# Let $R$ be a commutative ring and $A,B,C$ be $R$-modules, $A$ finitely presented, $C$ flat. Then Hom$(A,B\otimes C)\cong\text{Hom}(A,B)\otimes C$

Let $$R$$ be a commutative ring with $$A$$, $$B$$, and $$C$$ all $$R$$--modules. Suppose that $$A$$ is finitely presented and $$C$$ is flat (that is, the functor $${-} \otimes C$$ preserves short exact sequences). Show that $$\text{Hom}_R(A, B \otimes_R C) \cong \text{Hom}_R(A, B) \otimes_R C.$$

Here is what I have so far:

Since $$A$$ is finitely presented, there exists a short exact sequence $$R^m \to R^n \to A \to 0$$ And by the fact that $$\text{Hom}(-,B)$$ is left exact, we have an exact sequence $$0 \to \text{Hom}(A,B) \to \text{Hom}(R^n,B) \to \text{Hom}(R^m,B)$$ which, after tensoring with the flat module $$C$$, gives us the exactness of the sequence $$0 \to \text{Hom}_R(A,B) \otimes_R C \to \text{Hom}_R(R^n, B) \otimes_R C \to \text{Hom}(R^m, B) \otimes_R C.$$

However, I am not sure how to go from here. I know the tensor-hom adjunction $$\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, \text{Hom}(B,C))$$, but I don't know if the adjunction will help.

First, there is always a natural $$R$$-linear map \begin{align*} \Phi_{X,Y,Z}: & \text{Hom}_R(X,Y)\otimes Z \to \text{Hom}_R(X,Y\otimes Z) \\ & \varphi\otimes z \qquad \qquad \mapsto \quad (x \mapsto \varphi(x)\otimes z). \end{align*} For all $$k \in \mathbb{N}$$, the homomorphism $$\Phi_{R^k,Y,Z}:\text{Hom}_R(R^k,Y)\otimes Z \to \text{Hom}_R(R^k,Y\otimes Z)$$ is an isomorphism. For $$\Phi_{R,Y,Z}:\text{Hom}_R(R,Y)\otimes Z \to \text{Hom}_R(R,Y\otimes Z)$$ is an isomorphism, because there is a natural isomorphism of functors $$\text{Hom}_R(R,-) \cong \text{id}_{R-\text{Mod}}$$, and the following diagram commutes: $$\require{AMScd} \begin{CD} \text{Hom}_R(R^k,Y)\otimes Z @>>> \Pi_{i=1}^k\text{Hom}_R(R,Y)\otimes Z \\ @V\Phi_{R^k,Y,Z}VV @V\Pi_{i=1}^k\Phi_{R,Y,Z}VV \\ \text{Hom}_R(R^k,Y\otimes Z) @>>> \Pi_{i=1}^k\text{Hom}_R(R,Y\otimes Z)\\ \end{CD}$$
The following diagram is commutative: $$\require{AMScd} \begin{CD} 0 @>>> 0 @>>> \text{Hom}_R(A,B)\otimes C @>>> \text{Hom}_R(R^n,B)\otimes C @>>> \text{Hom}_R(R^m,B)\otimes C \\ @V0VV @V0VV @V\Phi_{A,B,C}VV @V\Phi_{R^n,B,C}VV @V\Phi_{R^m,B,C}VV \\ 0 @>>> 0 @>>> \text{Hom}_R(A,B\otimes C) @>>> \text{Hom}_R(R^n,B\otimes C) @>>> \text{Hom}_R(R^m,B\otimes C) \\ \end{CD}$$ By the five lemma the homomorphism $$\Phi_{A,B,C}$$ is an isomorphism.
• thanks for your answer. I am wondering if you could provide further explanation on why $\Phi_{R^k, Y, Z}$ is an isomorphism; I also don't get how $\text{Hom}_R(R,-) \cong \text{id}_{R-Mod}$. Commented Feb 28 at 2:37
• @Squirrel-Power There is a natural isomorphism $\text{Hom}_R(R,X) \to X, \phi \mapsto \phi(1)$ with inverse $x \mapsto (r \mapsto rx)$, so $\text{Hom}_R(R,-) \cong \text{id}_{R-Mod}$. The other thing is that there is a natural isomorphism $\text{Hom}_R(\bigoplus_{i \in I} X_i,-) \cong \Pi_{i \in I}\text{Hom}_R(X_i,-)$ (with the obvious natural maps). You can find such things for example in Rotman, An Introduction to Homological Algebra, Chapter 2. Commented Feb 28 at 10:51