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then $\phi(I)=[s'\in R'|s=\phi(s)\space \forall\space s\in I]$ is an ideal in R'
So I know that for something to be an ideal, it needs to be closed under subtraction and it must absorb products. I guess I get overwhelmed when there is a lot going on.
I want to say something along the lines of
Let $r,s\in \phi(I)$, then $\phi(r-s)=\phi(r+(-s))=\phi(r)+\phi(-s)=r+(-s)=r-s\in I$ since $I$ is an ideal...
Then let $t\in R'$ and $s\in \phi(I)$, so $t(s)=t(\phi(s))$ .... Not sure if I'm headed down the right path