Image of a maximal ideal

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 $I=S$. $\mathrm{QED}$

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.