Showing the inverse map of a ring homomorphism of a prime ideal is again a prime ideal

Question

Let $\phi : A \rightarrow B$ be a ring homomorphism and $I$ be a prime ideal of $B$.
(i) Show that $\phi^{-1}(I)$ is a prime ideal of $A$, and
(ii) find an example of $A$, $B$ and $I$ so that $I$ is a maximal ideal, but $\phi^{-1}(I)$ is not (a maximal ideal).

(i)

Let $x,y \in A$ with $xy \in \phi^{-1}(I)$ $\Rightarrow \exists z \in I: \phi(x) \phi(y) = \phi(xy) = z \in I$ $\Leftrightarrow x \in \phi^{-1}(I)$ or $y \in \phi^{-1}(I) \Rightarrow \phi^{-1}(I)$ is a prime ideal of A.

Can you please check my answer, is it (formally) correct?

(ii)

I have no clue how the example could look like, can you please help me with this?

Answer 1

(i) looks correct.

(ii): Consider $\varphi : \mathbb Z \to \mathbb Q$ the inclusion map. This is a ring homomorphism. $0$ is a maximal ideal in $\mathbb Q$, but $\varphi^{-1}(0) = 0$ is not a maximal ideal in $\mathbb Z$: pick any prime $p$ and notice that $0 \subsetneq (p) \subsetneq \mathbb Z$.

Answer 2

(ii): Take any integral domain $D$ which is not a field and let $k$ be it's field of fractions. Consider the natual inclusion $i :D[t] \to k[t].$ The ideal $M = tk[t]$ is a maximal ideal in $k[t].$ But $i^{-1}(M) = tD[t]$ is not a maximal ideal in $D[t].$