“Simplifying” an extension of scalars

Let $A$ be a commutative $\mathbb Z$-algebra and $M$ be a $\mathbb Z\oplus \mathbb Z$-module. Then $A\otimes_{\mathbb Z} M$ is an $A\oplus A$-module. Is it true that $(A\oplus A)\otimes_{\mathbb Z\oplus \mathbb Z} M \cong A\otimes_{\mathbb Z} M$ as $A\oplus A$-modules? It seems to me that the map given by $$(a_1, a_2)\times m \mapsto a_1\otimes_{\mathbb Z} (1,0)m + a_2\otimes_{\mathbb Z} (0,1) m$$ defines a map in one direction, and that the map $$a\times m\rightarrow (a,a)\otimes_{\mathbb Z\oplus \mathbb Z} m$$ defines its inverse.

More generally, is it true/ is there a slick reason that given commutative unital $R$-algebras $A$, $B$ and $C$ such that $B\cong A\otimes_{R} C$, and a $C$-module $M$, we have $B\otimes_C M\cong A\otimes_{R} M$ as $B$-modules?

Yes. You ask if in general $$(A \otimes_R C) \otimes_C M = A \otimes_R M,$$ when both sides make sense. This is true; the map is $(a \otimes c) \otimes m\mapsto a\otimes cm$ with inverse map $$a \otimes m \mapsto (a \otimes 1) \otimes m.$$ Of course there's a lot to check here before you believe me.
• (alternatively, prove the tensor product is associative and reduce to $C \otimes_C M = M$.) – user29743 Jul 27 '12 at 19:51