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Let $f :R \to S$ be a surjective ring homomorphism , $M$ be a maximal ideal of $S$ , I am writing a proof showing $f^{-1}(M)$ is a maximal ideal of $R$ , Please verify whether it is correct or not .

Proof :- Let $J$ be an ideal such that $f^{-1}(M) \subset J \subseteq R$ , we want to show $J=R$ i.e. $R \subseteq J$. Now as $f$ is surjective , $M=f(f^{-1}(M)) \subseteq f(J) \subseteq S$ . As $f$ is surjective , $f(J)$ is an ideal of $S$ . Now if it were possible that $M=f(J)$ , then $x \in J \implies f(x) \in f(J)=M \implies x \in f^{-1}(M)$ , so $J \subseteq f^{-1} (M)$ ,contarry to our assumption $f^{-1}(M) \subset J$ . Thus $M \ne f(J)$ , $f(J)$ is an ideal of $S$ containing $M$ ; since $M$ is maximal ideal in $S$ , so $f(J)=S$ . Then $x \in R \implies f(x) \in S=f(J) \implies \exists j \in J : f(x)=f(j) \implies f(x-j)=0_S \in M $

$\implies x-j \in f^{-1} (M) \subset J \implies x=x-j+j \in J$ , so $R \subseteq J$

Is the proof correct . Please comment .

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    $\begingroup$ I like the "pull back image" term. I always hate to say "inverse image". I'll use that from now on. $\endgroup$ – user207710 Mar 20 '15 at 14:12
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Your proof does indeed seem to be correct. It is written in a very convoluted way, however. Perhaps you should start with:

Let $J$ be an ideal containing $f^{-1}(M)$. As $f$ is surjective, $f(J)$ is an ideal, and it contains $M$. As $M$ is maximal, either $f(J)=M$ or $f(J)=S$. If $f(J)=M$, then...

Notice that this avoids your imprecision with the "contrary with our assumption", which actually is not contrary to any assumption you made. Also, it cleans up the proof, by making the chain of deduction clearer.

If I may suggest, a cleaner way of proving this is by an altogether different method, bypassing elements. Since $M$ is maximal, $k:=S/M$ is a field. Since $f$ is surjective, its composition with the canonical projection $\bar{f}:R\to k$ is a surjection. This means that $\ker \bar{f}$ is a maximal ideal. Can you compute it?

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  • $\begingroup$ But $J \subseteq f^{-1}(M)$ is a contradiction because I assumed $f^{-1}(M)$ is a "proper subset" of $J$ ... $\endgroup$ – Souvik Dey Mar 20 '15 at 14:54
  • $\begingroup$ Ah sorry,my bad. I usually don't use the inclusion sign to mean proper inclusion. Despite that, I think it still helps to write it along those lines. $\endgroup$ – Artur Araujo Mar 20 '15 at 15:00
  • $\begingroup$ @ArturAraujo Why is $\ker \overline{f}$ a maximal ideal? $\endgroup$ – user193319 Dec 21 '17 at 19:26
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    $\begingroup$ @user193319 given that it is a surjection onto a field, what would happen if the kernel were not maximal? $\endgroup$ – Artur Araujo Dec 21 '17 at 22:30
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    $\begingroup$ Points now raised at math.stackexchange.com/questions/2577094/… $\endgroup$ – Gerry Myerson Dec 22 '17 at 19:20
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I think it would be better to deduce this from the more general correspondence theorem.

This says that there is a one-to-one, inclusion-preserving correspondence between the ideals of $S$ and the ideals of $R$ which contain the kernel $K$ of $f$.

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