# Trying to Prove equivalent definition of Nakayama's Lemma

I have proven the following form of Nakayama's:

Suppose $$M$$ is a finitely-generated module over $$R, J(R)$$ is the Jacobson radical of $$R$$ and $$I\subseteq J(R)$$ where $$I$$ is an ideal. If $$I M=M$$, then $$M=0$$.

However, I want to show that if $$I$$ is an ideal of $$R$$ such that $$IM=M$$ then there exist $$i\in I$$ such that $$im=m$$ for all $$m\in M$$.

Any ideas? Im stuck at this point.

Let $$\mathfrak{m}$$ be a maximal ideal containing $$I$$, then we can see that $$M_{\mathfrak{m}}=0$$. So the support of $$M$$ (which is $$V(I_0)$$ where $$I_0$$ is the annihilator of $$M$$) is disjoint from $$V(I)$$. It means that $$(I,I_0)=R$$ so there is some $$a \in I$$ with $$(1-a) \in I_0$$ ie $$(1-a)M=0$$. Then $$a$$ works.
• $M_{\mathfrak{m}}$ is a finitely generated $R_{\mathfrak{m}}$ module and $\mathfrak{m}M_{\mathfrak{m}} =M_{\mathfrak{m}}$ so (as $\mathfrak{m}R_{\mathfrak{m}}$ is the Jacobson radical of $R_{\mathfrak{m}}$) by Nakayama $M_{\mathfrak{m}}=0$. $V(I_0)$ is the set of prime ideals of $R$ containing $I_0$ (endowed with the Zariski topology). Nov 3, 2021 at 13:42
• No, maximal works fine here: the above means that no maximal ideal $\mathfrak{m}$ containing $I$ may contain $I_0$, and thus $(I,I_0)=R$, hence the conclusion. Nov 3, 2021 at 14:12