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

Question

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 $R$.

Answer 1

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.