# Corollary to Lemma of Nakayama

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.

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?
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
• I cannot see why the first equality holds: if $I=0$, it seems to me that it does not hold. Apr 27, 2016 at 13:09
• 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. May 4, 2016 at 16:17