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In Corollary 1.2.3, in order to apply Nakayama's lemma, how to show $M$ is a finitely generated $A$-module?

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    $\begingroup$ $M$ is not necessarily f.g. as $A$-module! But if $M=mM$, then $M=nM$. $\endgroup$ – user26857 Sep 11 at 10:14

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