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

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?

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.
• I think so, multiplication and subtraction follow similarly, right? $\phi(rs)=\phi(r)\phi(s)=xy$, so $xy\in\operatorname{im}(\phi)$ $\phi(r-s)=\phi(r+(-s))=\phi(r)+\phi(-s)=x+(-y)=x-y$, so $x-y\in\operatorname{im}(\phi)$? – Mather Guy May 15 '19 at 22:23
So, in other words this image is not necessarly a subring, but, instead, is an ideal of $$\mathit{R'}$$.
• I don't think so because I'm not necessarily saying that it will absorb product from $R'$, just that subtraction holds. – Mather Guy May 15 '19 at 22:28