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.
