorphism. Uniqueness of lifting makes these isomorphisms compatible as $q$ varies. Their inverse limit gives the asserted completion isomorphism and identifies the two maximal-ideal filtrations. This is also the pointed same-residue-field comparison in the henselization construction below. ∎

Corollary 5.2. Suppose $R$ is Noetherian, $\mathfrak p$ is a prime and $R_{\mathfrak p}$ is a G-ring. A formal solution over $\widehat{R_{\mathfrak p}}$ admits approximation at every prescribed order in an étale $R$-algebra at a prime $\mathfrak p'$ with residue field $\kappa(\mathfrak p)$.

Proof. Apply Theorem 5.1 over $R_{\mathfrak p}$. Only finitely many coefficients occur in the resulting étale presentation and its Jacobian inverse data. Clearing their denominators descends the algebra to some $R_s$, with $s\notin\mathfrak p$, hence to an étale $R$-algebra. A further principal localization away from the selected prime clears all denominators in the solution and all zero witnesses for its finitely many equations. The local ring at that prime is unchanged, as are its residue marking, completion and congruences. The prime-localized approximation proof gives the same finite-data descent. ∎

B.6. Approximation of finite algebraic data

B.6.1. A marked family and its associated graded algebra

Theorem 6.1 (family approximation). Let $S$ be locally Noetherian and let $\mathcal X\to(\mathrm{Sch}/S)_{fppf}$ be a category fibred in groupoids that is limit preserving on objects. Let $x\in\mathcal X(R)$, where $(R,\mathfrak m_R)$ is complete Noetherian local and its residue field $k$ is of finite type over $S$. Write $s$ for the image of its closed point, and assume $\mathcal O_{S,s}$ is a G-ring. For each $N\geq1$ there are a finite-type $S$-algebra $A$, a maximal ideal $\mathfrak m_A$, an object $x_A\in\mathcal X(A)$, and

as $S$-algebras, together with a specified isomorphism of the restricted objects and an isomorphism

of graded $k$-algebras. Neither a perfect residue field nor characteristic zero is assumed.

Proof. Choose an affine open $\operatorname{Spec}\Lambda\subset S$ containing $s$. The map from the local scheme $\operatorname{Spec}R$ factors through it: its inverse image is an open containing the closed point of a local spectrum, hence the entire spectrum. The finite-type hypothesis on the residue field says that $k$ is a finitely generated $\Lambda$-algebra on this chart. Since $S$ is locally Noetherian, $\Lambda$ is Noetherian.

Write $R$ as the filtered colimit of its finite-type $\Lambda$-subalgebras. Limit preservation on objects gives a finite-type algebra $C$, a map $C\to R$, and an object $x_C$, with a chosen isomorphism $x_C|_R\simeq x$. No descent assertion about all arrows of $\mathcal X$ is being assumed. Choose a finite presentation

and write $\bar a_i\in R$ for the images of its generators.

Choose finitely many generators of $k$ as a $\Lambda$-algebra and lift them to $R$. Add lifts of a finite generating set of $\mathfrak m_R$. These choices define a polynomial ring $T=\Lambda[z_1,\ldots,z_e]$ and a map $T\to R$ whose reduction to $k$ is surjective and whose image contains generators of $\mathfrak m_R$. Thus $\mathfrak n=\ker(T\to k)$ is maximal. Elements outside $\mathfrak n$ map to units of $R$, so the map extends to the local ring $Q=T_{\mathfrak n}$ and then continuously to

Here is a direct proof of its surjectivity. The chosen elements generate $\mathfrak m_R^j/\mathfrak m_R^{j+1}$ by their degree-$j$ monomials with coefficients in $k$. Those coefficients lift through $Q\to k$. Starting with any element of $R$, lift its residue and then its successive errors by such monomials. At the $j$th correction the added element lies in $\mathfrak m_Q^j$. The partial lifts are Cauchy in $Q$, hence define an element of $P$, and their images converge to the prescribed element of the complete ring $R$. This proves surjectivity; it also proves the usual surjective-cotangent-space criterion in this instance.

The complete ring $P$ is Noetherian. Choose generators $b_1,\ldots,b_r$ of the kernel, and lifts $a_i\in P$ of the $\bar a_i$. Choose coefficients $c_{ji}$ such that

The syzygy module of $(b_1,\ldots,b_r)$ is finite. Choose generators $k_\ell=(k_{\ell1},\ldots,k_{\ell r})$, giving an exact presentation

The map $K$ sends the $\ell$th basis vector to $k_\ell$; in particular $\sum_i k_{\ell i}b_i=0$.

Let $c$ be one Artin–Rees constant for both displayed maps relative to $\mathfrak m_P$, and choose $M\geq\max(N,c+1)$. The common constant exists by Artin–Rees for finite modules, and increasing a constant preserves its defining inclusions. The finite-complex perturbation and canonical graded-quotient statements are the written Lesson 7, Appendix A, Lemmas A.1–A.2. We use their full statements here and do not reproduce their proof.

Put $\mathfrak p=\mathfrak n\cap\Lambda$. The hypothesis says that $\Lambda_{\mathfrak p}$ is a G-ring. The permanence theorem proved below makes its essentially finite-type algebra $Q$ a G-ring. Apply the pointed étale-neighbourhood approximation theorem of Section 5 simultaneously to the following finite polynomial system:

The tuple $(a_i,b_i,c_{ji},k_{\ell i})$ is an exact formal solution in $P$. We obtain an étale $Q$-algebra $B$, a maximal ideal $\mathfrak q$ above $\mathfrak n$ with residue field exactly $k$, and a solution $(a'_i,b'_i,c'_{ji},k'_{\ell i})$ whose entries differ from the old ones by $\mathfrak m_P^M$. The induced identification $\widehat{B_{\mathfrak q}}=P$ preserves the map from $Q$, its residue field and every finite-order quotient. There is no need to inject the entire possibly disconnected algebra $B$ into $P$.

Because $b_i$ lies in $\mathfrak m_P$ and $M\geq1$, each $b'_i$ lies in $\mathfrak q$. Set

The exact polynomial relations define $C\to A^*$ by $y_i\mapsto a'_i$, so they give $x^*=x_C|_{A^*}$. Completion of a quotient of a Noetherian local ring gives

This uses exactness of Noetherian completion for finite modules, supplied by the written completion prerequisite. As $b'_i-b_i\in\mathfrak m_P^M$, the ideals generated by the two lists become identical after adding $\mathfrak m_P^M$. Consequently

Localization at the selected maximal ideal does not change these quotients: every element outside that ideal is already a unit modulo any of its powers. The two maps from $C$ to the displayed common quotient agree, because $a'_i-a_i\in\mathfrak m_P^M$. Restricting the chosen $x_C|_R\simeq x$ therefore gives the required marked comparison of objects at order $M$, hence at order $N$.

The matrix $K'$ and row $(b'_i)$ still form a complex, since all the syzygy equations were preserved exactly. Their entries differ from those of the original exact presentation by $\mathfrak m_P^M\subset\mathfrak m_P^{c+1}$. Lesson 7, Lemma A.2 now identifies the initial ideals inside $\operatorname{gr}_{\mathfrak m_P}P$. Its canonical identity on that graded algebra yields

This is an isomorphism of graded algebras. The quotient filtration is the maximal-ideal filtration, so completion identifies its right side with $\operatorname{gr}_{\mathfrak m^*}A^*$. This explains why the entire relation complex was approximated: approximating only the ideal generators would not provide the reverse initial-ideal containment.

It remains to obtain a finite-type algebra over the original affine chart. The algebra $A^*$ is of finite presentation over $Q=T_{\mathfrak n}$. Its finitely many generators, relations and denominators descend to a finite-type $T$-algebra $A_0$ with

Products of denominators direct this system. Limit preservation on objects descends $x^*$ to an object $x_A$ at some stage $A=(A_0)_f$, with a specified isomorphism after base change to $A^*$. This algebra is finite type over $\Lambda$ and hence over $S$. Let $\mathfrak m_A$ be the inverse image of $\mathfrak m^*$. The image of $A\to k$ contains the image of $T$, which already surjects onto $k$ by the original residue-field generators. Thus $A/\mathfrak m_A=k$: the selected prime is maximal and has exactly the prescribed residue field.

Since $A^*$ is a localization of $A$ away from elements outside the selected point,

Localization at a maximal ideal preserves every quotient by its powers and therefore its associated graded algebra. All the ring comparisons constructed above consequently descend to $A$. Restricting the specified isomorphism $x_A|_{A^*}\simeq x^*$ to their common finite-order quotient supplies the required marking on objects; no finite-stage descent of an arbitrary arrow was needed. This proves all assertions. $\square$

The mathematical source is the Stacks Project, Tag 07XB. The complete argument above uses the independently written desingularization and G-ring proofs in this appendix, the pointed approximation result in B.5, and the exact graded comparison in Appendix A. In particular, the graded-algebra conclusion and finite-type return are both part of this proof.

B.7. Jacobian, parameter and cotangent calculations

Here are the calculations and corrections used in the proof chain. Native labels identify the corresponding mathematical source statements; the arguments in this lesson include the details described here.

1. For lemma-final-solve, the last step of B.3 proves smoothness on the cover $D(f_i)$. On its $i$th member the equation $\sum f_jZ_j=1$ eliminates $Z_i$, leaving a polynomial algebra over the smooth algebra localized at $f_i$.

2. For lemma-product, suppose $\Lambda_i=\varinjlim_\alpha B_{i,\alpha}$ with each $B_{i,\alpha}$ smooth over $R_i$. The product of the two index categories is filtered. It presents $\Lambda_1\times\Lambda_2$ as the colimit of $B_{1,\alpha}\times B_{2,\beta}$. Each stage is smooth over $R_1\times R_2$: the two idempotents split its spectrum and the base into open-and-closed pieces, on which the morphisms are the two given smooth maps. The structural and transition maps are product algebra maps.

3. In lemma-desingularize-lifting-apply, consider a prime $\eta$ of $\Lambda$ over the smooth locus of $\bar C$. If $\pi\notin\eta$, the lifting construction is already smooth there. If $\pi\in\eta$, its compatible maps $\bar C/\pi^4\to D/\pi^4\to\Lambda/\pi^4$ place the relevant prime of $D$ in its smooth locus. In either case $H_{D/R}\Lambda\nsubseteq\eta$. The inclusion $H_{D/R}B\subseteq H_{B/R}$ gives the same conclusion for $B$. On $\operatorname{Spec}(\Lambda/\pi^8)$ this is the inclusion of the smooth open for $\bar C$ in that for $B$; taking closed complements gives the required radical-ideal inclusion. Outside $V(\pi)$ the whole construction is smooth.

4. In lemma-ogoma, let $(s^n\pi)^2m=0$ with $n>0$. Localization kills $\pi^2m$, so $m\in K'$. The choice $s(K'/K)=0$ implies $s\pi m=0$, hence $s^n\pi m=0$. The reverse implication follows by multiplication. This proves the annihilator equality for every positive $n$.

5. In lemma-enlarge-solution-modulo, the exponent must be a power $q=p^r$ at least as large as the nilpotence order. Then $u^q=0$ for the relevant nilpotent $u$, and $(1+u)^q=1$ by Frobenius. Divisibility of $q$ by $p$ alone does not suffice. For an algebraic element $\alpha/F$, its irreducible polynomial has the form $h(T^{p^r})$ with $h'\ne0$; hence $\alpha^{p^r}$ is separable over $F$. Increase $r$ to meet both bounds, and lift the separable residue extension by finite étale lifting.

6. In lemma-resolve-general, after setting $e=8c$, use

The bound $N+dc$ by itself does not ensure $\mathfrak p^n\subseteq(\pi_1^e,\ldots,\pi_d^e)$. After localization the parameters generate $\mathfrak p$, and a monomial of degree $d(e-1)+1$ must have an exponent at least $e$. This proves the source inclusion. The identity $\mathfrak p\Lambda_{\mathfrak q}=\mathfrak q\Lambda_{\mathfrak q}$ gives its target version. Choose the Artinian approximation at this larger $n$, leaving $N,c,e$ fixed. The earlier estimate $\mathfrak q^n\subseteq\mathfrak q^{n-N}H_{A/k}\Lambda_{\mathfrak q}$ still applies. The order of the later operations is: solve modulo $\mathfrak p^n$, then pass to the quotient by $J$. This also corrects the source's fifth/sixth-step cross-references.

7. During the induction producing $\delta_i$, the finitely many localized coefficients have the form $v_j=\pi_{t+1}^N(c_j+c'_j)$. Write $v_j=a'_j/s_0$ with one $s_0\notin\mathfrak q$. The equality $\pi_{t+1}^{2N}=\sum_j a_jv_j$ may initially hold only in $\Lambda_{\mathfrak q}$. Choose $u\notin\mathfrak q$ killing the error after multiplication by $s_0$, and choose $s$ divisible by $us_0$. Then

hold in $\Lambda$. In $\bar\Lambda$, the coefficient image is $s^{2N}\pi_{t+1}^Nd_j$. The factor $\pi_{t+1}^N$ must be retained. After adjoining the allowed power of $s$, this image belongs to $D'$; the proof needs that membership, not equality with $s^{2N}d_j$. Replacing $s$ by a further power and scaling all $\mu_j$ accordingly preserves the equality, so the enlargement lemma can also arrange that the image of $s$ belongs to $D'$.

8. For the passage from $R/JR$ to $(R/I)_{\mathfrak r}$, all the $t_i$ are units at the selected prime: their images are $\delta_i\notin\mathfrak q$. Consequently $I_{\mathfrak r}=JR_{\mathfrak r}$ and $I\Lambda_{\mathfrak q}=J\Lambda_{\mathfrak q}$. Localize the existing solution over $(R/JR)_{\mathfrak p}$ at the remaining elements outside $\mathfrak r$. These identities give the solution over $(R/I)_{\mathfrak r}$ with the specified source and target maps.

9. In the final evaluation, $D'\to\bar\Lambda$ is flat and local, hence faithfully flat. The image of $\delta_i\in D'$ has nonzero residue, so $\delta_i$ is a unit in the Artinian local ring $D'$. The substitution

is therefore defined. The identity $(\pi_i\delta_i)^{2N}=\sum_j a_j\lambda_{ij}$ verifies its relations. Evaluating $t_i$ at $\delta_i$ recovers the original coefficients, which proves both required commutative triangles.

10. When $d=0$, the parameter-lifting lemma with hypothesis $r\geq1$ is not applicable and is not needed. In the separable case the localized target is a separable field extension, hence a filtered colimit of smooth algebras; height-zero delocalization gives the resolution. In the inseparable case take $c=1$, use the Artinian approximation without any $\pi_i$ or $t_i$, and keep the localization and height-zero argument. The solution-modulo construction gives an essentially smooth factorization at order $n\geq1$; finite presentation of its source descends the map to a smooth stage.

11. The projective-module argument in more-algebra-lemma-symmetric-algebra-smooth uses $K\to A^n$, tensored with $C$, with cokernel $M\otimes_A C$. It does not use the complex $K\to M$. For projective $M$ this sequence splits. The general conormal assertion requires a further correction: with $P=\operatorname{Sym}_A(A^n)$, $C=\operatorname{Sym}_A(M)$, $J=\ker(P\to C)$ and $S=\ker(K\otimes_A P\to P)$, the correct sequence is

The symmetric-algebra proof establishes this formula at its full scope. For $A=\mathbf Z$, $M=(\mathbf Z/2)^2$ and $K=2\mathbf Z^2$, the nonzero vector $(y,-x)$ lies in the kernel of $K\otimes_A C\to J/J^2$. Thus its unrestricted identification with the conormal module is false. The projective specialization is valid and supplies the smooth-presentation consumers. The projective-lifting argument also corrects the adjacent $R/I$ ring-name typo.

12. To prove composition of regular maps, fix a source prime, pass to its residue fibre and then to any finite purely inseparable extension of that residue field. The intermediate ring is Noetherian and regular. The target is Noetherian, and the base-changed second map is flat with regular fibres. The local regularity ascent theorem therefore makes the target regular. This proves geometric regularity of each composite fibre; flatness of the composite follows from composition of flat maps.

These corrections concern particular inferences, not counterexamples to Popescu's theorem. The separate comparison lemma keeps all six claims and explains why assertion 6 still needs a proof: scaling by an element of the presented algebra need not preserve the chosen invertible Jacobian minor. Assertion 5, used in the desingularization chain, is proved completely. The consulted native source files remain unchanged.

B.8. Commutative-algebra prerequisites

The following map leads to the complete proof at the scope used here. The local proof entry identifies any exact programme proof it reuses, together with the additional argument needed in this lesson. The component record retains the corresponding source and reader identities.

The supporting sections also prove the conormal and Koszul transitivity statements, lifting of finite projective modules and smooth sections, geometric regularity, the finite-over-regular complete-local construction, and G-ring permanence. The first-homology quotient argument is proved at its own scope; a comparison of cotangent presentations alone does not replace it. Arbitrary separable field extensions receive a full formal-smoothness proof. The coefficient-ring lesson is used only for the prime-field claim that it proves.

B.6.1 gives every part of the marked-family approximation used in Section 2.3: the presentation, completion and point comparisons, marked objects and finite-stage descent. Its Artin–Rees comparison is precisely Appendix A's finite-complex and graded-algebra argument. Imported scheme and deformation results outside this supporting algebra chain retain the prerequisite scope stated before the appendix.

B.9. Desingularization from singularity ideals

The proof below proceeds through singularity ideals, improved presentations, flat nilpotent lifting, lifting and desingularization, reduction to fields, localization and the two residue-field cases. It then proves Popescu's theorem and the two approximation statements, including the prime-localized form. The Néron-DVR interlude in the source is independent of this chain and is not included; the theorem for arbitrary regular maps of Noetherian rings keeps its full scope. B.7 records the concrete corrections applied throughout.

Singular ideals

The proof will measure progress by the open subset on which a finite presentation is smooth. We need a way to express that open subset using equations that can be transported to another ring. Jacobian minors provide such equations, provided we also control the relations omitted from the chosen minor. The constructions below make both requirements explicit.

The relevant mathematical references are the Stacks Project, Tags 07C5, 07C6, 07C7, 07ET, 07CA, 07CC and 07EU, together with the corresponding results in the AI Integrated Stacks Project, Smoothing Ring Maps.

Definition. The singularity ideal

For a homomorphism $R\to A$, let $U$ consist of the prime ideals of $A$ at which this homomorphism is smooth. Smoothness at a prime means smoothness on a principal neighbourhood of that prime, so $U$ is open; see Smoothness at a prime ideal. Define $H_{A/R}$ by

We use the convention that the intersection of an empty family of ideals is $A$. Since the complement of $U$ is closed, $H_{A/R}$ is its defining radical ideal. In particular,

This characterizes the singularity ideal uniquely. It does not assert that the ideal is finitely generated or that $U$ is quasi-compact.

Lemma. A Jacobian presentation near a smooth point

Fix a presentation $A=R[x_1,\ldots,x_n]/I$, with $I=(f_1,\ldots,f_m)$. Suppose that $R\to A$ is smooth at $\mathfrak q$. One can select $c$ equations and $c$ variables, where $0\leq c\leq\min(m,n)$, whose Jacobian minor $\Delta$ has the following property: there is an element $a\notin\mathfrak q$ divisible by $\Delta$ in $A$ such that

Here multiplication by an element of $A$ means multiplication on $I/I^2$; the displayed assertion is independent of a lift of $a$ to the polynomial ring. Equivalently, for the selected sets $U,V$ of column and row indices,

Proof. Work first over $A_{\mathfrak q}$. The presentation of the naive cotangent complex is

By smoothness, this map becomes a split injection near $\mathfrak q$, with finite projective cokernel. Its source is therefore finite projective there as well. Choose from the classes of the $f_j$ a basis of $(I/I^2)\otimes_A\kappa(\mathfrak q)$; call its size $c$. Choose $n-c$ of the coordinate differentials whose images form a basis of $\Omega_{A/R}\otimes_A\kappa(\mathfrak q)$. The remaining $c$ coordinates give the desired minor: projection onto them identifies the span of the selected relation differentials with a $c$-dimensional vector space. Thus $\Delta\notin\mathfrak q$.

Finite projectivity permits both chosen bases and the splitting to be used after inverting one element $h\notin\mathfrak q$. In particular, the selected equations generate $(I/I^2)_h$. For each of the finitely many $f_\ell$, clear the denominators in an expression for its class in this generating set. There is consequently a single exponent $N$ with

Take $a=h^N\Delta$. This element stays outside $\mathfrak q$, is divisible by the specified minor, and satisfies all the required containments. Localization of the presentation used above is justified by Localization of the naive cotangent complex. If $c=0$, the minor is the empty determinant $1$; the same argument gives $h^N I\subseteq I^2$. ∎

Definition. Strict standard smoothness

Let $A$ be finitely presented over $R$. The following definitions concern an element $a\in A$, but allow the choice of a presentation. Write such a presentation as

and choose an integer $c$ between $0$ and $\min(m,n)$. Both definitions require the first $c$ equations to control the remaining relations after multiplication by $a$:

An element is elementary standard if these data can be chosen so that it is a multiple of the first $c$-by-$c$ Jacobian minor:

It is strictly standard if, with the same condition on the relations, it belongs to the ideal of all maximal minors of the $c$ selected relation differentials:

We order the column indices in each determinant increasingly. Elementary standard implies strictly standard by assigning coefficient zero to the unused minors. Empty minors have determinant $1$; thus the definitions also cover $c=0$. The preceding lemma shows that every smooth prime has an elementary standard element outside it, after renumbering the selected equations and variables.

Lemma. The Jacobian relation in a strict presentation

Keep the presentation and the integer $c$ above. Let

If $a$ is an $A$-linear combination of the $c$-by-$c$ minors of $du$, there is an $A$-linear map $\psi:A^n\to A^c$ satisfying $\psi du=a\,1_{A^c}$. Conversely, the existence of such a map implies that $a^c$ belongs to the ideal of these minors.

Proof. Put $J=du$, regarded as an $n$-by-$c$ matrix. For a set $U$ of $c$ row indices, let $P_U:A^n\to A^c$ be coordinate projection and let $J_U=P_UJ$. The adjugate identity gives

If $a=\sum_U a_U\det(J_U)$, use

This has the asserted source and target, and its composite with $J$ is multiplication by $a$. For the converse, take determinants in $\psi J=a1_{A^c}$. Expanding the determinant of a product by its minors gives

The minor expansion follows directly from the multilinear determinant formula: terms with a repeated intermediate index cancel by alternation, and the remaining terms group according to $U$. This works over every commutative ring. For $c=0$, both the determinant of the identity on the zero module and the empty minor are $1$, so the assertion has the same interpretation. This is also the calculation underlying A left inverse for a matrix. ∎

Lemma. Controlling a power of the Jacobian determinant (Elkik)

For a finitely presented homomorphism $R\to A$, the elementary standard elements generate an ideal whose radical is $H_{A/R}$. The same is true if one uses all strictly standard elements.

Proof. We first check what happens after inverting a strictly standard element $a$. The relation condition says that the map

is surjective. The preceding matrix calculation gives $\psi_a d_a u_a=a1$. As $a$ is a unit, $u_a$ is also injective. Moreover $a^{-1}u_a\psi_a$ is a left inverse for $d_a$: this can be checked after precomposition with the surjection $u_a$. Hence $d_a$ is split injective. Its cokernel is a direct summand of the finite free module $A_a^n$. The finite-presentation cotangent criterion therefore proves that $R\to A_a$ is smooth. In particular, every strictly standard element belongs to $H_{A/R}$.

Let $J_e$ and $J_s$ be the ideals generated by elementary and strictly standard elements, respectively. We have

At every prime in the smooth locus, the Jacobian presentation lemma supplies an elementary standard element which is not in that prime. Thus $D(J_e)$ contains the entire smooth locus. The previous inclusion gives the reverse containment. Taking complements and then defining radical ideals yields

Neither this argument nor the preceding cotangent calculation requires a Noetherian base. ∎

Example. A smooth locus that is not quasi-compact

Let $k$ be a field and consider

The smooth locus of $\operatorname{Spec}A\to\operatorname{Spec}R$ is exactly $\bigcup_{i\geq1}D(y_i)$. Indeed, where $y_i$ is a unit the relation $xy_i=0$ forces $x=0$, so the morphism is an isomorphism. At the prime $(y_1,y_2,\ldots)$, its inverse image in $R$ is $(x,y_1,y_2,\ldots)$. In that local ring the ideal $(x)$ is nonzero: its annihilator is generated by all the $y_i$, so no element outside the prime kills $x$. This principal ideal is proper and cannot be idempotent, by Nakayama's lemma. The quotient by it is therefore not flat. To see the last implication directly, if $R/(x)$ were flat, tensoring $(x)\hookrightarrow R$ with the quotient would be injective; its resulting map is zero, forcing $(x)/(x)^2=0$. Smoothness consequently fails at this point.

Every prime of $A$ outside that one omits some $y_i$, proving the asserted description. No finite subcollection of these principal opens covers their union: for a finite set $E$ of indices, the prime $(y_i:i\in E)$ belongs to $D(y_j)$ for every $j\notin E$, while belonging to none of the selected $D(y_i)$. Thus even a finitely presented ring homomorphism can have a smooth locus which is not quasi-compact.

Lemma. Base change of a strict standard presentation

For a homomorphism $R\to R'$ and a finitely presented $R$-algebra $A$, write $A'=A\otimes_R R'$. If $a\in A$ is elementary standard, its image $a\otimes1$ is elementary standard over $R'$. The corresponding assertion holds for strictly standard elements.

Proof. Use a presentation witnessing the property of $a$. Tensoring the presentation with $R'$ gives

Formal differentiation of a polynomial commutes with mapping its coefficients from $R$ to $R'$. Consequently every relevant Jacobian minor maps to the corresponding minor of the new presentation. The equalities expressing $a$ as a multiple, or a linear combination, of these minors remain equalities after this coefficient map. So do the finitely many containments $a f_{c+j}\in(f_1,\ldots,f_c)+I^2$: map expressions witnessing those containments to the new polynomial ring. These are exactly the required conditions for $a\otimes1$. No flatness of $R'$ over $R$ is needed. ∎

Lemma. Solving the strict-standard Jacobian relations

Suppose $R\to A\to\Lambda$ is given, $A$ is finitely presented over $R$, and $H_{A/R}\Lambda=\Lambda$. There is a smooth $R$-algebra $B$ through which the specified map $A\to\Lambda$ factors.

Proof. The equality of ideals means that finitely many elements $f_i\in H_{A/R}$ and $\lambda_i\in\Lambda$ satisfy $\sum_i f_i\lambda_i=1$. Introduce variables $Z_i$ and set

Evaluation $Z_i\mapsto\lambda_i$ defines an $A$-algebra homomorphism $B\to\Lambda$. In $B$ the elements $f_i$ generate the unit ideal, so their principal opens cover $\operatorname{Spec}B$. On the $i$th open, eliminate $Z_i$ using the displayed equation:

The definition of $H_{A/R}$ says that $A_{f_i}$ is smooth over $R$. A polynomial algebra over it is smooth as well, so these covering opens establish smoothness of $B/R$. The presentation of $B$ is finite, as required. If the target is the zero ring, the smooth zero algebra supplies the factorization directly. ∎

Presentations of algebras

A presentation records more than generators of an algebra: its conormal module records which equations remain independent to first order. Adding the vector bundle associated with this module makes the conormal data free while retaining a section back to the original algebra. This gives a convenient route from arbitrary smooth or syntomic algebras to explicit equations.

We use the convention that a relative global complete intersection is a quotient of a polynomial algebra by $c$ equations which is flat over the base and whose nonempty fibres have pure dimension equal to the number of variables minus $c$. The corresponding local criteria, including the conormal statements used below, are supplied by Local criteria for a syntomic algebra, Local criteria for complete intersections, and Finite projectivity of a quasi-regular conormal module.

Lemma. Improving a finite presentation

Let $A$ be a finitely presented algebra over a ring $R$. There is a finite-type $A$-algebra $C$ with an $A$-algebra augmentation $C\to A$ such that the following assertions hold.

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

The composite $A\to C\to A$ is the identity. If $R$ is Noetherian, $C$ is also finitely presented over $R$.

Proof. Choose $P=R[x_1,\ldots,x_n]\twoheadrightarrow A$ with kernel $I=(f_1,\ldots,f_m)$, and put $M=I/I^2$. Define

Projection to degree zero supplies the required augmentation. The images of the $m$ chosen equations generate $M$, so they give $m$ algebra generators for $C/A$. Thus $C$ has finite type over $A$ and over $R$. Over a Noetherian base its relation ideal in a finite polynomial presentation is finitely generated, giving the last assertion.

To check the first assertion, fix $a$ with $A_a/R$ a local complete intersection and work with

The conormal criterion says that $M_a$ is finite projective. The sequence therefore splits, and $K_a\oplus M_a\simeq A_a^m$. In particular both summands are finite projective. Since $C_a=\operatorname{Sym}_{A_a}(M_a)$, it is a polynomial algebra on a free module locally on $\operatorname{Spec}A_a$. It follows that $C_a/A_a$ is smooth; see also Smoothness of a symmetric algebra.

Choose a polynomial $h\in P$ representing $a$. A polynomial presentation over $R$ for $A_a$ is obtained by adjoining $x_0$ and imposing $hx_0-1$ in addition to $I$. A presentation for $C_a$ further adjoins variables $y_1,\ldots,y_m$ representing the chosen generators of $M$. Write its relation ideal as $J'$. Cotangent transitivity for these presentations gives

Here the first summand is the relation $hx_0-1$. The right-hand term comes from the linear relations among the $y_j$. To see why there is no extra term on the left, locally split $A_a^m=M_a\oplus K_a$: the quotient defining its symmetric algebra then sets precisely the coordinates of $K_a$ equal to zero, a regular sequence. The complete-intersection conormal sequence is therefore left exact. These local identifications describe the same presentation maps and hence glue. This is the instance of Cotangent transitivity with a complete-intersection terminal map and The cotangent complex of a symmetric algebra used here.

The right-hand module is projective, so the sequence splits. Consequently

This proves the first assertion, including a presentation over $R$ itself.

If $A_a/R$ is smooth, the differential sequence for $R\to A_a\to C_a$ is split exact:

The right-hand identification is the differential calculation for a symmetric algebra; projectivity gives a splitting. The original smooth presentation also gives $M_a\oplus\Omega_{A_a/R}\simeq A_a^n$. Thus $\Omega_{C_a/R}\simeq C_a^n$. This is an existence of a free-module identification, not a claimed canonical choice of its basis. ∎

Proposition. Lifting a smooth algebra over a quotient

Let $R\twoheadrightarrow R_0=R/I$ be any quotient. Every syntomic $R_0$-algebra $A_0$ is the reduction of a syntomic $R$-algebra: there are $A/R$ syntomic and an $R_0$-algebra isomorphism $A/IA\simeq A_0$. If $A_0/R_0$ is smooth, $A$ can be chosen smooth. There is no nilpotence assumption on $I$.

Proof. First suppose $A_0/R_0$ is syntomic. For a finite polynomial presentation of $A_0$, its conormal module $P_0$ is finite projective. The preceding construction gives

smooth over $A_0$, with a presentation over $R_0$ whose conormal module is free. The presentation lemma for a conormal basis allows the equations to be chosen as lifts of a basis. Hence

Lift these finitely many polynomials to $R[y_1,\ldots,y_m]$. Their quotient reduces to $C_0$. The localization criterion for relative complete intersections permits a localization by an element equal to $1$ modulo $I$ after which the quotient is syntomic over $R$. Denote this quotient by $C$, retaining the specified identification $C/IC=C_0$.

Choose a finite projective complement $Q_0$ with $P_0\oplus Q_0\simeq A_0^n$. The projective-module lifting lemma gives an étale map $C\to C'$ whose reduction modulo $I$ is an isomorphism and a finite projective $C'$-module $Q$ lifting $Q_0\otimes_{A_0}C_0$. Put $D=\operatorname{Sym}_{C'}(Q)$. Then $D/C'$ is smooth, and composition shows that $D/R$ is syntomic. The chosen identifications give

All these maps are algebra maps over $A_0$; the last depends on the chosen complement and basis. Choose lifts $d_i\in D$ of the coordinates $t_i$, and set $A'=D/(d_1,\ldots,d_n)$. Its reduction is $A_0$, with its specified $R_0$-algebra structure.

The map $A'/R$ is syntomic in an open neighbourhood of $V(IA')$. Indeed, around any such point write the syntomic algebra $D$ as a relative complete intersection. In its reduction, the additional equations are the polynomial coordinates $t_i$, a regular sequence with quotient $A_0$ flat over $R_0$. Applying the same relative complete-intersection localization criterion to the combined equations proves the assertion at that point. This argument uses the composition criterion and local syntomic presentations, and therefore retains flatness as well as the complete-intersection condition.

Here and below a localization can preserve the entire closed fibre. If an open $U\subseteq\operatorname{Spec}A'$ contains $V(IA')$, write its closed complement as $V(J)$. The containment says $IA'+J=A'$. Choose $g\in J$ with $g\equiv1\pmod{IA'}$. Then $D(g)\subseteq U$ and $(A'_g)/I(A'_g)\simeq A'/IA'$. Apply this to the syntomic neighbourhood to obtain the desired syntomic $A$.

For the smooth assertion, start with this syntomic lift. It is flat and finitely presented over $R$. At every point of $V(IA)$, its fibre is a fibre of the given smooth algebra $A_0/R_0$. By flatness and smooth fibres, $A/R$ is smooth at all these points. The smooth locus is open. A further localization equal to $1$ on the closed fibre, chosen as above, therefore gives a smooth lift without changing $A/IA$. ∎

Lemma. Complete-intersection presentations of syntomic algebras

For every syntomic $R$-algebra $A$ there is a smooth $A$-algebra $C$ with an $A$-algebra augmentation such that $C/R$ is a relative global complete intersection. In particular it has a presentation

which is flat over $R$ and has fibres of pure dimension $n-c$ wherever they are nonempty.

Proof. Apply the improving-presentation lemma with $a=1$. Syntomic implies local complete intersection, so the resulting $C/A$ is smooth and its conormal module has a finite basis. The conormal-basis presentation lemma gives a presentation in which the defining equations lift that basis. Since smooth algebras are syntomic and syntomic maps compose, $C/R$ is syntomic, in particular flat. On each fibre the equations are a local complete-intersection sequence of length $c$ in a polynomial algebra on $n$ variables. The complete-intersection dimension criterion therefore gives dimension $n-c$ at every component. The augmentation is the degree-zero map in the original symmetric-algebra construction and has not been changed by choosing the presentation. ∎

Lemma. Standard presentations of smooth algebras

If $A/R$ is smooth, there is a smooth $A$-algebra $B$ with an $A$-algebra augmentation $B\to A$ such that $B/R$ is standard smooth: it admits a finite polynomial presentation with an invertible square Jacobian minor of size equal to the number of equations.

Proof. The preceding lemma gives a smooth $A$-algebra $C$, with augmentation, and a relative complete-intersection presentation

Because $C/R$ is smooth, the matrix

has a right inverse $T=(t_{i\ell})$ over $C$, so $JT=1_c$. This is the splitting of the conormal differential sequence. Introduce $c$ variables $y_\ell$ and set $B'=C[y_1,\ldots,y_c]$. Choose polynomial lifts of the $t_{i\ell}$ and define a map

We verify directly that this map is étale near the zero section $y=0$. Regard $B'$ as an algebra over $R[z_1,\ldots,z_n]$ with variables $y_1,\ldots,y_c,x_1,\ldots,x_n$ and equations

The square Jacobian matrix in those variables reduces modulo $(y_1,\ldots,y_c)$ to

Its determinant is $\det(JT)=1$. Let $g\in B'$ be the determinant before reduction. Then $g\equiv1\pmod{(y_1,\ldots,y_c)}$. On $B=B'_g$ the square minor is invertible, so this is a standard étale algebra over $R[z_1,\ldots,z_n]$. This follows from the standard smooth Jacobian criterion: the relative dimension is zero. If one writes localization as a polynomial presentation by adjoining $v$ with $gv-1=0$, the enlarged minor is the old one times $g$, hence is still invertible. Viewing the $z_i$ as additional free variables exhibits $B$ as standard smooth over $R$.

The map $C\to B$ is a polynomial extension followed by localization, hence smooth. The evaluation $y_\ell\mapsto0$ extends through that localization because $g$ evaluates to $1$. It gives a retraction $B\to C$. Composing with the original maps between $A$ and $C$ proves all the assertions about $A\to B\to A$. In particular, the localization has not removed the section needed for the return to $A$. ∎

Lemma. Standard smooth presentations in a filtered colimit

An $R$-algebra which is a filtered colimit of smooth $R$-algebras can also be expressed as a filtered colimit of standard smooth $R$-algebras.

Proof. Use the finite factorization criterion. A map $P\to\Lambda$ from a finitely presented $R$-algebra first factors through a smooth algebra $C$. Choose the standard smooth algebra $B$ with maps $C\to B\to C$ whose composite is the identity, as in the preceding lemma. The composite

has the original value on $P$. Thus every finite presentation mapping into $\Lambda$ factors through a standard smooth algebra. The same finite factorization proof applies to this class: tensor products and coequalizers need only be finitely presented before one applies the factorization property. It therefore constructs a filtered indexing category of standard smooth algebras with colimit $\Lambda$. ∎

Lemma. Including prescribed generators in a smooth presentation

Let $A/R$ be standard smooth and let $E=\{a_1,\ldots,a_n\}$ be a finite subset. There is a standard smooth presentation whose first $n$ variables represent these elements and whose invertible Jacobian minor includes these $n$ variables. In particular the number $c$ of equations can be chosen at least $n$.

Proof. Start with

where the minor in $y_1,\ldots,y_d$ is a unit. Choose polynomials $h_i(y)$ representing $a_i$. Adjoin $x_1,\ldots,x_n$ and impose $x_i-h_i(y)=0$ as well as the $g_j$. Eliminating the $x_i$ identifies the new quotient with $A$, with the required specified images. Taking derivatives with respect to $x_1,\ldots,x_n,y_1,\ldots,y_d$ gives a block triangular matrix with diagonal blocks $1_n$ and the original Jacobian minor. Its determinant is a unit. Thus the new presentation is standard smooth and has $c=n+d$ equations. ∎

Lemma. Comparing standard smooth presentations

Fix a finitely presented homomorphism $R\to A$ and an element $a\in A$. We distinguish six properties, retaining the numbering used by the source:

1. The localization $A_a$ is smooth over $R$.
2. It is smooth, and its module of relative differentials is stably free.
3. It is smooth, and its module of relative differentials is free.
4. It admits a standard smooth presentation over $R$.
5. The element $a$ is strictly standard for $A/R$.
6. The element $a$ is elementary standard for $A/R$.

The last two conditions refer to the presentation and relation conditions, with the empty determinant interpreted as $1$. A module is stably free here if adjoining a finite free summand makes it finite free.

The comparison assertions are as follows.

1. Property 4 implies 3, which implies 2, which implies 1.
2. Property 6 implies 5.
3. Property 6 implies 4.
4. Property 5 implies 2.
5. Under property 2 there is an integer $e_0$ such that $a^e$ is strictly standard whenever $e\geq e_0$.
6. The assertion that property 4 makes every sufficiently large power $a^e$ elementary standard is false over arbitrary bases. The integral cusp counterexample and the corrected differential-basis criterion below give its precise scope.

Proof of assertions 1–5. The first assertion follows by eliminating the differentials belonging to an invertible Jacobian minor in a standard presentation. The remaining coordinate differentials freely generate the differential module; the complete calculation is in Standard smooth algebras. A free module is stably free, and forgetting the module condition gives property 1. For the second assertion, a multiple of one minor is a linear combination of all the minors, with zero coefficients on the others. The relation condition is the same in properties 5 and 6.

For assertion 3, choose an elementary standard presentation

After renumbering, the chosen minor is $\Delta=\det(\partial f_j/\partial x_i)_{1\leq j,i\leq c}$. We have $a=a'\Delta$ in $A$, and multiplication by $a$ puts every relation class in the span of $[f_1],\ldots,[f_c]$. Lift $a$ to $h\in P$, and set

Then $Q/J=A_a$. Because $a$ is invertible in this quotient, the relation condition says $J=L+J^2$. The finitely generated $Q$-module $M=J/L$ consequently satisfies $M=JM$. Expressing a finite set of generators of $M$ as $J$-linear combinations of themselves and taking the determinant of the resulting identity-minus-coefficient matrix produces $g\in1+J$ with $gM=0$. Thus $J_g=L_g$. Since the image of $g$ in $Q/J$ is $1$, we obtain

This is the determinant construction in A presentation realizing a basis of the conormal module. Differentiate the displayed equations with respect to $x_0,x_1,\ldots,x_c,t$. The resulting square determinant is $h\Delta g$: the middle equations have no $x_0$ or $t$ terms, and the final column has just the entry $g$ in its last row. In the quotient $h$, $g$ and $\Delta$ are all units. We have therefore obtained a standard smooth presentation. The argument includes $c=0$.

For assertion 4, let a strict presentation have $c$ distinguished relations, and write

The Jacobian matrix identity supplies $\psi:A^n\to A^c$ with $\psi du=a\,1$. Over $A_a$, the relation condition makes $u_a$ surjective; the identity makes it injective. It also splits $d_a$. The strict-standard smoothness argument gives smoothness, and the split conormal sequence gives

This proves the required stable freeness.

For assertion 5, begin with any finite presentation $A=P/I$. Smoothness gives the split sequence

If $\Omega_{A_a/R}\oplus A_a^r\simeq A_a^s$, then $(I/I^2)_a\oplus A_a^s\simeq A_a^{n+r}$. Adjoin $s$ polynomial variables to $P$ and add those variables to the defining ideal. This adds a free summand of rank $s$ to the conormal module and leaves the algebra $A$ unchanged. Denote the new polynomial ring by $P'$ and its ideal by $J$. Now $(J/J^2)_a$ is free, of some finite rank $c$.

Choose a basis of this localized module. Multiplying individual basis vectors by powers of the unit $a$ makes them images of elements $f_1,\ldots,f_c\in J$; this still gives a basis. Extend this list to a finite generating list for $J$. The finite module $J/(J^2+(f_1,\ldots,f_c))$ becomes zero upon localization, so a single power $a^r$ kills it. Consequently the relation condition for strict standardness holds for every $a^e$ with $e\geq r$.

Let $D:A^c\to A^{n+s}$ be the matrix of these distinguished relation differentials. Over $A_a$ it has a left inverse. First clear denominators in the finitely many entries of that inverse. Its composite with $D$ agrees with a power of $a$ times the identity after localization; clear the finitely many remaining zero witnesses as well. This yields a map $\psi:A^{n+s}\to A^c$ and an integer $v\geq1$ for which

already over $A$. The minor expansion places $a^{cv}$ in the ideal of the $c$-by-$c$ minors of $D$. Multiplying its expression by $a^{e-cv}$ proves the minor condition for every $e\geq cv$. Hence $e_0=\max(r,cv)$ works. For $c=0$ the minor ideal is the unit ideal, so only the relation bound is needed. The argument also applies when $A_a$ is the zero ring: all vanishing witnesses still involve finitely many generators. This finishes assertions 1–5. ∎

The unresolved inference in assertion 6. The proposed argument starts with a standard presentation of $A_a$, includes prescribed algebra generators, and multiplies the other coordinates by powers of $a$ so that their images come from $A$. Clearing denominators in the equations then produces relations for a surjection from a polynomial ring onto $A$. To conclude, one needs a single chosen Jacobian minor that is invertible after inverting $a$, as well as generation of the localized conormal module. The coordinate change just described does not by itself establish that the chosen minor remains invertible: $a$ belongs to the presented algebra, so its relative differential can contribute new terms.

For a concrete failure, work over a field $k$ of characteristic $p>0$. Present $k[x,x^{-1}]$ with coordinates $Z=x$, $Y=x^{-1}$, $X=x$ and equations $Z-X$, $XY-1$. The determinant in columns $Z,Y$ is the unit $X$. Keep $Z$, and scale the other coordinates to $U=x^{N+1}$ and $V=x^{N-1}$, for $N\geq1$. After inverting $Z$, the equations

describe the same algebra. Their determinant in columns $Z,V$ is

It vanishes whenever $p\mid N+1$. Thus preservation of that minor cannot be inferred from the stated scaling operation. This scaling example alone does not disprove the existence assertion in part 6. The integral cusp below does disprove it, independently of the chosen presentation. The desingularization proofs below use the established strict-standard conclusion in assertion 5. The elementary-standard conclusion for localization by base scalars has its own argument, in which the scalars have zero relative differential.

Mathematical source: the Stacks Project authors, Tag 07EZ, smoothing.tex, lemma-compare-standard, in the pinned AI Integrated Stacks Project edition. The six comparison assertions, the failed scaling inference and the distinction between strict and elementary standardness are explicit above.

Elementary standard powers and exact differential coordinates

The assertion and an integral counterexample. The implication from standard smoothness of $A_a$ to eventual elementary standardness of $a^e$ is false for general base rings. Put

The map $\mathbf Z[X,Y]\to A$, $X\mapsto t^2$, $Y\mapsto t^3$, identifies

Indeed reduction by the monic relation in $Y$ gives a unique expression $P(X)+YQ(X)$. Its image has even exponents from $P(t^2)$ and odd exponents at least three from $t^3Q(t^2)$, so it vanishes only when $P=Q=0$. In particular this is a finite presentation over $R$.

After inverting $a$, the elements $Y/X=t$ and $Y/X^2=t^{-1}$ belong to $A_a$. Consequently

This is standard smooth over $\mathbf Z$: the derivative of $TV-1$ with respect to $V$ is the unit $T$. Its differential module is $\Omega_{B/\mathbf Z}=B\,dt$.

For every $u\in A$ there are integers $c_n$, only finitely many nonzero, such that

No such $du$ is a basis of $B\,dt$. To see this, the units of $\mathbf Z[t,t^{-1}]$ are exactly $\pm t^k$, $k\in\mathbf Z$. For a proof, the smallest and largest exponents of a nonzero Laurent polynomial add under multiplication, since $\mathbf Z$ is a domain. If a product is $1$, both factors must therefore have only one exponent, and their integer coefficients must be units. But a coefficient $nc_n$ with $n\geq2$ cannot be $1$ or $-1$. Distinct exponents in (ESC4) cannot cancel. Thus its parenthesized Laurent polynomial is never a unit.

This rules out elementary standardness in every finite polynomial presentation of $A$. Suppose, to the contrary, that some $a^e$, $e\geq1$, is elementary standard. Choose a witnessing presentation $A=\mathbf Z[z_1,\ldots,z_n]/I$ and distinguished relations $f_1,\ldots,f_c$. After localizing at $a$, their classes generate $(I/I^2)_a$, and their selected $c$-rowed Jacobian minor is a unit: the defining conditions are $a^e(I/I^2)\subset\sum B[f_j]$ after restriction and $a^e\in(\Delta)$ before restriction. The conormal sequence therefore presents $\Omega_{B/\mathbf Z}$ as the cokernel of the $c$ rows $df_j$. Elimination using the invertible minor shows that the differentials of the $n-c$ unselected variables are a basis of $\Omega_{B/\mathbf Z}$.

Tensoring with $\mathbf Q(t)$ gives $n-c=1$. Thus one variable image $u\in A$ has $du$ as a basis of $B\,dt$, contradicting (ESC4). There is no elementary standard power $a^e$ for any $e\geq1$. This is a counterexample to the implication itself, rather than only to one denominator-clearing argument.

The strictly standard conclusion remains valid in this example. In presentation (ESC2), the two one-rowed minors are $-3X^2$ and $2Y$, and in $A$

There are no remaining relations beyond the distinguished one. Hence $a^e$ is strictly standard for every $e\geq3$.

A corrected necessary and sufficient condition. Let $R\to A$ be finitely presented, let $a\in A$, and suppose $B=A_a\ne0$ is smooth over $R$. The following conditions are equivalent:

The length $r$ may be zero. In particular, freeness of $\Omega_{B/R}$ or standard smoothness of $B$ alone does not impose the requirement that a differential basis come from elements of $A$.

Proof. Condition (ii) implies (i). For (i) implies (iii), use a witnessing presentation and localize. The relation classes are generated by the distinguished $c$ relations, and the selected minor becomes invertible. The same conormal elimination as above gives a basis consisting of the differentials of the unselected variable images, all belonging to $A$. This argument allows an empty minor.

For (iii) implies (ii), choose a finite polynomial presentation

Adjoining these prescribed images gives such a presentation with finitely generated kernel. Explicitly, start with $A=R[w_1,\ldots,w_s]/(h_1,\ldots,h_l)$, represent $u_i$ by a polynomial $U_i(w)$, and take the variables $z_1,\ldots,z_r,w_1,\ldots,w_s$ with relations $h_j(w)$ and $z_i-U_i(w)$. Renumber these as the $n$ variables in (ESC7). This proves the finite generation without any Noetherian assumption. Choose $H\in P$ mapping to $a$. The conormal sequence over $B$ is short exact:

Here injectivity follows directly from formal smoothness. The square-zero surjection $P_H/I_H^2\to B$ admits an $R$-algebra section. Subtracting that section from $P_H\to P_H/I_H^2$ gives an $R$-derivation $P_H\to I_H/I_H^2$ restricting to the identity on $I_H/I_H^2$. The induced map from $\Omega_{P_H/R}\otimes B=B^n$ is a left inverse to the conormal map, proving the asserted injectivity.

By the basis condition, the first $r$ coordinate vectors of $B^n$ map isomorphically to $\Omega_{B/R}$. Consequently projection onto the last $c=n-r$ coordinates restricts to an isomorphism from the kernel in (ESC8) to $B^c$: each prescribed last-coordinate vector has a unique first-coordinate vector making its differential image zero. Let $v_1,\ldots,v_c$ be the resulting basis of $(I/I^2)_a$. There are $g_j\in I$ and integers $N_j\geq0$ with

Indeed every element of the localized conormal module has a numerator represented by one element of $I$ and a power of $a$ as denominator.

The Jacobian minor of $g_1,\ldots,g_c$ in the last $c$ variables therefore satisfies

It is a unit. Its inverse has the form $b/a^s$, $b\in A$, so for some $t\geq0$ the equality $a^t(b\Delta-a^s)=0$ holds in $A$. Thus $a^{s+t}\in(\Delta)$ in $A$.

Extend $g_1,\ldots,g_c$ by finitely many elements to a generating list of $I$. The finitely generated quotient $(I/I^2)/\sum A[g_j]$ vanishes after inverting $a$, since the classes in (ESC9) are a basis after localization. Clearing a denominator for each of its finitely many generators gives an integer $q\geq0$ with

For every $e\geq\max(1,q,s+t)$, both elementary-standard conditions hold. Renumber the variables to put the selected last $c$ variables first if necessary. For $c=0$, take $\Delta=1$ and use the same finite generation argument for the conormal module. This proves (ii) and the equivalence. If $A_a=0$, some power of $a$ is zero, and all its later powers are elementary standard with $c=0$.$\square$

Source correspondence. The assertion contradicted by (ESC1)--(ESC4) is part (f) of the Stacks Project, Tag 07EZ. The elementary and strictly standard conditions are those of Tag 07C7. The equivalence (ESC6) supplies the missing coordinate requirement over an arbitrary base; it does not change the valid strictly standard assertion or its applications.

Equations (ESC1)–(ESC5). A unit minor in any presentation would supply one differential basis element from $A$. The Laurent-unit calculation rules this out, while the two-minor identity proves strict standardness. The editable vector figure is integral-cusp-standardness.svg.

The lifting problem

We next show that ind-smoothness survives a flat nilpotent deformation. Here ind-smooth means a filtered colimit of smooth algebras. Two different operations are needed. First one lifts a finite diagram, allowing an error ideal inside the nilpotent ideal. Then flatness expresses the coefficients of that error through relations over the base, making it possible to eliminate the error without losing smoothness.

Lemma. A first lifting step

Suppose $R\to\Lambda$ is a ring map and $I\subseteq R$ satisfies $I^2=0$. Assume that $\Lambda/I\Lambda$ is ind-smooth over $R/I$. For every finitely presented $R$-algebra $A$ and every $R$-algebra map $\varphi:A\to\Lambda$, there are a smooth $R$-algebra $B$, a finitely generated ideal $J\subseteq IB$, and a factorization

of $\varphi$. Flatness of $\Lambda$ is not required for this first step.

Proof. The finite factorization criterion, together with standard smooth replacements, supplies a factorization of the reduction of $\varphi$ through a standard smooth $(R/I)$-algebra $\bar B$. The map $A/IA\to\bar B$ has finite presentation: both algebras have finite presentation over $R/I$, and one obtains a relative presentation by adjoining their finitely many generators and the equations identifying the images of the generators of $A/IA$. Choose such a presentation

Lift the image of each $t_i$ to an element $\lambda_i\in\Lambda$, and lift $\bar g_j$ to $g_j\in A[t_1,\ldots,t_r]$. Evaluating a lifted relation gives an element of $I\Lambda$. For each $j$, choose a finite expression

Introduce one variable $u_{j\ell}$ for each chosen summand and define

This is finitely presented over $R$. Evaluation at $t_i=\lambda_i$ and $u_{j\ell}=\mu_{j\ell}$ gives $A'\to\Lambda$ extending $\varphi$. Reduction modulo $I$ identifies $A'/IA'$ with the polynomial algebra $\bar B[(u_{j\ell})]$, which is standard smooth over $R/I$.

Choose a standard smooth presentation of $A'/IA'$ and lift its equations to $R$. Localize the lifted quotient at the same Jacobian minor. The Jacobian criterion gives a smooth $R$-algebra $B$ and an isomorphism $B/IB\simeq A'/IA'$. Concretely, if the equations are $f_1,\ldots,f_c$ and the chosen minor is $\Delta$, one can take

The last equation makes the enlarged Jacobian minor invertible. Its reduction gives exactly the chosen algebra because $\bar\Delta$ was already invertible there.

By formal smoothness, the map $B\to A'/IA'$ lifts through the square-zero quotient $A'\to A'/IA'$ to $\theta:B\to A'$. We check its surjectivity without a finite-generation assumption on its cokernel. Since the reduced map is surjective, write any $x\in A'$ as

For each of the finitely many $x_k$, write $x_k=\theta(b_k)+z_k$ with $z_k\in IA'$. Then $\epsilon_kz_k=0$, so $x=\theta(b+\sum_k\epsilon_kb_k)$. Thus $\theta$ is surjective. Its kernel $J$ is contained in $IB$ because the reduction of $\theta$ is an isomorphism. Finally $J$ is finitely generated: a surjection between finitely presented $R$-algebras has a finitely generated kernel, by independence of finite presentation from the chosen generators. Now $A'=B/J$, and the maps already constructed give the required factorization. ∎

Lemma. A second lifting step

Retain $I^2=0$ and the ind-smoothness of $\Lambda/I\Lambda$ over $R/I$, and assume in addition that $\Lambda$ is flat over $R$. Suppose $B/R$ is smooth, $\varphi:B\to\Lambda$ is an $R$-algebra map, and $J\subseteq IB$ is a finitely generated ideal annihilated by $\varphi$. There exist a smooth $R$-algebra $B'$ and maps

Proof. It suffices to eliminate one generator $h$ of $J$. Indeed, after eliminating it, the images of the remaining generators still lie in the ideal generated by $I$ and still vanish in $\Lambda$. Repetition therefore handles any finite generating set.

Write $h=\sum_{i=1}^q\epsilon_i b_i$ with $\epsilon_i\in I$ and $b_i\in B$. Its image is the relation $\sum_i\epsilon_i\varphi(b_i)=0$. The equational criterion for flatness gives finitely many $\lambda_j\in\Lambda$ and scalars $a_{ij}\in R$ such that

Thus the relation among the images of the $b_i$ is generated by relations among the coefficients in $R$. Encode these identities in the finitely presented algebra

It maps to $\Lambda$ by $x_j\mapsto\lambda_j$. The first lifting step factors this map through $B'/J'$, where $B'/R$ is smooth and $J'\subseteq IB'$. In particular $(J')^2=0$. Formal smoothness of $B/R$ lifts $B\to C\to B'/J'$ to a map $\alpha:B\to B'$. Let $\beta$ be the composite $B'\to B'/J'\to\Lambda$; then $\beta\alpha=\varphi$.

Choose $\xi_j\in B'$ lifting the image of $x_j$ in $B'/J'$. The defining equations of $C$ say that

Consequently

The first sum vanishes by the relations over $R$, and the second vanishes by $I^2=0$. This eliminates $h$, and the finite iteration described at the start proves the lemma. ∎

Proposition. Lifting an algebraic factorization

Let $\Lambda$ be a flat $R$-algebra, and let $I\subseteq R$ be nilpotent. If $\Lambda/I\Lambda$ is ind-smooth over $R/I$, then $\Lambda$ is ind-smooth over $R$.

Proof. Suppose first that $I^2=0$. Apply the first lifting step to an arbitrary map $A\to\Lambda$ with $A/R$ finitely presented. We obtain $A\to B/J\to\Lambda$, with $B/R$ smooth and $J\subseteq IB$ finitely generated. Apply the second lifting step to the induced map $B\to\Lambda$. The resulting map $B\to B'$ annihilates $J$, hence descends to $B/J\to B'$. Thus

factors the original map through a smooth $R$-algebra. The finite factorization criterion proves the square-zero case.

For the general case choose $N$ with $I^N=0$. Starting with the assumed ind-smooth algebra $\Lambda/I\Lambda$, successively pass from the base $R/I^r$ to $R/I^{r+1}$ for $r=1,\ldots,N-1$. The kernel $I^r/I^{r+1}$ has square zero, because $2r\geq r+1$. Flatness of $\Lambda/I^{r+1}\Lambda$ over $R/I^{r+1}$ follows by base change from the flatness of $\Lambda/R$. The square-zero case therefore applies at each step. At $r+1=N$ it yields the desired conclusion for $R$ and $\Lambda$. ∎

The lifting lemma

The next construction lifts a diagram modulo $\pi^2$ to an actual map into the target. It deliberately allows the new algebra to be singular over the singular part of the original diagram. Over its smooth part, extra variables record the divided errors in the equations, and compatibility relations make the different local presentations agree.

Lemma. The lifting lemma

Let $R$ be Noetherian, let $\Lambda$ be an $R$-algebra, and let $\pi\in R$ satisfy

Suppose $\bar C$ is finitely presented over $R/\pi^2R$ and is equipped with an $(R/\pi^2R)$-algebra map to $\Lambda/\pi^2\Lambda$. Then there is a finitely presented $R$-algebra $D$ and compatible maps

Compatibility means that the two resulting maps $\bar C\to\Lambda/\pi\Lambda$ coincide. They also commute with the structure maps from $R/\pi^2R$ and $R/\pi R$. These maps have the following properties.

1. $D[1/\pi]$ is smooth over $R$.
2. At every prime of $D$ containing $\pi$ whose image in $\operatorname{Spec}\bar C$ is smooth over $R/\pi^2R$, the map $R\to D$ is smooth.
3. The map $\bar C/\pi\bar C\to D/\pi D$ is smooth at every prime above that same smooth locus.

Proof. Put $P=R[x_1,\ldots,x_n]$ and choose a surjection $P\to\bar C$. Its kernel $I=(f_1,\ldots,f_m)$ contains $\pi^2$. Write $\bar P=P/\pi^2P$ and $\bar I=I/\pi^2P$. We first choose finitely many local presentations of the smooth locus.

Since $\bar C$ is Noetherian, that open locus is quasi-compact. The Jacobian presentation lemma therefore supplies finitely many elements $a_k\in P$ and subsets $E_k\subseteq\{1,\ldots,m\}$ such that the opens $D(a_k)$ cover precisely this locus and, on each of them, the classes of $f_j$ for $j\in E_k$ form a basis of $\bar I/\bar I^2$. We can also choose $|E_k|$ of the $x$-coordinates so that the corresponding Jacobian minor of these equations is invertible there. In fact the construction of that lemma chooses $a_k$ divisible by such a minor.

Let $I_k=(f_j:j\in E_k)$. We may arrange

Here is the needed refinement. Before this equality is imposed, the finite module $(\bar I/\bar I_k)_{a_k}$ equals its product with $\bar I_{a_k}$, because the chosen classes generate the conormal module. The determinant form of Nakayama's lemma gives an element $1+b'_k$, with $b'_k\in\bar I_{a_k}$, annihilating this module. Write $b'_k=b_k/a_k^N$ after increasing $N$. Replacing $a_k$ by $a_k(a_k^N+b_k)$ imposes the desired equality. Its image in $\bar C$ is $a_k^{N+1}$, so it defines the same open subset there and retains the invertibility of the chosen minor.

After a further power of each $a_k$, all the finitely many relations can be written in $P$ as

For $\ell\in E_k$, choose $h_{k\ell}^{j}=a_k\delta_{\ell j}$ and $g_{k\ell}=0$. Define polynomials in $P[z_1,\ldots,z_m]$ by

and set

All index sets are finite, so $D/R$ is finitely presented. Notice that $p_{k\ell}=0$ when $\ell\in E_k$. An identity that will control both the generic and the closed fibre is

It follows by substituting (L1) into the definitions, without dividing by $\pi$.

Choose $\lambda_i\in\Lambda$ lifting the specified images of $x_i$ modulo $\pi^2$. For each $j$, choose $\mu_j$ with $f_j(\lambda)=\pi^2\mu_j$. Send $x_i$ to $\lambda_i$ and $z_j$ to $\pi\mu_j$. The polynomials $q_j$ vanish under this assignment. The image of $p_{k\ell}$ is $\pi$ times

Equation (L1) says that $\pi^2$ annihilates this expression. The annihilator hypothesis in $\Lambda$ says that $\pi$ annihilates it as well. Hence we obtain $D\to\Lambda$. Modulo $\pi$, the relations $q_j$ reduce to $f_j$, giving the required map $\bar C/\pi\bar C\to D/\pi D$. Its compatibility with the specified target map is immediate from the chosen lifts of the $x_i$.

After inverting $\pi$, (L2) makes every $p_{k\ell}$ redundant, and the equations $q_j$ eliminate $z_j$ as $f_j/\pi$. Thus

which proves the first smoothness assertion.

For the assertions near the closed fibre, fix $k$ and retain only the equations belonging to that chart:

There is a surjection $D_k\to D$. Modulo $\pi$ and after inverting $a_k$, the equations $p_{k\ell}$ eliminate the $z_\ell$ with $\ell\notin E_k$. The equality of the localized ideals chosen above therefore gives

In particular this is smooth over $(\bar C/\pi\bar C)_{a_k}$.

We can verify smoothness of $D_k/R$ directly at a prime containing $\pi$ but not $a_k$. Put $c=|E_k|$. Take derivatives of its $m$ equations with respect to the $c$ chosen $x$-coordinates and the $m-c$ variables $z_\ell$ with $\ell\notin E_k$. Modulo $\pi$, the resulting square matrix has block form

where $\det J_k$ is the chosen invertible minor. Its determinant is therefore a unit at the prime. The Jacobian criterion proves that $D_k/R$ is smooth there. This also proves flatness at that prime.

It remains to show that the additional equations defining $D$ do not alter this local ring. Let $\mathfrak q\subset D$ contain $\pi$, with image in the smooth locus of $\bar C$, choose $k$ with $a_k\notin\mathfrak q$, and let $\mathfrak q_k$ be its inverse image in $D_k$. Set $T=(D_k)_{\mathfrak q_k}$. Equation (L2), together with the already imposed $p_{k\ell}$ and $q_j$ for $j\in E_k$, shows that $a_kq_\ell=0$ in $T$. Hence every $q_\ell$ is zero there.

For another chart $k'$, compare the two ways of expressing the class of $a_ka_{k'}f_\ell$ in $(\bar I/\bar I^2)_{a_k}$. Since the classes indexed by $E_k$ are a basis, this comparison gives, for every $j\in E_k$,

Thus $\delta_j$ belongs to $(I_k+\pi^2P)_{a_k}$, because $I_{a_k}=(I_k+\pi^2P)_{a_k}$. In $T$ the elements $f_j$ for $j\in E_k$ equal $\pi z_j$, so every $\delta_j$ lies in $\pi T$. Reducing the equations for chart $k$ modulo $\pi$ and substituting them into $p_{k'\ell}$ gives

Since $a_k$ is a unit, $p_{k'\ell}\in\pi T$. Meanwhile (L2) for chart $k'$ shows $\pi p_{k'\ell}=0$, because all the $q_j$ have already vanished in $T$.

Finally $T$ is flat over $R$. Tensoring the kernels of multiplication by $\pi$ and by $\pi^2$ with $T$ therefore transfers the annihilator equality in $R$ to $T$. If $p_{k'\ell}=\pi t$, the equality $\pi p_{k'\ell}=0$ says $\pi^2t=0$, and hence $\pi t=0$. Thus every $p_{k'\ell}$ vanishes in $T$. The surjection $D_k\to D$ induces an isomorphism $T\simeq D_{\mathfrak q}$.

The Jacobian calculation for $T/R$ and the polynomial description (L3) now give the second and third assertions. Both annihilator hypotheses have been used explicitly: the one in $\Lambda$ constructs the actual target map, and the one in $R$ removes the remaining compatibility equations in the smooth local model. ∎

The desingularization lemma

A section known to order $\pi^4$ allows a strict Jacobian presentation to be replaced by equations with an invertible minor along $\pi=0$. The construction below also tracks the precise obstruction when the annihilators of $\pi$ and $\pi^2$ in the base are unequal.

Lemma. The desingularization lemma

Let $R$ be Noetherian, let $\Lambda$ be an $R$-algebra, and assume

Suppose a finitely presented $R$-algebra $A$ maps to $\Lambda$, the image of $\pi$ is strictly standard for $A/R$, and there is an $R/\pi^4R$-algebra retraction

whose composite with $R/\pi^4R\to\Lambda/\pi^4\Lambda$ is the prescribed map from $A/\pi^4A$. Define

Then the map $A\to\Lambda$ factors through a finitely presented $R$-algebra $B$ such that $\mathfrak aB\subseteq H_{B/R}$.

Proof. Choose a strict presentation $A=R[x_1,\ldots,x_n]/(f_1,\ldots,f_m)$ and its distinguished equations $f_1,\ldots,f_c$. Translate the $x_i$ by lifts of their images under $\rho$. Thus $\rho(x_i)=0$, the constant term of every $f_j$ belongs to $\pi^4R$, and the image of each $x_i$ in $\Lambda$ has the form $\pi^4\lambda_i$.

Write $r_{ji}$ for the coefficient of $x_i$ in $f_j$. Apply $\rho$ to the strict Jacobian identity expressing $\pi$ as a combination of the $c$-row minors. The resulting identity over $R/\pi^4R$ lifts to an identity over $R$ expressing $u\pi$ as a combination of the minors of $(r_{ji})$, for some $u\in1+\pi^3R$. The adjugate calculation therefore gives an $n$-by-$c$ matrix $S=(s_{ik})$ satisfying

Introduce variables $v_1,\ldots,v_c,w_1,\ldots,w_n$ and substitute

For $j\leq c$, the linear part of $f_j$ becomes $u\pi^3v_j+\pi^3\sum_i r_{ji}w_i$. Its constant term and every term of degree at least two are divisible by $\pi^4$. Consequently we can choose actual polynomials $g_j\in R[v,w]$ such that, after (D1),

This is an identity obtained by factoring the displayed coefficients, so it does not require cancellation of $\pi$. Let $h_i=x_i-\pi^2\sum_k s_{ik}v_k-\pi^3w_i$ and put

The map $A\to B$ is induced by the $x$-coordinates. Define its prospective map to $\Lambda$ by

The $f_j$ and $h_i$ vanish. By (D2), the image $t$ of each $g_j$ belongs to $\pi\Lambda$ and satisfies $\pi^3t=0$. The assumed equality of annihilators implies

if $\pi^{e+1}z=0$, apply the equality for $\pi$ and $\pi^2$ to $\pi^{e-1}z$, then induct. Writing $t=\pi z$ now gives $\pi^4z=0$, hence $t=\pi z=0$. Thus $B\to\Lambda$ is well-defined.

Away from $\pi=0$, (D1) eliminates the $w_i$, and the equations $g_j$ follow from $f_j$. Hence $B_\pi\simeq A_\pi[v_1,\ldots,v_c]$. Strictness makes $A_\pi/R$ smooth by Elkik's singularity-ideal criterion, so $B_\pi/R$ is smooth.

To work along $\pi=0$, let $B'=R[v,w]/(g_1,\ldots,g_c)$. The Jacobian matrix of these equations in the variables $v_j$ is the identity modulo $\pi$. The Jacobian criterion therefore shows that $B'/R$ is smooth at every prime containing $\pi$. The map $B'\to B$ is surjective, with kernel generated by the substituted equations $f_{c+1},\ldots,f_m$.

Take a prime $\mathfrak q\subset B$ containing $\pi$ and an element $r\in\mathfrak a$ outside $\mathfrak q$. Let $T$ be the local ring of $B'$ at the inverse image of $\mathfrak q$, and let $K\subseteq T$ be the ideal generated by those remaining equations. We will prove $K=0$.

Every substituted $f_j$ belongs to $\pi^2T$: its constant term lies in $\pi^4R$, and each substituted $x_i$ is divisible by $\pi^2$. The relation condition in the strict presentation says that $\pi f_{c+j}$ belongs to $(f_1,\ldots,f_c)+(f_1,\ldots,f_m)^2$. Since the first $c$ equations vanish in $T$ by (D2), we get

The finite $T$-module $M=\pi K$ thus satisfies $M=\pi M$. Because $\pi$ is in the maximal ideal, Nakayama's lemma gives $\pi K=0$. Equivalently, one can iterate the last containment and use Krull intersection in the Noetherian local ring $T$.

Smoothness makes $T$ flat over $R$. Hence

The left side is zero because it is annihilated by $r$, which is a unit in $T$. Thus the annihilators in $T$ are equal, and their powers stabilize as above. For $f\in K$, write $f=\pi^2z$. Since $\pi f=0$, we have $\pi^3z=0$, hence $\pi z=0$ and $f=0$. This proves $K=0$.

We have shown that $B_{\mathfrak q}\simeq T$ is smooth over $R$ at every prime in $D(\mathfrak aB)$ lying over $\pi=0$, and we already proved smoothness away from $\pi=0$. Therefore $D(\mathfrak aB)$ is contained in the smooth locus, which is exactly the ideal containment $\mathfrak aB\subseteq H_{B/R}$. ∎

Lemma. Desingularization of a strict presentation

Let $R$ be Noetherian and let $\Lambda$ be an $R$-algebra. Suppose

Let $A,D$ be finitely presented $R$-algebras with maps to $\Lambda$. Assume $\pi$ is strictly standard in $A/R$ and there is a compatible $R$-algebra map $A/\pi^4A\to D/\pi^4D$. Then there are maps $A\to B$, $D\to B$, and $B\to\Lambda$, all compatible with the given target maps, such that $B/R$ is finitely presented and

Proof. Use $D$ as the base ring and $A\otimes_RD$ as the algebra to which the preceding lemma is applied. The base $D$ is Noetherian, strictness survives arbitrary base change, and the prescribed map modulo $\pi^4$ defines a $D/\pi^4D$-algebra retraction

Multiplication of the two given maps into $\Lambda$ supplies the map from the tensor product. The compatibility hypothesis gives exactly the required compatibility of the retraction. The preceding lemma yields $A\otimes_RD\to B\to\Lambda$ and

If $D_{\mathfrak p}$ is flat over $R$, the equality of annihilators in $R$ transfers to $D_{\mathfrak p}$. The finite $D$-module inside the last annihilator is therefore zero after localization at $\mathfrak p$. Finiteness implies that some element outside $\mathfrak p$ annihilates it, so $\mathfrak b\nsubseteq\mathfrak p$. Every smooth prime of $D/R$ is such a flat prime. Thus every prime of $B$ above the smooth locus of $D/R$ belongs to the smooth locus of $B/D$. In ideal language this is the first asserted containment. Smooth maps compose, so the same primes are smooth for $B/R$, proving the second. Finite presentation over $R$ follows by composing the finite presentations over $D$ and over $R$. ∎

Lemma. Combining lifting with desingularization

Let $R$ be Noetherian and let $\Lambda$ be an $R$-algebra. Assume the annihilators of $\pi$ and $\pi^2$ agree both in $R$ and in $\Lambda$. Let $A/R$ be finitely presented, with a map to $\Lambda$, and suppose $\pi$ is strictly standard for $A/R$. Given a factorization

through a finitely presented $(R/\pi^8R)$-algebra, there exists $A\to B\to\Lambda$ with $B/R$ finitely presented, $B_\pi/R_\pi$ smooth, and

Proof. The annihilator equalities for $\pi$ imply those for all its positive powers, by the induction used in the preceding proof. Apply the lifting lemma with $\pi^4$ in place of its parameter. It gives a finitely presented $D/R$, a map $D\to\Lambda$, and compatible maps

The algebra $D$ is smooth over $R$ away from $\pi=0$, and it is smooth along the inverse image of the smooth locus of $\bar C/(R/\pi^8R)$. Compose the given map from $A$ with this diagram. The preceding strict-desingularization lemma then gives compatible maps $A\to B\to\Lambda$ and $D\to B\to\Lambda$, with $H_{D/R}B\subseteq H_{B/R}$. It follows immediately that $B_\pi/R_\pi$ is smooth.

For the remaining assertion, let $\eta\subset\Lambda$ contain $\pi$, and suppose its image in $\operatorname{Spec}\bar C$ is smooth over $R/\pi^8R$. The compatibility modulo $\pi^4$ and the lifting lemma show that its inverse image in $D$ is smooth over $R$. Therefore $H_{D/R}\Lambda\nsubseteq\eta$, and the containment just proved implies $H_{B/R}\Lambda\nsubseteq\eta$. This proves the required containment of smooth opens in $\operatorname{Spec}(\Lambda/\pi^8\Lambda)$, hence the corresponding radical containment of ideals.

To identify that radical with the right side displayed in the statement, note also that smoothness of $B_\pi/R_\pi$ implies $\pi\in H_{B/R}$. Thus $\pi^8\Lambda$ already lies in $H_{B/R}\Lambda$, and passage to the quotient commutes with taking the radical of this ideal. This gives precisely the stated formula. ∎

Reduction to the field case

We now reduce the global theorem to a regular algebra over a field. The two return steps matter: a smooth factorization over the total ring of fractions must first be expressed over the original base, and a factorization modulo a power of a nonzerodivisor must then be lifted back to that base.

Situation. The global desingularization problem

In this section $R\to\Lambda$ is a regular homomorphism between Noetherian rings. We write $\operatorname{PT}(R,\Lambda)$ for the assertion that $\Lambda$ is ind-smooth over $R$. Equivalently, every map to $\Lambda$ from a finitely presented $R$-algebra factors through a smooth $R$-algebra, by the finite factorization criterion.

Lemma. Products of ind-smooth maps

For $i=1,2$, let $R_i\to\Lambda_i$ be regular maps of Noetherian rings satisfying $\operatorname{PT}(R_i,\Lambda_i)$. Then their product map satisfies $\operatorname{PT}(R_1\times R_2,\Lambda_1\times\Lambda_2)$.

Proof. Express $\Lambda_i$ as a filtered colimit of smooth $R_i$-algebras $B_{i,\alpha}$. Index the algebras $B_{1,\alpha}\times B_{2,\beta}$ by the product of the two indexing categories. This category is filtered: common targets and equalizers of parallel arrows can be chosen separately in its two factors. An element of the product of the two colimits comes from some pair of stages, and equality can also be checked at a pair of later stages. Thus the colimit is $\Lambda_1\times\Lambda_2$. Each stage is smooth over $R_1\times R_2$, because the two complementary idempotents split its spectrum into the two given smooth morphisms. This proves the assertion. ∎

Lemma. Returning from a localized base

Let $R\to A\to\Lambda$ be ring maps, with $A/R$ finitely presented, and let $S\subseteq R$ be multiplicative. Suppose $S^{-1}A\to C\to S^{-1}\Lambda$ factors the localized map and $C/S^{-1}R$ is smooth. Then $A\to\Lambda$ factors through a finitely presented $R$-algebra $B$ in which the image of some $s\in S$ is elementary standard.

Proof. Replace $C$ by a standard smooth algebra with a retraction to $C$, using the standard presentation lemma. Composing with that retraction preserves the required map to $S^{-1}\Lambda$. Present $A=R[x_1,\ldots,x_n]/(g_1,\ldots,g_t)$, and let $\lambda_i\in\Lambda$ be the images of its generators. The prescribed-generators lemma gives

with the prescribed first $n$ images and with the minor $\Delta=\det(\partial f_j/\partial x_i)_{1\leq j,i\leq c}$ invertible in $C$.

For each additional variable, multiply it by a suitable element of $S$ so that its target image is represented by an element $\lambda_i\in\Lambda$. These changes of variables are invertible diagonal changes over $S^{-1}R$. Their scaling factors are base scalars, so their differentials relative to the base vanish; consequently the chosen Jacobian minor changes only by a unit. Clear the coefficients in the finitely many equations. Multiplying an equation by an element of $S$ again changes its Jacobian row only by a base unit. We now have $f_j\in R[x_1,\ldots,x_{n+m}]$ presenting $C$ after localization, with the indicated generator images.

The inverse of $\Delta$ in the localized quotient can be represented by a polynomial. Clear its coefficients and the coefficients of the finitely many equations witnessing this inverse. Clearing also any remaining equality in the localized polynomial ring gives polynomials $b_0,b_1,\ldots,b_c$ and $s_0\in S$ with

Since each $g_\ell$ vanishes in $C$, choose $s_\ell\in S$ with $s_\ell g_\ell\in(f_1,\ldots,f_c)$ in the polynomial ring over $R$. Since each $f_j(\lambda)$ vanishes in $S^{-1}\Lambda$, choose $u_j\in S$ with $u_jf_j(\lambda)=0$ in $\Lambda$. Set

Evaluation at the $\lambda_i$ gives a factorization $A\to B\to\Lambda$. Let

The first $c$ equations of $B$ have minor $(\prod_j u_j)\Delta$. Multiplying the identity for $s_0$ by $\prod_j u_j$ and reducing in $B$ shows that $s$ is a multiple of this minor. Moreover, $sg_\ell\in(u_1f_1,\ldots,u_cf_c)$ for each $\ell$: multiply its earlier expression by the factors $u_j$, which supply every needed coefficient. These are the two conditions for $s$ to be elementary standard, with an even stronger relation containment than the definition requires. ∎

Lemma. Reducing desingularization to field bases

If $\operatorname{PT}(k,\Lambda)$ holds for every regular homomorphism from a field to a Noetherian ring, then $\operatorname{PT}(R,\Lambda)$ holds for every regular homomorphism of Noetherian rings.

Proof. Suppose a counterexample exists. For its fixed map $R\to\Lambda$, consider the ideals $I\subseteq R$ for which $R/I\to\Lambda/I\Lambda$ fails the assertion. Such quotient maps remain regular by base change of regular maps. The ascending-chain condition supplies a maximal failing ideal. Replace the map by its quotient by that ideal. Now every quotient by a nonzero ideal satisfies the assertion.

The new base $R$ is reduced. Otherwise its nonzero nilradical is nilpotent, since $R$ is Noetherian, and the quotient by that ideal satisfies $\operatorname{PT}$. Flatness of the regular map and nilpotent lifting would imply $\operatorname{PT}(R,\Lambda)$, a contradiction.

Let $S$ be the nonzerodivisors of this reduced Noetherian ring. Its total ring of fractions $S^{-1}R$ is a finite product of fields: there are finitely many minimal primes, no embedded associated primes in a reduced Noetherian ring, and localization at the nonzerodivisors gives their fraction fields. The precise algebraic statements are the total quotient ring without embedded primes and finiteness of the irreducible components. Localization preserves regularity. The assumed field case and the product lemma therefore give $\operatorname{PT}(S^{-1}R,S^{-1}\Lambda)$.

Take any finitely presented $A/R$ mapping to $\Lambda$. There is a smooth factorization after localization at $S$. The preceding return lemma replaces it by an actual factorization over $R$ whose intermediate algebra has an elementary standard element $\pi\in S$. We may replace $A$ by that intermediate algebra, since a smooth factorization of the new map will also factor the original one. Thus $\pi$ is strictly standard in $A/R$ and is a nonzerodivisor of $R$. It remains a nonzerodivisor of $\Lambda$, because $\Lambda/R$ is flat.

The ideal $(\pi^8)$ is nonzero unless $R=0$, whose conclusion is immediate. Hence maximality of the failing ideal gives $\operatorname{PT}(R/\pi^8R,\Lambda/\pi^8\Lambda)$. Factor the reduced map from $A/\pi^8A$ through a smooth algebra $\bar C$ over $R/\pi^8R$. Apply the combined lifting and desingularization lemma. It gives $A\to B\to\Lambda$, with $B/R$ finitely presented and $B_\pi/R_\pi$ smooth, and its radical-ideal conclusion reads

because $\bar C$ is smooth everywhere. Thus $H_{B/R}\Lambda=\Lambda$. The smooth factorization construction replaces $B$ by a smooth $R$-algebra still mapping to $\Lambda$. This factors every finite presentation into $\Lambda$ through a smooth algebra. The finite factorization criterion contradicts the assumed failure. ∎

Localization and descent of resolutions

The next arguments preserve two kinds of progress at once: they remove a specified prime from the nonsmooth locus, while retaining every point already known to be smooth. This second requirement explains the additional variables in the height-zero return construction.

Situation. The local desingularization problem

Fix a Noetherian ring $R$, a finitely presented $R$-algebra $A$, a map $A\to\Lambda$, and a prime $\mathfrak q\subset\Lambda$. No Noetherian hypothesis on $\Lambda$ is included in this setup. Put

A resolution at $\mathfrak q$ is a factorization $A\to B\to\Lambda$ with $B/R$ finitely presented and

Thus the smooth open in the target grows, and the new open contains $\mathfrak q$. The word resolution here refers to this factorization property, not to a proper birational resolution of a variety.

Lemma. Lifting a local desingularization solution

In the local setup, let $r\geq1$ and choose $\pi_1,\ldots,\pi_r\in R$ mapping into $\mathfrak q$. Assume each $\pi_i$ is strictly standard for $A/R$. For each $i$, assume also that the annihilators of $\pi_i$ and $\pi_i^2$ agree both in

If the induced local problem modulo $(\pi_1^8,\ldots,\pi_r^8)$ admits a resolution, then the original local problem admits a resolution.

Proof. Consider first one element $\pi$. Let $A/\pi^8A\to\bar C\to\Lambda/\pi^8\Lambda$ be a resolving factorization. The combined lifting lemma supplies $A\to B\to\Lambda$, with $B_\pi/R_\pi$ smooth and with the image of $H_{\bar C/(R/\pi^8R)}$ contained in the reduction of $\mathfrak h_B$. Since the reduced factorization resolves the selected prime, this containment gives $\mathfrak h_B\nsubseteq\mathfrak q$.

It also retains the original smooth locus. Smoothness is preserved by base change, so the reduction of $H_{A/R}$ belongs to $H_{(A/\pi^8A)/(R/\pi^8R)}$. The resolving property of $\bar C$ and the radical containment therefore place its target image inside $\mathfrak h_B$ modulo $\pi^8$. Since $B_\pi$ is smooth, $\pi\in H_{B/R}$; thus this containment lifts to $H_{A/R}\Lambda\subseteq\mathfrak h_B$. Taking radicals gives $\mathfrak h_A\subseteq\mathfrak h_B$, as required.

For several elements, first apply induction to the last $r-1$ elements over $R/\pi_1^8R$. The stated annihilator conditions are precisely those needed for this quotient setup, and strictness survives arbitrary base change. We obtain a resolution modulo $\pi_1^8$, to which the one-element argument applies. ∎

Lemma. Returning from a localized target

In the local setup, set $\mathfrak p=\mathfrak q\cap R$. Suppose $\mathfrak q$ is minimal over $\mathfrak h_A$ and the localized problem

admits a resolution at its maximal ideal. Then $A\to\Lambda$ factors through a finitely presented $R$-algebra $B$ for which $H_{B/R}\Lambda\nsubseteq\mathfrak q$. This conclusion alone does not assert $\mathfrak h_A\subseteq\mathfrak h_B$.

Proof. In the local target, a resolving algebra has singularity ideal generating the unit ideal: an ideal avoiding the maximal ideal contains a unit. The smooth factorization lemma therefore replaces it by a smooth $R_{\mathfrak p}$-algebra. The standard presentation lemma and the prescribed-generators lemma further give a factorization through

where $A=R[x_1,\ldots,x_n]/(g_1,\ldots,g_t)$, the first $n$ generators retain their given images, and a specified $c$-by-$c$ minor is invertible in $C$. Clearing base denominators makes the $f_j$ polynomials over $R$. Choose $s_0\notin\mathfrak p$ so that this minor is invertible in their quotient after inverting $s_0$. For each $g_\ell$, choose $s_\ell\notin\mathfrak p$ with $s_\ell g_\ell\in(f_1,\ldots,f_c)$.

Write the target images of the $y_i$ with one common denominator $\delta\in\Lambda\setminus\mathfrak q$, say $y_i\mapsto\nu_i/\delta$. Homogenize each $f_j(x,y)$ in the $y$-variables, using one extra variable $w$: let $F_j(x,Y,w)$ have degree $d_j$ in $(Y,w)$ and satisfy $F_j(x,y,1)=f_j(x,y)$. If $\lambda_i$ are the given images of $x_i$, then

There is a single $\epsilon\notin\mathfrak q$ annihilating all these finitely many values in $\Lambda$. Define

It receives $A$ and maps to $\Lambda$ by $x_i\mapsto\lambda_i$, $Y_i\mapsto\nu_i$, $w\mapsto\delta$, and $u\mapsto\epsilon$. Set $s=s_0s_1\cdots s_t$ and $b=suw\in B$. Its image avoids $\mathfrak q$.

After inverting $b$, all three factors $s,u,w$ are units. Use the coordinates $y_i=Y_i/w$ to remove the homogenization factors, which are powers of $w$. The equations $g_\ell$ are redundant because $s_\ell$ is invertible. We obtain the explicit isomorphism

The algebra inside parentheses is standard smooth by the choice of $s_0$, and adjoining two invertible variables preserves smoothness. Hence $b\in H_{B/R}$, proving the claimed avoidance of $\mathfrak q$. ∎

Lemma. Returning from a height-zero localization

In the local setup, assume $\mathfrak q$ is minimal over $\mathfrak h_A$, the localized problem over $R_{\mathfrak p}\to\Lambda_{\mathfrak q}$ admits a resolution, and $\dim\Lambda_{\mathfrak q}=0$. Then the original problem admits a resolution.

Proof. Since $A$ is Noetherian, choose finite generators $a_1,\ldots,a_r$ of $H_{A/R}$. Their images lie in $\mathfrak q$. A zero-dimensional local ring has only its maximal ideal as a prime, so that maximal ideal is its nilradical. Each $a_i/1$ is consequently nilpotent in $\Lambda_{\mathfrak q}$. Choose a common positive power $N$ killing these finitely many elements there, and then choose $\lambda\notin\mathfrak q$ with

This uses only finitely many nilpotent elements; it does not require $\Lambda_{\mathfrak q}$ to be Artinian.

The preceding lemma gives $A\to C\to\Lambda$, with $C/R$ finitely presented, and $c\in H_{C/R}$ whose target image avoids $\mathfrak q$. Choose a finite relative presentation $C=A[x_1,\ldots,x_n]/(f_1,\ldots,f_m)$. Form

The map to $\Lambda$ sends the $x$-coordinates through $C$, sends $t_{ij}$ to zero, sends $y_i$ to the image of $a_i^N$, and sends $z$ to $\lambda$. The identities just chosen make this well-defined.

Inverting $z$ forces all the $y_i$ to zero, so

Choose a polynomial in the original variables $A[x_1,\ldots,x_n]$ representing $c$, and denote its image in $B$ by $\widetilde c$. Under the displayed isomorphism it is precisely the coefficient $c\in C$, and its image in $\Lambda$ is the given image of $c$. Thus $B_{z\widetilde c}$ is smooth over $R$, and the image of $z\widetilde c$ in $\Lambda$ avoids $\mathfrak q$.

On the other hand, after inverting $y_\ell$, the equation $zy_\ell=0$ gives $z=0$, and the remaining relations solve uniquely for all $t_{\ell j}$. Thus $B_{y_\ell}$ is a polynomial algebra over $A$ with $y_\ell$ inverted. After also inverting $a_\ell$, it is smooth over $R$. Hence $a_\ell y_\ell\in H_{B/R}$. Its image in $\Lambda$ is $a_\ell^{N+1}$, showing $a_\ell\in\mathfrak h_B$ for every $\ell$. Therefore $\mathfrak h_A\subseteq\mathfrak h_B$. The element $z\widetilde c$ shows $\mathfrak h_B\nsubseteq\mathfrak q$, so this factorization is a resolution. ∎

Separable residue fields

At a regular local target with separable residue field, a system of parameters reduces the resolution problem to a nilpotent thickening of that field. Before making this reduction, we must choose representatives of the parameters whose annihilator properties hold in the original ring, not only in its localization.

Lemma. The annihilator stabilization step (Ogoma)

Let $M$ be finite over a Noetherian ring $A$, let $S\subseteq A$ be multiplicative, and let $\pi\in A$. Assume multiplication by $\pi$ and by $\pi^2$ have the same kernel on $S^{-1}M$. There exists $s\in S$ such that, for every $n\geq1$, multiplication by $s^n\pi$ and by $(s^n\pi)^2$ have the same kernel on $M$.

Proof. Let $K$ be the kernel of multiplication by $\pi$ on $M$, and let $K'$ be the inverse image in $M$ of the kernel of multiplication by $\pi^2$ on $S^{-1}M$. The assumption says $(K'/K)_S=0$. This quotient is finite, since it is a quotient of submodules of the finite module $M$ over a Noetherian ring. Some $s\in S$ therefore annihilates $K'/K$.

Fix $n\geq1$ and suppose $s^{2n}\pi^2m=0$. After localization, $s$ is a unit, so $m\in K'$. Our choice of $s$ gives $sm\in K$, hence $s\pi m=0$. Multiplying by $s^{n-1}$ gives $s^n\pi m=0$. The converse implication follows by multiplying once more by $s^n\pi$. Thus this single choice of $s$ works for every $n$. ∎

Lemma. Parameters adapted to the singularity ideal

Let $\Lambda$ be Noetherian, let $I\subseteq\mathfrak q\subset\Lambda$ with $\mathfrak q$ prime, and suppose $\Lambda_{\mathfrak q}$ is a regular local ring of dimension $d$. Given positive integers $n,e$ such that $\mathfrak q^n\Lambda_{\mathfrak q}\subseteq I\Lambda_{\mathfrak q}$, one can choose $\pi_1,\ldots,\pi_d\in\Lambda$ with

and, for each $i$, with equal annihilators of $\pi_i$ and $\pi_i^2$ in $\Lambda/(\pi_1^e,\ldots,\pi_{i-1}^e)\Lambda$.

Proof. Choose a regular system of parameters in $\Lambda_{\mathfrak q}$ and clear the denominators of its finitely many entries to obtain representatives in $\Lambda$. Multiplying an entry by an element outside $\mathfrak q$ does not change the parameter ideal in the localization. The hypothesis on $I$ allows such a multiplication for each entry so that its $n$th power belongs to $I$: if $s\pi_i^n\in I$, replace $\pi_i$ by $s\pi_i$ and use $s^n\pi_i^n\in I$.

Now adjust the entries successively. Suppose entries before $i$ have already been fixed. A regular local ring is Cohen–Macaulay, and powers of the initial terms of a regular sequence remain regular; see regular rings and powers of regular sequences. Thus $\pi_i$ is a nonzerodivisor on

Apply Ogoma's lemma to the finite module $\Lambda/(\pi_1^e,\ldots,\pi_{i-1}^e)\Lambda$, with $S=\Lambda\setminus\mathfrak q$. Multiplying $\pi_i$ by the resulting element of $S$ gives the desired annihilator equality before localization. Its $n$th-power containment persists, as does the parameter ideal. Earlier equalities are unaffected because their quotient ideals involve only earlier entries. After $d$ steps all the requirements hold. For $d=0$ the empty sequence satisfies them. ∎

Lemma. Desingularization with separable residue fields

Let $k\to A\to\Lambda$ be maps with $k$ a field, $A/k$ finitely presented, and $\Lambda$ Noetherian. Suppose $\mathfrak q$ is minimal over $\sqrt{H_{A/k}\Lambda}$, the local ring $\Lambda_{\mathfrak q}$ is regular, and $\kappa(\mathfrak q)/k$ is separable. Then the local desingularization problem at $\mathfrak q$ admits a resolution.

Proof. Set $d=\dim\Lambda_{\mathfrak q}$. If $d=0$, the localized target is the separable field extension $\kappa(\mathfrak q)/k$. Such an extension is ind-smooth over $k$, as in the field case of the smooth-colimit criterion. A map from the finite presentation $A$ therefore factors through a smooth algebra after localization. The height-zero return lemma gives a resolution before localization.

Assume $d>0$, and choose generators $a_1,\ldots,a_r$ of $H_{A/k}$. Put $I=H_{A/k}\Lambda$. Minimality of $\mathfrak q$ means that $I\Lambda_{\mathfrak q}$ has radical the maximal ideal. Since the local ring is Noetherian, choose $n\geq1$ with

Using the actual ideal $I$ here will ensure expressions in the generators $a_j$, rather than only membership in their radical.

Let $R=k[x_1,\ldots,x_d]$ and define an $R$-algebra

On $D(a_j)$ one can eliminate $z_{ij}$ for each $i$, leaving a polynomial algebra over $A_{a_j}\otimes_kR$. It is smooth over $R$, since $A_{a_j}/k$ is smooth. The opens $D(a_j)$ cover $D(x_i)$, by the displayed relations. Hence $B_{x_i}/R$ is smooth for every $i$.

Apply improvement of presentations to obtain $B\to C\to B$ with composite the identity. Since $R$ is Noetherian, $C/R$ is finitely presented. Each $C_{x_i}/R$ is smooth with free differentials. The strict part of the comparison lemma supplies a common $c\geq1$ such that every $x_i^c$ is strictly standard for $C/R$.

Apply the parameter lemma with the ideal $I$, the chosen $n$, and exponent $e=8c$. We obtain $\pi_1,\ldots,\pi_d\in\Lambda$ forming a parameter system after localization, with $\pi_i^n\in I$ and the required successive annihilator equalities. Choose coefficients $\lambda_{ij}$ with $\pi_i^n=\sum_j\lambda_{ij}a_j$ in $\Lambda$. Sending $x_i\mapsto\pi_i$ and $z_{ij}\mapsto\lambda_{ij}$ gives $B\to\Lambda$. Compose with the augmentation $C\to B$ to get the map from $C$.

The successive lifting lemma, applied over $R$ to the parameters $x_i^c$, reduces the problem for $C$ to its quotient by $(x_1^e,\ldots,x_d^e)$. In the polynomial base the required elements are nonzerodivisors in the successive quotients, since each new variable is independent of its predecessors. In $\Lambda$ the parameter lemma gives the equalities for $\pi_i$ modulo the preceding $e$th powers, and stabilization of annihilators gives those for $\pi_i^c$ as well.

We verify that the quotient problem can be resolved. Write $\mathfrak p=(x_1,\ldots,x_d)\subset R$. Its local ring is regular, and the images of its parameters form a regular sequence in $\Lambda_{\mathfrak q}$. The flatness criterion over a regular local ring makes $R_{\mathfrak p}\to\Lambda_{\mathfrak q}$ flat. Therefore

is a flat map of Artinian local rings. Reducing by the maximal ideal of its base gives precisely the separable extension $k\to\kappa(\mathfrak q)$, since the $\pi_i$ generate the localized maximal ideal. The base maximal ideal is nilpotent. The field smooth-colimit criterion followed by nilpotent lifting therefore makes this Artinian target ind-smooth over that base. This supplies a smooth factorization of the localized quotient problem.

The selected prime in the quotient target has height zero, because the localized quotient is Artinian. If it is already outside the singularity ideal of the quotient of $C$, the identity factorization resolves it. Otherwise its height-zero property makes it minimal over that ideal, and the height-zero return lemma applies to the smooth localized factorization just obtained. In either case the quotient problem is resolved. Successive lifting now yields $C\to D\to\Lambda$ resolving the problem over $R$.

Finally this also resolves the original problem over $k$. Indeed $C_{a_j}/R$ is smooth: $B_{a_j}/R$ is smooth, and the improvement construction is smooth over $B$ there. Hence the images of all the $a_j$ belong to $H_{C/R}\Lambda$. The resolution over $R$ retains their radical and avoids $\mathfrak q$. As $R/k$ is a polynomial algebra, composition of smooth maps gives $H_{D/R}\subseteq H_{D/k}$. Thus

Together with $A\to B\to C\to D\to\Lambda$, these are the required conclusions. ∎

Inseparable residue fields

In positive characteristic, the residue field need not be separable over the ground field, even for a geometrically regular local algebra. The construction therefore first captures the required infinitesimal data in a finite-dimensional local algebra. The key finiteness hypothesis concerns the first homology of the cotangent complex of the residue extension.

Lemma. Compatibility of the local approximation data

Let $k$ have characteristic $p>0$, and let $(\Lambda,\mathfrak m,K)$ be an Artinian local $k$-algebra. Suppose $H_1(L_{K/k})$ is finite-dimensional over $K$. Then $\Lambda$ is a filtered colimit of local Artinian $k$-algebras $A$ essentially of finite type over $k$, with $A\to\Lambda$ flat and $\mathfrak m_A\Lambda=\mathfrak m$. The maps can be taken to be inclusions, so this is a directed union of subalgebras. In particular, any finite subset of $\Lambda$ is contained in one such $A$.

Proof. Choose generators $\lambda_1,\ldots,\lambda_d$ of $\mathfrak m$, and let $n$ be the least positive integer with $\mathfrak m^n=0$. Every field of characteristic $p$ is formally smooth over its prime field, by formal smoothness over a perfect field. Thus the residue map admits a coefficient-field section $\sigma:K\to\Lambda$. Such a section is an $\mathbf F_p$-algebra map; it need not respect the given $k$-algebra structure.

For a chosen section, evaluation gives a surjection

Surjectivity follows by successively expressing an element modulo $\mathfrak m$, then modulo $\mathfrak m^2$, and so on, using coefficients from $\sigma(K)$. Nilpotence makes this a finite procedure.

We claim that one can choose $\sigma$ and a finitely generated subextension $k\subseteq F\subseteq K$ such that the image of the given map $k\to\Lambda$ is contained in $\Psi_\sigma(F[x])$. This is the step that retains the actual $k$-structure. We prove it by induction on $n$. For $n=1$, use $\Lambda=K$ and $F=k$.

For $n>1$, set $J=\mathfrak m^{n-1}$ and $\Lambda'=\Lambda/J$. By induction choose a coefficient section $\sigma':K\to\Lambda'$ and a finite subextension $F'/k$ whose polynomial image contains the image of $k$. Let $A'$ denote that polynomial image, with induced map $\tau':k\to A'$. Lift $\sigma'$ to a coefficient section $\sigma:K\to\Lambda$. The truncated polynomial algebra

maps to $\Lambda$, and its composite onto $A'$ has nilpotent kernel: on residue fields it is the inclusion of $F'$ in $K$, so its kernel is contained in the nilpotent ideal generated by the $x_i$. Formal smoothness of $k/\mathbf F_p$ lifts $\tau'$ to a map $\tau:k\to T$.

Let $i:k\to\Lambda$ be the given structure map. The two maps $i$ and $\Psi_\sigma\tau$ agree modulo $J$. Their difference

is therefore a derivation, with the $k$-action on $J$ given through its residue field. Here $\mathfrak mJ=0$ and $J^2=0$.

The Jacobi–Zariski sequence for fields, using formal smoothness over $\mathbf F_p$, gives the exact sequence

Choose a basis $\{\mathrm dy_j:j\in S\}$ of $\Omega_{k/\mathbf F_p}$ from its generating differentials. The finite-dimensional kernel in this sequence is contained in the span of finitely many of these basis vectors, indexed by $S_0\subset S$. The images of the remaining $\mathrm dy_j$ in $\Omega_{K/\mathbf F_p}$ are linearly independent over $K$: a relation would lie both in their span and in the span indexed by $S_0$.

Extend this independent set to a basis of $\Omega_{K/\mathbf F_p}$ and choose a $K$-linear map to $J$ taking $\mathrm dy_j$ to $\theta(y_j)$ for $j\notin S_0$. It corresponds to a derivation $D:K\to J$. Since $J^2=0$, $\sigma+D$ is again a coefficient-field section. This change replaces $\theta$ by $\theta-D|_k$. Indeed, only the constant term of $\tau(a)$ contributes to the change: every positive-degree monomial maps into $\mathfrak m$, which annihilates $J$. The residue of that constant term is $a$.

After making this change, $\theta$ vanishes on every $y_j$ outside $S_0$. Thus it is determined by the finitely many elements $\theta(y_j)\in J$ for $j\in S_0$. The degree-$n-1$ monomials in the $\lambda_i$ span $J$ over $K$, so write these elements as finite linear combinations of those monomials. Adjoin their finitely many coefficients to $F'$, obtaining $F$. Every value $\theta(a)$ then belongs to $\Psi_\sigma(F[x])$, as do all values of $\Psi_\sigma\tau$. This proves the claim for the actual map $i$.

Fix such a section $\sigma$. The kernel of $\Psi_\sigma$ is finitely generated, since $K[x]$ is Noetherian. Choose generators $g_1,\ldots,g_t$, including all degree-$n$ monomials if needed, and enlarge $F$ to contain their coefficients. For every further finitely generated intermediate field $F\subseteq F_1\subseteq K$, set

Extension of scalars identifies $K\otimes_{F_1}A_{F_1}$ with $\Lambda$. Hence $A_{F_1}\to\Lambda$ is faithfully flat and injective. All the $g_j$ have zero constant term, and the ideal $(x_1,\ldots,x_d)$ is nilpotent. Thus $A_{F_1}$ is local Artinian with residue field $F_1$, and its maximal ideal generates $\mathfrak m$ in $\Lambda$. The claim shows that its image contains the given image of $k$, making it a $k$-subalgebra.

For completeness, its finite-type assertion uses this actual $k$-structure, not the possibly different embedding of $k$ in the coefficient field. Choose finitely many elements of $A_{F_1}$ lifting field generators of $F_1/k$, and adjoin the images of $x_1,\ldots,x_d$. Localize the resulting finitely generated $k$-subalgebra at the inverse image of the maximal ideal of $A_{F_1}$. Its residue map is onto $F_1$. For any element of $A_{F_1}$, subtract an element of this local subalgebra with the same residue. The difference is a combination of the $x_i$. Replace its finitely many coefficients in the same way and repeat. After $n$ repetitions the remaining error is zero. Therefore this local subalgebra surjects onto $A_{F_1}$, proving that $A_{F_1}$ is essentially of finite type over $k$. This is the finite nilpotent-expansion argument behind the Artinian finite-type criterion.

Finally, the finitely generated fields $F_1$ containing the fixed $F$ form a directed family: use their compositum for a common enlargement. Every element of $\Lambda$ has a polynomial expression in the $\lambda_i$ with finitely many coefficients from $K$, so it belongs to one of these subalgebras. Their union is $\Lambda$, completing the proof. ∎

Lemma. A factorization modulo a prescribed ideal

Let $k$ be a field of characteristic $p>0$, let $\Lambda$ be a Noetherian geometrically regular $k$-algebra, and let $\mathfrak q\subset\Lambda$ be prime. For every $n\geq1$ and finite subset $E\subset\Lambda_{\mathfrak q}/\mathfrak q^n\Lambda_{\mathfrak q}$ there is a map

for some $m\geq0$ with the following properties. If $\mathfrak p=\varphi^{-1}(\mathfrak q)$, then $P_{\mathfrak p}\to\Lambda_{\mathfrak q}$ is flat and $\mathfrak p\Lambda_{\mathfrak q}=\mathfrak q\Lambda_{\mathfrak q}$. There is a factorization of local Artinian rings

in which the first map is essentially smooth, the second is flat, and the image of $D$ contains $E$.

Proof. Put $L=\Lambda_{\mathfrak q}/\mathfrak q^n\Lambda_{\mathfrak q}$ and $K=\kappa(\mathfrak q)$. We first establish a coefficient-enlargement fact, independently of the desired polynomial factorization.

Suppose $A\subset L$ is local Artinian, $A\to L$ is flat, and $\mathfrak m_AL=\mathfrak m_L$. Given a unit $u\in L$, there is an essentially smooth local Artinian extension $A\to A'\subset L$ such that $A'\to L$ remains flat, $\mathfrak m_{A'}L=\mathfrak m_L$, and $u^q\in A'$ for some integer $q>0$. Write $F$ for the residue field of $A$ and $\alpha$ for the residue of $u$.

If $\alpha\in F$, choose a unit $x\in A$ whose residue is $\alpha^{-1}$. Then $xu=1+v$ with $v$ nilpotent. For a power $q=p^r$ at least the nilpotence order, $(1+v)^q=1$. Hence $u^q=x^{-q}\in A$, and no extension is necessary.

If $\alpha$ is transcendental over $F$, use

This map to $L$ is defined because every inverted polynomial has nonzero residue at the transcendental element $\alpha$. The ring $A'$ is local Artinian with residue field $F(\alpha)$, and it is essentially smooth over $A$. The Noetherian fibrewise flatness criterion, applied to $A\to A'\to L$ and the finite $L$-module $L$, proves $L$ flat over $A'$: it is flat over $A$ by assumption, and its closed fibre $K$ is flat over the field $F(\alpha)$. The map is local, hence faithfully flat and injective. This identifies $A'$ with a subring of $L$ containing $u$, so take $q=1$.

If $\alpha$ is algebraic over $F$, its irreducible polynomial has the form $h(T^{p^r})$ with $h'\ne0$. Thus $\alpha^{p^r}$ is separable over $F$. Since Artinian local rings are henselian, lift the finite separable residue extension $F(\alpha^{p^r})/F$ to a finite étale local $A$-algebra $A'$, and lift its residue embedding into $K$ to a map $A'\to L$. These are the henselian lifting and finite étale classification statements. The same fibrewise flatness criterion shows this local map is faithfully flat. Apply the first case to $u^{p^r}$ and this enlarged coefficient ring to obtain a further power lying in $A'$. In all three cases $\mathfrak m_{A'}=\mathfrak m_AA'$, so the asserted maximal-ideal equality is preserved. This proves the enlargement fact. Successive enlargements handle any finite list of units.

Geometric regularity gives an injection $H_1(L_{K/k})\to\mathfrak q\Lambda_{\mathfrak q}/\mathfrak q^2\Lambda_{\mathfrak q}$, by the cotangent criterion. In particular this homology is finite-dimensional. Choose representatives $\tau_1,\ldots,\tau_d\in\Lambda$ of a regular system of parameters of $\Lambda_{\mathfrak q}$, clearing denominators if necessary. Apply the Artinian approximation lemma to a finite subset consisting of $E$ and the images of these representatives. We obtain a local Artinian $k$-subalgebra $A\subset L$, essentially of finite type, flatly embedded, with $\mathfrak m_AL=\mathfrak m_L$. Let $F$ be its residue field.

We next choose the polynomial coordinates in $\Lambda$ itself. Choose elements $b_1,\ldots,b_t\in A$ whose residues have differentials forming an $F$-basis of $\Omega_{F/k}$. Express each image in $L$ as $\ell_i/s_i$ with $\ell_i,s_i\in\Lambda$ and $s_i\notin\mathfrak q$. Apply the enlargement fact to the finitely many units represented by the $s_i$. This yields $A'\subset L$ containing powers $s_i^{q_i}$ and hence containing the images of the global elements

It is still essentially of finite type over $k$, since each enlargement is essentially smooth and therefore essentially of finite type, and it still contains $E$ and the parameter images. Write $F'$ for its residue field. In each transcendental enlargement the new residue-field differential is that of the adjoined global unit; in each finite separable enlargement no new relative differential is needed. It follows that $\Omega_{F'/k}$ is generated by the old $\mathrm db_i$ and the differentials of the adjoined powers of the $s_i$. Since in $F'$ we have $v_i=b_i s_i^{q_i}$, the identity

and invertibility of $s_i^{q_i}$ show that it is generated by differentials of finitely many global elements of $\Lambda$ whose images belong to $A'$. Select a subset whose differentials form a basis. Denote these global elements by $\lambda_1,\ldots,\lambda_a$.

Apply geometric regularity over a field to these elements and the finite subextension $F'/k$ of $K$. For

and $\mathfrak p'=(P'\to\Lambda)^{-1}(\mathfrak q)$, the map $P'_{\mathfrak p'}\to\Lambda_{\mathfrak q}$ is flat with regular closed fibre. Its proof also shows that a regular system of parameters of $P'_{\mathfrak p'}$ maps to part of one for $\Lambda_{\mathfrak q}$. Complete it by a suitable subset of the $\tau_i$: their images generate the maximal ideal, so a subset supplies a basis of the remaining cotangent space. Adjoin these selected $\tau_i$ as further polynomial coordinates. The resulting $P\to\Lambda$ has $\mathfrak p\Lambda_{\mathfrak q}=\mathfrak q\Lambda_{\mathfrak q}$, and a parameter system of the regular local ring $P_{\mathfrak p}$ maps to a parameter system of $\Lambda_{\mathfrak q}$. The regular-local flatness criterion proves the required flatness. All its coordinate images in $L$ belong to $A'$.

Consequently $Q=P_{\mathfrak p}/\mathfrak p^nP_{\mathfrak p}$ maps to $A'$: elements outside $\mathfrak p$ have nonzero residue and are units there, while $\mathfrak p^n$ vanishes in $L$ and hence in its subring $A'$. The map $Q\to L$ is flat, by the flatness just proved and the equality of the extended maximal ideals. Since $A'\to L$ is faithfully flat, descent of flatness gives $A'/Q$ flat.

Moreover $\mathfrak m_QA'=\mathfrak m_{A'}$: both ideals become $\mathfrak m_L$ after the faithfully flat extension. The extension $F'/\kappa(\mathfrak p)$ is finitely generated and has zero differentials, since the selected coordinates span $\Omega_{F'/k}$ and the additional parameter coordinates have residue zero. The field differential criterion makes it finite separable. Lifting a finite basis of this residue extension generates $A'$ as a $Q$-module: subtract a linear combination modulo $\mathfrak m_Q$, then repeat on the coefficients; nilpotence of $\mathfrak m_Q$ terminates the process. Thus $A'/Q$ is finite flat with finite separable closed fibre. The étale criterion makes it finite étale, in particular essentially smooth. Taking $D=A'$ proves every assertion, including containment of the prescribed set $E$. ∎

Lemma. Enlarging a factorization in positive characteristic

Retain a factorization supplied by the preceding lemma,

For every $\lambda\in\Lambda\setminus\mathfrak q$, there is an integer $q>0$ and a local Artinian factorization $D\to D'\to L$ such that $D'/D$ is essentially smooth, $D'\to L$ is flat, and the image of $\lambda^q$ lies in $D'$.

Proof. The flat local map $D\to L$ is injective, so identify $D$ with its image. Since the source $P_{\mathfrak p}/\mathfrak p^n$ maps essentially smoothly to the Artinian local ring $D$, its fibre is a zero-dimensional local essentially smooth algebra, hence a field. Its maximal ideal therefore generates $\mathfrak m_D$. The equality $\mathfrak p\Lambda_{\mathfrak q}=\mathfrak q\Lambda_{\mathfrak q}$ gives $\mathfrak m_DL=\mathfrak m_L$. The image of $\lambda$ is a unit in $L$. Apply the coefficient-enlargement fact proved at the start of the preceding proof with $A=D$ and this unit. It gives exactly the asserted factorization and power, with no change to the original maps from $P_{\mathfrak p}/\mathfrak p^n$. ∎

Lemma. Desingularization with inseparable residue fields

Let $k$ have characteristic $p>0$, let $A/k$ be finitely presented, and let $A\to\Lambda$ be a $k$-algebra map with $\Lambda$ Noetherian and geometrically regular over $k$. If $\mathfrak q$ is minimal over $\mathfrak h_A=\sqrt{H_{A/k}\Lambda}$, then the local desingularization problem at $\mathfrak q$ admits a resolution.

Proof. Put $d=\dim\Lambda_{\mathfrak q}$. If $d=0$, this local ring is a geometrically regular field extension of $k$, hence separable, and the preceding separable-residue-field argument applies. One may also use the finite-jet factorization at order one: its polynomial local base then has dimension zero and is a rational function field over $k$; its essentially smooth intermediate algebra supplies a smooth stage containing the finitely many source generators. The height-zero return lemma gives the desired global factorization. In particular, no positive-length parameter-lifting argument is needed in this case.

Assume $d>0$. We organize the construction around a strict presentation, an Artinian approximation, and the return from that approximation.

Preparing a strict presentation. Choose $N\geq1$ with

which is possible by minimality and Noetherianity. Write $H_{A/k}=(a_1,\ldots,a_s)$, set $P_0=k[x_1,\ldots,x_d]$, and form

Each $B_{a_j}/P_0$ is smooth, by eliminating the corresponding $z_{ij}$ and using smoothness of $A_{a_j}/k$. The equations imply that the opens $D(a_j)$ cover every $D(x_i)$, so $B_{x_i}/P_0$ is smooth. Improve this presentation to $B\to C\to B$ with composite the identity. Then $C/P_0$ is finitely presented, and $C_{x_i}/P_0$ is smooth with free differentials. The strict part of the comparison lemma gives a common integer $c\geq1$ for which every $x_i^c$ is strictly standard for $C/P_0$.

Set $e=8c$ and choose

Apply the finite-jet factorization at this order, requiring its intermediate algebra to contain the images of generators of $A$. We obtain a polynomial algebra $P=k[y_1,\ldots,y_m]$, a map $P\to\Lambda$, a prime $\mathfrak p$ over $\mathfrak q$, and

The first map is essentially smooth, the second is flat local, and $\mathfrak p\Lambda_{\mathfrak q}=\mathfrak q\Lambda_{\mathfrak q}$. Since the flat map $P_{\mathfrak p}\to\Lambda_{\mathfrak q}$ has a field as closed fibre, the dimension formula gives $\dim P_{\mathfrak p}=d$. Choose $\pi_1,\ldots,\pi_d\in\mathfrak p$ representing a regular system of parameters there. They also form a parameter system in $\Lambda_{\mathfrak q}$. Identify $D$ with its image in $L$, using faithful flatness. Containment of the source generators defines a map $A\to D$: all their polynomial relations already vanish in the containing ring $L$.

Arranging annihilators over the polynomial base. Put $R=P[t_1,\ldots,t_d]$ and $\gamma_i=\pi_it_i$. Let $S=P\setminus\mathfrak p$, so $S^{-1}R=P_{\mathfrak p}[t]$. The parameters $\pi_i$ are a permutable regular sequence in $P_{\mathfrak p}$. Therefore the products $\pi_it_i$, and the successive powers needed here, are regular sequences in this polynomial algebra, by regular sequences with separate polynomial variables and powers of regular sequences.

Use Ogoma's lemma successively on the finite $R$-modules $R/(\gamma_1^e,\ldots,\gamma_{i-1}^e)$. Multiplying $\pi_i$ by an appropriate element of $S$ ensures

These successive changes leave earlier equalities intact and do not alter either localized parameter system. They also leave every $\pi_i$ in the image of $P$ in $D$.

Arranging coefficients and annihilators in the target. We construct units $\delta_i\in\Lambda\setminus\mathfrak q$ and coefficients $\lambda_{ij}\in\Lambda$, together with an essentially smooth local Artinian enlargement $D\to D'\subset L$, such that

all the images of $\delta_i,\lambda_{ij}$ belong to $D'$, and the annihilators of $\delta_i\pi_i$ and its square agree modulo the preceding $(\delta_h\pi_h)^e$. The map $D'\to L$ will remain flat throughout.

Suppose the construction has been completed for indices before $i$. The element $\pi_i^N$ belongs to $(a_1,\ldots,a_s)L$, so faithful flatness contracts this membership to the current coefficient ring: write

Lift the $d_j$ to $b_j\in\Lambda_{\mathfrak q}$. The error lies in $\mathfrak q^n\Lambda_{\mathfrak q}$, which is contained in $\mathfrak q^{n-N}(a_1,\ldots,a_s)\Lambda_{\mathfrak q}$. Hence choose $b'_j\in\mathfrak q^{n-N}\Lambda_{\mathfrak q}$ with

After multiplying by $\pi_i^N$, put $v_j=\pi_i^N(b_j+b'_j)$. Its image in $L$ is $\pi_i^Nd_j$, since $\pi_i^Nb'_j\in\mathfrak q^n\Lambda_{\mathfrak q}$. Write all $v_j=w_j/s_0$ with $w_j\in\Lambda$ and $s_0\notin\mathfrak q$. Some $u\notin\mathfrak q$ annihilates the error in the equality $s_0\pi_i^{2N}=\sum_j a_jw_j$. Choose $s\notin\mathfrak q$ divisible by $us_0$ and set

This is an element of $\Lambda$, and the chosen divisibility gives the exact equality

Its coefficient images in $L$ are $s^{2N}\pi_i^Nd_j$. By the enlargement lemma, replace $s$ by a positive power and enlarge $D'$ so its image lies in $D'$. Scale the $\mu_j$ by the corresponding $2N$th power; both the equality and the displayed coefficient formula persist. Thus the new coefficient images lie in $D'$ as well.

In $\Lambda_{\mathfrak q}$, the preceding $(\delta_h\pi_h)^e$ followed by $s\pi_i$ form a regular sequence. Apply Ogoma's lemma to the finite module $\Lambda/((\delta_h\pi_h)^e:h<i)$ to find $s'\notin\mathfrak q$ such that every $(s')^q s\pi_i$, $q>0$, has equal first and second annihilators there. Enlarge $D'$ once more so that some $(s')^q$ belongs to it. Taking

establishes all requirements for index $i$. The finitely many successive essentially smooth enlargements compose, completing this induction.

Constructing the compatible finite-jet map. Send $x_i\mapsto\delta_i\pi_i$ and $z_{ij}\mapsto\lambda_{ij}$ to obtain $B\to\Lambda$, then use the augmentation $C\to B$ to define $C\to\Lambda$. Give $R$ its map to $\Lambda$ by $t_i\mapsto\delta_i$, and map $P_0\to R$ by $x_i\mapsto\gamma_i$. Let $C_R=C\otimes_{P_0}R$ and $B_R=B\otimes_{P_0}R$.

Write $Q=P_{\mathfrak p}/\mathfrak p^nP_{\mathfrak p}$. Define a $Q[t]$-algebra map

using the given map $A\to D'$ and the formulas

Each $\delta_i$ is a unit in $D'$: its image in the residue field of $L$ is nonzero and the map is local. Equation (I2), multiplied by $t_i^{2N}/\delta_i^{2N}$, verifies every relation of $B$. Evaluation $t_i\mapsto\delta_i$ gives the specified map to $L$, including the original values of $z_{ij}$. Thus both the base map and the target map commute. Compose with the retraction from $C_R$ to obtain the same factorization for $C_R\otimes_RQ[t]$.

The algebra $D'/Q$ is essentially smooth, so $D'[t]$ is a filtered colimit of smooth $Q[t]$-algebras, by writing its localization as a filtered colimit of principal localizations. Since $C_R\otimes_RQ[t]$ is finitely presented, its map factors through one smooth stage $T\to L$. This gives an actual smooth factorization over $Q[t]$.

Returning to the original rings. Set $J=(\pi_1^e,\ldots,\pi_d^e)\subset P$ and $I=(\gamma_1^e,\ldots,\gamma_d^e)\subset R$. Because the parameters generate $\mathfrak pP_{\mathfrak p}$, every monomial of degree $d(e-1)+1$ in them contains an $e$th power. Our choice of $n$ therefore gives

Base-change the smooth stage $T$ from $Q[t]$ to $P_{\mathfrak p}[t]/J$. Its target is now $\Lambda_{\mathfrak q}/J\Lambda_{\mathfrak q}$. Let $\mathfrak r$ be the inverse image of $\mathfrak q$ in $R$. Every $t_i$ is a unit in $R_{\mathfrak r}$, because its target image is $\delta_i\notin\mathfrak q$. Thus

Localizing the smooth factorization accordingly gives one for $C_R\otimes_R(R/I)_{\mathfrak r}$ into $\Lambda_{\mathfrak q}/I\Lambda_{\mathfrak q}$. This target is Artinian, since $J$ consists of positive powers of a full parameter system. The height-zero return lemma therefore resolves the quotient problem before localization. If its selected prime is already smooth, use the identity factorization; otherwise height zero makes it minimal over the singularity ideal, as required by that lemma.

By arbitrary base change, every $\gamma_i^c$ is strictly standard for $C_R/R$. Equalities (I1) and the target equalities arranged above imply the corresponding annihilator equalities for their $c$th powers, since first-power stabilization implies stabilization of all positive powers. Apply successive lifting with these elements. Their eighth powers generate $I$, so it gives a resolution $C_R\to V\to\Lambda$ over $R$.

Finally the images of $a_j$ lie in $H_{C_R/R}$: they already make $B/P_0$ smooth, the improvement is smooth there, and base change preserves smoothness. The resolution over $R$ retains their radical and avoids $\mathfrak q$. Since $R/k$ is polynomial, $H_{V/R}\subseteq H_{V/k}$ by composition of smooth maps. The composite $A\to B\to C\to C_R\to V\to\Lambda$ therefore satisfies $\mathfrak h_A\subseteq\mathfrak h_V\nsubseteq\mathfrak q$, proving the lemma. ∎

The main theorem

The local constructions now give a terminating global procedure: unless the target is already covered by the smooth locus of the current presentation, resolve a minimal prime of its remaining nonsmooth locus. Each step strictly enlarges a radical ideal in the Noetherian target.

Theorem. General Néron desingularization (Popescu)

For a regular homomorphism $R\to\Lambda$ between Noetherian rings, the $R$-algebra $\Lambda$ is a filtered colimit of smooth $R$-algebras.

Proof. By the field reduction, it is enough to take $R=k$ a field and $\Lambda/k$ Noetherian and geometrically regular. Fix a finitely presented $k$-algebra $A$ mapping to $\Lambda$, and set $\mathfrak h_A=\sqrt{H_{A/k}\Lambda}$.

If this radical is proper, choose a prime minimal over it. In characteristic zero its residue extension is separable, so the separable resolution lemma applies. In positive characteristic use the inseparable resolution lemma. Either gives a finitely presented algebra $B$ through which the map factors, with $\mathfrak h_A\subseteq\mathfrak h_B$ and $\mathfrak h_B$ avoiding the selected prime. This containment is strict, because the selected prime contains $\mathfrak h_A$.

Repeat if the new radical is still proper. An infinite repetition would be an infinite strictly ascending chain of ideals in the Noetherian ring $\Lambda$, which is impossible. Thus a finite sequence of factorizations reaches $B$ with $\sqrt{H_{B/k}\Lambda}=\Lambda$, equivalently $H_{B/k}\Lambda=\Lambda$. The smooth factorization lemma replaces it by a smooth algebra still factoring the original map $A\to\Lambda$. Every finite presentation has therefore been factored through a smooth algebra. The finite factorization criterion proves the theorem, including the filteredness of the resulting presentation. The zero target is covered by the smooth zero algebra. ∎

The approximation property for G-rings

For a Noetherian local ring $(R,\mathfrak m)$, the G-ring condition can be tested on the completion map $R\to\widehat R$: it is equivalent to regularity of that map. See the maximal-ideal test for G-rings. Henselization and strict henselization preserve this condition, by henselian permanence.

The examples used here include algebras essentially of finite type over a field, over a complete Noetherian local ring, over $\mathbf Z$, or over a characteristic-zero Dedekind ring. Their G-ring property is supplied by examples and permanence. These examples are not an extra hypothesis on the following results: the stated G-ring condition is sufficient.

For polynomials $f_1,\ldots,f_m\in R[X_1,\ldots,X_n]$, a solution in an $R$-algebra $S$ is a tuple $a\in S^n$ with $f_j(a)=0$ for every $j$. Approximation asks for exact solutions with specified finite-order agreement with a solution in the completion.

Theorem. Artin approximation for henselian local G-rings

Let $(R,\mathfrak m)$ be a henselian Noetherian local G-ring. Given $f_1,\ldots,f_m\in R[X_1,\ldots,X_n]$ and a solution $a\in\widehat R^n$, for every $N\geq0$ there is a solution $b\in R^n$ satisfying

Proof. First reduce finite-order approximation to existence of an exact solution. Choose $c_i\in R$ representing $a_i$ modulo $\mathfrak m^N\widehat R$, choose generators $d_1,\ldots,d_s$ of $\mathfrak m^N$, and write $a_i=c_i+\sum_\ell d_\ell u_{i\ell}$ in $\widehat R$. Substitute

in each $f_j$. The resulting finite polynomial system has the solution $(u_{i\ell})$ in $\widehat R$. Any solution of that new system in $R$ gives an exact solution of the original system with the desired congruence. Thus it remains to prove existence in $R$ for any finite system solvable in $\widehat R$.

Let $C\subseteq\widehat R$ be the $R$-subalgebra generated by its finitely many solution coordinates. It is finitely presented, because $R$ is Noetherian. The G-ring condition makes $R\to\widehat R$ regular, so Popescu's theorem and finite factorization give

with $B/R$ smooth. Composing the second map with the residue map gives a point $B\to R/\mathfrak m$. The smooth-section lifting lemma gives an étale $R$-algebra $R'$ with $R'/\mathfrak mR'\simeq R/\mathfrak m$ and a map $B\to R'$ lifting this point. The henselian section criterion supplies an $R$-algebra map $R'\to R$. The composite $C\to B\to R'\to R$ sends the coordinates to a solution in $R$. Apply this existence assertion to the substituted system to obtain the required $b$. ∎

Order zero imposes no congruence restriction. If negative orders are included with the convention $\mathfrak m^N=R$, the same existence assertion covers them as well.

A pointed étale neighbourhood also has a canonical comparison with the completion. If $R\to R'$ is étale and $\mathfrak m'\subset R'$ lies over $\mathfrak m$ with unchanged residue field, then

are injective local maps, and the completions of the first two local rings are canonically isomorphic. These are the pointed étale and henselization comparisons in the henselian construction and Noetherian henselization. The ring being localized is $R'$, so the second term is $R'_{\mathfrak m'}$.

Theorem. Artin approximation in an étale neighbourhood

Let $(R,\mathfrak m)$ be a Noetherian local G-ring, and suppose a finite polynomial system over $R$ has a solution $a\in\widehat R^n$. For every $N\geq0$ there are an étale $R$-algebra $R'$, a maximal ideal $\mathfrak m'$ over $\mathfrak m$ with $\kappa(\mathfrak m')=\kappa(\mathfrak m)$, and a solution $b\in(R')^n$ such that

under the pointed completion comparison just described.

Proof. Perform the same substitution $X_i=c_i+\sum_\ell d_\ell U_{i\ell}$ with $d_\ell$ generating $\mathfrak m^N$. Apply Popescu factorization and smooth-section lifting to the new system, obtaining $C\to B\to R'$ with $R'$ étale over $R$ and $R'/\mathfrak mR'\simeq R/\mathfrak m$. This gives an exact solution of the substituted system in $R'$, without needing a henselian section back to $R$.

Let $\mathfrak m'$ be the kernel of its given residue map. The resulting original coordinates $b_i=c_i+\sum_\ell d_\ell v_{i\ell}$ solve all the equations. Under $R'\to R'_{\mathfrak m'}\to\widehat R$, the difference from $a_i=c_i+\sum_\ell d_\ell u_{i\ell}$ belongs to $\mathfrak m^N\widehat R=(\mathfrak m')^N\widehat R$. This proves the assertion. As an alternative, one can apply the henselian theorem to $R^h$, which is a G-ring, and descend the finitely many solution coordinates and equations to a pointed étale stage in its filtered presentation. ∎

Lemma. Approximation after localization at a prime

Let $R$ be Noetherian, let $\mathfrak p\subset R$ be prime, and assume $R_{\mathfrak p}$ is a G-ring. Suppose polynomials over $R$ have a solution $a\in\widehat{R_{\mathfrak p}}^{n}$. For each $N\geq0$ there are an étale $R$-algebra $R'$, a prime $\mathfrak p'$ above $\mathfrak p$ with the same residue field, and a solution $b\in(R')^n$ satisfying

after identifying this completion with $\widehat{R_{\mathfrak p}}$.

Proof. Apply the preceding theorem over $R_{\mathfrak p}$. It gives a pointed étale algebra $R''/R_{\mathfrak p}$ and a solution there. The finite presentation of $R''$, its étale presentation data, and the inverse of the chosen Jacobian determinant involve only finitely many elements of $R_{\mathfrak p}$. Clear their denominators to descend the algebra to an étale algebra over $R_s$ for some $s\notin\mathfrak p$. Since $R_s/R$ is étale, this is an étale $R$-algebra $R'$; see also finite-presentation descent of étale algebras.

The finitely many solution coordinates descend after one further localization away from the chosen prime. Clear the finitely many relations asserting that they solve the equations at the same time. Such localizations preserve the selected local ring and the residue identification. If $\mathfrak p'$ is the inverse image of the chosen prime of $R''$, the local rings $R'_{\mathfrak p'}$ and the selected localization of $R''$ are identical. Their completions and the order-$N$ congruences are therefore the ones already obtained. ∎

B.10. Predecessor constructions used in the proof

The following statements and proofs supply the algebraic constructions used above. Native labels identify their mathematical correspondence with the cited Stacks sources. Each entry gives its argument here or identifies an exact complete programme proof with the same hypotheses; that proof's own prerequisites retain their stated scope.

B.10.1. More on Algebra

The lifting arguments below preserve the entire quotient by the given ideal. The ideal need not be nilpotent or lie in the Jacobson radical. The permitted change of base is an étale map whose reduction is the identity on that quotient.

Lemma. Localizing an algebra while preserving its closed fibre

Suppose $A\to B$ is étale at all primes of $B$ containing an ideal $J$. There is an element $g\in B$, with $g\equiv1\pmod J$, for which $B_g$ is étale over $A$. In particular localization at $g$ leaves $B/J$ unchanged.

Proof. The étale locus is open; this is part of the local characterization in Étale ring maps. Write its complement as $V(L)$. The hypothesis says $V(L+J)=\varnothing$, so $L+J=B$. Choose $g\in L$ with $1-g\in J$. Then $D(g)$ lies in the étale locus and contains $V(J)$. These are exactly the two required properties. ∎

Lemma. Lifting a monic polynomial factorization

Let $I\subset A$ be any ideal, and let $f\in A[t]$ be monic. Suppose its reduction has a factorization $\overline f=\overline g\,\overline h$ into monic polynomials generating $(A/I)[t]$ as an ideal. There are an étale $A$-algebra $A'$, an isomorphism $A'/IA'\simeq A/I$ compatible with $A$, and monic factors $g',h'\in A'[t]$ of $f$ reducing to $\overline g,\overline h$.

Proof. Put $n=\deg\overline g$ and $m=\deg\overline h$. Introduce coefficients for two monic polynomials $G,H$ of these degrees. Equating the $n+m$ remaining coefficients of $GH$ with those of $f$ defines a finitely presented $A$-algebra $B$. The specified factors define an $A$-algebra surjection $\epsilon:B\to A/I$.

At these coefficients, the derivative of the multiplication map is

This map is an isomorphism. Indeed, for a polynomial $w$ on the right, invert $\overline h$ modulo the monic polynomial $\overline g$ to obtain the unique $u$ of degree less than $n$ with $\overline h u\equiv w\pmod{\overline g}$. Then $v=(w-\overline h u)/\overline g$ has degree less than $m$. This construction is linear and also proves uniqueness. It includes the degree-zero cases, with the corresponding polynomial module equal to zero.

Consequently the determinant $d$ of the coefficient Jacobian maps to a unit under $\epsilon$. The standard étale criterion makes $B_d$ étale over $A$; this is also the construction in Étale algebras from polynomial factorizations. Put $D=B_d/IB_d$. The surjection $D\to A/I$ is a map between étale $A/I$-algebras, hence is étale by Morphisms between étale algebras. A surjective flat map of finite presentation selects an open and closed component: by Finite presentation and flatness, there is an idempotent $e\in D$ with $D_e\simeq A/I$. Lift $e$ to $v\in B_d$ and set $A'=(B_d)_v$. This remains étale over $A$, has the specified quotient, and carries the required images of $G,H$. ∎

Lemma. Lifting a coprime factorization

The same conclusion holds when the two factors of $\overline f$ are coprime and the leading coefficient of $\overline g$ is a unit, without requiring the factors themselves to be monic.

Proof. Lift that leading coefficient to $u\in A$. Since $u$ becomes a unit in $A/I$, the localization $A\to A_u$ is étale and induces an isomorphism on quotients by $I$. Over this localization replace the prescribed factors by $\overline u^{-1}\overline g$ and $\overline u\,\overline h$. Both are monic: the degree and leading coefficient of a product with a monic factor are determined without cancellation. Apply the preceding lemma, then multiply the lifted factors by $u$ and $u^{-1}$ respectively. Their product is still $f$ and their reductions are the original factors. ∎

Lemma. Separating a closed image from another closed subset

For a map $R\to S$ and ideals $I\subset R$, $J\subset S$, assume that the closure of the image of $V(J)$ in $\operatorname{Spec}R$ avoids $V(I)$. Then some $f\in R$ satisfies $f\equiv1\pmod I$ and has image in $J$.

Proof. Express that closure as $V(L)$. The disjointness gives $I+L=R$, so choose $b\in L$ congruent to $1$ modulo $I$. Every prime of $S$ containing $J$ contains the image of $b$. Thus that image lies in $\sqrt J$, and a power $f=b^r$ has image in $J$. The same power remains congruent to $1$ modulo $I$. This uses the ideal–closed-set correspondence in the affine Zariski topology. ∎

Lemma. Integral elements compatible with a lifted factorization

Let $B$ be integral over $A$, and suppose the image of $b\in B$ in $B/IB$ is idempotent. Then a monic polynomial $f\in A[t]$ annihilates $b$ and satisfies

Proof. The element $z=b^2-b$ belongs to $IB$. Choose a finite expression for $z$ with coefficients in $I$ and elements of $B$. Adjoining those elements and $b$ to $A$ gives a finite $A$-algebra $B_0\subset B$, since each adjoined element is integral. Moreover $zB_0\subset IB_0$. On a finite set of module generators for $B_0$, multiplication by $z$ is therefore represented by a matrix with entries in $I$. The determinant trick produces a monic polynomial

that annihilates multiplication by $z$, and hence annihilates $z$ itself. Take a nonempty generating set so that $d\geq1$. Now $f(t)=q(t^2-t)$ has all the asserted properties. This is the integral-over-an-ideal argument of Integral extensions, applied inside the finite subalgebra containing the data. ∎

Lemma. Lifting an idempotent after localization

If $B$ is an integral $A$-algebra and $\overline e\in B/IB$ is idempotent, then after an étale base change $A\to A'$ inducing $A'/IA'\simeq A/I$, the element $\overline e$ lifts to an idempotent of $B\otimes_A A'$.

Proof. Choose $y\in B$ reducing to $\overline e$ and use the preceding lemma to obtain $f(y)=0$ with $\overline f=t^d(t-1)^d$. Lift this coprime factorization by an étale change of base preserving $A/I$. Temporarily rename the new rings $A,B$; then

Set $b_1=g(y)$, $b_2=h(y)$. Their product is zero, and their reductions are $\overline e$ and $(-1)^d(1-\overline e)$. Thus $V(b_1,b_2)$ avoids $V(IB)$.

An integral morphism of spectra is closed, by going up and the closed-map criterion. Its image here avoids $V(I)$. The preceding separation lemma supplies $a\equiv1\pmod I$ whose image lies in $(b_1,b_2)B$. Localizing $A$ at $a$ is another permitted étale change. In the resulting $B$, choose $u,v$ with $ub_1+vb_2=1$. The element $e=ub_1$ is idempotent because

Reducing the displayed Bézout identity modulo $IB$ and multiplying it by $\overline e$ gives $\overline u\,\overline e=\overline e$. Therefore $e$ reduces to the prescribed idempotent. The composite base change is étale and still induces the identity quotient by $I$. ∎

Lemma. Lifting a finite projective module

A finite projective module $\overline P$ over $A/I$ lifts to a finite projective module over some étale $A$-algebra $A'$ satisfying $A'/IA'\simeq A/I$.

Proof. Realize $\overline P$ as the image of an idempotent endomorphism $\overline p$ of $(A/I)^n$. If $n=0$, there is nothing to lift. Otherwise choose a matrix $\varphi\in M_n(A)$ lifting $\overline p$ and let $f(t)$ be its characteristic polynomial. The finite integral algebra $B=A[t]/(f)$ acts on $A^n$ by $t\mapsto\varphi$, by Cayley–Hamilton.

In each residue field of $A/I$, the characteristic polynomial of $\overline p$ is $t^{n-r}(t-1)^r$, where $r$ is its rank at that point. It follows that the image of $t(1-t)$ in $B/IB$ belongs to every prime: for a prime of $B/IB$, pass to the residue field at its contraction in $A/I$ and then use this factorization. Choose $N\geq1$ for which $t^N(1-t)^N=0$ in $B/IB$. The two terms $t^N$ and $(1-t)^N$ generate the unit ideal, so their sum is a unit when their product is zero. Consequently

is idempotent. Under the action on $(A/I)^n$, it maps to $\overline p$, since $\overline p^N=\overline p$ and $(1-\overline p)^N=1-\overline p$.

Apply the idempotent-lifting lemma to $B/A$. The lifted idempotent in $B\otimes_A A'$ acts as a projector $p'$ on $(A')^n$. Its image $P'$ is a finite projective direct summand, and the direct-sum decomposition remains split after reduction modulo $IA'$. Thus $P'/IP'\simeq\overline P$, as required. ∎

Lemma. The cotangent complex of a symmetric algebra

Write $F=A^m$, choose an exact sequence $0\to K\to F\to M\to0$, and put

For $k=(k_1,\ldots,k_m)\in K$, let $\ell_k=\sum_i k_i y_i\in P$. Define the $P$-module of relations among these linear equations by

There is a right-exact sequence

Thus the naive cotangent complex of this presentation, in degrees $-1,0$, is

In every case there is a canonical identification

If $M$ is projective, the second arrow in (SA1) is an isomorphism, so (SA2) simplifies to $[K\otimes_A C\to C^m]$.

Proof. The universal property of the symmetric algebra identifies $C$ with the quotient of $P$ by the linear forms $\ell_k$. Hence $K\otimes_A P\to J$ is surjective with kernel $S$. Tensor this presentation with $C=P/J$. Right exactness gives (SA1), because $J\otimes_P C=J/J^2$. Differentiating a relation $\sum p_i\ell_{k_i}=0$ and then reducing modulo $J$ gives $\sum \overline p_i k_i=0$ in $C^m$. Consequently the displayed differential descends to the quotient in (SA2). This is exactly the defining conormal presentation of the naive cotangent complex.

For any $C$-module $N$, an $A$-derivation $C\to N$ is uniquely determined by its restriction to $M\subset C$, and that restriction can be any $A$-linear map $M\to N$. The universal properties of differentials and tensor products therefore give $\Omega_{C/A}\simeq C\otimes_A M$. This argument also applies when $M$ has no finite generating set.

When $M$ is projective, $F\to M$ splits. Both $M$ and $K$ are then finite projective. Locally on $\operatorname{Spec}A$, choose bases for the two summands in $F=M\oplus K$. The resulting polynomial coordinates identify $J$ with the ideal generated by the coordinates corresponding to $K$. Its conormal module is free on those coordinates, and the canonical map $K\otimes_A C\to J/J^2$ is an isomorphism. These are local checks of a single globally defined map, so they prove the assertion over $A$. ∎

Why the relation term is necessary. The unrestricted formula $J/J^2\simeq K\otimes_A C$ in Stacks, Tag 07EV fails already over $A=\mathbb Z$. Take

Using $(2,0)$ and $(0,2)$ as a basis of $K$, the natural map is

It kills $(y,-x)$. This vector is nonzero, as is seen after reducing to $\mathbb F_2[x,y]^2$. Thus the proposed canonical map is not injective. More explicitly, in graded degree two its domain has dimension four over $\mathbb F_2$, whereas its codomain has dimension three. The missing relation is $y(2x)-x(2y)=0$. Replacing the image of a map by its domain before taking a cokernel loses precisely such relations.

The corrected statement retains the arbitrary-module case through (SA1)–(SA2), rather than using the projective case as a substitute for it. The improved smooth presentation uses the projective case after localization at the specified smooth locus.

Lemma. Smoothness of a symmetric algebra

For any $A$-module $M$, its symmetric algebra $C=\operatorname{Sym}_A(M)$ is smooth over $A$ exactly when $M$ is finite projective.

Proof. Suppose first that $C/A$ is smooth. Then $\Omega_{C/A}$ is a finite projective $C$-module. Pull it back along the augmentation $C\to A$ and use the preceding differential calculation:

Base change preserves finite projectivity, so $M$ has the required property. Equivalently, the augmentation ideal has conormal module $M$, and the same conclusion follows from the conormal module of a smooth section.

Conversely, choose a finite free module surjecting onto the finite projective module $M$. Its kernel $K$ is a finite projective direct summand. Thus $C$ has a finite presentation, with equations given by a finite generating set of $K$. Locally on $\operatorname{Spec}A$, the module $M$ is free of finite rank and $C$ is a polynomial algebra. Smoothness is local on the base, which proves the claim. One may also use the projective case of (SA2): the inclusion $K\otimes_A C\to C^m$ splits and its cokernel is the finite projective module $M\otimes_A C$. The cotangent criterion for smoothness gives the same conclusion. ∎

Lemma. Lifting a section of a smooth morphism

Let $A\to B$ be smooth, let $I\subset A$ be an ideal, and fix an $A$-algebra map $\epsilon:B\to A/I$. There exist an étale $A$-algebra $A'$ with $A'/IA'\simeq A/I$ and an $A$-algebra map $B\to A'$ whose reduction is $\epsilon$.

Proof. Put $J=\ker\epsilon$. The chosen map is surjective because it extends $A\to A/I$. The augmentation $B/IB\to A/I$ identifies

For completeness, this is the conormal identity for an augmented algebra: modulo the square of its augmentation ideal, the map taking an element to that element minus its constant term is a derivation. It gives the inverse to the universal differential map on the augmentation ideal. Thus (SL1) does not require a nilpotence assumption on $I$. Smoothness makes $P_0$ finite projective.

Choose a finite projective complement $K_0$ with $P_0\oplus K_0\simeq(A/I)^n$. By Lifting a finite projective module, an étale change $A\to A_1$ preserving $A/I$ lifts $K_0$ to a finite projective $A_1$-module $K$. Replace $B$ by

This is smooth over $A_1$, by Smoothness of a symmetric algebra, base change, and composition of smooth maps. The original section and the symmetric-algebra augmentation define $B_1\to A/I$. Its module (SL1) is $P_0\oplus K_0$, because relative differentials of the tensor product split as the sum of the two pulled-back differential modules. Every map from $B_1$ extending this section also gives the required map from the original $B$. We can therefore rename $A_1,B_1$ as $A,B$ and assume that $P_0$ is free of rank $n$.

Choose $f_1,\ldots,f_n\in J$ whose classes form a basis of $P_0$, and let $C=B/(f_1,\ldots,f_n)$. We claim that $C/A$ is étale along the closed subset defined by $JC$. Let $\mathfrak q\supset J$ be a prime of $B$, and let $\mathfrak r$ be its image in $C$. Since $\Omega_{B/A}$ is finite projective and has rank $n$ at this section, its localization at $\mathfrak q$ is free of rank $n$. The differentials $df_i$ reduce to a basis modulo $J$, so Nakayama's lemma makes the map

an isomorphism: it is a surjection between free modules of the same finite rank over a local ring.

Write $L=(f_1,\ldots,f_n)\subset B$. The conormal map

factors the preceding isomorphism through the surjection $C_{\mathfrak r}^n\to(L/L^2)_{\mathfrak r}$. Both arrows in this factorization are consequently isomorphisms. The cotangent transitivity sequence, with $B/A$ smooth and $\Omega_{B/A}$ projective, now gives

The algebra $C$ is finitely presented over $A$. The pointwise smoothness criterion therefore gives smoothness at $\mathfrak r$, and the vanishing of relative differentials gives étaleness there.

By Localizing an algebra while preserving its closed fibre, choose a localization $C_g$ that is étale over $A$ and still maps onto $C/JC=A/I$. The quotient map $C_g/IC_g\to A/I$ is a surjection between étale $A/I$-algebras. As in Lifting a monic polynomial factorization, isolate its open and closed component by localizing at a lift of the corresponding idempotent. The result is an étale $A$-algebra $A'$ with $A'/IA'\simeq A/I$ and a map $B\to C_g\to A'$ reducing to $\epsilon$.

Finally compose with the initial base change $A\to A_1$ if one was made. Étaleness and the asserted quotient identification are preserved by that composition, so this proves the claim for the original rings and section. ∎

Lemma. The conormal sequence for a first-homology regular sequence

Suppose $I\subset J$ are ideals of $A$, and $J/I$ is generated by a finite sequence with vanishing first Koszul homology over $A/I$. Then

Proof. Lift the sequence to $g_1,\ldots,g_m\in J$ and put $G=(g_1,\ldots,g_m)$, so $J=I+G$. First observe that $I\cap G=IG$. Indeed, the short exact sequence of Koszul complexes

is exact term by term because Koszul terms are free. The hypothesis on $H_1$ makes $I/IG\to A/G$ injective in its homology sequence. This gives the asserted intersection equality; see also First cotangent homology after a regular quotient.

Now $J^2=I^2+IG+G^2$. If $x\in I\cap J^2$, subtract its terms in $I^2+IG$ to obtain an element of $I\cap G^2\subset I\cap G=IG$. Thus $x\in I^2+IG=IJ$. The reverse inclusion follows directly from $I\subset J$. ∎

Lemma. The conormal sequence for a first-homology regular ideal

The equality $I\cap J^2=IJ$ also holds if $J/I$ is an $H_1$-regular ideal of $A/I$.

Proof. At a prime containing $J$, the definition supplies a neighborhood where $J/I$ has the finite sequence required in the preceding lemma. At a prime outside $V(J)$, the localized ideal $J$ is the whole ring, and both sides become $I$. Intersections, products and squares of these ideals commute with localization. The two submodules therefore agree at every prime and hence agree globally. ∎

Lemma. Finite projectivity of a quasi-regular conormal module

If $I$ is a quasi-regular ideal of $R$, then $I/I^2$ is finite projective over $R/I$.

Proof. On the neighborhoods in the definition of a quasi-regular ideal, its chosen generators give a basis of $I/I^2$: this is the degree-one part of the quasi-regularity isomorphism with the associated graded algebra. Thus $I/I^2$ is locally free of finite rank on $\operatorname{Spec}(R/I)$. Quasi-compactness permits a finite such cover. The finite-projective criterion then applies, without requiring that the rank be the same on different connected components. ∎

Lemma. Syntomic algebras and local complete intersections

A ring map $R\to S$ is syntomic if and only if it is flat and a local complete intersection.

Proof. A syntomic map is flat by definition. Its local presentations are relative global complete intersections, by Local criteria for a syntomic algebra. Their defining equations are Koszul-regular by Koszul complexes of global complete intersections. The locality of the complete-intersection condition therefore makes $R\to S$ a local complete intersection.

For the converse, choose a finite polynomial presentation $S=R[x_1,\ldots,x_n]/I$. The local complete-intersection hypothesis says that $I$ is locally generated by finite Koszul-regular sequences around its zero set. Away from that set it is locally the unit ideal. A finite principal-open cover of the polynomial spectrum consequently shows that $I$ is finitely generated, so $S$ is finitely presented over $R$.

It remains to check the fibres. For a field $k$ over $R$ and a prime of $S\otimes_R k$, choose a neighborhood in $R[x]$ where $I$ is generated by a Koszul-regular sequence $f_1,\ldots,f_r$. Write the localized polynomial ring as $Q$ and its quotient as $T$. The ring $Q$ is flat over $R$, and $T$ is flat over $R$ by the assumed flatness of $S$. The Koszul resolution of $T$ consists of finite free $Q$-modules. Its successive kernels are flat over $R$: start with the exact sequence ending in the flat module $T$ and proceed through the resolution, using closure of flat modules under kernels of surjections between flat modules. Tensoring the resolution with $k$ is therefore still exact in positive degrees. Hence the images of the $f_i$ are Koszul-regular in $Q\otimes_R k$.

At the chosen prime this is a sequence in the maximal ideal of a Noetherian local ring. The Noetherian equivalence of regularity conditions makes it a regular sequence. The local complete-intersection criterion over a field gives a complete-intersection neighborhood of this prime in the fibre. Since the prime and field were arbitrary, all fibres are local complete intersections. Together with flatness and finite presentation, this is the syntomic criterion.

Alternatively, the same fibre step follows from Relative regular immersions in affine algebra: it preserves $H_1$-regularity under this base change because the quotient is flat. The same Noetherian local criterion then gives the regular sequence in the fibre. ∎

Lemma. Cotangent transitivity with a complete-intersection terminal map

Let $A\to B\to C$ be ring maps with $B\to C$ a local complete intersection. Choose presentations

whose kernels are $I$ and $J$. Put $Q=P[y_1,\ldots,y_m]$ and $K=\ker(Q\to C)$. The presentation maps give the following commutative diagram with exact rows; all tensor products in the differential row are along the indicated polynomial maps to $C$:

Consequently the cotangent transitivity sequence begins with zero:

Here $\mathrm{NL}_{B/A}\otimes_B C$ means the ordinary tensor product of the two-term presentation complex; the tensor symbol in this term is not being replaced by a derived tensor product.

Proof. Let $I_Q=IQ$. Then $Q/I_Q=B[y]$ and $K/I_Q=J$. The complete-intersection hypothesis makes $J$ Koszul-regular, and therefore $H_1$-regular. The preceding intersection lemma gives

It follows that the usual conormal sequence is left exact, because its left-hand term and kernel identify as

Its cokernel is $J/J^2$. The differential row is split exact by separating the $dx_s$ and $dy_i$ coordinates. Applying universal derivations makes the three squares commute, so (CT1) is proved. Taking homology of this short exact sequence of two-term complexes gives (CT2), including the initial zero. This refines the general cotangent transitivity sequence precisely at its left end. ∎

Lemma. Cotangent transitivity for filtered complete intersections

The conormal diagram and exact sequence just proved remain valid when $C$ is a filtered colimit of local complete-intersection $B$-algebras. Polynomial presentations may use arbitrary sets of variables when finite sets do not suffice.

Proof. Write $C=\varinjlim C_\lambda$. For presentations chosen compatibly with this system, formation of the two-term cotangent complexes and their maps commutes with the filtered colimit; see Filtered colimits of naive cotangent complexes. Filtered colimits of modules are exact, so the preceding result at each stage gives (CT2).

One can also check directly the only potentially missing injection in (CT1), for any chosen polynomial presentation $Q\twoheadrightarrow C$. An element of $I_Q\cap K^2$ is witnessed by a finite expression as a sum of products of elements of $K$. Only finitely many polynomial variables, coefficients in $C$, and relations equal to zero in $C$ occur in this expression. Lift the coefficients to some $C_\lambda$ and move to a later stage where those finitely many relations vanish. Present that stage over $B$ with the selected elements among its polynomial generators. Apply the preceding intersection equality there. Map its additional polynomial generators back to any chosen polynomial representatives of their images in $C$, leaving the original selected variables fixed. Its relation ideal maps into $K$, and its equality expresses the original element as an element of $I_QK$. Hence $I_Q\cap K^2=I_QK$ also in the chosen presentation of $C$. The rest of (CT1) and its homology sequence follows exactly as above. ∎

Lemma. Cartier's equality for differentials (Cartier equality)

For a finitely generated extension of fields $K/k$, both $\Omega_{K/k}$ and $H_1(L_{K/k})$ have finite dimension over $K$, and

Proof. Let $t_1,\ldots,t_d$ be a transcendence basis and write $E_0=k(t_1,\ldots,t_d)$. Choose a finite tower $E_i=E_{i-1}(\alpha_i)$ ending in $E_r=K$. No separability is assumed. At step $i$, express the coefficients of the monic minimal polynomial of $\alpha_i$ as polynomials in $\alpha_1,\ldots,\alpha_{i-1}$ with coefficients in $E_0$. This is possible because a finite algebraic field extension is generated as an algebra by these elements. Lift those coefficient expressions to obtain triangular polynomials $f_i(T_1,\ldots,T_i)$, monic in $T_i$.

A single nonzero $D\in k[t_1,\ldots,t_d]$ clears all coefficients' denominators. Over $R=k[t_1,\ldots,t_d]_D$, form

Each successive quotient is finite free over the preceding one because the next equation is monic. After tensoring with $E_0$ these quotients are exactly the fields $E_i$. Their freeness over the domain $R$ makes the map to this localization injective. Thus $A$ is a domain with fraction field $K$.

As a finitely presented $k$-algebra, $A$ has variables $t_1,\ldots,t_d,U,T_1,\ldots,T_r$ and equations $UD-1,f_1,\ldots,f_r$. The first equation is a nonzerodivisor in the polynomial ring; after taking its quotient, each subsequent monic equation is a nonzerodivisor in the corresponding polynomial variable. This explicitly produces a global complete-intersection model with $d+r+1$ variables and $r+1$ equations.

The conormal module is free on those equations by The conormal module of a global complete intersection. Localize the presentation at the fraction field, using Localization of the naive cotangent complex. The resulting two-term complex is

Its kernel is $H_1(L_{K/k})$ and its cokernel is $\Omega_{K/k}$. Both are finite dimensional, and rank-nullity gives (FC1). Equivalently, the complete-intersection dimension calculation gives $\dim A=d$, by dimension in a polynomial ring, and the same Euler characteristic equals that dimension. ∎

Lemma. Transitivity of first cotangent homology

For any tower of fields $K\subset L\subset M$, the sequence

is exact, including its first zero. The extensions need not be finitely generated.

Proof. Every finitely generated subextension of $M/L$ is the fraction field of a complete-intersection $L$-algebra by the preceding construction. A fraction field is the filtered colimit of localizations at single nonzero elements. Each such localization is again a complete intersection: adjoin a variable $v$ and the equation $vs-1$ after the existing regular sequence. For any two of these subalgebras, choose a finitely generated subfield containing both and then localize its complete-intersection model enough to contain their finite sets of generators. This proves the required directedness under inclusion. Their union is $M$, so $M$ is a filtered colimit of local complete-intersection $L$-algebras. Apply Cotangent transitivity for filtered complete intersections to $K\to L\to M$. Since tensoring an $L$-vector space with $M$ is exact, the leftmost homology term there is $H_1(L_{L/K})\otimes_L M$. This gives (FC2). ∎

Lemma. Compatibility of cotangent homology with a quotient

Consider field inclusions $k\subset K\subset K'$ and $k\subset k'\subset K'$ with the same composite map $k\to K'$. Assume that $k'/k$ and $K'/K$ are finitely generated. For the natural comparison maps

their four kernel and cokernel spaces are finite dimensional over $K'$, and

Proof. Factor both maps through the cotangent spaces for $K'/k$:

These factorizations follow from functoriality of derivations and of the presentation cotangent complex.

For the tower $k\subset K\subset K'$, sequence (FC2) shows that $\beta_1$ is injective and that its cokernel embeds in $H_1(L_{K'/K})$. It also identifies $\operatorname{coker}\alpha_1$ with $\Omega_{K'/K}$ and gives

All these dimensions are finite by (FC1). If $\operatorname{ind}(u)=\dim\ker u-\dim\operatorname{coker}u$ for a map with finite kernel and cokernel, we obtain

For the tower $k\subset k'\subset K'$, sequence (FC2) identifies $\ker\beta_2$ with $H_1(L_{k'/k})\otimes_{k'}K'$ and embeds $\operatorname{coker}\beta_2$ into $\Omega_{k'/k}\otimes_{k'}K'$. The map $\alpha_2$ is surjective, with

Again the spaces involved are finite dimensional, and (FC1) now gives

Finally, finite kernel and cokernel are preserved under composition, and the index is additive. Both assertions follow from the exact sequence

Apply this to the two factorizations and add the index identities to obtain (FC3). This argument never subtracts dimensions of the potentially infinite spaces $\Omega_{K/k}$ or $H_1(L_{K/k})$ themselves. ∎

Proposition. Characterizations of geometric regularity

Let $k$ have characteristic $p>0$, and let $(A,\mathfrak m,K)$ be a Noetherian local $k$-algebra. The following conditions are equivalent:

1. $A$ is geometrically regular over $k$.
2. $A\otimes_k k'$ is regular for every finite intermediate field $k\subset k'\subset k^{1/p}$.
3. $A$ is regular and the boundary $H_1(L_{K/k})\to\mathfrak m/\mathfrak m^2$ is injective.
4. $A$ is regular and the natural map $\Omega_{k/\mathbb F_p}\otimes_k K\to\Omega_{A/\mathbb F_p}\otimes_A K$ is injective.

Proof. We first prove that (3) implies (1). It suffices to test finite purely inseparable extensions $k'/k$, by Criteria for geometric regularity. For such an extension put $A'=A\otimes_k k'$. The finite faithfully flat map $A\to A'$ is a universal homeomorphism, so $A'$ is local and $\dim A'=\dim A=d$. Write its maximal ideal and residue field as $\mathfrak m'$ and $K'$. The extension $K'/K$ is finite.

For this paragraph all vector spaces are over $K'$. Set

The quotient cotangent sequences and base change for differentials give a commutative diagram

Only the upper row is asserted to be left exact at $H$; that is hypothesis (3). In particular $H$ is finite dimensional. The middle isomorphism makes $\alpha$ surjective. The field comparison formula gives finite kernel and cokernel for $\beta$ and gives

Thus $H'$ is also finite dimensional. Since $W\simeq W'$ and $\alpha$ is surjective, the kernels of the maps from these middle spaces fit into

The first kernel has dimension $d-\dim H$. The lower row of (GR1) consequently bounds

The embedding dimension of a Noetherian local ring is at least its Krull dimension. Hence $A'$ has both dimensions equal to $d$ and is regular. This proves the required purely inseparable tests. Notice that no dimension of $W$ or $V$ was assumed finite.

Next compare (3) and (4). Every field over $\mathbb F_p$ is separable, so its first cotangent homology over $\mathbb F_p$ vanishes; see Characterizations of separable field extensions. Apply cotangent transitivity to $\mathbb F_p\to A\to K$ and $\mathbb F_p\to k\to K$. It gives exact rows and their natural vertical maps:

The kernel of the middle vertical map lies in the image of the upper left injection, because the next vertical map is the identity. The lower left injection then identifies it with the kernel of the first vertical map. Thus those kernels are canonically isomorphic. With the shared regularity hypothesis on $A$, (3) and (4) are equivalent.

The implication (2) to (4) follows Faltings's argument, credited in the Stacks Project under the citation key Faltings-einfacher. We now prove (2) implies (4), by adjoining $p$th roots. Taking $k'=k$ first shows that $A$ is regular. Choose $a_1,\ldots,a_n\in k$ such that $da_1,\ldots,da_n$ are linearly independent in $\Omega_{k/\mathbb F_p}$. By Degrees of extensions obtained by adjoining p-th roots,

Set $A'=A\otimes_k k'$ and use the same notation $\mathfrak m',K'$ as above. Both $A$ and $A'$ are regular of dimension $d$. The equations are a regular sequence, successively monic in distinct variables. For the conormal basis choose $g_i=a_i-x_i^p$, the negatives of the displayed equations. Their differentials relative to $A$ vanish. Hence

In transitivity over $\mathbb F_p$, the class of $g_i$ maps to $da_i$ in $\Omega_{A/\mathbb F_p}\otimes_A A'$. This specifies the sign of the boundary with the chosen equations. Since $\Omega_{A'/A}$ is free, the terminal short exact sequence splits and remains exact after tensoring with $K'$. Right exactness on the preceding terms gives

In particular $\dim\operatorname{coker}v=n$.

The quotient cotangent sequences over $\mathbb F_p$ are short exact:

The two spaces for $u$ have dimension $d$, so $u$ has index zero. For $w$, sequence (FC2) for $\mathbb F_p\subset K\subset K'$ and Cartier's equality for the finite extension $K'/K$ show that its kernel and cokernel are finite dimensional of equal dimension. The snake lemma applied to (GR4) therefore gives finite kernel and cokernel for $v$, with index zero. Combining this with (GR3) yields $\dim\ker v=n$. The first map in (GR3) is consequently an isomorphism onto this kernel. In particular the $da_i$ remain independent after mapping into $\Omega_{A/\mathbb F_p}\otimes_A K'$ and hence after mapping into $\Omega_{A/\mathbb F_p}\otimes_A K$.

The differentials $da$, $a\in k$, span $\Omega_{k/\mathbb F_p}$. Every finite-dimensional subspace generated by them has a basis chosen from them. The independence just proved therefore establishes the full injection in (4). Finally (1) immediately implies (2). All four conditions are now equivalent. ∎

Lemma. Geometric regularity over a field

Let $(A,\mathfrak m,K)$ be a Noetherian local algebra geometrically regular over a field $k$ of characteristic $p>0$. Let $k\subset F\subset K$ with $F/k$ finitely generated. Suppose a $k$-algebra map $\varphi:k[y_1,\ldots,y_m]\to A$ has residues $\overline y_i\in F$ whose differentials form an $F$-basis of $\Omega_{F/k}$. If $\mathfrak p=\varphi^{-1}(\mathfrak m)$, then

is flat, and $A/\mathfrak pA$ is regular.

Proof. Write $A_0=k[y]_{\mathfrak p}$, with maximal ideal $\mathfrak m_0$ and residue field $K_0=k(\overline y_1,\ldots,\overline y_m)\subset F$. The natural surjection

is an isomorphism. Indeed, after extending scalars to $F$ and mapping to $\Omega_{F/k}$, its coordinate vectors become the given basis, so its kernel is zero. The regular local ring $A_0$ is geometrically regular over $k$. The proposition and the quotient cotangent sequence therefore identify

In the natural square

the left arrow is injective by field cotangent transitivity, and the bottom arrow is injective by geometric regularity of $A$. Thus the right arrow is injective.

A regular system of parameters of $A_0$ consequently maps to linearly independent cotangent vectors in the regular local ring $A$. Extend those vectors to a basis of $\mathfrak m/\mathfrak m^2$; any lifts form a regular system of parameters of $A$. The chosen images are therefore an $A$-regular sequence, and their quotient is regular. The flatness criterion over a regular local ring, together with regular rings being Cohen–Macaulay, proves flatness of $A_0\to A$. Its parameter ideal is $\mathfrak pA$, so the quotient just considered is precisely $A/\mathfrak pA$. ∎

Lemma. Base change of regular ring maps (Regular maps and base change)

If $R\to\Lambda$ is regular and $R'$ is a finite type $R$-algebra, then $R'\to\Lambda\otimes_R R'$ is regular.

Proof. Flatness survives the base change. For $\mathfrak p'\in\operatorname{Spec}R'$ over $\mathfrak p\in\operatorname{Spec}R$, its fibre is

The first factor is Noetherian and geometrically regular by hypothesis. The residue extension is finitely generated, since $R'/R$ is finite type. The fibre is consequently Noetherian by Noetherianity under extension of the ground field. Every further finitely generated field extension of $\kappa(\mathfrak p')$ is still finitely generated over $\kappa(\mathfrak p)$, so the same fibre is geometrically regular by the field criterion. This checks every fibre and proves regularity of the base-changed map. ∎

Lemma. Composition of regular ring maps (Composition of regular maps)

For regular maps $A\to B\to C$, the composite is regular provided all its fibre rings are Noetherian.

Proof. The composite is flat. Fix $\mathfrak p\in\operatorname{Spec}A$ and a finite purely inseparable extension $k/\kappa(\mathfrak p)$. Put

The ring $B'$ is Noetherian and regular by the geometric regularity of the fibre of $B/A$. The ring $C'$ is Noetherian by the assumed Noetherianity of the composite fibre and the finite field extension. The map $B'\to C'$ is flat.

Its fibres are regular as well. In fact, for a prime $\mathfrak q'$ of $B'$ contracting to $\mathfrak q$ of $B$, the residue extension $\kappa(\mathfrak q')/\kappa(\mathfrak q)$ is finite. To see this, obtain $B'$ by first passing to $B\otimes_A\kappa(\mathfrak p)$, a localization of $B/\mathfrak pB$, and then making the finite extension $k/\kappa(\mathfrak p)$. The first operation preserves the residue field at the corresponding prime; the second gives a finite ring extension. Thus

is regular by the regular-map hypothesis on $B\to C$.

The flat regular-base and regular-fibre criterion now makes $C'$ regular. Since these tests cover every finite purely inseparable extension of every composite residue field, the composite fibres are geometrically regular. Together with flatness and their stated Noetherianity, this proves the claim. The argument does not assume that $A\to\kappa(\mathfrak p)$ is of finite type. ∎

Lemma. Permanence of regular ring maps

Suppose $A\to C$ is regular and $B\to C$ is faithfully flat, where $A\to B\to C$ is a factorization. Then $A\to B$ is regular. Equivalently, it is enough that $B\to C$ be flat and surjective on spectra.

Proof. Flatness of $A\to B$ follows by faithfully flat descent from flatness of $A\to C$; see Permanence of flat ring maps. At each $\mathfrak p\in\operatorname{Spec}A$ the fibre map

is faithfully flat. Its target is Noetherian, so its source is Noetherian by descent of Noetherianity. After any finite purely inseparable extension of $\kappa(\mathfrak p)$, the fibre map remains faithfully flat and its target is regular. Descent of regularity gives regularity of the corresponding source. The field criterion proves that the original source fibre is geometrically regular. This is also the fibrewise descent of geometric regularity. ∎

B.10.2. Commutative Algebra

These algebraic facts supply the matrix, flatness and finite-generation steps used above. All ring maps preserve the identity, and a regular sequence is required to generate a proper ideal.

Lemma. A left inverse for a matrix

Let $A$ be an $n\times m$ matrix over a ring $R$, with $n\geq m$, and let $J$ be the ideal of its maximal minors. Then:

1. Every $f\in J$ admits an $m\times n$ matrix $B$ with $BA=fI_m$.
2. Conversely, such an identity implies $f^m\in J$.

Proof. For each set $E$ of $m$ row indices, let $p_E:R^n\to R^m$ be the coordinate projection in increasing index order. Put $A_E=p_EA$. The adjugate identity gives

Write $f=\sum_E c_E\det A_E$. Then $B=\sum_E c_E\operatorname{adj}(A_E)p_E$ has the property in (1).

For (2), take determinants in $BA=fI_m$ and use Cauchy–Binet:

where $B^E$ selects the columns with indices in $E$. This lies in $J$. If $m=0$, the empty minor is $1$, so $J=R$; the empty matrices and the convention $f^0=1$ make both assertions valid. ∎

Lemma. Flatness over a regular local ring

Let $R\to S$ be a homomorphism of Noetherian local rings. Suppose $R$ is regular and a regular system of parameters $x_1,\ldots,x_d$ of $R$ maps to an $S$-regular sequence. Then $S$ is flat over $R$.

Proof. The last quotient $S/(x_1,\ldots,x_d)S$ is flat over the field $R/(x_1,\ldots,x_d)$. Now lift flatness one parameter at a time. At step $i$, multiplication by $x_i$ is injective on both $R/(x_1,\ldots,x_{i-1})$ and $S/(x_1,\ldots,x_{i-1})S$, and the quotient of the latter by $x_i$ is already flat over the corresponding quotient of the former. The local criterion for flatness therefore gives flatness before that last quotient. Apply this for $i=d,d-1,\ldots,1$.

All these quotient rings are Noetherian local, and the parameters lie in their maximal ideals: properness of the regular-sequence ideal in the local ring $S$ ensures this on the target. Thus the local criterion applies at every step. For $d=0$, the initial field argument is already the conclusion. ∎

Lemma. Regular sequences in a polynomial ring

Let $f_1,\ldots,f_r\in R$ generate a proper ideal. The following conditions are equivalent:

1. Every permutation of the sequence is regular.
2. Every subsequence, in its original order, is regular.
3. The sequence $f_1x_1,\ldots,f_rx_r$ is regular in $R[x_1,\ldots,x_r]$.

Proof. Under (1), place any chosen subsequence first in a permutation and take that initial segment. This proves (2).

To recover (1) from (2), first note a two-element interchange rule. If $a,b$ are nonzerodivisors and $a,b$ is regular, then $b,a$ is regular. Indeed, if $au=bv$, reduction modulo $a$ gives $v=aw$ because $b$ is a nonzerodivisor modulo $a$. Cancelling $a$ in $a(u-bw)=0$ gives $u=bw$. Thus $a$ is a nonzerodivisor modulo $b$.

Proceed by induction on $r$. All proper subsequences may be permuted by induction. To put $f_s$ last, only its injectivity modulo the other terms remains to be checked. If $s=r$, this is already known. Otherwise quotient by the terms with indices outside $\{s,r\}$. Induction shows that both remaining terms are nonzerodivisors there, and that they occur as the regular pair $f_s,f_r$: permute the first $r-1$ terms to put $f_s$ last among them, without changing their generated ideal. The interchange rule makes $f_r,f_s$ regular in that quotient. This proves injectivity of the desired final term. Any order of its predecessors is regular by induction, proving (1).

For the polynomial assertion, after quotienting by the first $i$ elements $f_jx_j$, the underlying $R$-module splits by monomials:

Multiplication by $f_{i+1}x_{i+1}$ sends the summand for $e$ to that for $e+(0,\ldots,1,\ldots,0)$, by multiplication by $f_{i+1}$ on its coefficient module. The coefficient ideal depends only on the first $i$ exponents and hence is unchanged. Distinct source monomials have distinct targets. The multiplication is therefore injective exactly when $f_{i+1}$ is a nonzerodivisor modulo every ideal generated by a subset of $f_1,\ldots,f_i$. These are precisely the successive injectivity conditions for all ordered subsequences in (2). Every such subset occurs by taking its exponents positive and the others zero. This proves (2) iff (3). The relevant quotient ideals are proper, by the hypothesis in $R$ and by evaluation at $x_1=\cdots=x_r=0$ in the polynomial ring. ∎

Lemma. Finite generation in an Artinian local target

Let $R\to S$ be a ring map, with $S$ Artinian local and residue field $K=S/\mathfrak m$. The map to $S$ is finite, finite type, or essentially finite type, respectively, if and only if the composite map to $K$ has that property.

Proof. Each of the three properties passes to a quotient, so consider the converse directions.

If $K$ is a finite $R$-module, take a composition series of $S$ as an $S$-module. Such a finite series exists by Finite length over an Artinian ring, and every factor is $K$. Viewed as a series of $R$-modules, it shows that $S$ is finite over $R$: extensions of finite modules are finite.

If $K$ is a finite type $R$-algebra, lift a finite set of algebra generators to $s_1,\ldots,s_n\in S$. The map $R[X_1,\ldots,X_n]\to K$ defined by their residues is surjective. The finite-module case applied to $R[X]\to S$ makes $S$ finite over this polynomial ring, and hence finite type over $R$.

Finally, suppose $K$ is a localization of a finite type $R$-algebra. Choose lifts to $S$ of a finite generating set for the image of that algebra in $K$, and let $A\subset S$ be the $R$-subalgebra they generate. Put $\mathfrak p=A\cap\mathfrak m$. Then $A/\mathfrak p$ is that same image, and its fraction field is $K$: the given localization is already a field. Elements of $A\setminus\mathfrak p$ map to units of $S$, so the inclusion extends to $A_{\mathfrak p}\to S$. Its map to $K$ is surjective, and the first case makes $S$ finite over $A_{\mathfrak p}$. Since $A$ is finite type over $R$, composition of essentially finite-type maps now shows that $S/R$ is essentially finite type. No finite-generation assumption on the ring $R$ is used. ∎

Lemma. Syntomic algebras in a filtered colimit

Any field extension $K/k$ is a filtered colimit of global complete-intersection $k$-algebras. If the extension is separable, smooth $k$-algebras suffice.

Proof. We construct such subalgebras containing any given finite subset $E\subset K$. Let $L=k(E)$. The triangular monic construction in the proof of Cartier's equality gives a domain $A\subset L$ that is a global complete intersection over $k$ and has fraction field $L$. Express each element of $E$ as a fraction in $A$, and invert the product of its finitely many nonzero denominators. The resulting algebra contains $E$. It remains a global complete intersection: for a localization at $s$, adjoin a variable $v$ and append the equation $vs-1$ to the regular sequence presenting $A$.

In the separable case, the finitely generated extension $L/k$ has a separating transcendence basis $t_1,\ldots,t_d$, and $L/k(t)$ is finite separable. Choose a primitive element $\alpha$ with monic minimal polynomial $f(T)\in k(t)[T]$. Clear its coefficients' denominators by a nonzero $D\in k[t]$. The ring $A=k[t]_D[T]/(f)$ embeds in $L$ and has fraction field $L$. Since $f'(\alpha)\ne0$, localizing $A$ at $f'(\alpha)$ makes it étale over the smooth algebra $k[t]_D$ and therefore smooth over $k$. Further localization at the finitely many denominators of the elements of $E$ preserves smoothness and puts $E$ in the algebra. This is the explicit construction behind Smooth localizations of separable extensions.

The family is directed under inclusion: for two constructed finite type subalgebras, apply the appropriate construction to the union of their finite generating sets. Its union contains every element of $K$ and is contained in $K$, so its filtered colimit is exactly $K$. ∎

B.11. The additional G-ring permanence argument

Artin approximation over the local polynomial algebra requires permanence of the G-ring condition under essentially finite-type extensions. We prove that permanence through formal-fibre calculations, beginning with the reductions used to test those fibres.

G-rings

For a Noetherian local ring $A$ and a prime $\mathfrak q\subset A$, the formal fibre at $\mathfrak q$ is the $\kappa(\mathfrak q)$-algebra

Here completion uses the maximal ideal of $A$. Completion of a Noetherian ring commutes with quotient by a finitely generated ideal, so this is also

The completion map is flat and its fibres are Noetherian. It is therefore regular exactly when these formal fibres are geometrically regular over their indicated residue fields. In particular, a formal fibre is generally a localization of a quotient of $\widehat A$; the quotient cannot be omitted when $\mathfrak q\ne0$.

Definition. G-rings

A G-ring is a Noetherian ring $R$ such that, for every prime $\mathfrak p$, the local completion map

is regular. Equivalently, every local ring of $R$ has geometrically regular formal fibres. If $R$ contains $\mathbb Q$, geometric regularity in this test is equivalent to regularity, since all the residue fields have characteristic zero; apply the field criterion and regularity under separable field extension.

Lemma. Recognizing a G-ring from a completion

A Noetherian ring $R$ is a G-ring if and only if, for every chain $\mathfrak q\subset\mathfrak p$ of primes, the ring

is geometrically regular over $\kappa(\mathfrak q)$. The completion on the left is at the maximal ideal of $(R/\mathfrak q)_{\mathfrak p}$.

Proof. Quotient compatibility of Noetherian completion gives

Localizing away from $\mathfrak q$ identifies the stated ring with $\widehat{R_{\mathfrak p}}\otimes_R\kappa(\mathfrak q)$. As $\mathfrak q$ runs through the primes contained in $\mathfrak p$, these are exactly the fibres of the completion map of $R_{\mathfrak p}$. The definition now gives both implications. ∎

Lemma. G-rings under quasi-finite extensions

Suppose $R\to R'$ is a finite type map of Noetherian rings. Let $\mathfrak q'\subset\mathfrak p'$ be primes of $R'$ contracting to $\mathfrak q\subset\mathfrak p$ in $R$, and assume the map is quasi-finite at $\mathfrak p'$. Then:

1. Geometric regularity of $\widehat{R_{\mathfrak p}}\otimes_R\kappa(\mathfrak q)$ over $\kappa(\mathfrak q)$ implies geometric regularity of $\widehat{R'_{\mathfrak p'}}\otimes_{R'}\kappa(\mathfrak q')$ over $\kappa(\mathfrak q')$.
2. If every formal fibre of $R_{\mathfrak p}$ is geometrically regular, the same holds for $R'_{\mathfrak p'}$.
3. If $R\to R'$ is quasi-finite everywhere and $R$ is a G-ring, then $R'$ is a G-ring.

Proof. By Completion at a quasi-finite prime, the completed local ring at $\mathfrak p'$ is a direct factor of the indicated base change:

Tensor this decomposition, as one of $R'$-algebras, with $\kappa(\mathfrak q')$. Its left-hand side becomes

The residue extension is finitely generated, so this is Noetherian and geometrically regular under the hypothesis of (1), by Noetherianity under field extension and the geometric-regularity criterion. A direct factor has the same property, since it is localization at an idempotent, before and after every finite field extension. This proves (1). Applying it to every prime $\mathfrak q'\subset\mathfrak p'$ proves (2), and then applying (2) at every $\mathfrak p'$ proves (3). ∎

Lemma. Testing geometric regularity of formal fibres

For a Noetherian ring $R$, the G-ring condition is equivalent to the following test: every finite free $R$-algebra $S$ has regular formal fibres at all its local rings.

Proof. If $R$ is a G-ring, a finite free algebra is a finite, hence quasi-finite, algebra over $R$. The preceding lemma makes $S$ a G-ring, so its formal fibres are geometrically regular and in particular regular.

Conversely, assume the test. Fix primes $\mathfrak q\subset\mathfrak p$ of $R$ and a finite purely inseparable extension $L/\kappa(\mathfrak q)$. By Finite free algebras with a prescribed residue extension, choose a finite free $R$-algebra $R'$ for which $\mathfrak q'=\mathfrak qR'$ is prime and $\kappa(\mathfrak q')\simeq L$ over $\kappa(\mathfrak q)$.

The finite-extension completion decomposition gives

with a finite index set; see Completion of a finite ring extension. After tensoring over $R'$ with $\kappa(\mathfrak q')=L$, its left side is

On the right, a factor is zero unless $\mathfrak q'\subset\mathfrak p_i'$: otherwise some element of $\mathfrak q'$ is already invertible in $R'_{\mathfrak p_i'}$ and is killed by the tensor product. Each remaining factor is the formal fibre of $R'_{\mathfrak p_i'}$ at $\mathfrak q'R'_{\mathfrak p_i'}$, hence regular by the test. The finite product is therefore regular. Every finite purely inseparable extension $L$ has now been checked, so the field criterion makes the original formal fibre geometrically regular. Apply the preceding characterization for all $\mathfrak q\subset\mathfrak p$ to conclude that $R$ is a G-ring. ∎

Lemma. Geometric regularity of generic formal fibres in positive characteristic

For a field $k$ of characteristic $p>0$, put $A=k[[x_1,\ldots,x_n]][y_1,\ldots,y_m]$ and $K=\operatorname{Frac}A$. For every prime $\mathfrak p$ of $A$, the generic formal fibre $\widehat{A_{\mathfrak p}}\otimes_A K$ is geometrically regular over $K$.

Proof. For a finite purely inseparable extension $L/K$, we prove regularity of $\widehat{A_{\mathfrak p}}\otimes_A L$ by induction on its degree. At degree one, $A$ is regular, its local ring and completion are regular, and the ring in question is a localization of that completion. Use Regularity and completion.

For the induction step choose $K\subset M\subset L$ with $[L:M]=p$, and write $L=M(z)$ with $z^p=f\in M\setminus M^p$. Choose a finite domain $B\subset M$ over $A$, with fraction field $M$, such that a $p$-power of every element of $B$ belongs to $A$. Here is a construction: take finitely many generators $\alpha_i$ of $M/K$, choose powers $\alpha_i^{p^{e_i}}=a_i/b_i$ in $K$, and put $B=A[b_i\alpha_i\mid i]$. Each displayed generator has a $p$-power in $A$, so it is integral; a common Frobenius power works for every polynomial in them. The resulting finite algebra has fraction field $M$.

Write $f=b/c$ with $b,c\in B$, $c\ne0$. Replacing $z$ by $cz$ replaces $f$ by $c^pf=c^{p-1}b\in B$ and gives the same extension $L/M$. The new $f$ is still not a $p$th power in $M$. Thus we may assume $f\in B$. Put $C=B[z]/(z^p-f)\subset L$. It is a finite domain over $B$, and every element of $C$ also has a $p$-power in $A$. These integral radicial extensions have unique primes $\mathfrak r\subset B$ and $\mathfrak q\subset C$ above $\mathfrak p$.

The finite-extension completion formula and uniqueness of these primes identify

The induction hypothesis says that $T$ is regular. The domain $B$ is finite type over the complete local ring $k[[x_1,\ldots,x_n]]$. The derivation lemma in positive characteristic supplies a derivation $D:B\to B$ with $D(f)\ne0$. By Extending a derivation, it extends first to $B_{\mathfrak r}$, then to its completion, and finally to $T$. In the last ring, the nonzero element $D(f)\in B$ is invertible because we have tensored with the fraction field $M$.

Extend $D$ to $T[z]$ by $D(z)=0$. It sends $z^p-f$ to the unit $-D(f)$. The regular-quotient criterion by a derivation therefore makes $T[z]/(z^p-f)$ regular. This completes the induction. The purely inseparable field criterion proves geometric regularity of the original generic formal fibre. ∎

Proposition. Complete Noetherian rings are G-rings

Every Noetherian complete local ring is a G-ring.

Proof. Let $A$ be such a ring and fix $\mathfrak q\subset\mathfrak p$ in $\operatorname{Spec}A$. By quotient compatibility of completion, the formal fibre to be tested is the generic formal fibre of the complete local domain $B=A/\mathfrak q$ at $\mathfrak p/\mathfrak q$. Thus it suffices to establish geometric regularity of all generic formal fibres of every complete local domain.

For such a domain $B$, the complete-domain structure theorem gives a regular complete local subring $B_0\subset B$, finite in $B$, where $B_0$ is a power-series ring over a field or over a Cohen ring. The generic prime of $B$ contracts to the generic prime of $B_0$. Part (1) of G-rings under quasi-finite extensions, applied at each chosen upper prime, therefore transfers geometric regularity of the generic formal fibres of $B_0$ to those of $B$. This use requires only the indicated generic fibres of $B_0$, not the whole G-ring assertion for $B_0$.

Let $K_0=\operatorname{Frac}B_0$. The ring $\widehat{(B_0)_{\mathfrak p_0}}\otimes_{B_0}K_0$ is regular: it is a localization of the completion of a regular local ring. If $K_0$ has characteristic zero, this is already geometric regularity. If its characteristic is positive, $B_0$ is a power-series ring over a field, and the preceding lemma, with no polynomial variables, proves geometric regularity. Every required generic fibre is now covered. Returning through $B=A/\mathfrak q$ proves the G-ring condition for all pairs $\mathfrak q\subset\mathfrak p$ of the original $A$. ∎

Lemma. Testing the G-ring property at maximal ideals

A Noetherian ring $R$ is a G-ring if and only if every maximal localization $R_{\mathfrak m}$ has geometrically regular formal fibres.

Proof. The forward implication is part of the definition. For the converse fix $\mathfrak p\subset\mathfrak m$, with $\mathfrak m$ maximal, and write $S=\widehat{R_{\mathfrak m}}$. Faithful flatness of completion supplies a prime $\mathfrak p'\subset S$ over $\mathfrak pR_{\mathfrak m}$. Put $T=S_{\mathfrak p'}$ and $D=\widehat T$.

By hypothesis $R_{\mathfrak m}\to S$ is regular. The localization $S\to T$ is regular, and $T\to D$ is regular because the preceding proposition makes the complete local ring $S$ a G-ring. Composition gives a regular map $R_{\mathfrak m}\to D$; its fibres are Noetherian since $D$ is Noetherian. The map factors through $R_{\mathfrak p}$, and its fibres and flatness at primes of that localization are unchanged. Hence $R_{\mathfrak p}\to D$ is regular.

There is a compatible local map $\widehat{R_{\mathfrak p}}\to D$. We verify that it is faithfully flat before applying descent. More generally, if $U\to V$ is a flat local map of Noetherian local rings and $W=\widehat V$, then the induced map $\widehat U\to W$ is flat. Indeed $W$ is flat over $U$. A free resolution of the residue field $\kappa_U$ over $U$, tensored with the flat ring $\widehat U$, is a free resolution of the same residue field over $\widehat U$. Consequently

Apply the local flatness criterion to the finite $W$-module $W$, the local map $\widehat U\to W$, and the maximal ideal of $\widehat U$. The residue quotient is a vector space over $\kappa_U$ and hence flat, so $W$ is flat over $\widehat U$. As a local flat map of nonzero local rings, the map is faithfully flat.

Use this observation with $U=R_{\mathfrak p}$ and $V=T$. The original map $U\to T$ is flat and local, as follows from the completion and localization construction. Thus $\widehat{R_{\mathfrak p}}\to D$ is faithfully flat. Permanence of regular ring maps applied to

now proves regularity of the first map. Since $\mathfrak p$ was arbitrary, $R$ is a G-ring. ∎

Lemma. The G-ring property under henselization

If $R$ is a Noetherian local G-ring, both its henselization $R^h$ and its strict henselization $R^{sh}$ are G-rings.

Proof. Let $H$ be either henselization and put $D=\widehat H$. The Noetherian henselization theorem says that $H$ is Noetherian local. In the ordinary case it identifies $D$ with $\widehat R$. In the strict case it gives a formally smooth local map $\widehat R\to D$, for the maximal-ideal topology on $D$. By Formal smoothness and regularity, this map is regular. Since $R\to\widehat R$ is regular, composition shows in both cases that $R\to D$ is regular; the required Noetherian fibre condition holds because $D$ is Noetherian.

Fix $\mathfrak p\in\operatorname{Spec}R$. The henselization fibre description gives a finite product

where the $\mathfrak q_i$ are the primes of $H$ over $\mathfrak p$ and each residue extension is separable algebraic. Tensoring this equality over $H$ with $D$ yields

The left-hand side is geometrically regular over $\kappa(\mathfrak p)$, since $R\to D$ is regular. Each factor has that same property. The coefficient-field change from $\kappa(\mathfrak p)$ to the separable algebraic field $\kappa(\mathfrak q_i)$ preserves geometric regularity in this situation, by Geometric regularity under separable algebraic extensions. Thus every formal fibre of the local ring $H$ is geometrically regular over its own residue field. The maximal-ideal test proves that $H$ is a G-ring, in both cases. ∎

Lemma. Geometric regularity of polynomial formal fibres in positive characteristic

Let $A$ be a Noetherian complete local domain whose fraction field $K$ has characteristic $p>0$. Let $\mathfrak q\subset A[x]$ be maximal, with contraction the maximal ideal of $A$. If $0\ne\mathfrak r\subset\mathfrak q$ and $\mathfrak r\cap A=0$, then

is geometrically regular over $\kappa(\mathfrak r)$.

Proof. Localizing by the nonzero elements of $A$ sends $\mathfrak r$ to a nonzero prime of the principal ideal domain $K[x]$. Thus $\kappa(\mathfrak r)$ is finite over $K$. Fix a finite purely inseparable extension $L/\kappa(\mathfrak r)$; we will show that $F\otimes_{\kappa(\mathfrak r)}L$ is regular.

First choose a domain $B$, finite over $A$, contained in $L$, with fraction field $L$. To construct it, take finitely many algebraic generators of $L/K$ and multiply each by a nonzero element of $A$ that clears the denominators of its monic equation. The scaled generators are integral over $A$ and still generate $L$ as a field. Their $A$-algebra is the required $B$. A finite algebra over a complete Noetherian local ring is a product of complete local rings; because $B$ is a domain, it has just one factor. Hence $B$ is itself complete local.

Map $B[x]$ to $L$ by sending $x$ to its image in $\kappa(\mathfrak r)\subset L$, and denote the kernel by $\mathfrak r'$. The fraction field of the image is $L$, so $\kappa(\mathfrak r')=L$. Moreover $\mathfrak r'\cap B=0$ and $\mathfrak r'\ne0$: over $\operatorname{Frac}B=L$ its ideal is generated by a linear polynomial. Write $\mathfrak q_i$ for the finitely many primes of $B[x]$ above $\mathfrak q$. By Completion of a finite ring extension,

Terms with $\mathfrak r'\not\subset\mathfrak q_i$ vanish, since some element killed in $\kappa(\mathfrak r')$ is already invertible in that factor. Every remaining $\mathfrak q_i$ is maximal and lies over the maximal ideal of $B$. Thus it suffices to prove ordinary regularity in the special case $K=\kappa(\mathfrak r)$: the replacement $(B,\mathfrak q_i,\mathfrak r')$ has precisely that property.

In this special case $\mathfrak rK[x]=(x-f)$ for some $f\in K$. Set

The map $A\to A[x]_{\mathfrak q}$ is formally smooth: polynomial algebras have the lifting property, and localization preserves it because a lift of a unit modulo a nilpotent ideal is a unit. Its composite with completion is formally smooth for the maximal-ideal topology, by Formal smoothness and completion. Formal smoothness and regularity makes this composite regular; hence its generic fibre $T$ is regular.

The $A$-derivation $\partial/\partial x$ extends through localization and completion and then to $T$, by Extending a derivation. It kills $K$ and sends $x-f$ to $1$. Therefore Regularity of a quotient shows that

is regular. Applying this argument to all the surviving factors above proves regularity after every finite purely inseparable extension $L$. The field criterion now gives the asserted geometric regularity. ∎

Proposition. The G-ring property for finite-type algebras

If $R$ is a G-ring and $S$ is an essentially finite-type $R$-algebra, then $S$ is a G-ring.

Proof. A localization of a G-ring satisfies the definition, since its local rings are local rings of the original ring. Quotients preserve the G-ring property by G-rings under quasi-finite extensions. Presenting a finite-type algebra as a quotient of a polynomial algebra, and then adding its variables one at a time, reduces the assertion to $S=R[x]$.

By Testing the G-ring property at maximal ideals, fix a maximal $\mathfrak q\subset R[x]$ and prove regularity of the completion map of $T=R[x]_{\mathfrak q}$. Put $\mathfrak p=\mathfrak q\cap R$. Replacing $R$ by $R_{\mathfrak p}$ does not change $T$; thus $R$ is now local, with maximal ideal $\mathfrak m=\mathfrak pR_{\mathfrak p}$, and $\mathfrak q$ lies over $\mathfrak m$. In $\widehat R[x]$ there is a unique prime $\mathfrak q'$ over $\mathfrak q$: modulo $\mathfrak m$ the extension is the identity on $(R/\mathfrak m)[x]$. This prime is maximal. Put $T'=\widehat R[x]_{\mathfrak q'}$.

The map $T\to T'$ is regular, since it is obtained from the regular map $R\to\widehat R$ by polynomial base change and localization. It is also local. The induced map $\widehat T\to\widehat{T'}$ is faithfully flat, by the completion argument proved in the maximal-ideal test. Consequently, if $T'\to\widehat{T'}$ is regular, composition and faithfully flat descent of regularity give regularity of $T\to\widehat T$. We have reduced the problem to a complete Noetherian local base $R$.

Fix a prime $\mathfrak r\subset\mathfrak q$. We must prove geometric regularity of

Let $\mathfrak p=\mathfrak r\cap R$. Quotient compatibility of completion identifies this ring with the corresponding formal fibre for $(R/\mathfrak p)[x]$, at the images of $\mathfrak r$ and $\mathfrak q$; its residue field is unchanged. We may therefore assume that $R$ is a complete local domain and $\mathfrak r\cap R=0$.

Choose a regular complete local subring $R_0\subset R$, finite in $R$, using A complete local domain finite over a regular ring. Contraction sends our chain to $\mathfrak r_0\subset\mathfrak q_0\subset R_0[x]$, with $\mathfrak r_0\cap R_0=0$ and $\mathfrak q_0$ maximal over the maximal ideal of $R_0$. Part (1) of the quasi-finite extension lemma transfers geometric regularity of this particular formal fibre of $R_0[x]$ to the desired fibre of $R[x]$. Thus we need only treat a regular complete local base. This reduction uses the stated fibre-transfer result, without assuming the G-ring property for the polynomial algebra being proved.

For such a base, put $K=\operatorname{Frac}R$ and $C=\widehat{R[x]_{\mathfrak q}}$. The ring $R[x]$ is regular, as are its local ring at $\mathfrak q$ and the completion $C$; use Regularity ascends along a regular ring map and Regularity and completion. There are the following cases.

If $\operatorname{char}K=0$ and $\mathfrak r=0$, the formal fibre is a localization of $C$, hence regular and geometrically regular over the characteristic-zero field $K(x)$.

If $\operatorname{char}K=0$ and $\mathfrak r\ne0$, write $\mathfrak rK[x]=(f)$ for a monic irreducible polynomial $f\in K[x]$. Then $\kappa(\mathfrak r)=K[x]/(f)$, and the formal fibre is

The ambient ring $C\otimes_R K$ is regular. Extend $\partial/\partial x$ from $R[x]$ to it by localization and completion. Since $f$ is separable, its derivative has nonzero image in the field $K[x]/(f)$, so this derivative becomes a unit in the displayed quotient. The regular-quotient criterion proves regularity of the fibre. Its coefficient field has characteristic zero, which gives geometric regularity. This quotient argument is essential: for nonzero $\mathfrak r$ the formal fibre is not merely a localization of $C$.

Finally suppose $\operatorname{char}K=p>0$. The regular complete local ring has the form $R=k[[x_1,\ldots,x_n]]$. If $\mathfrak r=0$, apply Geometric regularity of generic formal fibres in positive characteristic, with the one polynomial variable $x$. If $\mathfrak r\ne0$, apply the preceding lemma on polynomial formal fibres. Thus all the formal fibres of $T$ are geometrically regular. The completion map is flat, so it is regular. Returning through the reductions proves the proposition. ∎

Remark. Failure of the G-ring property under completion

The G-ring property does not persist under completion with respect to an arbitrary ideal. In particular, there are a G-ring $R$ and an ideal $I$ for which the $I$-adic completion is not a G-ring. A construction is given by Jun-ichi Nishimura, On ideal-adic completion of Noetherian rings, J. Math. Kyoto Univ. 21 (1981), no. 1, 153–169. The final section of Tiberiu Dumitrescu, On some examples of atomic domains and of $G$-rings, Comm. Algebra 28 (2000), no. 3, 1115–1123, develops this example further. These are references for the counterexamples; no counterexample construction is supplied in this remark.

Proposition. Examples and permanence of G-rings

Each of the following is a G-ring:

1. a field;
2. a Noetherian complete local ring;
3. $\mathbb Z$;
4. a Dedekind domain whose fraction field has characteristic zero;
5. a finite-type algebra over any ring in the preceding classes.

Proof. A field is a complete Noetherian local ring. The second assertion was proved in Complete Noetherian rings are G-rings.

For a Dedekind domain $R$ as in (4), localization at the zero prime is a field. At a nonzero prime it is a discrete valuation ring $V$, and $\widehat V$ is again a discrete valuation ring. The closed formal fibre is the residue field of $V$, over itself. The generic formal fibre is $\operatorname{Frac}(\widehat V)$ over $\operatorname{Frac}V$; it is geometrically regular because the latter field has characteristic zero. Both formal fibres are therefore geometrically regular, proving (4). The case $\mathbb Z$ follows since it is such a Dedekind domain. The finite-type assertion is the preceding proposition. ∎

Additional proofs of the supporting constructions

The following results give the supporting arguments for the approximation theorem. Where a complete proof already appears earlier in this lesson, the reference identifies that result and explains the match of hypotheses and conclusions. The remaining supporting units retain their individual source references.

Versality and algebraicity criteria

Lemma. Approximation of a marked family with its associated graded algebra

Let $S$ be locally Noetherian and let $\mathcal X\to(\mathrm{Sch}/S)_{fppf}$ be a category fibred in groupoids. Suppose $x\in\mathcal X(R)$, where $(R,\mathfrak m_R)$ is a complete Noetherian local ring and its residue field $k$ defines a finite-type morphism $\operatorname{Spec}k\to S$. Denote the image point by $s$. Assume that $\mathcal O_{S,s}$ is a G-ring and that $\mathcal X$ is limit preserving on objects.

For every $N\geq1$ there are an $S$-algebra $A$ of finite type, a maximal ideal $\mathfrak m_A$, an object $x_A\in\mathcal X(A)$, and an $S$-algebra isomorphism

under which the restricted objects $x$ and $x_A$ are isomorphic. One can require, in addition, an isomorphism of graded $k$-algebras

using the residue-field identification induced by the first map.

Proof. This is Theorem 6.1 in B.6.1, whose full proof applies to precisely this category, base, object and integer $N$. To identify every step of the construction: it first descends $x$ to a finite-type algebra $C$ on an affine chart $\operatorname{Spec}\Lambda\subset S$. Residue-field generators and generators of $\mathfrak m_R$ then give a surjection

The successive-order lifting argument in that proof establishes this surjection directly. Finite generators of its kernel and of their relation module turn the presentation and the map from $C$ into one finite polynomial system.

The G-ring theorem and pointed approximation solve this system in an étale neighbourhood, retaining its exact relations and matching the formal coefficients to any chosen sufficiently high order. Appendix A, Lemmas A.1–A.2 then give the associated graded algebra comparison. Finally the last two paragraphs of B.6.1 descend the algebra and object to a finite-type $\Lambda$-algebra, prove that the selected point is maximal with residue field exactly $k$, and carry back both the finite-order marking and the graded isomorphism. Thus none of the six requested data is lost in the passage from the essentially finite-type intermediate ring to $A$. ∎

Commutative algebra and regularity

Definition. Smoothness at a prime ideal

For a homomorphism $R\to S$ and $\mathfrak q\in\operatorname{Spec}S$, smoothness at $\mathfrak q$ means that $R\to S_g$ is smooth for some $g\in S\setminus\mathfrak q$. Thus the definition requires a smooth principal neighbourhood of the point.

Lemma. Localization of the naive cotangent complex

For $A\to B$ and a multiplicative subset $U\subset B$, localization gives a quasi-isomorphism

Proof. The principal localizations $B_u$, indexed by $u\in U$ with common later terms obtained by taking products, have colimit $U^{-1}B$. For each $u$, The cotangent complex of a principal localization gives the required comparison over $B_u$. Take their filtered colimit. Tensor product commutes with this colimit on the left; Filtered colimits of naive cotangent complexes identifies the right-hand side. Filtered colimits of modules are exact, so they preserve the homology isomorphisms of these two-term complexes. This proves the assertion for any multiplicative set, without a finiteness hypothesis on it. ∎

Definition. Smooth ring maps

A homomorphism $R\to S$ is smooth if it is finitely presented and its naive cotangent complex has zero first homology and finite projective degree-zero homology. Equivalently,

and $S$ is finitely presented over $R$. This says that the complex is quasi-isomorphic to a finite projective module concentrated in degree zero.

Lemma. The transitivity sequence for the naive cotangent complex (Jacobi-Zariski sequence)

Let $A\to B\to C$ be homomorphisms. Choose polynomial presentations

with kernels $I$ and $J$. The variable sets may be infinite. Put $Q=P[y_t\mid t\in T]$ and $K=\ker(Q\to C)$. There is a canonical commutative diagram with exact rows

It gives the exact sequence

The first tensor product is the ordinary tensor product of a two-term presentation complex. If

its first homology identifies with $H_1(L_{B/A})\otimes_B C$.

Proof. The differential row separates the free basis $\{dx_s,dy_t\}$ into its two groups. Its first map includes the $dx_s$ summands, and its last map kills those summands and carries each $dy_t$ to the corresponding relative differential.

For the conormal row, put $I_Q=IQ$. The surjection $Q\to B[y]$ has kernel $I_Q$, so $K/I_Q=J$. The map $K/K^2\to J/J^2$ is onto and has kernel $(I_Q+K^2)/K^2$. The left-hand module is canonically

and inclusion of $I_Q$ into $K$ induces its surjection onto that kernel. These are the asserted maps. Universal derivations commute with the polynomial maps and with passage to $C$, proving commutativity.

To obtain the homology sequence without assuming that $E\to K/K^2$ is injective, let $D$ be its image. Its kernel consists of elements represented in $I_Q\cap K^2$, whose differential is zero after tensoring with $C$. Therefore the left vertical differential factors through $D$. Replacing $E$ by $D$ gives a short exact sequence of two-term complexes, with the middle and right complexes representing $\mathrm{NL}_{C/A}$ and $\mathrm{NL}_{C/B}$. The homology sequence of that short exact sequence has the claimed form with $H_1(D\to\Omega_{P/A}\otimes_P C)$ at the left.

Every degree-one cycle of this last complex lifts to a degree-one cycle of $E\to\Omega_{P/A}\otimes_P C$, because the map in degree zero is the identity. Their first homologies thus map surjectively. Their degree-zero homologies agree, and are $\Omega_{B/A}\otimes_B C$. Substitution gives the displayed exact sequence with its stated first term. The comparison of polynomial presentations identifies all these homology modules with the indicated cotangent homology groups.

For the last assertion, write a presentation complex for $B/A$ as $N^{-1}\to N^0$, with $N^0$ free, and denote its image by $M$. The exact sequence

remains left exact after tensoring with $C$ when $\operatorname{Tor}_1^B(\Omega_{B/A},C)=0$. Since $N^0$ is free, its Tor sequence also gives

Tensoring $0\to H_1(L_{B/A})\to N^{-1}\to M\to0$ consequently remains exact on the left as well. Combining the two sequences identifies the kernel of $N^{-1}\otimes_B C\to N^0\otimes_B C$ with $H_1(L_{B/A})\otimes_B C$, as required. ∎

Lemma. The conormal module of a syntomic presentation

Suppose $S=R[x_1,\ldots,x_n]/I$ with $I$ finitely generated, and let $g\in S$. If $S_g$ is syntomic over $R$, then $(I/I^2)_g$ is finite projective as an $S_g$-module.

Proof. The local syntomic criterion covers $\operatorname{Spec}S_g$ by principal opens on which the algebra is a relative global complete intersection. On each such open, compare its complete-intersection presentation with the given presentation. Localization of a conormal module gives the comparison after adding finite free summands, and The conormal module of a global complete intersection makes the complete-intersection conormal finite free. Hence the conormal for the given presentation is finite projective on each open. Take a finite subcover and apply the local finite-projective criterion. This proves the claim over the whole ring $S_g$. ∎

Lemma. A presentation realizing a basis of the conormal module

Let $S$ be a finitely presented $R$-algebra. If it has a presentation $S=R[x_1,\ldots,x_n]/I$ with $I/I^2$ free over $S$, then it has a finite presentation whose defining equations themselves give a basis of its conormal module.

Proof. Put $P=R[x_1,\ldots,x_n]$. Finite presentation of $S$ makes $I$ finitely generated, by Independence of finite presentation. Choose $f_1,\ldots,f_c\in I$ whose classes are a basis of the finite free module $I/I^2$, and set $F=(f_1,\ldots,f_c)$. Then $I=F+I^2$, so the finite $P$-module $I/F$ satisfies $I(I/F)=I/F$. The determinant form of Nakayama's lemma gives $g\in1+I$ with $gI\subset F$. Thus $I_g=F_g$, while $g$ has image $1$ in $S$.

It follows that

Write $J=(f_1,\ldots,f_c,gt-1)\subset P[t]$. To verify the promised basis explicitly, consider

For $h\in J$, evaluation at $1$ lies in $I$. Evaluation of a product of two elements of $J$ lies in $I^2$, and its derivative evaluates into $I$, so the map is well-defined. It is $S$-linear by the product rule. The images of the defining equations are $([f_i],0)$ and $([g-1],1)$. These form a basis of $(I/I^2)\oplus S$. Since their classes also generate $J/J^2$, the map is an isomorphism and their classes are a basis there. The new presentation has $c+1$ equations and $n+1$ variables, proving the assertion. ∎

Lemma. Localization of a relative complete intersection

Write $S=R[x_1,\ldots,x_n]/(f_1,\ldots,f_c)$. In each situation below, there is a polynomial $h$ with image $g\in S$ such that

is a relative global complete intersection, with the stated control on $g$:

1. If $I\subset R$ and every nonempty fibre of $S/IS$ over $R/I$ has dimension $n-c$, one can arrange $g\equiv1\pmod{IS}$.
2. If $\mathfrak p\subset R$ and $\dim(S\otimes_R\kappa(\mathfrak p))=n-c$, one can arrange that $g$ is a unit on this entire fibre.
3. If $\mathfrak q\subset S$ lies over $\mathfrak p$ and $\dim_{\mathfrak q}(S/R)=n-c$, one can arrange $g\notin\mathfrak q$.

Proof. By The open fibre-dimension bound, the points where the fibre dimension is at most $n-c$ form an open set $W\subset\operatorname{Spec}S$. Write its complement as $V(J)$.

In (1), $W$ contains $V(IS)$, so $J+IS=S$. Choose $g\in J$ with $g\equiv1$ modulo $IS$. In (2), the extension of $J$ to $S\otimes_R\kappa(\mathfrak p)$ is the unit ideal. Express $1$ as a finite combination of images of elements of $J$, and clear the denominators from $R\setminus\mathfrak p$. This gives $g\in J$ whose image in the fibre is the image of an element of $R\setminus\mathfrak p$, hence a unit. In (3), choose $g\in J\setminus\mathfrak q$, since $\mathfrak q\in W$.

In every case $D(g)\subset W$. Lift $g$ to $h\in R[x]$. The displayed presentation has $n+1$ variables and $c+1$ equations, and every nonempty fibre has dimension at most $n-c$ by its construction. Conversely a nonzero quotient of a polynomial ring in $n+1$ variables over a field by $c+1$ equations has dimension at least $n-c$, by the height bound for a finitely generated ideal. Thus each such fibre has dimension exactly $n-c$. This is the required relative global complete-intersection presentation, with all the claimed conditions on $g$. ∎

Lemma. Composition of syntomic ring maps

For homomorphisms $R\to S\to T$:

1. If both maps are syntomic, so is $R\to T$.
2. If both maps are relative global complete intersections, so is $R\to T$.

Proof. First suppose

are presentations of the kind in (2). Lifting the coefficients of the $h_j$ to $R[x]$ gives

For a residue field $k$ of $R$, the nonempty base fibre has dimension $n-c$, and the nonempty fibres above it of the second map have dimension $m-d$. The base–fibre dimension bound therefore gives dimension at most $(n-c)+(m-d)$ for $T\otimes_R k$. When this algebra is nonzero, its displayed presentation gives the reverse inequality by the height bound. Hence each nonempty fibre has exactly that dimension, proving (2).

Now assume the maps are syntomic and choose $\mathfrak q'\in\operatorname{Spec}T$, with image $\mathfrak q$ in $S$. The local syntomic criterion supplies $g'\in T\setminus\mathfrak q'$ and $g\in S\setminus\mathfrak q$ for which $S\to T_{g'}$ and $R\to S_g$ are relative global complete intersections. Base change makes $S_g\to T_{gg'}$ one as well. By (2), $R\to T_{gg'}$ is a relative global complete intersection, hence syntomic. These neighbourhoods cover $\operatorname{Spec}T$; choose a finite subcover and apply Locality of syntomic ring maps. This proves (1). ∎

Lemma. Smooth algebras are syntomic

If $R\to S$ is smooth, then $\operatorname{Spec}S$ admits a cover by principal opens $D(g)$ for which $S_g$ is standard smooth over $R$. In particular, every smooth ring map is syntomic.

Proof. The full local-standard-form argument is Formally smooth, unramified and étale ring maps, Theorem 5.1. It applies to an arbitrary base ring. Section 4 of that lesson proves that its definition of smoothness agrees with the naive-cotangent definition used here. The theorem's proof starts with the finite projective conormal module, chooses a basis at the given prime, uses Nakayama in the localized polynomial ring to generate the actual ideal, clears finitely many denominators, and adjoins an inverse variable to give a standard smooth presentation. Thus it supplies the asserted principal neighbourhood at every prime, including rank zero.

By Standard smooth algebras, each resulting chart is a relative global complete intersection. The complete-intersection criterion makes every chart syntomic. A finite subcover and locality then make $R\to S$ syntomic. For the zero algebra the spectrum is empty and the same local assertion is vacuous. ∎

Lemma. Local criteria for a syntomic algebra

Let $R\to S$, and let $\mathfrak q\subset S$ contract to $\mathfrak p\subset R$. The following conditions are equivalent:

1. Some $g\in S\setminus\mathfrak q$ makes $R\to S_g$ syntomic.
2. Some $g\in S\setminus\mathfrak q$ makes $S_g$ a relative global complete intersection over $R$.
3. There is a finitely presented neighbourhood $R\to S_g$ with $g\notin\mathfrak q$, the map $R_{\mathfrak p}\to S_{\mathfrak q}$ is flat, and the fibre local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is a complete intersection over $\kappa(\mathfrak p)$.

Proof. Condition (1) gives (3) by Complete intersections at a prime. The implication (2) to (1) is the relative global complete-intersection criterion. It remains to start with (3) and construct the presentation in (2).

First replace $S$ by its finitely presented neighbourhood, and write $S=P/I$, where $P=R[x_1,\ldots,x_n]$ and $I$ is finitely generated. Let $\mathfrak q'$ be the inverse image of $\mathfrak q$ in $P$, and put $k=\kappa(\mathfrak p)$. The fibre presentation has ideal $\overline I\subset k[x]$, and let $\overline{\mathfrak q}'$ be its indicated prime. By The field complete-intersection criterion, choose $f_1,\ldots,f_c\in I$ whose images form a minimal generating set of $\overline I_{\overline{\mathfrak q}'}$. They generate that ideal and are a regular sequence there. Such lifts can be chosen in $I$ because the images of $I$ span its minimal-generator vector space.

Set $S'=P/(f_1,\ldots,f_c)$ and $J=\ker(S'\to S)$. This kernel is finitely generated. Localize the exact sequence $0\to J\to S'\to S\to0$ at $\mathfrak q'$. Its last term is $S_{\mathfrak q}$, which is flat over $R$ by (3) and localization of the base. Tensoring with $k$ is therefore still injective at its first term. By the chosen generators, the map between the two fibre local rings is an isomorphism. Hence

Nakayama applies to this finite module over the local ring $S'_{\mathfrak q'}$, since $\mathfrak p$ is contained in its maximal ideal. It gives $J_{\mathfrak q'}=0$. A product of annihilating denominators for a finite generating set of $J$ now gives $h\in P\setminus\mathfrak q'$ with $J_h=0$, so $S'_h\simeq S_h$.

The regular sequence in the fibre polynomial local ring gives fibre dimension $n-c$ at the chosen point. Apply the preceding localization lemma to $S'$ and that point, and combine its denominator with $h$. The resulting neighbourhood is a relative global complete intersection and still contains $\mathfrak q$. This proves (2). ∎

Lemma. Smoothness from flatness and smooth fibres

For $R\to S$ and $\mathfrak q\subset S$ over $\mathfrak p\subset R$, suppose that $S$ is finitely presented over $R$ on a neighbourhood of $\mathfrak q$, that $R_{\mathfrak p}\to S_{\mathfrak q}$ is flat, and that $S\otimes_R\kappa(\mathfrak p)$ is smooth at the fibre point defined by $\mathfrak q$. Then $R\to S$ is smooth at $\mathfrak q$.

Proof. A smooth algebra over a field has complete-intersection local rings, by the field Jacobian criterion. The preceding syntomic criterion therefore replaces $S$, near $\mathfrak q$, by a relative global complete-intersection presentation

For each set $E$ of $c$ variable indices, let $\Delta_E$ be the corresponding Jacobian minor. On passing to the residue field $\kappa(\mathfrak p)$, the same coefficient map sends $\Delta_E$ to the Jacobian minor of the images of the $f_i$: differentiation and determinants commute with coefficient change.

Apply Smoothness of a global complete intersection first to the fibre. Its smoothness at the selected point gives an $E$ for which $\Delta_E\notin\mathfrak q$. Apply the same result to the displayed presentation over $R$. It says that this principal neighbourhood is smooth. Thus the original map is smooth at $\mathfrak q$. ∎

Lemma. Local criteria for complete intersections

Let $S$ be a finite-type algebra over a field $k$, fix $\mathfrak q\in\operatorname{Spec}S$, and choose any polynomial presentation $S=P/I$ with $P=k[x_1,\ldots,x_n]$. Denote the inverse image of $\mathfrak q$ by $\mathfrak q'$. Put

Here dimension at a point means the minimum dimension of its open neighbourhoods; the equality is the finite-type field dimension formula. The following are equivalent:

1. There is $g\in S\setminus\mathfrak q$ such that $S_g$ is a global complete intersection over $k$.
2. The ideal $I_{\mathfrak q'}$ in $P_{\mathfrak q'}$ admits a generating list of $c$ elements.
3. The conormal module $(I/I^2)_{\mathfrak q}$ admits a generating list of $c$ elements over $S_{\mathfrak q}$.
4. The module $(I/I^2)_{\mathfrak q}$ is free of rank $c$.
5. The ideal $I_{\mathfrak q'}$ is generated by a regular sequence in $P_{\mathfrak q'}$.

When these conditions hold, any $c$ elements of $I_{\mathfrak q'}$ that generate $I_{\mathfrak q'}/\mathfrak q'I_{\mathfrak q'}$ form a regular sequence.

Proof. Write $Q=P_{\mathfrak q'}$ and $H=I_{\mathfrak q'}$. The ring $Q$ is regular local, hence Cohen–Macaulay, and

If $c$ elements generate $H/H^2$, Nakayama applied to their quotient in the finite $Q$-module $H$ shows that they generate $H$: the remaining module equals its product with $H\subset\mathfrak q'Q$. Thus (3) implies (2); the converse follows by taking the quotient by $H^2$.

In a Cohen–Macaulay local ring, $c$ generators of an ideal whose quotient has dimension $\dim Q-c$ form a regular sequence, by The Cohen–Macaulay parameter criterion. Consequently (2) implies (5). A regular sequence has a free conormal module on its classes, by Regular sequences are quasi-regular. If its length is $e$, successive regular quotients have dimension $\dim Q-e$, so the displayed equality forces $e=c$. This proves (5) implies (4), which immediately implies (3). The same Nakayama and parameter argument proves the final assertion for every generating list of the specified residue vector space.

To obtain (1) from (2), clear the denominators of the local generators and choose $f_1,\ldots,f_c\in I$ that still generate $H$. Since $I$ is finitely generated, there is $h\in P\setminus\mathfrak q'$ with $I_h=(f_1,\ldots,f_c)_h$. The rings $P/(f_1,\ldots,f_c)$ and $S$ therefore agree near the chosen point, where their dimension at the point is $n-c$. Localization of a relative complete intersection, case (3), supplies a further principal neighbourhood that is a global complete intersection over $k$. Combining its denominator with $h$ gives (1).

Conversely, suppose (1) and present

as a global complete intersection. Its conormal module $J/J^2$ is free of rank $t$, where $J=(h_1,\ldots,h_t)$, and every component has dimension $m-t$. At the specified point this gives $m-t=\dim_{\mathfrak q}\operatorname{Spec}S$. The comparison of conormal modules for two presentations gives

Localizing at $\mathfrak q$ makes $(I/I^2)_{\mathfrak q}$ finite projective over a local ring, hence free, of rank

Thus (1) implies (4), completing the equivalences. ∎

Definition. Unramified ring maps

For a homomorphism $R\to S$:

1. Unramified means finite type with $\Omega_{S/R}=0$.
2. G-unramified means finite presentation with $\Omega_{S/R}=0$.
3. It is unramified at $\mathfrak q$, for a prime $\mathfrak q$ of $S$, if some $g\notin\mathfrak q$ makes $R\to S_g$ unramified.
4. It is G-unramified at $\mathfrak q$ if some such principal neighbourhood is G-unramified.

The two finiteness requirements are kept distinct over a non-Noetherian base.

Lemma. Flatness from Cohen--Macaulayness over a regular base

Let $(R,\mathfrak m_R)\to(S,\mathfrak m_S)$ be a local homomorphism of Noetherian local rings. If $R$ is regular, $S$ is Cohen–Macaulay, and

then $S$ is flat over $R$.

Proof. Induct on $d=\dim R$. At $d=0$, the regular local ring $R$ is a field, so flatness is automatic. Suppose $d>0$. For every minimal prime $\mathfrak q_i$ of $S$, Cohen–Macaulayness gives $\dim(S/\mathfrak q_i)=\dim S$, by The Cohen–Macaulay chain theorem. Thus $\mathfrak m_RS$ cannot be contained in $\mathfrak q_i$: otherwise the quotient $S/\mathfrak q_i$ would have dimension at most $\dim(S/\mathfrak m_RS)$, contrary to the displayed equality and $d>0$.

Each contraction $\mathfrak p_i=R\cap\mathfrak q_i$ is therefore a proper subprime of $\mathfrak m_R$. Prime avoidance with the square of the maximal ideal provides

Since $R$ is regular, $x$ is part of a regular system of parameters. Hence $R/xR$ is regular of dimension $d-1$. The associated primes of the Cohen–Macaulay local ring $S$ are its minimal primes, and $x$ avoids all of them. Thus $x$ is also a nonzerodivisor on $S$, and $S/xS$ is Cohen–Macaulay of dimension $\dim S-1$.

The closed fibre of $R/xR\to S/xS$ is still $S/\mathfrak m_RS$. Subtracting one from both ring dimensions preserves the required dimension equality, so induction makes $S/xS$ flat over $R/xR$. The free resolution $0\to R\xrightarrow{x}R\to R/xR\to0$ shows

Apply The local flatness criterion for an ideal with the ideal $(x)$ and the finite $S$-module $S$. Its two hypotheses are exactly this Tor vanishing and the established flatness of $S/xS$ over $R/xR$. It follows that $S$ is flat over $R$. ∎

Lemma. Smoothness after an algebraic closure of the ground field

Let $k$ be algebraically closed, $S$ a finite-type $k$-algebra, and $\mathfrak m$ a maximal ideal of $S$. The following are equivalent:

1. $S_{\mathfrak m}$ is regular local.
2. $\dim_{\kappa(\mathfrak m)}(\Omega_{S/k}\otimes_S\kappa(\mathfrak m))\leq\dim S_{\mathfrak m}$.
3. The two dimensions in (2) are equal.
4. Some $g\in S\setminus\mathfrak m$ makes $S_g$ smooth over $k$.

Proof. The Nullstellensatz gives $\kappa(\mathfrak m)=k$. The exact cotangent calculation and the closed-point smoothness criterion are proved in Smooth algebras over a field and the Jacobian criterion, Proposition 3.1, using the full field criterion of its Theorem 2.1. In the present notation the calculation identifies

Its dimension is the embedding dimension of $S_{\mathfrak m}$, which is at least its Krull dimension. Therefore (2) is equivalent to (3), and (3) is precisely regularity, proving equivalence with (1). Proposition 3.1 identifies that regularity with smoothness at this closed point, meaning the principal neighbourhood in (4). Its hypotheses are exactly a finite-type algebra, an algebraically closed field and a closed point; thus all four assertions follow at their stated scope. ∎

Lemma. The fibrewise criterion for flatness (Critère de platitude par fibres)

Let $R\to S\to S'$ be local homomorphisms of local rings, and let $\mathfrak m$ be the maximal ideal of $R$. Suppose that both $S$ and $S'$ are essentially finitely presented over $R$, and that $M\ne0$ is a finitely presented $S'$-module. If $M$ is flat over $R$ and $M/\mathfrak mM$ is flat over $S/\mathfrak mS$, then $S$ is flat over $R$ and $M$ is flat over $S$.

Proof. We reduce the data, including both flatness conditions, to Noetherian local models. Express $R$ as a filtered colimit of local rings $R_\lambda$ essentially of finite type over $\mathbb Z$, with local transition maps and local maps to $R$. Denote their maximal ideals by $\mathfrak m_\lambda$. One construction takes finite-type subrings of $R$ and localizes at the inverse images of $\mathfrak m$. The essential finite-presentation model lemma permits a common later index at which both ring maps and their composite have compatible finite presentations.

More explicitly, after fixing finite lists of coefficients we obtain local rings

with colimits $S$ and $S'$. The indicated primes are the inverse images of the final maximal ideals. Lift a finite presentation matrix for $M$ at a further index and set $M_\lambda$ to be its cokernel over $S'_\lambda$. Then

for all sufficiently late indices. Every $M_\lambda$ is nonzero, since a zero one would make its base change $M$ zero.

For $\mu\geq\lambda$, the maps

are localizations. The same is true of $S'_\lambda\otimes_{R_\lambda}R_\mu\to S'_\mu$. Apply Eventual flatness in a filtered colimit to the models $R_\lambda\to S'_\lambda,M_\lambda$. The flatness of $M$ over $R$ gives flatness of $M_\lambda$ over $R_\lambda$ at all sufficiently late indices.

The maximal ideals have colimit $\mathfrak m$, so quotienting gives

and likewise for $S'$. These quotient models still have the required localization and module base-change properties. Indeed, quotienting the preceding localized tensor products by $\mathfrak m_\mu$ identifies their source with

and its map to $S'_\mu/\mathfrak m_\mu S'_\mu$ is a localization. The presentation matrix gives the analogous base-change isomorphism for the quotient modules. Thus these are essentially finitely presented module models. Applying eventual flatness a second time, now to the assumed flatness of $M/\mathfrak mM$, makes

at all sufficiently late indices.

Choose one index at which both conclusions hold. The rings $R_\lambda,S_\lambda,S'_\lambda$ are Noetherian local, and $M_\lambda$ is finite and nonzero. The Noetherian fibrewise criterion gives flatness of $S_\lambda$ over $R_\lambda$ and of $M_\lambda$ over $S_\lambda$. Finally $S$ is a localization of $S_\lambda\otimes_{R_\lambda}R$, and $M$ is obtained from $M_\lambda\otimes_{S_\lambda}S$ by the corresponding further localization. Flatness survives both base change and localization, yielding the two required conclusions over the original, possibly non-Noetherian rings. ∎

Lemma. Characterizations of étale algebras

Let $R\to S$ be finitely presented, and let $\mathfrak q\subset S$ lie over $\mathfrak p\subset R$. Suppose $R_{\mathfrak p}\to S_{\mathfrak q}$ is flat, $\mathfrak pS_{\mathfrak q}$ is the maximal ideal of $S_{\mathfrak q}$, and $\kappa(\mathfrak q)/\kappa(\mathfrak p)$ is finite separable. Then $R\to S$ is étale at $\mathfrak q$.

Proof. The fibre local ring is the field $\kappa(\mathfrak q)$. By An isolated point of a fibre, a principal neighbourhood of $\mathfrak q$ has no other point in that fibre. Replace $S$ by this neighbourhood. Its fibre algebra is local and equals its localization at that one point, hence equals $\kappa(\mathfrak q)$. The finite separable field extension is étale, so this fibre is smooth.

Smoothness from flatness and smooth fibres now gives a smooth principal neighbourhood. Refine it by the standard smooth neighbourhood theorem to a presentation

with an invertible $c$-column Jacobian minor. Its fibre at the retained point is still the same field. A nonempty fibre of this standard presentation has dimension $n-c$, by Standard smooth algebras, so $n=c$. The presentation therefore has invertible full square Jacobian and zero differentials. It is smooth and unramified, hence étale, proving the assertion at $\mathfrak q$. ∎

Lemma. Étale algebras in standard smooth form

Every étale $R$-algebra $S$ admits a presentation

whose full Jacobian determinant is invertible in $S$. In particular, the algebra is standard smooth globally.

Proof. Choose any finite polynomial presentation $S=P/I$. Étaleness says that its conormal differential

is an isomorphism. In particular, the conormal module is free. Apply A presentation realizing a conormal basis to obtain a presentation whose defining equations are a basis of that module. In this new presentation the conormal differential is still an isomorphism. For $S\ne0$, equality of the ranks of finite free modules gives as many equations as variables, and the matrix of the differential is the full Jacobian. Its determinant is a unit, as asserted. The zero algebra has the presentation $R[t]/(1)$; its Jacobian image is the identity element of the zero ring and hence is a unit there. This also covers that case. ∎

Lemma. Recognizing a filtered colimit of finite presentations

Let $R\to\Lambda$ be a homomorphism and let $\mathcal E$ be a set of finitely presented $R$-algebras. Then $\Lambda$ is a filtered colimit of algebras in $\mathcal E$ if and only if every homomorphism $A\to\Lambda$ from a finitely presented $R$-algebra factors as $A\to B\to\Lambda$ with $B\in\mathcal E$.

Proof. In the forward direction, a map from a finitely presented algebra into a filtered colimit factors through a stage, by The finite-presentation criterion.

Conversely, form the category of pairs $(B,\phi)$ with $B\in\mathcal E$ and $\phi:B\to\Lambda$; arrows are $R$-algebra maps compatible with these structure maps. This category is small. Applying the factorization hypothesis to $R\to\Lambda$ makes it nonempty. Two objects have a common target: their maps give $B_1\otimes_R B_2\to\Lambda$, and this finitely presented algebra factors through an object of the category.

To equalize two parallel arrows $u,v:B_1\to B_2$, choose a finite list of algebra generators $b_i$ of $B_1$. The quotient

is still finitely presented over $R$, and its map to $\Lambda$ factors through an object of $\mathcal E$. The resulting arrow out of $B_2$ equalizes $u$ and $v$ on the generators, hence on all of $B_1$. The category is therefore filtered.

Its colimit maps to $\Lambda$. This map is surjective: for any $\lambda\in\Lambda$, apply the hypothesis to $R[t]\to\Lambda$, $t\mapsto\lambda$. It is injective as well. If an element $b$ represented at a stage $B$ maps to zero, the finitely presented quotient $B/(b)$ maps to $\Lambda$ and factors through another stage; that transition kills $b$. Filteredness first puts any two representatives at a common stage, so this kernel calculation proves injectivity in general. Thus the colimit is $\Lambda$, with exactly the required kind of stages. ∎

Lemma. Standard smooth algebras

Suppose $S=R[x_1,\ldots,x_n]/(f_1,\ldots,f_c)=P/I$ is standard smooth, with its invertible Jacobian minor in the first $c$ variable columns. Then:

1. $R\to S$ is smooth.
2. $\Omega_{S/R}$ is free with basis $dx_{c+1},\ldots,dx_n$.
3. $I/I^2$ is free with basis the classes of $f_1,\ldots,f_c$.
4. Every principal localization $S_g$ is standard smooth over $R$.
5. Every base change $R'\to R'\otimes_R S$ is standard smooth.
6. If $f\in R$ is invertible in $S$, the induced map $R_f\to S$ is standard smooth.
7. $S$ is a relative global complete intersection over $R$.

Proof. Parts (1)–(5) are proved, with these exact bases and the polynomial-quotient convention, in Formally smooth, unramified and étale ring maps, Theorem 4.1. Its square-zero correction solves the equations by the invertible Jacobian block. Projection onto that block proves independence of the conormal generators, and eliminating those differential coordinates gives the stated basis of $\Omega$. Its localization proof adjoins a variable $z$ with equation $hz-1$, where $h$ lifts $g$, and uses the block determinant $h\Delta$. Its base-change proof applies the coefficient map to the equations and to $\Delta$. Thus the full proof includes both preservation assertions, not just smoothness itself.

For (6), apply (5) with $R'=R_f$. Since $f$ already has an inverse in $S$, the canonical map $R_f\otimes_R S\to S$ is an isomorphism. This gives the asserted standard smooth presentation over $R_f$.

For (7), take a residue field $k$ of $R$. Part (5) gives the same standard smooth presentation for $S\otimes_R k$, and part (2) gives differential rank $n-c$ at every point of a nonempty fibre. The full field criterion in Smooth algebras over a field and the Jacobian criterion, Theorem 2.1 identifies that rank with dimension at the point. Hence the fibre has dimension $n-c$. Every nonempty fibre has now been checked, which is exactly the relative global complete-intersection condition for the given presentation. ∎

Proposition. Formal smoothness of smooth algebras

For any ring homomorphism $R\to S$, the following are equivalent:

1. The algebra $S$ is finitely presented over $R$ and is formally smooth over $R$.
2. The map $R\to S$ is smooth.

Proof. Choose a finite polynomial presentation $S=P/I$. Its conormal sequence is

The complete equivalence between formal smoothness and this sequence being split exact with an initial zero is Formally smooth, unramified and étale ring maps, Theorem 3.1. Its proof constructs the lift from a section of $P/I^2\to S$ and constructs that section from the splitting by subtracting the induced derivation.

If (1) holds, the sequence is split exact. The middle term is finite free, so its quotient $\Omega_{S/R}$ is finite projective, and injectivity on the left says $H_1(\mathrm{NL}_{S/R})=0$. These are the smoothness conditions. Conversely, smoothness makes that conormal map injective and its cokernel finite projective. The sequence therefore splits, so Theorem 3.1 gives formal smoothness. Finite presentation is already included in smoothness. This proves both implications. ∎

Lemma. Nakayama's lemma

The lemma is associated with Nakayama, Azumaya and Krull; see the historical attribution accompanying Stacks, Tag 00DV.

Fix a commutative ring $R$, an ideal $I$, and an $R$-module $M$. Write $J(R)$ for the Jacobson radical. The following forms will be used:

1. A finite module satisfying $M=IM$ is annihilated by some $f\in1+I$.
2. In (1), the additional containment $I\subseteq J(R)$ forces $M=0$.
3. Suppose $N,N'\subseteq M$, the module $N'$ is finite, and $M=N+IN'$. Some $f\in1+I$ then satisfies $fM\subseteq N$; in particular, $N_f=M_f$.
4. Under the hypotheses of (3), if $I\subseteq J(R)$, then $N=M$.
5. Let $u:N\to M$ be linear, with $M$ finite. Surjectivity of $N/IN\to M/IM$ implies surjectivity of $u_f:N_f\to M_f$ for some $f\in1+I$.
6. Under the hypotheses of (5), $u$ itself is surjective when $I\subseteq J(R)$.
7. If $M$ is finite and the images of $x_1,\ldots,x_n\in M$ span $M/IM$, these elements span $M_f$ over $R_f$ for a suitable $f\in1+I$.
8. With the hypotheses of (7) and $I\subseteq J(R)$, the elements already span $M$ over $R$.
9. For nilpotent $I$, the equality $M=IM$ implies $M=0$, without a finiteness condition on $M$.
10. For nilpotent $I$, any equality $M=N+IN'$ with submodules $N,N'\subseteq M$ implies $M=N$; neither submodule needs to be finite.
11. For nilpotent $I$, a linear map $u:N\to M$ is surjective whenever its reduction $N/IN\to M/IM$ is surjective.
12. For nilpotent $I$, an arbitrary family $(x_\alpha)_{\alpha\in A}$ spanning $M/IM$ spans $M$ itself.

Proof. The determinant argument for a finite module is proved in Localization, local properties and support, Lemma 4.1 and Theorem 4.2. Apply that lemma to the identity endomorphism of $M$, whose image lies in $IM$. Evaluation of its monic annihilating polynomial at $1$ gives an annihilator in $1+I$, proving (1). When $I\subseteq J(R)$, such an element is a unit: it belongs to no maximal ideal. This proves (2).

Here are the quotient arguments that give the remaining forms. Under (3), put $Q=M/N$. The map $N'\to Q$ is surjective, since $M=N+IN'\subseteq N+N'$, so $Q$ is finite. Also $Q=IQ$. Part (1) supplies $fQ=0$, which is exactly $fM\subseteq N$. Localizing makes $Q_f=0$, hence $M_f=N_f$. Part (2) instead gives $Q=0$ under (4).

For (5) and (6), use $Q=\operatorname{coker}(u)$. It is finite as a quotient of $M$, and the surjectivity modulo $I$ says $Q/IQ=0$. Thus (1) and (2) give the respective conclusions. Apply these conclusions to the map $R^n\to M$ taking its standard basis to the $x_i$ to obtain (7) and (8).

If $I^a=0$ and $Q=IQ$, iteration gives $Q=I^aQ=0$ for every module $Q$. This proves (9). Apply it to $M/N$, to $\operatorname{coker}(u)$, and to the cokernel of $R^{(A)}\to M$, respectively. Their reductions modulo $I$ vanish under (10), (11), and (12). The direct sum $R^{(A)}$ permits any indexing set, so no finite-generation assumption has entered these last four assertions. ∎

Lemma. Finite presentation and finite algebras

If $S$ is a finitely presented $R$-algebra, the kernel of every surjection $\alpha:R[x_1,\ldots,x_n]\to S$ is finitely generated as an ideal.

Proof. Take one finite presentation $S=R[y_1,\ldots,y_m]/(f_1,\ldots,f_s)$. Choose polynomials $g_i(y)$ representing $\alpha(x_i)$, and choose polynomials $h_j(x)$ whose images under $\alpha$ are the classes of $y_j$. In $R[x]$, form the finite ideal

Each displayed generator maps to zero in $S$, so $\alpha$ descends to $\bar\alpha:R[x]/L\to S$. In the other direction, the substitution $y_j\mapsto h_j(x)\bmod L$ kills every $f_\ell$ and defines $\beta:S\to R[x]/L$.

On the generators $y_j$ of $S$, the composite $\bar\alpha\beta$ is the identity by the choice of $h_j$. On the generators $x_i$ of $R[x]/L$, the composite $\beta\bar\alpha$ takes $x_i$ to $g_i(h(x))=x_i\bmod L$. Thus these maps are inverse $R$-algebra isomorphisms. It follows that $\ker\alpha=L$, giving the required finite generators over an arbitrary base ring. ∎

Lemma. The equational criterion for flatness (Equational criterion of flatness)

For any ring $R$ and any $R$-module $M$, flatness is equivalent to the following condition. Whenever a finite list satisfies $\sum_{i=1}^n a_i x_i=0$ in $M$, there are a finite list $y_1,\ldots,y_s\in M$ and a matrix $(b_{ij})$ over $R$ such that

A relation admitting this factorization is called trivial.

Proof. This is the full generality of Tor and flat modules, Theorem 5.1. Its forward argument lifts the relation through $\ker(R^n\to(a_1,\ldots,a_n))\otimes_R M$, producing exactly the matrix above. In the reverse direction that factorization kills every kernel tensor for $I\otimes_R M\to M$. Theorem 2.1 in the same lesson, Section 2, supplies the complete ideal criterion: it passes from finitely generated ideals to all ideals, then to submodules of finite free modules, arbitrary free modules, and finally arbitrary inclusions. These proofs impose no finite-presentation condition on the ideal or module and no Noetherian condition on $R$. ∎

Lemma. Krull's intersection theorem

If $(R,\mathfrak m)$ is Noetherian local, $M$ is a finite $R$-module, and $I\subsetneq R$ is an ideal, then

Proof. The more general Noetherian and Artinian rings, Theorem 6.1, proves that this intersection $K$ satisfies $IK=K$ and is annihilated by one element of $1+I$. Its preceding Section 5 proves the Artin–Rees identity used to obtain $IK=K$; thus the theorem includes the proof of the intersection assertion. Here $I\subseteq\mathfrak m=J(R)$, so its annihilator is a unit. Consequently $K=0$. ∎

Lemma. Composition of smooth ring maps

Given smooth maps $R\to S$ and $S\to T$, the composite $R\to T$ is smooth.

Proof by the cotangent sequence. First, $T$ is finitely presented over $R$: combine a finite presentation of $S$ over $R$ with a finite presentation of $T$ over $S$, lifting the coefficients of the latter equations to the former polynomial ring. The module $\Omega_{S/R}$ is projective, so its first and second Tor groups with $T$ vanish. The transitivity sequence, including its comparison of the first homology after tensoring, therefore gives

Smoothness of the two given maps makes both outside $H_1$ terms zero. Hence $H_1(\mathrm{NL}_{T/R})=0$, and the differential terms form a short exact sequence with initial term $\Omega_{S/R}\otimes_S T$. That term is finite projective by scalar extension, and the last term $\Omega_{T/S}$ is finite projective. Projectivity of the last term splits the sequence, so the middle term is finite projective as well. These are precisely the smoothness conditions for $R\to T$.

Proof using standard presentations. This also makes the local geometric mechanism explicit. Let $\mathfrak q\in\operatorname{Spec}(T)$ and $\mathfrak p$ its inverse image in $S$. The standard-neighborhood theorem for a smooth algebra gives a principal neighborhood $S_f$ of $\mathfrak p$ standard smooth over $R$. The map $S_f\to T_f$ is smooth by localization, and a further principal neighborhood $(T_f)_g$ of $\mathfrak q$ is standard smooth over $S_f$. The composition calculation for standard smooth presentations makes $(T_f)_g$ standard smooth over $R$: its Jacobian has the two invertible diagonal blocks. If $g=t/f^a$, this is a principal neighborhood $T_{ft}$ of $\mathfrak q$. Such neighborhoods cover $\operatorname{Spec}(T)$, so locality of smoothness proves the assertion. This route uses the previously proved standard-neighborhood and locality statements, with their finite-presentation hypotheses retained. ∎

Lemma. Total rings of fractions without embedded primes

Suppose that a ring $R$ has a finite set of minimal primes $\{\mathfrak q_1,\ldots,\mathfrak q_t\}$ and that its zero divisors are exactly $\bigcup_i\mathfrak q_i$. Then the localization maps induce an isomorphism

No Noetherian assumption is required.

Proof. Write $U$ for the multiplicative set of nonzerodivisors. Every $u\in U$ lies outside each $\mathfrak q_i$, giving the indicated maps from $U^{-1}R=Q(R)$. By the prime correspondence for localization, a prime of $Q(R)$ comes from a prime $\mathfrak p$ of $R$ disjoint from $U$. Such a $\mathfrak p$ is contained in $\bigcup_i\mathfrak q_i$, and finite prime avoidance puts it inside one $\mathfrak q_i$. Minimality then gives $\mathfrak p=\mathfrak q_i$. Conversely, each $\mathfrak q_i$ is disjoint from $U$.

Thus all primes of $Q(R)$ are the finitely many points corresponding to the $\mathfrak q_i$, and each is maximal. Each singleton is closed; its finite complement is also closed, so the spectrum is discrete. The idempotent decomposition for a disjoint spectrum expresses $Q(R)$ as a product with one local factor at each of these points. Localizing that product at its $i$th prime keeps exactly its $i$th factor. Transitivity of localization identifies it with $R_{\mathfrak q_i}$. These identifications intertwine the factor projections with the original localization maps, proving the stated canonical isomorphism. If $R=0$, the spectrum is empty and the assertion is the empty product, namely the zero ring. ∎

Lemma. Irreducible components of a Noetherian spectrum

A Noetherian ring has only finitely many minimal prime ideals. Equivalently, its spectrum has finitely many irreducible components and therefore finitely many component generic points.

Proof. For $X=\operatorname{Spec}(R)$, a descending sequence of closed sets gives an ascending sequence of their radical ideals. The Noetherian condition makes that sequence stabilize. Thus $X$ satisfies the descending chain condition on closed sets.

Every closed subset of such a space is a finite union of irreducible closed subsets. Indeed, otherwise choose a minimal closed counterexample $Z$ by the descending chain condition. It is nonempty and cannot itself be irreducible. Write $Z=Z_1\cup Z_2$ with both $Z_i$ proper closed subsets of $Z$. Minimality gives finite irreducible decompositions of both $Z_i$, which together give one of $Z$, a contradiction. Apply this to $X$ and discard members contained in others. The remaining sets are exactly its irreducible components: any irreducible subset of a finite union of closed sets lies in one member.

For the affine irreducibility criterion, an irreducible closed set is $V(\mathfrak p)$ for a prime $\mathfrak p$, with generic point $\mathfrak p$. A component corresponds to a prime minimal under inclusion. Hence the finite component decomposition gives the asserted finite set of minimal primes. The zero ring gives the empty decomposition. ∎

Lemma. Regular rings are Cohen--Macaulay

Let $(R,\mathfrak m)$ be regular local of dimension $d$, and choose any minimal generating list $x_1,\ldots,x_d$ for $\mathfrak m$. This list is $R$-regular. For each $0\leq c\leq d$, the local ring $R/(x_1,\ldots,x_c)$ is regular of dimension $d-c$. In particular, $R$ is Cohen–Macaulay.

Proof. Use the complete proof of Regular sequences, depth and Cohen–Macaulay modules, Theorem 6.1 and Corollary 6.2. The theorem begins with any minimal generating list. It proves that $\operatorname{gr}_{\mathfrak m}R$ is the polynomial ring on the initial forms of that list, then uses Krull separation to obtain injectivity successively in the quotients. It therefore proves regularity of the given list, not merely the existence of some regular list, and proves that $R$ is Cohen–Macaulay.

For completeness, the quotient conclusion follows within that same proof route as follows. Corollary 6.2 gives dimension $d-c$ after a prefix of length $c$. The remaining $d-c$ elements generate its maximal ideal, so its embedding dimension is at most $d-c$. Embedding dimension is at least Krull dimension for a Noetherian local ring. Equality follows, which is exactly regularity. At $c=d$ the quotient is the residue field, and $d=0$ is already the field case of Theorem 6.1. ∎

Lemma. Powers of a regular sequence

For an arbitrary ring $R$, an arbitrary $R$-module $M$, and positive integers $e_1,\ldots,e_r$, one has

Regularity includes nonvanishing of the final quotient.

Proof. This is Regular sequences, depth and Cohen–Macaulay modules, Proposition 1.3, with exactly the arbitrary-ring and arbitrary-module hypotheses above. Its proof treats both directions by the finite filtration of $M/f_1^{e_1}M$ whose factors are $M/f_1M$. After each later equation, it proves exactness of the quotient filtration; in the reverse direction the embedded first factor detects the next injectivity condition. The final filtration also proves equivalence of nonvanishing of the last quotient. Induction then changes the other exponents. The proof includes the empty list and does not use permutation of a regular sequence, which would require additional hypotheses. ∎

Lemma. Elementary formally smooth extensions

For a field extension $K/k$, each of the following hypotheses implies formal smoothness of $k\to K$:

1. $K$ is purely transcendental over $k$, with any cardinality of transcendence basis.
2. $K/k$ is separable algebraic, without a finite-degree assumption.
3. $K/k$ is separable, in the sense that every finitely generated intermediate extension is separably generated.

Proof. For (1), write $K=k(t_j\mid j\in J)$. Polynomial algebras on an arbitrary set of variables are formally smooth, and localization preserves formal smoothness, by the complete proof of Formally smooth, unramified and étale ring maps, Theorem 1.2. Its localization step applies here because every nonzero polynomial has a unit image under any test map from the field; units lift across a square-zero ideal. Thus it treats arbitrary $J$ and every required denominator.

For (2), every finite intermediate field $E/k$ is formally étale by the same lesson, Proposition 6.3, whose proof lifts a separable generating tower by unique root corrections. The fields $E$ form a directed system with union $K$. Given a square-zero lifting problem for $K$, lift its restriction to each $E$. Uniqueness makes these lifts agree under every inclusion, so they define a lift on the union. Restriction to each $E$ also proves uniqueness. Hence $K/k$ is in fact formally étale.

For (3), let $E$ range over all intermediate fields finitely generated over $k$. Separability supplies a transcendence basis $\mathbf t$ of $E/k$ with $E/k(\mathbf t)$ separable algebraic. Parts (1) and (2), followed by the composition assertion of Theorem 1.2, give formal smoothness of each $E/k$.

To pass to their union, use the cotangent criterion for a field extension: it gives $H_1(L_{E/k})=0$. The filtered-colimit comparison for naive cotangent complexes then gives $H_1(L_{K/k})=0$. More explicitly, tensor each presentation complex with the field $K$; this is exact on $E$-modules, and the resulting filtered colimit is the presentation complex for $K/k$. Exactness of filtered colimits therefore gives the stated homology vanishing. Apply the field criterion again to conclude formal smoothness. This argument does not require arbitrary choices of smooth lifts at different stages to be compatible. ∎

Lemma. Descent of flatness

Let $R$ be a ring, let $S\to S'$ be a flat homomorphism of $R$-algebras, and let $M$ be an $S$-module. Set $M'=M\otimes_S S'$.

1. If $M$ is flat as an $R$-module, then so is $M'$.
2. If $S\to S'$ is faithfully flat, then $M$ is flat over $R$ exactly when $M'$ is flat over $R$.

Proof. For any inclusion $U\hookrightarrow V$ of $R$-modules, let

This is an $S$-module. Exactness of $-\otimes_S S'$ and the associative tensor identifications give

If $M$ is $R$-flat, then $K=0$ for every inclusion, so the right-hand kernels vanish; this proves (1). Conversely, if $M'$ is $R$-flat, the right-hand kernel always vanishes. Under faithful flatness this implies $K=0$. Tensoring with $M$ therefore preserves every inclusion of $R$-modules, which proves (2). No finiteness hypothesis on any of these rings or modules is used. ∎

Lemma. Henselianity in local dimension zero

Every local ring $(R,\mathfrak m)$ of dimension zero is henselian, including rings that are not Noetherian.

Proof by root correction. The only prime of $R$ is $\mathfrak m$, so $\mathfrak m$ is its nilradical. Take a monic polynomial $f\in R[T]$ with a simple root $\bar a\in R/\mathfrak m$, and choose a lift $a_0\in R$. The element $e=f(a_0)$ is nilpotent, while $f'(a_0)$ is a unit. Define

All $a_n$ have residue $\bar a$, so all displayed inverses exist. Polynomial expansion shows

For sufficiently large $n$ this error is zero. The resulting $a_n$ is a root lifting $\bar a$, as required. Only the single element $e$ has been assumed nilpotent; a common nilpotence exponent for $\mathfrak m$ is unnecessary.

Proof by finite algebras. If $S$ is finite over $R$, each of its primes contracts to $\mathfrak m$. The fibre $S/\mathfrak mS$ is finite-dimensional over $R/\mathfrak m$, so there are finitely many such primes. Incomparability for an integral extension makes them all maximal. Its finite spectrum is therefore discrete, and the idempotent decomposition writes $S$ as a finite product of local rings. The complete finite-algebra criterion in Henselian local rings and henselization, Theorem 2.2, now also gives henselianity of $R$. ∎

Lemma. Finite étale algebras over a henselian ring

For a henselian local ring $(R,\mathfrak m,\kappa)$, the functor

from finite étale $R$-algebras to finite étale $\kappa$-algebras is an equivalence of categories. Its action on morphisms is reduction modulo $\mathfrak m$.

Proof. Apply Étale neighbourhoods, henselization and quasi-finite morphisms, Theorem 5.2. Its complete proof has exactly this arbitrary-local-ring scope. For essential surjectivity it lifts a separable monic polynomial for each residue-field factor and proves that its derivative is a unit throughout the finite lift. For full faithfulness it decomposes the target into henselian local factors, then lifts each specified residue map by the unique section through the corresponding rational fibre point. Combining those lifts recovers every morphism and proves its uniqueness. The theorem even proves this morphism assertion when the target is any finite $R$-algebra.

The algebraic inputs to that argument are the complete finite-algebra criterion and permanence for finite local algebras. Thus the local factors used in the proof are finite over $R$ and henselian; in the finite étale case they remain étale over $R$. The equivalence retains residue-field automorphisms, rather than identifying all lifts by an unspecified unique isomorphism. ∎

Lemma. Dimension of a flat family

Suppose $R\to S$ is a homomorphism of Noetherian rings with going down. Let $\mathfrak q\subset S$ contract to $\mathfrak p\subset R$. Then

In particular this holds for a flat map, by going down for flat ring maps.

Proof. Put $A=R_{\mathfrak p}$, $B=S_{\mathfrak q}$, with maximal ideals $\mathfrak m,\mathfrak n$, and let $d=\dim A$, $e=\dim(B/\mathfrak mB)$. These dimensions are finite. Choose $d$ parameters in $A$ and lift a system of $e$ parameters of $B/\mathfrak mB$ to $B$. The ideal they together generate in $B$ has radical $\mathfrak n$. Indeed, a power of $\mathfrak m$ is contained in the ideal of the first parameters, and the second parameters cut the fibre down to its closed point. The height bound for an ideal generated by $d+e$ elements therefore gives $\dim B\leq d+e$. This is the parameter proof of the base–fibre dimension bound.

For the reverse inequality, take a chain of length $e$ in $\operatorname{Spec}(B/\mathfrak mB)$ ending at its maximal ideal. Its inverse images are

all contracting to $\mathfrak m$ in $A$. Choose a chain of length $d$ in $A$ ending at $\mathfrak m$. Going down, which persists under these localizations, successively extends the chain below $\mathfrak q_0$ by one prime over each earlier member of the chosen base chain. Those $d$ new inclusions are strict because their contractions are distinct. The resulting chain in $B$ has length $d+e$. Hence $\dim B\geq d+e$, proving equality, including $d=0$ or $e=0$. ∎

Lemma. Universal injectivity of a faithfully flat ring map

For a faithfully flat homomorphism $R\to S$, the map $R\to S$ remains injective after tensoring with any $R$-module. In particular it identifies $R$ with a subring of $S$, and for every ideal $I\subset R$ one has $R\cap IS=I$.

Proof. These are precisely the conclusions of Faithful flatness and the local criterion for flatness, Theorem 2.2. Its proof tensors $N\to N\otimes_R S$ once more with $S$, exhibits multiplication on the two $S$-factors as a retraction, and reflects injectivity by faithful flatness. Taking $N=R/I$ gives the asserted contraction, for every ideal without a finite-generation condition. ∎

Lemma. Characterizations of henselian local rings

Let $(R,\mathfrak m,\kappa)$ be a local ring. Bars denote reduction to $\kappa$. The following thirteen conditions are equivalent:

1. $R$ is henselian: every simple residue root of a monic polynomial over $R$ lifts to a root in $R$.
2. For every $f\in R[T]$, with no monicity requirement, each $a_0\in\kappa$ satisfying $\bar f(a_0)=0$ and $\overline{f'}(a_0)\ne0$ lifts to a root of $f$.
3. For monic $f\in R[T]$, every factorization $\bar f=g_0h_0$ with $g_0,h_0\in\kappa[T]$ coprime lifts to $f=gh$ with $\bar g=g_0$, $\bar h=h_0$.
4. The factors in (3) can always be chosen with $\deg g=\deg g_0$.
5. The factorization assertion of (3) holds for every $f\in R[T]$, not necessarily monic.
6. The factors in (5) can always be chosen with $\deg g=\deg g_0$ whenever $g_0\ne0$.
7. Whenever an étale $R$-algebra $S$ has a prime $\mathfrak q$ above $\mathfrak m$ with canonical residue extension $\kappa\to\kappa(\mathfrak q)$ an isomorphism, $R\to S$ has an $R$-algebra retraction $S\to R$.
8. For each $(S,\mathfrak q)$ in (7), there is exactly one such retraction $\tau$ with $\tau^{-1}(\mathfrak m)=\mathfrak q$.
9. Every finite $R$-algebra is a product of local rings.
10. Every finite $R$-algebra is a finite product of local rings.
11. Each finite type $R$-algebra $S$ has a decomposition $S=A\times B$, with $A$ finite over $R$, such that $R\to B$ is not quasi-finite at any point above $\mathfrak m$.
12. Each finite type $R$-algebra $S$ has a decomposition $S=A\times B$, with $A$ finite over $R$, such that every irreducible component of $\operatorname{Spec}(B\otimes_R\kappa)$ has dimension at least one.
13. Each quasi-finite $R$-algebra $S$ has a decomposition $S=A\times B$, with $A$ finite over $R$ and $B\otimes_R\kappa=0$.

Empty products are allowed for the zero algebra. No Noetherian condition is imposed on $R$.

Proof. We first connect the root, section and finite-algebra conditions, then treat arbitrary polynomial factorizations.

The equivalence of (1) and (8), including uniqueness through the specified point, is the complete section criterion in Henselian local rings and henselization, Theorem 1.2. Its proof uses a standard étale neighborhood to turn the section problem into the specified simple-root problem. Lemma 1.1 there proves uniqueness by applying Nakayama to the finite ideal of differences of two evaluations. The equivalence of (1) and (10) is the complete Theorem 2.2 of that lesson: coprime factorization lifts idempotents in every finite algebra, and a lifted rank-one factor recovers a simple root.

Certainly (8) implies (7). Conversely, suppose (7) and fix its point $\mathfrak q$. The closed fibre of $S$ is a finite product of finite separable fields. Choose an element of $S$ whose image is one in the factor belonging to $\mathfrak q$ and zero in all other factors. Inverting it gives an étale algebra with exactly one point above $\mathfrak m$, still with residue field $\kappa$. Apply (7) to that algebra. Any resulting retraction has inverse image of $\mathfrak m$ equal to its sole closed-fibre point. Restrict to $S$ to get the required retraction through $\mathfrak q$. Uniqueness is Lemma 1.1 just cited. Thus (7) implies (8).

Conditions (9) and (10) are equivalent. A finite algebra over a local ring has only finitely many maximal ideals: all contract to $\mathfrak m$ by integrality and correspond to maximal ideals in its finite-dimensional residue algebra. A product of nonzero local rings has at least as many distinct maximal ideals as factors, using the coordinate projections. Therefore any product decomposition as in (9) already has finitely many factors. The reverse implication is immediate.

Under (8), let $f$ and $a_0$ be as in (2). The algebra $R[T]_{f'}/(f)$ is étale, by its invertible one-by-one Jacobian; evaluation at $a_0$ defines a closed-fibre point of residue field $\kappa$. The retraction supplied by (8) sends $T$ to the desired root. Hence (8) implies (2), and (2) plainly implies (1).

We next connect the finite-type conditions. The full finite-branch decomposition, Étale neighbourhoods, henselization and quasi-finite morphisms, Lemma 4.1, proves (1) implies (11). Its argument isolates each quasi-finite point by algebraic Zariski's Main Theorem, chooses a finite algebra of integral numerators, and uses its henselian local factors to obtain the actual idempotent decomposition of $S$. It applies to finite type algebras without assuming finite presentation. A finite type scheme over a field has a zero-dimensional irreducible component exactly at an isolated point, and its quasi-finite points are exactly those isolated points. Thus the conditions on $B$ in (11) and (12) agree. When $S$ is quasi-finite everywhere, (11) forces its residual closed fibre to be empty, giving (13).

The étale-neighborhood route to (11) and (10) is also useful. Under (8), apply Making a quasi-finite algebra finite étale locally to $S$ and $\mathfrak m$. It gives an étale $R\to R'$ with a point $\mathfrak m'$ of residue field $\kappa$ and

where each $A'_i$ is finite with one point over $\mathfrak m'$, while $B'$ has no quasi-finite point there. The retraction $\tau:R'\to R$ through $\mathfrak m'$ gives

The residue map of $\tau$ is the prescribed identity on $\kappa$. Consequently the closed fibres of these factors identify with their old fibres at $\mathfrak m'$. Finiteness is preserved by this base change, so their product is the required $A$ and the last factor is the required $B$. If $S$ was finite, the last factor is finite with empty closed fibre and hence zero by Nakayama. Each remaining finite factor has one maximal ideal, giving (10) by this second route as well.

To prove (13) implies (1), start with monic $f$ and a simple root $a_0$ of $\bar f$. Choose a polynomial $u\in R[T]$ reducing to $\bar f/(T-a_0)$, and put

This algebra is quasi-finite, being a principal localization of a finite algebra, and its closed fibre is $\kappa$: the polynomial $u$ removes all factors except the simple root $a_0$. It is flat over $R$, since $R[T]/(f)$ is finite free and localization is flat. Under (13), write $S=A\times B$ with $A$ finite and $B/\mathfrak mB=0$. Then $A/\mathfrak mA=\kappa$. The factor $A$ is flat over $R$, hence finite free by the finite-flat local theorem, and its rank is one. Its unit is a basis: it spans the reduction, so Nakayama makes it a generator, and a generator of a rank-one free module is a basis. The map $R\to A$ is therefore an algebra isomorphism. The image of $T$ in $A=R$ is the requested root. This completes the equivalence of (1), (2), (7)–(13).

It remains to include the four factorization conditions. There are immediate implications

We will prove (3) implies (1) and (1) implies (6).

Assume (3), and write $\bar f=(T-a_0)h_0$ for a monic simple-root problem. Lift this factorization to $f=gh$, without imposing degree or monicity on the lifted factors. In the finite free algebra $D=R[T]/(f)$, the images of $g$ and $h$ generate the unit ideal: the quotient by both is finite and has zero reduction, so Nakayama applies. Since their product is zero, their generated ideals have zero intersection. Chinese remainders give

The first factor is finite projective over $R$, hence free because $R$ is local. Its reduction is $\kappa[T]/(T-a_0)=\kappa$, so it has rank one. The unit-basis argument of the preceding paragraph identifies it with $R$ and gives a root of $f$ lifting $a_0$. Thus (3) implies (1).

Finally assume (1) and take $\bar f=g_0h_0$ with coprime factors, allowing $f$ to be nonmonic. If either residue factor vanishes, coprimality makes the other a nonzero scalar. Choose a unit of $R$ representing that scalar and use it as the corresponding factor of $f$; division of $f$ by this unit supplies the remaining factor. Their residues are the prescribed pair. In the case $g_0\ne0$ this construction makes $g$ a unit constant, so its degree is the required zero. This proves (6) for all vanishing-factor cases.

Now both residue factors are nonzero. If the leading coefficient of $g_0$ is $\lambda\in\kappa^\times$, choose a unit $v\in R$ lifting $\lambda$ and replace the prescribed pair by $(\lambda^{-1}g_0,\lambda h_0)$. Once this normalized pair is lifted, multiplication of its first lift by $v$ and its second by $v^{-1}$ restores the original residues and degrees. We may thus construct the lift with $g_0$ monic.

Some coefficient of $f$ is a unit of $R$, because $\bar f\ne0$. Hence the image of $f$ in $\kappa(\mathfrak p)[T]$ is nonzero for every $\mathfrak p\subset R$. It follows that $S=R[T]/(f)$ has finite fibres and is quasi-finite. It is also flat over $R$: at each prime of $R[T]$ containing $f$, apply the fibrewise nonzerodivisor criterion to the essentially finitely presented flat local map from the corresponding localization of $R$. Its fibre is a localization of a polynomial ring over a field, where this nonzero polynomial is a nonzerodivisor. The criterion gives flatness of the quotient at every prime, and locality of flatness gives the assertion for $S$.

By (13) and (10), decompose $S$ into finite local factors and a factor with empty closed fibre. The coprime factorization gives

The finite local factors of $S$ reduce to exactly the local factors of this Artinian algebra. Let $A$ be the product of those belonging to $\kappa[T]/(g_0)$. It is finite flat over $R$, hence free, of rank $r=\deg g_0$. Let $g$ be the characteristic polynomial of multiplication by the image of $T$ on $A$. It is monic of degree $r$ and reduces to $g_0$, since multiplication by $T$ on $\kappa[T]/(g_0)$ has characteristic polynomial $g_0$.

Cayley–Hamilton gives a surjection $R[T]/(g)\to A$. Both modules are free of rank $r$, and reduction makes this map an isomorphism. Its matrix has unit determinant, so it is an isomorphism over $R$; the rank-zero case is the isomorphism of zero modules. Since $f$ vanishes in $A$, we obtain $f=gh$ in $R[T]$. Reducing and cancelling the nonzero polynomial $g_0$ in $\kappa[T]$ gives $\bar h=h_0$. The normalization step then restores any original leading coefficient. This proves (6) and finishes all thirteen equivalences. ∎

Lemma. Extending a henselian lifting problem to a finite algebra

Let $R\to S$ be a local homomorphism of local rings, and write $S\to S^h$ for the henselization. Suppose $R\to A$ is étale and $\mathfrak q\subset A$ lies above $\mathfrak m_R$ with canonical isomorphism $R/\mathfrak m_R\cong\kappa(\mathfrak q)$. There is exactly one $R$-algebra map $f:A\to S^h$ with

In particular $R\to A\xrightarrow f S^h$ agrees with $R\to S\to S^h$.

Proof. The residue field of $S^h$ is canonically that of $S$. Locality of $R\to S$ and the given residue isomorphism specify the evaluation

Apply the general evaluation assertion in Henselian local rings and henselization, Theorem 1.2, to the henselian target $S^h$. It gives the unique $R$-algebra lift of this evaluation. Its residue kernel is $\mathfrak q$, proving the asserted inverse-image condition. Conversely, every $R$-algebra map with that inverse image induces this same evaluation, because $\kappa(\mathfrak q)=\kappa(R)$ and its action on $R$ is fixed. The theorem's uniqueness therefore proves uniqueness in the exact scope stated here. ∎

Lemma. Étale morphisms

The following properties hold for étale ring maps.

1. Every principal localization $R\to R_f$ is étale, including $f=0$.
2. The composite of two étale maps is étale.
3. Any base change of an étale map is étale.
4. Suppose $g_1,\ldots,g_m\in S$ generate the unit ideal. If every $R\to S_{g_i}$ is étale, then $R\to S$ is étale.
5. If $R\to S$ is finitely presented and $R\to R'$ is flat, put $S'=R'\otimes_R S$. The étale locus of $S'/R'$ is exactly the inverse image of the étale locus of $S/R$.
6. Every étale map is syntomic and hence flat.
7. For a finite type algebra $S$ over a field $k$, étaleness is equivalent to $\Omega_{S/k}=0$.
8. Every étale $R$-algebra $S$ is obtained by base change from an étale $R_0$-algebra $S_0$, where $R_0\subseteq R$ is a finite type $\mathbf Z$-algebra.
9. If $A=\operatorname{colim}_i A_i$ is a filtered colimit of rings and $B$ is étale over $A$, some stage has an étale algebra $B_i/A_i$ with $A\otimes_{A_i}B_i\cong B$.
10. If $U\subset A$ is multiplicative and $B'$ is étale over $U^{-1}A$, there is an étale $A$-algebra $B$ such that $U^{-1}B\cong B'$.
11. For $B=B'\times B''$ as $A$-algebras, $B/A$ is étale if and only if both factors are étale over $A$.