Let $R$ and $R'$ be two rings and $f:R\to R'$ be a ring homomorphism. Let $I$ be an ideal of $R.$ Is $f(I)$ an ideal of $R'?$ Justify.
I can prove that $f(I)$ is a subgroup of $R'.$ This is because, say, $a,b\in f(I)$ then, $\exists x,y \in I$ such that $f(x)=a,f(y)=b$. Now, $$f(x+y)=f(x)+f(y)=a+b\implies a+b\in f(I).$$
Also, $f(-x)=-f(x)=-a\in f(I).$ This proves that $f(I)$ is a subgroup of $R'$ under addition.
However, $f$ is not onto so, we can really say $f(I)$ is an ideal. But then, I am unable to find a counter example as well.