$\text{Hom}_{A \otimes_K L}(P \otimes_K L, P \otimes_K L) \cong \text{Hom}_{A}(P,P) \otimes_K \text{Hom}_{L}(L,L)$?

Question

Let $K$ be a field and $L$ an extensional field of $K$. Fix $A$ a finite $K$-algebra and $P$ a right $A$-module. Whether $\text{Hom}_{A \otimes_K L}(P \otimes_K L, P \otimes_K L) \cong \text{Hom}_{A}(P,P) \otimes_K \text{Hom}_{L}(L,L)$?

Answer 1

Not a definite answer, but at least partial:

First note that $\mathrm{Hom}_L(L, L)=L,$ so what one wants to prove is that $$\mathrm{Hom}_{A \otimes_K L}(P \otimes_K L, P \otimes_K L) \simeq \mathrm{Hom}_A(P, P)\otimes_K L.$$

First, observe that there is a natural (in $P$ at least, but in $Q$ as well) morphism $$\varphi_{P, Q}:\mathrm{Hom}_A(P, Q)\otimes_K L \rightarrow \mathrm{Hom}_{A \otimes_K L}(P \otimes_K L, Q \otimes_K L)$$ given by $f \otimes \lambda \mapsto h_{f, \lambda},$ where $h_{f, \lambda}$ is a homomorphism $P \otimes_K L \rightarrow Q \otimes_K L$ define on generators by $(p \otimes \mu)\mapsto f(p)\otimes(\lambda \mu)$.

I claim that this morphism is isomorphism for all finitely presented modules $P$.

First of all, observe that it is an isomorphism for $P=A$, the regular module. Symbolically, we have

$$\mathrm{Hom}_{A \otimes_K L}(A \otimes_K L, Q \otimes_K L) \simeq Q \otimes_K L \simeq \mathrm{Hom}_{A}(A, Q)\otimes_K L$$ and going through the two obvious isomorphism quickly shows that the inverse of their composite is precisely $\varphi_{A, Q}$.

Next, observe that if $\varphi_{P, Q}$ and $\varphi_{P', Q}$ are isomorphisms, then so is $\varphi_{P \oplus P', Q}:$ again, symbolically, we have

$$\mathrm{Hom}_{A \otimes_K L}((P\oplus P') \otimes_K L, Q \otimes_K L)\simeq \mathrm{Hom}_{A \otimes_K L}((P \otimes_K L)\oplus (P' \otimes_K L), Q \otimes_K L)$$ $$\simeq \mathrm{Hom}_{A \otimes_K L}(P \otimes_K L, Q \otimes_K L) \oplus \mathrm{Hom}_{A \otimes_K L}(P' \otimes_K L, Q \otimes_K L)$$ $$\simeq^{*} \mathrm{Hom}_A(P, Q)\otimes_K L \oplus \mathrm{Hom}_A(P', Q)\otimes_K L $$ $$\simeq(\mathrm{Hom}_A(P, Q) \oplus \mathrm{Hom}_A(P', Q))\otimes_K L \simeq \mathrm{Hom}_A(P\oplus P', Q) \otimes_K L$$ and again, the painful checking of the used standard isomorphism would verify that the composite (from right to left) is $\varphi_{P \oplus P', Q}$. (Note: the isomorphism $\simeq^{*}$ is induced by the isomorphisms $\varphi_{P, Q}, \varphi_{P', Q}$.)

Thus, we know know that $\varphi_{P, Q}$ is an isomorphism whenever $P$ is finitely generated and free (i.e. $P=A^n=A \oplus A \oplus A \dots \oplus A$, by obvious induction).

Now, if $P$ is finitely presented, we have an exact sequence of the form $$A^n \rightarrow A^m \rightarrow P \rightarrow 0$$. Apply to it the functors $\mathrm{Hom}_{A\otimes_K L}(-\otimes L, Q \otimes L)$ and $\mathrm{Hom}_A(-, Q)\otimes L$. Both these functors are left exact - for the second one, it is because the field $L$ is a flat $K$-module. Thus, we obtain a comm. diagram

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllll} 0 & \ra{} & \mathrm{Hom}_A(P, Q)\otimes L & \ra{} & \mathrm{Hom}_A(A^m, Q)\otimes L & \ra{} & \mathrm{Hom}_A(A^n, Q)\otimes L & \\ & &\da{\varphi_{P, Q}} & & \da{\varphi_{A^m, Q},\; \simeq} & & \da{\varphi_{A^n, Q},\; \simeq} & & \\ 0 & \ra{} & \mathrm{Hom}_{A\otimes_K L}(P\otimes L, Q \otimes L) & \ra{} & \mathrm{Hom}_{A\otimes_K L}(A^m\otimes L, Q \otimes L)& \ra{} & \mathrm{Hom}_{A\otimes_K L}(A^n\otimes L, Q \otimes L) & \\ \end{array} $$

with exact rows. Finally, using five lemma, one can deduce that $\varphi_{P, Q}$ is an isomorphism.

Thus, if $P$ is finitely presented (e.g. if $\mathrm{dim}_K A < \infty $ and $\mathrm{dim}_K P < \infty $), setting $Q :=P$ yields the desired isomorphism.