Weakly regular ideals
Written by Claude Opus 5.5 (Anthropic) on 22 October 2026. Self-checked by the writing AI. Public domain (CC0).
Let \(I\) be an ideal of a ring \(A\) of a topos and \(B=A/I\). The sheaves \(\operatorname{Tor}^A_q(B,B)\) form a strictly graded commutative \(B\)-algebra, \(\operatorname{Tor}^A_1(B,B)\) is the conormal module \(I/I^2\), and the inverse of this isomorphism extends to a homomorphism \(\theta\) from the exterior algebra of \(I/I^2\). The ideal \(I\) is weakly regular when \(I/I^2\) is flat over \(B\) and \(\theta\) is an isomorphism. A weakly regular ideal behaves like an ideal generated by a regular sequence: its associated graded ring is the symmetric algebra of \(I/I^2\), and the Koszul complex of \(I/I^2\) computes the sheaves \(\operatorname{Tor}^A_q(B,A/I^{n+1})\). The lesson proves:
- the homotopy of a simplicial ring is strictly graded commutative: classes of odd degree have square zero;
- \(\operatorname{Tor}^A_\bullet(B,B)\) is a strictly graded commutative \(B\)-algebra, independent of the flat resolution used to compute it, and the inverse of \(\operatorname{Tor}^A_1(B,B)\cong I/I^2\) extends to \(\theta\colon\Lambda_B(I/I^2)\to\operatorname{Tor}^A_\bullet(B,B)\);
- the sheaves \(\operatorname{Tor}^A_q(B,I^n/I^{n+1})\) form a bigraded algebra on which the connecting maps act as a derivation, and \(\theta\) extends to a homomorphism from \(\Lambda_B(I/I^2)\otimes_BS_B(I/I^2)\) that carries the Koszul differential to the connecting maps;
- if \(I\) is weakly regular, this homomorphism is an isomorphism; so \(S^n_B(I/I^2)\cong I^n/I^{n+1}\), and for \(q\geq1\) the Koszul complex resolves \(\operatorname{Tor}^A_q(B,A/I^{n+1})\);
- the ideal of the variables of the symmetric algebra of a flat module is weakly regular, and so is every Koszul-regular ideal.
Prerequisites. Simplicial rings and modules in a topos (Proposition 2.1); Derived tensor products of simplicial modules (Definition 3.1, Propositions 2.2, 3.2, 3.3 and 3.4, Lemma 5.1); Non-additive functors of complexes in a ringed topos (Sections 2 and 5, Proposition 2.3, Theorem 4.4, Example 4.7); Koszul complexes and the shift formulas (Proposition 1.1, Theorems 2.1 and 3.2, Proposition 3.4, Exercise 4.4); Filtered derived categories (the triangle of a short exact sequence, in the conventions); The cotangent complex (Theorems 1.2 and 1.3). From the AI-integrated Stacks project: the standard resolution of a sheaf of rings [Stacks, Tag 08SR]; flat modules and exactness [Tag 03EX]; bounded above complexes of flat modules are K-flat [Tag 06YQ]; the triangle of a short exact sequence [Tag 0152]; Koszul-regular sequences [Tag 062E].
Conventions
\(T\) is a topos and rings are commutative. For a simplicial module \(E\) over a simplicial ring, \(H_n(E)=\pi_n(E)\) is the \(n\)-th homology sheaf of its normalized complex \(NE\). For a ring \(A\) of \(T\) and \(A\)-modules \(M\), \(N\), \(\operatorname{Tor}^A_q(M,N)=H_q(M\otimes^\ell_AN)\), the derived tensor product of the lesson on derived tensor products (Definition 3.1 there). It is the classical one (Proposition 3.4 there): if \(L\to M\) is a quasi-isomorphism from a simplicial \(A\)-module with flat terms, or from a complex of flat \(A\)-modules in nonnegative homological degrees, then \(\operatorname{Tor}^A_q(M,N)=H_q(L\otimes_AN)\) (Proposition 3.2 (1) there, and [Tag 06YQ] for complexes). A short exact sequence \(0\to N'\to N\to N''\to0\) gives the exact sequence \(0\to L\otimes_AN'\to\) \(L\otimes_AN\to L\otimes_AN''\to0\) and hence the long exact sequence of \(\operatorname{Tor}\) sheaves, with the connecting maps of the snake lemma: lift a cycle, apply the differential, pull back. These are the maps on homology of the image under \(M\otimes^\ell_A-\) of the triangle of the sequence [Tag 0152] (Proposition 3.3 of the lesson on derived tensor products, and the conventions of the lesson on filtered derived categories), so they do not depend on \(L\); the same holds in the first variable. For a module \(F\) over a ring \(B\), \(S_B(F)\) and \(\Lambda_B(F)\) are the symmetric and exterior algebras, \(S^{\geq m}_B(F)=\bigoplus_{k\geq m}S^k_B(F)\) and \(S^+_B(F)=S^{\geq1}_B(F)\). If \(F\) is flat, so are the \(S^k_B(F)\) and \(\Lambda^k_B(F)\) (Example 4.7 of the lesson on non-additive functors).
For local sections \(f_1,\dots,f_r\) of \(A\) over an object \(U\) of \(T\), the Koszul complex \(K(f)\) is the exterior algebra \(\Lambda_{A_U}(A_U^r)\) on \(e_1,\dots,e_r\) with the derivation \(d\) of degree \(-1\) that maps \(e_i\) to \(f_i\): \[ \begin{aligned}&d(e_{i_1}\wedge\cdots\wedge e_{i_k})\\&\quad=\sum_{j=1}^k(-1)^{j-1}f_{i_j}\,e_{i_1}\wedge\cdots\widehat{e_{i_j}}\cdots\wedge e_{i_k}.\end{aligned} \] It satisfies \(d\circ d=0\) by the argument of Proposition 1.1 of the lesson on Koszul complexes, so it is a differential graded \(A_U\)-algebra, augmented to \(A_U/(f_1,\dots,f_r)\). The sequence is Koszul-regular if \(K(f)\) has no homology in positive degrees [Tag 062E]. An ideal \(I\) of \(A\) is Koszul-regular if every object of \(T\) has a covering by objects \(U\) over which \(I\) is generated by a Koszul-regular sequence.
1. Strict graded commutativity
Lemma 1.1. Let \(R\) be a simplicial ring of \(T\), and let \(NR\) carry the product of Proposition 2.1 of the lesson on simplicial rings and modules, which is graded commutative.
- For \(x\in N_nR\) with \(n\) odd, \(x\cdot x=0\).
- Every local section \(y\) of \(\pi_n(R)\), \(n\) odd, satisfies \(y^2=0\).
- Let \(M\) be a \(\pi_0(R)\)-module and \(f\colon M\to\pi_1(R)\) a \(\pi_0(R)\)-linear map. There is a unique homomorphism of graded \(\pi_0(R)\)-algebras \(\Lambda_{\pi_0(R)}(M)\to\pi_\bullet(R)\) extending \(f\).
Proof. (1) The product is the shuffle map followed by the multiplication of \(R_{2n}\). By Theorem 1.3 of The cotangent complex, \[ x\cdot y=\sum_{(\mu,\nu)}(-1)^{\varepsilon(\mu,\nu)}\,(s_\nu x)(s_\mu y) \] for \(x,y\in R_n\), where \((\mu,\nu)\) runs over the partitions of \(\{0,\dots,2n-1\}\) into two increasing sequences of length \(n\), \(s_\nu=s_{\nu_n}\cdots s_{\nu_1}\), \(s_\mu=s_{\mu_n}\cdots s_{\mu_1}\), and \(\varepsilon(\mu,\nu)\) is the sign of the permutation \((\mu_1,\dots,\mu_n,\nu_1,\dots,\nu_n)\). Exchanging \(\mu\) and \(\nu\) is an involution of this set of partitions without fixed points, since \(\mu\) and \(\nu\) are disjoint and nonempty, and \(\varepsilon(\nu,\mu)\equiv\varepsilon(\mu,\nu)+n^2\) modulo \(2\), the two permutations differing by the exchange of two blocks of \(n\) entries. For \(y=x\) the terms of \((\mu,\nu)\) and \((\nu,\mu)\) are \((-1)^{\varepsilon(\mu,\nu)}a\) and \((-1)^{\varepsilon(\mu,\nu)+n^2}a\) with the same \(a=(s_\nu x)(s_\mu x)\), because \(R_{2n}\) is commutative. For \(n\) odd they cancel.
(2) Locally \(y\) is the class of a cycle \(x\in N_nR\), and \(y^2\) is the class of \(x\cdot x\).
(3) The homomorphism of \(\pi_0(R)\)-algebras from the tensor algebra of \(M\) induced by \(f\) maps \(m\otimes m\) to \(f(m)^2=0\), by (2). So it factors through \(\Lambda_{\pi_0(R)}(M)\), the quotient of the tensor algebra by the ideal generated by these sections; uniqueness holds because \(M\) generates \(\Lambda_{\pi_0(R)}(M)\). \(\square\)
2. The Tor algebra and weakly regular ideals
Let \(A\) be a ring of \(T\) and \(B\) an \(A\)-algebra. By Lemma 5.1 of the lesson on derived tensor products, there is a quasi-isomorphism of simplicial \(A\)-algebras \(P\to B\), natural in \(B\), with each \(P_n\) a polynomial \(A\)-algebra \(A[X_n]\) on a sheaf of sets \(X_n\); for the constant algebra \(B\) it is the standard resolution [Tag 08SR]. Put \(Q=P\otimes_AB\), a simplicial \(B\)-algebra through the second factor.
Definition 2.1. The Tor algebra of \(B\) over \(A\) is \(\operatorname{Tor}^A_\bullet(B,B)=\pi_\bullet(Q)\), with the product of Proposition 2.1 of the lesson on simplicial rings and modules. By that proposition and Lemma 1.1 it is a strictly graded commutative \(\pi_0(Q)\)-algebra, and \(\pi_0(Q)=B\otimes_AB\).
Proposition 2.2.
- \(\pi_q(Q)=\operatorname{Tor}^A_q(B,B)\) for every \(q\).
- Let \(W\to B\) be a quasi-isomorphism of simplicial \(A\)-algebras with each \(W_n\) flat over \(A\). Then there is an isomorphism of graded rings \(\pi_\bullet(W\otimes_AB)\cong\pi_\bullet(Q)\) which, under (1) and the analogous identification of \(\pi_\bullet(W\otimes_AB)\) with \(\operatorname{Tor}^A_\bullet(B,B)\), is the identity.
- A commutative square of rings from \(A\to B\) to \(A'\to B'\) induces a homomorphism of graded rings \(\operatorname{Tor}^A_\bullet(B,B)\to\operatorname{Tor}^{A'}_\bullet(B',B')\), compatible with composition of squares and, under (1), with the maps of \(\operatorname{Tor}\) sheaves.
Proof. (1) \(P\) is flat over \(A\), so \(Q\) represents \(B\otimes^\ell_AB\) (Proposition 3.2 (1) of the lesson on derived tensor products).
(2) Let \(P'\to W\) be the resolution of Lemma 5.1 of the lesson on derived tensor products, applied to the \(A\)-algebra \(W\), and let \(P'\to P\) be the map that its naturality gives for \(W\to B\); both are maps of simplicial \(A\)-algebras over \(B\), and quasi-isomorphisms. In \[ P\otimes_AB\longleftarrow P'\otimes_AB\longrightarrow W\otimes_AB \] both maps are maps of simplicial rings and quasi-isomorphisms, by Proposition 2.2 there, since \(P\), \(P'\) and \(W\) are flat. So they induce isomorphisms of graded rings on \(\pi_\bullet\). They are transition maps of the essentially constant pro-object of Proposition 3.2 (2) there, so under the identifications of Proposition 3.2 (1) there they induce the identity of \(\operatorname{Tor}^A_\bullet(B,B)\).
(3) The free-algebra construction is functorial in the square, so it gives a map of simplicial rings \(P\to P'\) over \(B\to B'\), hence \(Q\to Q'=P'\otimes_{A'}B'\). On \(\pi_\bullet\) it is a homomorphism of graded rings, and on underlying modules it is the map \(B\otimes^\ell_AB\to B'\otimes^\ell_{A'}B'\). \(\square\)
2.3. The conormal isomorphism. Let \(I\subseteq A\) be an ideal and \(B=A/I\). Then \(B\otimes_AB=B\). Since \(P\) is flat, \(0\to P\otimes_AI\to P\to P\otimes_AB\to0\) is an exact sequence of simplicial modules, and \(\operatorname{Tor}^A_q(B,N)=\pi_q(P\otimes_AN)\) for every \(A\)-module \(N\). Its long exact sequence contains \[ \begin{aligned}&0=\pi_1(P)\to\operatorname{Tor}^A_1(B,B)\\ &\quad\xrightarrow{\partial}B\otimes_AI\to B\otimes_AA,\end{aligned} \] in which \(B\otimes_AI=I/I^2\) and the last map, induced by \(I\to A\), is zero. So the connecting map is an isomorphism \(\partial\colon\operatorname{Tor}^A_1(B,B)\to I/I^2\). It is natural in the pair \((A,I)\): a ring homomorphism \(A\to A'\) mapping \(I\) into an ideal \(I'\) gives a map \(P\to P'\) as in the proof of Proposition 2.2 (3), and hence a map of these exact sequences.
Definition 2.4. Let \(B=A/I\). The homomorphism of graded \(B\)-algebras \[ \theta=\theta_{A,I}\colon\Lambda_B(I/I^2)\to\operatorname{Tor}^A_\bullet(B,B) \] is the one extending \(\partial^{-1}\) (Lemma 1.1 (3), with \(\pi_0(Q)=B\)). The ideal \(I\) is weakly regular if \(I/I^2\) is a flat \(B\)-module and \(\theta\) is an isomorphism.
By Proposition 2.2 (3), \(\theta\) is natural: a ring homomorphism \(A\to A'\) mapping \(I\) into an ideal \(I'\) commutes with \(\theta_{A,I}\) and \(\theta_{A',I'}\), both being homomorphisms of algebras out of an algebra generated in degree one on which they agree. By Proposition 2.2 (2), \(\theta\) can be computed with any resolution \(W\) as there, and whether \(I\) is weakly regular does not depend on that choice. The resolution \(P\) is built from free algebras on sheaves of sets, whose formation commutes with restriction to the objects of \(T\); so \(\operatorname{Tor}^A_\bullet(B,B)\), \(\partial\) and \(\theta\) commute with restriction, and weak regularity is a local property.
3. The bigraded algebra of the powers of an ideal
In this section \(I\) is an ideal of a ring \(A\) of \(T\), \(B=A/I\), \(N=I/I^2\), and \(P\to B\) is the resolution of Section 2. Put \(I^k=A\) for \(k\leq0\) and \(G_n=I^n/I^{n+1}\), so that \(G_n=0\) for \(n<0\), and let \(G=\bigoplus_{n\geq0}G_n=\operatorname{gr}_I(A)\), a graded commutative \(A\)-algebra with \(G_0=B\) and \(G_1=N\). Since \(P\) is flat over \(A\), \(P\otimes_AG\) is a simplicial graded commutative ring, and \[ E_{q,n}=\pi_q(P\otimes_AG_n)=\operatorname{Tor}^A_q(B,G_n). \] By Proposition 2.1 of the lesson on simplicial rings and modules and Lemma 1.1, applied to the simplicial ring \(P\otimes_AG\), \(E=\bigoplus_{q,n}E_{q,n}\) is a bigraded ring, strictly graded commutative for the degree \(q\); the degree \(n\) plays no part in the signs. Moreover \(E_{0,\bullet}=B\otimes_AG=G\) and \(E_{\bullet,0}=\operatorname{Tor}^A_\bullet(B,B)\), so \(E\) is a bigraded \(B\)-algebra. Let \[ \delta\colon E_{q,n}\to E_{q-1,n+1} \] be the connecting map of the exact sequence \(0\to G_{n+1}\to I^n/I^{n+2}\to G_n\to0\).
Lemma 3.1. For \(x\in E_{q,n}\) and \(y\in E_{q',n'}\), \(\delta(xy)=\delta(x)\,y+(-1)^qx\,\delta(y)\). Moreover \(\delta\circ\delta=0\).
Proof. Let \(\tilde G=\bigoplus_{n\geq0}I^n/I^{n+2}\), a graded ring under the multiplication \(I^n/I^{n+2}\times I^{n'}/I^{n'+2}\to I^{n+n'}/I^{n+n'+2}\), and let \(K\subseteq\tilde G\) be the ideal with \(K_n=I^{n+1}/I^{n+2}\). Then \(\tilde G/K=G\), \(K\cdot K=0\), and \(K\) is the \(G\)-module \(G\) with its degrees lowered by one. Since \(P\) is flat, \(0\to P\otimes_AK\to\) \(P\otimes_A\tilde G\to P\otimes_AG\to0\) is an exact sequence of simplicial modules, \(P\otimes_A\tilde G\) is a simplicial ring and \(P\otimes_AK\) is an ideal of square zero. Let \(x\) and \(y\) be represented, locally, by cycles \(x_0\in N_q(P\otimes_AG_n)\) and \(y_0\in N_{q'}(P\otimes_AG_{n'})\), with lifts \(\tilde x\) and \(\tilde y\) in \(N(P\otimes_A\tilde G)\). Then \(d\tilde x\in N_{q-1}(P\otimes_AK_n)\) represents \(\delta(x)\), and \(\tilde x\tilde y\) lifts \(x_0y_0\). By the Leibniz rule of the differential graded ring \(N(P\otimes_A\tilde G)\) (Proposition 2.1 of the lesson on simplicial rings and modules), \[ d(\tilde x\tilde y)=(d\tilde x)\,\tilde y+(-1)^q\,\tilde x\,(d\tilde y), \] and \((d\tilde x)\tilde y=(d\tilde x)\,y_0\), \(\tilde x(d\tilde y)=x_0\,(d\tilde y)\) for the module structure of \(P\otimes_AK\) over \(P\otimes_AG\), because \(K\cdot K=0\). This is the first formula. For the second: by naturality, the connecting map of \(0\to G_{n+1}\to I^n/I^{n+2}\to G_n\to0\) is the connecting map of \(0\to I^{n+1}/I^{n+3}\to I^n/I^{n+3}\to G_n\to0\) followed by the projection to \(G_{n+1}\), and that projection followed by the connecting map of \(0\to G_{n+2}\to I^{n+1}/I^{n+3}\to G_{n+1}\to0\) is zero. \(\square\)
Proposition 3.2. There is a unique homomorphism of bigraded \(B\)-algebras \[ \theta\colon\Lambda_B(N)\otimes_BS_B(N)\to E, \] with \(\Lambda^q_BN\otimes_BS^n_BN\) in bidegree \((q,n)\), that is \(\partial^{-1}\) on \(\Lambda^1_BN=N\) and the identity of \(N=G_1=E_{0,1}\) on \(S^1_BN\). Its components \(\theta_{q,0}\) form the homomorphism \(\theta\) of Definition 2.4, its components \(\theta_{0,n}\colon S^n_BN\to G_n=I^n/I^{n+1}\) are the multiplication maps, it is natural in the pair \((A,I)\), and \(\delta\circ\theta=\theta\circ d_K\), where \(d_K\) is the Koszul differential of \(1_N\) (Proposition 1.1 of the lesson on Koszul complexes): \[ \begin{aligned}&d_K(x_1\wedge\cdots\wedge x_q\otimes s)\\&\quad=\sum_{j=1}^q(-1)^{j-1}\,x_1\wedge\cdots\widehat{x_j}\cdots\wedge x_q\\&\qquad\otimes x_js.\end{aligned} \]
Proof. The map \(\partial^{-1}\colon N\to E_{1,0}\) extends to the homomorphism \(\theta\) of Definition 2.4 from \(\Lambda_B(N)\) to \(E_{\bullet,0}\), and the identity of \(N\) extends to the homomorphism of commutative \(B\)-algebras \(S_B(N)\to G=E_{0,\bullet}\), given by the multiplication maps. Elements of \(E_{0,\bullet}\) have even degree, so they commute with all of \(E\), and the two homomorphisms combine to a homomorphism \(\theta\) on the tensor product; it is unique because \(N\) in the bidegrees \((1,0)\) and \((0,1)\) generates \(\Lambda_B(N)\otimes_BS_B(N)\). A homomorphism \((A,I)\to(A',I')\) gives maps \(P\to P'\) and \(G\to G'\), hence a homomorphism of bigraded algebras \(E\to E'\) commuting with \(\delta\); its composite with \(\theta\) and the composite of \(\theta'\) with the induced map agree on generators, by the naturality of \(\partial\), so they are equal.
The maps \(D=\delta\circ\theta\) and \(D'=\theta\circ d_K\) are additive of bidegree \((-1,1)\), and both satisfy \(D(ab)=D(a)\theta(b)+(-1)^q\theta(a)D(b)\) for \(a\) of degree \(q\), by Lemma 3.1 and because \(d_K\) is a derivation. Such a map is determined by its values on \(B\) and on the generators. On \(B\) and on \(S^1_BN\) both vanish, their values lying in \(E_{-1,\bullet}=0\). For \(x\in N=\Lambda^1_BN\), \(D'(x)=\theta(1\otimes x)=x\in E_{0,1}\). The inclusions of \(0\to I\to A\to B\to0\) into \(0\to I/I^2\to A/I^2\to B\to0\) show that \(\delta\colon E_{1,0}\to E_{0,1}\) is \(\partial\) followed by the identity of \(I/I^2=\pi_0(P\otimes_AI)=\pi_0(P\otimes_AI/I^2)\). So \(D(x)=\delta(\partial^{-1}x)=x\), and \(D=D'\). \(\square\)
Theorem 3.3. Let \(I\) be weakly regular.
- \(\theta\) is an isomorphism. In particular the multiplication maps \(S^n_B(I/I^2)\to I^n/I^{n+1}\) are isomorphisms, the \(B\)-modules \(I^n/I^{n+1}\) are flat, and \(\operatorname{Tor}^A_q(B,I^n/I^{n+1})\cong\) \(\Lambda^q_BN\otimes_BS^n_BN\).
- For \(q\geq1\) and \(k\geq0\), the map \(\operatorname{Tor}^A_q(B,A/I^{k+1})\to\) \(\operatorname{Tor}^A_q(B,A/I^k)\) induced by the projection is zero.
Proof. Write \(T_q(M)=\operatorname{Tor}^A_q(B,M)\). For \(k\geq1\) let \[ \begin{aligned}&0\to G_{k-1}\xrightarrow{i_k}A/I^k\\&\quad\xrightarrow{j_k}A/I^{k-1}\to0\end{aligned}\tag{3.1} \] be the exact sequence of the inclusion and the projection, and \(\partial_k\colon T_{q+1}(A/I^{k-1})\to T_q(G_{k-1})\) its connecting map. Three facts are used.
(a) For \(k\geq1\), \(\delta=\partial_{k+1}\circ(i_k)_*\) as maps \(T_{q+1}(G_{k-1})\to T_q(G_k)\): the inclusions map the sequence \(0\to G_k\to I^{k-1}/I^{k+1}\to G_{k-1}\to0\) to (3.1) for \(k+1\), with the identity on \(G_k\). Moreover \((i_k)_*\circ\partial_k=0\), as consecutive maps of a long exact sequence.
(b) If \(G_k\) is flat over \(B\), the product \(E_{q,0}\otimes_BG_k\to E_{q,k}\) is an isomorphism. Indeed \(P\otimes_AG_k=Q\otimes_BG_k\), so \(E_{q,k}=\pi_q(Q)\otimes_BG_k\) because \(G_k\) is flat [Tag 03EX], and the shuffle product of \(x\in N_q(Q)\) with the element \(1\otimes g\) of degree \(0\) is \(x\otimes g\). Hence, if \(\theta_{0,k}\) is an isomorphism, then \(G_k\cong S^k_BN\) is flat, and \(\theta_{q,k}\), which is \(\theta_{q,0}\otimes\theta_{0,k}\) followed by this product, is an isomorphism for every \(q\).
(c) For \(k\geq1\), \(\partial_{k+1}\colon T_1(A/I^k)\to T_0(G_k)=G_k\) is an isomorphism. Indeed the sequence \(0\to I\to A\to B\to0\) in the first variable identifies \(T_1(M)\) with the kernel of \(I\otimes_AM\to M\), naturally in \(M\). For \(M=A/I^k\) this kernel is \(I^k/I^{k+1}\subseteq I/I^{k+1}\), and the map \(T_1(A/I^{k+1})\to T_1(A/I^k)\) becomes the restriction of the projection \(I/I^{k+2}\to I/I^{k+1}\) to \(I^{k+1}/I^{k+2}\), which is zero. In the long exact sequence of (3.1) for \(k+1\), \(\partial_{k+1}\) is therefore injective on \(T_1(A/I^k)\), and it is surjective because the next map \(G_k\to T_0(A/I^{k+1})=B\) is zero.
For \(m\geq1\) consider the statements:
- \((A_m)\): \(\theta_{0,k}\colon S^k_BN\to G_k\) is an isomorphism for \(k\leq m\);
- \((B_m)\): \((j_k)_*\colon T_q(A/I^k)\to T_q(A/I^{k-1})\) is zero for \(q\geq1\) and \(1\leq k\leq m\).
\((A_1)\) holds because \(\theta_{0,0}\) and \(\theta_{0,1}\) are identities, and \((B_1)\) because \(A/I^0=0\). Under \((B_m)\), for \(1\leq k\leq m\) and \(q\geq1\) the map \((i_k)_*\colon T_q(G_{k-1})\to T_q(A/I^k)\) is surjective. Under \((A_m)\), by (b), \(\theta_{q,k}\) is an isomorphism for every \(q\) and every \(k\leq m\).
\((A_m)\) and \((B_m)\) imply \((B_{m+1})\). Let \(q\geq0\) and let \(x\in T_{q+1}(A/I^m)\) be in the image of \((j_{m+1})_*\). Then \(\partial_{m+1}(x)=0\), and locally \(x=(i_m)_*(y)\) with \(y\in T_{q+1}(G_{m-1})\). By (a), \(\delta(y)=0\). Through \(\theta_{q+2,m-2}\), \(\theta_{q+1,m-1}\) and \(\theta_{q,m}\), which are isomorphisms, Proposition 3.2 identifies \[ T_{q+2}(G_{m-2})\xrightarrow{\delta}T_{q+1}(G_{m-1})\xrightarrow{\delta}T_q(G_m) \] with a part of weight \(q+m\) of the Koszul complex of \(1_N\), which is exact because \(N\) is flat (Theorem 2.1 of the lesson on Koszul complexes). So locally \(y=\delta(z)\) with \(z\in T_{q+2}(G_{m-2})\), and \(x=(i_m)_*(\delta(z))\). This is zero: for \(m=1\) because \(G_{-1}=0\), and for \(m\geq2\) because \(\delta(z)=\partial_m((i_{m-1})_*z)\) by (a) and \((i_m)_*\circ\partial_m=0\).
\((A_m)\) and \((B_{m+1})\) imply \((A_{m+1})\). In the sequence \[ T_2(G_{m-1})\xrightarrow{\delta}T_1(G_m)\xrightarrow{\delta}T_0(G_{m+1})\to0 \] the second map is \(\partial_{m+2}\circ(i_{m+1})_*\) by (a), the composite of a surjection and, by (c), an isomorphism; so it is surjective. If \(\delta(y)=0\), then \((i_{m+1})_*(y)=0\) by (c), so \(y=\partial_{m+1}(w)\) with \(w\in T_2(A/I^m)\), locally \(w=(i_m)_*(z)\), and \(y=\delta(z)\) by (a). With \(\delta\circ\delta=0\) the sequence is exact. Through \(\theta\) it receives the exact sequence \(\Lambda^2_BN\otimes_BS^{m-1}_BN\to N\otimes_BS^m_BN\) \(\to S^{m+1}_BN\to0\) of the Koszul complex, and \(\theta_{2,m-1}\) and \(\theta_{1,m}\) are isomorphisms; so \(\theta_{0,m+1}\) is an isomorphism.
By induction \((A_m)\) and \((B_m)\) hold for every \(m\). With (b) this gives (1), and (2) is \((B_m)\). \(\square\)
Corollary 3.4. Let \(I\) be weakly regular, \(q\geq1\) and \(n\geq0\). The sequence \[ \begin{aligned}&0\to\Lambda^{q+n}_BN\to\cdots\to\Lambda^{q+1}_BN\otimes_BS^{n-1}_BN\\ &\quad\to\Lambda^q_BN\otimes_BS^n_BN\\ &\quad\to\operatorname{Tor}^A_q(B,A/I^{n+1})\to0,\end{aligned} \] whose maps are Koszul differentials and \((i_{n+1})_*\circ\theta_{q,n}\), is exact. It is natural in the pair \((A,I)\).
Proof. We use the notation of the proof of Theorem 3.3. By Theorem 3.3 (2) and the long exact sequence of (3.1) for \(k=n+1\), \((i_{n+1})_*\colon T_q(G_n)\to T_q(A/I^{n+1})\) is surjective with kernel the image of \(\partial_{n+1}\). For \(n=0\) this image is zero, \(A/I^0\) being \(0\). For \(n\geq1\), \((i_n)_*\colon T_{q+1}(G_{n-1})\to T_{q+1}(A/I^n)\) is surjective by the same theorem, applied with \(k=n-1\), and \(\partial_{n+1}\circ(i_n)_*=\delta\) by naturality, as in the proof of the theorem; so the kernel is the image of \(\delta\). Through the isomorphisms \(\theta\) (Proposition 3.2 and Theorem 3.3), it is the image of the Koszul differential \(\Lambda^{q+1}_BN\otimes S^{n-1}_BN\to\Lambda^q_BN\otimes S^n_BN\). The rest of the sequence is the part of weight \(q+n\) of the Koszul complex of \(1_N\) in the degrees \(\geq q\), which is exact (Theorem 2.1 of the lesson on Koszul complexes). The maps \(\theta\), \(\delta\) and the connecting maps are natural in \((A,I)\). \(\square\)
4. Examples
Proposition 4.1. Let \(B\) be a ring of \(T\), \(F\) a flat \(B\)-module, \(R=S_B(F)\), augmented to \(B\) by \(F\mapsto0\), and \(J=S^+_B(F)\). The ideal \(J\) of \(R\) is weakly regular.
Proof. Here \(R/J=B\) and \(J/J^2=F\). Let \(C\) be the simplicial \(B\)-module that corresponds, under the Dold–Kan correspondence (Theorem 1.2 of The cotangent complex), to the complex \(K_C=(F\xrightarrow{1}F)\) in homological degrees \(1\) and \(0\), and let \(F\subseteq C\) be the constant submodule corresponding to the subcomplex \(F\) in degree \(0\). In each degree \(C_n=F\oplus C'_n\) with \(C'_n\) a finite sum of copies of \(F\), and \(\Sigma F=C/F\) corresponds to \(F\) placed in degree \(1\).
Put \(W=S_B(C)\), with the map \(R=S_B(F)\to S_B(C)\) induced by \(F\subseteq C\), and the augmentation \(W\to S^0_B(C)=B\). Each \(W_n=S_B(F)\otimes_BS_B(C'_n)\) is flat over \(R\). For \(k\geq1\), the normalized complex of \(S^k_B(C)\) is \(S^k(K_C)\) in the notation of Section 2 of the lesson on non-additive functors. The complex \(K_C\) is homotopy equivalent to \(0\) and \(S^k(0)=0\), so \(S^k(K_C)\) is contractible (Proposition 2.3 there). Hence \(W\to B\) is a quasi-isomorphism of simplicial \(R\)-algebras, and Proposition 2.2 (2) gives an isomorphism of graded rings \[ c\colon\operatorname{Tor}^R_\bullet(B,B)\to\pi_\bullet(W\otimes_RB), \] compatible with the identifications with \(\operatorname{Tor}^R_\bullet(B,B)\). Here \(W\otimes_RB=S_B(C)/F\,S_B(C)=S_B(\Sigma F)\). In degree \(0\), \(c\) is the identity of \(B\), so \(c\) is \(B\)-linear.
The normalized complex of \(\Sigma F\) is \(F[1]\) (\(F\) in cohomological degree \(-1\)), a complex of flat modules. By Theorem 4.4 and Section 2 of the lesson on non-additive functors, the normalized complex of \(S^k_B(\Sigma F)\) is \(S^k(F[1])\), which represents \(LS^k(F[1])\); by Theorem 3.2 of the lesson on Koszul complexes, \(LS^k(F[1])\cong L\Lambda^k(F)[k]=\Lambda^k_B(F)[k]\) for \(k\geq1\). So \(\pi_q(S^k_B(\Sigma F))\) is \(\Lambda^q_B(F)\) for \(k=q\) and \(0\) otherwise. Let \(\sigma_q\colon\Lambda^q_B(F)\to\pi_q(S_B(\Sigma F))\), for \(q\geq1\), be the inverse of the isomorphism \(\pi_q(S^q_B(\Sigma F))\to\Lambda^q_B(F)\) induced by the shift isomorphism, and \(\sigma_0\) the identity of \(B\); \(\sigma_1\) is the identity of \(F=\pi_1(\Sigma F)\) (Exercise 4.4 there).
The product of \(\pi_\bullet(S_B(\Sigma F))\), the shuffle map followed by multiplication, is on \(\pi_a\otimes\pi_b\) the map induced by the product \(S^a(F[1])\otimes S^b(F[1])\to S^{a+b}(F[1])\) of Section 5 of the lesson on non-additive functors, because for complexes in nonpositive cohomological degrees the map (5.1) there is the shuffle map. By Proposition 3.4 of the lesson on Koszul complexes, for \(a,b\geq1\) the shift isomorphisms carry this product to the product of \(\Lambda_B(F)\), shifted and composed with \(\tau\), and \(\tau\) multiplies \(x\otimes y\) by \((-1)^{ab}\) for \(x\in\Lambda^a_BF\) and \(y\in\Lambda^b_BF\), because \(x\) has degree \(-a\) in \(\Lambda^a_B(F)[a]\). So \(\sigma_a(x)\sigma_b(y)=(-1)^{ab}\sigma_{a+b}(x\wedge y)\), also when \(a\) or \(b\) is \(0\). Put \(\psi_q=(-1)^{q(q-1)/2}\sigma_q\). Since \[ \tfrac{(a+b)(a+b-1)}2=\tfrac{a(a-1)}2+\tfrac{b(b-1)}2+ab, \] the maps \(\psi_q\) form an isomorphism of graded \(B\)-algebras \(\psi\colon\Lambda_B(F)\to\pi_\bullet(S_B(\Sigma F))\).
Both \(c\circ\theta\) and \(\psi\) are homomorphisms of graded \(B\)-algebras \(\Lambda_B(F)\to\pi_\bullet(S_B(\Sigma F))\) that are isomorphisms in degree \(1\). Put \(\varphi=\psi_1^{-1}\circ c\circ\theta_1\), an automorphism of \(F\). The homomorphisms \(c\circ\theta\) and \(\psi\circ\Lambda(\varphi)\) agree on \(F\), which generates \(\Lambda_B(F)\), so they are equal, and \(\theta=c^{-1}\circ\psi\circ\Lambda(\varphi)\) is an isomorphism. Since \(J/J^2=F\) is flat, \(J\) is weakly regular. \(\square\)
By Theorem 3.3 and Corollary 3.4, \(J^n/J^{n+1}=S^n_BF\), as it is by construction, and for \(q\geq1\) the sheaf \(\operatorname{Tor}^R_q(B,R/J^{n+1})\) is the cokernel of the Koszul differential \(\Lambda^{q+1}_BF\otimes_BS^{n-1}_BF\to\Lambda^q_BF\otimes_BS^n_BF\). Exercise 5.5 obtains this directly from the Koszul resolution of \(B\).
Proposition 4.2. Every Koszul-regular ideal \(I\) of \(A\) is weakly regular. If \(I\) is generated over \(U\) by a Koszul-regular sequence \(f_1,\dots,f_r\), then \(I/I^2\) is free over \(U\), with the classes of the \(f_i\) as a basis.
Proof. Both statements are local, so we may assume that \(f_1,\dots,f_r\) are global sections generating \(I\). Locally, there are \(x_i\in N_1P\) with \(dx_i=f_i\), since the image of \(f_i\) in \(\pi_0(P)=B\) is zero; work over an object where they exist. The normalized complex \(NP\) is a differential graded \(A\)-algebra under the shuffle product, graded commutative (Proposition 2.1 of the lesson on simplicial rings and modules), and every \(v\in N_1P\) satisfies \(v^2=0\) (Lemma 1.1 (1)). So, as in Lemma 1.1 (3), there is a homomorphism of graded \(A\)-algebras \[ \Phi\colon K(f)=\Lambda_A(A^r)\to NP,\quad e_i\mapsto x_i. \] The maps \(d\circ\Phi\) and \(\Phi\circ d\) satisfy the same Leibniz rule along \(\Phi\) and agree on \(A\) and on the \(e_i\), so \(\Phi\) commutes with the differentials. In degree \(0\) it is the structure map \(A\to P_0\), so \(\Phi\) is a map of resolutions of \(B\) by complexes of flat \(A\)-modules, and \(\Phi\otimes_AB\colon K(f)\otimes_AB\to\) \(NP\otimes_AB=N(Q)\) induces an isomorphism on homology, both sides computing \(\operatorname{Tor}^A_\bullet(B,B)\) compatibly with \(\Phi\). The differential of \(K(f)\otimes_AB=\Lambda_B(B^r)\) is zero, since the \(f_i\) vanish in \(B\). So \(H(\Phi\otimes_AB)\colon\Lambda_B(B^r)\to\pi_\bullet(Q)\) is an isomorphism of graded \(B\)-algebras.
In degree one it maps \(e_i\) to the class of \(x_i\otimes1\). Its image under \(\partial\) is the class \(\bar f_i\) of \(f_i\): the cycle \(x_i\otimes1\) of \(N_1(P\otimes_AB)\) lifts to \(x_i\in N_1P\), whose differential \(f_i\) is the image of \(1\otimes f_i\in P_0\otimes_AI\), with class \(\bar f_i\) in \(\pi_0(P\otimes_AI)=I/I^2\). Let \(\phi\colon B^r\to I/I^2\) map \(e_i\) to \(\bar f_i\). Then \(\phi=\partial\circ H_1(\Phi\otimes_AB)\) is an isomorphism, so \(I/I^2\) is free with basis \(\bar f_1,\dots,\bar f_r\). The homomorphisms \(H(\Phi\otimes_AB)\) and \(\theta\circ\Lambda(\phi)\) of graded \(B\)-algebras agree in degree one, so they are equal, and \(\theta=H(\Phi\otimes_AB)\circ\Lambda(\phi)^{-1}\) is an isomorphism. \(\square\)
For a regular sequence in a ring, Section 7 of The cotangent complex reads the Tor algebra off the Koszul complex in the same way.
5. Exercises
Exercise 5.1. Show that the graded ring \(\mathbb F_2[x]\) with \(x\) in degree \(1\) is graded commutative but not strictly graded commutative, and conclude that it is not \(\pi_\bullet\) of a simplicial ring.
Exercise 5.2. Let \(k\neq0\) be a ring of \(T\), \(A=k[t]/(t^2)\), \(I=(t)\) and \(B=k\). Compute \(\operatorname{Tor}^A_q(k,k)\) for all \(q\) from the periodic resolution \(\cdots\to A\xrightarrow{t}A\xrightarrow{t}A\to k\), and show in two ways that \(I\) is not weakly regular although \(I/I^2\) is free over \(B\): through \(\theta\), and through Theorem 3.3.
Exercise 5.3. Let \(k\) be a field, \(A=k[x,y]/(xy)\) and \(I=(x,y)\), so that \(B=k\). Show that \(I/I^2\) is free of rank \(2\) over \(k\), that the multiplication \(S^2_k(I/I^2)\to I^2/I^3\) is not injective, and conclude from Theorem 3.3 that \(I\) is not weakly regular. Then check directly that \(\operatorname{Tor}^A_2(k,k)\) has dimension \(2\), using the complex \(A^2\to A^2\to A\to k\) with the maps \((a,b)\mapsto(ya,xb)\) and \((a,b)\mapsto ax+by\).
Exercise 5.4. Let \(R=B[t]\) and \(J=(t)\), so that \(R=S_B(B)\). Compute \(\operatorname{Tor}^R_q(B,R/J^m)\) from the resolution \(0\to R\xrightarrow{t}R\to B\to0\), and compare with Corollary 3.4.
Exercise 5.5. Let \(F\) be a flat \(B\)-module, \(R=S_B(F)\) and \(J=S^+_B(F)\). The Koszul complex \(K=(\Lambda_BF\otimes_BS_BF,d)\) of \(1_F\) is a resolution of \(B\) by the flat \(R\)-modules \(\Lambda^i_BF\otimes_BR\) (Proposition 1.1 and Theorem 2.1 of the lesson on Koszul complexes). Use it to show directly that, for \(q\geq1\) and \(m\geq1\), \(\operatorname{Tor}^R_q(B,R/J^m)\) is the cokernel of the Koszul differential \(\Lambda^{q+1}_BF\otimes S^{m-2}_BF\to\Lambda^q_BF\otimes S^{m-1}_BF\).
6. Complete solutions
Solution 5.1. For homogeneous \(a\), \(b\) of degrees \(p\), \(q\), \(ab=ba=(-1)^{pq}ba\) because \(-1=1\) in \(\mathbb F_2\); so the ring is graded commutative, and \(x\cdot x=x^2\neq0\) with \(x\) of odd degree. By Lemma 1.1 (2), every class of odd degree in \(\pi_\bullet(R)\) has square zero, so no simplicial ring has this homotopy ring.
Solution 5.2. The resolution is exact, \(t\) having kernel and image \(tA\). Tensoring with \(k\) gives \(\cdots\xrightarrow{0}k\xrightarrow{0}k\), so \(\operatorname{Tor}^A_q(k,k)=k\) for every \(q\geq0\). Here \(I/I^2=tA/0=kt\) is free of rank one, so \(\Lambda_k(I/I^2)\) is zero in degrees \(\geq2\), while \(\operatorname{Tor}^A_2(k,k)=k\neq0\); so \(\theta\) is not surjective in degree \(2\). Alternatively, \(I^2=0\), so \(I^2/I^3=0\), while \(S^2_k(I/I^2)=k\,t^2\neq0\); by Theorem 3.3 (1), \(I\) is not weakly regular.
Solution 5.3. The ring \(A\) has the \(k\)-basis \(1\), \(x^i\), \(y^j\) (\(i,j\geq1\)), and \(xy=0\). So \(I^2=(x^2,y^2)\), \(I^3=(x^3,y^3)\), \(I/I^2\) has the basis \(x\), \(y\), and \(I^2/I^3\) has the basis \(x^2\), \(y^2\). The symmetric square \(S^2_k(I/I^2)\) has the basis \(x^2\), \(xy\), \(y^2\), and \(xy\) maps to \(0\). By Theorem 3.3 (1), \(I\) is not weakly regular.
For the direct check: if \(ax+by=0\) in \(A\), write \(a=a_0+a'(x)+a''(y)\) with \(a_0\in k\), \(a'\in xk[x]\), \(a''\in yk[y]\), and similarly \(b\); then \(ax+by=(a_0+a')x+(b_0+b'')y\), so \(a_0=a'=0\) and \(b_0=b''=0\), that is \(a\in yA\) and \(b\in xA\). Hence \(A^2\to A^2\to A\to k\to0\) is exact at the middle \(A^2\) and at \(A\), and it is the beginning of a free resolution whose next map, onto the kernel \(xA\oplus yA\) of \((a,b)\mapsto(ya,xb)\), is \((a,b)\mapsto(xa,yb)\). All these matrices have entries in \((x,y)\), so after tensoring with \(k\) all differentials vanish and \(\operatorname{Tor}^A_2(k,k)=k^2\), while \(\Lambda^2_k(I/I^2)=k\).
Solution 5.4. Tensoring the resolution with \(R/J^m=B[t]/(t^m)\) gives \(t\colon B[t]/(t^m)\to B[t]/(t^m)\), with kernel \(t^{m-1}B[t]/(t^m)\cong B\) and cokernel \(B\). So \(\operatorname{Tor}^R_0=B\), \(\operatorname{Tor}^R_1(B,R/J^m)\cong B\), generated by the class of \(t^{m-1}\), and \(\operatorname{Tor}^R_q=0\) for \(q\geq2\). In Corollary 3.4, with \(n=m-1\), \(N=B\) and \(\Lambda^iN=0\) for \(i\geq2\): for \(q\geq2\) every term vanishes, and for \(q=1\) the sequence reduces to \(0\to N\otimes S^{m-1}N\to\) \(\operatorname{Tor}^R_1(B,R/J^m)\to0\). This is the same answer: \(N\otimes S^{m-1}N=B\cdot e\otimes t^{m-1}\), and \(e\otimes t^{m-1}\) is the generator of the resolution tensored with \(R/J^m\) in degree \(1\), mapping to \(t^m=0\).
Solution 5.5. For an \(R\)-module \(M\), \(\operatorname{Tor}^R_i(B,M)=H_i(K\otimes_RM)\). The complex \(K\otimes_RR/J^m=\Lambda_BF\otimes_BS^{<m}_BF\) is the sum over the weights \(w\) of the parts \(\Lambda^iF\otimes S^{w-i}F\) with \(w-i<m\), with the Koszul differential, terms landing in \(S^{\geq m}F\) being zero. The weight-\(w\) part of \(K\) is exact for \(w\geq1\), because \(K\to B\) is a quasi-isomorphism and \(B\) has weight \(0\). Its truncation to the degrees \(i\geq w-m+1\) therefore has homology only in degree \(i_0=w-m+1\) when \(i_0\geq1\), equal to the cokernel of \(\Lambda^{i_0+1}F\otimes S^{m-2}F\to\Lambda^{i_0}F\otimes S^{m-1}F\); for \(i_0\leq0\) it is the whole weight-\(w\) part and has no homology in degrees \(\geq1\). So \(\operatorname{Tor}^R_q(B,R/J^m)\), for \(q\geq1\), is the homology in degree \(q\) of the part of weight \(q+m-1\), which is the stated cokernel.
Sources
The algebra structure on \(\operatorname{Tor}^A_\bullet(B,B)\) through simplicial resolutions, the ideals for which it is the exterior algebra on \(I/I^2\), the bigraded algebra \(\operatorname{Tor}^A_\bullet(B,\operatorname{gr}_I(A))\) with its homomorphism from the Koszul complex, Theorem 3.3, Corollary 3.4 and the examples of Section 4 are due to Quillen for rings, in Section 8 of his MIT notes on the homology of commutative rings (1968), available from Haynes Miller's page, where these ideals are called quasi-regular. Their extension to rings of a topos, and the name weakly regular, are due to Illusie, in his work on the cotangent complex.