Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.