Is a semisimple A-module semisimple over its endomorphism ring? Let $A$ be a ring, $M$ be a semisimple $A$-module and let $B=End_A(M)$. Show that $M$ is semisimple as a $B$-module. 
My thoughts so far are: if I can show that $B$ is a semisimple ring, then it would immediately follow that $M$ is $B$-semisimple. However I haven't had much luck proving that (in fact I think this statement may be false).
Another approach I've tried is showing that the $A$-submodules and the $B$-submodules of $M$ are the same and thus showing that every $B$-submodule of $M$ is a direct summand, but since $A$ is not necessarily commutative I'm not sure if this may be true.
 A: Nice job including your thoughts: that gives us a lot to start with!


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*if I can show that B is a semisimple ring, then it would immediately follow that M is B-semisimple. Very true! Unfortunately, your instincts were right that this is false in general. The standard example to have would be the ring $R=End(_kV)$ where $k$ is a field and $V$ has countably infinite dimension over $k$. This ring is not Artinian or Noetherian, so it definitely can't be semisimple (but $_kV$ is surely a semisimple $k$ module). (Factoid: the endomorphism ring of a semisimple module is always a von Neumann regular ring. )

*Another approach I've tried is showing that the A-submodules and the B-submodules of M are the same This is another good first thought; however, the example of $R$ again shows that this is not the case. While $_kV$ is definitely a semisimple $k$ module, and it can be written as a sum of countably many simple $k$ submodules, it's easy to show using linear algebra that $V_R$ is a simple $R$ module!
Those two ideas didn't pan out directly, but they are full of information that points us in the right direction (and besides, they bring up a lot of things you might use in future problems), so it's good they came up :) 
The first thing to notice about our example of $R$ is that all of its simple submodules were isomorphic. Our module $M$ may not have that property. So what happens if we look at $M$ broken into homogeneous pieces? 
What I mean is this: let's consider a fixed simple submodule $S_\alpha$ of $M$, and then look at $M_\alpha:=\sum\{ S \mid S\cong S_\alpha\}$ (isomorphisms of $A$ modules). Exercise: Show that this is a $B$ submodule of $M$, and that $M=\oplus M_\alpha$ where the $\alpha$ have been chosen to index a representative set of isomorphism types of simple submodules of $M$.
But now this is starting to look like our $V$ in the $R$ example! Is $M_\alpha$ a simple $B$ module? If so, then $M$ is certainly a semisimple $B$ module. So finally I can hand the problem back to you:
Hint: argue that $M_\alpha$ is a simple $B$ module. (The easiest way for me to think about this was to show that for any nonzero $x,y\in M_\alpha$ there exists an $A$ linear map that sends $x$ to $y$.)
