Prove that $M_{nm}(A \otimes B) \cong M_{n}(A) \otimes M_m(B)$ Prove that $M_{nm}(A \otimes B) \cong M_{n}(A) \otimes M_m(B)$
There's a similar question floating around but I was merely wondering if the result holds in the same way when we let A and B be fields, Rings, K-algebras, etc.
 A: Let $\varphi : M_n(A) \times M_m (B) \to M_{mn}(A \otimes B), \ \ \varphi : (a_{ij}) \times (b_{ij}) \mapsto $ let $m \gt n$ wlog, then fill in the $mn \times mn$ matrix with $m^2$ copies of the $n^2$ matrix $(a_{ij})$.  At copy $(i,j)$ dot each entry with $\otimes b_{ij}$.  So for instance $\begin{pmatrix} a \end{pmatrix} \times \begin{pmatrix} b & c \\ d & e \end{pmatrix} \mapsto \begin{pmatrix} a \otimes b & a \otimes c \\ a \otimes d & a \otimes e\end{pmatrix}$.  Then  check that $\varphi$ is an $R$-bilinear map.  
Let $M$ be a right $R$-module and $N$ a left $R$-module and let $L$ be a left $R$-module.  Then there's a bijective correspondence between $R$-bilinear maps from $\phi: M\times N \to L$ and $R$-module homs $\Phi : M \otimes_R N \to L$ and for corresponding $\phi, \Phi$, we have $\Phi  = \phi \circ \iota$ for $\iota : M \times N \to M \otimes_R N, \ \ m \times n \mapsto m \otimes n$ the usual $R$-balanced map into $M \otimes_R N$.
Thus there is a hom $\Phi$ corresponding to $\varphi$ sending $M_n(A) \otimes M_n(B) \to M_{mn} (A \otimes B)$  where the tensor product is taken over the common shared subfield $M \cap N$, or $R$ for $R$-modules, $k$ for $k$-algebras.
Now you just have to prove that $\Phi$ is injective and bijective and that the proof does not depend on what structure your taking the product with.
All you need for the tensor product to be defined is an $R$-module structure for a ring $R$.  For fields $K, L \supset F$, the're $F$-modules, a $k$-algebra is a $k$-module.  And so on...  So it's my guess that the result holds for any structure your tensor multiplying.
$\varphi$ is surjective.  Let $(c_{ij}) = \begin{pmatrix} \sum_k a_{ij,k} \otimes b_{ij,k} \end{pmatrix} \in M_{mn}( A\otimes B)$.  I.e. it represents any element since it's filled with general sums of simple tensors.
Then
$$
(c_{ij}) = \sum_{i,j} C_{ij}
$$
where each $C_{ij} = $ a matrix of zeros except for entry $i,j$ which equals the entry of $(c_{ij})$.  Then further decompose $C_{ij}$ into
$$
C_{ij} = \sum_{k=1}^{n_{ij}} B_{ij,k}
$$
where $B_{ij,k} = $ a matrix of zeros except for entry $i,j$ which equals $a_{ij,k} \otimes b_{ij,k}$.  Then by bilinearity of $\varphi$, we've shown that this tensor is the image of a sum of matrix pairs in $M_n(A) \times M_m(B)$.
