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.


1 Answer 1


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.

  • $\begingroup$ $M_{\mathfrak{m}}$ 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). $\endgroup$
    – Aphelli
    Nov 3, 2021 at 13:42
  • $\begingroup$ Thanks. Just a minor observation, i think m in your proof should be prime i.e. "m is a prime ideal containing I" since V(I) and V(I_0) are sets of prime ideals. $\endgroup$
    – Jhon Doe
    Nov 3, 2021 at 14:02
  • $\begingroup$ 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. $\endgroup$
    – Aphelli
    Nov 3, 2021 at 14:12

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