**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 10.19.1) 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 $M$.
$\square$

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