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Let $R$ and $S$ be commutative rings with $1$ and $\phi: R\rightarrow S$ be a surjective ring homomorphism. Then for an arbitrary maximal ideal $I$ of $R$, does $\phi(I)$ have to be maximal in $S$?

My idea: from the isomorphism theorem, maximal ideals of $R$ containing $Ker\phi$ are in 1-1 correspondence with maximal ideals of $S$. Hence there may exist some maximal ideals $I$ of $R$ such that $Ker \phi$ not contained in $I$, thus $\phi(I)$ is not a maximal ideal of $S$. But this is not valid proof (this is because that if we replace maximal with prime, we can use the same argument. However, the surjective homomorphism image of a prime ideal is prime, which I have proved using the definition). What should I do?

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The answer is NO! take $R=k[x]$, $S=k[x]/x$, and $I=(x-1)$.

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From $\phi$ you can construct $\varphi\colon R \to S/\phi(I)$. It is a surjective homomorphism with kernel $I$. Therefore, by the isomorphism theorem $R/I \cong S/\phi(I)$. If $I$ is maximal, then $R/I$ is a field, then $S/\phi(I)$ is a field and then $\phi(I)$ is maximal.

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    $\begingroup$ How do you know that the kernel of $\varphi$ is exactly $I$? It certainly contains $I$, but the reverse inclusion fails. Look at adrido's answer: in his example, $\phi(I) = S$, so $\ker(\varphi) = R$. $\endgroup$ – André 3000 Jul 11 '15 at 3:07
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It is, note that if $J \subseteq S$ is an ideal of $S$, then $\phi^{-1}(J)$ is an ideal of $R$.

If $\phi(I) \subset J$ and $\phi(I) \not = J$, then $\phi^{-1}(J)$ must equal $ R$ since $I$ is maximal. Hence $J = \phi(R)$ and since $\phi$ is surjective we have $J = S$.

Therefore $\phi(I)$ is maximal.

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    $\begingroup$ How do you know that $\varphi(I)$ is a proper ideal? You didn't show that $\varphi(I) \neq S$. Look at adrido's counterexample. $\endgroup$ – André 3000 Jul 11 '15 at 3:05
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This is true if and only if the kernel of $ϕ$ is a subset of I. then you could use the isomorphism theorem and the fact that this is an epimorphism

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