Give an example to show that in general $h(P)$ is not an ideal of $S$.

Question

Let $h: R\to S$ be a ring homomorphism. Let $P\subset R$ be a prime ideal. Give an example to show that in general $h(P)$ is not an ideal of $S$

The first thing I think is to take $R=\mathbb{Z}$ and $P=(2)$ but I do not know how to take $S$ or if this works in this way, any help is appreciated.

Note: $R$ and $S$ are commutative rings with unit.

Answer 1

Consider $S=\mathbb{R}$. With the injection map $h(x)=x$, $R=\mathbb{Z}$ and any prime ideal of $\mathbb{Z}$, say $P=(2)$ as you suggested.

Answer 2

You can consider $R=\mathbb{R}[x]$ and $S=\mathbb{C}$ and $h:R \to S$ as in $h(p)=p(1)$.

You can see that $<x>$ is prime in $R$. And $h(R)=\mathbb{R}$, which is not an ideal in $\mathbb{C}$.