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.


1 Answer 1


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.

  • $\begingroup$ 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 $\text{Hom}_R(R,-) \cong \text{id}_{R-Mod}$. $\endgroup$ Commented Feb 28 at 2:37
  • 1
    $\begingroup$ @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. $\endgroup$
    – psl2Z
    Commented Feb 28 at 10:51

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