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Simple Module

What isomorphism shall I use to show that any simple module V (over a ring R) is isomorphic to R/M, where M is a maximal ideal?
Is this not simialr to things like in group and ring theory where you use what is called the canonical (or natural) homomorphism to show very similar types of theorems?

Do you have something you call a canonical or natural homomorphism/isomorphism for use on modules? I've never done any work specifically on modules, so I'm not sure even how you define one, never mind how to do about defining the canonical homomorphism on one, but I guess it would be similar to groups.
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J.F.N
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Roger Kirk
Is this not simialr to things like in group and ring theory where you use what is called the canonical (or natural) homomorphism to show very similar types of theorems?

Do you have something you call a canonical or natural homomorphism/isomorphism for use on modules? I've never done any work specifically on modules, so I'm not sure even how you define one, never mind how to do about defining the canonical homomorphism on one, but I guess it would be similar to groups.


Never mind, I got it. I used an isomorphism from R to V, showed that R/ker is a field, and that ker is hence maximal. Then using the first iso. theorem, R/ker is isomorphic to Im, but Im=V since V is simple, so the result is proved.
J.F.N
Never mind, I got it. I used an isomorphism from R to V, showed that R/ker is a field, and that ker is hence maximal. Then using the first iso. theorem, R/ker is isomorphic to Im, but Im=V since V is simple, so the result is proved.

here is a way to do it.
there is a bijection between the ideals J of R st IcJCR and sub-modules of R/I given by J ->J/I. from this we can deduce R/I is simple iff I is maximal.
to show every simple module is iso to R/I with I maximal:
let M be a simple module.
let x be in M
define f:R ->M by r->rx
then M =im f since M is simple
by first iso
R/ker(f)=imf=M
J.F.N
Never mind, I got it. I used an isomorphism from R to V, showed that R/ker is a field, and that ker is hence maximal. Then using the first iso. theorem, R/ker is isomorphic to Im, but Im=V since V is simple, so the result is proved.
I guess that is similar to what I was talking about then, as you generally create the natural hom to prove first iso theorems I think.

Glad you're sorted anyway :smile:
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J.F.N
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While we're at it, what are the finitely generated modules over the ring Z/nZ?

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