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.
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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?
egreg
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awesome, thanks. This correspondence between R and S eluded me. – tmpys May 03 '18 at 08:10
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Is this homomorphism a module homomorphism or ring homomorphism? – tmpys May 04 '18 at 04:40
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@tmpys Module hom – egreg May 04 '18 at 06:09
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So we are considering R as an R-module here? – tmpys May 04 '18 at 06:45
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@tmpys Yes, of course: we're dealing with the simple modules, aren't we? – egreg May 04 '18 at 06:55
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yes but we also talking about rings and quotient rings. I don't ask questions I know the answer to. – tmpys May 04 '18 at 06:59
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@tmpys OK, I added some details. Dealing with commutative rings may indeed raise doubts about the matter. – egreg May 04 '18 at 07:26