1
$\begingroup$

Let $S$ be a commutative ring, $R$ a subring of $S$, and $M$ a non-zero $S$-module. If $M$ finitely generated as an $R$-module, do we have that $M$ is finitely generated as an $S$-module, and $S$ is finitely generated as an $R$-module?

I have proven the converse to this statement (i.e. $M$ finitely generated as an $S$-module and $S$ finitely generated as an $R$-module together imply that $M$ is finitely generated as an $R$-module), but I have no idea whether the other statement is true or not. My guess would be no, but I cannot think of a counterexample!

Can anyone provide some tips please?

$\endgroup$
2
$\begingroup$

If $M$ is finitely generated as an $R$-module, then since $R$ is a subring of $S$ we have that $M$ is finitely generated as an $S$-module (we just happen to be able to restrict the coefficients to be only elements of $R$ if we want, which are still elements of $S$). But $S$ need not be finitely generated as an $R$-module.

For example, we could use $S=\mathbb Z^\omega$, $R$ the subring isomorphic to $\mathbb{Z}$ where the elements in each coordinate are the same, $M = \mathbb{Z}$ and $S$ acts on $M$ by multiplying by the first coordinate. Then $M$ is finitely generated as an $R$-module (and hence as an $S$-module) but clearly $S$ is not finitely generated as an $R$-module.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.