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Let $f:R\rightarrow S$ be a surjective homomorphism of commutative rings with unity. I want to prove that if $M$ is a maximal ideal then $f(M)$ is either $S$ or it is a maximal ideal of $S$. I get the feeling I should somehow use the correspondence theorem, but I just can't see how to exactly use it. Thank you in advance.

I also was wondering if the same statement holds for prime ideals?

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  • $\begingroup$ $f(M)$ need not even be an ideal. Do you mean the ideal generated by $f(M)$ or do you want $f$ to be surjective? $\endgroup$ – Tobias Kildetoft Sep 25 '13 at 7:10
  • $\begingroup$ You are right! f should be surjective. I'll edit that now. Thank you. $\endgroup$ – Tim Sep 25 '13 at 7:11
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    $\begingroup$ In that case, if $f(M)$ is a proper non-maximal ideal of $S$, it will be contained in a maximal ideal. What happens if you pull back this maximal ideal? $\endgroup$ – Tobias Kildetoft Sep 25 '13 at 7:11
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    $\begingroup$ You could also think in terms of quotients. $\endgroup$ – Alex Youcis Sep 25 '13 at 7:17
  • $\begingroup$ When you say "pull back" the maximal ideal which contains $f(M)$, do you mean take its inverse image? $\endgroup$ – Tim Sep 25 '13 at 7:18
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If $f(M) \subseteq I \subseteq S$ is an ideal, then $M \subseteq f^{-1}(I) \subseteq R$. Since $M$ is maximal, we get $M=f^{-1}(I)$ or $f^{-1}(I)=R$, i.e. $f(M)=I$ or $I=S$. $\mathrm{QED}$

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  • $\begingroup$ Hi, Just wondering if i consider q: Z to Z/2Z. Then 3Z is a maximal ideal of Z but q(3Z) = Z/2Z is not a maximal ideal in Z/2Z since it is not a proper ideal. What am I misunderstanding here? Your proof seems to be correct but i don' think there is anything wrong with my counterexample. $\endgroup$ – user10024395 Apr 29 '15 at 4:31
  • $\begingroup$ @Aha What's wrong it's your interpretation of question! The claim to prove in the question is that the image of ideal is an ideal or all the codomain $S$, please don't delete your comment, it will be instructive, I got stuck in the same thing. $\endgroup$ – Santropedro Jul 15 '17 at 16:10
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Let $f: R\rightarrow S$ be a surjective homomorphism.

Suppose $M$ be a maximal ideal of $R$ ans suppose $f(M)$ is not a maximal ideal.

Then we should have $f(M)\subseteq N$ for a maximal ideal $N$ of $S$.

As $f$ is surjective we can consider $f^{-1}(N)$.

As inverse image of maximal ideal is maximal ideal we see that $f^{-1}(N)$ is maximal ideal.

$M\subseteq f^{-1}(N)$ But, $M$ is maximal ideal and thus $M=f^{-1}(N)$ and so, $f(M)=N$

Thus, $f(M)$ is maximal ideal.

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  • $\begingroup$ Inverse image of maximal ideal needn't be maximal in general. For example, $\mathbb{Z}\hookrightarrow\mathbb{Q}$. Or more intuitively, subrings of fields needn't be fields. This shows why the same question for prime ideals is true though, since subrings of domains are domains. $\endgroup$ – Alex Youcis Sep 25 '13 at 7:33
  • $\begingroup$ @AlexYoucis But since the map is surjective it does hold in this case. $\endgroup$ – Tobias Kildetoft Sep 25 '13 at 7:37
  • $\begingroup$ @TobiasKildetoft I am well aware of that :) I just wanted Praphulla to acknowledge this fact, since this is the crux of his argument. $\endgroup$ – Alex Youcis Sep 25 '13 at 7:38
  • $\begingroup$ @AlexYoucis : I have already said that it is surjective and i have used it :O What is the point of downvote(if you have done that :P) $\endgroup$ – user87543 Sep 25 '13 at 7:40
  • $\begingroup$ @PraphullaKoushik I have not downvoted. What you have written would seem to suggest that you can only consider the preimage $N$ because it's surjective. It certainly does not suggest the surjectivity is being used (and especially is pivotal) for the fact that preimage of maximal is maximal. $\endgroup$ – Alex Youcis Sep 25 '13 at 7:41
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Let $f:R \to S$ be surjective, as above, and let $\mathfrak m$ be a maximal ideal. Then, since $f$ is surjective, the image of $\mathfrak m$ is an ideal also, which we denote by $f(\mathfrak m)$. We get an induced map on quotient rings: $f:R/\mathfrak m \to S/f(\mathfrak m)$. Now, the claim is equivalent to $S/f(\mathfrak m)$ being either a field or the zero ring.

If it is neither, it has a proper ideal, say $\mathfrak a$, whose inverse image is a proper ideal of $R/\mathfrak m$. But this is a contradiction, as $R/\mathfrak m$ is a field, thus having no proper ideals.

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  • $\begingroup$ @downvoter: I'm not able to find a mistake in my answer. Why the downvote? $\endgroup$ – Fredrik Meyer May 31 '16 at 10:47

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