# A local ring with unique maximal ideal M and M' be a finitely generated A-module

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.

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 $$M'/N=0$$.