Adjunction of $A$-Module Morphisms

Question

Let $K$ a field, $A$ a $K$-algebra and $L$ a finite field extension of $K$. Futhermore let $M, N$ be $A$-modules.

I want to know why & how to see that we have following identifications between morphisms:

1. $$\hom_A(M,N) \otimes_K L \cong \hom_A(M,N \otimes_K L) $$

2. $$\hom_A(M,N \otimes_K L) \cong \hom_{A \otimes_K L}(M \otimes_K L,N \otimes_K L) $$

My ideas: Regarding the first one we have: $L \cong K^d$ for $d= dim_K(L)$. Therefore $N \otimes_K L \cong N^d$, so it would suffice to show $\hom_A(M,N)^d \cong \hom_A(M,N^d)$, but this is the universal property of product. But here I "forget" the $A$ module structure of $N$, so I'm not sure if I can argue in this way.

Concerning the second identity I have no idea.

Answer 1

1. Your identification works as $A$-modules. It is not the $A$-module structure we forget here (in neither steps) but the multiplication of $L$.
   $\def\t{\underset K\otimes}$ Suppose $e_1,\dots,e_d$ is a basis of $L$ over $K$ (usually it's taken to be $1,\alpha,\alpha^2,\dots,\alpha^{d-1}$ for a single $\alpha$), then an element $u$ of $N\t L$ can be written as $\ u=\sum_in_i\otimes e_i$ with $n_i\in N$, then we simply have $$au\ =\ a\cdot\left(\sum_in_i\otimes e_i\right)=\sum_i(an_i)\otimes e_i$$ So the action of $A$ is working coordinatewise, hence the identification $L\cong K^d$ does induce an $A$-module isomorphism $N\t L\cong N^d$.
   As you say, $\hom_A(M,N)^d\cong\hom_A(M,N^d)$ basically by definition of the product, which is taken in the category of $A$-modules, so there's no problem here neither. If $A$ is not commutative, the homsets will carry only $K$-vector space structure.
   Tensoring by $L$ will lift it to an $L$-vector space.

2. Define a pair of homomorphisms and prove they are inverses to each other.
   Suppose $\varphi\in\hom_A(M,\,N\t L)$. Its correspondent will be $$m\otimes\lambda\,\longmapsto\,\varphi(m)\cdot\lambda $$ It preserves the $A\t L$-action [$(a\otimes\lambda_1)\cdot(m\otimes\lambda_2):=(am\otimes\lambda_1\lambda_2)$].

   For the other direction, if $\psi\in\hom_{A\t L}(M\t L,\,N\t L)$, map it to $$m\mapsto \psi(m\otimes1)$$