# Relationship of Hom and tensor for vector spaces

Let $$U,V$$ be two vector spaces over $$K$$ and $$L$$ is a field extension of $$K$$.

Is it true that for any finite dimension $$U,V,L$$ over $$K$$ that we have natural isomorphisms

$$\mathrm{Hom}_K(U,V)\otimes L\cong \mathrm{Hom}_K(U\otimes L,V)\cong\mathrm{Hom}_K(U,V\otimes L)?$$

It looks that it is related to the Hom-tensor adjunction, but slightly different.

Is this still true for infinite dimensional?

Bacially, I want to show this on page 26 lemma 1.2.6.

$$\mathrm{Hom}_\mathbb{R}(V,\mathbb{R})\otimes \mathbb{C}\cong \mathrm{Hom}_\mathbb{R}(V\otimes \mathbb{C},\mathbb{R})\cong\mathrm{Hom}_\mathbb{R}(V,\mathbb{R}\otimes \mathbb{C}).$$

The natural isomorphism between the first and third is clear via $$V^*\otimes \mathbb{C}\cong\mathrm{Hom}_{\mathbb{R}}(V,\mathbb{C})$$. But what about the second one?

• I think I can see a natural isomorphism between the first and third term, based on the isomorphism $A^{*} \otimes B \to \mathrm{Hom}(A, B)$ given by $f \otimes b \mapsto (a \mapsto f(a) b)$ . This holds when the vector spaces are finite-dimensional. Jan 1 '21 at 15:18

Not a full answer, but in a more general context (modules over a commutative ring $$R$$), we have a natural homomorphism: $$\DeclareMathOperator{\Hom}{Hom}$$ $$\Hom_R(U,V)\otimes_R\Hom(M,L)\longrightarrow\Hom_R(U\otimes_R M,V\otimes_R L)$$ which is an isomorphism is any of the pairs $$(U,M)$$ or $$(U,V)$$ or$$(M,L)$$ is made up of projective modules of finite type.
Considering the particular case when $$M=R$$, which implies $$\Hom_R(R,L)\simeq L$$, we have that $$\Hom_R(U,V)\otimes_R L\longrightarrow\Hom_R(U,V\otimes_R L)$$ is an isomorphism is $$U$$ or $$L$$ is a projective module of finite type – which, if the base ring is a field $$K$$, means $$U$$ or $$L$$ is a finite dimensional vector space.
• Thank you! So somehow $\mathrm{Hom}_\mathbb{R}(V,\mathbb{C})\otimes\mathbb{C}$ will not be isomorphic to $\mathrm{Hom}_\mathbb{R}(V\otimes\mathbb{C},\mathbb{R})$ naturally I guess? Because $\mathrm{Hom}_\mathbb{R}(\mathbb{C},\mathbb{R})$ and $\mathbb{C}$ are not naturally isomorphic.
• I didn't have the time to think about it. Of course, if I find an answer later, I'll update my answer. You're probably right, as there's a canonical isomorphism $\operatorname{Hom}(V\otimes \mathbf C,\mathbf R)\simeq\operatorname{Hom}\bigl(V,\operatorname{Hom}(\mathbf C,\mathbf R)\bigr)$. Jan 1 '21 at 15:47