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How can I prove that if $R$ is a finite dimensional algebra over a field then $R$ is simple as a ring if and only if it has a faithful simple left $R$-module?

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How much theory are you allowing yourself to use? –  rschwieb May 4 '12 at 18:27
    
I don't know how to explain, there are no restriction I think, the standard theory of simple modules. –  Alex M May 5 '12 at 22:14
    
Is it OK to use Artin–Wedderburn structure theorem for semisimple rings for instance? I can supply an answer in that case. –  Cihan May 6 '12 at 5:36
    
^Well I also need basic theory of the Jacobson radical for the answer I have in mind. –  Cihan May 6 '12 at 5:46
    
yeah yeah, it's ok –  Alex M May 6 '12 at 9:38

1 Answer 1

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OK: if you count Artin-Wedderburn among the simple results you can use, then let's try this.

In one direction, every (unital, nonzero) module over a simple ring with unity is faithful.

In the other direction, the Jacobson radical $rad(R)$ is clearly zero if you look at it as "The intersection of annihilators of simple right $R$-modules." Since an Artinian ring with $rad(R)=\{0\}$ is semisimple, then $R$ is at least semisimple.

Since $R$ is semisimple you can embed your simple module in $R$ as a minimal right ideal $S$. Since nonisomorphic minimal right ideals would annihilate your simple module, there is only one isotype of minimal right ideal. From Artin-Wedderburn theory, $R=\Sigma${minimal right ideals isomorphic to $S$}$\cong M_n(D)$ for a division ring $D$. So, $R$ would be simple.

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