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Let $R$ and $S$ be rings, $I$ a maximal ideal of $R$, and let $f: R\rightarrow S$ be a surjective ring homomorphism such that $\ker f \subseteq I$. I’d like to show that the image $f(I)$ will be a maximal ideal of $S$.

My idea is to construct $g\colon R \to S/f(I)$, which is a surjective homomorphism such that $g(x)=f(x)+f(I)$. Then, if I could show that $\ker g = I$, I could use the first isomorphism theorem to conclude that $R/I \cong S/f(I)$. Finally, if $I$ is maximal, then $R/I$ would be a field, $S/f(I)$ would also be a field and then $f(I)$ would be maximal.

However, I’m stuck at showing that $\ker g = I$ (I guess here is where I should use that $\ker f \subseteq I$?). Is my reasoning correct so far?

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    $\begingroup$ Perhaps you already know this, but I'd like to point out that the image of an ideal need not be an ideal unless the map is onto. Since your map is onto everything is fine, but if you haven't shown this before it's worth doing. $\endgroup$ Apr 18, 2021 at 16:10
  • $\begingroup$ @paulblartmathcop Thanks! I did know about that, but I forgot writing that step down so it’s a good reminder! $\endgroup$
    – dahemar
    Apr 18, 2021 at 18:12

2 Answers 2

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If you consider two points $x,y\in R$ such that $x-y\in I$, then

$f(x)-f(y)\in f(I)$

And so

$F: R/I\to S/f(I)$ is well defined.

The map is injective:

If $F(x)\in f(I)$ then $x\in f^{-1}(f(I))=I+ker(f)=I$

This means $x+I=I$

Until this step you don’t need to have $I$ maximal.

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You're doing great.

Hint: $x \in \ker g$ iff $f(x) \in f(I)$ iff $x \in f^{-1}(f(I)) = I + \ker f = \cdots$.

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  • $\begingroup$ Thanks! Why is it that $f^{-1}(f(I)) = I + \ker f$? $\endgroup$
    – dahemar
    Apr 18, 2021 at 18:11
  • $\begingroup$ @dahemar, try proving it yourself in the usual way: $\subseteq$ and $\supseteq$. $\endgroup$
    – lhf
    Apr 18, 2021 at 22:38
  • $\begingroup$ I think I got the $\supseteq$ part but I kinda got stuck on the $\subseteq$ one... I don’t know how to make that kernel “appear”. $\endgroup$
    – dahemar
    Apr 20, 2021 at 14:08
  • $\begingroup$ @dahemar, $f^{-1}(f(I)) \supseteq f^{-1}(0) = \ker f$ $\endgroup$
    – lhf
    Apr 20, 2021 at 14:19
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    $\begingroup$ @dahemar, indeed, sorry about that. Try $x \in f^{-1}(f(I)) \implies f(x) = f(i)$ for some $i \in I$. Then $x-i \in \ker f$. $\endgroup$
    – lhf
    Apr 20, 2021 at 15:50

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