Corollary to Lemma of Nakayama

Question

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.

Answer 1

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?

Answer 2

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