Prove: If $\phi:R\rightarrow R'$ is a ring homomorphism, then $image(\phi)$ is a subring of $R'$

Question

Do I have to use the regular axioms for proving something is a subring? i.e. closed under subtraction and multiplication. If so, can I say $$im(r-s)=r-s\in R$$ $$im(rs)=rs\in R$$ Therefore $im(\phi)$ is a subgroup of $R'$? Do I have some notation mixed up? Or am I just thinking about it completely wrong?

Answer 1

You have your notation a little mixed up. I'll get you started by showing that $\operatorname{im}(\phi)$ is closed under addition.

Assume $x, y \in \operatorname{im}(\phi) \subseteq R'$. Then $\exists r, s \in R \text{ such that } \phi(r)=x, \phi (s)=y$. Since $\phi$ is a homomorphism, $\phi(r+s)=\phi(r)+\phi(s)=x+y$, so $x+y \in \operatorname{im}(\phi)$. Thus, $\operatorname{im}(\phi)$ is closed under addition.

Can you finish off the proof from there?

Answer 2

So, in other words this image is not necessarly a subring, but, instead, is an ideal of $\mathit{R'}$.