Lemma 10.51.1. 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
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Comment #3028 by Brian Lawrence on
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