Let ${A}$ be a regular local (noetherian) ring with maximal ideal ${\mathfrak{m}}$ and residue field ${k}$. The purpose of this post is to construct an equivalence (in fact, a duality)

$\displaystyle \mathbb{D}: \mathrm{Mod}_{\mathrm{sm}}(A) \simeq \mathrm{Mod}_{\mathrm{sm}}(A)^{op}$

between the category ${\mathrm{Mod}_{\mathrm{sm}}(A)}$ of finite length ${A}$-modules (i.e., finitely generated modules annihilated by a power of ${\mathfrak{m}}$) and its opposite. Such an anti-equivalence holds in fact for any noetherian local ring ${A}$, but in this post we will mostly stick to the regular case. In the next post, we’ll use this duality to give a description of the local cohomology groups of a noetherian local ring. Most of this material can be found in the first couple of sections of SGA 2 or in Hartshorne’s Local Cohomology.

1. Duality in the derived category

Let ${A}$ be any commutative ring, and let ${\mathrm{D}_{\mathrm{perf}}(A)}$ be the perfect derived category of ${A}$. This is the derived category (or preferably, derived ${\infty}$-category) of perfect complexes of ${A}$-modules: that is, complexes containing a finite number of projectives. ${\mathrm{D}_{\mathrm{perf}}(A)}$ is the smallest stable subcategory of the derived category containing the complex ${A}$ in degree zero, and closed under retracts. It can also be characterized abstractly: ${\mathrm{D}_{\mathrm{perf}}(A)}$ consists of the compact objects in the derived category of ${A}$. That is, a complex ${X}$ is quasi-isomorphic to something in ${\mathrm{D}_{\mathrm{perf}}(A)}$ if and only if the functor

$\displaystyle \hom(X, \cdot) : \mathrm{D}(A) \rightarrow \mathbf{Spaces}$

commutes with homotopy colimits. (“Chain complexes” could replace “spaces.”)

Now, ${\mathrm{D}_{\mathrm{perf}}(A)}$ is a symmetric monoidal ${\infty}$-category under (derived) tensor product, and as it happens here we have dual objects. There is a duality

$\displaystyle D: \mathrm{D}_{\mathrm{perf}}(A) \simeq \mathrm{D}_{\mathrm{perf}}(A)^{op}$

sending a complex ${X}$ to its dual ${\mathbf{Hom}(X, A)}$ (where the bolded ${\mathbf{Hom}}$ means the internal mapping object: that is, the mapping chain complex). Given a symmetric monoidal stable ${\infty}$-category, this sort of duality on compact objects is a frequent occurrence. For instance, in the case of spectra (i.e., modules over the sphere spectrum), the compact objects are the finite spectra, and the relevant duality is Spanier-Whitehead duality. The analog in the 1-categorical case might be duality for finitely generated projective modules over a ring.

The rough idea is that there is a natural map

$\displaystyle X \rightarrow \mathbf{Hom}(\mathbf{Hom}(X, A)) ,$

for ${X}$ an arbitrary complex, adjoint to the evaluation map ${X \otimes \mathbf{Hom}(X, A) \rightarrow A}$. This is an equivalence for ${X = A}$, and thus on the smallest ${\infty}$-category containing ${A}$ closed under finite limits and colimits and retracts: that is, on the perfect derived category.

2.The duality functor as an Ext

Now let’s return to commutative algebra. We saw in the previous section that there is an equivalence

$\displaystyle \mathrm{D}_{\mathrm{perf}}(A) \simeq \mathrm{D}_{\mathrm{perf}}(A)^{op}$

for any commutative ring ${A}$. We want to turn this into a statement about the ordinary 1-categories ${\mathrm{Mod}_{\mathrm{sm}}(A)}$, when ${A}$ is a regular local ring.

So suppose that ${A}$ is regular local. Then every ${A}$-module determines an object in the derived category of ${A}$, and a finitely generated ${A}$-module in fact lives in ${\mathrm{D}_{\mathrm{perf}}(A)}$. This is not formal: it follows from the fact that ${A}$ has finite global dimension. In particular, we have a (fully faithful) inclusion of categories

$\displaystyle \mathrm{Mod}_{\mathrm{sm}}(A) \subset \mathrm{D}_{\mathrm{perf}}(A).$

A complex ${X_\bullet}$ belongs to ${\mathrm{Mod}_{\mathrm{sm}}(A)}$ if and only if it is homologically concentrated in degree zero and the group ${\pi_0 X_\bullet}$ is annihilated by a power of ${\mathfrak{m} }$.

The point of the present post is that the functor

$\displaystyle \mathrm{Mod}_{\mathrm{sm}}(A) \subset \mathrm{D}_{\mathrm{perf}}(A) \stackrel{D}{\rightarrow} \mathrm{D}_{\mathrm{perf}}(A)^{op}$

has its image contained in a shift of ${\mathrm{Mod}_{\mathrm{sm}}(A)}$. In fact, let ${M \in \mathrm{Mod}_{\mathrm{sm}}(A)}$; we have to show that the complex ${\mathbf{Hom}(M, A)}$ is homologically concentrated in one degree and is ${\mathfrak{m}}$-power torsion there. However, we observe that

$\displaystyle \pi_i( \mathbf{Hom}(M, A)) = \pi_0 \hom ( M[i], A) = \mathrm{Ext}^{-i}(M, A),$

for all ${i}$.

We now invoke:

Proposition 1 Let ${A}$ be regular local of dimension ${d}$. Then if ${M \in \mathrm{Mod}_{\mathrm{sm}}(A)}$, we have

$\displaystyle \mathrm{Ext}^i( M, A) = 0 \quad \text{if} \ i \neq d.$

Moreover, ${\mathrm{Ext}^d(M, A) \in \mathrm{Mod}_{\mathrm{sm}}(A)}$.

When ${i < d}$, this is a consequence of the fact that ${A}$ has depth ${d}$, and the description of depth in terms of vanishing of ${\mathrm{Ext}}$ groups. When ${i > d}$, the vanishing is a consequence of the fact that ${A}$ has global dimension ${d}$. The fact that ${\mathrm{Ext}^d(M, A)}$ is annihilated by a power of ${\mathfrak{m}}$ follows from the fact that ${M}$ is.

It follows that the duality functor ${\mathbf{Hom}(\cdot, A)}$ sends ${\mathrm{Mod}_{\mathrm{sm}}(A)}$ into the category ${\mathrm{Mod}_{\mathrm{sm}}(A) [-d] \subset \mathrm{D}_{\mathrm{perf}}(A)}$.

Combining these observations, we get a fully faithful duality functor ${\mathbb{D}}$ fitting into the above diagram

Concretely, it is given by

$\displaystyle \mathbb{D}: M \mapsto \mathrm{Ext}^d(M, A).$

The last thing we need to check is that ${\mathbb{D}}$ actually induces an equivalence. To see this, we observe that its inverse is again ${\mathbb{D}}$: this follows from the fact that ${\mathbf{Hom}(\cdot, A)}$ was a duality functor.

We can conclude:

Theorem 2 The duality functor ${\mathbb{D}: M \mapsto \mathrm{Ext}^d(M, A)}$ induces an equivalence (in fact, a duality) ${\mathrm{Mod}_{\mathrm{sm}}(A) \simeq \mathrm{Mod}_{\mathrm{sm}}(A)^{op}}$.

Our next goal is to describe this functor in another way.

3. The dualizing module

Observe that this functor ${\mathbb{D}}$ is actually ind-representable, because it is exact (being an equivalence). In other words, there is an ${\mathrm{Ind} }$-object ${Q}$ of ${\mathrm{Mod}_{\mathrm{sm}}(A)}$ (that is, an ${\mathfrak{m}}$-torsion module which is not necessarily finitely generated) such that

$\displaystyle \mathbb{D} M = \hom(M, Q) , \quad M \in \mathrm{Mod}_{\mathrm{sm}}(A).$

This follows from a suitable version of the adjoint functor theorem and ${\mathrm{Ind}}$-ization. We can actually write down this module very explicitly. Namely, we can take

$\displaystyle Q = \varinjlim \mathbb{D} A/\mathfrak{m}^i.$

This is an ${\mathfrak{m}}$-torsion module. Once one knows that ${\mathbb{D}}$ is ${\mathrm{Ind}}$-representable, it is straightforward that the representing object must be as written.

Observe that it satisfies ${\hom_A(k, Q) = k}$, because ${\hom_A(k, Q) = \mathbb{D} k}$ must be a simple object of ${\mathrm{Mod}_{\mathrm{sm}}(A)}$ (i.e., ${k}$). Below, we will give a characterization of ${Q}$ in terms of ${k}$.

We will need:

Proposition 3 ${Q}$ is an injective ${A}$-module: in fact, it is the injective hull of ${k}$.

Proof: We know that ${\mathbb{D}}$ is exact on the category ${\mathrm{Mod}_{\mathrm{sm}}(A)}$, from which we can see that ${Q}$ is an injective in the category ${\mathrm{Ind}(\mathrm{Mod}_{\mathrm{sm}}(A))}$ of ${\mathfrak{m}}$-torsion ${A}$-modules. In fact, injectivity of ${Q}$ will follow from the following result: the functor

$\displaystyle \mathrm{Ind}(\mathrm{Mod}_{\mathrm{sm}}(A)) \rightarrow \mathrm{Mod}(A)$

preserves injective objects. That is, an ${\mathfrak{m}}$-torsion module which has the relevant extension property for monomorphisms in ${\mathrm{Mod}_{\mathrm{sm}}(A)}$ is actually an injective ${A}$-module.

This is not (as far as I can tell) completely formal, and I’ll just sketch the proof. Given a module ${I}$ which is injective in ${\mathrm{Ind}(\mathrm{Mod}_{\mathrm{sm}}(A))}$, and given an injection of finitely generated ${A}$-modules ${E \hookrightarrow F}$, any map

$\displaystyle E \rightarrow I$

factors through ${E/(\mathfrak{m}^t F \cap E) \rightarrow I}$ for some ${t \gg 0}$, in view of the Artin-Rees lemma. Now we can extend the map

$\displaystyle E/(\mathfrak{m}^t F \cap E) \rightarrow I$

to a map ${F/\mathfrak{m}^t F \rightarrow I}$, by hypothesis. This proves injectivity.

Next we need to see that ${Q}$ is actually the injective hull of ${k}$. In other words, we have to see that the map ${k \rightarrow Q}$ (which is well-defined up to scaling) is essential: that is, any nonzero submodule of ${Q}$ intersects ${k}$ nontrivially. We can reduce to the finitely generated (hence small) case, so we need to show that for an injection ${W \rightarrow \varinjlim \mathbb{D} A/\mathfrak{m}^i}$, the fibered product

is nonempty. Suppose ${W \rightarrow Q}$ factors through some ${\mathbb{D} A/\mathfrak{m}^N}$, ${N \gg 0}$. Then we might as well take the fibered product

which is dual, under the anti-equivalence ${\mathbb{D}}$, to the push-out square

Now, by duality, ${A/\mathfrak{m}^i \rightarrow \mathbb{D} W}$ is a surjection, and by defintion the map ${A/\mathfrak{m}^i \rightarrow \mathbb{D} k = k}$ is the reduction map. It follows that the push-out is a quotient of ${A/\mathfrak{m}^i}$ by an ideal contained in ${\mathfrak{m}}$, so that ${\mathbb{D}(W \cap k)}$ is not zero (in fact, it is ${k}$). This means ${W \cap k \neq 0}$, as desired. $\Box$

So this means that we’ve gotten a fairly concrete result. Given a regular local ring ${A}$, we can consider the injective envelope ${Q}$ of the residue field; then ${\hom(\cdot, Q)}$ determines an anti-equivalence between ${\mathrm{Mod}_{\mathrm{sm}}(A) }$ and itself. We could have, in fact, proved this directly, and in fact:

Proposition 4 Let ${A}$ be a noetherian local ring (not necessarily regular) and let ${Q}$ be the injective envelope of the residue field ${k}$. Then ${\hom(\cdot, Q)}$ defines a duality ${\mathrm{Mod}_{\mathrm{sm}}(A) \simeq \mathrm{Mod}_{\mathrm{sm}}(A)^{op}}$.

In fact, we need to check that the canonical biduality map

$\displaystyle t_M: M \rightarrow \hom(\hom(M, Q), Q)$

is an isomorphism if ${M \in \mathrm{Mod}_{\mathrm{sm}}(A)}$. Both functors are exact, as ${Q}$ is injective, and the collection of ${M}$ such that ${t_M}$ is an isomorphism is stable under extensions. Consequently it suffices to check that ${t_M}$ is an isomorphism when ${M = k}$. Then it follows from the definition of an injective hull: ${\hom(k, Q) \simeq k}$. (The same sort of argument can be used to show that $\mathbb{D}$ actually takes values in finite length modules.)

In the next post, we’ll apply this to a calculation of the local cohomology groups of ${A}$.