# Left Artinian Rings

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

$R$ 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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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.