# $\operatorname{Hom}_{K}(K \otimes_k U, K \otimes_k V) \cong K \otimes_k \operatorname{Hom}_{k}(U, V)$

Let $$K$$ be a field extension of $$k$$, $$U$$ and $$V$$ finite dimensional (just in case) vector spaces over $$k$$.

How do I construct an isomorphism like this?

$$\operatorname{Hom}_{K}(K \otimes_k U, K \otimes_k V) \cong K \otimes_k \operatorname{Hom}_{k}(U, V)$$

I have a lot of problems similar to this and I think it'll start to go easy as soon as I figure this one out. Normally I would try to check the universal property, but it is an implicit way.

And also is this true for infinite dimensional case?

• I don't know what you mean by "normally I would try to check the universal property, but it is an implicit way". Are you saying that you would normally use the universal property but can't find a way to do so in this case, or are you saying that you would like to find another approach because using the universal property is too "implicit" (whatever it is you mean by that). – Omnomnomnom May 21 at 18:59
• @Omnomnomnom both. By implicit I mean that it proves the existence of an isomorphism without explicit formula. – Vladislav May 21 at 19:04

We are looking for an isomorphism $$\phi: \operatorname{Hom}_K(K \otimes_k U, K \otimes_k V) \to K \otimes \operatorname{Hom}_k(U,V).$$

I find that a nice approach to these kinds of problems is to first figure out what should happen to the "nice" elements of the spaces in question. For the tensor products, the nice elements are the pure tensors, and for the Hom spaces, the nice elements are the rank-1 elements. Note that the rank-1 elements of $$\operatorname{Hom}_k(U,V)$$ are of the form $$vf$$, where $$v \in V$$ and $$f \in U^*$$.

So, in order to produce this $$\phi$$, we consider what should happen to elements of the form $$\bar v \bar f$$, with $$\bar v \in K \otimes_k V$$ and $$\bar f \in (K \otimes_k U)^*$$. We'll go further and start with "nice" elements $$\bar v, \bar f$$, namely $$a_V \otimes_k v$$ and $$a_U \otimes_k f$$ with $$a_U,a_V \in K$$. The most "obvious" way to rearrange these elements to produce and element of the codomain is to take $$\phi[(a_V \otimes_k v)(a_U \otimes_k f)] = (a_Va_U) \otimes_k (vf).$$ Now, extending this map linearly uniquely defines $$\phi$$ to all of $$\operatorname{Hom}_k(K \otimes_k U, K \otimes_k V)$$. To argue that this map has a unique linear extension, it suffices to use the universal property on the corresponding map $$\bar \phi$$ from $$\operatorname{Hom}_K(K \otimes_k U,K \otimes_k V) \cong (K \otimes _k U)^* \otimes_K (K \otimes_k V)$$.

Note also that if we fix basis of $$K,U,V$$, then we describe this map more concretely by saying what $$\phi$$ does to the coordinate vectors of elements of $$\operatorname{Hom}_K(K \otimes_k U, K \otimes_k V)$$ relative to the basis corresponding to the chosen bases of $$K,U,V$$.

With that, it now suffices to define $$\phi^{-1}$$ similarly and show that the compositions $$\phi \circ \phi^{-1}$$ and $$\phi^{-1} \circ \phi$$ yield the identity over the "nice" elements, and therefore over the whole space.

Alternatively, it suffices to show that the map is bijective. To show that $$\phi$$ is surjective, note that the "nice" elements $$a \otimes_k (vf)$$ with $$a \in K$$, $$v \in V,$$ and $$f \in U^*$$ span the codomain, which means that it suffices to show that these elements lie in the image of $$\phi$$ (which is easy given the above construction). Showing that the map is injective is a bit trickier; I suspect that there is a straightforward approach via the universal property.

Actually, if you have established that $$\operatorname{Hom}_k(U,V) \cong U^* \otimes_k V$$ and $$(K \otimes_k U) \otimes_K (K \otimes_k V)$$ $$\cong K \otimes (U \otimes_k V)$$, then we have the following chain of isomorphisms: $$\operatorname{Hom}_K(K \otimes_k U, K \otimes_k V) \cong\\ (K \otimes_k U)^*\otimes_K (K \otimes_k V) \cong\\ (K \otimes_k U^*) \otimes_K (K \otimes_k V) \cong\\ K \otimes_k (U^* \otimes_k V) \cong\\ K \otimes_k \operatorname{Hom}_k(U,V).$$

• For the bijectivity part I think one could try the following: $U$ and $V$ are both finite-dimensional $k$-vector spaces and $K \otimes U$ resp. $K \otimes V$ resp. $K \otimes \text{Hom}_k(U,V)$ can all be seen as $K$-vector spaces of the same dimension over $K$ as $U$ resp. $V$ resp. $\text{Hom}_k(U,V)$ over $k$. I think then it suffices to show either surjectivity or injectivity of this $K$-linear map $\phi$ since both sides would have dimension dim$(U)$dim$(V)$. – M. Wang May 21 at 19:59
• @Brian It does matter in this context and it should; I'll edit accordingly – Omnomnomnom May 21 at 20:03
• @Brian In fact, both make sense. For a concrete example where this matters: if $K = U = V = \Bbb C$ and $k = \Bbb R$, then the conjugation map $a + bi \mapsto a - bi$ lies in the $\operatorname{Hom}_k$ version but not the $\operatorname{Hom}_K$ version. – Omnomnomnom May 21 at 20:13
• @M.Wang That is correct, and your approach makes sense given that the surjectivity of this map is easier to show. However, because of the algebraic nature of the question, I was hoping to answer this question as generically as possible, which is to say in a way that allows for arbitrary vector spaces. – Omnomnomnom May 21 at 21:33