For positive integers $m$ and $n$ and a field $k$, write $\mathbf{M}_{m\times n}(k)$ for the $k$-vector space of $m\times n$ matrices with entries in $k$. In particular $\mathbf{M}_n(k):=\mathbf{M}_{n\times n}(k)$ is a $k$-algebra under matrix multiplication.
The vector space $\mathbf{M}_{p\times n}(k)$ has a natural structure of a right $\mathbf{M}_n(k)$-module, while $\mathbf{M}_{n\times q}(k)$ has a natural structure of left $\mathbf{M}_n(k)$-module, so we can construct the tensor product $$ T := \mathbf{M}_{p\times n}(k) \otimes_{\mathbf{M}_{n}(k)} \mathbf{M}_{n\times q}(k). $$
Now, there is a natural $\mathbf{M}_{n}(k)$-balanced map $$ \iota:\mathbf{M}_{p\times n}(k)\times \mathbf{M}_{n\times q}(k)\to \mathbf{M}_{p\times q}(k) $$ given by $$ \iota(a,b) = ab, \quad a\in \mathbf{M}_{p\times n}(k), b\in \mathbf{M}_{n\times q}(k), $$ and I wondered if the corresponding induced map $$ i:T\to \mathbf{M}_{p\times q}(k) $$ is an isomorphism.
I was trying to prove this by see if the pair $(\mathbf{M}_{p\times q}(k),\iota)$ satisfies the universal property for tensor product, but I was unable to see it.
I have to clarify: I have no evidence that supports the fact that $i$ is an isomorphism. It is just that the map $\iota$ is so "natural" that makes me think so.
So my question is: Is $i$ an isomorphism? If not, is there a simple description of $T$ (for example, as a matrix vector space)?
Yes, good find! $(\mathbf{M}_{p\times q}(k),\iota)$ does satisfy the tensor product's universal property.
When I write $E_{j,\ell}$ I'm referring to a matrix with a $1$ in the $(j,\ell)$ entry and zeros elsewhere. This notation suppresses the dimensions of the matrix, but hopefully this doesn't hurt your understanding. Do let me know if this causes any confusion, though.
This proof is somewhat prototypical for tensor product calculations IMO. It's not hard to make a guess for what a relevant extension of a given balanced map might be, but there's a sickening sense that it's not unique enough to be correct. But if we are bold enough to simply write something down, we discover that uniqueness is actually obvious, and that the interesting part is actually to show that the choice actually works at all. And at some crucial step in that proof the intuitive "non-uniqueness" is instead seen to be helping us, formally expressed in the balanced-ness of the original map.
But enough heuristics. Let $h:\mathbf{M}_{p\times n}(k)\times \mathbf{M}_{n\times q}(k)\to V$ be an $\mathbf{M}_{n\times n}(k)$-balanced map. We claim that the $k$-linear map $\tilde h: \mathbf{M}_{p\times q}(k)\to V$ defined by $E_{a,d}\mapsto h(E_{a,1},E_{1,d})$ is the unique $k$-linear map such that $h=\tilde h\circ \iota$. Uniqueness is clear because $E_{a,d}=E_{a,1}E_{1,d}$ imposes this constraint on any purported extension $\tilde h$. It remains to show that this choice actually satisfies the equation.
Let $(E_{a,b},E_{c,d})\in \mathbf{M}_{p\times n}(k)\times \mathbf{M}_{n\times q}(k)$; we must show that $h(E_{a,b},E_{c,d})=\tilde h(\delta_{b,c} E_{a,d}) = \delta_{b,c}h(E_{a,1},E_{1,d})$. Indeed, since $h$ is balanced we can write $$h(E_{a,b},E_{c,d})=h(E_{a,1}E_{1,b},E_{c,1}E_{1,d})=h(E_{a,1},E_{1,b}E_{c,1}E_{1,d})$$ When $b\neq c$ the second input on the RHS is zero, and so bilinearity implies the whole thing is zero. Conversely, if $b=c$ then $E_{1,b}E_{c,1}E_{1,d}=E_{1,d}$. In either case we got the desired result, that is, $h(E_{a,b},E_{c,d})=\delta_{b,c}h(E_{a,1},E_{1,d})$.