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

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?

It's false: Take $$R=\mathbb{Z}$$, $$S=\mathbb{Q}$$ and $$\phi$$ to be the inclusion.
• Or more generally, $R$ any integral domain that is not a field, $S$ its field of fractions, and $\phi$ to be again the inclusion. – Travis Willse Oct 29 '18 at 4:48
• @numericalorange sure, take $n\in \mathbb{Z}$ to $n\in \mathbb{Q}$ – qbert Oct 29 '18 at 5:00
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).