# How to prove that $(M \oplus N) \otimes P$ is isomorphic to $(M \otimes P) \oplus (N \otimes P)$?

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.

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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 $f$.
So, you know what you need to send $(x \otimes z, y \otimes z)$ to, but every element isn't of this form. However, your map will have to be additive, so for an arbitrary pair of simple tensors $$(m \otimes p_1, n \otimes p_2)$$ we write as $$(0 \otimes p_2, n \otimes p_2) + (m \otimes p_1, 0 \otimes p_1)$$ which we had better map to $$(0, n)\otimes p_2 + (m, 0) \otimes p_1$$ if this map is going to be inverse to the map you already defined. There's thus only one thing the map can be. You still must check that this is well-defined (i.e. the corresponding map out of the products is bilinear).
Hint: Find module homomorphisms $M\otimes P\rightarrow (M\oplus N)\otimes P$ and $N\otimes P\rightarrow (M\oplus N)\otimes P$.