1. ## Semisimple algebra

Show that an algebra $\displaystyle A$ is semisimple if and only if every $\displaystyle A$-module is completely reducible.

I know that if $\displaystyle A$ is semisimple then every irreducible $\displaystyle A$-module is isomorphic to a submodule of $\displaystyle A^{*}$ ($\displaystyle A^{*}$ is $\displaystyle A$ viewed as a module over itself). Then the submodules of $\displaystyle A$ are direct summands (there exists another submodule such that the direct sum equals $\displaystyle A$).
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?

2. Originally Posted by robeuler
Show that an algebra $\displaystyle A$ is semisimple if and only if every $\displaystyle A$-module is completely reducible.
Definition. The algebra A is semisimple if every nonzero A-module is semisimple.

If your definition of the semisimple algebra is here, then I think your question is probably,

Show that an algebra $\displaystyle A$ is semisimple if and only if the $\displaystyle A$-module A is completely reducible (semisimple).

Proof.
Suppose that the A-module A is semisimple, and let M be an arbitrary A-module generated by $\displaystyle \{b_1, \cdots , b_m\}$. We shall show that M is a semisimple A-module. Let $\displaystyle A^m$ denote the direct sum of m copies of A. Let $\displaystyle \phi:A^m \rightarrow M$ be a map, which sends $\displaystyle (x_1, \cdots, x_m)$ to $\displaystyle x_1b_1 + \cdots +x_mb_m$. We see that $\displaystyle \phi$ is an A-module epimorphism. It follows that M is isomorphic with a quotient module of the semisimple module $\displaystyle A^m$. Thus it is semisimple by Lemma 1 below. We conclude that A is a semisimple algebra by the definition given above.

I know that if $\displaystyle A$ is semisimple then every irreducible $\displaystyle A$-module is isomorphic to a submodule of $\displaystyle A^{*}$ ($\displaystyle A^{*}$ is $\displaystyle A$ viewed as a module over itself). Then the submodules of $\displaystyle A$ are direct summands (there exists another submodule such that the direct sum equals $\displaystyle A$).
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?
Lemma 1. Quotient modules and submodules of semisimple modules are also semisimple.

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:

$\displaystyle M=S_1 \oplus \cdots \oplus S_n$.

The quotient module of M has the form M/N, where N is an A-module. Let $\displaystyle \phi$ be an A-module epimorphism such that $\displaystyle \phi:M \rightarrow M/N$. Then,

$\displaystyle M/N=\phi(M)=\phi(S_1) \oplus \cdots \oplus \phi(S_n)$.

Can you conclude that M/N is semisimple, which implies that a quotient module of a semisimple module is semisimple?