## 47.7 Injective hull of the residue field

Most of our results will be for Noetherian local rings in this section.

Lemma 47.7.1. 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 $E[I] = \mathop{\mathrm{Hom}}\nolimits _ R(S, E)$ as $S = R/I$. Hence $E[I]$ is an injective $S$-module by Lemma 47.3.4. 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. $\square$

Lemma 47.7.2. 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 47.2.4) we conclude that the kernel of $M \to E^{\oplus n}$ is zero. $\square$

Lemma 47.7.3. 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 47.7.1. We have $E = \bigcup E_ n$ because $\bigcup E_ n = E[\mathfrak m^\infty ]$ is an injective $R$-submodule which contains $\kappa$, see Lemma 47.3.9. $\square$

The following lemma tells us the injective hull of the residue field of a Noetherian local ring only depends on the completion.

Lemma 47.7.4. 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. 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 47.3.2. Choose an extension $E_ R \to E_ S$ of the identification of residue fields. This map is an isomorphism by Lemma 47.7.3 because $R \to S$ induces an isomorphism $R/\mathfrak m_ R^ n \to S/\mathfrak m_ S^ n$ for all $n$. $\square$

Lemma 47.7.5. Let $(R, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $E$ be an injective hull of $\kappa$ over $R$. Then $\mathop{\mathrm{Hom}}\nolimits _ R(E, E)$ is canonically isomorphic to the completion of $R$.

Proof. Write $E = \bigcup E_ n$ with $E_ n = E[\mathfrak m^ n]$ as in Lemma 47.7.3. Any endomorphism of $E$ preserves this filtration. Hence

$\mathop{\mathrm{Hom}}\nolimits _ R(E, E) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ R(E_ n, E_ n)$

The lemma follows as $\mathop{\mathrm{Hom}}\nolimits _ R(E_ n, E_ n) = \mathop{\mathrm{Hom}}\nolimits _{R/\mathfrak m^ n}(E_ n, E_ n) = R/\mathfrak m^ n$ by Lemma 47.6.2. $\square$

Lemma 47.7.6. 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

$\mathop{\mathrm{Hom}}\nolimits _ R(E, E) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_1, E) \to \mathop{\mathrm{Hom}}\nolimits _ R(M_2, E) \to \ldots$

is sequence of surjections. By Lemma 47.7.5 each of these is a module over the completion $R^\wedge = \mathop{\mathrm{Hom}}\nolimits _ R(E, E)$. Since $R^\wedge$ is Noetherian (Algebra, Lemma 10.97.6) the sequence stabilizes: $\mathop{\mathrm{Hom}}\nolimits _ R(M_ n, E) = \mathop{\mathrm{Hom}}\nolimits _ R(M_{n + 1}, E) = \ldots$. Since $E$ is injective, this can only happen if $\mathop{\mathrm{Hom}}\nolimits _ R(M_ n/M_{n + 1}, E)$ 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 47.7.3. In this case $M_ n/M_{n + 1}$ has a nonzero map into $E$, contradicting the assumed vanishing. This finishes the proof. $\square$

Lemma 47.7.7. Let $(R, \mathfrak m, \kappa )$ be a Noetherian local ring. Let $E$ be an injective hull of $\kappa$.

1. For an $R$-module $M$ the following are equivalent:

1. $M$ satisfies the ascending chain condition,

2. $M$ is a finite $R$-module, and

3. there exist $n, m$ and an exact sequence $R^{\oplus m} \to R^{\oplus n} \to M \to 0$.

2. For an $R$-module $M$ the following are equivalent:

1. $M$ satisfies the descending chain condition,

2. $M$ is $\mathfrak m$-power torsion and $\dim _\kappa (M[\mathfrak m]) < \infty$, and

3. 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 10.20.1) 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 47.7.2 we see that $M$ is a submodule of $E^{\oplus n}$ for some $n$. Consider the quotient $N = E^{\oplus n}/M$. By Lemma 47.7.6 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 47.7.2 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 47.7.6 so does $M$. $\square$

Proposition 47.7.8 (Matlis duality). Let $(R, \mathfrak m, \kappa )$ be a complete local Noetherian ring. Let $E$ be an injective hull of $\kappa$ over $R$. The functor $D(-) = \mathop{\mathrm{Hom}}\nolimits _ R(-, E)$ 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 47.7.5 we have $R = \mathop{\mathrm{Hom}}\nolimits _ R(E, E) = D(E)$. Of course we have $E = \mathop{\mathrm{Hom}}\nolimits _ R(R, E) = D(R)$. Since $E$ is injective the functor $D$ is exact. The result now follows immediately from the description of the categories in Lemma 47.7.7. $\square$

Remark 47.7.9. 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 $M = \mathop{\mathrm{Hom}}\nolimits _ R(N, E)$ 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 $N \supset N' \mapsto M' = \mathop{\mathrm{Hom}}\nolimits _ R(N/N', E) \subset M$ is an inclusion preserving map from submodules of $N$ to submodules of $M$, whence the conclusion.

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