If A is semisimple algebra then every A-module is completely reducible

Question

A semisimple Algebra means the regular module is completely reducible. Let $A= \sum_{i}A_{i}$ where $A_{i}$ is irreducible sub modules of A. Suppose V is A-module. I tried to define a map between the $A_{i}$ and V. $$\theta_{i}: A_{i} \longrightarrow V$$ $$ x_{i}\mapsto vx_{i} $$ the map is homomorphism so $A_{i}\setminus ker(\theta_{i})$ is isomorphic to $\theta (A_{i}) \subseteq V$ the kernel is $\{0\}$ since $A_{i}$ is irreducible, so $A_{i}$ is isomorphic to $\theta_{i} (A_{i}) \subseteq V$.

• $\theta_{i} (A_{i})$ is irreducible because of the isomorphic. Now, $\oplus \sum_{i} \theta_{i} (A_{i})$ is the direct sum of irreducible V submodules.

$\oplus \sum_{i}A_{i}$ is isomorphic to $\oplus \sum_{i} \theta_{i} (A_{i})$, i.e $A$ is isomorphic to $\oplus \sum_{i} \theta_{i} (A_{i})$

If the steps are right, how can I show that $\oplus \sum_{i} \theta_{i} (A_{i})= V$

Answer 1

$A$ is semisimple as (right) $A$-module, $V$ is a (right) $A$-module and $v\in V$.

$\ker(A\to vA)$ is a submodule of $A$ thus it has a complementary module $A =\ker(A\to vA)\oplus B$ $$B\cong \ker(A\to vA)\backslash A\cong vA$$ so that $vA\cong B$ is a submodule of $A$, it is semisimple.

For another cyclic submodule $wA\subset V$ and $U\subset V$ semisimple we have that $$U= (wA\cap U)\oplus S \qquad \implies wA+U = wA\oplus S$$ is semisimple.

Thus every (finitely generated) $A$-module is semisimple.

Answer 2

I tried to define a map between the $A$ and V.

For some v $\in$ V define the following map:

$$\theta: A \longrightarrow V$$ $$ x\mapsto vx $$ the map is homomorphism so $A\setminus ker(\theta)$ is isomorphic to $\theta (A) \subseteq V$

$\theta (A) = vA$ $\subset V$

Then I proved that $A\setminus ker(\theta)$ is completely reducible so is $vA$

So $vA$ can be written as sum of irreducible submodules $V_{\alpha}$

$V= \sum_{v \in\ V\setminus {0}} vA = \sum_{s=1}^{n-1} \sum V_{\alpha}$

So $V$ is written as sum of irreducible submodules so $V$ is completely reducible