TITLE: How to prove that $(M \oplus N) \otimes P$ is isomorphic to $(M \otimes P) \oplus (N \otimes P)$?
QUESTION [10 upvotes]: Let $M, N, P$ be $A$-modules, where $A$ is a commutative ring with identity. I want to prove that $(M \oplus N) \otimes P$ is isomorphic to $(M \otimes P) \oplus (N \otimes P)$. 
I start by defining a map $(M \oplus N) \times P \rightarrow (M \otimes P) \oplus (N \otimes P)$ by $(x+y,z) \mapsto (x \otimes z, y \otimes z)$. This map is $A$-bilinear, thus it induces a homomorphism $f:(M \oplus N) \otimes P \rightarrow (M \otimes P) \oplus (N \otimes P)$ such that $(x,y) \otimes z \mapsto (x \otimes z, y \otimes z)$. 
Now i want to construct a homomorphism the other way, i.e. 
$g: (M \otimes P) \oplus (N \otimes P) \rightarrow (M \oplus N) \otimes P$ and show that $g \circ f, g \circ g$ are the identity maps. I am having difficulty in constructing a suitable bilinear map that will induce $g$.
Any insights?
Thanks.

REPLY [10 votes]: Consider the two maps $$\varphi: M \times P \longrightarrow (M \oplus N) \otimes P,$$ $$(x,z) \mapsto (x,0) \otimes z$$ and $$\psi: N \times P \longrightarrow (M \oplus N) \otimes P,$$ $$(y,z) \mapsto (0,y) \otimes z.$$ These maps are easily checked to be $A$-bilinear, so they define homomorphisms $$\Phi: M \otimes P \longrightarrow (M \oplus N) \otimes P$$ and $$\Psi: N \otimes P \longrightarrow (M \oplus N) \otimes P.$$ Now check that $g = \Phi \oplus \Psi$ is an inverse to $f$.