Image of a maximal ideal 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?
 A: 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.
A: 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 $I=S$. $\mathrm{QED}$
A: 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.
