If $\phi:R\rightarrow R'$ is a surjective ring homomorphism and I is an ideal in R… continued below [duplicate]

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

marked as duplicate by Lord Shark the Unknown abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 19 at 4:34

• $\phi(I) = \{ r \in R^{'} | r = \phi(s) \ \exists s\in I \}$ – dcolazin May 15 at 22:32

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 $$R^\prime$$.