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This is an exercise of Jacobson's algebra volume II, page $155$:

Let $R$ and $S$ be rings. Let $P$ be a finitely generated projective left $R$-module, $M$ an $R-S$ bimodule, $N$ a left $S$-module. Then there is a group isomorhpism:

$\alpha: hom_{R}(P,M) \otimes_{S} N \rightarrow hom_{R}(P,M \otimes_{S}N)$

such that for $f \in hom_{R}(P,M)$ and $y \in N$ then $\alpha(f \otimes y)$ is the homomorphism $x \mapsto f(x) \otimes {y}$.

I can see why this is a group homomorphism, but why is it a bijection?

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up vote 3 down vote accepted

Hint: Keeping $M$ and $N$ fixed, call the statement $A(P)$. It suffices to show $$ A(R) $$ and $$ A(P)\text{ and }A(Q)\iff A(P\oplus Q), $$ which are clear.

Answer to a comment:

Again, we keep $M$ and $N$ fixed. Write your map as $$ \alpha_P:X(P)\to Y(P), $$ and call $A(P)$ the statement "$\alpha_P$ is an abelian group isomorphism".

Let's agree that $\simeq$ means "canonical isomorphism", and that $P$ and $Q$ are ("variable") finitely generated projective left $R$-modules.

I claim:

(a) $A(R)$ holds,

(b) $X(P\oplus Q)\simeq X(P)\oplus X(Q)$,

(c) $Y(P\oplus Q)\simeq Y(P)\oplus Y(Q)$,

(d) $\alpha_{P\oplus Q}=\alpha_P\oplus\alpha_Q$,

(e) $A(P)$.

Note that (d) implies: "$\alpha_{P\oplus Q}$ is an isomorphism if and only if $\alpha_P$ and $\alpha_Q$ are isomorphisms".

I hope this is clearer.

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Dear @user10: I tried to answer your comment by editing the answer. – Pierre-Yves Gaillard Apr 2 '12 at 11:58

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