Simple modules over $F[x]$ are the $F[x]/I$ where $I$ is a prime ideal

Question

I am trying to compare some theorems about general groups to some theorems about modules. In the document given below, it states that simple modules over $F[x]$ are the $F[x]/I$ where $I$ is a prime ideal. Can this be proved easily? If so, please provide a proof.

http://www.ucl.ac.uk/~ucahaya/SemisimpleModules.pdf

Answer 1

If $R$ is a commutative ring (with identity), then a simple module over $R$ is of the form $R/M$, where $M$ is a maximal ideal of $R$.

If we're dealing with noncommutative rings, a simple left $R$-module is of the form $R/M$, where $M$ is a maximal left ideal of $R$.

Indeed, if $S$ is a simple left $R$-module and $x\in S$, $x\ne0$, then $S=Rx$ (because $S$ is simple) and therefore the map $\mu_x\colon R\to S$, $\mu_x(r)=rx$ is a surjective module homomorphism. By the correspondence theorem, $M=\ker\mu_x$ is a (left) maximal ideal and $S\cong R/M$ by the homomorphism theorem.

Now, what are the maximal ideals in a principal ideal domain?