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I have an attempt to prove the claim that $R$ has finitely many nonisomorphic simple $R$-modules if $R$ is left artinian.
I would like to know if it's a good attempt. Helpful hints are very much welcome.

Attempt

Since $R$ is left artinian, $R$ has minimal left ideals. Also, $R$ is left semisimple, so $R$ can written as a direct sum of the minimal ideals, which can be grouped according to their isomorphic types as left $R-$modules. So, $$R\cong n_{1}L_{1}\oplus\cdots\oplus n_{r}L_{r}$$ where the $L_{r}$ are mutually nonisomorphic simple left $R-$modules. Let $N$ be any simple left $R-$module. Then $N\cong$a quotient of $R$ and thus $\cong L_{i}$ for some $i.$ Thus, $\left\{ L_{1},\cdots,L_{r}\right\} $ is the set of nonisomorphic left simple $R-$modules.

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From your first paragraph it seems that your only hypothesis in $R$ is that it is left artinian. Why are you using then in your argument that it is semisimple? There are plenty of non semisimple left artinian rings! –  Mariano Suárez-Alvarez May 8 '11 at 20:31
    
-Thanks Mariano...I'll try again. –  Godwin May 8 '11 at 23:38

2 Answers 2

up vote 3 down vote accepted

$R$ has finite length as a left $R$-module, and using the uniqueness (up to isomorphism and permutation) of a composition series, you get that the simple $R$-modules all appear in a given composition series of $R$.

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No, this attempt does not work.

For instance, take $R=\mathbb{Z}/8\mathbb{Z}$, which is left Artinian but is not a direct sum of minimal left ideals. The only minimal left ideal is $R(4) = 4\mathbb{Z}/8\mathbb{Z}$. No simple module of R is a quotient of R by a direct sum of minimal left ideals.

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Can you please give hints on how to begin...Thanks –  Godwin May 8 '11 at 23:39
    
I like Plop's argument, but it requires the Hopkins-Levitzki theorem, so I don't know if it is good for your course. –  Jack Schmidt May 9 '11 at 0:02
    
Thanks, Jack. I'm self studying, so I'll look for the theorem and try to apply it. –  Godwin May 9 '11 at 1:21
    
@Godwin: I tried to write another solution using the nilpotence of the Jacobson radical, but I think I need to know that R is noetherian for that too, and then I like Plop's answer better. At any rate, a very important fact about artinian rings (probably equivalent to what you want to prove) is that R/Jac(R) is artinian semisimple, so a direct product of finitely many matrix rings over division rings. There is one direct factor for each simple module. In other words, you use your proof after you know R/J(R) is semisimple. –  Jack Schmidt May 9 '11 at 15:27
    
Thanks. I'd like to see the proof supposing R is noetherian. –  Godwin May 9 '11 at 19:38

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