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Postmodern Algebra

Sam Macdonald

Section 20.1 Local Duality

Let \(\mathfrak{m}\) be the maximal ideal of \(R\text{.}\) We know that:
  • If \(M\) is a finitely generated \(R\)-module, then \(\mathrm{H}_{\mathfrak{m}}^{i}(M)\) is an artinian module;
  • If \(R=M\) is a polyonomial ring over a field \(k\text{,}\) then \(\mathrm{H}_{\mathfrak{m}}^{d}(M) \cong E_{R}(k)\text{.}\)
The Matlis duals of \(\mathrm{H}_{\mathfrak{m}}^{\bullet}(M)\) are evidently finitely generated modules (if \(R\) is complete), and we might hope that they can be realized by formulas in terms of finitely generated elements.
We will see that this turns out to be true. We will establish formulas of this form for an increasingly general class of rings. Along the way, to understand the new class of rings we will meet, we will pursue a more refined study of injective resolutions.
We will give two proofs of this fact.

Proof.

First, we note that both \(\mathrm{H}_{\mathfrak{m}}^{d}(R)\) and \(E_{R}(k)\) stay the same if \(R\) is replaced by its completion, so we may assume that \(R\) is complete. By Cohen’s Structure Theorem, \(R=K \llbracket \underline{x} \rrbracket, V \llbracket \underline{x} \rrbracket\text{,}\) or \(V \llbracket \underline{x} \rrbracket /(f)\) where \(K\) is a field and \(V\) a DVR.
We have seen this by explicit computation for \(R=k \llbracket \underline{x} \rrbracket\) (worksheet) or \(R=V \llbracket \underline{x} \rrbracket\text{,}\) where \(V\) is a DVR (#2 on HW #2). Also, if \(f \in V \llbracket \underline{x} \rrbracket\text{,}\) then, from the SES \(0 \rightarrow R \rightarrow R \rightarrow R / f R \rightarrow 0\text{,}\) we get the LES
\begin{equation*} \cdots \longrightarrow 0 \longrightarrow \mathrm{H}_{\mathfrak{m}}^{d-1}(V \llbracket \underline{x} \rrbracket /(f)) \longrightarrow \mathrm{H}_{\mathfrak{m}}^{d}(V \llbracket \underline{x} \rrbracket) \stackrel{f}{\longrightarrow} \mathrm{H}_{\mathfrak{m}}^{d}(V \llbracket \underline{x} \rrbracket) \longrightarrow 0, \end{equation*}
where the last terms is zero since \(\operatorname{dim}(R / f R)=d-1\text{.}\) Consequently,
\begin{equation*} \mathrm{H}_{\mathfrak{m}}^{d-1}(V \llbracket \underline{x} \rrbracket /(f))=\operatorname{ann}_{\mathrm{H}_{\mathfrak{m}}^{d}(V \llbracket \underline{x} \rrbracket)}(f)=\operatorname{ann}_{E_{V \llbracket \underline{x} \rrbracket}}(f)=E_{V \llbracket \underline{x} \rrbracket /(f)}(k) . \end{equation*}
Thus, this holds for all complete regular rings.

Proof.

When \(R\) is regular, the following is an injective resolution of \(R\) (by #6 HW #2):
\begin{equation*} 0(\rightarrow R) \rightarrow E_{R}(R) \rightarrow \bigoplus_{\mathrm{ht} \mathfrak{p}=1} E_{R}(R / \mathfrak{p}) \rightarrow \bigoplus_{\mathrm{ht} \mathfrak{p}=2} E_{R}(R / \mathfrak{p}) \rightarrow \cdots \rightarrow \bigoplus_{\mathrm{ht} \mathfrak{p}=d} E_{R}(R / \mathfrak{p}) \rightarrow 0 \end{equation*}
By applying \(\Gamma_{\mathfrak{m}}\) to this resolution, we are only left with \(E(R / \mathfrak{m})\) in the \(d\)-th spot.
If you don’t believe in homework, don’t worry; we will generalize this soon.

Exercise 20.2. Exercise 4.2..

  1. If \(P\) is a flat \(R\)-module, then \(\operatorname{Tor}_{i}^{R}(P, M)=0\) for all \(i>0\) and all \(R\)-modules \(M\text{.}\)
  2. If
    \begin{equation*} \cdots \longrightarrow P_{n} \longrightarrow \cdots \longrightarrow P_{1} \longrightarrow P_{0} \longrightarrow(M \longrightarrow) 0 \end{equation*}
    is exact and each \(P_{i}\) is flat (we say this is a flat resolution of \(M\) ), then \(\operatorname{Tor}_{i}^{R}(M, N)=\) \(\mathrm{H}_{i}\left(P_{\bullet} \otimes_{R} N\right)\) for all \(i\text{.}\)

Proof.

Let \(x_{1}, \ldots, x_{d}\) be an SOP for \(R\text{.}\) The augmented Čech complex
Table 20.4.
flat resolution \(\mathrm{d}\) \(\mathrm{d}-1\) 0
\(\begin{array}{lllll}\text { Čech indexing } & 0 & 1 & \mathrm{~d}\end{array}\)
\begin{equation*} 0 \longrightarrow R \longrightarrow \bigoplus_{i} R_{x_{i}} \longrightarrow \cdots \longrightarrow R_{x_{1} \cdots x_{d}} \longrightarrow\left(\mathrm{H}_{\mathfrak{m}}^{d}(R) \longrightarrow\right) 0 \end{equation*}
is exact, since the lower local cohomologies vanish. Thus, \(\check{C}^{\bullet}(\underline{x} ; R)\) is a flat resolution of \(\mathrm{H}_{\mathfrak{m}}^{d}(R)\text{.}\) Then,
\begin{equation*} \mathrm{H}_{\mathfrak{m}}^{i}(M)=\check{H}^{i}\left(\check{C}^{\bullet}(\underline{x} ; M)\right)=\check{H}^{i}\left(\check{C}^{\bullet}(\underline{x} ; R) \otimes_{R} M\right)=\operatorname{Tor}_{d-i}^{R}\left(\mathrm{H}_{\mathfrak{m}}^{d}(R), M\right) \end{equation*}
We note that there is a switch from cohomological indexing of Čech complex and homological indexing of a flat resolution.

Proof.

Let \(P_{\bullet}(\longrightarrow M)\) be a projective resolution of \(M\text{.}\)
\begin{equation*} \operatorname{Tor}_{i}^{R}(M, N)^{\vee} \cong \mathrm{H}_{i}\left(P_{\bullet} \otimes N\right)^{\vee} \cong \mathrm{H}_{i}\left(\left(P_{\bullet} \otimes N\right)^{\vee}\right) \cong \mathrm{H}_{i}\left(\operatorname{Hom}\left(P_{\bullet}, N^{\vee}\right)\right) \cong \operatorname{Ext}_{R}^{i}\left(M, N^{\vee}\right) \end{equation*}
Given a map \(M \rightarrow M^{\prime}\text{,}\) we can lift to a map on projective resolutions, and it is easy to see that we get commuting maps through all of the isomorphisms above.
  1. By taking a minimal resolution, we may assume that \(P_{\bullet}(\longrightarrow M)\) is a complex of finitely generated free \(R\)-modules. In this case, there is a natural isomorphism (exercise!)
\begin{equation*} P_{\bullet} \otimes \operatorname{Hom}_{R}(N, E) \cong \operatorname{Hom}_{R}\left(\operatorname{Hom}_{R}\left(P_{\bullet}, N\right), E\right) \end{equation*}
Taking homology gives the isomorphism.
To help remember which hypothesis goes with which situation in the following, one might keep in mind that Matlis duals of artinian modules are complete.

Proof.

\begin{equation*} \mathrm{H}_{\mathfrak{m}}^{i}(M) \cong \operatorname{Tor}_{d-i}^{R}\left(M, \mathrm{H}_{\mathfrak{m}}^{d}(R)\right) \cong \operatorname{Tor}_{d-i}^{R}\left(M, R^{\vee}\right) \cong \operatorname{Ext}_{R}^{d-i}(M, R)^{\vee} . \end{equation*}
  1. Since \(R\) is complete, \(\mathrm{H}_{\mathfrak{m}}^{d}(R)^{\vee} \cong R^{\vee \vee} \cong R\text{.}\) Then,
\begin{equation*} \mathrm{H}_{\mathfrak{m}}^{i}(M)^{\vee} \cong \operatorname{Tor}_{d-i}^{R}\left(M, \mathrm{H}_{\mathfrak{m}}^{d}(R)\right)^{\vee} \cong \operatorname{Ext}_{R}^{d-i}\left(M, \mathrm{H}_{\mathfrak{m}}^{d}(R)^{\vee}\right) \cong \operatorname{Ext}_{R}^{d-i}(M, R) \end{equation*}
One often finds the just the first statement as local duality, but the second "opposite of local duality" is quite useful as well.

Remark 20.7.

Remark 4.6. By the exact same proof, if \(R\) is Cohen-Macaulay and \(W\) is some finitely generated module such that \(W^{\vee} \cong \mathrm{H}_{\mathfrak{m}}^{d}(R)\text{,}\) then
  • If \(M\) is a finitely generated \(R\)-module, then \(\mathrm{H}_{\mathfrak{m}}^{i}(M)=\operatorname{Ext}_{R}^{d-i}(M, W)^{\vee}\text{.}\)
  • If \(R\) is complete, then \(\mathrm{H}_{\mathfrak{m}}^{i}(M)^{\vee}=\operatorname{Ext}_{R}^{d-i}(M, W)\) by the same proof.
We will return to this later.
Local duality is a powerful tool to study local cohomology modules. Part of its power comes from its flexibility: there are many different rings, modules, and indices to plug in. Here are some special cases:

Example 20.8.

Example 4.7. Let \((R, \mathfrak{m})\) be a local ring, and \((S, \mathfrak{n})\) another local ring such that \(S\) is regular, and \(R\) is a quotient of \(S\text{.}\) (Such an \(S\) exists whenever \(R\) is complete, or essentially of finite type over a field or \(\mathbb{Z}\text{.}\)) Then, since \(R\) is a finitely generated \(S\)-module, \(\mathrm{H}_{\mathfrak{m}}^{i}(R) \cong \mathrm{H}_{\mathfrak{n}}^{i}(R) \cong \operatorname{Ext}_{S}^{\operatorname{dim}(S)-i}(R, S)^{\vee}\) for all \(i\text{.}\)

Example 20.9.

Example 4.8. Let \((R, \mathfrak{m}) \rightarrow(S, \mathfrak{n})\) be a map of complete local rings, and suppose that \(R\) is regular, e.g., a power series ring over a field \(K\text{.}\) Then, \(\mathrm{H}_{\mathfrak{m}}^{\operatorname{dim}(R)}(S)^{\vee} \cong \operatorname{Hom}_{R}(S, R)\text{.}\)
Here is an important application of Local Duality. Note that we are not assuming that \(R\) is regular in the statement.

Proof.

We already know \(\leqslant\text{.}\) Without loss of generality, we can assume that \(M\) is a faithful module, by invariance of base. Complete; \(\widehat{M}\) is \(\widehat{R}\)-faithful, \(\operatorname{dim}(M)=\operatorname{dim}(\widehat{M})\text{,}\) and \(\mathrm{H}_{\mathfrak{m}}^{\bullet}(M) \cong \mathrm{H}_{\mathfrak{m} \widehat{R}}^{\bullet}(\widehat{M})\text{.}\) Therefore, we may assume without loss of generality that \(R\) and \(M\) are complete.
Now \(R\) is a quotient of a power series ring, so by invariance of base we may assume without loss of generality that \(R\) is a regular ring. Note, however, that \(M\) is not necessarily faithful anymore.
We claim that if the depth of \(\operatorname{ann}(M)\) on an \(R\)-module \(N\) is \(t\text{,}\) then \(\operatorname{Ext}_{R}^{<t}(M, N)=0\text{.}\) This is essentially the same as Rees’s Theorem, but we include the argument anyway. Indeed, by induction on the depth (if positive), we can assume that \(\operatorname{Ext}_{R}^{<t-1}(M, N / x N)=0\) for \(x\) an \(N\)-regular element in \(\operatorname{ann}(M)\text{.}\) From the usual LES, we see that, for \(i<t, x\) acts injectively on \(\operatorname{Ext}_{R}^{i}(M, N)=0\text{,}\) but \(x\) annihilates this module as well, so it must be the zero module. This establishes the claim.
Now, by Local Duality, \(\mathrm{H}_{\mathfrak{m}}^{i}(M) \cong \operatorname{Ext}_{R}^{\operatorname{dim}(R)-i}(M, R)^{\vee}\text{.}\) Since Matlis Duality is faithful, the indicated Ext is nonzero if and only if its dual is. Thus,
\begin{equation*} \begin{aligned} \operatorname{cd}(\mathfrak{m}, M)=\max \left\{j \mid \mathrm{H}_{m}^{j}(M) \neq 0\right\} & =\max \left\{\operatorname{dim}(R)-i \mid \operatorname{Ext}_{R}^{\operatorname{dim}(R)-i}(M, R) \neq 0\right\} \\ & =\operatorname{dim}(R)-\min \left\{i \mid \operatorname{Ext}_{R}^{i}(M, R) \neq 0\right\} \\ & \geq \operatorname{dim}(R)-\operatorname{depth}_{\operatorname{ann}(M)}(R) \\ & \geq \operatorname{dim}(R)-\operatorname{height}(\operatorname{ann}(M)) \\ & \geq \operatorname{dim}(R / \operatorname{ann}(M))=\operatorname{dim}(M) . \end{aligned} \end{equation*}
Since \(\operatorname{dim}(M) \geq \operatorname{cd}(\mathfrak{m}, M)\text{,}\) equality holds throughout.

Remark 20.11.

Remark 4.10. A consequence of this argument, by taking \(M=R / I\text{,}\) is that if \(R\) is a regular local ring (or any ring for which Local Duality holds!) there are equalities for any ideal \(I \subset R\) :
\begin{equation*} \operatorname{depth}_{I}(R)=\operatorname{height}(I)=\operatorname{dim}(R)-\operatorname{dim}(R / I) \end{equation*}
Of course, these can be obtained more directly (for Cohen-Macaulay rings).
Following Remark ??, we should be able to generalize local duality, so that we can compute local cohomology as duals of Ext modules over rings that are more relevant. The best generalization would be in the case of a Cohen-Macaulay local ring \(R\) such that \(R^{\vee} \cong \mathrm{H}_{\mathfrak{m}}^{d}(R)\text{.}\) Failing that, we would like to understand when a f.g. module \(K\) such that \(K^{\vee} \cong H_{\mathfrak{m}}^{d}(R)\) exists over a CM local ring, and how else we might recognize such a \(K\text{.}\)
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