The exercise is:
Given that $ \phi : R \to S $ is an onto ring homomorphism, let $ B $ be a maximal ideal of $ S $. Prove that $ A = \phi^{-1} (B) $ is a maximal ideal of $R$.
Ok, I previously proved that $A$ is an ideal of $R$, now to prove that it is maximal:
Let $I$ be an ideal of $R$ such that $ A \subset I$. From this follows that $B \subset \phi (I)$, and since $B$ is maximal $ \phi (I) = S$, and $I = R$, which means that $A$ is maximal.
Is it correct? Because I'm not too sure.
You're a little too fast. Suppose $A\subset I$ (proper subset). Then $\phi(A)\subset\phi(I)$, which is an ideal because $\phi$ is surjective; since $\phi$ is surjective, $\phi(A)=B$. By maximality, either $\phi(I)=B$ or $\phi(I)=S$.
Now note that $\phi^{-1}(\phi(I))=I$, so we must have $I=R$.
Better technique: the homomorphism theorem provides an isomorphism $$ R/\phi^{-1}(B)\to S/B $$ The ideal $B$ is maximal if and only if $S/B$ is…