Q:

Prove that if $M$ is maximal ideal of $S$ and $\varphi$ is surjective then $\varphi^{-1}\left( M\right) $ is maximal ideal of $R$.

I'm thinking of Dummit Abstract Algebra, Chap 7, sectin 4, exercise 13b. I've tried to show that 
$R/\varphi^{-1}\left(  M\right)  $ is division ring, but was not successful. Can anybody help me?
Let $\varphi:R\rightarrow S$ be a homomorphism of commutative rings.
(a) If $P$ is a prime ideal of $S$, then either $\varphi^{-1}\left(  P\right)
=R$ or $\varphi^{-1}\left(  P\right)  $ is prime. OK!
(b) Prove that if $M$ is maximal ideal of $S$ and $\varphi$ is surjective, then
$\varphi^{-1}\left(  M\right)  $ is a maximal ideal of $R$.

A:

If $\pi:S\to S/M$ is the natural map, then $\pi\circ\phi$ is surjective, and
$$ \ker(\pi\circ\phi)=\{x\in R:\pi(\phi(x))=0\}=\{x\in R:\phi(x)\in M\}=\phi^{-1}(M)$$
Therefore $R/\phi^{-1}(M)\simeq S/M$ by the first isomorphism theorem, so if $M$ is a maximal ideal then $S/M$ is a field, hence $R/\phi^{-1}(M)$ is a field, hence $\phi^{-1}(M)$ is a maximal ideal.

