Why is this ring semisimple? 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?
 A: 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.
A: 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.
A: 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.
