Lemma 10.50.1. Let $R$ be a Noetherian ring. Any finite $R$-module is of finite presentation. Any submodule of a finite $R$-module is finite. The ascending chain condition holds for $R$-submodules of a finite $R$-module.

Proof. We first show that any submodule $N$ of a finite $R$-module $M$ is finite. We do this by induction on the number of generators of $M$. If this number is $1$, then $N = J/I \subset M = R/I$ for some ideals $I \subset J \subset R$. Thus the definition of Noetherian implies the result. If the number of generators of $M$ is greater than $1$, then we can find a short exact sequence $0 \to M' \to M \to M'' \to 0$ where $M'$ and $M''$ have fewer generators. Note that setting $N' = M' \cap N$ and $N'' = \mathop{\mathrm{Im}}(N \to M'')$ gives a similar short exact sequence for $N$. Hence the result follows from the induction hypothesis since the number of generators of $N$ is at most the number of generators of $N'$ plus the number of generators of $N''$.

To show that $M$ is finitely presented just apply the previous result to the kernel of a presentation $R^ n \to M$.

It is well known and easy to prove that the ascending chain condition for $R$-submodules of $M$ is equivalent to the condition that every submodule of $M$ is a finite $R$-module. We omit the proof. $\square$

## Comments (2)

Comment #3028 by Brian Lawrence on

Suggested slogan: Over a Noetherian ring: Any finite module is of finite presentation, any submodule of a finite module is finite, and the ascending chain condition holds for any finite module.

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