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Prove the following result: Let $A$ be a local ring with unique maximal ideal $\mathfrak{m}$. Let $M'$ be a finitely generated A-module. If $N$ is a submodule of $M'$ such that $M'= N+ \mathfrak{m}M'$, then show that $N=M'$.

I tried by considering $M'/N$ and using Nakayama Lemma but I was unable to do it as I don't think it can be proved. $\mathfrak{m}(M'/N) = ((N+ \mathfrak{m}M') /N) = (\mathfrak{m}N + \mathfrak{m}M')/N = (N+\mathfrak{m}M' )/N.$

This question is from my Commutative Algebra assignment and I was unable to solve this question. So, asking for help here. I have been following Atiyah and Macdonald.

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Well you have the argument. Just observe that $\mathfrak{m}(M'/N)=M'/N$.

To see this observe that every element $m'\in M'$, $m'=n+a$ for some $a\in \mathfrak{m}M'$.

So $N+m'=N+n+a=N+a$. Hence $\mathfrak{m}(M'/N)=M'/N$.

Now applying the Nakayama's lemma to $M'/N$, we get $M'/N=0$.

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