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Let $M$ be a module over a commutative ring $R$. Let $\Omega$ be the set of all maximal ideals of $R$. Prove that $\operatorname{Rad}(M)=\bigcap_{\mathfrak m\in \Omega}\mathfrak mM$, where $\operatorname{Rad}(M)$ is the intersection of maximal submodules of $M$. (This is exercise 15.5, p. 174, from Anderson and Fuller, Rings and Categories of Modules.)

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1 Answer 1

By definition, $\operatorname{Rad}(M)$ equals the intersection of all submodules $N$ of $M$ such that $M/N$ is simple.

Let $N$ be a submodule of $M$ such that $M/N$ is simple. Then there exists $\mathfrak m\in\operatorname{Max}(R)$ such that $M/N\simeq R/\mathfrak m$. In particular, $\mathfrak m(M/N)=0$ and thus $\mathfrak mM\subseteq N$. This shows that $\bigcap_{\mathfrak m\in \Omega}\mathfrak mM\subseteq \operatorname{Rad}(M)$.

For the converse note that $\operatorname{Rad}(M)$ also equals the intersection of all submodules $N$ of $M$ such that $M/N$ is semisimple. Since $M/\mathfrak mM$ is an $R/\mathfrak m$-vectorspace, then $M/\mathfrak mM$ is semisimple and we are done.

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