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Let $f: R \to S$ be a surjective ring homomorphism.

Let $M \subset S$ be maximal, and let $f^{-1}(M) \subset I$ for some ideal $I \subset R$.

Then $M=f(f^{-1}(M)) \subset f(I)$ since $f$ is surjective.

Since $M$ is maximal in $S$, then either $f(I)=M$ or $f(I)=S$.

If $f(I)=M$, then $I \subset f^{-1}(f(I)) =f^{-1}(M)$, hence $I=f^{-1}(M)$.

Now, I'm having trouble showing that if $f(I)=S$, then $I=R$.

user5826
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  • Would you accept an answer that proves the statement a different (but still simple) way, or are you specifically looking to complete this argument? – Sebastian Monnet May 23 '20 at 20:51
  • @Qwertiops I know how to prove it via $S \cong R/ \ker f$ and using correspondence theorem. I also know the argument where we have $R/f^{-1}(M) \cong S/M$. I figured these out after spending quite a while with my original attempt above. So, mainly just looking for a way to finish up the argument in the post. – user5826 May 23 '20 at 22:16

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