Question

Let R be a k-algebra where k is a field, and M,N be left Rmodules, with dimkM <∞. It is known that, for any field extension K ⊇ k, the natural map θ : (HomR(M,N))K → HomRK(MK,NK) is an isomorphism of K-vector spaces. Replacing the hypothesis dimkM < ∞ by “M is a finitely presented R-module,” give a basis-free proof for the fact that θ is a K-isomorphism.

Answer 1

First consider the special case when $M \sim R^n$ . In this case, we have the canonical isomorphisms

$\left(\operatorname{Hom}_R(M,N)\right)^K \sim \left(\operatorname{Hom}_R(R^n,N)\right)^K \sim (nN)^K \sim n \cdot N^K,$

$\operatorname{Hom}_{R^K}(M^K, N^K) \sim \operatorname{Hom}^{R^K}((R^K)^n, N^K) \sim n \cdot N^K.$

From this, we can safely conclude that $\theta$ is an isomorphism. (Some commutative diagrams must be checked, but it is mostly routine work.) Next we assume $M$ is a finitely presented $R$ -module, which means that there exists an exact sequence of $R$ -modules $M1 \to M2 \to M \to 0$ where $M_1 = R^n$ and $M_2 = R^m$ . Applying the left-exact Hom - functors into $N$ and into $N^K$ , we have the following commutative diagram:

0 $\rightarrow$ $\left(Hom_{R}\left(M,N\right)\right)^{K}\rightarrow \left(Hom_{R}\left(M_{2},N\right)\right)^{K}$ $\left(Hom_{R}\left(M_{1},N\right)\right)^{K}$

$\downarrow \theta$

$\downarrow \theta_{2}$

$\downarrow \theta_{1}$

0 $\rightarrow$ $Hom_{R^K}\left(M^K,N^K\right)\rightarrow Hom_{R^K}\left(M_2^K,N^K\right)\rightarrow Hom_{R^K}\left(M_1^K,N^K\right)$

Since $\theta_{1},\theta_{2}$ are both isomorphisms, an easy diagram chase shows that $\theta$ is also an isomorphism.