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 $S$.

## 47.7 Injective hull of the residue field

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

**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.**
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 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

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

is a 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 $.

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 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

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.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #7412 by Hanlin on

Comment #7413 by Hanlin on

Comment #7421 by Alex Scheffelin on