Let $f:R\rightarrow S$ be an epimorphism of commutative rings with unit $1$ and let $I\unlhd R$ be a maximal ideal that does not contain $ker(f)$. I'm trying to show if it's true or not that $f(I)$ is maximal in S (probably not).
Can you please give me a direction for a counterexample?
On
Under these assumptions, one can actually show that always $f(I)=S$.
Since $f$ is surjective, it is easy to see that $f(I)$ is an ideal: it is clearly an additive subgroup of $S$ and whenever $j \in f(I)$ and $s \in S$, one can find $i \in I$ and $r \in R$ such that $f(i)=j, f(r)=s$. Then $sj=f(r)f(i)=f(ri) \in f(I),$ since $ri \in I$.
Now: From the standard correspondence theorem, we have that $J=f^{-1}(f(I))$ is an ideal of $R$ containing both $I$ and $\ker f$. From the fact that $\ker f \not \subseteq I$ we see that $I \subsetneq J$. Since $I$ was maximal, it follows that $J=R$. But then $f(I)=f(f^{-1}(f(I)))=f(J)=f(R)=S$ from the surjectivity assumption.
How does it work in general: If one omits the assumption that $I$ is maximal, the fact that $f(I)$ is an ideal still holds. Then $J=f^{-1}(f(I))$ is again an ideal containing $I\cup \ker f$ and with the property that $f(J)=f(I)$. The same holds for the ideal $I+\ker f$ and from the fact that the correspondence theorem talks about one to one correspondence of ideals of $R$ containing $\ker f$ and ideals of $S$, we have $J=I+\ker f$. Hence the image of $I$ is the same as when one considers the image of $I+\ker f$, i.e. takes the smallest ideal above $I$ such that it contains $\ker f$ as well.
Consider $\Bbb Z\times \Bbb Z$ and the homomorphism $f:\Bbb Z\times \Bbb Z\to \Bbb Z$ given by $f(a,b)=a$, and the ideal $\Bbb Z\times 2\Bbb Z$.