# If $M$ finitely generated as an $R$-module, is $M$ is finitely generated as an $S$-module, and $S$ is finitely generated as an $R$-module?

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?

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.