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Let $R$ be a simple ring (i.e. a ring with no nontrivial two-sided ideals) which contains a left ideal which is simple as a left $R$-module. How can I prove that $R$ is semisimple?

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@Geoff Robinson Not every simple ring is semisimple, the ones that are seki-simple, are necessarily, by Artin-Wedderburn, isomorphic to a matrix ring over a division ring. –  Olivier Bégassat May 6 '12 at 13:51
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You can look at the sum of all simple submodules (i.e. all simple left ideals) isomorphic to that given simple submodule of $_RR$... this turns out to be a (two-sided) ideal of $R$, irrespective of what kind of ring $R$ is. However, if $R$ is simple you don't have a choice. Then $_RR$ is a sum of simple submodules (i.e. simple left ideals), thus a direct (why?) sum of simple left ideals, thus a finite (why?) direct sum of simple left ideals all isomorphic to the simple left ideal you started with .

You have a division ring at hand, the endomorphisms of that simple left submodule you started with. You finish off by saying $R$ is the ring of endomorphisms of $_RR$ (acting from the right) and this ring is also a matrix ring because of the previously mentioned direct sum decomposition of $_RR$ into a finite number of simple left ideals all isomorphic to the first one. This finishes the proof, and shows that all (left) artinian simple rings are matrix rings over some division ring.

This is very well explained in Lam's book A First Course in Noncommutative Rings.

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I didn 't undestand what is $_RR$, could you explain it, please? –  Alex M May 7 '12 at 0:42
    
It's how you call the canonical left $R$-module structure on $R$. In other words, $_RR$ is the left $R$ module on the abelian group $M=(R,+)$ where for all $r\in R, m\in M,~r.m=r\times m$. Here "$.$" defines the left $R$-module structure on $M$, and "$\times$" is the ring product of $R$. –  Olivier Bégassat May 7 '12 at 0:59
    
Let's see if I understood: let $I$ be the simple left ideal. Define $J:=\sum_{I_i\cong I} I_i$ (where the sum is over the left ideals isomorphic to $I$). Then $J=R$ and so $R=I_1\oplus\cdots\oplus I_n$ where $I_j\cong I$. This is enough to say that $R$ is semisimple, right? I don't understand your discussion in the second paragraph. –  Alex M May 7 '12 at 2:29
    
That's it. You then say some thing like $\mathrm{End}(_RR)\simeq R^{op}$ where for any $x\in R$, you define the endomorphism $\rho_x$ of the left $R$ module $_RR$ by $\rho_x(r):=r\times x$. Thus $R^{op}\simeq \mathrm{End}_R(R)\simeq \mathrm{End}_R(\oplus^n I)\simeq\mathrm{Mat}_n(\mathrm{End}_R(I))$. but the ring of endomorphisms of $I$ is a division ring by Schur's lemma nd you're done. –  Olivier Bégassat May 7 '12 at 10:16
    
What I'm not understanding is this: the discussion in your second paragraph (the fact that $R^{op}\cong\mathrm{Mat}_n(\mathrm{End}_R(I))$) is extra or I need it? because it seems to me that the exercise is solved once we note that $R=I_1\oplus\cdots\oplus I_n$ with the $I_j$'s simple. –  Alex M May 7 '12 at 19:42
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Given a left $R$-module $M$, let $\text{soc}(M)$ be the sum of all simple submodules of $M$ (if $M$ has no simple submodule, then of course $\text{soc}(M) = 0$). The submodule $\text{soc}(M)$ is called the socle of $M$. Note that $M$ is semisimple if and only if $\text{soc}(M) = M$.

Lemma: Let $\varphi:M \rightarrow N$ be an $R$-module homomorphism. Then $\varphi(\text{soc}(M)) \subseteq \text{soc}(N)$.
Proof: Use Schur's lemma.

Consider $S = \text{soc}(_R R)$. By construction $S$ is a left ideal. I claim that $S$ is actually a two-sided ideal: Indeed for every $r \in R$ right multiplication by $r$ is an endomorphism of $_R R$; hence by the Lemma above $Sr \subseteq S$.

By hypothesis $S \neq 0$. But $R$ is a simple ring, thus $S=R$, i.e. $R$ is semisimple.

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As Olivier nicely explains, the key fact here is as follows:

Lemma: For a minimal left ideal $U$ of a ring $R$, let $B_U$ be the left ideal generated by all left ideals $I$ of $R$ which are isomorphic to $U$ as left $R$-modules. Then $B_U$ is a two-sided ideal of $R$.

A proof of this can be found in $\S 2.3$ of my noncommutative algebra notes.

Later on in $\S 2.4$ this lemma is used to prove a result which, in particular, answers your question.

Theorem: For a simple ring $R$, the following are equivalent:
(i) $R$ is left Artinian.
(ii) $R$ has a minimal left ideal.
(iii) $R$ is left semismple.
(iv) For some $n \in \mathbb{Z}^+$, $R$ is isomorphic to the ring of $n \times n$ matrices over a division ring.

Note that a ring satisfies condition (iv) iff its opposite ring does, so all of the instances of "left" in conditions (i) through (iii) can be replaced by "right".

Let me also add that these conditions are nonvacuous: a simple ring need not be left Artinian. (Further, there are simple rings which are left Noetherian but not left Artinian and also simple rings which are not left Noetherian.) Some examples are given (following Lam) in $\S 1.10$. Thus -- confusingly for the unwary! -- a simple ring need not be semisimple.

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Could you maybe take a look at one of my recent questions (if you haven't already) about the interactions between ring theory and Lie algebra theory? It is kind of vague but some guiding principle behind the similarities of both theories would help me a lot! –  Olivier Bégassat May 6 '12 at 20:25
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