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If M irreducible R-module, then M is isomorphic to R/I for I a maximal ideal of R. [If M is irreducible there is a natural map $R \rightarrow M$ defined by $r \mapsto rm$, where m is any fixed nonzero element of M.]

I am self-studying Abstract algebra I, and my attempted proof so far is this. However I am not sure if the proof is correct.

Let $I_{k}$ be any maximal ideal of R if any. Define $\varphi_{k}: R \rightarrow M$ where $\varphi_{k}(r) = 0$ if $r \in I_{k}$ else $\varphi_{k}(r) = rm$ where m is any fixed non-zero element of M. Hence $I_{k}$ is the annihilator of M in R. Now $\varphi_{k}$ is an R-module homomorphism since if $r_{1}, r_{2} \in R$

$\varphi_{k}(r_{1}+r_{2}) = (r_{1}+r_{2})m =r_{1}m+r_{2}m = \varphi_{k}(r_{1}) + \varphi_{k}(r_{2})$

$\varphi_{k}(r_{1}r_{2}) = (r_{1}r_{2})m = r_{1}(r_{2}m) = r_{1}\varphi_{k}(r_{2})$

Since R is an R-module itself, by the First isomorphism theorem for modules, kernel of $\varphi_{k} = I_{k}$ is a submodule of R and $R/I_{k} \cong \varphi_{k}(R)$. I am assuming $\varphi_{k}(R) = M$ but how would I show this?

metalder9
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1 Answers1

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You don't need the case by case definition of $\varphi_k$: $\varphi_k(r)=rm$ suffices in all cases.

$\varphi_k(R)$ is the image of a module homomorphism, so is a submodule of $M$. As $M$ is irreducible, it only submodules are $0$ and $M$, and $\varphi_k$ has been set up to have $m$ in its image. Therefore, $\varphi_k(R)=M$.

Angina Seng
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  • Thanks, could you explain why you do not need the case by case definition. Is it because the first isomorphism theorem already establishes the isomorphism R/I to M, hence requiring that $\varphi_{k}(r) = 0$ for $r \in I_{k}$ is unnecessary? – metalder9 Jul 02 '17 at 16:14
  • @metalder9 It's just straightforward definition chasing that $r\mapsto rm$ is always a module homomorphism. Then $I$ will be its kernel ... you do have to check then that $I$ is maximal. – Angina Seng Jul 02 '17 at 16:18
  • @metalder9 Don't forget that $M$ is irreducible. – Angina Seng Jul 02 '17 at 16:25
  • What would happen if I was not maximal and any ideal of R. What would break in the above proof? Thanks for the help. I wasn't sure if you replied this in your above comment. – metalder9 Jul 02 '17 at 16:50
  • Oh I think I get it: since M is irreducible, I can build a contradiction with an ideal contained in the maximal ideal $I_{k}$, where the image is a submodule 'smaller' than M. Or that show that the quotient R/smaller ideal contained in $I_{k}$ is also isomorphic to M, which is impossible. Thanks! – metalder9 Jul 03 '17 at 02:32