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 .
Best Answer
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?