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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?

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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?

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  • $\begingroup$ 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 $x-y\in\operatorname{im}(\phi)$? $\endgroup$ – Mather Guy May 15 '19 at 22:23
  • $\begingroup$ All correct. You've got it. $\endgroup$ – Robert Shore May 15 '19 at 23:02
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So, in other words this image is not necessarly a subring, but, instead, is an ideal of $\mathit{R'}$.

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  • $\begingroup$ I don't think so because I'm not necessarily saying that it will absorb product from $R'$, just that subtraction holds. $\endgroup$ – Mather Guy May 15 '19 at 22:28

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