Say $0 \to M_1 \to M_2 \to M_3 \to 0$ is a short exact sequence and each $M_i$ is an $R$-module. If $M_2$ is finitely generated, then so also is $M_1$?
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No – unless $M_2$ is a noetherian $R$-module, e.g. if $R$ is a noetherian ring. – Bernard Oct 10 '16 at 08:44
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Your question can be rephrased to the question: Is a submodule of a finitely generated module again finitely generated, because for $M_1\subset M_2$ the sequence $$ 0\rightarrow M_1 \rightarrow M_2 \rightarrow M_1/ M_2\rightarrow 0 $$ is always exact.
This is known to be false in general. See the following question which has stronger assumptions on the ring and the module and it is still false.
Is the submodule of a finitely generated free module finitely generated?
