Prove that a commutative ring is primitive if and only if it is a field.

Question

I want to prove that a commutative ring $ R $ is primitive if and only if it is a field.

Here is my proof of "$\impliedby$":

Let $ R $ be a field. Then $ 0 $ and $ R $ is its only ideals. Since ann$_R(R) $ is an ideal of $ R $, ann$ _R(R)=R $ or ann$_R(R)=0 $. If ann$_R(R)=R $, then $ rR=0 $ for all $ r\in R $. In particular, $ 1\cdot R=0 $, a contradiction. Hence ann$ _R(R)=0 $. Since $ R $ is a simple faithful module of $ R $, $ R $ is primitive.

Can anyone help me to get started with "$\implies$"?

Answer 1

Let $R$ be primitive and commutative. Then for some maximal ideal $M$, $R/M$ is a faithful module.

But $M$ is the annihilator of $R/M$ so...