Introduction
In this chapter we discuss dualizing complexes in commutative algebra. A reference is [RD].
We begin with a discussion of essential surjections and essential injections, projective covers, injective hulls, duality for Artinian rings, and study injective hulls of residue fields, leading quickly to a proof of Matlis duality. See Sections 08XI, 08XN, 08XX, 08Y1, 08YW, and 08Z1 and Proposition 08Z9.
This is followed by three sections discussing local cohomology in great generality, see Sections 0BJA, 0952, and 0BJD. We apply some of this to a discussion of depth in Section 0AVY. In another application we show how, given a finitely generated ideal \(I\) of a ring \(A\), the “\(I\)-complete” and “\(I\)-torsion” objects of the derived category of \(A\) are equivalent, see Section 0A6V. To learn more about local cohomology, for example the finiteness theorem (which relies on local duality – see below) please visit Local Cohomology, Section 0DWP.
The bulk of this chapter is devoted to duality for a ring map and dualizing complexes. See Sections 0A6Z, 0E28, 0A7A, 0A7M, 0A7W, 0A81, 0DW3, 0DW4, 0DW6, 0DWB, and 0BJM. The key definition is that of a dualizing complex \(\omega_A^\bullet\) over a Noetherian ring \(A\) as an object \(\omega_A^\bullet \in D^{+}(A)\) whose cohomology modules \(H^i(\omega_A^\bullet)\) are finite \(A\)-modules, which has finite injective dimension, and is such that the map \[A \longrightarrow R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)\] is a quasi-isomorphism. After establishing some elementary properties of dualizing complexes, we show a dualizing complex gives rise to a dimension function. Next, we prove Grothendieck’s local duality theorem. After briefly discussing dualizing modules and Cohen-Macaulay rings, we introduce Gorenstein rings and we show many familiar Noetherian rings have dualizing complexes. In a last section we apply the material to show there is a good theory of Noetherian local rings whose formal fibres are Gorenstein or local complete intersections.
In the last few sections, we describe an algebraic construction of the “upper shriek functors” used in algebraic geometry, for example in the book [RD]. This topic is continued in the chapter on duality for schemes. See Duality for Schemes, Section 0DWF.
Essential surjections and injections
We will mostly work in categories of modules, but we may as well make the definition in general.
Definition
Let \(\mathcal{A}\) be an abelian category.
An injection \(A \subset B\) of \(\mathcal{A}\) is essential, or we say that \(B\) is an essential extension of \(A\), if every nonzero subobject \(B' \subset B\) has nonzero intersection with \(A\).
A surjection \(f : A \to B\) of \(\mathcal{A}\) is essential if for every proper subobject \(A' \subset A\) we have \(f(A') \not = B\).
Some lemmas about this notion.
Lemma
Let \(\mathcal{A}\) be an abelian category.
If \(A \subset B\) and \(B \subset C\) are essential extensions, then \(A \subset C\) is an essential extension.
If \(A \subset B\) is an essential extension and \(C \subset B\) is a subobject, then \(A \cap C \subset C\) is an essential extension.
If \(A \to B\) and \(B \to C\) are essential surjections, then \(A \to C\) is an essential surjection.
Given an essential surjection \(f : A \to B\) and a surjection \(A \to C\) with kernel \(K\), the morphism \(C \to B/f(K)\) is an essential surjection.
Proof
Omitted.
Lemma
Let \(R\) be a ring. Let \(M\) be an \(R\)-module. Let \(E = \colim E_i\) be a filtered colimit of \(R\)-modules. Suppose given a compatible system of essential injections \(M \to E_i\) of \(R\)-modules. Then \(M \to E\) is an essential injection.
Proof
Immediate from the definitions and the fact that filtered colimits are exact (Algebra, Lemma 00DB).
Lemma
Let \(R\) be a ring. Let \(M \subset N\) be \(R\)-modules. The following are equivalent
\(M \subset N\) is an essential extension,
for all \(x \in N\) nonzero there exists an \(f \in R\) such that \(fx \in M\) and \(fx \not = 0\).
Proof
Assume (1) and let \(x \in N\) be a nonzero element. By (1) we have \(Rx \cap M \not = 0\). This implies (2).
Assume (2). Let \(N' \subset N\) be a nonzero submodule. Pick \(x \in N'\) nonzero. By (2) we can find \(f \in R\) with \(fx \in M\) and \(fx \not = 0\). Thus \(N' \cap M \not = 0\).
Injective modules
Some results about injective modules over rings.
Lemma
Let \(R\) be a ring. Any product of injective \(R\)-modules is injective.
Proof
Special case of Homology, Lemma 0137.
Lemma
Let \(R \to S\) be a flat ring map. If \(E\) is an injective \(S\)-module, then \(E\) is injective as an \(R\)-module.
Proof
This is true because \(\Hom_R(M, E) = \Hom_S(M \otimes_R S, E)\) by Algebra, Lemma 05DQ and the fact that tensoring with \(S\) is exact.
Lemma
Let \(R \to S\) be an epimorphism of rings. Let \(E\) be an \(S\)-module. If \(E\) is injective as an \(R\)-module, then \(E\) is an injective \(S\)-module.
Proof
This is true because \(\Hom_R(N, E) = \Hom_S(N, E)\) for any \(S\)-module \(N\), see Algebra, Lemma 08YS.
Lemma
Let \(R \to S\) be a ring map. If \(E\) is an injective \(R\)-module, then \(\Hom_R(S, E)\) is an injective \(S\)-module.
Proof
This is true because \(\Hom_S(N, \Hom_R(S, E)) = \Hom_R(N, E)\) by Algebra, Lemma 08YP.
Lemma
Let \(R\) be a ring. Let \(I\) be an injective \(R\)-module. Let \(E \subset I\) be a submodule. The following are equivalent
\(E\) is injective, and
for all \(E \subset E' \subset I\) with \(E \subset E'\) essential we have \(E = E'\).
In particular, an \(R\)-module is injective if and only if every essential extension is trivial.
Proof
The final assertion follows from the first and the fact that the category of \(R\)-modules has enough injectives (More on Algebra, Section 01D8).
Assume (1). Let \(E \subset E' \subset I\) as in (2). Then the map \(\text{id}_E : E \to E\) can be extended to a map \(\alpha : E' \to E\). The kernel of \(\alpha\) has to be zero because it intersects \(E\) trivially and \(E'\) is an essential extension. Hence \(E = E'\).
Assume (2). Let \(M \subset N\) be \(R\)-modules and let \(\varphi : M \to E\) be an \(R\)-module map. In order to prove (1) we have to show that \(\varphi\) extends to a morphism \(N \to E\). Consider the set \(\mathcal{S}\) of pairs \((M', \varphi')\) where \(M \subset M' \subset N\) and \(\varphi' : M' \to E\) is an \(R\)-module map agreeing with \(\varphi\) on \(M\). We define an ordering on \(\mathcal{S}\) by the rule \((M', \varphi') \leq (M'', \varphi'')\) if and only if \(M' \subset M''\) and \(\varphi''|_{M'} = \varphi'\). It is clear that we can take the maximum of a totally ordered subset of \(\mathcal{S}\). Hence by Zorn’s lemma we may assume \((M, \varphi)\) is a maximal element.
Choose an extension \(\psi : N \to I\) of \(\varphi\) composed with the inclusion \(E \to I\). This is possible as \(I\) is injective. If \(\psi(N) \subset E\), then \(\psi\) is the desired extension. If \(\psi(N)\) is not contained in \(E\), then by (2) the inclusion \(E \subset E + \psi(N)\) is not essential. hence we can find a nonzero submodule \(K \subset E + \psi(N)\) meeting \(E\) in \(0\). This means that \(M' = \psi^{-1}(E + K)\) strictly contains \(M\). Thus we can extend \(\varphi\) to \(M'\) using \[M' \xrightarrow{\psi|_{M'}} E + K \to (E + K)/K = E\] This contradicts the maximality of \((M, \varphi)\).
Example
Let \(R\) be a reduced ring. Let \(\mathfrak p \subset R\) be a minimal prime so that \(K = R_\mathfrak p\) is a field (Algebra, Lemma 00EU). Then \(K\) is an injective \(R\)-module. Namely, we have \(\Hom_R(M, K) = \Hom_K(M_\mathfrak p, K)\) for any \(R\)-module \(M\). Since localization is an exact functor and taking duals is an exact functor on \(K\)-vector spaces we conclude \(\Hom_R(-, K)\) is an exact functor, i.e., \(K\) is an injective \(R\)-module.
Lemma
Let \(R\) be a Noetherian ring. A direct sum of injective modules is injective.
Proof
Let \(E_i\) be a family of injective modules parametrized by a set \(I\). Set \(E = \bigoplus E_i\). To show that \(E\) is injective we use Injectives, Lemma 05NU. Thus let \(\varphi : J \to E\) be a module map from an ideal of \(R\) into \(E\). As \(J\) is a finite \(R\)-module (because \(R\) is Noetherian) we can find finitely many elements \(i_1, \ldots, i_r \in I\) such that \(\varphi\) maps into \(\bigoplus_{j = 1, \ldots, r} E_{i_j}\). Then we can extend \(\varphi\) into \(\bigoplus_{j = 1, \ldots, r} E_{i_j}\) using the injectivity of the modules \(E_{i_j}\).
Lemma
Let \(R\) be a Noetherian ring. Let \(S \subset R\) be a multiplicative subset. If \(E\) is an injective \(R\)-module, then \(S^{-1}E\) is an injective \(S^{-1}R\)-module.
Proof
Since \(R \to S^{-1}R\) is an epimorphism of rings, it suffices to show that \(S^{-1}E\) is injective as an \(R\)-module, see Lemma 08YV. To show this we use Injectives, Lemma 05NU. Thus let \(I \subset R\) be an ideal and let \(\varphi : I \to S^{-1} E\) be an \(R\)-module map. As \(I\) is a finitely presented \(R\)-module (because \(R\) is Noetherian) we can find an \(f \in S\) and an \(R\)-module map \(I \to E\) such that \(f\varphi\) is the composition \(I \to E \to S^{-1}E\) (Algebra, Lemma 0583). Then we can extend \(I \to E\) to a homomorphism \(R \to E\). Then the composition \[R \to E \to S^{-1}E \xrightarrow{f^{-1}} S^{-1}E\] is the desired extension of \(\varphi\) to \(R\).
Lemma
Let \(R\) be a Noetherian ring. Let \(I\) be an injective \(R\)-module.
Let \(f \in R\). Then \(E = \bigcup I[f^n] = I[f^\infty]\) is an injective submodule of \(I\).
Let \(J \subset R\) be an ideal. Then the \(J\)-power torsion submodule \(I[J^\infty]\) is an injective submodule of \(I\).
Proof
We will use Lemma 08XS to prove (1). Suppose that \(E \subset E' \subset I\) and that \(E'\) is an essential extension of \(E\). We will show that \(E' = E\). If not, then we can find \(x \in E'\) and \(x \not \in E\). Let \(J = \{ a \in R \mid ax \in E\}\). Since \(R\) is Noetherian, we may write \(J = (g_1, \ldots, g_t)\) for some \(g_i \in R\). By definition \(E\) is the set of elements of \(I\) annihilated by powers of \(f\), so we may choose integers \(n_i\) so that \(f^{n_i}g_ix = 0\). Set \(n = \mathrm{max}\{ n_i \}\). Then \(x' = f^n x\) is an element of \(E'\) not in \(E\) and is annihilated by \(J\). Set \(J' = \{ a \in R \mid ax' \in E \}\) so \(J \subset J'\). Conversely, we have \(a \in J'\) if and only if \(ax' \in E\) if and only if \(f^m a x' = 0\) for some \(m \geq 0\). But then \(f^m a x' = f^{m + n} a x\) implies \(ax \in E\), i.e., \(a \in J\). Hence \(J = J'\). Thus \(J = J' = \text{Ann}(x')\), so \(Rx' \cap E = 0\). Hence \(E'\) is not an essential extension of \(E\), a contradiction.
To prove (2) write \(J = (f_1, \ldots, f_t)\). Then \(I[J^\infty]\) is equal to \[(\ldots((I[f_1^\infty])[f_2^\infty])\ldots)[f_t^\infty]\] and the result follows from (1) and induction.
Lemma
Let \(A\) be a Noetherian ring. Let \(E\) be an injective \(A\)-module. Then \(E \otimes_A A[x]\) has injective-amplitude \([0, 1]\) as an object of \(D(A[x])\). In particular, \(E \otimes_A A[x]\) has finite injective dimension as an \(A[x]\)-module.
Proof
Let us write \(E[x] = E \otimes_A A[x]\). Consider the short exact sequence of \(A[x]\)-modules \[0 \to E[x] \to \Hom_A(A[x], E[x]) \to \Hom_A(A[x], E[x]) \to 0\] where the first map sends \(p \in E[x]\) to \(f \mapsto fp\) and the second map sends \(\varphi\) to \(f \mapsto \varphi(xf) - x\varphi(f)\). The second map is surjective because \(\Hom_A(A[x], E[x]) = \prod_{n \geq 0} E[x]\) as an abelian group and the map sends \((e_n)\) to \((e_{n + 1} - xe_n)\) which is surjective. As an \(A\)-module we have \(E[x] \cong \bigoplus_{n \geq 0} E\) which is injective by Lemma 08XV. Hence the \(A[x]\)-module \(\Hom_A(A[x], E[x])\) is injective by Lemma 08XR and the proof is complete.
Projective covers
In this section we briefly discuss projective covers.
Definition
Let \(R\) be a ring. A surjection \(P \to M\) of \(R\)-modules is said to be a projective cover, or sometimes a projective envelope, if \(P\) is a projective \(R\)-module and \(P \to M\) is an essential surjection.
Projective covers do not always exist. For example, if \(k\) is a field and \(R = k[x]\) is the polynomial ring over \(k\), then the module \(M = R/(x)\) does not have a projective cover. Namely, for any surjection \(f : P \to M\) with \(P\) projective over \(R\), the proper submodule \((x - 1)P\) surjects onto \(M\). Hence \(f\) is not essential.
Lemma
Let \(R\) be a ring and let \(M\) be an \(R\)-module. If a projective cover of \(M\) exists, then it is unique up to isomorphism.
Proof
Let \(P \to M\) and \(P' \to M\) be projective covers. Because \(P\) is a projective \(R\)-module and \(P' \to M\) is surjective, we can find an \(R\)-module map \(\alpha : P \to P'\) compatible with the maps to \(M\). Since \(P' \to M\) is essential, we see that \(\alpha\) is surjective. As \(P'\) is a projective \(R\)-module we can choose a direct sum decomposition \(P = \Ker(\alpha) \oplus P'\). Since \(P' \to M\) is surjective and since \(P \to M\) is essential we conclude that \(\Ker(\alpha)\) is zero as desired.
Here is an example where projective covers exist.
Lemma
Let \((R, \mathfrak m, \kappa)\) be a local ring. Any finite \(R\)-module has a projective cover.
Proof
Let \(M\) be a finite \(R\)-module. Let \(r = \dim_\kappa(M/\mathfrak m M)\). Choose \(x_1, \ldots, x_r \in M\) mapping to a basis of \(M/\mathfrak m M\). Consider the map \(f : R^{\oplus r} \to M\). By Nakayama’s lemma this is a surjection (Algebra, Lemma 00DV). If \(N \subset R^{\oplus r}\) is a proper submodule, then \(N/\mathfrak m N \to \kappa^{\oplus r}\) is not surjective (by Nakayama’s lemma again) hence \(N/\mathfrak m N \to M/\mathfrak m M\) is not surjective. Thus \(f\) is an essential surjection.
Injective hulls
In this section we briefly discuss injective hulls.
Definition
Let \(R\) be a ring. An injection \(M \to I\) of \(R\)-modules is said to be an injective hull if \(I\) is an injective \(R\)-module and \(M \to I\) is an essential injection.
Injective hulls always exist.
Lemma
Let \(R\) be a ring. Any \(R\)-module has an injective hull.
Proof
Let \(M\) be an \(R\)-module. By More on Algebra, Section 01D8 the category of \(R\)-modules has enough injectives. Choose an injection \(M \to I\) with \(I\) an injective \(R\)-module. Consider the set \(\mathcal{S}\) of submodules \(M \subset E \subset I\) such that \(E\) is an essential extension of \(M\). We order \(\mathcal{S}\) by inclusion. If \(\{E_\alpha\}\) is a totally ordered subset of \(\mathcal{S}\), then \(\bigcup E_\alpha\) is an essential extension of \(M\) too (Lemma 08XL). Thus we can apply Zorn’s lemma and find a maximal element \(E \in \mathcal{S}\). We claim \(M \subset E\) is an injective hull, i.e., \(E\) is an injective \(R\)-module. This follows from Lemma 08XS.
Lemma
Let \(R\) be a ring. Let \(M\), \(N\) be \(R\)-modules and let \(M \to E\) and \(N \to E'\) be injective hulls. Then
for any \(R\)-module map \(\varphi : M \to N\) there exists an \(R\)-module map \(\psi : E \to E'\) such that \[\xymatrix{ M \ar[r] \ar[d]_\varphi & E \ar[d]^\psi \\ N \ar[r] & E' }\] commutes,
if \(\varphi\) is injective, then \(\psi\) is injective,
if \(\varphi\) is an essential injection, then \(\psi\) is an isomorphism,
if \(\varphi\) is an isomorphism, then \(\psi\) is an isomorphism,
if \(M \to I\) is an embedding of \(M\) into an injective \(R\)-module, then there is an isomorphism \(I \cong E \oplus I'\) compatible with the embeddings of \(M\),
In particular, the injective hull \(E\) of \(M\) is unique up to isomorphism.
Proof
Part (1) follows from the fact that \(E'\) is an injective \(R\)-module. Part (2) follows as \(\Ker(\psi) \cap M = 0\) and \(E\) is an essential extension of \(M\). Assume \(\varphi\) is an essential injection. Then \(E \cong \psi(E) \subset E'\) by (2) which implies \(E' = \psi(E) \oplus E''\) because \(E\) is injective. Since \(E'\) is an essential extension of \(M\) (Lemma 08XK) we get \(E'' = 0\). Part (4) is a special case of (3). Assume \(M \to I\) as in (5). Choose a map \(\alpha : E \to I\) extending the map \(M \to I\). Arguing as before we see that \(\alpha\) is injective. Thus as before \(\alpha(E)\) splits off from \(I\). This proves (5).
Example
Let \(R\) be a domain with fraction field \(K\). Then \(R \subset K\) is an injective hull of \(R\). Namely, by Example 08XT we see that \(K\) is an injective \(R\)-module and by Lemma 08XM we see that \(R \subset K\) is an essential extension.
Definition
An object \(X\) of an additive category is called indecomposable if it is nonzero and if \(X = Y \oplus Z\), then either \(Y = 0\) or \(Z = 0\).
Lemma
Let \(R\) be a ring. Let \(E\) be an indecomposable injective \(R\)-module. Then
\(E\) is the injective hull of any nonzero submodule of \(E\),
the intersection of any two nonzero submodules of \(E\) is nonzero,
\(\text{End}_R(E)\) is a noncommutative local ring with maximal ideal those \(\varphi : E \to E\) whose kernel is nonzero, and
the set of zerodivisors on \(E\) is a prime ideal \(\mathfrak p\) of \(R\) and \(E\) is an injective \(R_\mathfrak p\)-module.
Proof
Part (1) follows from Lemma 08Y4. Part (2) follows from part (1) and the definition of injective hulls.
Proof of (3). Set \(A = \text{End}_R(E)\) and \(I = \{\varphi \in A \mid \Ker(\varphi) \not = 0\}\). The statement means that \(I\) is a two sided ideal and that any \(\varphi \in A\), \(\varphi \not \in I\) is invertible. Suppose \(\varphi\) and \(\psi\) are not injective. Then \(\Ker(\varphi) \cap \Ker(\psi)\) is nonzero by (2). Hence \(\varphi + \psi \in I\). It follows that \(I\) is a two sided ideal. If \(\varphi \in A\), \(\varphi \not \in I\), then \(E \cong \varphi(E) \subset E\) is an injective submodule, hence \(E = \varphi(E)\) because \(E\) is indecomposable.
Proof of (4). Consider the ring map \(R \to A\) and let \(\mathfrak p \subset R\) be the inverse image of the maximal ideal \(I\). Then it is clear that \(\mathfrak p\) is a prime ideal and that \(R \to A\) extends to \(R_\mathfrak p \to A\). Thus \(E\) is an \(R_\mathfrak p\)-module. It follows from Lemma 08YV that \(E\) is injective as an \(R_\mathfrak p\)-module.
Lemma
Let \(\mathfrak p \subset R\) be a prime of a ring \(R\). Let \(E\) be the injective hull of \(R/\mathfrak p\). Then
\(E\) is indecomposable,
\(E\) is the injective hull of \(\kappa(\mathfrak p)\),
\(E\) is the injective hull of \(\kappa(\mathfrak p)\) over the ring \(R_\mathfrak p\).
Proof
By Lemma 08XM the inclusion \(R/\mathfrak p \subset \kappa(\mathfrak p)\) is an essential extension. Then Lemma 08Y4 shows (2) holds. For \(f \in R\), \(f \not \in \mathfrak p\) the map \(f : \kappa(\mathfrak p) \to \kappa(\mathfrak p)\) is an isomorphism hence the map \(f : E \to E\) is an isomorphism, see Lemma 08Y4. Thus \(E\) is an \(R_\mathfrak p\)-module. It is injective as an \(R_\mathfrak p\)-module by Lemma 08YV. Finally, let \(E' \subset E\) be a nonzero injective \(R\)-submodule. Then \(J = (R/\mathfrak p) \cap E'\) is nonzero. After shrinking \(E'\) we may assume that \(E'\) is the injective hull of \(J\) (see Lemma 08Y4 for example). Observe that \(R/\mathfrak p\) is an essential extension of \(J\) for example by Lemma 08XM. Hence \(E' \to E\) is an isomorphism by Lemma 08Y4 part (3). Hence \(E\) is indecomposable.
Lemma
Let \(R\) be a Noetherian ring. Let \(E\) be an indecomposable injective \(R\)-module. Then there exists a prime ideal \(\mathfrak p\) of \(R\) such that \(E\) is the injective hull of \(\kappa(\mathfrak p)\).
Proof
Let \(\mathfrak p\) be the prime ideal found in Lemma 08Y7. Say \(\mathfrak p = (f_1, \ldots, f_r)\). Pick a nonzero element \(x \in \bigcap \Ker(f_i : E \to E)\), see Lemma 08Y7. Then \((R_\mathfrak p)x\) is a module isomorphic to \(\kappa(\mathfrak p)\) inside \(E\). We conclude by Lemma 08Y7.
Proposition
Let \(R\) be a Noetherian ring. Every injective module is a direct sum of indecomposable injective modules. Every indecomposable injective module is the injective hull of the residue field at a prime.
Proof
The second statement is Lemma 08Y9. For the first statement, let \(I\) be an injective \(R\)-module. We will use transfinite recursion to construct \(I_\alpha \subset I\) for ordinals \(\alpha\) which are direct sums of indecomposable injective \(R\)-modules \(E_{\beta + 1}\) for \(\beta < \alpha\). For \(\alpha = 0\) we let \(I_0 = 0\). Suppose given an ordinal \(\alpha\) such that \(I_\alpha\) has been constructed. Then \(I_\alpha\) is an injective \(R\)-module by Lemma 08XV. Hence \(I \cong I_\alpha \oplus I'\). If \(I' = 0\) we are done. If not, then \(I'\) has an associated prime by Algebra, Lemma 0587. Thus \(I'\) contains a copy of \(R/\mathfrak p\) for some prime \(\mathfrak p\). Hence \(I'\) contains an indecomposable submodule \(E\) by Lemmas 08Y4 and 08Y8. Set \(I_{\alpha + 1} = I_\alpha \oplus E_\alpha\). If \(\alpha\) is a limit ordinal and \(I_\beta\) has been constructed for \(\beta < \alpha\), then we set \(I_\alpha = \bigcup_{\beta < \alpha} I_\beta\). Observe that \(I_\alpha = \bigoplus_{\beta < \alpha} E_{\beta + 1}\). This concludes the proof.
Duality over Artinian local rings
Let \((R, \mathfrak m, \kappa)\) be an artinian local ring. Recall that this implies \(R\) is Noetherian and that \(R\) has finite length as an \(R\)-module. Moreover an \(R\)-module is finite if and only if it has finite length. We will use these facts without further mention in this section. Please see Algebra, Sections 00IU and 00J4 and Algebra, Proposition 00KJ for more details.
Lemma
Let \((R, \mathfrak m, \kappa)\) be an artinian local ring. Let \(E\) be an injective hull of \(\kappa\). For every finite \(R\)-module \(M\) we have \[\text{length}_R(M) = \text{length}_R(\Hom_R(M, E))\] In particular, the injective hull \(E\) of \(\kappa\) is a finite \(R\)-module.
Proof
Because \(E\) is an essential extension of \(\kappa\) we have \(\kappa = E[\mathfrak m]\) where \(E[\mathfrak m]\) is the \(\mathfrak m\)-torsion in \(E\) (notation as in More on Algebra, Section 0ALX). Hence \(\Hom_R(\kappa, E) \cong \kappa\) and the equality of lengths holds for \(M = \kappa\). We prove the displayed equality of the lemma by induction on the length of \(M\). If \(M\) is nonzero there exists a surjection \(M \to \kappa\) with kernel \(M'\). Since the functor \(M \mapsto \Hom_R(M, E)\) is exact we obtain a short exact sequence \[0 \to \Hom_R(\kappa, E) \to \Hom_R(M, E) \to \Hom_R(M', E) \to 0.\] Additivity of length for this sequence and the sequence \(0 \to M' \to M \to \kappa \to 0\) and the equality for \(M'\) (induction hypothesis) and \(\kappa\) implies the equality for \(M\). The final statement of the lemma follows as \(E = \Hom_R(R, E)\).
Lemma
Let \((R, \mathfrak m, \kappa)\) be an artinian local ring. Let \(E\) be an injective hull of \(\kappa\). For any finite \(R\)-module \(M\) the evaluation map \[M \longrightarrow \Hom_R(\Hom_R(M, E), E)\] is an isomorphism. In particular \(R = \Hom_R(E, E)\).
Proof
Observe that the displayed arrow is injective. Namely, if \(x \in M\) is a nonzero element, then there is a nonzero map \(Rx \to \kappa\) which we can extend to a map \(\varphi : M \to E\) that doesn’t vanish on \(x\). Since the source and target of the arrow have the same length by Lemma 08YX we conclude it is an isomorphism. The final statement follows on taking \(M = R\).
To state the next lemma, denote \(\text{Mod}^{fg}_R\) the category of finite \(R\)-modules over a ring \(R\).
Lemma
Let \((R, \mathfrak m, \kappa)\) be an artinian local ring. Let \(E\) be an injective hull of \(\kappa\). The functor \(D(-) = \Hom_R(-, E)\) induces an exact anti-equivalence \(\text{Mod}^{fg}_R \to \text{Mod}^{fg}_R\) and \(D \circ D \cong \text{id}\).
Proof
We have seen that \(D \circ D = \text{id}\) on \(\text{Mod}^{fg}_R\) in Lemma 08YY. It follows immediately that \(D\) is an anti-equivalence.
Lemma
Assumptions and notation as in Lemma 08YZ. Let \(I \subset R\) be an ideal and \(M\) a finite \(R\)-module. Then \[D(M[I]) = D(M)/ID(M) \quad\text{and}\quad D(M/IM) = D(M)[I]\]
Proof
Say \(I = (f_1, \ldots, f_t)\). Consider the map \[M^{\oplus t} \xrightarrow{f_1, \ldots, f_t} M\] with cokernel \(M/IM\). Applying the exact functor \(D\) we conclude that \(D(M/IM)\) is \(D(M)[I]\). The other case is proved in the same way.
Injective hull of the residue field
Most of our results will be for Noetherian local rings in this section.
Lemma
Let \(R \to S\) be a surjective map of local rings with kernel \(I\). Let \(E\) be the injective hull of the residue field of \(R\) over \(R\). Then \(E[I]\) is the injective hull of the residue field of \(S\) over \(S\).
Proof
Observe that \(E[I] = \Hom_R(S, E)\) as \(S = R/I\). Hence \(E[I]\) is an injective \(S\)-module by Lemma 08XR. Since \(E\) is an essential extension of \(\kappa = R/\mathfrak m_R\) it follows that \(E[I]\) is an essential extension of \(\kappa\) as well. The result follows.
Lemma
Let \((R, \mathfrak m, \kappa)\) be a local ring. Let \(E\) be the injective hull of \(\kappa\). Let \(M\) be a \(\mathfrak m\)-power torsion \(R\)-module with \(n = \dim_\kappa(M[\mathfrak m]) < \infty\). Then \(M\) is isomorphic to a submodule of \(E^{\oplus n}\).
Proof
Observe that \(E^{\oplus n}\) is the injective hull of \(\kappa^{\oplus n} = M[\mathfrak m]\). Thus there is an \(R\)-module map \(M \to E^{\oplus n}\) which is injective on \(M[\mathfrak m]\). Since \(M\) is \(\mathfrak m\)-power torsion the inclusion \(M[\mathfrak m] \subset M\) is an essential extension (for example by Lemma 08XM) we conclude that the kernel of \(M \to E^{\oplus n}\) is zero.
Lemma
Let \((R, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(E\) be an injective hull of \(\kappa\) over \(R\). Let \(E_n\) be an injective hull of \(\kappa\) over \(R/\mathfrak m^n\). Then \(E = \bigcup E_n\) and \(E_n = E[\mathfrak m^n]\).
Proof
We have \(E_n = E[\mathfrak m^n]\) by Lemma 08Z2. We have \(E = \bigcup E_n\) because \(\bigcup E_n = E[\mathfrak m^\infty]\) is an injective \(R\)-submodule which contains \(\kappa\), see Lemma 08XW.
The following lemma tells us the injective hull of the residue field of a Noetherian local ring only depends on the completion.
Lemma
Let \(R \to S\) be a flat local homomorphism of local Noetherian rings such that \(R/\mathfrak m_R \cong S/\mathfrak m_R S\). Then the injective hull of the residue field of \(R\) is the injective hull of the residue field of \(S\).
Proof
Note that \(\mathfrak m_RS = \mathfrak m_S\) as the quotient by the former is a field. Set \(\kappa = R/\mathfrak m_R = S/\mathfrak m_S\). Let \(E_R\) be the injective hull of \(\kappa\) over \(R\). Let \(E_S\) be the injective hull of \(\kappa\) over \(S\). Observe that \(E_S\) is an injective \(R\)-module by Lemma 08XQ. Choose an extension \(E_R \to E_S\) of the identification of residue fields. This map is an isomorphism by Lemma 08Z4 because \(R \to S\) induces an isomorphism \(R/\mathfrak m_R^n \to S/\mathfrak m_S^n\) for all \(n\).
Lemma
Let \((R, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(E\) be an injective hull of \(\kappa\) over \(R\). Then \(\Hom_R(E, E)\) is canonically isomorphic to the completion of \(R\).
Proof
Write \(E = \bigcup E_n\) with \(E_n = E[\mathfrak m^n]\) as in Lemma 08Z4. Any endomorphism of \(E\) preserves this filtration. Hence \[\Hom_R(E, E) = \lim \Hom_R(E_n, E_n)\] The lemma follows as \(\Hom_R(E_n, E_n) = \Hom_{R/\mathfrak m^n}(E_n, E_n) = R/\mathfrak m^n\) by Lemma 08YY.
Lemma
Let \((R, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(E\) be an injective hull of \(\kappa\) over \(R\). Then \(E\) satisfies the descending chain condition.
Proof
If \(E \supset M_1 \supset M_2 \supset \ldots\) is a sequence of submodules, then \[\Hom_R(E, E) \to \Hom_R(M_1, E) \to \Hom_R(M_2, E) \to \ldots\] is a sequence of surjections. By Lemma 08Z6 each of these is a module over the completion \(R^\wedge = \Hom_R(E, E)\). Since \(R^\wedge\) is Noetherian (Algebra, Lemma 0316) the sequence stabilizes: \(\Hom_R(M_n, E) = \Hom_R(M_{n + 1}, E) = \ldots\). Since \(E\) is injective, this can only happen if \(\Hom_R(M_n/M_{n + 1}, E)\) is zero. However, if \(M_n/M_{n + 1}\) is nonzero, then it contains a nonzero element annihilated by \(\mathfrak m\), because \(E\) is \(\mathfrak m\)-power torsion by Lemma 08Z4. In this case \(M_n/M_{n + 1}\) has a nonzero map into \(E\), contradicting the assumed vanishing. This finishes the proof.
Lemma
Let \((R, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(E\) be an injective hull of \(\kappa\).
For an \(R\)-module \(M\) the following are equivalent:
\(M\) satisfies the ascending chain condition,
\(M\) is a finite \(R\)-module, and
there exist \(n, m\) and an exact sequence \(R^{\oplus m} \to R^{\oplus n} \to M \to 0\).
For an \(R\)-module \(M\) the following are equivalent:
\(M\) satisfies the descending chain condition,
\(M\) is \(\mathfrak m\)-power torsion and \(\dim_\kappa(M[\mathfrak m]) < \infty\), and
there exist \(n, m\) and an exact sequence \(0 \to M \to E^{\oplus n} \to E^{\oplus m}\).
Proof
We omit the proof of (1).
Let \(M\) be an \(R\)-module with the descending chain condition. Let \(x \in M\). Then \(\mathfrak m^n x\) is a descending chain of submodules, hence stabilizes. Thus \(\mathfrak m^nx = \mathfrak m^{n + 1}x\) for some \(n\). By Nakayama’s lemma (Algebra, Lemma 00DV) this implies \(\mathfrak m^n x = 0\), i.e., \(x\) is \(\mathfrak m\)-power torsion. Since \(M[\mathfrak m]\) is a vector space over \(\kappa\) it has to be finite dimensional in order to have the descending chain condition.
Assume that \(M\) is \(\mathfrak m\)-power torsion and has a finite dimensional \(\mathfrak m\)-torsion submodule \(M[\mathfrak m]\). By Lemma 08Z3 we see that \(M\) is a submodule of \(E^{\oplus n}\) for some \(n\). Consider the quotient \(N = E^{\oplus n}/M\). By Lemma 08Z7 the module \(E\) has the descending chain condition hence so do \(E^{\oplus n}\) and \(N\). Therefore \(N\) satisfies (2)(a) which implies \(N\) satisfies (2)(b) by the second paragraph of the proof. Thus by Lemma 08Z3 again we see that \(N\) is a submodule of \(E^{\oplus m}\) for some \(m\). Thus we have a short exact sequence \(0 \to M \to E^{\oplus n} \to E^{\oplus m}\).
Assume we have a short exact sequence \(0 \to M \to E^{\oplus n} \to E^{\oplus m}\). Since \(E\) satisfies the descending chain condition by Lemma 08Z7 so does \(M\).
Proposition
Let \((R, \mathfrak m, \kappa)\) be a complete local Noetherian ring. Let \(E\) be an injective hull of \(\kappa\) over \(R\). The functor \(D(-) = \Hom_R(-, E)\) induces an anti-equivalence \[\left\{ \begin{matrix} R\text{-modules with the} \\ \text{descending chain condition} \end{matrix} \right\} \longleftrightarrow \left\{ \begin{matrix} R\text{-modules with the} \\ \text{ascending chain condition} \end{matrix} \right\}\] and we have \(D \circ D = \text{id}\) on either side of the equivalence.
Proof
By Lemma 08Z6 we have \(R = \Hom_R(E, E) = D(E)\). Of course we have \(E = \Hom_R(R, E) = D(R)\). Since \(E\) is injective the functor \(D\) is exact. The result now follows immediately from the description of the categories in Lemma 08Z8.
Remark
Let \((R, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(E\) be an injective hull of \(\kappa\) over \(R\). Here is an addendum to Matlis duality: If \(N\) is an \(\mathfrak m\)-power torsion module and \(M = \Hom_R(N, E)\) is a finite module over the completion of \(R\), then \(N\) satisfies the descending chain condition. Namely, for any submodules \(N'' \subset N' \subset N\) with \(N'' \not = N'\), we can find an embedding \(\kappa \subset N''/N'\) and hence a nonzero map \(N' \to E\) annihilating \(N''\) which we can extend to a map \(N \to E\) annihilating \(N''\). Thus \(N \supset N' \mapsto M' = \Hom_R(N/N', E) \subset M\) is an inclusion preserving map from submodules of \(N\) to submodules of \(M\), whence the conclusion.
Deriving torsion
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal (if \(I\) is not finitely generated perhaps a different definition should be used). Let \(Z = V(I) \subset \Spec(A)\). Recall that the category \(I^\infty\text{-torsion}\) of \(I\)-power torsion modules only depends on the closed subset \(Z\) and not on the choice of the finitely generated ideal \(I\) such that \(Z = V(I)\), see More on Algebra, Lemma 0953. In this section we will consider the functor \[H^0_{I} : \text{Mod}_A \longrightarrow I^\infty\text{-torsion},\quad M \longmapsto M[I^\infty] = \bigcup M[I^n]\] which sends \(M\) to the submodule of \(I\)-power torsion.
Let \(A\) be a ring and let \(I\) be a finitely generated ideal. Note that \(I^\infty\text{-torsion}\) is a Grothendieck abelian category (direct sums exist, filtered colimits are exact, and \(\bigoplus A/I^n\) is a generator by More on Algebra, Lemma 05E8). Hence the derived category \(D(I^\infty\text{-torsion})\) exists, see Injectives, Remark 079Q. Our functor \(H^0_I\) is left exact and has a derived extension which we will denote \[R\Gamma_I : D(A) \longrightarrow D(I^\infty\text{-torsion}).\] Warning: this functor does not deserve the name local cohomology unless the ring \(A\) is Noetherian. The functors \(H^0_I\), \(R\Gamma_I\), and the satellites \(H^p_I\) only depend on the closed subset \(Z \subset \Spec(A)\) and not on the choice of the finitely generated ideal \(I\) such that \(V(I) = Z\). However, we insist on using the subscript \(I\) for the functors above as the notation \(R\Gamma_Z\) is going to be used for a different functor, see (0A6Q), which agrees with the functor \(R\Gamma_I\) only (as far as we know) in case \(A\) is Noetherian (see Lemma 0955).
Lemma
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. The functor \(R\Gamma_I\) is right adjoint to the functor \(D(I^\infty\text{-torsion}) \to D(A)\).
Proof
This follows from the fact that taking \(I\)-power torsion submodules is the right adjoint to the inclusion functor \(I^\infty\text{-torsion} \to \text{Mod}_A\). See Derived Categories, Lemma 09T5.
Lemma
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. For any object \(K\) of \(D(A)\) we have \[R\Gamma_I(K) = \text{hocolim}\ R\Hom_A(A/I^n, K)\] in \(D(A)\) and \[R^q\Gamma_I(K) = \colim_n \Ext_A^q(A/I^n, K)\] as modules for all \(q \in \mathbf{Z}\).
Proof
Let \(J^\bullet\) be a K-injective complex representing \(K\). Then \[R\Gamma_I(K) = J^\bullet[I^\infty] = \colim J^\bullet[I^n] = \colim \Hom_A(A/I^n, J^\bullet)\] where the first equality is the definition of \(R\Gamma_I(K)\). By Derived Categories, Lemma 0949 we obtain the first displayed equality in the statement of the lemma. The second displayed equality in the statement of the lemma then follows because \(H^q(\Hom_A(A/I^n, J^\bullet)) = \Ext^q_A(A/I^n, K)\) and because filtered colimits are exact in the category of abelian groups.
Lemma
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. Let \(K^\bullet\) be a complex of \(A\)-modules such that \(f : K^\bullet \to K^\bullet\) is an isomorphism for some \(f \in I\), i.e., \(K^\bullet\) is a complex of \(A_f\)-modules. Then \(R\Gamma_I(K^\bullet) = 0\).
Proof
Namely, in this case the cohomology modules of \(R\Gamma_I(K^\bullet)\) are both \(f\)-power torsion and \(f\) acts by automorphisms. Hence the cohomology modules are zero and hence the object is zero.
Let \(A\) be a ring and \(I \subset A\) a finitely generated ideal. By More on Algebra, Lemma 0A6K the category of \(I\)-power torsion modules is a Serre subcategory of the category of all \(A\)-modules. Hence there is a functor [0A6N]\[\begin{equation} D(I^\infty\text{-torsion}) \longrightarrow D_{I^\infty\text{-torsion}}(A) \end{equation}\] where the right hand side is the full subcategory of \(D(A)\) consising of objects whose cohomologies are \(I\)-power torsion modules. Both the functor and the category are discussed in Derived Categories, Section 06UP.
Lemma
Let \(A\) be a ring and let \(I\) be a finitely generated ideal. Let \(M\) and \(N\) be \(I\)-power torsion modules.
\(\Hom_{D(A)}(M, N) = \Hom_{D({I^\infty\text{-torsion}})}(M, N)\),
\(\Ext^1_{D(A)}(M, N) = \Ext^1_{D({I^\infty\text{-torsion}})}(M, N)\),
\(\Ext^2_{D({I^\infty\text{-torsion}})}(M, N) \to \Ext^2_{D(A)}(M, N)\) is not surjective in general,
(0A6N) is not an equivalence in general.
Proof
Parts (1) and (2) follow immediately from the fact that \(I\)-power torsion forms a Serre subcategory of \(\text{Mod}_A\). Part (4) follows from part (3).
For part (3) let \(A\) be a ring with an element \(f \in A\) such that \(A[f]\) contains a nonzero element \(x\) annihilated by \(f\) and \(A\) contains elements \(x_n\) with \(f^nx_n = x\). Such a ring \(A\) exists because we can take \[A = \mathbf{Z}[f, x, x_n]/(fx, f^nx_n - x)\] Given \(A\) set \(I = (f)\). Then the exact sequence \[0 \to A[f] \to A \xrightarrow{f} A \to A/fA \to 0\] defines an element in \(\Ext^2_A(A/fA, A[f])\). We claim this element does not come from an element of \(\Ext^2_{D(f^\infty\text{-torsion})}(A/fA, A[f])\). Namely, if it did, then there would be an exact sequence \[0 \to A[f] \to M \to N \to A/fA \to 0\] where \(M\) and \(N\) are \(f\)-power torsion modules defining the same \(2\) extension class. Since \(A \to A\) is a complex of free modules and since the \(2\) extension classes are the same we would be able to find a map \[\xymatrix{ 0 \ar[r] & A[f] \ar[r] \ar[d] & A \ar[r] \ar[d]_\varphi & A \ar[r] \ar[d]_\psi & A/fA \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A[f] \ar[r] & M \ar[r] & N \ar[r] & A/fA \ar[r] & 0 }\] (some details omitted). Then we could replace \(M\) by the image of \(\varphi\) and \(N\) by the image of \(\psi\). Then \(M\) would be a cyclic module, hence \(f^n M = 0\) for some \(n\). Considering \(\varphi(x_{n + 1})\) we get a contradiction with the fact that \(f^{n + 1}x_n = x\) is nonzero in \(A[f]\).
Local cohomology
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. Set \(Z = V(I) \subset \Spec(A)\). We will construct a functor [0A6Q]\[\begin{equation} R\Gamma_Z : D(A) \longrightarrow D_{I^\infty\text{-torsion}}(A). \end{equation}\] which is right adjoint to the inclusion functor. For notation see Section 0BJA. The cohomology modules of \(R\Gamma_Z(K)\) are the local cohomology groups of \(K\) with respect to \(Z\). By Lemma 0A6P this functor will in general not be equal to \(R\Gamma_I( - )\) even viewed as functors into \(D(A)\). In Section 0BJD we will show that if \(A\) is Noetherian, then the two agree.
We will continue the discussion of local cohomology in the chapter on local cohomology, see Local Cohomology, Section 0DWP. For example, there we will show that \(R\Gamma_Z\) computes cohomology with support in \(Z\) for the associated complex of quasi-coherent sheaves on \(\Spec(A)\). See Local Cohomology, Lemma 0A6T.
Lemma
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. There exists a right adjoint \(R\Gamma_Z\) (0A6Q) to the inclusion functor \(D_{I^\infty\text{-torsion}}(A) \to D(A)\). In fact, if \(I\) is generated by \(f_1, \ldots, f_r \in A\), then we have \[R\Gamma_Z(K) = (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \otimes_A^\mathbf{L} K\] functorially in \(K \in D(A)\).
Proof
Say \(I = (f_1, \ldots, f_r)\) is an ideal. Let \(K^\bullet\) be a complex of \(A\)-modules. There is a canonical map of complexes \[(A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \longrightarrow A.\] from the extended Čech complex to \(A\). Tensoring with \(K^\bullet\), taking associated total complex, we get a map \[\text{Tot}\left( K^\bullet \otimes_A (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r})\right) \longrightarrow K^\bullet\] in \(D(A)\). We claim the cohomology modules of the complex on the left are \(I\)-power torsion, i.e., the LHS is an object of \(D_{I^\infty\text{-torsion}}(A)\). Namely, we have \[(A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) = \colim K(A, f_1^n, \ldots, f_r^n)\] by More on Algebra, Lemma 0913. Moreover, multiplication by \(f_i^n\) on the complex \(K(A, f_1^n, \ldots, f_r^n)\) is homotopic to zero by More on Algebra, Lemma 0663. Since \[H^q\left( LHS \right) = \colim H^q(\text{Tot}(K^\bullet \otimes_A K(A, f_1^n, \ldots, f_r^n)))\] we obtain our claim. On the other hand, if \(K^\bullet\) is an object of \(D_{I^\infty\text{-torsion}}(A)\), then the complexes \(K^\bullet \otimes_A A_{f_{i_0} \ldots f_{i_p}}\) have vanishing cohomology. Hence in this case the map \(LHS \to K^\bullet\) is an isomorphism in \(D(A)\). The construction \[R\Gamma_Z(K^\bullet) = \text{Tot}\left( K^\bullet \otimes_A (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r})\right)\] is functorial in \(K^\bullet\) and defines an exact functor \(D(A) \to D_{I^\infty\text{-torsion}}(A)\) between triangulated categories. It follows formally from the existence of the natural transformation \(R\Gamma_Z \to \text{id}\) given above and the fact that this evaluates to an isomorphism on \(K^\bullet\) in the subcategory, that \(R\Gamma_Z\) is the desired right adjoint.
Lemma
Let \(A \to B\) be a ring homomorphism and let \(I \subset A\) be a finitely generated ideal. Set \(J = IB\). Set \(Z = V(I)\) and \(Y = V(J)\). Then \[R\Gamma_Z(M_A) = R\Gamma_Y(M)_A\] functorially in \(M \in D(B)\). Here \((-)_A\) denotes the restriction functors \(D(B) \to D(A)\) and \(D_{J^\infty\text{-torsion}}(B) \to D_{I^\infty\text{-torsion}}(A)\).
Proof
This follows from uniqueness of adjoint functors as both \(R\Gamma_Z((-)_A)\) and \(R\Gamma_Y(-)_A\) are right adjoint to the functor \(D_{I^\infty\text{-torsion}}(A) \to D(B)\), \(K \mapsto K \otimes_A^\mathbf{L} B\). Alternatively, one can use the description of \(R\Gamma_Z\) and \(R\Gamma_Y\) in terms of alternating Čech complexes (Lemma 0A6R). Namely, if \(I = (f_1, \ldots, f_r)\) then \(J\) is generated by the images \(g_1, \ldots, g_r \in B\) of \(f_1, \ldots, f_r\). Then the statement of the lemma follows from the existence of a canonical isomorphism \[\begin{align*} & M_A \otimes_A (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \\ & = M \otimes_B (B \to \prod\nolimits_{i_0} B_{g_{i_0}} \to \prod\nolimits_{i_0 < i_1} B_{g_{i_0}g_{i_1}} \to \ldots \to B_{g_1\ldots g_r}) \end{align*}\] for any \(B\)-module \(M\).
Lemma
Let \(A \to B\) be a ring homomorphism and let \(I \subset A\) be a finitely generated ideal. Set \(J = IB\). Let \(Z = V(I)\) and \(Y = V(J)\). Then \[R\Gamma_Z(K) \otimes_A^\mathbf{L} B = R\Gamma_Y(K \otimes_A^\mathbf{L} B)\] functorially in \(K \in D(A)\).
Proof
Write \(I = (f_1, \ldots, f_r)\). Then \(J\) is generated by the images \(g_1, \ldots, g_r \in B\) of \(f_1, \ldots, f_r\). Then we have \[(A \to \prod A_{f_{i_0}} \to \ldots \to A_{f_1\ldots f_r}) \otimes_A B = (B \to \prod B_{g_{i_0}} \to \ldots \to B_{g_1\ldots g_r})\] as complexes of \(B\)-modules. Represent \(K\) by a K-flat complex \(K^\bullet\) of \(A\)-modules. Since the total complexes associated to \[K^\bullet \otimes_A (A \to \prod A_{f_{i_0}} \to \ldots \to A_{f_1\ldots f_r}) \otimes_A B\] and \[K^\bullet \otimes_A B \otimes_B (B \to \prod B_{g_{i_0}} \to \ldots \to B_{g_1\ldots g_r})\] represent the left and right hand side of the displayed formula of the lemma (see Lemma 0A6R) we conclude.
Lemma
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. Let \(K^\bullet\) be a complex of \(A\)-modules such that \(f : K^\bullet \to K^\bullet\) is an isomorphism for some \(f \in I\), i.e., \(K^\bullet\) is a complex of \(A_f\)-modules. Then \(R\Gamma_Z(K^\bullet) = 0\).
Proof
Namely, in this case the cohomology modules of \(R\Gamma_Z(K^\bullet)\) are both \(f\)-power torsion and \(f\) acts by automorphisms. Hence the cohomology modules are zero and hence the object is zero.
Lemma
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. For \(K, L \in D(A)\) we have \[R\Gamma_Z(K \otimes_A^\mathbf{L} L) = K \otimes_A^\mathbf{L} R\Gamma_Z(L) = R\Gamma_Z(K) \otimes_A^\mathbf{L} L = R\Gamma_Z(K) \otimes_A^\mathbf{L} R\Gamma_Z(L)\] If \(K\) or \(L\) is in \(D_{I^\infty\text{-torsion}}(A)\) then so is \(K \otimes_A^\mathbf{L} L\).
Proof
By Lemma 0A6R we know that \(R\Gamma_Z\) is given by \(C \otimes^\mathbf{L} -\) for some \(C \in D(A)\). Hence, for \(K, L \in D(A)\) general we have \[R\Gamma_Z(K \otimes_A^\mathbf{L} L) = K \otimes^\mathbf{L} L \otimes_A^\mathbf{L} C = K \otimes_A^\mathbf{L} R\Gamma_Z(L)\] The other equalities follow formally from this one. This also implies the last statement of the lemma.
Lemma
Let \(A\) be a ring and let \(I, J \subset A\) be finitely generated ideals. Set \(Z = V(I)\) and \(Y = V(J)\). Then \(Z \cap Y = V(I + J)\) and \(R\Gamma_Y \circ R\Gamma_Z = R\Gamma_{Y \cap Z}\) as functors \(D(A) \to D_{(I + J)^\infty\text{-torsion}}(A)\). For \(K \in D^+(A)\) there is a spectral sequence \[E_2^{p, q} = H^p_Y(H^q_Z(K)) \Rightarrow H^{p + q}_{Y \cap Z}(K)\] as in Derived Categories, Lemma 015N.
Proof
There is a bit of abuse of notation in the lemma as strictly speaking we cannot compose \(R\Gamma_Y\) and \(R\Gamma_Z\). The meaning of the statement is simply that we are composing \(R\Gamma_Z\) with the inclusion \(D_{I^\infty\text{-torsion}}(A) \to D(A)\) and then with \(R\Gamma_Y\). Then the equality \(R\Gamma_Y \circ R\Gamma_Z = R\Gamma_{Y \cap Z}\) follows from the fact that \[D_{I^\infty\text{-torsion}}(A) \to D(A) \xrightarrow{R\Gamma_Y} D_{(I + J)^\infty\text{-torsion}}(A)\] is right adjoint to the inclusion \(D_{(I + J)^\infty\text{-torsion}}(A) \to D_{I^\infty\text{-torsion}}(A)\). Alternatively one can prove the formula using Lemma 0A6R and the fact that the tensor product of extended Čech complexes on \(f_1, \ldots, f_r\) and \(g_1, \ldots, g_m\) is the extended Čech complex on \(f_1, \ldots, f_n. g_1, \ldots, g_m\). The final assertion follows from this and the cited lemma.
The following lemma is the analogue of More on Algebra, Lemma 0925 for complexes with torsion cohomologies.
Lemma
Let \(A \to B\) be a flat ring map and let \(I \subset A\) be a finitely generated ideal such that \(A/I = B/IB\). Then base change and restriction induce quasi-inverse equivalences \(D_{I^\infty\text{-torsion}}(A) = D_{(IB)^\infty\text{-torsion}}(B)\).
Proof
More precisely the functors are \(K \mapsto K \otimes_A^\mathbf{L} B\) for \(K\) in \(D_{I^\infty\text{-torsion}}(A)\) and \(M \mapsto M_A\) for \(M\) in \(D_{(IB)^\infty\text{-torsion}}(B)\). The reason this works is that \(H^i(K \otimes_A^\mathbf{L} B) = H^i(K) \otimes_A B = H^i(K)\). The first equality holds as \(A \to B\) is flat and the second by More on Algebra, Lemma 05E9.
The following lemma was shown for \(\Hom\) and \(\Ext^1\) of modules in More on Algebra, Lemmas 05EC and 05EG.
Lemma
Let \(A \to B\) be a flat ring map and let \(I \subset A\) be a finitely generated ideal such that \(A/I \to B/IB\) is an isomorphism. For \(K \in D_{I^\infty\text{-torsion}}(A)\) and \(L \in D(A)\) the map \[R\Hom_A(K, L) \longrightarrow R\Hom_B(K \otimes_A B, L \otimes_A B)\] is a quasi-isomorphism. In particular, if \(M\), \(N\) are \(A\)-modules and \(M\) is \(I\)-power torsion, then the canonical map \[\Ext^i_A(M, N) \longrightarrow \Ext^i_B(M \otimes_A B, N \otimes_A B)\] is an isomorphism for all \(i\).
Proof
Let \(Z = V(I) \subset \Spec(A)\) and \(Y = V(IB) \subset \Spec(B)\). Since the cohomology modules of \(K\) are \(I\) power torsion, the canonical map \(R\Gamma_Z(L) \to L\) induces an isomorphism \[R\Hom_A(K, R\Gamma_Z(L)) \to R\Hom_A(K, L)\] in \(D(A)\). Similarly, the cohomology modules of \(K \otimes_A B\) are \(IB\) power torsion and we have an isomorphism \[R\Hom_B(K \otimes_A B, R\Gamma_Y(L \otimes_A B)) \to R\Hom_B(K \otimes_A B, L \otimes_A B)\] in \(D(B)\). By Lemma 0ALZ we have \(R\Gamma_Z(L) \otimes_A B = R\Gamma_Y(L \otimes_A B)\). Hence it suffices to show that the map \[R\Hom_A(K, R\Gamma_Z(L)) \to R\Hom_B(K \otimes_A B, R\Gamma_Z(L) \otimes_A B)\] is a quasi-isomorphism. This follows from Lemma 0AM0.
Local cohomology for Noetherian rings
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. Set \(Z = V(I) \subset \Spec(A)\). Recall that (0A6N) is the functor \[D(I^\infty\text{-torsion}) \to D_{I^\infty\text{-torsion}}(A)\] In fact, there is a natural transformation of functors [0A6U]\[\begin{equation} (\href{dualizing.html#dualizing-equation-compare-torsion}{0A6N}) \circ R\Gamma_I(-) \longrightarrow R\Gamma_Z(-) \end{equation}\] Namely, given a complex of \(A\)-modules \(K^\bullet\) the canonical map \(R\Gamma_I(K^\bullet) \to K^\bullet\) in \(D(A)\) factors (uniquely) through \(R\Gamma_Z(K^\bullet)\) as \(R\Gamma_I(K^\bullet)\) has \(I\)-power torsion cohomology modules (see Lemma 0A6L). In general this map is not an isomorphism (we’ve seen this in Lemma 0A6P).
Lemma
Let \(A\) be a Noetherian ring and let \(I \subset A\) be an ideal.
the adjunction \(R\Gamma_I(K) \to K\) is an isomorphism for \(K \in D_{I^\infty\text{-torsion}}(A)\),
the functor (0A6N) \(D(I^\infty\text{-torsion}) \to D_{I^\infty\text{-torsion}}(A)\) is an equivalence,
the transformation of functors (0A6U) is an isomorphism, in other words \(R\Gamma_I(K) = R\Gamma_Z(K)\) for \(K \in D(A)\).
Proof
A formal argument, which we omit, shows that it suffices to prove (1).
Let \(M\) be an \(I\)-power torsion \(A\)-module. Choose an embedding \(M \to J\) into an injective \(A\)-module. Then \(J[I^\infty]\) is an injective \(A\)-module, see Lemma 08XW, and we obtain an embedding \(M \to J[I^\infty]\). Thus every \(I\)-power torsion module has an injective resolution \(M \to J^\bullet\) with \(J^n\) also \(I\)-power torsion. It follows that \(R\Gamma_I(M) = M\) (this is not a triviality and this is not true in general if \(A\) is not Noetherian). Next, suppose that \(K \in D_{I^\infty\text{-torsion}}^+(A)\). Then the spectral sequence \[R^q\Gamma_I(H^p(K)) \Rightarrow R^{p + q}\Gamma_I(K)\] (Derived Categories, Lemma 015J) converges and above we have seen that only the terms with \(q = 0\) are nonzero. Thus we see that \(R\Gamma_I(K) \to K\) is an isomorphism.
Suppose \(K\) is an arbitrary object of \(D_{I^\infty\text{-torsion}}(A)\). We have \[R^q\Gamma_I(K) = \colim \Ext^q_A(A/I^n, K)\] by Lemma 0954. Choose \(f_1, \ldots, f_r \in A\) generating \(I\). Let \(K_n^\bullet = K(A, f_1^n, \ldots, f_r^n)\) be the Koszul complex with terms in degrees \(-r, \ldots, 0\). Since the pro-objects \(\{A/I^n\}\) and \(\{K_n^\bullet\}\) in \(D(A)\) are the same by More on Algebra, Lemma 0921, we see that \[R^q\Gamma_I(K) = \colim \Ext^q_A(K_n^\bullet, K)\] Pick any complex \(K^\bullet\) of \(A\)-modules representing \(K\). Since \(K_n^\bullet\) is a finite complex of finite free modules we see that \[\Ext^q_A(K_n, K) = H^q(\text{Tot}((K_n^\bullet)^\vee \otimes_A K^\bullet))\] where \((K_n^\bullet)^\vee\) is the dual of the complex \(K_n^\bullet\). See More on Algebra, Lemma 0A66. As \((K_n^\bullet)^\vee\) is a complex of finite free \(A\)-modules sitting in degrees \(0, \ldots, r\) we see that the terms of the complex \(\text{Tot}((K_n^\bullet)^\vee \otimes_A K^\bullet)\) are the same as the terms of the complex \(\text{Tot}((K_n^\bullet)^\vee \otimes_A \tau_{\geq q - r - 2} K^\bullet)\) in degrees \(q - 1\) and higher. Hence we see that \[\Ext^q_A(K_n, K) = \text{Ext}^q_A(K_n, \tau_{\geq q - r - 2}K)\] for all \(n\). It follows that \[R^q\Gamma_I(K) = R^q\Gamma_I(\tau_{\geq q - r - 2}K) = H^q(\tau_{\geq q - r - 2}K) = H^q(K)\] Thus we see that the map \(R\Gamma_I(K) \to K\) is an isomorphism.
Lemma
Let \(A\) be a Noetherian ring and let \(I = (f_1, \ldots, f_r)\) be an ideal of \(A\). Set \(Z = V(I) \subset \Spec(A)\). There are canonical isomorphisms \[R\Gamma_I(A) \to (A \to \prod\nolimits_{i_0} A_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_r}) \to R\Gamma_Z(A)\] in \(D(A)\). If \(M\) is an \(A\)-module, then we have similarly \[R\Gamma_I(M) \cong (M \to \prod\nolimits_{i_0} M_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} M_{f_{i_0}f_{i_1}} \to \ldots \to M_{f_1\ldots f_r}) \cong R\Gamma_Z(M)\] in \(D(A)\).
Proof
This follows from Lemma 0955 and the computation of the functor \(R\Gamma_Z\) in Lemma 0A6R.
Lemma
If \(A \to B\) is a homomorphism of Noetherian rings and \(I \subset A\) is an ideal, then in \(D(B)\) we have \[R\Gamma_I(A) \otimes_A^\mathbf{L} B = R\Gamma_Z(A) \otimes_A^\mathbf{L} B = R\Gamma_Y(B) = R\Gamma_{IB}(B)\] where \(Y = V(IB) \subset \Spec(B)\).
Proof
Depth
In this section we revisit the notion of depth introduced in Algebra, Section 00LE.
Lemma
Let \(A\) be a Noetherian ring, let \(I \subset A\) be an ideal, and let \(M\) be a finite \(A\)-module such that \(IM \not = M\). Then the following integers are equal:
\(\text{depth}_I(M)\),
the smallest integer \(i\) such that \(\Ext_A^i(A/I, M)\) is nonzero, and
the smallest integer \(i\) such that \(H^i_I(M)\) is nonzero.
Moreover, we have \(\Ext^i_A(N, M) = 0\) for \(i < \text{depth}_I(M)\) for any finite \(A\)-module \(N\) annihilated by a power of \(I\).
Proof
We prove the equality of (1) and (2) by induction on \(\text{depth}_I(M)\) which is allowed by Algebra, Lemma 0AUJ.
Base case. If \(\text{depth}_I(M) = 0\), then \(I\) is contained in the union of the associated primes of \(M\) (Algebra, Lemma 00LD). By prime avoidance (Algebra, Lemma 00DS) we see that \(I \subset \mathfrak p\) for some associated prime \(\mathfrak p\). Hence \(\Hom_A(A/I, M)\) is nonzero. Thus equality holds in this case.
Assume that \(\text{depth}_I(M) > 0\). Let \(f \in I\) be a nonzerodivisor on \(M\) such that \(\text{depth}_I(M/fM) = \text{depth}_I(M) - 1\). Consider the short exact sequence \[0 \to M \to M \to M/fM \to 0\] and the associated long exact sequence for \(\Ext^*_A(A/I, -)\). Note that \(\Ext^i_A(A/I, M)\) is a finite \(A/I\)-module (Algebra, Lemmas 08YR and 00LV). Hence we obtain \[\Hom_A(A/I, M/fM) = \Ext^1_A(A/I, M)\] and short exact sequences \[0 \to \Ext^i_A(A/I, M) \to \text{Ext}^i_A(A/I, M/fM) \to \Ext^{i + 1}_A(A/I, M) \to 0\] Thus the equality of (1) and (2) by induction.
Observe that \(\text{depth}_I(M) = \text{depth}_{I^n}(M)\) for all \(n \geq 1\) for example by Algebra, Lemma 07DV. Hence by the equality of (1) and (2) we see that \(\Ext^i_A(A/I^n, M) = 0\) for all \(n\) and \(i < \text{depth}_I(M)\). Let \(N\) be a finite \(A\)-module annihilated by a power of \(I\). Then we can choose a short exact sequence \[0 \to N' \to (A/I^n)^{\oplus m} \to N \to 0\] for some \(n, m \geq 0\). Then \(\Hom_A(N, M) \subset \Hom_A((A/I^n)^{\oplus m}, M)\) and \(\Ext^i_A(N, M) \subset \text{Ext}^{i - 1}_A(N', M)\) for \(i < \text{depth}_I(M)\). Thus a simply induction argument shows that the final statement of the lemma holds.
Finally, we prove that (3) is equal to (1) and (2). We have \(H^p_I(M) = \colim \Ext^p_A(A/I^n, M)\) by Lemma 0954. Thus we see that \(H^i_I(M) = 0\) for \(i < \text{depth}_I(M)\). For \(i = \text{depth}_I(M)\), using the vanishing of \(\Ext_A^{i - 1}(I/I^n, M)\) we see that the map \(\Ext_A^i(A/I, M) \to H_I^i(M)\) is injective which proves nonvanishing in the correct degree.
Lemma
Let \(A\) be a Noetherian ring. Let \(0 \to N' \to N \to N'' \to 0\) be a short exact sequence of finite \(A\)-modules. Let \(I \subset A\) be an ideal.
\(\text{depth}_I(N) \geq \min\{\text{depth}_I(N'), \text{depth}_I(N'')\}\)
\(\text{depth}_I(N'') \geq \min\{\text{depth}_I(N), \text{depth}_I(N') - 1\}\)
\(\text{depth}_I(N') \geq \min\{\text{depth}_I(N), \text{depth}_I(N'') + 1\}\)
Proof
Assume \(IN \not = N\), \(IN' \not = N'\), and \(IN'' \not = N''\). Then we can use the characterization of depth using the Ext groups \(\Ext^i(A/I, N)\), see Lemma 0AVZ, and use the long exact cohomology sequence \[\begin{matrix} 0 \to \Hom_A(A/I, N') \to \Hom_A(A/I, N) \to \Hom_A(A/I, N'') \\ \phantom{0\ } \to \Ext^1_A(A/I, N') \to \Ext^1_A(A/I, N) \to \Ext^1_A(A/I, N'') \to \ldots \end{matrix}\] from Algebra, Lemma 00LU. This argument also works if \(IN = N\) because in this case \(\Ext^i_A(A/I, N) = 0\) for all \(i\). Similarly in case \(IN' \not = N'\) and/or \(IN'' \not = N''\).
Lemma
Let \(A\) be a Noetherian ring, let \(I \subset A\) be an ideal, and let \(M\) a finite \(A\)-module with \(IM \not = M\).
If \(x \in I\) is a nonzerodivisor on \(M\), then \(\text{depth}_I(M/xM) = \text{depth}_I(M) - 1\).
Any \(M\)-regular sequence \(x_1, \ldots, x_r\) in \(I\) can be extended to an \(M\)-regular sequence in \(I\) of length \(\text{depth}_I(M)\).
Proof
Part (2) is a formal consequence of part (1). Let \(x \in I\) be as in (1). By the short exact sequence \(0 \to M \to M \to M/xM \to 0\) and Lemma 0BUV we see that \(\text{depth}_I(M/xM) \geq \text{depth}_I(M) - 1\). On the other hand, if \(x_1, \ldots, x_r \in I\) is a regular sequence for \(M/xM\), then \(x, x_1, \ldots, x_r\) is a regular sequence for \(M\). Hence (1) holds.
Lemma
Let \(R\) be a Noetherian local ring. If \(M\) is a finite Cohen-Macaulay \(R\)-module and \(I \subset R\) a nontrivial ideal. Then \[\text{depth}_I(M) = \dim(\text{Supp}(M)) - \dim(\text{Supp}(M/IM)).\]
Proof
We will prove this by induction on \(\text{depth}_I(M)\).
If \(\text{depth}_I(M) = 0\), then \(I\) is contained in one of the associated primes \(\mathfrak p\) of \(M\) (Algebra, Lemma 00LL). Then \(\mathfrak p \in \text{Supp}(M/IM)\), hence \(\dim(\text{Supp}(M/IM)) \geq \dim(R/\mathfrak p) = \dim(\text{Supp}(M))\) where equality holds by Algebra, Lemma 0BUS. Thus the lemma holds in this case.
If \(\text{depth}_I(M) > 0\), we pick \(x \in I\) which is a nonzerodivisor on \(M\). Note that \((M/xM)/I(M/xM) = M/IM\). On the other hand we have \(\text{depth}_I(M/xM) = \text{depth}_I(M) - 1\) by Lemma 0BUW and \(\dim(\text{Supp}(M/xM)) = \dim(\text{Supp}(M)) - 1\) by Algebra, Lemma 0B52. Thus the result by induction hypothesis.
Lemma
Let \(R \to S\) be a flat local ring homomorphism of Noetherian local rings. Denote \(\mathfrak m \subset R\) the maximal ideal. Let \(I \subset S\) be an ideal. If \(S/\mathfrak mS\) is Cohen-Macaulay, then \[\text{depth}_I(S) \geq \dim(S/\mathfrak mS) - \dim(S/\mathfrak mS + I)\]
Proof
By Algebra, Lemma 00MG any sequence in \(S\) which maps to a regular sequence in \(S/\mathfrak mS\) is a regular sequence in \(S\). Thus it suffices to prove the lemma in case \(R\) is a field. This is a special case of Lemma 0BUX.
Lemma
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. Let \(M\) be an \(A\)-module. Let \(Z = V(I)\). Then \(H^0_I(M) = H^0_Z(M)\). Let \(N\) be the common value and set \(M' = M/N\). Then
\(H^0_I(M') = 0\) and \(H^p_I(M) = H^p_I(M')\) and \(H^p_I(N) = 0\) for all \(p > 0\),
\(H^0_Z(M') = 0\) and \(H^p_Z(M) = H^p_Z(M')\) and \(H^p_Z(N) = 0\) for all \(p > 0\).
Proof
By definition \(H^0_I(M) = M[I^\infty]\) is \(I\)-power torsion. By Lemma 0A6R we see that \[H^0_Z(M) = \Ker(M \longrightarrow M_{f_1} \times \ldots \times M_{f_r})\] if \(I = (f_1, \ldots, f_r)\). Thus \(H^0_I(M) \subset H^0_Z(M)\) and conversely, if \(x \in H^0_Z(M)\), then it is annihilated by a \(f_i^{e_i}\) for some \(e_i \geq 1\) hence annihilated by some power of \(I\). This proves the first equality and moreover \(N\) is \(I\)-power torsion. By Lemma 0A6L we see that \(R\Gamma_I(N) = N\). By Lemma 0A6R we see that \(R\Gamma_Z(N) = N\). This proves the higher vanishing of \(H^p_I(N)\) and \(H^p_Z(N)\) in (1) and (2). The vanishing of \(H^0_I(M')\) and \(H^0_Z(M')\) follow from the preceding remarks and the fact that \(M'\) is \(I\)-power torsion free by More on Algebra, Lemma 05EB. The equality of higher cohomologies for \(M\) and \(M'\) follow immediately from the long exact cohomology sequence.
Torsion versus complete modules
Let \(A\) be a ring and let \(I\) be a finitely generated ideal. In this case we can consider the derived category \(D_{I^\infty\text{-torsion}}(A)\) of complexes with \(I\)-power torsion cohomology modules (Section 0952) and the derived category \(D_{comp}(A, I)\) of derived complete complexes (More on Algebra, Section 091N). In this section we show these categories are equivalent. A more general statement can be found in [Dwyer-Greenlees].
Lemma
Let \(A\) be a ring and let \(I\) be a finitely generated ideal. Let \(R\Gamma_Z\) be as in Lemma 0A6R. Let \({\ }^\wedge\) denote derived completion as in More on Algebra, Lemma 091V. For an object \(K\) in \(D(A)\) we have \[R\Gamma_Z(K^\wedge) = R\Gamma_Z(K) \quad\text{and}\quad (R\Gamma_Z(K))^\wedge = K^\wedge\] in \(D(A)\).
Proof
Choose \(f_1, \ldots, f_r \in A\) generating \(I\). Recall that \[K^\wedge = R\Hom_A\left((A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_r}), K\right)\] by More on Algebra, Lemma 091V. Hence the cone \(C = \text{Cone}(K \to K^\wedge)\) is given by \[R\Hom_A\left((\prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_r}), K\right)\] which can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to \[R\Hom_A(A_{f_{i_0} \ldots f_{i_p}}, K), \quad p > 0\] These complexes vanish on applying \(R\Gamma_Z\), see Lemma 0A6S. Applying \(R\Gamma_Z\) to the distinguished triangle \(K \to K^\wedge \to C \to K[1]\) we see that the first formula of the lemma is correct.
Recall that \[R\Gamma_Z(K) = K \otimes^\mathbf{L} (A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_r})\] by Lemma 0A6R. Hence the cone \(C = \text{Cone}(R\Gamma_Z(K) \to K)\) can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to \[K \otimes_A A_{f_{i_0} \ldots f_{i_p}}, \quad p > 0\] These complexes vanish on applying \({\ }^\wedge\), see More on Algebra, Lemma 0A6D. Applying derived completion to the distinguished triangle \(R\Gamma_Z(K) \to K \to C \to R\Gamma_Z(K)[1]\) we see that the second formula of the lemma is correct.
The following result is a special case of a very general phenomenon concerning admissible subcategories of a triangulated category.
Proposition
Let \(A\) be a ring and let \(I \subset A\) be a finitely generated ideal. The functors \(R\Gamma_Z\) and \({\ }^\wedge\) define quasi-inverse equivalences of categories \[D_{I^\infty\text{-torsion}}(A) \leftrightarrow D_{comp}(A, I)\]
Proof
Follows immediately from Lemma 0A6W.
The following addendum of the proposition above makes the correspondence on morphisms more precise.
Lemma
With notation as in Lemma 0A6W. For objects \(K, L\) in \(D(A)\) there is a canonical isomorphism \[R\Hom_A(K^\wedge, L^\wedge) \longrightarrow R\Hom_A(R\Gamma_Z(K), R\Gamma_Z(L))\] in \(D(A)\).
Proof
Say \(I = (f_1, \ldots, f_r)\). Denote \(C = (A \to \prod A_{f_i} \to \ldots \to A_{f_1 \ldots f_r})\) the alternating Čech complex. Then derived completion is given by \(R\Hom_A(C, -)\) (More on Algebra, Lemma 091V) and local cohomology by \(C \otimes^\mathbf{L} -\) (Lemma 0A6R). Combining the isomorphism \[R\Hom_A(K \otimes^\mathbf{L} C, L \otimes^\mathbf{L} C) = R\Hom_A(K, R\Hom_A(C, L \otimes^\mathbf{L} C))\] (More on Algebra, Lemma 0A65) and the map \[L \to R\Hom_A(C, L \otimes^\mathbf{L} C)\] (More on Algebra, Lemma 0A6B) we obtain a map \[\gamma : R\Hom_A(K, L) \longrightarrow R\Hom_A(K \otimes^\mathbf{L} C, L \otimes^\mathbf{L} C)\] On the other hand, the right hand side is derived complete as it is equal to \[R\Hom_A(C, R\Hom_A(K, L \otimes^\mathbf{L} C)).\] Thus \(\gamma\) factors through the derived completion of \(R\Hom_A(K, L)\) by the universal property of derived completion. However, the derived completion goes inside the \(R\Hom_A\) by More on Algebra, Lemma 0A6E and we obtain the desired map.
To show that the map of the lemma is an isomorphism we may assume that \(K\) and \(L\) are derived complete, i.e., \(K = K^\wedge\) and \(L = L^\wedge\). In this case we are looking at the map \[\gamma : R\Hom_A(K, L) \longrightarrow R\Hom_A(R\Gamma_Z(K), R\Gamma_Z(L))\] By Proposition 0A6X we know that the cohomology groups of the left and the right hand side coincide. In other words, we have to check that the map \(\gamma\) sends a morphism \(\alpha : K \to L\) in \(D(A)\) to the morphism \(R\Gamma_Z(\alpha) : R\Gamma_Z(K) \to R\Gamma_Z(L)\). We omit the verification (hint: note that \(R\Gamma_Z(\alpha)\) is just the map \(\alpha \otimes \text{id}_C : K \otimes^\mathbf{L} C \to L \otimes^\mathbf{L} C\) which is almost the same as the construction of the map in More on Algebra, Lemma 0A6B).
Lemma
Let \(I\) and \(J\) be ideals in a Noetherian ring \(A\). Let \(M\) be a finite \(A\)-module. Set \(Z =V(J)\). Consider the derived \(I\)-adic completion \(R\Gamma_Z(M)^\wedge\) of local cohomology. Then
we have \(R\Gamma_Z(M)^\wedge = R\lim R\Gamma_Z(M/I^nM)\), and
there are short exact sequences \[0 \to R^1\lim H^{i - 1}_Z(M/I^nM) \to H^i(R\Gamma_Z(M)^\wedge) \to \lim H^i_Z(M/I^nM) \to 0\]
In particular \(R\Gamma_Z(M)^\wedge\) has vanishing cohomology in negative degrees.
Proof
Suppose that \(J = (g_1, \ldots, g_m)\). Then \(R\Gamma_Z(M)\) is computed by the complex \[M \to \prod M_{g_{j_0}} \to \prod M_{g_{j_0}g_{j_1}} \to \ldots \to M_{g_1g_2\ldots g_m}\] by Lemma 0A6R. By More on Algebra, Lemma 0EEU the derived \(I\)-adic completion of this complex is given by the complex \[\lim M/I^nM \to \prod \lim (M/I^nM)_{g_{j_0}} \to \ldots \to \lim (M/I^nM)_{g_1g_2\ldots g_m}\] of usual completions. Since \(R\Gamma_Z(M/I^nM)\) is computed by the complex \(M/I^nM \to \prod (M/I^nM)_{g_{j_0}} \to \ldots \to (M/I^nM)_{g_1g_2\ldots g_m}\) and since the transition maps between these complexes are surjective, we conclude that (1) holds by More on Algebra, Lemma 091D. Part (2) then follows from More on Algebra, Lemma 0CQE.
Lemma
With notation and hypotheses as in Lemma 0EEW assume \(A\) is \(I\)-adically complete. Then \[H^0(R\Gamma_Z(M)^\wedge) = \colim H^0_{V(J')}(M)\] where the filtered colimit is over \(J' \subset J\) such that \(V(J') \cap V(I) = V(J) \cap V(I)\).
Proof
Since \(M\) is a finite \(A\)-module, we have that \(M\) is \(I\)-adically complete. The proof of Lemma 0EEW shows that \[H^0(R\Gamma_Z(M)^\wedge) = \Ker(M^\wedge \to \prod M_{g_j}^\wedge) = \Ker(M \to \prod M_{g_j}^\wedge)\] where on the right hand side we have usual \(I\)-adic completion. The kernel \(K_j\) of \(M_{g_j} \to M_{g_j}^\wedge\) is \(\bigcap I^n M_{g_j}\). By Algebra, Lemma 00IQ for every \(\mathfrak p \in V(IA_{g_j})\) we find an \(f \in A_{g_j}\), \(f \not \in \mathfrak p\) such that \((K_j)_f = 0\).
Let \(s \in H^0(R\Gamma_Z(M)^\wedge)\). By the above we may think of \(s\) as an element of \(M\). The support \(Z'\) of \(s\) intersected with \(D(g_j)\) is disjoint from \(D(g_j) \cap V(I)\) by the arguments above. Thus \(Z'\) is a closed subset of \(\Spec(A)\) with \(Z' \cap V(I) \subset V(J)\). Then \(Z' \cup V(J) = V(J')\) for some ideal \(J' \subset J\) with \(V(J') \cap V(I) \subset V(J)\) and we have \(s \in H^0_{V(J')}(M)\). Conversely, any \(s \in H^0_{V(J')}(M)\) with \(J' \subset J\) and \(V(J') \cap V(I) \subset V(J)\) maps to zero in \(M_{g_j}^\wedge\) for all \(j\). This proves the lemma.
Trivial duality for a ring map
Let \(A \to B\) be a ring homomorphism. Consider the functor \[\Hom_A(B, -) : \text{Mod}_A \longrightarrow \text{Mod}_B,\quad M \longmapsto \Hom_A(B, M)\] This functor is left exact and has a derived extension \(R\Hom(B, -) : D(A) \to D(B)\).
Lemma
Let \(A \to B\) be a ring homomorphism. The functor \(R\Hom(B, -)\) constructed above is right adjoint to the restriction functor \(D(B) \to D(A)\).
Proof
This is a consequence of the fact that restriction and \(\Hom_A(B, -)\) are adjoint functors by Algebra, Lemma 08YP. See Derived Categories, Lemma 09T5.
Lemma
Let \(A \to B \to C\) be ring maps. Then \(R\Hom(C, -) \circ R\Hom(B, -) : D(A) \to D(C)\) is the functor \(R\Hom(C, -) : D(A) \to D(C)\).
Proof
Follows from uniqueness of right adjoints and Lemma 0A70.
Lemma
Let \(\varphi : A \to B\) be a ring homomorphism. For \(K\) in \(D(A)\) we have \[\varphi_*R\Hom(B, K) = R\Hom_A(B, K)\] where \(\varphi_* : D(B) \to D(A)\) is restriction. In particular \(R^q\Hom(B, K) = \Ext_A^q(B, K)\).
Proof
Choose a K-injective complex \(I^\bullet\) representing \(K\). Then \(R\Hom(B, K)\) is represented by the complex \(\Hom_A(B, I^\bullet)\) of \(B\)-modules. Since this complex, as a complex of \(A\)-modules, represents \(R\Hom_A(B, K)\) we see that the lemma is true.
Let \(A\) be a Noetherian ring. We will denote \[D_{\textit{Coh}}(A) \subset D(A)\] the full subcategory consisting of those objects \(K\) of \(D(A)\) whose cohomology modules are all finite \(A\)-modules. This makes sense by Derived Categories, Section 06UP because as \(A\) is Noetherian, the subcategory of finite \(A\)-modules is a Serre subcategory of \(\text{Mod}_A\).
Lemma
With notation as above, assume \(A \to B\) is a finite ring map of Noetherian rings. Then \(R\Hom(B, -)\) maps \(D^+_{\textit{Coh}}(A)\) into \(D^+_{\textit{Coh}}(B)\).
Proof
We have to show: if \(K \in D^+(A)\) has finite cohomology modules, then the complex \(R\Hom(B, K)\) has finite cohomology modules too. This follows for example from Lemma 0A71 if we can show the ext modules \(\Ext^i_A(B, K)\) are finite \(A\)-modules. Since \(K\) is bounded below there is a convergent spectral sequence \[\Ext^p_A(B, H^q(K)) \Rightarrow \text{Ext}^{p + q}_A(B, K)\] This finishes the proof as the modules \(\Ext^p_A(B, H^q(K))\) are finite by Algebra, Lemma 08YR.
Remark
Let \(A\) be a ring and let \(I \subset A\) be an ideal. Set \(B = A/I\). In this case the functor \(\Hom_A(B, -)\) is equal to the functor \[\text{Mod}_A \longrightarrow \text{Mod}_B,\quad M \longmapsto M[I]\] which sends \(M\) to the submodule of \(I\)-torsion.
Situation
Let \(R \to A\) be a ring map. We will give an alternative construction of \(R\Hom(A, -)\) which will stand us in good stead later in this chapter. Namely, suppose we have a differential graded algebra \((E, d)\) over \(R\) and a quasi-isomorphism \(E \to A\) where we view \(A\) as a differential graded algebra over \(R\) with zero differential. Then we have commutative diagrams \[\vcenter{ \xymatrix{ D(E, \text{d}) \ar[rd] & & D(A) \ar[ll] \ar[ld] \\ & D(R) } } \quad\text{and}\quad \vcenter{ \xymatrix{ D(E, \text{d}) \ar[rr]_{- \otimes_E^\mathbf{L} A} & & D(A) \\ & D(R) \ar[lu]^{- \otimes_R^\mathbf{L} E} \ar[ru]_{- \otimes_R^\mathbf{L} A} } }\] where the horizontal arrows are equivalences of categories (Differential Graded Algebra, Lemma 09S6). It is clear that the first diagram commutes. The second diagram commutes because the first one does and our functors are their left adjoints (Differential Graded Algebra, Example 0BYX) or because we have \(E \otimes^\mathbf{L}_E A = E \otimes_E A\) and we can use Differential Graded Algebra, Lemma 0BZ3.
Lemma
In Situation 0BZB the functor \(R\Hom(A, -)\) is equal to the composition of \(R\Hom(E, -) : D(R) \to D(E, \text{d})\) and the equivalence \(- \otimes^\mathbf{L}_E A : D(E, \text{d}) \to D(A)\).
Proof
This is true because \(R\Hom(E, -)\) is the right adjoint to \(- \otimes^\mathbf{L}_R E\), see Differential Graded Algebra, Lemma 09LT. Hence this functor plays the same role as the functor \(R\Hom(A, -)\) for the map \(R \to A\) (Lemma 0A70), whence these functors must correspond via the equivalence \(- \otimes^\mathbf{L}_E A : D(E, \text{d}) \to D(A)\).
Lemma
In Situation 0BZB assume that
\(E\) viewed as an object of \(D(R)\) is compact, and
\(N = \Hom^\bullet_R(E^\bullet, R)\) computes \(R\Hom(E, R)\).
Then \(R\Hom(E, -) : D(R) \to D(E)\) is isomorphic to \(K \mapsto K \otimes_R^\mathbf{L} N\).
Proof
Special case of Differential Graded Algebra, Lemma 0BZ0.
Lemma
In Situation 0BZB assume \(A\) is a perfect \(R\)-module. Then \[R\Hom(A, -) : D(R) \to D(A)\] is given by \(K \mapsto K \otimes_R^\mathbf{L} M\) where \(M = R\Hom(A, R) \in D(A)\).
Proof
We apply Divided Power Algebra, Lemma 0BZ9 to choose a Tate resolution \((E, \text{d})\) of \(A\) over \(R\). Note that \(E^i = 0\) for \(i > 0\), \(E^0 = R[x_1, \ldots, x_n]\) is a polynomial algebra, and \(E^i\) is a finite free \(E^0\)-module for \(i < 0\). It follows that \(E\) viewed as a complex of \(R\)-modules is a bounded above complex of free \(R\)-modules. We check the assumptions of Lemma 0BZD. The first holds because \(A\) is perfect (hence compact by More on Algebra, Proposition 07LT) and the second by More on Algebra, Lemma 0A66. From the lemma conclude that \(K \mapsto R\Hom(E, K)\) is isomorphic to \(K \mapsto K \otimes_R^\mathbf{L} N\) for some differential graded \(E\)-module \(N\). Observe that \[(R \otimes_R E) \otimes_E^\mathbf{L} A = R \otimes_E E \otimes_E A\] in \(D(A)\). Hence by Differential Graded Algebra, Lemma 0BZ5 we conclude that the composition of \(- \otimes_R^\mathbf{L} N\) and \(- \otimes_R^\mathbf{L} A\) is of the form \(- \otimes_R M\) for some \(M \in D(A)\). To finish the proof we apply Lemma 0BZC.
Lemma
Let \(R \to A\) be a surjective ring map whose kernel \(I\) is an invertible \(R\)-module. The functor \(R\Hom(A, -) : D(R) \to D(A)\) is isomorphic to \(K \mapsto K \otimes_R^\mathbf{L} N[-1]\) where \(N\) is inverse of the invertible \(A\)-module \(I \otimes_R A\).
Proof
Since \(A\) has the finite projective resolution \[0 \to I \to R \to A \to 0\] we see that \(A\) is a perfect \(R\)-module. By Lemma 0BZE it suffices to prove that \(R\Hom(A, R)\) is represented by \(N[-1]\) in \(D(A)\). This means \(R\Hom(A, R)\) has a unique nonzero cohomology module, namely \(N\) in degree \(1\). As \(\text{Mod}_A \to \text{Mod}_R\) is fully faithful it suffice to prove this after applying the restriction functor \(i_* : D(A) \to D(R)\). By Lemma 0A71 we have \[i_*R\Hom(A, R) = R\Hom_R(A, R)\] Using the finite projective resolution above we find that the latter is represented by the complex \(R \to I^{\otimes -1}\) with \(R\) in degree \(0\). The map \(R \to I^{\otimes -1}\) is injective and the cokernel is \(N\).
Base change for trivial duality
In this section we consider a cocartesian square of rings \[\xymatrix{ A \ar[r]_\alpha & A' \\ R \ar[u]^\varphi \ar[r]^\rho & R' \ar[u]_{\varphi'} }\] In other words, we have \(A' = A \otimes_R R'\). If \(A\) and \(R'\) are tor independent over \(R\) then there is a canonical base change map [0E29]\[\begin{equation} R\Hom(A, K) \otimes_A^\mathbf{L} A' \longrightarrow R\Hom(A', K \otimes_R^\mathbf{L} R') \end{equation}\] in \(D(A')\) functorial for \(K\) in \(D(R)\). Namely, by the adjointness of Lemma 0A70 such an arrow is the same thing as a map \[\varphi'_*\left(R\Hom(A, K) \otimes_A^\mathbf{L} A'\right) \longrightarrow K \otimes_R^\mathbf{L} R'\] in \(D(R')\) where \(\varphi'_* : D(A') \to D(R')\) is the restriction functor. We may apply More on Algebra, Lemma 0661 to the left hand side to get that this is the same thing as a map \[\varphi_*(R\Hom(A, K)) \otimes_R^\mathbf{L} R' \longrightarrow K \otimes_R^\mathbf{L} R'\] in \(D(R')\) where \(\varphi_* : D(A) \to D(R)\) is the restriction functor. For this we can choose \(can \otimes^\mathbf{L} \text{id}_{R'}\) where \(can : \varphi_*(R\Hom(A, K)) \to K\) is the counit of the adjunction between \(R\Hom(A, -)\) and \(\varphi_*\).
Lemma
In the situation above, the map (0E29) is an isomorphism if and only if the map \[R\Hom_R(A, K) \otimes_R^\mathbf{L} R' \longrightarrow R\Hom_R(A, K \otimes_R^\mathbf{L} R')\] of More on Algebra, Lemma 0BYN is an isomorphism.
Proof
To see that the map is an isomorphism, it suffices to prove it is an isomorphism after applying \(\varphi'_*\). Applying the functor \(\varphi'_*\) to (0E29) and using that \(A' = A \otimes_R^\mathbf{L} R'\) we obtain the base change map \(R\Hom_R(A, K) \otimes_R^\mathbf{L} R' \to R\Hom_{R'}(A \otimes_R^\mathbf{L} R', K \otimes_R^\mathbf{L} R')\) for derived hom of More on Algebra, Equation (0E1X). Unwinding the left and right hand side exactly as in the proof of More on Algebra, Lemma 0A6A and in particular using More on Algebra, Lemma 0E1W gives the desired result.
Lemma
Let \(R \to A\) and \(R \to R'\) be ring maps and \(A' = A \otimes_R R'\). Assume
\(A\) is pseudo-coherent as an \(R\)-module,
\(R'\) has finite tor dimension as an \(R\)-module (for example \(R \to R'\) is flat),
\(A\) and \(R'\) are tor independent over \(R\).
Then (0E29) is an isomorphism for \(K \in D^+(R)\).
Proof
Follows from Lemma 0E2A and More on Algebra, Lemma 0ATK part (4).
Lemma
Let \(R \to A\) and \(R \to R'\) be ring maps and \(A' = A \otimes_R R'\). Assume
\(A\) is perfect as an \(R\)-module,
\(A\) and \(R'\) are tor independent over \(R\).
Then (0E29) is an isomorphism for all \(K \in D(R)\).
Proof
Follows from Lemma 0E2A and More on Algebra, Lemma 0ATK part (1).
Dualizing complexes
In this section we define dualizing complexes for Noetherian rings.
Definition
Let \(A\) be a Noetherian ring. A dualizing complex is a complex of \(A\)-modules \(\omega_A^\bullet\) such that
\(\omega_A^\bullet\) has finite injective dimension,
\(H^i(\omega_A^\bullet)\) is a finite \(A\)-module for all \(i\), and
\(A \to R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)\) is a quasi-isomorphism.
This definition takes some time getting used to. It is perhaps a good idea to prove some of the following lemmas yourself without reading the proofs.
Lemma
Let \(A\) be a Noetherian ring. Let \(K, L \in D_{\textit{Coh}}(A)\) and assume \(L\) has finite injective dimension. Then \(R\Hom_A(K, L)\) is in \(D_{\textit{Coh}}(A)\).
Proof
Pick an integer \(n\) and consider the distinguished triangle \[\tau_{\leq n}K \to K \to \tau_{\geq n + 1}K \to \tau_{\leq n}K[1]\] see Derived Categories, Remark 08J5. Since \(L\) has finite injective dimension we see that \(R\Hom_A(\tau_{\geq n + 1}K, L)\) has vanishing cohomology in degrees \(\geq c - n\) for some constant \(c\). Hence, given \(i\), we see that \(\Ext^i_A(K, L) \to \Ext^i_A(\tau_{\leq n}K, L)\) is an isomorphism for some \(n \gg - i\). By Derived Categories of Schemes, Lemma 0D0C applied to \(\tau_{\leq n}K\) and \(L\) we see conclude that \(\Ext^i_A(K, L)\) is a finite \(A\)-module for all \(i\). Hence \(R\Hom_A(K, L)\) is indeed an object of \(D_{\textit{Coh}}(A)\).
Lemma
Let \(A\) be a Noetherian ring. If \(\omega_A^\bullet\) is a dualizing complex, then the functor \[D : K \longmapsto R\Hom_A(K, \omega_A^\bullet)\] is an anti-equivalence \(D_{\textit{Coh}}(A) \to D_{\textit{Coh}}(A)\) which exchanges \(D^+_{\textit{Coh}}(A)\) and \(D^-_{\textit{Coh}}(A)\) and induces an anti-equivalence \(D^b_{\textit{Coh}}(A) \to D^b_{\textit{Coh}}(A)\). Moreover \(D \circ D\) is isomorphic to the identity functor.
Proof
Let \(K\) be an object of \(D_{\textit{Coh}}(A)\). From Lemma 0G4H we see \(R\Hom_A(K, \omega_A^\bullet)\) is an object of \(D_{\textit{Coh}}(A)\). By More on Algebra, Lemma 0A69 and the assumptions on the dualizing complex we obtain a canonical isomorphism \[K = R\Hom_A(\omega_A^\bullet, \omega_A^\bullet) \otimes_A^\mathbf{L} K \longrightarrow R\Hom_A(R\Hom_A(K, \omega_A^\bullet), \omega_A^\bullet)\] Thus our functor has a quasi-inverse and the proof is complete.
Let \(R\) be a ring. Recall that an object \(L\) of \(D(R)\) is invertible if it is an invertible object for the symmetric monoidal structure on \(D(R)\) given by derived tensor product. In More on Algebra, Lemma 0FNT we have seen this means \(L\) is perfect, \(L = \bigoplus H^n(L)[-n]\), this is a finite sum, each \(H^n(L)\) is finite projective, and there is an open covering \(\Spec(R) = \bigcup D(f_i)\) such that \(L \otimes_R R_{f_i} \cong R_{f_i}[-n_i]\) for some integers \(n_i\).
Lemma
Let \(A\) be a Noetherian ring. Let \(F : D^b_{\textit{Coh}}(A) \to D^b_{\textit{Coh}}(A)\) be an \(A\)-linear equivalence of categories. Then \(F(A)\) is an invertible object of \(D(A)\).
Proof
Let \(\mathfrak m \subset A\) be a maximal ideal with residue field \(\kappa\). Consider the object \(F(\kappa)\). Since \(\kappa = \Hom_{D(A)}(\kappa, \kappa)\) we find that all cohomology groups of \(F(\kappa)\) are annihilated by \(\mathfrak m\). We also see that \[\Ext^i_A(\kappa, \kappa) = \text{Ext}^i_A(F(\kappa), F(\kappa)) = \Hom_{D(A)}(F(\kappa), F(\kappa)[i])\] is zero for \(i < 0\). Say \(H^a(F(\kappa)) \not = 0\) and \(H^b(F(\kappa)) \not = 0\) with \(a\) minimal and \(b\) maximal (so in particular \(a \leq b\)). Then there is a nonzero map \[F(\kappa) \to H^b(F(\kappa))[-b] \to H^a(F(\kappa))[-b] \to F(\kappa)[a - b]\] in \(D(A)\) (nonzero because it induces a nonzero map on cohomology). This proves that \(b = a\). We conclude that \(F(\kappa) = \kappa[-a]\).
Let \(G\) be a quasi-inverse to our functor \(F\). Arguing as above we find an integer \(b\) such that \(G(\kappa) = \kappa[-b]\). On composing we find \(a + b = 0\). Let \(E\) be a finite \(A\)-module which is annihilated by a power of \(\mathfrak m\). Arguing by induction on the length of \(E\) we find that \(G(E) = E'[-b]\) for some finite \(A\)-module \(E'\) annihilated by a power of \(\mathfrak m\). Then \(E[-a] = F(E')\). Next, we consider the groups \[\Ext^i_A(A, E') = \text{Ext}^i_A(F(A), F(E')) = \Hom_{D(A)}(F(A), E[-a + i])\] The left hand side is nonzero if and only if \(i = 0\) and then we get \(E'\). Applying this with \(E = E' = \kappa\) and using Nakayama’s lemma this implies that \(H^j(F(A))_\mathfrak m\) is zero for \(j > a\) and generated by \(1\) element for \(j = a\). On the other hand, if \(H^j(F(A))_\mathfrak m\) is not zero for some \(j < a\), then there is a map \(F(A) \to E[-a + i]\) for some \(i < 0\) and some \(E\) (More on Algebra, Lemma 0A7D) which is a contradiction. Thus we see that \(F(A)_\mathfrak m = M[-a]\) for some \(A_\mathfrak m\)-module \(M\) generated by \(1\) element. However, since \[A_\mathfrak m = \Hom_{D(A)}(A, A)_\mathfrak m = \Hom_{D(A)}(F(A), F(A))_\mathfrak m = \Hom_{A_\mathfrak m}(M, M)\] we see that \(M \cong A_\mathfrak m\). We conclude that there exists an element \(f \in A\), \(f \not \in \mathfrak m\) such that \(F(A)_f\) is isomorphic to \(A_f[-a]\). This finishes the proof.
Lemma
Let \(A\) be a Noetherian ring. If \(\omega_A^\bullet\) and \((\omega'_A)^\bullet\) are dualizing complexes, then \((\omega'_A)^\bullet\) is quasi-isomorphic to \(\omega_A^\bullet \otimes_A^\mathbf{L} L\) for some invertible object \(L\) of \(D(A)\).
Proof
By Lemmas 0A7C and 0A7E the functor \(K \mapsto R\Hom_A(R\Hom_A(K, \omega_A^\bullet), (\omega_A')^\bullet)\) maps \(A\) to an invertible object \(L\). In other words, there is an isomorphism \[L \longrightarrow R\Hom_A(\omega_A^\bullet, (\omega_A')^\bullet)\] Since \(L\) has finite tor dimension, this means that we can apply More on Algebra, Lemma 0A69 to see that \[R\Hom_A(\omega_A^\bullet, (\omega'_A)^\bullet) \otimes_A^\mathbf{L} K \longrightarrow R\Hom_A(R\Hom_A(K, \omega_A^\bullet), (\omega_A')^\bullet)\] is an isomorphism for \(K\) in \(D^b_{\textit{Coh}}(A)\). In particular, setting \(K = \omega_A^\bullet\) finishes the proof.
Lemma
Let \(A\) be a Noetherian ring. Let \(B = S^{-1}A\) be a localization. If \(\omega_A^\bullet\) is a dualizing complex, then \(\omega_A^\bullet \otimes_A B\) is a dualizing complex for \(B\).
Proof
Let \(\omega_A^\bullet \to I^\bullet\) be a quasi-isomorphism with \(I^\bullet\) a bounded complex of injectives. Then \(S^{-1}I^\bullet\) is a bounded complex of injective \(B = S^{-1}A\)-modules (Lemma 0A6I) representing \(\omega_A^\bullet \otimes_A B\). Thus \(\omega_A^\bullet \otimes_A B\) has finite injective dimension. Since \(H^i(\omega_A^\bullet \otimes_A B) = H^i(\omega_A^\bullet) \otimes_A B\) by flatness of \(A \to B\) we see that \(\omega_A^\bullet \otimes_A B\) has finite cohomology modules. Finally, the map \[B \longrightarrow R\Hom_A(\omega_A^\bullet \otimes_A B, \omega_A^\bullet \otimes_A B)\] is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 0A6A.
Lemma
Let \(A\) be a Noetherian ring. Let \(f_1, \ldots, f_n \in A\) generate the unit ideal. If \(\omega_A^\bullet\) is a complex of \(A\)-modules such that \((\omega_A^\bullet)_{f_i}\) is a dualizing complex for \(A_{f_i}\) for all \(i\), then \(\omega_A^\bullet\) is a dualizing complex for \(A\).
Proof
Consider the double complex \[\prod\nolimits_{i_0} (\omega_A^\bullet)_{f_{i_0}} \to \prod\nolimits_{i_0 < i_1} (\omega_A^\bullet)_{f_{i_0}f_{i_1}} \to \ldots\] The associated total complex is quasi-isomorphic to \(\omega_A^\bullet\) for example by Descent, Remark 023O or by Derived Categories of Schemes, Lemma 08D2. By assumption the complexes \((\omega_A^\bullet)_{f_i}\) have finite injective dimension as complexes of \(A_{f_i}\)-modules. This implies that each of the complexes \((\omega_A^\bullet)_{f_{i_0} \ldots f_{i_p}}\), \(p > 0\) has finite injective dimension over \(A_{f_{i_0} \ldots f_{i_p}}\), see Lemma 0A6I. This in turn implies that each of the complexes \((\omega_A^\bullet)_{f_{i_0} \ldots f_{i_p}}\), \(p > 0\) has finite injective dimension over \(A\), see Lemma 08XQ. Hence \(\omega_A^\bullet\) has finite injective dimension as a complex of \(A\)-modules (as it can be represented by a complex endowed with a finite filtration whose graded parts have finite injective dimension). Since \(H^n(\omega_A^\bullet)_{f_i}\) is a finite \(A_{f_i}\) module for each \(i\) we see that \(H^i(\omega_A^\bullet)\) is a finite \(A\)-module, see Algebra, Lemma 00EO. Finally, the (derived) base change of the map \(A \to R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)\) to \(A_{f_i}\) is the map \(A_{f_i} \to R\Hom_A((\omega_A^\bullet)_{f_i}, (\omega_A^\bullet)_{f_i})\) by More on Algebra, Lemma 0A6A. Hence we deduce that \(A \to R\Hom_A(\omega_A^\bullet, \omega_A^\bullet)\) is an isomorphism and the proof is complete.
Lemma
Let \(A \to B\) be a finite ring map of Noetherian rings. Let \(\omega_A^\bullet\) be a dualizing complex. Then \(R\Hom(B, \omega_A^\bullet)\) is a dualizing complex for \(B\).
Proof
Let \(\omega_A^\bullet \to I^\bullet\) be a quasi-isomorphism with \(I^\bullet\) a bounded complex of injectives. Then \(\Hom_A(B, I^\bullet)\) is a bounded complex of injective \(B\)-modules (Lemma 08XR) representing \(R\Hom(B, \omega_A^\bullet)\). Thus \(R\Hom(B, \omega_A^\bullet)\) has finite injective dimension. By Lemma 0A72 it is an object of \(D_{\textit{Coh}}(B)\). Finally, we compute \[\Hom_{D(B)}(R\Hom(B, \omega_A^\bullet), R\Hom(B, \omega_A^\bullet)) = \Hom_{D(A)}(R\Hom(B, \omega_A^\bullet), \omega_A^\bullet) = B\] and for \(n \not = 0\) we compute \[\Hom_{D(B)}(R\Hom(B, \omega_A^\bullet), R\Hom(B, \omega_A^\bullet)[n]) = \Hom_{D(A)}(R\Hom(B, \omega_A^\bullet), \omega_A^\bullet[n]) = 0\] which proves the last property of a dualizing complex. In the displayed equations, the first equality holds by Lemma 0A70 and the second equality holds by Lemma 0A7C.
Lemma
Let \(A \to B\) be a surjective homomorphism of Noetherian rings. Let \(\omega_A^\bullet\) be a dualizing complex. Then \(R\Hom(B, \omega_A^\bullet)\) is a dualizing complex for \(B\).
Proof
Special case of Lemma 0AX0.
Lemma
Let \(A\) be a Noetherian ring. If \(\omega_A^\bullet\) is a dualizing complex, then \(\omega_A^\bullet \otimes_A A[x]\) is a dualizing complex for \(A[x]\).
Proof
Set \(B = A[x]\) and \(\omega_B^\bullet = \omega_A^\bullet \otimes_A B\). It follows from Lemma 0A6J and More on Algebra, Lemma 0A5V that \(\omega_B^\bullet\) has finite injective dimension. Since \(H^i(\omega_B^\bullet) = H^i(\omega_A^\bullet) \otimes_A B\) by flatness of \(A \to B\) we see that \(\omega_A^\bullet \otimes_A B\) has finite cohomology modules. Finally, the map \[B \longrightarrow R\Hom_B(\omega_B^\bullet, \omega_B^\bullet)\] is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 0A6A.
Proposition
Let \(A\) be a Noetherian ring which has a dualizing complex. Then any \(A\)-algebra essentially of finite type over \(A\) has a dualizing complex.
Proof
This follows from a combination of Lemmas 0A7G, 0A7I, and 0A7J.
Lemma
Let \(A\) be a Noetherian ring. Let \(\omega_A^\bullet\) be a dualizing complex. Let \(\mathfrak m \subset A\) be a maximal ideal and set \(\kappa = A/\mathfrak m\). Then \(R\Hom_A(\kappa, \omega_A^\bullet) \cong \kappa[n]\) for some \(n \in \mathbf{Z}\).
Proof
This is true because \(R\Hom_A(\kappa, \omega_A^\bullet)\) is a dualizing complex over \(\kappa\) (Lemma 0A7I), because dualizing complexes over \(\kappa\) are unique up to shifts (Lemma 0A7F), and because \(\kappa\) is a dualizing complex over \(\kappa\).
Dualizing complexes over local rings
In this section \((A, \mathfrak m, \kappa)\) will be a Noetherian local ring endowed with a dualizing complex \(\omega_A^\bullet\) such that the integer \(n\) of Lemma 0A7L is zero. More precisely, we assume that \(R\Hom_A(\kappa, \omega_A^\bullet) = \kappa[0]\). In this case we will say that the dualizing complex is normalized. Observe that a normalized dualizing complex is unique up to isomorphism and that any other dualizing complex for \(A\) is isomorphic to a shift of a normalized one (Lemma 0A7F).
Lemma
Let \((A, \mathfrak m, \kappa) \to (B, \mathfrak m', \kappa')\) be a finite local map of Noetherian local rings. Let \(\omega_A^\bullet\) be a normalized dualizing complex. Then \(\omega_B^\bullet = R\Hom(B, \omega_A^\bullet)\) is a normalized dualizing complex for \(B\).
Proof
By Lemma 0AX0 the complex \(\omega_B^\bullet\) is dualizing for \(B\). We have \[R\Hom_B(\kappa', \omega_B^\bullet) = R\Hom_B(\kappa', R\Hom(B, \omega_A^\bullet)) = R\Hom_A(\kappa', \omega_A^\bullet)\] by Lemma 0A70. Since \(\kappa'\) is isomorphic to a finite direct sum of copies of \(\kappa\) as an \(A\)-module and since \(\omega_A^\bullet\) is normalized, we see that this complex only has cohomology placed in degree \(0\). Thus \(\omega_B^\bullet\) is a normalized dualizing complex as well.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Let \(A \to B\) be surjective. Then \(\omega_B^\bullet = R\Hom_A(B, \omega_A^\bullet)\) is a normalized dualizing complex for \(B\).
Proof
Special case of Lemma 0AX1.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(F\) be an \(A\)-linear self-equivalence of the category of finite length \(A\)-modules. Then \(F\) is isomorphic to the identity functor.
Proof
Since \(\kappa\) is the unique simple object of the category we have \(F(\kappa) \cong \kappa\). Since our category is abelian, we find that \(F\) is exact. Hence \(F(E)\) has the same length as \(E\) for all finite length modules \(E\). Since \(\Hom(E, \kappa) = \Hom(F(E), F(\kappa)) \cong \Hom(F(E), \kappa)\) we conclude from Nakayama’s lemma that \(E\) and \(F(E)\) have the same number of generators. Hence \(F(A/\mathfrak m^n)\) is a cyclic \(A\)-module. Pick a generator \(e \in F(A/\mathfrak m^n)\). Since \(F\) is \(A\)-linear we conclude that \(\mathfrak m^n e = 0\). The map \(A/\mathfrak m^n \to F(A/\mathfrak m^n)\) has to be an isomorphism as the lengths are equal. Pick an element \[e \in \lim F(A/\mathfrak m^n)\] which maps to a generator for all \(n\) (small argument omitted). Then we obtain a system of isomorphisms \(A/\mathfrak m^n \to F(A/\mathfrak m^n)\) compatible with all \(A\)-module maps \(A/\mathfrak m^n \to A/\mathfrak m^{n'}\) (by \(A\)-linearity of \(F\) again). Since any finite length module is a cokernel of a map between direct sums of cyclic modules, we obtain the isomorphism of the lemma.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Let \(E\) be an injective hull of \(\kappa\). Then there exists a functorial isomorphism \[R\Hom_A(N, \omega_A^\bullet) = \Hom_A(N, E)[0]\] for \(N\) running through the finite length \(A\)-modules.
Proof
By induction on the length of \(N\) we see that \(R\Hom_A(N, \omega_A^\bullet)\) is a module of finite length sitting in degree \(0\). Thus \(R\Hom_A(-, \omega_A^\bullet)\) induces an anti-equivalence on the category of finite length modules. Since the same is true for \(\Hom_A(-, E)\) by Proposition 08Z9 we see that \[N \longmapsto \Hom_A(R\Hom_A(N, \omega_A^\bullet), E)\] is an equivalence as in Lemma 0A7P. Hence it is isomorphic to the identity functor. Since \(\Hom_A(-, E)\) applied twice is the identity (Proposition 08Z9) we obtain the statement of the lemma.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Let \(M\) be a finite \(A\)-module and let \(d = \dim(\text{Supp}(M))\). Then
if \(\Ext^i_A(M, \omega_A^\bullet)\) is nonzero, then \(i \in \{-d, \ldots, 0\}\),
the dimension of the support of \(\Ext^i_A(M, \omega_A^\bullet)\) is at most \(-i\),
\(\text{depth}(M)\) is the smallest integer \(\delta \geq 0\) such that \(\Ext^{-\delta}_A(M, \omega_A^\bullet) \not = 0\).
Proof
We prove this by induction on \(d\). If \(d = 0\), this follows from Lemma 0A7Q and Matlis duality (Proposition 08Z9) which guarantees that \(\Hom_A(M, E)\) is nonzero if \(M\) is nonzero.
Assume the result holds for modules with support of dimension \(< d\) and that \(M\) has depth \(> 0\). Choose an \(f \in \mathfrak m\) which is a nonzerodivisor on \(M\) and consider the short exact sequence \[0 \to M \to M \to M/fM \to 0\] Since \(\dim(\text{Supp}(M/fM)) = d - 1\) (Algebra, Lemma 0B52) we may apply the induction hypothesis. Writing \(E^i = \Ext^i_A(M, \omega_A^\bullet)\) and \(F^i = \Ext^i_A(M/fM, \omega_A^\bullet)\) we obtain a long exact sequence \[\ldots \to F^i \to E^i \xrightarrow{f} E^i \to F^{i + 1} \to \ldots\] By induction \(E^i/fE^i = 0\) for \(i + 1 \not \in \{-\dim(\text{Supp}(M/fM)), \ldots, -\text{depth}(M/fM)\}\). By Nakayama’s lemma (Algebra, Lemma 00DV) and Algebra, Lemma 090R we conclude \(E^i = 0\) for \(i \not \in \{-\dim(\text{Supp}(M)), \ldots, -\text{depth}(M)\}\). Moreover, in the boundary case \(i = - \text{depth}(M)\) we deduce that \(E^i\) is nonzero as \(F^{i + 1}\) is nonzero by induction. Since \(E^i/fE^i \subset F^{i + 1}\) we get \[\dim(\text{Supp}(F^{i + 1})) \geq \dim(\text{Supp}(E^i/fE^i)) \geq \dim(\text{Supp}(E^i)) - 1\] (see lemma used above) we also obtain the dimension estimate (2).
If \(M\) has depth \(0\) and \(d > 0\) we let \(N = M[\mathfrak m^\infty]\) and set \(M' = M/N\) (compare with Lemma 0AW0). Then \(M'\) has depth \(> 0\) and \(\dim(\text{Supp}(M')) = d\). Thus we know the result for \(M'\) and since \(R\Hom_A(N, \omega_A^\bullet) = \Hom_A(N, E)\) (Lemma 0A7Q) the long exact cohomology sequence of \(\Ext\)’s implies the result for \(M\).
Remark
Let \((A, \mathfrak m)\) and \(\omega_A^\bullet\) be as in Lemma 0A7U. By More on Algebra, Lemma 0A5T we see that \(\omega_A^\bullet\) has injective-amplitude in \([-d, 0]\) because part (3) of that lemma applies. In particular, for any \(A\)-module \(M\) (not necessarily finite) we have \(\Ext^i_A(M, \omega_A^\bullet) = 0\) for \(i \not \in \{-d, \ldots, 0\}\).
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Let \(M\) be a finite \(A\)-module. The following are equivalent
\(M\) is Cohen-Macaulay,
\(\Ext^i_A(M, \omega_A^\bullet)\) is nonzero for at most one \(i\),
\(\Ext^{-i}_A(M, \omega_A^\bullet)\) is zero for \(i \not = \dim(\text{Supp}(M))\).
Denote \(CM_d\) the category of finite Cohen-Macaulay \(A\)-modules of depth \(d\). Then \(M \mapsto \Ext^{-d}_A(M, \omega_A^\bullet)\) defines an anti-auto-equivalence of \(CM_d\).
Proof
We will use the results of Lemma 0A7U without further mention. Fix a finite module \(M\). If \(M\) is Cohen-Macaulay, then only \(\Ext^{-d}_A(M, \omega_A^\bullet)\) can be nonzero, hence (1) \(\Rightarrow\) (3). The implication (3) \(\Rightarrow\) (2) is immediate. Assume (2) and let \(N = \Ext^{-\delta}_A(M, \omega_A^\bullet)\) be the nonzero \(\Ext\) where \(\delta = \text{depth}(M)\). Then, since \[M[0] = R\Hom_A(R\Hom_A(M, \omega_A^\bullet), \omega_A^\bullet) = R\Hom_A(N[\delta], \omega_A^\bullet)\] (Lemma 0A7C) we conclude that \(M = \Ext_A^{-\delta}(N, \omega_A^\bullet)\). Thus \(\delta \geq \dim(\text{Supp}(M))\). However, since we also know that \(\delta \leq \dim(\text{Supp}(M))\) (Algebra, Lemma 00LK) we conclude that \(M\) is Cohen-Macaulay.
To prove the final statement, it suffices to show that \(N = \Ext^{-d}_A(M, \omega_A^\bullet)\) is in \(CM_d\) for \(M\) in \(CM_d\). Above we have seen that \(M[0] = R\Hom_A(N[d], \omega_A^\bullet)\) and this proves the desired result by the equivalence of (1) and (3).
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). If \(\dim(A) = 0\), then \(\omega_A^\bullet \cong E[0]\) where \(E\) is an injective hull of the residue field.
Proof
Immediate from Lemma 0A7Q.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex. Let \(I \subset \mathfrak m\) be an ideal of finite length. Set \(B = A/I\). Then there is a distinguished triangle \[\omega_B^\bullet \to \omega_A^\bullet \to \Hom_A(I, E)[0] \to \omega_B^\bullet[1]\] in \(D(A)\) where \(E\) is an injective hull of \(\kappa\) and \(\omega_B^\bullet\) is a normalized dualizing complex for \(B\).
Proof
Use the short exact sequence \(0 \to I \to A \to B \to 0\) and Lemmas 0A7Q and 0A7N.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Let \(f \in \mathfrak m\) be a nonzerodivisor. Set \(B = A/(f)\). Then there is a distinguished triangle \[\omega_B^\bullet \to \omega_A^\bullet \to \omega_A^\bullet \to \omega_B^\bullet[1]\] in \(D(A)\) where \(\omega_B^\bullet\) is a normalized dualizing complex for \(B\).
Proof
Use the short exact sequence \(0 \to A \to A \to B \to 0\) and Lemma 0A7N.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Let \(\mathfrak p\) be a minimal prime of \(A\) with \(\dim(A/\mathfrak p) = e\). Then \(H^i(\omega_A^\bullet)_\mathfrak p\) is nonzero if and only if \(i = -e\).
Proof
Since \(A_\mathfrak p\) has dimension zero, there exists an integer \(n > 0\) such that \(\mathfrak p^nA_\mathfrak p\) is zero. Set \(B = A/\mathfrak p^n\) and \(\omega_B^\bullet = R\Hom_A(B, \omega_A^\bullet)\). Since \(B_\mathfrak p = A_\mathfrak p\) we see that \[(\omega_B^\bullet)_\mathfrak p = R\Hom_A(B, \omega_A^\bullet) \otimes_A^\mathbf{L} A_\mathfrak p = R\Hom_{A_\mathfrak p}(B_\mathfrak p, (\omega_A^\bullet)_\mathfrak p) = (\omega_A^\bullet)_\mathfrak p\] The second equality holds by More on Algebra, Lemma 0A6A. By Lemma 0A7N we may replace \(A\) by \(B\). After doing so, we see that \(\dim(A) = e\). Then we see that \(H^i(\omega_A^\bullet)_\mathfrak p\) can only be nonzero if \(i = -e\) by Lemma 0A7U parts (1) and (2). On the other hand, since \((\omega_A^\bullet)_\mathfrak p\) is a dualizing complex for the nonzero ring \(A_\mathfrak p\) (Lemma 0A7G) we see that the remaining module has to be nonzero.
Dualizing complexes and dimension functions
Our results in the local setting have the following consequence: a Noetherian ring which has a dualizing complex is a universally catenary ring of finite dimension.
Lemma
Let \(A\) be a Noetherian ring. Let \(\mathfrak p\) be a minimal prime of \(A\). Then \(H^i(\omega_A^\bullet)_\mathfrak p\) is nonzero for exactly one \(i\).
Proof
The complex \(\omega_A^\bullet \otimes_A A_\mathfrak p\) is a dualizing complex for \(A_\mathfrak p\) (Lemma 0A7G). The dimension of \(A_\mathfrak p\) is zero as \(\mathfrak p\) is minimal. Hence the result follows from Lemma 0A7R.
Let \(A\) be a Noetherian ring and let \(\omega_A^\bullet\) be a dualizing complex. Lemma 0A7L allows us to define a function \[\delta = \delta_{\omega_A^\bullet} : \Spec(A) \longrightarrow \mathbf{Z}\] by mapping \(\mathfrak p\) to the integer of Lemma 0A7L for the dualizing complex \((\omega_A^\bullet)_\mathfrak p\) over \(A_\mathfrak p\) (Lemma 0A7G) and the residue field \(\kappa(\mathfrak p)\). To be precise, we define \(\delta(\mathfrak p)\) to be the unique integer such that \[(\omega_A^\bullet)_\mathfrak p[-\delta(\mathfrak p)]\] is a normalized dualizing complex over the Noetherian local ring \(A_\mathfrak p\).
Lemma
Let \(A\) be a Noetherian ring and let \(\omega_A^\bullet\) be a dualizing complex. Let \(A \to B\) be a surjective ring map and let \(\omega_B^\bullet = R\Hom(B, \omega_A^\bullet)\) be the dualizing complex for \(B\) of Lemma 0A7I. Then we have \[\delta_{\omega_B^\bullet} = \delta_{\omega_A^\bullet}|_{\Spec(B)}\]
Proof
This follows from the definition of the functions and Lemma 0A7N.
Lemma
Let \(A\) be a Noetherian ring and let \(\omega_A^\bullet\) be a dualizing complex. The function \(\delta = \delta_{\omega_A^\bullet}\) defined above is a dimension function (Topology, Definition 02I9).
Proof
Let \(\mathfrak p \subset \mathfrak q\) be an immediate specialization. We have to show that \(\delta(\mathfrak p) = \delta(\mathfrak q) + 1\). We may replace \(A\) by \(A/\mathfrak p\), the complex \(\omega_A^\bullet\) by \(\omega_{A/\mathfrak p}^\bullet = R\Hom(A/\mathfrak p, \omega_A^\bullet)\), the prime \(\mathfrak p\) by \((0)\), and the prime \(\mathfrak q\) by \(\mathfrak q/\mathfrak p\), see Lemma 0A7Y. Thus we may assume that \(A\) is a domain, \(\mathfrak p = (0)\), and \(\mathfrak q\) is a prime ideal of height \(1\).
Then \(H^i(\omega_A^\bullet)_{(0)}\) is nonzero for exactly one \(i\), say \(i_0\), by Lemma 0A7X. In fact \(i_0 = -\delta((0))\) because \((\omega_A^\bullet)_{(0)}[-\delta((0))]\) is a normalized dualizing complex over the field \(A_{(0)}\).
On the other hand \((\omega_A^\bullet)_\mathfrak q[-\delta(\mathfrak q)]\) is a normalized dualizing complex for \(A_\mathfrak q\). By Lemma 0A7V we see that \[H^e((\omega_A^\bullet)_\mathfrak q[-\delta(\mathfrak q)])_{(0)} = H^{e - \delta(\mathfrak q)}(\omega_A^\bullet)_{(0)}\] is nonzero only for \(e = -\dim(A_\mathfrak q) = -1\). We conclude \[-\delta((0)) = -1 - \delta(\mathfrak q)\] as desired.
Lemma
Let \(A\) be a Noetherian ring which has a dualizing complex. Then \(A\) is universally catenary of finite dimension.
Proof
Because \(\Spec(A)\) has a dimension function by Lemma 0A7Z it is catenary, see Topology, Lemma 02IA. Hence \(A\) is catenary, see Algebra, Lemma 02IH. It follows from Proposition 0A7K that \(A\) is universally catenary.
Because any dualizing complex \(\omega_A^\bullet\) is in \(D^b_{\textit{Coh}}(A)\) the values of the function \(\delta_{\omega_A^\bullet}\) in minimal primes are bounded by Lemma 0A7X. On the other hand, for a maximal ideal \(\mathfrak m\) with residue field \(\kappa\) the integer \(i = -\delta(\mathfrak m)\) is the unique integer such that \(\Ext_A^i(\kappa, \omega_A^\bullet)\) is nonzero (Lemma 0A7L). Since \(\omega_A^\bullet\) has finite injective dimension these values are bounded too. Since the dimension of \(A\) is the maximal value of \(\delta(\mathfrak p) - \delta(\mathfrak m)\) where \(\mathfrak p \subset \mathfrak m\) are a pair consisting of a minimal prime and a maximal prime we find that the dimension of \(\Spec(A)\) is bounded.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Let \(d = \dim(A)\) and \(\omega_A = H^{-d}(\omega_A^\bullet)\). Then
the support of \(\omega_A\) is the union of the irreducible components of \(\Spec(A)\) of dimension \(d\),
\(\omega_A\) satisfies \((S_2)\), see Algebra, Definition 031P.
Proof
We will use Lemma 0A7U without further mention. By Lemma 0A7V the support of \(\omega_A\) contains the irreducible components of dimension \(d\). Let \(\mathfrak p \subset A\) be a prime. By Lemma 0A7Z the complex \((\omega_A^\bullet)_{\mathfrak p}[-\dim(A/\mathfrak p)]\) is a normalized dualizing complex for \(A_\mathfrak p\). Hence if \(\dim(A/\mathfrak p) + \dim(A_\mathfrak p) < d\), then \((\omega_A)_\mathfrak p = 0\). This proves the support of \(\omega_A\) is the union of the irreducible components of dimension \(d\), because the complement of this union is exactly the primes \(\mathfrak p\) of \(A\) for which \(\dim(A/\mathfrak p) + \dim(A_\mathfrak p) < d\) as \(A\) is catenary (Lemma 0A80). On the other hand, if \(\dim(A/\mathfrak p) + \dim(A_\mathfrak p) = d\), then \[(\omega_A)_\mathfrak p = H^{-\dim(A_\mathfrak p)}\left( (\omega_A^\bullet)_{\mathfrak p}[-\dim(A/\mathfrak p)] \right)\] Hence in order to prove \(\omega_A\) has \((S_2)\) it suffices to show that the depth of \(\omega_A\) is at least \(\min(\dim(A), 2)\). We prove this by induction on \(\dim(A)\). The case \(\dim(A) = 0\) is trivial.
Assume \(\text{depth}(A) > 0\). Choose a nonzerodivisor \(f \in \mathfrak m\) and set \(B = A/fA\). Then \(\dim(B) = \dim(A) - 1\) and we may apply the induction hypothesis to \(B\). By Lemma 0A7T we see that multiplication by \(f\) is injective on \(\omega_A\) and we get \(\omega_A/f\omega_A \subset \omega_B\). This proves the depth of \(\omega_A\) is at least \(1\). If \(\dim(A) > 1\), then \(\dim(B) > 0\) and \(\omega_B\) has depth \(> 0\). Hence \(\omega_A\) has depth \(> 1\) and we conclude in this case.
Assume \(\dim(A) > 0\) and \(\text{depth}(A) = 0\). Let \(I = A[\mathfrak m^\infty]\) and set \(B = A/I\). Then \(B\) has depth \(\geq 1\) and \(\omega_A = \omega_B\) by Lemma 0A7S. Since we proved the result for \(\omega_B\) above the proof is done.
The local duality theorem
The main result in this section is due to Grothendieck.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(\omega_A^\bullet\) be a normalized dualizing complex. Let \(Z = V(\mathfrak m) \subset \Spec(A)\). Then \(E = R^0\Gamma_Z(\omega_A^\bullet)\) is an injective hull of \(\kappa\) and \(R\Gamma_Z(\omega_A^\bullet) = E[0]\).
Proof
By Lemma 0955 we have \(R\Gamma_{\mathfrak m} = R\Gamma_Z\). Thus \[R\Gamma_Z(\omega_A^\bullet) = R\Gamma_{\mathfrak m}(\omega_A^\bullet) = \text{hocolim}\ R\Hom_A(A/\mathfrak m^n, \omega_A^\bullet)\] by Lemma 0954. Let \(E'\) be an injective hull of the residue field. By Lemma 0A7Q we can find isomorphisms \[R\Hom_A(A/\mathfrak m^n, \omega_A^\bullet) \cong \Hom_A(A/\mathfrak m^n, E')[0]\] compatible with transition maps. Since \(E' = \bigcup E'[\mathfrak m^n] = \colim \Hom_A(A/\mathfrak m^n, E')\) by Lemma 08Z4 we conclude that \(E \cong E'\) and that all other cohomology groups of the complex \(R\Gamma_Z(\omega_A^\bullet)\) are zero.
Remark
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with a normalized dualizing complex \(\omega_A^\bullet\). By Lemma 0A82 above we see that \(R\Gamma_Z(\omega_A^\bullet)\) is an injective hull of the residue field placed in degree \(0\). In fact, this gives a “construction” or “realization” of the injective hull which is slightly more canonical than just picking any old injective hull. Namely, a normalized dualizing complex is unique up to isomorphism, with group of automorphisms the group of units of \(A\), whereas an injective hull of \(\kappa\) is unique up to isomorphism, with group of automorphisms the group of units of the completion \(A^\wedge\) of \(A\) with respect to \(\mathfrak m\).
Here is the main result of this section.
Theorem
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(\omega_A^\bullet\) be a normalized dualizing complex. Let \(E\) be an injective hull of the residue field. Let \(Z = V(\mathfrak m) \subset \Spec(A)\). Denote \({}^\wedge\) derived completion with respect to \(\mathfrak m\). Then \[R\Hom_A(K, \omega_A^\bullet)^\wedge \cong R\Hom_A(R\Gamma_Z(K), E[0])\] for \(K\) in \(D(A)\).
Proof
Observe that \(E[0] \cong R\Gamma_Z(\omega_A^\bullet)\) by Lemma 0A82. By More on Algebra, Lemma 0A6E completion on the left hand side goes inside. Thus we have to prove \[R\Hom_A(K^\wedge, (\omega_A^\bullet)^\wedge) = R\Hom_A(R\Gamma_Z(K), R\Gamma_Z(\omega_A^\bullet))\] This follows from the equivalence between \(D_{comp}(A, \mathfrak m)\) and \(D_{\mathfrak m^\infty\text{-torsion}}(A)\) given in Proposition 0A6X. More precisely, it is a special case of Lemma 0A6Y.
Here is a special case of the theorem above.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(\omega_A^\bullet\) be a normalized dualizing complex. Let \(E\) be an injective hull of the residue field. Let \(K \in D_{\textit{Coh}}(A)\). Then \[\Ext^{-i}_A(K, \omega_A^\bullet)^\wedge = \Hom_A(H^i_{\mathfrak m}(K), E)\] where \({}^\wedge\) denotes \(\mathfrak m\)-adic completion.
Proof
By Lemma 0A7C we see that \(R\Hom_A(K, \omega_A^\bullet)\) is an object of \(D_{\textit{Coh}}(A)\). It follows that the cohomology modules of the derived completion of \(R\Hom_A(K, \omega_A^\bullet)\) are equal to the usual completions \(\Ext^i_A(K, \omega_A^\bullet)^\wedge\) by More on Algebra, Lemma 0A06. On the other hand, we have \(R\Gamma_{\mathfrak m} = R\Gamma_Z\) for \(Z = V(\mathfrak m)\) by Lemma 0955. Moreover, the functor \(\Hom_A(-, E)\) is exact hence factors through cohomology. Hence the lemma is consequence of Theorem 0A84.
Dualizing modules
If \((A, \mathfrak m, \kappa)\) is a Noetherian local ring and \(\omega_A^\bullet\) is a normalized dualizing complex, then we say the module \(\omega_A = H^{-\dim(A)}(\omega_A^\bullet)\), described in Lemma 0AWE, is a dualizing module for \(A\). This module is a canonical module of \(A\). It seems generally agreed upon to define a canonical module for a Noetherian local ring \((A, \mathfrak m, \kappa)\) to be a finite \(A\)-module \(K\) such that \[\Hom_A(K, E) \cong H^{\dim(A)}_\mathfrak m(A)\] where \(E\) is an injective hull of the residue field. A dualizing module is canonical because \[\Hom_A(H^{\dim(A)}_\mathfrak m(A), E) = (\omega_A)^\wedge\] by Lemma 0AAK and hence applying \(\Hom_A(-, E)\) we get \[\begin{align*} \Hom_A(\omega_A, E) & = \Hom_A((\omega_A)^\wedge, E) \\ & = \Hom_A(\Hom_A(H^{\dim(A)}_\mathfrak m(A), E), E) \\ & = H^{\dim(A)}_\mathfrak m(A) \end{align*}\] the first equality because \(E\) is \(\mathfrak m\)-power torsion, the second by the above, and the third by Matlis duality (Proposition 08Z9). The utility of the definition of a canonical module given above lies in the fact that it makes sense even if \(A\) does not have a dualizing complex.
Cohen-Macaulay rings
Cohen-Macaulay modules and rings were studied in Algebra, Sections 00N2 and 00N7.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\). Then \(\text{depth}(A)\) is equal to the smallest integer \(\delta \geq 0\) such that \(H^{-\delta}(\omega_A^\bullet) \not = 0\).
Proof
This follows immediately from Lemma 0A7U. Here are two other ways to see that it is true.
First alternative. By Nakayama’s lemma we see that \(\delta\) is the smallest integer such that \(\Hom_A(H^{-\delta}(\omega_A^\bullet), \kappa) \not = 0\). In other words, it is the smallest integer such that \(\Ext_A^{-\delta}(\omega_A^\bullet, \kappa)\) is nonzero. Using Lemma 0A7C and the fact that \(\omega_A^\bullet\) is normalized this is equal to the smallest integer such that \(\Ext_A^\delta(\kappa, A)\) is nonzero. This is equal to the depth of \(A\) by Algebra, Lemma 00LW.
Second alternative. By the local duality theorem (in the form of Lemma 0AAK) \(\delta\) is the smallest integer such that \(H^\delta_\mathfrak m(A)\) is nonzero. This is equal to the depth of \(A\) by Lemma 0AVZ.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring with normalized dualizing complex \(\omega_A^\bullet\) and dualizing module \(\omega_A = H^{-\dim(A)}(\omega_A^\bullet)\). The following are equivalent
\(A\) is Cohen-Macaulay,
\(\omega_A^\bullet\) is concentrated in a single degree, and
\(\omega_A^\bullet = \omega_A[\dim(A)]\).
In this case \(\omega_A\) is a maximal Cohen-Macaulay module.
Proof
Follows immediately from Lemma 0B5A.
Lemma
Let \(A\) be a Noetherian ring. If there exists a finite \(A\)-module \(\omega_A\) such that \(\omega_A[0]\) is a dualizing complex, then \(A\) is Cohen-Macaulay.
Proof
We may replace \(A\) by the localization at a prime (Lemma 0A7G and Algebra, Definition 00NC). In this case the result follows immediately from Lemma 0AWS.
Lemma
Let \(A\) be a Noetherian ring with dualizing complex \(\omega_A^\bullet\). Let \(M\) be a finite \(A\)-module. Then \[U = \{\mathfrak p \in \Spec(A) \mid M_\mathfrak p\text{ is Cohen-Macaulay}\}\] is an open subset of \(\Spec(A)\) whose intersection with \(\text{Supp}(M)\) is dense.
Proof
If \(\mathfrak p\) is a generic point of \(\text{Supp}(M)\), then \(\text{depth}(M_\mathfrak p) = \dim(M_\mathfrak p) = 0\) and hence \(\mathfrak p \in U\). This proves denseness. If \(\mathfrak p \in U\), then we see that \[R\Hom_A(M, \omega_A^\bullet)_\mathfrak p = R\Hom_{A_\mathfrak p}(M_\mathfrak p, (\omega_A^\bullet)_\mathfrak p)\] has a unique nonzero cohomology module, say in degree \(i_0\), by Lemma 0B5A. Since \(R\Hom_A(M, \omega_A^\bullet)\) has only a finite number of nonzero cohomology modules \(H^i\) and since each of these is a finite \(A\)-module, we can find an \(f \in A\), \(f \not \in \mathfrak p\) such that \((H^i)_f = 0\) for \(i \not = i_0\). Then \(R\Hom_A(M, \omega_A^\bullet)_f\) has a unique nonzero cohomology module and reversing the arguments just given we find that \(D(f) \subset U\).
Lemma
Let \(A\) be a Noetherian ring. If \(A\) has a dualizing complex \(\omega_A^\bullet\), then \(\{\mathfrak p \in \Spec(A) \mid A_\mathfrak p\text{ is Cohen-Macaulay}\}\) is a dense open subset of \(\Spec(A)\).
Proof
Immediate consequence of Lemma 0EHS and the definitions.
Gorenstein rings
So far, the only explicit dualizing complex we’ve seen is \(\kappa\) on \(\kappa\) for a field \(\kappa\), see proof of Lemma 0A7L. By Proposition 0A7K this means that any finite type algebra over a field has a dualizing complex. However, it turns out that there are Noetherian (local) rings which do not have a dualizing complex. Namely, we have seen that a ring which has a dualizing complex is universally catenary (Lemma 0A80) but there are examples of Noetherian local rings which are not catenary, see Examples, Section 02JE.
Nonetheless many rings in algebraic geometry have dualizing complexes simply because they are quotients of Gorenstein rings. This condition is in fact both necessary and sufficient. That is: a Noetherian ring has a dualizing complex if and only if it is a quotient of a finite dimensional Gorenstein ring. This is Sharp’s conjecture ([Sharp]) which can be found as [Kawasaki, Corollary 1.4] in the literature. Returning to our current topic, here is the definition of Gorenstein rings.
Definition
Gorenstein rings.
Let \(A\) be a Noetherian local ring. We say \(A\) is Gorenstein if \(A[0]\) is a dualizing complex for \(A\).
Let \(A\) be a Noetherian ring. We say \(A\) is Gorenstein if \(A_\mathfrak p\) is Gorenstein for every prime \(\mathfrak p\) of \(A\).
This definition makes sense, because if \(A[0]\) is a dualizing complex for \(A\), then \(S^{-1}A[0]\) is a dualizing complex for \(S^{-1}A\) by Lemma 0A7G. A finite dimensional Noetherian ring is Gorenstein if it has finite injective dimension as a module over itself, see Lemma 0HB2.
Lemma
A Gorenstein ring is Cohen-Macaulay.
Proof
Follows from Lemma 0AWS.
An example of a Gorenstein ring is a regular ring.
Lemma
A regular local ring is Gorenstein. A regular ring is Gorenstein.
Proof
Let \(A\) be a regular ring of finite dimension \(d\). Then \(A\) has finite global dimension \(d\), see Algebra, Lemma 00OE. Hence \(\Ext^{d + 1}_A(M, A) = 0\) for all \(A\)-modules \(M\), see Algebra, Lemma 065R. Thus \(A\) has finite injective dimension as an \(A\)-module by More on Algebra, Lemma 0A5T. It follows that \(A[0]\) is a dualizing complex, hence \(A\) is Gorenstein by the remark following the definition.
Lemma
Let \(A\) be a Noetherian ring.
If \(A\) has a dualizing complex \(\omega_A^\bullet\), then
\(A\) is Gorenstein \(\Leftrightarrow\) \(\omega_A^\bullet\) is an invertible object of \(D(A)\),
\(A_\mathfrak p\) is Gorenstein \(\Leftrightarrow\) \((\omega_A^\bullet)_\mathfrak p\) is an invertible object of \(D(A_\mathfrak p)\),
\(\{\mathfrak p \in \Spec(A) \mid A_\mathfrak p\text{ is Gorenstein}\}\) is an open subset.
If \(A\) is Gorenstein, then \(A\) has a dualizing complex if and only if \(A[0]\) is a dualizing complex.
Proof
For invertible objects of \(D(A)\), see More on Algebra, Lemma 0FNT and the discussion in Section 0A7A.
By Lemma 0A7G for every \(\mathfrak p\) the complex \((\omega_A^\bullet)_\mathfrak p\) is a dualizing complex over \(A_\mathfrak p\). By definition and uniqueness of dualizing complexes (Lemma 0A7F) we see that (1)(b) holds.
To see (1)(c) assume that \(A_\mathfrak p\) is Gorenstein. Let \(n_x\) be the unique integer such that \(H^{n_{x}}((\omega_A^\bullet)_\mathfrak p)\) is nonzero and isomorphic to \(A_\mathfrak p\). Since \(\omega_A^\bullet\) is in \(D^b_{\textit{Coh}}(A)\) there are finitely many nonzero finite \(A\)-modules \(H^i(\omega_A^\bullet)\). Thus there exists some \(f \in A\), \(f \not \in \mathfrak p\) such that only \(H^{n_x}((\omega_A^\bullet)_f)\) is nonzero and generated by \(1\) element over \(A_f\). Since dualizing complexes are faithful (by definition) we conclude that \(A_f \cong H^{n_x}((\omega_A^\bullet)_f)\). In this way we see that \(A_\mathfrak q\) is Gorenstein for every \(\mathfrak q \in D(f)\). This proves that the set in (1)(c) is open.
Proof of (1)(a). The implication \(\Leftarrow\) follows from (1)(b). The implication \(\Rightarrow\) follows from the discussion in the previous paragraph, where we showed that if \(A_\mathfrak p\) is Gorenstein, then for some \(f \in A\), \(f \not \in \mathfrak p\) the complex \((\omega_A^\bullet)_f\) has only one nonzero cohomology module which is invertible.
If \(A[0]\) is a dualizing complex then \(A\) is Gorenstein by part (1). Conversely, we see that part (1) shows that \(\omega_A^\bullet\) is locally isomorphic to a shift of \(A\). Since being a dualizing complex is local (Lemma 0A7H) the result is clear.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring. Then \(A\) is Gorenstein if and only if \(\Ext^i_A(\kappa, A)\) is zero for \(i \gg 0\).
Proof
Observe that \(A[0]\) is a dualizing complex for \(A\) if and only if \(A\) has finite injective dimension as an \(A\)-module (follows immediately from Definition 0A7B). Thus the lemma follows from More on Algebra, Lemma 0AVJ.
Lemma
Let \(A\) be a Noetherian ring. The following are equivalent
\(A\) has finite injective dimension as a module over itself,
\(A\) is Gorenstein and finite dimensional.
In this case \(A[0]\) is a dualizing complex.
Proof
If \(A\) has finite injective dimension as a module over \(A\), then \(A[0]\) is a dualizing complex over \(A\) (see Definition 0A7B) and we conclude \(A\) has finite dimension by Lemma 0A80 and is Gorenstein by Lemma 0DW9. Assume \(A\) is Gorenstein and \(\dim(A) \leq d < \infty\). Let \(I \subset A\) be any ideal. We claim \(\Ext^i_A(A/I, A) = 0\) for \(i > d\); this proves that \(A\) has finite injective dimension as a module over itself by More on Algebra, Lemma 0A5T. Forming the \(\Ext\) groups commutes with localization (see for example More on Algebra, Lemma 0A6A). Thus to prove vanishing we may assume \(A\) is local of dimension \(e < d\). As \(A\) is Gorenstein we see that \(\omega_A^\bullet = A[e]\) is a normalized dualizing complex. Hence the claimed vanishing by Lemma 0A7U for example.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local ring. Let \(f \in \mathfrak m\) be a nonzerodivisor. Set \(B = A/(f)\). Then \(A\) is Gorenstein if and only if \(B\) is Gorenstein.
Proof
If \(A\) is Gorenstein, then \(B\) is Gorenstein by Lemma 0A7T. Conversely, suppose that \(B\) is Gorenstein. Then \(\Ext^i_B(\kappa, B)\) is zero for \(i \gg 0\) (Lemma 0BJI). Recall that \(R\Hom(B, -) : D(A) \to D(B)\) is a right adjoint to restriction (Lemma 0A70). Hence \[R\Hom_A(\kappa, A) = R\Hom_B(\kappa, R\Hom(B, A)) = R\Hom_B(\kappa, B[1])\] The final equality by direct computation or by Lemma 0BZH. Thus we see that \(\Ext^i_A(\kappa, A)\) is zero for \(i \gg 0\) and \(A\) is Gorenstein (Lemma 0BJI).
Lemma
If \(A \to B\) is a local complete intersection homomorphism of rings and \(A\) is a Noetherian Gorenstein ring, then \(B\) is a Gorenstein ring.
Proof
By More on Algebra, Definition 07D0 we can write \(B = A[x_1, \ldots, x_n]/I\) where \(I\) is a Koszul-regular ideal. Observe that a polynomial ring over a Gorenstein ring \(A\) is Gorenstein: reduce to \(A\) local and then use Lemmas 0A7J and 0DW9. A Koszul-regular ideal is by definition locally generated by a Koszul-regular sequence, see More on Algebra, Section 07CU. Looking at local rings of \(A[x_1, \ldots, x_n]\) we see it suffices to show: if \(R\) is a Noetherian local Gorenstein ring and \(f_1, \ldots, f_c \in \mathfrak m_R\) is a Koszul regular sequence, then \(R/(f_1, \ldots, f_c)\) is Gorenstein. This follows from Lemma 0BJJ and the fact that a Koszul regular sequence in \(R\) is just a regular sequence (More on Algebra, Lemma 09CC).
Lemma
Let \(A \to B\) be a flat local homomorphism of Noetherian local rings. The following are equivalent
\(B\) is Gorenstein, and
\(A\) and \(B/\mathfrak m_A B\) are Gorenstein.
Proof
Below we will use without further mention that a local Gorenstein ring has finite injective dimension as well as Lemma 0BJI. By More on Algebra, Lemma 087R we have \[\Ext^i_A(\kappa_A, A) \otimes_A B = \Ext^i_B(B/\mathfrak m_A B, B)\] for all \(i\).
Assume (2). Using that \(R\Hom(B/\mathfrak m_A B, -) : D(B) \to D(B/\mathfrak m_A B)\) is a right adjoint to restriction (Lemma 0A70) we obtain \[R\Hom_B(\kappa_B, B) = R\Hom_{B/\mathfrak m_A B}(\kappa_B, R\Hom(B/\mathfrak m_A B, B))\] The cohomology modules of \(R\Hom(B/\mathfrak m_A B, B)\) are the modules \(\Ext^i_B(B/\mathfrak m_A B, B) = \Ext^i_A(\kappa_A, A) \otimes_A B\). Since \(A\) is Gorenstein, we conclude only a finite number of these are nonzero and each is isomorphic to a direct sum of copies of \(B/\mathfrak m_A B\). Hence since \(B/\mathfrak m_A B\) is Gorenstein we conclude that \(R\Hom_B(B/\mathfrak m_B, B)\) has only a finite number of nonzero cohomology modules. Hence \(B\) is Gorenstein.
Assume (1). Since \(B\) has finite injective dimension, \(\Ext^i_B(B/\mathfrak m_A B, B)\) is \(0\) for \(i \gg 0\). Since \(A \to B\) is faithfully flat we conclude that \(\Ext^i_A(\kappa_A, A)\) is \(0\) for \(i \gg 0\). We conclude that \(A\) is Gorenstein. This implies that \(\Ext^i_A(\kappa_A, A)\) is nonzero for exactly one \(i\), namely for \(i = \dim(A)\), and \(\Ext^{\dim(A)}_A(\kappa_A, A) \cong \kappa_A\) (see Lemmas 0AX1, 0AWS, and 0DW8). Thus we see that \(\Ext^i_B(B/\mathfrak m_A B, B)\) is zero except for one \(i\), namely \(i = \dim(A)\) and \(\Ext^{\dim(A)}_B(B/\mathfrak m_A B, B) \cong B/\mathfrak m_A B\). Thus \(B/\mathfrak m_A B\) is Gorenstein by Lemma 0AX1.
Lemma
Let \((A, \mathfrak m, \kappa)\) be a Noetherian local Gorenstein ring of dimension \(d\). Let \(E\) be the injective hull of \(\kappa\). Then \(\text{Tor}_i^A(E, \kappa)\) is zero for \(i \not = d\) and \(\text{Tor}_d^A(E, \kappa) = \kappa\).
Proof
Since \(A\) is Gorenstein \(\omega_A^\bullet = A[d]\) is a normalized dualizing complex for \(A\). Also \(E\) is the only nonzero cohomology module of \(R\Gamma_\mathfrak m(\omega_A^\bullet)\) sitting in degree \(0\), see Lemma 0A82. By Lemma 0ALY we have \[E \otimes_A^\mathbf{L} \kappa = R\Gamma_\mathfrak m(\omega_A^\bullet) \otimes_A^\mathbf{L} \kappa = R\Gamma_\mathfrak m(\omega_A^\bullet \otimes_A^\mathbf{L} \kappa) = R\Gamma_\mathfrak m(\kappa[d]) = \kappa[d]\] and the lemma follows.
The ubiquity of dualizing complexes
Many Noetherian rings have dualizing complexes.
Lemma
Let \(A \to B\) be a local homomorphism of Noetherian local rings. Let \(\omega_A^\bullet\) be a normalized dualizing complex. If \(A \to B\) is flat and \(\mathfrak m_A B = \mathfrak m_B\), then \(\omega_A^\bullet \otimes_A B\) is a normalized dualizing complex for \(B\).
Proof
It is clear that \(\omega_A^\bullet \otimes_A B\) is in \(D^b_{\textit{Coh}}(B)\). Let \(\kappa_A\) and \(\kappa_B\) be the residue fields of \(A\) and \(B\). By More on Algebra, Lemma 0A6A we see that \[R\Hom_B(\kappa_B, \omega_A^\bullet \otimes_A B) = R\Hom_A(\kappa_A, \omega_A^\bullet) \otimes_A B = \kappa_A[0] \otimes_A B = \kappa_B[0]\] Thus \(\omega_A^\bullet \otimes_A B\) has finite injective dimension by More on Algebra, Lemma 0AVJ. Finally, we can use the same arguments to see that \[R\Hom_B(\omega_A^\bullet \otimes_A B, \omega_A^\bullet \otimes_A B) = R\Hom_A(\omega_A^\bullet, \omega_A^\bullet) \otimes_A B = A \otimes_A B = B\] as desired.
Lemma
Let \(A \to B\) be a flat map of Noetherian rings. Let \(I \subset A\) be an ideal such that \(A/I = B/IB\) and such that \(IB\) is contained in the Jacobson radical of \(B\). Let \(\omega_A^\bullet\) be a dualizing complex. Then \(\omega_A^\bullet \otimes_A B\) is a dualizing complex for \(B\).
Proof
It is clear that \(\omega_A^\bullet \otimes_A B\) is in \(D^b_{\textit{Coh}}(B)\). By More on Algebra, Lemma 0A6A we see that \[R\Hom_B(K \otimes_A B, \omega_A^\bullet \otimes_A B) = R\Hom_A(K, \omega_A^\bullet) \otimes_A B\] for any \(K \in D^b_{\textit{Coh}}(A)\). For any ideal \(IB \subset J \subset B\) there is a unique ideal \(I \subset J' \subset A\) such that \(A/J' \otimes_A B = B/J\). Thus \(\omega_A^\bullet \otimes_A B\) has finite injective dimension by More on Algebra, Lemma 0DW2. Finally, we also have \[R\Hom_B(\omega_A^\bullet \otimes_A B, \omega_A^\bullet \otimes_A B) = R\Hom_A(\omega_A^\bullet, \omega_A^\bullet) \otimes_A B = A \otimes_A B = B\] as desired.
Lemma
Let \(A\) be a Noetherian ring and let \(I \subset A\) be an ideal. Let \(\omega_A^\bullet\) be a dualizing complex.
\(\omega_A^\bullet \otimes_A A^h\) is a dualizing complex on the henselization \((A^h, I^h)\) of the pair \((A, I)\),
\(\omega_A^\bullet \otimes_A A^\wedge\) is a dualizing complex on the \(I\)-adic completion \(A^\wedge\), and
if \(A\) is local, then \(\omega_A^\bullet \otimes_A A^h\), resp. \(\omega_A^\bullet \otimes_A A^{sh}\) is a dualzing complex on the henselization, resp. strict henselization of \(A\).
Proof
Immediate from Lemmas 0AWD and 0DWC. See More on Algebra, Sections 09XD, 07NU, and 07QL and Algebra, Sections 00M9 and 0BNH for information on completions and henselizations.
Lemma
The following types of rings have a dualizing complex:
fields,
Noetherian complete local rings,
\(\mathbf{Z}\),
Dedekind domains,
any ring which is obtained from one of the rings above by taking an algebra essentially of finite type, or by taking an ideal-adic completion, or by taking a henselization, or by taking a strict henselization.
Proof
Part (5) follows from Proposition 0A7K and Lemma 0DWD. By Lemma 0AWX a regular local ring has a dualizing complex. A complete Noetherian local ring is the quotient of a regular local ring by the Cohen structure theorem (Algebra, Theorem 032A). Let \(A\) be a Dedekind domain. Then every ideal \(I\) is a finite projective \(A\)-module (follows from Algebra, Lemma 00NX and the fact that the local rings of \(A\) are discrete valuation ring and hence PIDs). Thus every \(A\)-module has finite injective dimension at most \(1\) by More on Algebra, Lemma 0A5T. It follows easily that \(A[0]\) is a dualizing complex.
Formal fibres
This section is a continuation of More on Algebra, Section 0BIR. There we saw there is a (fairly) good theory of Noetherian rings \(A\) whose local rings have Cohen-Macaulay formal fibres. Namely, we proved (1) it suffices to check the formal fibres of localizations at maximal ideals are Cohen-Macaulay, (2) the property is inherited by rings of finite type over \(A\), (3) the fibres of \(A \to A^\wedge\) are Cohen-Macaulay for any completion \(A^\wedge\) of \(A\), and (4) the property is inherited by henselizations of \(A\). See More on Algebra, Lemma 0BIU, Proposition 0BIV, Lemma 0BK9, and Lemma 0BKA. Similarly, for Noetherian rings whose local rings have formal fibres which are geometrically reduced, geometrically normal, \((S_n)\), and geometrically \((R_n)\). In this section we will see that the same is true for Noetherian rings whose local rings have formal fibres which are Gorenstein or local complete intersections. This is relevant to this chapter because a Noetherian ring which has a dualizing complex is an example.
Lemma
Properties (A), (B), (C), (D), and (E) of More on Algebra, Section 0BIR hold for \(P(k \to R) =\)“\(R\) is a Gorenstein ring”.
Proof
Since we already know the result holds for Cohen-Macaulay instead of Gorenstein, we may in each step assume the ring we have is Cohen-Macaulay. This is not particularly helpful for the proof, but psychologically may be useful.
Part (A). Let \(K/k\) be a finitely generated field extension. Let \(R\) be a Gorenstein \(k\)-algebra. We can find a global complete intersection \(A = k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\) over \(k\) such that \(K\) is isomorphic to the fraction field of \(A\), see Algebra, Lemma 07BV. Then \(R \to R \otimes_k A\) is a relative global complete intersection. Hence \(R \otimes_k A\) is Gorenstein by Lemma 0DWA. Thus \(R \otimes_k K\) is too as a localization.
Proof of (B). This is clear because a ring is Gorenstein if and only if all of its local rings are Gorenstein.
Part (C). Let \(A \to B \to C\) be flat maps of Noetherian rings. Assume the fibres of \(A \to B\) are Gorenstein and \(B \to C\) is regular. We have to show the fibres of \(A \to C\) are Gorenstein. Clearly, we may assume \(A = k\) is a field. Then we may assume that \(B \to C\) is a regular local homomorphism of Noetherian local rings. Then \(B\) is Gorenstein and \(C/\mathfrak m_B C\) is regular, in particular Gorenstein (Lemma 0AWX). Then \(C\) is Gorenstein by Lemma 0BJL.
Part (D). This follows from Lemma 0BJL. Part (E) is immediate as the condition does not refer to the ground field.
Lemma
Let \(A\) be a Noetherian local ring. If \(A\) has a dualizing complex, then the formal fibres of \(A\) are Gorenstein.
Proof
Let \(\mathfrak p\) be a prime of \(A\). The formal fibre of \(A\) at \(\mathfrak p\) is isomorphic to the formal fibre of \(A/\mathfrak p\) at \((0)\). The quotient \(A/\mathfrak p\) has a dualizing complex (Lemma 0A7I). Thus it suffices to check the statement when \(A\) is a local domain and \(\mathfrak p = (0)\). Let \(\omega_A^\bullet\) be a dualizing complex for \(A\). Then \(\omega_A^\bullet \otimes_A A^\wedge\) is a dualizing complex for the completion \(A^\wedge\) (Lemma 0AWD). Then \(\omega_A^\bullet \otimes_A K\) is a dualizing complex for the fraction field \(K\) of \(A\) (Lemma 0A7G). Hence \(\omega_A^\bullet \otimes_A K\) is isomorphic to \(K[n]\) for some \(n \in \mathbf{Z}\). Similarly, we conclude a dualizing complex for the formal fibre \(A^\wedge \otimes_A K\) is \[\omega_A^\bullet \otimes_A A^\wedge \otimes_{A^\wedge} (A^\wedge \otimes_A K) = (\omega_A^\bullet \otimes_A K) \otimes_K (A^\wedge \otimes_A K) \cong (A^\wedge \otimes_A K)[n]\] as desired.
Here is the verification promised in Divided Power Algebra, Remark 09QC.
Lemma
Properties (A), (B), (C), (D), and (E) of More on Algebra, Section 0BIR hold for \(P(k \to R) =\)“\(R\) is a local complete intersection”. See Divided Power Algebra, Definition 09Q3.
Proof
Part (A). Let \(K/k\) be a finitely generated field extension. Let \(R\) be a \(k\)-algebra which is a local complete intersection. We can find a global complete intersection \(A = k[x_1, \ldots, x_n]/(f_1, \ldots, f_c)\) over \(k\) such that \(K\) is isomorphic to the fraction field of \(A\), see Algebra, Lemma 07BV. Then \(R \to R \otimes_k A\) is a relative global complete intersection. It follows that \(R \otimes_k A\) is a local complete intersection by Divided Power Algebra, Lemma 09Q7.
Proof of (B). This is clear because a ring is a local complete intersection if and only if all of its local rings are complete intersections.
Part (C). Let \(A \to B \to C\) be flat maps of Noetherian rings. Assume the fibres of \(A \to B\) are local complete intersections and \(B \to C\) is regular. We have to show the fibres of \(A \to C\) are local complete intersections. Clearly, we may assume \(A = k\) is a field. Then we may assume that \(B \to C\) is a regular local homomorphism of Noetherian local rings. Then \(B\) is a complete intersection and \(C/\mathfrak m_B C\) is regular, in particular a complete intersection (by definition). Then \(C\) is a complete intersection by Divided Power Algebra, Lemma 09Q7.
Part (D). This follows by the same arguments as in (C) from the other implication in Divided Power Algebra, Lemma 09Q7. Part (E) is immediate as the condition does not refer to the ground field.
Upper shriek algebraically
For a finite type homomorphism \(R \to A\) of Noetherian rings we will construct a functor \(\varphi^! : D(R) \to D(A)\) well defined up to nonunique isomorphism which as we will see in Duality for Schemes, Remark 0BV2 agrees up to isomorphism with the upper shriek functors one encounters in the duality theory for schemes. To motivate the construction we mention two additional properties:
\(\varphi^!\) sends a dualizing complex for \(R\) (if it exists) to a dualizing complex for \(A\), and
\(\omega_{A/R}^\bullet = \varphi^!(R)\) is a kind of relative dualizing complex: it lies in \(D^b_{\textit{Coh}}(A)\) and restricts to a dualizing complex on the fibres provided \(R \to A\) is flat.
These statements are Lemmas 0BZL and 0BZW.
Let \(\varphi : R \to A\) be a finite type homomorphism of Noetherian rings. We will define a functor \(\varphi^! : D(R) \to D(A)\) in the following way
If \(\varphi : R \to A\) is surjective we set \(\varphi^!(K) = R\Hom(A, K)\). Here we use the functor \(R\Hom(A, -) : D(R) \to D(A)\) of Section 0A6Z, and
in general we choose a surjection \(\psi : P \to A\) with \(P = R[x_1, \ldots, x_n]\) and we set \(\varphi^!(K) = \psi^!(K \otimes_R^\mathbf{L} P)[n]\). Here we use the functor \(- \otimes_R^\mathbf{L} P : D(R) \to D(P)\) of More on Algebra, Section 06Y5.
Note the shift \([n]\) by the number of variables in the polynomial ring. This construction is not canonical and the functor \(\varphi^!\) will only be well defined up to a (nonunique) isomorphism of functors1.
Lemma
Let \(\varphi : R \to A\) be a finite type homomorphism of Noetherian rings. The functor \(\varphi^!\) is well defined up to isomorphism.
Proof
Suppose that \(\psi_1 : P_1 = R[x_1, \ldots, x_n] \to A\) and \(\psi_2 : P_2 = R[y_1, \ldots, y_m] \to A\) are two surjections from polynomial rings onto \(A\). Then we get a commutative diagram \[\xymatrix{ R[x_1, \ldots, x_n, y_1, \ldots, y_m] \ar[d]^{x_i \mapsto g_i} \ar[rr]_-{y_j \mapsto f_j} & & R[x_1, \ldots, x_n] \ar[d] \\ R[y_1, \ldots, y_m] \ar[rr] & & A }\] where \(f_j\) and \(g_i\) are chosen such that \(\psi_1(f_j) = \psi_2(y_j)\) and \(\psi_2(g_i) = \psi_1(x_i)\). By symmetry it suffices to prove the functors defined using \(P \to A\) and \(P[y_1, \ldots, y_m] \to A\) are isomorphic. By induction we may assume \(m = 1\). This reduces us to the case discussed in the next paragraph.
Here \(\psi : P \to A\) is given and \(\chi : P[y] \to A\) induces \(\psi\) on \(P\). Write \(Q = P[y]\). Choose \(g \in P\) with \(\psi(g) = \chi(y)\). Denote \(\pi : Q \to P\) the \(P\)-algebra map with \(\pi(y) = g\). Then \(\chi = \psi \circ \pi\) and hence \(\chi^! = \psi^! \circ \pi^!\) as both are adjoint to the restriction functor \(D(A) \to D(Q)\) by the material in Section 0A6Z. Thus \[\chi^!\left(K \otimes_R^\mathbf{L} Q\right)[n + 1] = \psi^!\left(\pi^!\left(K \otimes_R^\mathbf{L} Q\right)[1]\right)[n]\] Hence it suffices to show that \(\pi^!(K \otimes_R^\mathbf{L} Q[1]) = K \otimes_R^\mathbf{L} P\) Thus it suffices to show that the functor \(\pi^!(-) : D(Q) \to D(P)\) is isomorphic to \(K \mapsto K \otimes_Q^\mathbf{L} P[-1]\). This follows from Lemma 0BZH.
Lemma
Let \(\varphi : R \to A\) be a finite type homomorphism of Noetherian rings.
\(\varphi^!\) maps \(D^+(R)\) into \(D^+(A)\) and \(D^+_{\textit{Coh}}(R)\) into \(D^+_{\textit{Coh}}(A)\).
if \(\varphi\) is perfect, then \(\varphi^!\) maps \(D^-(R)\) into \(D^-(A)\), \(D^-_{\textit{Coh}}(R)\) into \(D^-_{\textit{Coh}}(A)\), and \(D^b_{\textit{Coh}}(R)\) into \(D^b_{\textit{Coh}}(A)\).
Proof
Choose a factorization \(R \to P \to A\) as in the definition of \(\varphi^!\). The functor \(- \otimes_R^\mathbf{L} : D(R) \to D(P)\) preserves the subcategories \(D^+, D^+_{\textit{Coh}}, D^-, D^-_{\textit{Coh}}, D^b_{\textit{Coh}}\). The functor \(R\Hom(A, -) : D(P) \to D(A)\) preserves \(D^+\) and \(D^+_{\textit{Coh}}\) by Lemma 0A72. If \(R \to A\) is perfect, then \(A\) is perfect as a \(P\)-module, see More on Algebra, Lemma 068Y. Recall that the restriction of \(R\Hom(A, K)\) to \(D(P)\) is \(R\Hom_P(A, K)\). By More on Algebra, Lemma 07VI we have \(R\Hom_P(A, K) = E \otimes_P^\mathbf{L} K\) for some perfect \(E \in D(P)\). Since we can represent \(E\) by a finite complex of finite projective \(P\)-modules it is clear that \(R\Hom_P(A, K)\) is in \(D^-(P), D^-_{\textit{Coh}}(P), D^b_{\textit{Coh}}(P)\) as soon as \(K\) is. Since the restriction functor \(D(A) \to D(P)\) reflects these subcategories, the proof is complete.
Lemma
Let \(\varphi\) be a finite type homomorphism of Noetherian rings. If \(\omega_R^\bullet\) is a dualizing complex for \(R\), then \(\varphi^!(\omega_R^\bullet)\) is a dualizing complex for \(A\).
Proof
Lemma
Let \(R \to R'\) be a flat homomorphism of Noetherian rings. Let \(\varphi : R \to A\) be a finite type ring map. Let \(\varphi' : R' \to A' = A \otimes_R R'\) be the map induced by \(\varphi\). Then we have a functorial maps \[\varphi^!(K) \otimes_A^\mathbf{L} A' \longrightarrow (\varphi')^!(K \otimes_R^\mathbf{L} R')\] for \(K\) in \(D(R)\) which are isomorphisms for \(K \in D^+(R)\).
Proof
Choose a factorization \(R \to P \to A\) where \(P\) is a polynomial ring over \(R\). This gives a corresponding factorization \(R' \to P' \to A'\) by base change. Since we have \((K \otimes_R^\mathbf{L} P) \otimes_P^\mathbf{L} P' = (K \otimes_R^\mathbf{L} R') \otimes_{R'}^\mathbf{L} P'\) by More on Algebra, Lemma 08YU it suffices to construct maps \[R\Hom(A, K \otimes_R^\mathbf{L} P[n]) \otimes_A^\mathbf{L} A' \longrightarrow R\Hom(A', (K \otimes_R^\mathbf{L} P[n]) \otimes_P^\mathbf{L} P')\] functorial in \(K\). For this we use the map (0E29) constructed in Section 0E28 for \(P, A, P', A'\). The map is an isomorphism for \(K \in D^+(R)\) by Lemma 0BZM.
Lemma
Let \(R \to R'\) be a homomorphism of Noetherian rings. Let \(\varphi : R \to A\) be a perfect ring map (More on Algebra, Definition 067H) such that \(R'\) and \(A\) are tor independent over \(R\). Let \(\varphi' : R' \to A' = A \otimes_R R'\) be the map induced by \(\varphi\). Then we have a functorial isomorphism \[\varphi^!(K) \otimes_A^\mathbf{L} A' = (\varphi')^!(K \otimes_R^\mathbf{L} R')\] for \(K\) in \(D(R)\).
Proof
We may choose a factorization \(R \to P \to A\) where \(P\) is a polynomial ring over \(R\) such that \(A\) is a perfect \(P\)-module, see More on Algebra, Lemma 068Y. This gives a corresponding factorization \(R' \to P' \to A'\) by base change. Since we have \((K \otimes_R^\mathbf{L} P) \otimes_P^\mathbf{L} P' = (K \otimes_R^\mathbf{L} R') \otimes_{R'}^\mathbf{L} P'\) by More on Algebra, Lemma 08YU it suffices to construct maps \[R\Hom(A, K \otimes_R^\mathbf{L} P[n]) \otimes_A^\mathbf{L} A' \longrightarrow R\Hom(A', (K \otimes_R^\mathbf{L} P[n]) \otimes_P^\mathbf{L} P')\] functorial in \(K\). We have \[A \otimes_P^\mathbf{L} P' = A \otimes_R^\mathbf{L} R' = A'\] The first equality by More on Algebra, Lemma 0661 applied to \(R, R', P, P'\). The second equality because \(A\) and \(R'\) are tor independent over \(R\). Hence \(A\) and \(P'\) are tor independent over \(P\) and we can use the map (0E29) constructed in Section 0E28 for \(P, A, P', A'\) get the desired arrow. By Lemma 0BZP to finish the proof it suffices to prove that \(A\) is a perfect \(P\)-module which we saw above.
Lemma
Let \(R \to R'\) be a homomorphism of Noetherian rings. Let \(\varphi : R \to A\) be flat of finite type. Let \(\varphi' : R' \to A' = A \otimes_R R'\) be the map induced by \(\varphi\). Then we have a functorial isomorphism \[\varphi^!(K) \otimes_A^\mathbf{L} A' = (\varphi')^!(K \otimes_R^\mathbf{L} R')\] for \(K\) in \(D(R)\).
Proof
Lemma
Let \(A \xrightarrow{a} B \xrightarrow{b} C\) be finite type homomorphisms of Noetherian rings. Then there is a transformation of functors \(b^! \circ a^! \to (b \circ a)^!\) which is an isomorphism on \(D^+(A)\).
Proof
Choose a polynomial ring \(P = A[x_1, \ldots, x_n]\) over \(A\) and a surjection \(P \to B\). Choose elements \(c_1, \ldots, c_m \in C\) generating \(C\) over \(B\). Set \(Q = P[y_1, \ldots, y_m]\) and denote \(Q' = Q \otimes_P B = B[y_1, \ldots, y_m]\). Let \(\chi : Q' \to C\) be the surjection sending \(y_j\) to \(c_j\). Picture \[\xymatrix{ & Q \ar[r]_{\psi'} & Q' \ar[r]_\chi & C \\ A \ar[r] & P \ar[r]^\psi \ar[u] & B \ar[u] }\] By Lemma 0BZM for \(M \in D(P)\) we have an arrow \(\psi^!(M) \otimes_B^\mathbf{L} Q' \to (\psi')^!(M \otimes_P^\mathbf{L} Q)\) which is an isomorphism whenever \(M\) is bounded below. Also we have \(\chi^! \circ (\psi')^! = (\chi \circ \psi')^!\) as both functors are adjoint to the restriction functor \(D(C) \to D(Q)\) by Section 0A6Z. Then we see \[\begin{align*} b^!(a^!(K)) & = \chi^!(\psi^!(K \otimes_A^\mathbf{L} P)[n] \otimes_B^\mathbf{L} Q)[m] \\ & \to \chi^!((\psi')^!(K \otimes_A^\mathbf{L} P \otimes_P^\mathbf{L} Q))[n + m] \\ & = (\chi \circ \psi')^!(K\otimes_A^\mathbf{L} Q)[n + m] \\ & = (b \circ a)^!(K) \end{align*}\] where we have used in addition to the above More on Algebra, Lemma 08YU.
Lemma
Let \(\varphi : R \to A\) be a finite map of Noetherian rings. Then \(\varphi^!\) is isomorphic to the functor \(R\Hom(A, -) : D(R) \to D(A)\) from Section 0A6Z.
Proof
Suppose that \(A\) is generated by \(n > 1\) elements over \(R\). Then can factor \(R \to A\) as a composition of two finite ring maps where in both steps the number of generators is \(< n\). Since we have Lemma 0BZT and Lemma 0C0F we conclude that it suffices to prove the lemma when \(A\) is generated by one element over \(R\). Since \(A\) is finite over \(R\), it follows that \(A\) is a quotient of \(B = R[x]/(f)\) where \(f\) is a monic polynomial in \(x\) (Algebra, Lemma 00GK). Again using the lemmas on composition and the fact that we have agreement for surjections by definition, we conclude that it suffices to prove the lemma for \(R \to B = R[x]/(f)\). In this case, the functor \(\varphi^!\) is isomorphic to \(K \mapsto K \otimes_R^\mathbf{L} B\); you prove this by using Lemma 0BZH for the map \(R[x] \to B\) (note that the shift in the definition of \(\varphi^!\) and in the lemma add up to zero). For the functor \(R\Hom(B, -) : D(R) \to D(B)\) we can use Lemma 0BZE to see that it suffices to show \(\Hom_R(B, R) \cong B\) as \(B\)-modules. Suppose that \(f\) has degree \(d\). Then an \(R\)-basis for \(B\) is given by \(1, x, \ldots, x^{d - 1}\). Let \(\delta_i : B \to R\), \(i = 0, \ldots, d - 1\) be the \(R\)-linear map which picks off the coefficient of \(x^i\) with respect to the given basis. Then \(\delta_0, \ldots, \delta_{d - 1}\) is a basis for \(\Hom_R(B, R)\). Finally, for \(0 \leq i \leq d - 1\) a computation shows that \[x^i \delta_{d - 1} = \delta_{d - 1 - i} + b_1 \delta_{d - i} + \ldots + b_i \delta_{d - 1}\] for some \(c_1, \ldots, c_d \in R\)2. Hence \(\Hom_R(B, R)\) is a principal \(B\)-module with generator \(\delta_{d - 1}\). By looking at ranks we conclude that it is a rank \(1\) free \(B\)-module.
Lemma
Let \(R\) be a Noetherian ring and let \(f \in R\). If \(\varphi\) denotes the map \(R \to R_f\), then \(\varphi^!\) is isomorphic to \(- \otimes_R^\mathbf{L} R_f\). More generally, if \(\varphi : R \to R'\) is a map such that \(\Spec(R') \to \Spec(R)\) is an open immersion, then \(\varphi^!\) is isomorphic to \(- \otimes_R^\mathbf{L} R'\).
Proof
Choose the presentation \(R \to R[x] \to R[x]/(fx - 1) = R_f\) and observe that \(fx - 1\) is a nonzerodivisor in \(R[x]\). Thus we can apply using Lemma 0BZH to compute the functor \(\varphi^!\). Details omitted; note that the shift in the definition of \(\varphi^!\) and in the lemma add up to zero.
In the general case note that \(R' \otimes_R R' = R'\). Hence the result follows from the base change results above. Either Lemma 0BZN or Lemma 0BZR will do.
Lemma
Let \(\varphi : R \to A\) be a perfect homomorphism of Noetherian rings (for example \(\varphi\) is flat of finite type). Then \(\varphi^!(K) = K \otimes_R^\mathbf{L} \varphi^!(R)\) for \(K \in D(R)\).
Proof
(The parenthetical statement follows from More on Algebra, Lemma 067J.) We can choose a factorization \(R \to P \to A\) where \(P\) is a polynomial ring in \(n\) variables over \(R\) and then \(A\) is a perfect \(P\)-module, see More on Algebra, Lemma 068Y. Recall that \(\varphi^!(K) = R\Hom(A, K \otimes_R^\mathbf{L} P[n])\). Thus the result follows from Lemma 0BZE and More on Algebra, Lemma 08YU.
Lemma
Let \(\varphi : A \to B\) be a finite type homomorphism of Noetherian rings. Let \(\omega_A^\bullet\) be a dualizing complex for \(A\). Set \(\omega_B^\bullet = \varphi^!(\omega_A^\bullet)\). Denote \(D_A(K) = R\Hom_A(K, \omega_A^\bullet)\) for \(K \in D_{\textit{Coh}}(A)\) and \(D_B(L) = R\Hom_B(L, \omega_B^\bullet)\) for \(L \in D_{\textit{Coh}}(B)\). Then there is a functorial isomorphism \[\varphi^!(K) = D_B(D_A(K) \otimes_A^\mathbf{L} B)\] for \(K \in D_{\textit{Coh}}(A)\).
Proof
Observe that \(\omega_B^\bullet\) is a dualizing complex for \(B\) by Lemma 0BZL. Let \(A \to B \to C\) be finite type homomorphisms of Noetherian rings. If the lemma holds for \(A \to B\) and \(B \to C\), then the lemma holds for \(A \to C\). This follows from Lemma 0BZT and the fact that \(D_B \circ D_B \cong \text{id}\) by Lemma 0A7C. Thus it suffices to prove the lemma in case \(A \to B\) is a surjection and in the case where \(B\) is a polynomial ring over \(A\).
Assume \(B = A[x_1, \ldots, x_n]\). Since \(D_A \circ D_A \cong \text{id}\), it suffices to prove \(D_B(K \otimes_A B) \cong D_A(K) \otimes_A B[n]\) for \(K\) in \(D_{\textit{Coh}}(A)\). Choose a bounded complex \(I^\bullet\) of injectives representing \(\omega_A^\bullet\). Choose a quasi-isomorphism \(I^\bullet \otimes_A B \to J^\bullet\) where \(J^\bullet\) is a bounded complex of \(B\)-modules. Given a complex \(K^\bullet\) of \(A\)-modules, consider the obvious map of complexes \[\Hom^\bullet(K^\bullet, I^\bullet) \otimes_A B[n] \longrightarrow \Hom^\bullet(K^\bullet \otimes_A B, J^\bullet[n])\] The left hand side represents \(D_A(K) \otimes_A B[n]\) and the right hand side represents \(D_B(K \otimes_A B)\). Thus it suffices to prove this map is a quasi-isomorphism if the cohomology modules of \(K^\bullet\) are finite \(A\)-modules. Observe that the cohomology of the complex in degree \(r\) (on either side) only depends on finitely many of the \(K^i\). Thus we may replace \(K^\bullet\) by a truncation, i.e., we may assume \(K^\bullet\) represents an object of \(D^-_{\textit{Coh}}(A)\). Then \(K^\bullet\) is quasi-isomorphic to a bounded above complex of finite free \(A\)-modules. Therefore we may assume \(K^\bullet\) is a bounded above complex of finite free \(A\)-modules. In this case it is easy to that the displayed map is an isomorphism of complexes which finishes the proof in this case.
Assume that \(A \to B\) is surjective. Denote \(i_* : D(B) \to D(A)\) the restriction functor and recall that \(\varphi^!(-) = R\Hom(A, -)\) is a right adjoint to \(i_*\) (Lemma 0A70). For \(F \in D(B)\) we have \[\begin{align*} \Hom_B(F, D_B(D_A(K) \otimes_A^\mathbf{L} B)) & = \Hom_B((D_A(K) \otimes_A^\mathbf{L} B) \otimes_B^\mathbf{L} F, \omega_B^\bullet) \\ & = \Hom_A(D_A(K) \otimes_A^\mathbf{L} i_*F, \omega_A^\bullet) \\ & = \Hom_A(i_*F, D_A(D_A(K))) \\ & = \Hom_A(i_*F, K) \\ & = \Hom_B(F, \varphi^!(K)) \end{align*}\] The first equality follows from More on Algebra, Lemma 0A65 and the definition of \(D_B\). The second equality by the adjointness mentioned above and the equality \(i_*((D_A(K) \otimes_A^\mathbf{L} B) \otimes_B^\mathbf{L} F) = D_A(K) \otimes_A^\mathbf{L} i_*F\) (More on Algebra, Lemma 06Y6). The third equality follows from More on Algebra, Lemma 0A65. The fourth because \(D_A \circ D_A = \text{id}\). The final equality by adjointness again. Thus the result holds by the Yoneda lemma.
Relative dualizing complexes in the Noetherian case
Let \(\varphi : R \to A\) be a finite type homomorphism of Noetherian rings. Then we define the relative dualizing complex of \(A\) over \(R\) as the object \[\omega_{A/R}^\bullet = \varphi^!(R)\] of \(D(A)\). Here \(\varphi^!\) is as in Section 0BZI. From the material in that section we see that \(\omega_{A/R}^\bullet\) is well defined up to (non-unique) isomorphism.
Lemma
Let \(R \to R'\) be a homomorphism of Noetherian rings. Let \(R \to A\) be of finite type. Set \(A' = A \otimes_R R'\). If
\(R \to R'\) is flat, or
\(R \to A\) is flat, or
\(R \to A\) is perfect and \(R'\) and \(A\) are tor independent over \(R\),
then there is an isomorphism \(\omega_{A/R}^\bullet \otimes_A^\mathbf{L} A' \to \omega^\bullet_{A'/R'}\) in \(D(A')\).
Proof
Follows from Lemmas 0BZN, 0BZS, and 0BZR and the definitions.
Lemma
Let \(\varphi : R \to A\) be a finite type map of Noetherian rings.
If \(\varphi\) is perfect, then
\(\omega_{A/R}^\bullet\) is in \(D^b_{\textit{Coh}}(A)\),
\(\omega_{A/R}^\bullet\) has finite tor dimension over \(R\), and
\(A \to R\Hom_A(\omega_{A/R}^\bullet, \omega_{A/R}^\bullet)\) is an isomorphism.
If \(\varphi\) is flat, then (1)(a), (1)(b), and (1)(c) hold and
\(\omega_{A/R}^\bullet\) is \(R\)-perfect (More on Algebra, Definition 0DHS), and
for every map \(R \to k\) to a field the base change \(\omega_{A/R}^\bullet \otimes_A^\mathbf{L} (A \otimes_R k)\) is a dualizing complex for \(A \otimes_R k\).
Proof
Assume \(R \to A\) is perfect. Choose \(R \to P \to A\) as in the definition of \(\varphi^!\). Then \(A\) is perfect as a \(P\)-modue (More on Algebra, Lemma 068Y). This shows that \(\omega_{A/R}^\bullet\) is in \(D^b_{\textit{Coh}}(A)\) by Lemma 0BZK. This proves (1)(a). To show \(\omega_{A/R}^\bullet\) has finite tor dimension as a complex of \(R\)-modules, observe that \(\omega_{A/R}^\bullet = \varphi^!(R) = R\Hom(A, P)[n]\) maps to \(R\Hom_P(A, P)[n]\) in \(D(P)\), which is perfect in \(D(P)\) (More on Algebra, Lemma 07VI), hence has finite tor dimension in \(D(R)\) as \(R \to P\) is flat. This proves (1)(b). The object \(R\Hom_A(\omega_{A/R}^\bullet, \omega_{A/R}^\bullet)\) of \(D(A)\) maps in \(D(P)\) to \[\begin{align*} R\Hom_P(\omega_{A/R}^\bullet, R\Hom(A, P)[n]) & = R\Hom_P(R\Hom_P(A, P)[n], P)[n] \\ & = R\Hom_P(R\Hom_P(A, P), P) \end{align*}\] This is equal to \(A\) by the already used More on Algebra, Lemma 07VI. This proves (1)(c).
Assume \(\varphi\) is flat. Then \(R \to A\) is a perfect ring map (More on Algebra, Lemma 067J) and we see that (1)(a), (1)(b), and (1)(c) hold. Of course, then \(\omega_{A/R}^\bullet\) is \(R\)-perfect by (1)(a) and (1)(b) and the definitions. Let \(R \to k\) be as in (2)(b). By Lemma 0BZV there is an isomorphism \[\omega_{A/R}^\bullet \otimes_A^\mathbf{L} (A \otimes_R k) \cong \omega^\bullet_{A \otimes_R k/k}\] and the right hand side is a dualizing complex by Lemma 0BZL. This finishes the proof.
Lemma
Let \(K/k\) be an extension of fields. Let \(A\) be a finite type \(k\)-algebra. Let \(A_K = A \otimes_k K\). If \(\omega_A^\bullet\) is a dualizing complex for \(A\), then \(\omega_A^\bullet \otimes_A A_K\) is a dualizing complex for \(A_K\).
Proof
By the uniqueness of dualizing complexes, it doesn’t matter which dualizing complex we pick for \(A\); we omit the detailed proof. Denote \(\varphi : k \to A\) the algebra structure. We may take \(\omega_A^\bullet = \varphi^!(k[0])\) by Lemma 0BZL. We conclude by Lemma 0BZW.
Lemma
Let \(\varphi : R \to A\) be a local complete intersection homomorphism of Noetherian rings. Then \(\omega_{A/R}^\bullet\) is an invertible object of \(D(A)\) and \(\varphi^!(K) = K \otimes_R^\mathbf{L} \omega_{A/R}^\bullet\) for all \(K \in D(R)\).
Proof
Recall that a local complete intersection homomorphism is a perfect ring map by More on Algebra, Lemma 07EN. Hence the final statement holds by Lemma 0BZU. By More on Algebra, Definition 07D0 we can write \(A = R[x_1, \ldots, x_n]/I\) where \(I\) is a Koszul-regular ideal. The construction of \(\varphi^!\) in Section 0BZI shows that it suffices to show the lemma in case \(A = R/I\) where \(I \subset R\) is a Koszul-regular ideal. Checking \(\omega_{A/R}^\bullet\) is invertible in \(D(A)\) is local on \(\Spec(A)\) by More on Algebra, Lemma 0FNT. Moreover, formation of \(\omega_{A/R}^\bullet\) commutes with localization on \(R\) by Lemma 0BZN. Combining More on Algebra, Definition 07CV and Lemma 09CC and Algebra, Lemma 061L we can find \(g_1, \ldots, g_r \in R\) generating the unit ideal in \(A\) such that \(I_{g_j} \subset R_{g_j}\) is generated by a regular sequence. Thus we may assume \(A = R/(f_1, \ldots, f_c)\) where \(f_1, \ldots, f_c\) is a regular sequence in \(R\). Then we consider the ring maps \[R \to R/(f_1) \to R/(f_1, f_2) \to \ldots \to R/(f_1, \ldots, f_c) = A\] and we use Lemma 0BZT (and the final statement already proven) to see that it suffices to prove the lemma for each step. Finally, in case \(A = R/(f)\) for some nonzerodivisor \(f\) we see that the lemma is true since \(\varphi^!(R) = R\Hom(A, R)\) is invertible by Lemma 0BZH.
Lemma
Let \(\varphi : R \to A\) be a flat finite type homomorphism of Noetherian rings. The following are equivalent
the fibres \(A \otimes_R \kappa(\mathfrak p)\) are Gorenstein for all primes \(\mathfrak p \subset R\), and
\(\omega_{A/R}^\bullet\) is an invertible object of \(D(A)\), see More on Algebra, Lemma 0FNT.
Proof
If (2) holds, then the fibre rings \(A \otimes_R \kappa(\mathfrak p)\) have invertible dualizing complexes, and hence are Gorenstein. See Lemmas 0BZW and 0DW9.
For the converse, assume (1). Observe that \(\omega_{A/R}^\bullet\) is in \(D^b_{\textit{Coh}}(A)\) by Lemma 0BZK (since flat finite type homomorphisms of Noetherian rings are perfect, see More on Algebra, Lemma 067J). Take a prime \(\mathfrak q \subset A\) lying over \(\mathfrak p \subset R\). Then \[\omega_{A/R}^\bullet \otimes_A^\mathbf{L} \kappa(\mathfrak q) = \omega_{A/R}^\bullet \otimes_A^\mathbf{L} (A \otimes_R \kappa(\mathfrak p)) \otimes_{(A \otimes_R \kappa(\mathfrak p))}^\mathbf{L} \kappa(\mathfrak q)\] Applying Lemmas 0BZW and 0DW9 and assumption (1) we find that this complex has \(1\) nonzero cohomology group which is a \(1\)-dimensional \(\kappa(\mathfrak q)\)-vector space. By More on Algebra, Lemma 0BYP we conclude that \((\omega_{A/R}^\bullet)_f\) is an invertible object of \(D(A_f)\) for some \(f \in A\), \(f \not \in \mathfrak q\). This proves (2) holds.
The following lemma is useful to see how dimension functions change when passing to a finite type algebra over a Noetherian ring.
Lemma
Let \(\varphi : R \to A\) be a finite type homomorphism of Noetherian rings. Assume \(R\) local and let \(\mathfrak m \subset A\) be a maximal ideal lying over the maximal ideal of \(R\). If \(\omega_R^\bullet\) is a normalized dualizing complex for \(R\), then \(\varphi^!(\omega_R^\bullet)_\mathfrak m\) is a normalized dualizing complex for \(A_\mathfrak m\).
Proof
We already know that \(\varphi^!(\omega_R^\bullet)\) is a dualizing complex for \(A\), see Lemma 0BZL. Choose a factorization \(R \to P \to A\) with \(P = R[x_1, \ldots, x_n]\) as in the construction of \(\varphi^!\). If we can prove the lemma for \(R \to P\) and the maximal ideal \(\mathfrak m'\) of \(P\) corresponding to \(\mathfrak m\), then we obtain the result for \(R \to A\) by applying Lemma 0A7N to \(P_{\mathfrak m'} \to A_\mathfrak m\) or by applying Lemma 0A7Y to \(P \to A\). In the case \(A = R[x_1, \ldots, x_n]\) we see that \(\dim(A_\mathfrak m) = \dim(R) + n\) for example by Algebra, Lemma 00ON (combined with Algebra, Lemma 00OP to compute the dimension of the fibre). The fact that \(\omega_R^\bullet\) is normalized means that \(i = -\dim(R)\) is the smallest index such that \(H^i(\omega_R^\bullet)\) is nonzero (follows from Lemmas 0A7U and 0A7V). Then \(\varphi^!(\omega_R^\bullet)_\mathfrak m = \omega_R^\bullet \otimes_R A_\mathfrak m[n]\) has its first nonzero cohomology module in degree \(-\dim(R) - n\) and therefore is the normalized dualizing complex for \(A_\mathfrak m\).
Lemma
Let \(R \to A\) be a finite type homomorphism of Noetherian rings. Let \(\mathfrak q \subset A\) be a prime ideal lying over \(\mathfrak p \subset R\). Then \[H^i(\omega_{A/R}^\bullet)_\mathfrak q \not = 0 \Rightarrow - d \leq i\] where \(d\) is the dimension of the fibre of \(\Spec(A) \to \Spec(R)\) over \(\mathfrak p\) at the point \(\mathfrak q\).
Proof
Choose a factorization \(R \to P \to A\) with \(P = R[x_1, \ldots, x_n]\) as in Section 0BZI so that \(\omega_{A/R}^\bullet = R\Hom(A, P)[n]\). We have to show that \(R\Hom(A, P)_\mathfrak q\) has vanishing cohomology in degrees \(< n - d\). By Lemma 0A71 this means we have to show that \(\Ext_P^i(P/I, P)_{\mathfrak r} = 0\) for \(i < n - d\) where \(\mathfrak r \subset P\) is the prime corresponding to \(\mathfrak q\) and \(I\) is the kernel of \(P \to A\). We may rewrite this as \(\Ext_{P_\mathfrak r}^i(P_\mathfrak r/IP_\mathfrak r, P_\mathfrak r)\) by More on Algebra, Lemma 087R. Thus we have to show \[\text{depth}_{IP_\mathfrak r}(P_\mathfrak r) \geq n - d\] by Lemma 0AVZ. By Lemma 0BUY we have \[\text{depth}_{IP_\mathfrak r}(P_\mathfrak r) \geq \dim((P \otimes_R \kappa(\mathfrak p))_\mathfrak r) - \dim((P/I \otimes_R \kappa(\mathfrak p))_\mathfrak r)\] The two expressions on the right hand side agree by Algebra, Lemma 00P2.
Lemma
Let \(R \to A\) be a flat finite type homomorphism of Noetherian rings. Let \(\mathfrak q \subset A\) be a prime ideal lying over \(\mathfrak p \subset R\). Then \[H^i(\omega_{A/R}^\bullet)_\mathfrak q \not = 0 \Rightarrow - d \leq i \leq 0\] where \(d\) is the dimension of the fibre of \(\Spec(A) \to \Spec(R)\) over \(\mathfrak p\) at the point \(\mathfrak q\). If all fibres of \(\Spec(A) \to \Spec(R)\) have dimension \(\leq d\), then \(\omega_{A/R}^\bullet\) has tor amplitude in \([-d, 0]\) as a complex of \(R\)-modules.
Proof
The lower bound has been shown in Lemma 0E9P. Choose a factorization \(R \to P \to A\) with \(P = R[x_1, \ldots, x_n]\) as in Section 0BZI so that \(\omega_{A/R}^\bullet = R\Hom(A, P)[n]\). The upper bound means that \(\Ext^i_P(A, P)\) is zero for \(i > n\). This follows from More on Algebra, Lemma 068X which shows that \(A\) is a perfect \(P\)-module with tor amplitude in \([-n, 0]\).
Proof of the final statement. Let \(R \to R'\) be a ring homomorphism of Noetherian rings. Set \(A' = A \otimes_R R'\). Then \[\omega_{A'/R'}^\bullet = \omega_{A/R}^\bullet \otimes_A^\mathbf{L} A' = \omega_{A/R}^\bullet \otimes_R^\mathbf{L} R'\] The first isomorphism by Lemma 0BZV and the second, which takes place in \(D(R')\), by More on Algebra, Lemma 0661. By the first part of the proof (note that the fibres of \(\Spec(A') \to \Spec(R')\) have dimension \(\leq d\)) we conclude that \(\omega_{A/R}^\bullet \otimes_R^\mathbf{L} R'\) has cohomology only in degrees \([-d, 0]\). Taking \(R' = R \oplus M\) to be the square zero thickening of \(R\) by a finite \(R\)-module \(M\), we see that \(R\Hom(A, P) \otimes_R^\mathbf{L} M\) has cohomology only in the interval \([-d, 0]\) for any finite \(R\)-module \(M\). Since any \(R\)-module is a filtered colimit of finite \(R\)-modules and since tensor products commute with colimits we conclude.
Lemma
Let \(R \to A\) be a finite type homomorphism of Noetherian rings. Let \(\mathfrak p \subset R\) be a prime ideal. Assume
\(R_\mathfrak p\) is Cohen-Macaulay, and
for any minimal prime \(\mathfrak q \subset A\) we have \(\text{trdeg}_{\kappa(R \cap \mathfrak q)} \kappa(\mathfrak q) \leq r\).
Then \[H^i(\omega_{A/R}^\bullet)_\mathfrak p \not = 0 \Rightarrow - r \leq i\] and \(H^{-r}(\omega_{A/R}^\bullet)_\mathfrak p\) is \((S_2)\) as an \(A_\mathfrak p\)-module.
Proof
We may replace \(R\) by \(R_\mathfrak p\) by Lemma 0BZV. Thus we may assume \(R\) is a Cohen-Macaulay local ring and we have to show the assertions of the lemma for the \(A\)-modules \(H^i(\omega_{A/R}^\bullet)\).
Let \(R^\wedge\) be the completion of \(R\). The map \(R \to R^\wedge\) is flat and \(R^\wedge\) is Cohen-Macaulay (More on Algebra, Lemma 07NX). Observe that the minimal primes of \(A \otimes_R R^\wedge\) lie over minimal primes of \(A\) by the flatness of \(A \to A \otimes_R R^\wedge\) (and going down for flatness, see Algebra, Lemma 00HS). Thus condition (2) holds for the finite type ring map \(R^\wedge \to A \otimes_R R^\wedge\) by Morphisms, Lemma 02FY. Appealing to Lemma 0BZV once again it suffices to prove the lemma for \(R^\wedge \to A \otimes_R R^\wedge\). In this way, using Lemma 0BFR, we may assume \(R\) is a Noetherian local Cohen-Macaulay ring which has a dualizing complex \(\omega_R^\bullet\).
Let \(\mathfrak m \subset A\) be a maximal ideal. It suffices to show that the assertions of the lemma hold for \(H^i(\omega_{A/R}^\bullet)_\mathfrak m\). If \(\mathfrak m\) does not lie over the maximal ideal of \(R\), then we replace \(R\) by a localization to reduce to this case (small detail omitted).
We may assume \(\omega_R^\bullet\) is normalized. Setting \(d = \dim(R)\) we see that \(\omega_R^\bullet = \omega_R[d]\) for some \(R\)-module \(\omega_R\), see Lemma 0AWS. Set \(\omega_A^\bullet = \varphi^!(\omega_R^\bullet)\). By Lemma 0E9L we have \[\omega_{A/R}^\bullet = R\Hom_A(\omega_R[d] \otimes_R^\mathbf{L} A, \omega_A^\bullet)\] By the dimension formula we have \(\dim(A_\mathfrak m) \leq d + r\), see Morphisms, Lemma 0BAE and use that \(\kappa(\mathfrak m)\) is finite over the residue field of \(R\) by the Hilbert Nullstellensatz. By Lemma 0E9N we see that \((\omega_A^\bullet)_\mathfrak m\) is a normalized dualizing complex for \(A_\mathfrak m\). Hence \(H^i((\omega_A^\bullet)_\mathfrak m)\) is nonzero only for \(-d - r \leq i \leq 0\), see Lemma 0A7U. Since \(\omega_R[d] \otimes_R^\mathbf{L} A\) lives in degrees \(\leq -d\) we conclude the vanishing holds. Finally, we also see that \[H^{-r}(\omega_{A/R}^\bullet)_\mathfrak m = \Hom_A(\omega_R \otimes_R A, H^{-d - r}(\omega_A^\bullet))_\mathfrak m\] Since \(H^{-d - r}(\omega_A^\bullet)_\mathfrak m\) is \((S_2)\) by Lemma 0AWE we find that the final statement is true by More on Algebra, Lemma 0AV6.
More on dualizing complexes
Some lemmas which don’t fit anywhere else very well.
Lemma
Let \(A \to B\) be a faithfully flat map of Noetherian rings. If \(K \in D(A)\) and \(K \otimes_A^\mathbf{L} B\) is a dualizing complex for \(B\), then \(K\) is a dualizing complex for \(A\).
Proof
Since \(A \to B\) is flat we have \(H^i(K) \otimes_A B = H^i(K \otimes_A^\mathbf{L} B)\). Since \(K \otimes_A^\mathbf{L} B\) is in \(D^b_{\textit{Coh}}(B)\) we first find that \(K\) is in \(D^b(A)\) and then we see that \(H^i(K)\) is a finite \(A\)-module by Algebra, Lemma 03C4. Let \(M\) be a finite \(A\)-module. Then \[R\Hom_A(M, K) \otimes_A B = R\Hom_B(M \otimes_A B, K \otimes_A^\mathbf{L} B)\] by More on Algebra, Lemma 0A6A. Since \(K \otimes_A^\mathbf{L} B\) has finite injective dimension, say injective-amplitude in \([a, b]\), we see that the right hand side has vanishing cohomology in degrees \(> b\). Since \(A \to B\) is faithfully flat, we find that \(R\Hom_A(M, K)\) has vanishing cohomology in degrees \(> b\). Thus \(K\) has finite injective dimension by More on Algebra, Lemma 0A5T. To finish the proof we have to show that the map \(A \to R\Hom_A(K, K)\) is an isomorphism. For this we again use More on Algebra, Lemma 0A6A and the fact that \(B \to R\Hom_B(K \otimes_A^\mathbf{L} B, K \otimes_A^\mathbf{L} B)\) is an isomorphism.
Lemma
Let \(\varphi : A \to B\) be a homomorphism of Noetherian rings. Assume
\(A \to B\) is syntomic and induces a surjective map on spectra, or
\(A \to B\) is a faithfully flat local complete intersection, or
\(A \to B\) is faithfully flat of finite type with Gorenstein fibres.
Then \(K \in D(A)\) is a dualizing complex for \(A\) if and only if \(K \otimes_A^\mathbf{L} B\) is a dualizing complex for \(B\).
Proof
Observe that \(A \to B\) satisfies (1) if and only if \(A \to B\) satisfies (2) by More on Algebra, Lemma 07D3. Observe that in both (2) and (3) the relative dualzing complex \(\varphi^!(A) = \omega_{B/A}^\bullet\) is an invertible object of \(D(B)\), see Lemmas 0E4B and 0E4C. Moreover we have \(\varphi^!(K) = K \otimes_A^\mathbf{L} \omega_{B/A}^\bullet\) in both cases, see Lemma 0BZU for case (3). Thus \(\varphi^!(K)\) is the same as \(K \otimes_A^\mathbf{L} B\) up to tensoring with an invertible object of \(D(B)\). Hence \(\varphi^!(K)\) is a dualizing complex for \(B\) if and only if \(K \otimes_A^\mathbf{L} B\) is (as being a dualizing complex is local and invariant under shifts). Thus we see that if \(K\) is dualizing for \(A\), then \(K \otimes_A^\mathbf{L} B\) is dualizing for \(B\) by Lemma 0BZL. To descend the property, see Lemma 0E4A.
Lemma
Let \((A, \mathfrak m, \kappa) \to (B, \mathfrak n, l)\) be a flat local homorphism of Noetherian rings such that \(\mathfrak n = \mathfrak m B\). If \(E\) is the injective hull of \(\kappa\), then \(E \otimes_A B\) is the injective hull of \(l\).
Proof
Write \(E = \bigcup E_n\) as in Lemma 08Z4. It suffices to show that \(E_n \otimes_{A/\mathfrak m^n} B/\mathfrak n^n\) is the injective hull of \(l\) over \(B/\mathfrak n\). This reduces us to the case where \(A\) and \(B\) are Artinian local. Observe that \(\text{length}_A(A) = \text{length}_B(B)\) and \(\text{length}_A(E) = \text{length}_B(E \otimes_A B)\) by Algebra, Lemma 02M1. By Lemma 08YX we have \(\text{length}_A(E) = \text{length}_A(A)\) and \(\text{length}_B(E') = \text{length}_B(B)\) where \(E'\) is the injective hull of \(l\) over \(B\). We conclude \(\text{length}_B(E') = \text{length}_B(E \otimes_A B)\). Observe that \[\dim_l((E \otimes_A B)[\mathfrak n]) = \dim_l(E[\mathfrak m] \otimes_A B) = \dim_\kappa(E[\mathfrak m]) = 1\] where we have used flatness of \(A \to B\) and \(\mathfrak n = \mathfrak mB\). Thus there is an injective \(B\)-module map \(E \otimes_A B \to E'\) by Lemma 08Z3. By equality of lengths shown above this is an isomorphism.
Lemma
Let \(\varphi : A \to B\) be a flat homorphism of Noetherian rings such that for all primes \(\mathfrak q \subset B\) we have \(\mathfrak p B_\mathfrak q = \mathfrak qB_\mathfrak q\) where \(\mathfrak p = \varphi^{-1}(\mathfrak q)\), for example if \(\varphi\) is étale. If \(I\) is an injective \(A\)-module, then \(I \otimes_A B\) is an injective \(B\)-module.
Proof
Étale maps satisfy the assumption by Algebra, Lemma 00U4. By Lemma 08XV and Proposition 08YA we may assume \(I\) is the injective hull of \(\kappa(\mathfrak p)\) for some prime \(\mathfrak p \subset A\). Then \(I\) is a module over \(A_\mathfrak p\). It suffices to prove \(I \otimes_A B = I \otimes_{A_\mathfrak p} B_\mathfrak p\) is injective as a \(B_\mathfrak p\)-module, see Lemma 08XQ. Thus we may assume \((A, \mathfrak m, \kappa)\) is local Noetherian and \(I = E\) is the injective hull of the residue field \(\kappa\). Our assumption implies that the Noetherian ring \(B/\mathfrak m B\) is a product of fields (details omitted). Thus there are finitely many prime ideals \(\mathfrak m_1, \ldots, \mathfrak m_n\) in \(B\) lying over \(\mathfrak m\) and they are all maximal ideals. Write \(E = \bigcup E_n\) as in Lemma 08Z4. Then \(E \otimes_A B = \bigcup E_n \otimes_A B\) and \(E_n \otimes_A B\) is a finite \(B\)-module with support \(\{\mathfrak m_1, \ldots, \mathfrak m_n\}\) hence decomposes as a product over the localizations at \(\mathfrak m_i\). Thus \(E \otimes_A B = \prod (E \otimes_A B)_{\mathfrak m_i}\). Since \((E \otimes_A B)_{\mathfrak m_i} = E \otimes_A B_{\mathfrak m_i}\) is the injective hull of the residue field of \(\mathfrak m_i\) by Lemma 0E4E we conclude.
Relative dualizing complexes
For a finite type ring map \(\varphi : R \to A\) of Noetherian rings we have the relative dualizing complex \(\omega_{A/R}^\bullet = \varphi^!(R)\) considered in Section 0E9M. If \(R\) is not Noetherian, a similarly constructed complex will in general not have good properties. In this section, we give a definition of a relative dualizing complex for a flat and finitely presented ring maps \(R \to A\) of non-Noetherian rings. The definition is chosen to globalize to flat and finitely presented morphisms of schemes, see Duality for Schemes, Section 0E2S. We will show that relative dualizing complexes exist (when the definition applies), are unique up to (noncanonical) isomorphism, and that in the Noetherian case we recover the complex of Section 0E9M.
The Noetherian reader may safely skip this section!
Definition
Let \(R \to A\) be a flat ring map of finite presentation. A relative dualizing complex is an object \(K \in D(A)\) such that
\(K\) is \(R\)-perfect (More on Algebra, Definition 0DHS), and
\(R\Hom_{A \otimes_R A}(A, K \otimes_A^\mathbf{L} (A \otimes_R A))\) is isomorphic to \(A\).
To understand this definition you may have to read and understand some of the following lemmas. Lemmas 0E2E and 0E2D show this definition does not clash with the definition in Section 0E9M.
Lemma
Let \(R \to A\) be a flat ring map of finite presentation. Any two relative dualizing complexes for \(R \to A\) are isomorphic.
Proof
Let \(K\) and \(L\) be two relative dualizing complexes for \(R \to A\). Denote \(K_1 = K \otimes_A^\mathbf{L} (A \otimes_R A)\) and \(L_2 = (A \otimes_R A) \otimes_A^\mathbf{L} L\) the derived base changes via the first and second coprojections \(A \to A \otimes_R A\). By symmetry the assumption on \(L_2\) implies that \(R\Hom_{A \otimes_R A}(A, L_2)\) is isomorphic to \(A\). By More on Algebra, Lemma 0ATK part (3) applied twice we have \[A \otimes_{A \otimes_R A}^\mathbf{L} L_2 \cong R\Hom_{A \otimes_R A}(A, K_1 \otimes_{A \otimes_R A}^\mathbf{L} L_2) \cong A \otimes_{A \otimes_R A}^\mathbf{L} K_1\] Applying the restriction functor \(D(A \otimes_R A) \to D(A)\) for either coprojection we obtain the desired result.
Lemma
Let \(\varphi : R \to A\) be a flat finite type ring map of Noetherian rings. Then the relative dualizing complex \(\omega_{A/R}^\bullet = \varphi^!(R)\) of Section 0E9M is a relative dualizing complex in the sense of Definition 0E2C.
Proof
From Lemma 0BZW we see that \(\varphi^!(R)\) is \(R\)-perfect. Denote \(\delta : A \otimes_R A \to A\) the multiplication map and \(p_1, p_2 : A \to A \otimes_R A\) the coprojections. Then \[\varphi^!(R) \otimes_A^\mathbf{L} (A \otimes_R A) = \varphi^!(R) \otimes_{A, p_1}^\mathbf{L} (A \otimes_R A) = p_2^!(A)\] by Lemma 0BZN. Recall that \(R\Hom_{A \otimes_R A}(A, \varphi^!(R) \otimes_A^\mathbf{L} (A \otimes_R A))\) is the image of \(\delta^!(\varphi^!(R) \otimes_A^\mathbf{L} (A \otimes_R A))\) under the restriction map \(\delta_* : D(A) \to D(A \otimes_R A)\). Use the definition of \(\delta^!\) from Section 0BZI and Lemma 0A71. Since \(\delta^!(p_2^!(A)) \cong A\) by Lemma 0BZT we conclude.
Lemma
Let \(R \to A\) be a flat ring map of finite presentation. Then
there exists a relative dualizing complex \(K\) in \(D(A)\), and
for any ring map \(R \to R'\) setting \(A' = A \otimes_R R'\) and \(K' = K \otimes_A^\mathbf{L} A'\), then \(K'\) is a relative dualizing complex for \(R' \to A'\).
Moreover, if \[\xi : A \longrightarrow K \otimes_A^\mathbf{L} (A \otimes_R A)\] is a generator for the cyclic module \(\Hom_{D(A \otimes_R A)}(A, K \otimes_A^\mathbf{L} (A \otimes_R A))\) then in (2) the derived base change of \(\xi\) by \(A \otimes_R A \to A' \otimes_{R'} A'\) is a generator for the cyclic module \(\Hom_{D(A' \otimes_{R'} A')}(A', K' \otimes_{A'}^\mathbf{L} (A' \otimes_{R'} A'))\)
Proof
We first reduce to the Noetherian case. By Algebra, Lemma 02JO there exists a finite type \(\mathbf{Z}\) subalgebra \(R_0 \subset R\) and a flat finite type ring map \(R_0 \to A_0\) such that \(A = A_0 \otimes_{R_0} R\). By Lemma 0E2E there exists a relative dualizing complex \(K_0 \in D(A_0)\). Thus if we show (2) for \(K_0\), then we find that \(K_0 \otimes_{A_0}^\mathbf{L} A\) is a dualizing complex for \(R \to A\) and that it also satisfies (2) by transitivity of derived base change. The uniqueness of relative dualizing complexes (Lemma 0E2D) then shows that this holds for any relative dualizing complex.
Assume \(R\) Noetherian and let \(K\) be a relative dualizing complex for \(R \to A\). Given a ring map \(R \to R'\) set \(A' = A \otimes_R R'\) and \(K' = K \otimes_A^\mathbf{L} A'\). To finish the proof we have to show that \(K'\) is a relative dualizing complex for \(R' \to A'\). By More on Algebra, Lemma 0DHW we see that \(K'\) is \(R'\)-perfect in all cases. By Lemmas 0BZV and 0E2E if \(R'\) is Noetherian, then \(K'\) is a relative dualizing complex for \(R' \to A'\) (in either sense). Transitivity of derived tensor product shows that \(K \otimes_A^\mathbf{L} (A \otimes_R A) \otimes_{A \otimes_R A}^\mathbf{L} (A' \otimes_{R'} A') = K' \otimes_{A'}^\mathbf{L} (A' \otimes_{R'} A')\). Flatness of \(R \to A\) guarantees that \(A \otimes_{A \otimes_R A}^\mathbf{L} (A' \otimes_{R'} A') = A'\); namely \(A \otimes_R A\) and \(R'\) are tor independent over \(R\) so we can apply More on Algebra, Lemma 0661. Finally, \(A\) is pseudo-coherent as an \(A \otimes_R A\)-module by More on Algebra, Lemma 0E1T. Thus we have checked all the assumptions of More on Algebra, Lemma 0E1U. We find there exists a bounded below complex \(E^\bullet\) of \(R\)-flat finitely presented \(A \otimes_R A\)-modules such that \(E^\bullet \otimes_R R'\) represents \(R\Hom_{A' \otimes_{R'} A'}(A', K' \otimes_{A'}^\mathbf{L} (A' \otimes_{R'} A'))\) and these identifications are compatible with derived base change. Let \(n \in \mathbf{Z}\), \(n \not = 0\). Define \(Q^n\) by the sequence \[E^{n - 1} \to E^n \to Q^n \to 0\] Since \(\kappa(\mathfrak p)\) is a Noetherian ring, we know that \(H^n(E^\bullet \otimes_R \kappa(\mathfrak p)) = 0\), see remarks above. Chasing diagrams this means that \[Q^n \otimes_R \kappa(\mathfrak p) \to E^{n + 1} \otimes_R \kappa(\mathfrak p)\] is injective. Hence for a prime \(\mathfrak q\) of \(A \otimes_R A\) lying over \(\mathfrak p\) we have \(Q^n_\mathfrak q\) is \(R_\mathfrak p\)-flat and \(Q^n_\mathfrak p \to E^{n + 1}_\mathfrak q\) is \(R_\mathfrak p\)-universally injective, see Algebra, Lemma 00ME. Since this holds for all primes, we conclude that \(Q^n\) is \(R\)-flat and \(Q^n \to E^{n + 1}\) is \(R\)-universally injective. In particular \(H^n(E^\bullet \otimes_R R') = 0\) for any ring map \(R \to R'\). Let \(Z^0 = \Ker(E^0 \to E^1)\). Since there is an exact sequence \(0 \to Z^0 \to E^0 \to E^1 \to Q^1 \to 0\) we see that \(Z^0\) is \(R\)-flat and that \(Z^0 \otimes_R R' = \Ker(E^0 \otimes_R R' \to E^1 \otimes_R R')\) for all \(R \to R'\). Then the short exact sequence \(0 \to Q^{-1} \to Z^0 \to H^0(E^\bullet) \to 0\) shows that \[H^0(E^\bullet \otimes_R R') = H^0(E^\bullet) \otimes_R R' = A \otimes_R R' = A'\] as desired. This equality furthermore gives the final assertion of the lemma.
Lemma
Let \(R \to A\) be a flat ring map of finite presentation. Let \(K\) be a relative dualizing complex. Then \(A \to R\Hom_A(K, K)\) is an isomorphism.
Proof
By Algebra, Lemma 02JO there exists a finite type \(\mathbf{Z}\) subalgebra \(R_0 \subset R\) and a flat finite type ring map \(R_0 \to A_0\) such that \(A = A_0 \otimes_{R_0} R\). By Lemmas 0E2D, 0E2E, and 0E2F there exists a relative dualizing complex \(K_0 \in D(A_0)\) and its derived base change is \(K\). This reduces us to the situation discussed in the next paragraph.
Assume \(R\) Noetherian and let \(K\) be a relative dualizing complex for \(R \to A\). Given a ring map \(R \to R'\) set \(A' = A \otimes_R R'\) and \(K' = K \otimes_A^\mathbf{L} A'\). To finish the proof we show \(R\Hom_{A'}(K', K') = A'\). By Lemma 0BZW we know this is true whenever \(R'\) is Noetherian. Since a general \(R'\) is a filtered colimit of Noetherian \(R\)-algebras, we find the result holds by More on Algebra, Lemma 0DHX.
Lemma
Let \(R \to A \to B\) be a ring maps which are flat and of finite presentation. Let \(K_{A/R}\) and \(K_{B/A}\) be relative dualizing complexes for \(R \to A\) and \(A \to B\). Then \(K = K_{A/R} \otimes_A^\mathbf{L} K_{B/A}\) is a relative dualizing complex for \(R \to B\).
Proof
We will use reduction to the Noetherian case. Namely, by Algebra, Lemma 02JO there exists a finite type \(\mathbf{Z}\) subalgebra \(R_0 \subset R\) and a flat finite type ring map \(R_0 \to A_0\) such that \(A = A_0 \otimes_{R_0} R\). After increasing \(R_0\) and correspondingly replacing \(A_0\) we may assume there is a flat finite type ring map \(A_0 \to B_0\) such that \(B = B_0 \otimes_{R_0} R\) (use the same lemma). If we prove the lemma for \(R_0 \to A_0 \to B_0\), then the lemma follows by Lemmas 0E2D, 0E2E, and 0E2F. This reduces us to the situation discussed in the next paragraph.
Assume \(R\) is Noetherian and denote \(\varphi : R \to A\) and \(\psi : A \to B\) the given ring maps. Then \(K_{A/R} \cong \varphi^!(R)\) and \(K_{B/A} \cong \psi^!(A)\), see references given above. Then \[K = K_{A/R} \otimes_A^\mathbf{L} K_{B/A} \cong \varphi^!(R) \otimes_A^\mathbf{L} \psi^!(A) \cong \psi^!(\varphi^!(R)) \cong (\psi \circ \varphi)^!(R)\] by Lemmas 0BZU and 0BZT. Thus \(K\) is a relative dualizing complex for \(R \to B\).
It is possible to make the construction canonical: use \(\Omega^n_{P/R}[n]\) instead of \(P[n]\) in the construction and use this in Lemma 0BZJ. The material in this section becomes a lot more involved if one wants to do this.↩︎
If \(f = x^d + a_1 x^{d - 1} + \ldots + a_d\), then \(c_1 = -a_1\), \(c_2 = a_1^2 - a_2\), \(c_3 = -a_1^3 + 2a_1a_2 -a_3\), etc.↩︎