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This question already has an answer here:

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

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marked as duplicate by Lord Shark the Unknown abstract-algebra May 19 at 4:34

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    $\begingroup$ $\phi(I) = \{ r \in R^{'} | r = \phi(s) \ \exists s\in I \}$ $\endgroup$ – dcolazin May 15 at 22:32
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I take $I$ to be a two-sided ideal in $R$.

Given

$x, y \in \phi(I), \tag 1$

there exist

$r, s \in I \tag 2$

with

$\phi(r) = x, \; \phi(s) = y; \tag 3$

then

$x - y = \phi(r) - \phi(s) = \phi(r -s) \in \phi(I) \tag 4$

since

$r - s \in I. \tag 5$

Also, if

$b \in R^\prime, \tag 6$

then since $\phi:R \to R^\prime$ is surjective, there is some

$a \in R, \; \phi(a) = b; \tag 7$

then

$bx = \phi(a)\phi(r) = \phi(ar) \in \phi(I), \tag 8$

by virtue of the fact that

$ar \in I, \tag 9$

$I$ being an ideal in $R$; likewise,

$xb \in \phi(I) \tag{10}$

as well. Thus $\phi(I)$ is an ideal in $R^\prime$.

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