Example of Ring Homomorphism such that inverse image of maximal ideal is not maximal.

Question

Give an example of a ring homomorphism $f : R \rightarrow S$ and a maximal ideal $M \unlhd S$ such that $f^{-1}(M) \subset R$ is not a maximal ideal.
The example I came up with is $f : \mathbb{Z} \rightarrow \mathbb{Q}[[x]]$ by $f(x) = x$ and the maximal ideal of $\mathbb{Q}[[x]]$ is $(x)$. Then I obtained that $f^{-1}(M) = \{0\}$ which is not a maximal ideal of $\mathbb{Z}$, take for example $\{0 \} \subset (p) \subset \mathbb{Z}$. Is this a correct example?

Answer 1

Yes, the example is correct (although you should use a different letter for defining $f$, because $x$ is already a particular element of $\mathbb{Q}[[x]]$), but any domain $R$ (not a field) and its field of quotients $F$ provide for an example.

The simplest instance is $\mathbb{Z}\to\mathbb{Q}$. The inverse image of the maximal ideal $(0)$ is the ideal $(0)$ which is not maximal in $\mathbb{Z}$.

This is an instance of a more general situation. Any time you have a commutative ring $R$, having a nonmaximal prime ideal $P$, the localization of $R$ at $P$ has a unique maximal ideal, whose inverse image under the localization map is $P$.