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?

  • 1
    $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ Thank you this helps! Where can I find this? $\endgroup$
    – CO2
    Jan 1 '21 at 15:26
  • $\begingroup$ Bourbaki, Algebra, Ch. II, §4 Relations between tensor products and homomorphisms modules, n°4, p.274. $\endgroup$
    – Bernard
    Jan 1 '21 at 15:30
  • $\begingroup$ 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. $\endgroup$
    – CO2
    Jan 1 '21 at 15:40
  • $\begingroup$ 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 $\operatorname{Hom}(V\otimes \mathbf C,\mathbf R)\simeq\operatorname{Hom}\bigl(V,\operatorname{Hom}(\mathbf C,\mathbf R)\bigr)$. $\endgroup$
    – Bernard
    Jan 1 '21 at 15:47

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