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In Matsumura's Commutative Algebra there is the following Corollary to the Lemma of Nakayama:

Let $A$ be a ring, $M$ an $A$-module, $N$ and $N'$ submodules of $M$, and $I$ an ideal of $A$. Suppose that $M=N+IN'$, and that either:

  1. $I$ is nilpotent
  2. $I\subseteq \mathrm{rad}(A)$ and $N'$ is finitely generated.

Then $M=N$.

In both cases the idea is to prove that $M/N=I(M/N)$ and I am struggling with that.

We have that $M/N=(N+IN')/N=IN'/(IN'\cap N)$ and I do not know how to take it from here.

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2 Answers 2

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  1. Since $M=N+IN'\subseteq N+IM$ you get $M=N+IM$. Can you continue from here?

  2. Since $M=N+IN'\subseteq N+N'$ you get $M=N+N'$. Can you see now why $M/N$ is finitely generated?

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Hint: if $I$ is nilpotent, it is contained in the Jacobson radical, you deduce that $N+J(A)N'=N+IN'=M$. You have $I\subset J(A)$, thus $IN'\subset J(A)N'$ and $M=N+IN'\subset N+J(A)N'\subset M$. You can apply the Nakayama lemma. statement 3

https://en.wikipedia.org/wiki/Nakayama_lemma#Statement

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  • $\begingroup$ I cannot see why the first equality holds: if $I=0$, it seems to me that it does not hold. $\endgroup$
    – Iulia
    Apr 27, 2016 at 13:09
  • $\begingroup$ Have you proved $M=N+J(A)M$ in order to use the linked version of NAK? Moreover, in the first case $M/N$ is not necessarily finitely generated. $\endgroup$
    – user26857
    May 4, 2016 at 16:17

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