Let $R$ be a (commutative unitary) ring, and let $M,N\subset E$ be three $R$-modules. Assume that $N$ is finitely generated, and that there is an ideal $I\subset \operatorname{rad}(N)$ such that $N\subset M+IN$. Show that $N\subset M$.
Lemma. Let $L'\subset L$ be two $R$-modules, with $L/L'$ finitely generated. If an ideal $I\subset\operatorname{rad}L$ satisfies $L=L'+IL$, then $L'=L$.
Obviously we want to solve the exercise using the lemma, with $L:=N+M$ and $L':=M$. So we must check that:
1. $(N+M)/M\cong N/(N\cap M)$ is finitely generated; true because (any quotient of) $N$ is finitely generated.
2. $N+M=M+I(M+N)$, true as $M+I(M+N)=M+IN$; in fact $M+IN\subset M+N$ trivially, while $M+N\subset M+M+IN=M+IN$ by the hypothesis.
3. Finally, I don't understand why $I\subset \operatorname{rad}(M+N)$. If $N\subset M+N$, then $\operatorname{ann}(N)\supset\operatorname{ann}(M+N)$, so if a maximal ideal contains $\operatorname{ann}(N)$ it also contains $\operatorname{ann}(M+N)$, hence $\operatorname{rad}(N)\supset\operatorname{rad}(M+N)$. Since with the opposite inclusion the proof would work, I feel like I am wrong somewhere in this third check, but I can't see where. Thanks for any help