If $\phi : A \to B$ is a ring homomorphism then the image of a subring $S \subset A$ is a subring $\phi(A) \subset B$. Is the image of an ideal under a ring homomorphism also an ideal? The answer is negative. Let’s provide a simple counterexample.

Let’s take $A=\mathbb Z$ the ring of the integers and for $B$ the ring of the polynomials with integer coefficients $\mathbb Z[x]$. The inclusion $\phi : \mathbb Z \to \mathbb Z[x]$ is a ring homorphism. The subset $2 \mathbb Z \subset \mathbb Z$ of even integers is an ideal. However $2 \mathbb Z$ is not an ideal of $\mathbb Z[x]$ as for example $2x \notin 2\mathbb Z$.