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Postmodern Algebra

Sam Macdonald

Section 18.3 Mathis Duality

In the previous worksheet, we encountered the Matlis duality functor in a special setting. We define it the same way in general.

Definition 18.23.

Definition 2.34. Let \((R, \mathfrak{m}, k)\) be a local ring. The Matlis duality functor is \((-)^{\vee}=\operatorname{Hom}_{R}\left(-, E_{R}(k)\right)\text{.}\)
We expect this functor to behave like vector space duality \(\operatorname{Hom}_{k}(-, k)\text{.}\) A more homological motivation for this functor is given by the fact hom into an injective is exact, and the only indecomposable injective into which every module could admit a nonzero map is \(E_{R}(k)\text{.}\)
We hope/expect this to be a duality in some sense. The following lemma gives a few first ways in which this is true.

Proof.

For nonzero \(x \in M\) set \(L=R x \subseteq M\text{,}\) and consider the composed map \(L \rightarrow L / \mathfrak{m} L \cong\) \(K \rightarrow E_{R}(k)\text{.}\) This map extends to a homomorphism of \(R\)-modules \(\epsilon: M \rightarrow E_{R}(k)\) with \(\epsilon(x) \neq 0\text{.}\)
  1. First, we assume that \(\ell(M)\) is finite, and induce on length. Note that \(k\) is the only simple module over \(R\text{,}\) and thus the subquotients in any decomposition series are isomorphic to \(k\text{.}\) Therefore, if \(\ell(M)=1\text{,}\) then \(M \cong k\text{,}\) and the result follows from \(\operatorname{Hom}_{R}(k, E(k)) \cong k\text{.}\) When \(\ell(M) \geqslant 2\text{,}\) the composition series of \(M\) gives an exact sequence
    \begin{equation*} 0 \longrightarrow L \longrightarrow M \longrightarrow k \longrightarrow 0 \end{equation*}
    Since length is additive on exact sequences, \(\ell(L)=\ell(M)-1\text{.}\) By induction hypothesis, \(\ell\left(L^{\vee}\right)=\ell(L)\text{.}\)
  2. On the other hand, since \((-)^{\vee}\) is an exact functor, applying \((-)^{\vee}\) to the short exact sequence above yields a short exact sequence
    \begin{equation*} 0 \longrightarrow k \longrightarrow M^{\vee} \longrightarrow L^{\vee} \longrightarrow 0 \text {. } \end{equation*}
    By additivity of \(\ell\) on this short exact sequence, \(\ell\left(M^{\vee}\right)=\ell\left(L^{\vee}\right)+1\text{.}\) Thus \(\ell\left(M^{\vee}\right)=\ell(L)+1=\) \(\ell(M)\text{.}\)
  3. From the first two parts, we have that \(M \rightarrow M^{\vee \vee}\) is injective, and \(\ell(M)=\ell\left(M^{\vee \vee}\right)\text{,}\) so the map must be surjective as well.

Remark 18.25.

Remark 2.36. The natural map "evaluate at" from \(M\) to \(M^{\vee \vee}\) commmutes with double-duals: given \(f: M \rightarrow N\text{,}\) the diagram
commutes.
Our goal now is to bootstrap up this duality from the finite length case to local rings in general. The following statement on compatibility of injective hulls will be the key.

Proof.

We ahve that \(S / \mathfrak{m} S \cong S \otimes_{R} k\) has finite length, so \(\mathfrak{m} S\) is \(\mathfrak{n}\)-primary. Since some power of \(\mathfrak{n}\) is contained in \(\mathfrak{m} S\text{,}\) and \(E_{R}(k)\) is \(\mathfrak{m}\)-torsion, \(\operatorname{Hom}_{R}\left(S, E_{R}(k)\right)\) is \(\mathfrak{n}\)-torsion as an \(S\)-module. It is also injective by Lemma ??. Thus, by the structure theorem, it is isomorphic to some number of copies of \(E_{S}(\ell)\text{.}\) We know that this number of copies is the \(\ell\)-vector space dimension of \(\operatorname{Hom}_{S}\left(\ell, \operatorname{Hom}_{R}\left(S, E_{R}(k)\right)\right)\text{.}\) Consider the isomorphisms
\begin{equation*} \begin{aligned} \operatorname{Hom}_{S}\left(\ell, \operatorname{Hom}_{R}\left(S, E_{R}(k)\right)\right) & \cong \operatorname{Hom}_{R}\left(\ell \otimes_{S} S, E_{R}(k)\right) \\ & \cong \operatorname{Hom}_{R}\left(\ell \otimes_{k} k, E_{R}(k)\right) \\ & \cong \operatorname{Hom}_{k}\left(\ell, \operatorname{Hom}_{R}\left(k, E_{R}(k)\right)\right) \\ & \cong \operatorname{Hom}_{k}(\ell, k) . \end{aligned} \end{equation*}
The last hom module is a one-dimensional \(\ell\)-vector space, so \(\operatorname{Hom}_{R}\left(S, E_{R}(k)\right) \cong E_{S}(\ell)\text{.}\)

Remark 18.27.

Remark 2.38. Let \((R, \mathfrak{m}, k)\) be a local ring. For each ideal \(I\) in \(R\text{,}\) the theorem above implies that the injective hull of \(k\) over \(R / I\) is \(\left(0:_{E_{R}(k)} I\right)\text{.}\) In particular, since \(E_{R}(k)\) is \(\mathfrak{m}\)-torsion, one has
\begin{equation*} E_{R}(k)=\bigcup_{t}\left(0:_{E_{R}(k)} \mathfrak{m}^{t}\right)=\bigcup_{t} E_{R / \mathfrak{m}^{t}}(k) \end{equation*}

Remark 18.28.

Remark 2.39. Let \((R, \mathfrak{m})\) is local, and \(M\) is an \(\mathfrak{m}\)-torsion \(R\)-module, then \(M\) is naturally a \(\widehat{R}\) module: if \(\mathfrak{m}^{n} x=0\text{,}\) and \(\widehat{r} \in \widehat{R}\text{,}\) take some \(r \in R\) such that \(r-\widehat{r} \in \mathfrak{m}^{n}\text{,}\) and set \(\widehat{r} m=r m\text{.}\) One sees easily that this action is well-defined, and additionally, that any \(R\)-linear endomorphism of \(M\) is also an \(\widehat{R}\)-linear endomorphism.

Proof.

The containment \(k \subseteq E\) is an essential extension of \(R\)-modules, and thus of \(\widehat{R}\)-modules as well. Given an essential extension \(E \subseteq M\) of \(\widehat{R}\)-modules, \(M\) must be \(\mathfrak{m}\)-torsion, since it has the same associated primes (just \(\mathfrak{m}\) ). Therefore, \(R m=\widehat{R} m\) for each \(m \in M\text{,}\) which implies that \(E \subseteq M\) is also an essential extension of \(R\)-modules. Hence \(M=E\text{,}\) and we conclude that \(E=E_{\widehat{R}}(k)\text{.}\)
Since \(E\) is \(\mathfrak{m}\)-torsion, \(\operatorname{Hom}_{\widehat{R}}(E, E)=\operatorname{Hom}_{R}(E, E)\text{,}\) so we may assume that \(\widehat{R}\) is complete for the rest of the proof. Then, we can identify the given map \(R \rightarrow R^{\vee \vee}\) with the natural map \(r \mapsto(\phi \mapsto \phi(r))\text{.}\) Hence, this map is injective.
When \(R\) is artinian, \(\ell(R)<\infty\text{,}\) so the map \(R \rightarrow R^{\vee \vee}\) is an isomorphism.
For each \(i \geqslant 1\text{,}\) set \(R_{i}:=R / \mathfrak{m}^{i}\text{.}\) The \(R_{i}\)-module \(E_{i}=\left(0:_{E} \mathfrak{m}^{i}\right)\) is the injective hull of \(k\) as an \(R_{i}\)-module. For each \(\varphi \in \operatorname{Hom}_{R}(E, E), \varphi\left(E_{i}\right) \subseteq E_{i}\text{,}\) and thus \(\varphi\) restricts to an element of \(\operatorname{Hom}_{R_{i}}\left(E_{i}, E_{i}\right)=R_{i}\text{,}\) where the equality holds because \(R_{i}\) is artinian. Consequently, \(\varphi\) restricted to \(E_{i}\) is multiplication by an element \(\overline{r_{i}} \in R_{i}\text{,}\) with \(r_{i+1}-r_{i} \in \mathfrak{m}^{i}\text{.}\) Thus \(\varphi\) is multiplication by the element \(\varliminf_{\longleftarrow} r_{i}\) in \(R\text{.}\)

Remark 18.30.

Remark 2.41. We have a pretty concrete idea of what any injective module in any noetherian ring looks like now. They are direct sums of modules \(E_{R}(R / \mathfrak{p})\) for primes \(\mathfrak{p}\text{,}\) and \(E_{R}(R / \mathfrak{p}) \cong E_{R_{\mathfrak{p}}}\left(R_{\mathfrak{p}} / \mathfrak{p} R_{\mathfrak{p}}\right) \cong E_{\widehat{R_{\mathfrak{p}}}}\left(\widehat{R_{\mathfrak{p}}} / \mathfrak{p} \widehat{R_{\mathfrak{p}}}\right) \cong\left\{\begin{array}{l}E_{K \llbracket \underline{x} / I}(K) \cong \operatorname{ann}_{I}\left(E_{K \llbracket \rrbracket}(K)\right) \cong \operatorname{ann}_{I}\left(K\left[\underline{\left.\left.x^{-1}\right]\right)}\right.\right. \\ E_{V \llbracket \underline{x} \rrbracket / I}(V / p V) \cong \operatorname{ann}_{I}\left(E_{V \llbracket \underline{x} \rrbracket}(V / p V)\right) \cong \operatorname{ann}_{I}(\star),\end{array}\right.\) where \(K\left[\underline{x^{-1}}\right]\) is the continuous hom module you described in the worksheet, and \(\star\) is the analogous mixed characteristic thing you described in the homework. Note that \(K \cong R_{\mathfrak{p}} / \mathfrak{p} R_{\mathfrak{p}}\text{.}\)

Definition 18.31.

An \(R\)-module \(M\) is artinian if every descending chain of submodules of \(M\) eventually stabilizes.
It is evident that submodules and quotient modules of artinian modules are also artinian. We recall that artinian rings - rings that satisfy the DCC condition on ideals - are necessarily noetherian, and hence have finite length. This is not true for modules.

Proof.

Consider a chain of submodules \(E_{R}(k) \supseteq E_{1} \supseteq E_{2} \supseteq \cdots\text{.}\) Applying the functor \((-)^{\vee}\) yields surjections
\begin{equation*} \widehat{R} \longrightarrow E_{1}^{\vee} \longrightarrow E_{2}^{\vee} \longrightarrow \cdots \end{equation*}
The ideals \(\operatorname{ker}\left(\widehat{R} \longrightarrow E_{i}^{\vee}\right)\) form an ascending chain of ideals, and thus stabilize. Thus \(E_{i}^{\vee} \longrightarrow E_{i+1}^{\vee}\) is an isomorphism for \(i \gg 0\text{.}\) Since \((-)^{\vee}\) is faithful, it follows that \(E_{i}=E_{i+1}\) for \(i \gg 0\text{.}\)

Proof.

The implications \((1) \Rightarrow(2) \Rightarrow(3) \Rightarrow(4)\) are straightforward. To show \((4) \Rightarrow(1)\text{,}\) consider \(x \in M\text{,}\) and the descending chain \(R x \supseteq \mathfrak{m} x \supseteq \mathfrak{m}^{2} x \supseteq \cdots\text{,}\) which stabilizes. Therefore, \(\mathfrak{m}^{t+1} x=\mathfrak{m}^{t} x\) for some \(t\text{,}\) which by NAK implies that \(\mathfrak{m}^{t} x=0\text{.}\) Therefore, \(M\) is \(\mathfrak{m}\)-torsion. Finally, the socle of \(M\) is artinian and a \(k\)-vector space, so it must have finite rank.

Proof.

Write \(E=E_{R}(k)\text{.}\) If \(M\) is noetherian, consider a presentation
\begin{equation*} R^{m} \longrightarrow R^{n} \longrightarrow M \longrightarrow 0 \text {. } \end{equation*}
Applying \((-)^{\vee}\text{,}\) we get an exact sequence
\begin{equation*} 0 \longrightarrow M^{\vee} \longrightarrow E^{n} \longrightarrow E^{m} \text {. } \end{equation*}
The module \(\left(R^{n}\right)^{\vee} \cong E^{n}\) is artinian, hence so is the submodule \(M^{\vee}\text{.}\) Applying \((-)^{\vee}\) again, we get the commutative diagram
and thus \(M \cong M^{\vee \vee}\text{.}\)
Similarly, if \(M\) is artinian, we note that \(R \cong R^{\vee \vee}\) implies \(E \cong R^{\vee} \cong R^{\vee \vee \vee} \cong E^{\vee \vee}\text{.}\) Embed \(M\) into some \(E^{a}\text{;}\) since the cokernel is a quotient of an artinian module, hence artinian, we get a left-exact sequence
\begin{equation*} 0 \longrightarrow M \longrightarrow E^{a} \longrightarrow E^{b} \end{equation*}
As before, we obtain a commutative diagram
and thus \(M \cong M^{\vee \vee}\text{.}\)
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