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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?

Alex
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  • Are you sure you got the first definition correct? Isn't it $ \operatorname{Rad}(M)=\bigcap {\textbf{Ann}(N) \mid N \text{ is a maximal submodule of } M}$, so that it is also an ideal? (The question of whether they are equal is then still an issue of course) $ – Vincent Jan 11 '21 at 16:44
  • (bold face for emphasis, not part of standard notation) – Vincent Jan 11 '21 at 16:45
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    Yes, I'm sure that first definition is what I want. This definition can be found in many non-commutative ring theory books and also in wiki https://en.wikipedia.org/wiki/Radical_of_a_module – Alex Jan 11 '21 at 18:44
  • Hmm, interesting! I only knew radicals as ideals – Vincent Jan 11 '21 at 18:56

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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.

Atticus Stonestrom
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    I understand your example, $\mathbb{Q}$ has no maximal submodules because there are no maps $\mathbb{Q} \to \mathbb{F}_p$. But to be honest, I want to see more examples of this sort, where there are maximal submodules in $M$. – Alex Jan 11 '21 at 18:48
  • @Alex indeed, that argument works! re: your question, you can actually extend the counterexample $\mathbb{Q}$ to a counterexample that does have maximal submodules fairly easy; I've given an example above, hopefully it's helpful :) – Atticus Stonestrom Jan 11 '21 at 19:06