Let $k$ be a field and $A$ a finite-dimensional unital associative $k$-algebra. For a finite-dimensional left $A$-module $V$, the socle of $V$, written $\operatorname{soc}V$, is defined to be the sum of all simple submodules of $V$ (or equivalently the largest semisimple submodule of $V$).
Let $V$ be a finite-dimensional left $A$-module, and suppose that $W_1,W_2\subseteq V$ are submodules such that $W_1+W_2=V$. I want to know whether we have, in general $$\operatorname{soc}W_1+\operatorname{soc}W_2=\operatorname{soc}V.$$ Clearly $\operatorname{soc}W_1+\operatorname{soc}W_2\subseteq \operatorname{soc}V$.
To try to prove the opposite, I've considered a simple submodule $L\subseteq V$. Then if $L\cap W_1\neq 0$ or $L\cap W_2\neq 0$ then by simplicity we know $L\subseteq W_1$ or $L\subseteq W_2$, so let's suppose that $L\cap W_1=0$ and $L\cap W_2=0$. The hope is that isomorphic copies of $L$ lie inside both $W_1$ and $W_2$, and we are seeing a diagonal copy of it. However, all I can deduce from this assumption is that $L$ is a submodule of $W_1/(W_1\cap W_2)$ and $W_2/(W_1\cap W_2)$ (one can see this by quotienting $V$ by $W_1$ or $W_2$ and applying the diamond isomorphism thoerem). I don't know how to argue that a copy of $L$ actually lies in the socle of $W_1$ and $W_2$.
Any ideas/comments/counterexamples/proofs/references?
There is a simple counterexample explained to me. Consider the commutative ring $A=k[z]/(z^2)$. Then consider the three dimensional representation $k\langle v_1,v_2,v_3\rangle$ where $zv_1=zv_2=v_3$ and $zv_3=0$. Then the socle of this module is $k\langle v_1-v_2,v_3\rangle$. On the other hand $W_1=k\langle v_1,v_3\rangle$ and $W_2=k\langle v_2,v_3\rangle$ are submodules with socle $k\langle v_3\rangle$ that add up to $V$.