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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 .

Souvik Dey
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    I like the "pull back image" term. I always hate to say "inverse image". I'll use that from now on. –  Mar 20 '15 at 14:12
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    I like "preimage" more than either pullback image or inverse image. – user5826 May 25 '20 at 16:32

4 Answers4

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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?

Artur Araujo
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  • But $J \subseteq f^{-1}(M)$ is a contradiction because I assumed $f^{-1}(M)$ is a "proper subset" of $J$ ... – Souvik Dey Mar 20 '15 at 14:54
  • 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. – Artur Araujo Mar 20 '15 at 15:00
  • @ArturAraujo Why is $\ker \overline{f}$ a maximal ideal? – user193319 Dec 21 '17 at 19:26
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    @user193319 given that it is a surjection onto a field, what would happen if the kernel were not maximal? – Artur Araujo Dec 21 '17 at 22:30
  • Okay. Let $f : R \to F$ be a surjective ring homomorphism, where $R$ is some (commutative?) ring and $F$ is a field. But suppose that $\ker f$ is not maximal. Then there exists an ideal $I$ in $R$ such that $\ker f \subset I \subset R$ and therefore ${0} \subseteq f(I) \subseteq F$. This would be a contradiction if images generally preserved strict inclusions, but I believe this is only true of injective maps. So I don't see why $\ker f$ has to be maximal in $R$. – user193319 Dec 21 '17 at 23:02
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    Points now raised at https://math.stackexchange.com/questions/2577094/is-the-kernel-a-maximal-ideal – Gerry Myerson Dec 22 '17 at 19:20
  • It would not be hard to rewrite your answer in terms of simple rings (I didn’t don’t see any indications the user assumed commutativity, and the argument is the same otherwise.) – rschwieb Dec 22 '17 at 19:44
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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 $f$.

Andreas Caranti
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  • There are many ways to prove this and the correspondence theorem seems to be the most popular. But this way "set theoretic" kind of way is more elementary. See my comment in my post: https://math.stackexchange.com/questions/3688465/how-to-finish-this-argument-showing-preimage-of-maximal-ideal-is-maximal-under-s – user5826 May 25 '20 at 16:27
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This post answered my post here, but for the sake of completion with my post, I include my answer here. It's very similar to yours with some minor changes.

To answer your post: Yes, your proof is correct.

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

Let $M$ 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 $I=f^{-1}(M)$.

If $f(I)=S$, then $M \subset f(I)$.

Let $x \in R$. Then $f(x) \in S = f(I)$.

So, there exists $i \in I$ with $f(x)=f(i)$, hence $f(x-i) =0$.

Then $x-i \in f^{-1}(M) \subset I$, hence $x \in I$.

Thus, $I=R$.

So, $f^{-1}(M)$ is a maximal ideal.

user5826
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If $f\colon R\to S$ is surjective, then for any ideal $J\subset S$, the induced ring homomorphism $$\varphi\colon R/f^{-1}J\to S/J$$ is also surjective. In fact, the kernel of $\varphi$ is trivial, so $\varphi$ is an isomorphism. Recall that an ideal $J$ of a ring $S$ is maximal if and only if $S/J$ is a field. So, $J\subset S$ is maximal if and only if $S/J$ is a field if and only if $R/f^{-1}J$ is a field if and only if $f^{-1}J\subset R$ is maximal.

Karl Kroningfeld
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