Is $ mA \otimes B = m(A\otimes B)$?

Question

I am writing a proof which would simplify if the following was true.

Let $ A $ and $ B $ be two $ R $-modules, and suppose that $ B $ is flat over $ R $. Let $ m \in \mathbb{Z} $. Is it true that $$ mA \otimes _{ R } B = m(A\otimes _ { R} B) ? $$

Answer 1

By flatness, if $A'$ is a submodule of $A$ then one can identify $A'\otimes_R B$ with a submodule of $A\otimes_R B$, namely as that generated by all $a'\otimes b$ with $a'\in A'$.

With this observation it's surely obvious that $mA\otimes_R B$ is the submodule of $A\otimes_R B$ generates by the elements $(ma)\otimes b=m(a\otimes b)$ and that is $m(A\otimes_R B)$.