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 10.96.6) 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. $\square$

## 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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08Z7. Beware of the difference between the letter 'O' and the digit '0'.