Prove or disprove: The image of a ring homomorphism $\phi:R\to S$ is an ideal in $S$.

Question

Prove or disprove: The image of a ring homomorphism $\phi:R\to S$ is an ideal in $S$.

I only see examples where they use the image of an ideal, but I don't think this is the case for my question.

At first I tried using the ideal test, and I had a feeling that it wasn't working out the way it should, so now I am trying to find a counterexample. Any ideas for a simple counterexample?

Answer 1

It's false: Take $R=\mathbb{Z}$, $S=\mathbb{Q}$ and $\phi$ to be the inclusion.

Answer 2

Another example I like : Take $R$ any ring and the polynomial ring $R[X]$. Now consider the morphism $f:R[X]\rightarrow R[X]$ which sends $X$ to $X^2$ (extend it).