$Hom(A^n, A^m)$ is a free $mn$ module

Question

Prove $Hom(A^n, A^m) $ is a free module of dimension $mn$.

The fact that the dimension of $Hom(A^n, A^m) $ is $mn$ I think can be proven defining an explicit construction of the element, this is using a basis The following link has a proof I think can be adapted to this theorem https://www.ma.utexas.edu/users/kit/Algebra/Section_4.3--Dual_Spaces.pdf.

But is there a more refined way to do it?

Answer 1

Let $e_1, \dots, e_n$ be a basis for $A^n$ and $d_1, \dots, d_m$ a basis for $A^m$. An $A$-linear map $A^n \to A^m$ is determined exactly by where it sends each $e_i$ (do you see why? given where the map sends each $e_i$, where does it send the general element $\sum_i c_i e_i$?), and each $e_i$ is sent to some element of the form $$ \sum_{j=1}^m a_{ij}d_j \in A^m, $$ where the $a_{ij}$ are in $A$. Conversely, clearly any set of $a_{ij}$ also gives rise to a map $A^n \to A^m$ in the same way. In other words, each map can be expressed uniquely as a $m \times n$ matrix of elements of $A$, and certainly this is just a free $A$-module of rank $mn$.

Answer 2

A product of modules is characterized by the property that to give a homomorphism into the product is to give a collection of homomorphisms going into each component:

$$\textrm{Hom}_A(M, \prod\limits_i N_i) \cong \prod\limits_i \textrm{Hom}_A(M, N_i)$$

A coproduct (direct sum) of modules is characterized by the property that to give a homomorphism out of the coproduct is to give a collection of homomorphisms going out of each component:

$$\textrm{Hom}_A(\bigoplus\limits_i M_i, N) \cong \prod\limits_i\textrm{Hom}_A(M_i,N)$$

A priori, these are just bijections of sets, but all these guys naturally have module structures, and the bijections are easily seen to be module homomorphisms.

Putting these facts together, and using the fact that finite direct sums are the same thing as finite direct products, we get

$$\textrm{Hom}_A(A^n,A^m) \cong \prod\limits_{i=1}^n \textrm{Hom}_A(A,A^m) \cong \prod\limits_{i=1}^n \prod\limits_{j=1}^m \textrm{Hom}_A(A,A) \cong \prod\limits_{i=1}^n \prod\limits_{j=1}^m A \cong A^{nm}$$