Let $R$ be a commutative ring and $M$ is an $R$-module. There are two definitions of the radical of $M$ in the literature. First definition is
$$
\operatorname{Rad}(M)=\bigcap \{N \mid N \text{ is a maximal submodule of } M\},
$$
which gives us a submodule of $M$. The second definition is
$$
\operatorname{rad}(M) = \bigcap \{m \subset R \mid m \text{ is a maximal ideal and }\operatorname{Ann}(M) \subset m\},
$$
which is an ideal. This definition is used, for example, in the book "A term of Commutative Algebra" by Altman and Kleiman.
Is it true that $\operatorname{rad}(M) M = \operatorname{Rad}(M)$?
I see how to show that
$$
\bigcap_{m \text{ is maximal ideal}} mM=\operatorname{Rad}(M),
$$
but I have problems showing that $\operatorname{rad}(M)M = \bigcap_{m \text{ is maximal ideal}} mM$. If this is not correct, what is a counterexample?
Best Answer
As written, the desired equality does not hold, and can in fact fail very acutely! For instance, let $R=\mathbb{Z}$ and $M=\mathbb{Q}$. Then $M$ has no maximal submodules (why?), and so in particular we have $\text{Rad}(M)=M$. However, the annihilator of $M$ is clearly just the ideal $(0)<\mathbb{Z}$, so $\text{rad}(M)$ is the intersection of all maximal ideals of $\mathbb{Z}$, which is $(0)$. Hence $\text{rad}(M)M=\{0\}$, which is certainly not equal to $M$.
For an example where $M$ does have maximal submodules, let $M=\mathbb{Q}\times \mathbb{Z}$, with $R=\mathbb{Z}$ as above. By lemma 2 here, $$\text{Rad}(M)=\text{Rad}(\mathbb{Q})\times\text{Rad}(\mathbb{Z})=\mathbb{Q}\times\{0\},$$ which is not all of $M$. (In particular, $\mathbb{Q}\times(p)$ is a maximal submodule of $M$ for any $p\in\mathbb{Z}$ prime.) However, we again have $\text{Ann}(M)=(0)$ and thus $\text{rad}(M)=(0)$, so this gives another counterexample.