Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It's stated here that for a commutative ring $R$, every simple module over $R$ is isomorphic as an $R$-module to a quotient ring of $R$ by a maximal ideal.

Intuitively this seems likely, since a quotient ring is a field if any only if we quotient by a maximal ideal, and a field has no nonzero proper ideals. This makes me suspect that an $R$-module $M$ is simple if and only if $M\cong R/I$ (as $R$-modules) for a maximal ideal $I$ of $R$.

Is there a proof of whether this is true or not?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Let $M$ be a simple module and let $m \in M$ be nonzero. By simplicity, $\{ rm : r \in R \} = R$, so the map $r \mapsto rm$ defines a natural surjective homomorphism $R \to M$ of $R$-modules giving an isomorphism $M \cong R/I$ where $I$ is the kernel. If $I$ is not maximal, then letting $m$ be any maximal ideal containing it we have another surjection $R/I \to R/m$ with nontrivial kernel, and this contradicts simplicity.

share|improve this answer

If $M$ is a simple module for each $x\in M$ with $x\neq 0$, $Rx=M$, so take a $x\in M$ with $x\neq 0$, and consider the next morfism $f:R\longrightarrow M$ given by $f(a)=ax$, by the last observation is an epimorphism. Now apply the first isomorphism theorem and you get $M/\ker f\cong M$. Now the $\ker f=I$ is an ideal of $R$. To see that $I$ is maximal use the theorem of the bijective correspondence. Also the same theorem gives you a proof of the return.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.