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?
Best Answer
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.