completely reducible module iff intersection of maximal submodules is trivial

Question

Let $V$ be a finite dimensional $A$-module where $A$ is a finite dimensional algebra. I want to show that $V$ is completely reducible (semi-simple) if and only the intersection of all the maximal submodules (Jacobson radical?) is trivial.

My Thoughts...

Assuming $V$ is completely reducible we can write $V$ as

$$V=\dot\sum^n V_i$$

where the $V_i$ are irreducible for all $i$. I was thinking originally that all the maximal submodules could be generated by taking any $n-1$ components of the above sum and looking at their sum. However, this doesn't seem credible to me.

I can conclude that any maximal submodule must either completely contain or trivially intersect any of the irreducible submodules and each irreducible submodule must be contained in some maximal submodule (because duh). I do believe that any $n-1$ of the irreducible submodules must be maximal (as otherwise writing the module as an n-tuple I would find a proper submodule of an irreducible submodule). The intersection of these $n$ maximal submodules must be trivial as it can be written as

$$V_1 \cap V_2 \cap \dots \cap V_n \cap \text{some other stuff} = \{0\}$$

We can conclude then the intersection of all maximal submodules, which must be contained in the above, must be trivial.

I'm not sure about the converse direction. I'm not really looking for an answer more a kind of a push in the right direction.

Answer 1

Suppose that $V$ is completely reducible and $0\neq W$ be the intersection of all of the maximal submodules of $V$. Then let $W\oplus U=V$, from $W\cap U=0$ we know that $W\not\subset U$ hence $U$ is not maximal. So $U$ is contained in a maximal submodule by finite dimensionality, namely $U\subset V'$, hence $W\oplus V'\subsetneq V$, contradiction.
Conversely, suppose that the intersection of all of the maximal submodule of $V$ is trivial, then the natural homomorphism $V\rightarrow \oplus_{i=1}^nV/M_i$ is injective where $V/M_i$ is irreducible. So $V$ is completely reducible, as a submodule of a completely reducible module.

Answer 2

If $V$ is irreducible, we are done. Otherwise, let $W_1$ be a non-zero proper irreducible submodule of $V$. Since the intersection of all maximal submodules is zero, there must exist a maximal submodule $V_1$ such that $W_1 \not \subset V_1$. Then $V = W_1 \oplus V_1$.

If $V_1$ is irreducible, we are done. Otherwise, repeat the above process as follows: let $W_2$ be a non-zero proper irreducible submodule of $V_1$, and choose a maximal $U_2$ such that $W_2 \not \subset U_2$. Then check that $V_1 = W_2 \oplus V_2$ where $V_2 = U_2 \cap V_1$.

If $V_2$ is irreducible, we are done, else repeat as above. Since $V$ is finite dimensional, this process terminates and we conclude that $V$ is completely reducible.