If $V$ is a simple $R$-module and $R = S\oplus T$, then $V = SV \oplus TV$

Question

Let $V$ be a simple left $R$-module where $R$ is a unital ring.

Suppose we can write $R = S \oplus T$ for some ideals $S,T$. Is it true that

$$V = SV \oplus TV?$$

Attempt:

I can see that we certainly have $V= RV = (S+T)V \subseteq SV + TV \subseteq V$ so it suffices to check that $SV \cap TV = 0$.

Note that $SV \cap TV$ is a submodule of $V$. If it were non-zero, then $SV \cap TV = V$ implying that $SV = TV = V$, but I cannot find a contradiction.

Maybe the statement is false?

EDIT: I attempted to answer my own question below. Please have a look and share your feedback!

Answer 1

Answering my own question. Feedback is appreciated.

The claim is true.

Assume to the contrary that $SV \cap TV \neq 0$. Since $V$ is simple, we get $SV \cap TV = V$. In particular, also $SV = TV = V$. Then

$$V= SV = S(TV)= (ST) V$$

and thus there are elements $s \in S, t \in T$ such that $st \neq 0$ (otherwise we obtain that $ST = 0$ and then $V = 0$, contradicting that by definition of simple module $V \neq 0$).

Thus $$0 \neq st \in sT \cap St \subseteq T \cap S = \{0\}$$

which is a contradiction.