If your definition of the semisimple algebra is here, then I think your question is probably,
Show that an algebra is semisimple if and only if the -module A is completely reducible (semisimple).
Suppose that the A-module A is semisimple, and let M be an arbitrary A-module generated by . We shall show that M is a semisimple A-module. Let denote the direct sum of m copies of A. Let be a map, which sends to . We see that is an A-module epimorphism. It follows that M is isomorphic with a quotient module of the semisimple module . Thus it is semisimple by Lemma 1 below. We conclude that A is a semisimple algebra by the definition given above.
Lemma 1. Quotient modules and submodules of semisimple modules are also semisimple.I know that if is semisimple then every irreducible -module is isomorphic to a submodule of ( is viewed as a module over itself). Then the submodules of are direct summands (there exists another submodule such that the direct sum equals ).
Is it true that if a module is completely reducible, then all of its submodules are as well?
For the other direction I don't have any good ideas. How about you guys?
Let M be a semisimple A-module. Then every submodule of M is isomorphic with a quotient module of M by the first isomorphism theorem. It remains to show that the quotient modules of M are semisimple. Since M is a semisimple A-module, M has the following form:
The quotient module of M has the form M/N, where N is an A-module. Let be an A-module epimorphism such that . Then,
Can you conclude that M/N is semisimple, which implies that a quotient module of a semisimple module is semisimple?