Let $M$ be a commutative ring with a subring $R \subset M$ and let $(C_\bullet,\partial^C)$, $(D_\bullet,\partial^D)$ be two chain complexes that are freely and finitely generated $R$-modules in every degree with the boundary operators being $R$-linear.
Let us assume that $C_\bullet$ and $D_\bullet$ are chain homotopy equivalent over $R$, i.e. there exist two chain maps $$i \colon C_\bullet \to D_\bullet, \, j \colon D_\bullet \to C_\bullet$$ that are $R$-linear so that both $j \circ i$ and $i \circ j$ are chain homotopic (over $R$) to the respective identities $\mathrm{id}_{C}$ and $\mathrm{id}_D$.
After tensoring with $M$ over $R$ we obtain two new chain complexes $(C_\bullet \otimes_R M,\partial^C \otimes \mathrm{id}_M)$ and $(D_\bullet \otimes_R M,\partial^D \otimes \mathrm{id}_M)$. Note that these two are now also $M$-modules with the action given by $n \cdot (c \otimes m):=c \otimes n \cdot m$ for $c \in C_\bullet, \, m,n \in M$ (similarly for $D_\bullet \otimes_R M$).
I did a direct computation and believe to have shown that $$I:=i \otimes \mathrm{id}_M \text{ and } J:=j\otimes \mathrm{id}_M$$ define a $M$-linear chain homotopy equivalence.
Question: Is this true or have I gone wrong somewhere? If it's true, does one have a reference to such a result? It looks so standard and easy, but I couldn't find any similar statement which led to some serious doubts...