Kronecker product for matrices over noncommutative fields

Question

Suppose $A, B, C, D$ are matrices over a commutative field such that $AC$ and $BD$ are well defined. Then we know that

$$(A \otimes B)(C \otimes D) = (AC) \otimes (BD).$$

Are there clean formulas available over noncommutative fields (starting from the left-hand side) ?

Answer 1

There are two fundamental problems going on here:

1. Over a noncommutative ring, one cannot take a tensor product of two right modules, and
2. Over a noncommutative ring, the space of linear maps is not a module over the original ring.

The first problem is fairly obvious when you try to do it, you really need to be taking a tensor product of a left module with a right module to impose the equivariance condition. I find the second problem more interesting, since a bunch of stuff that you would otherwise think is meaningful (writing down diagonal scalar matrices for instance) seems to go horrifically wrong in the noncommutative case.

Let $R$ be a unital ring, not necessarily commutative. Then obviously we can form matrices with entries in $R$, and we still have matrix addition (commutative) and matrix multiplication (associative), we can scale them on the left or on the right by some scalar, and we can form Kronecker products. However, I argue that only matrix addition and multiplication are meaningful, while scalar multiplication and Kronecker products are meaningless.

Suppose that $U$ and $V$ are right $R$-modules, both free of finite rank, and we have chosen bases $(u_i)$ of $U$ and $(v_i)$ of $V$. Then a map of abelian groups $\varphi \colon U \to V$ is called $R$-linear if $\varphi(u \lambda) = \varphi(u) \lambda$ for all $u \in U$ and $\lambda \in R$. An $R$-linear map $\varphi$ determines a matrix $[\varphi]_{i, j}$ with entries in $R$ in the usual way, where the coefficients are defined by the equation $$ \varphi(v_j) = \sum_i u_i [\varphi]_{i, j}. $$ Addition of linear maps correpsonds to matrix addition, and composition of linear maps corresponds to matrix multiplication as usual. But scaling a linear map here is undefined, since the abelian group $\operatorname{Hom}_R(U, V)$ of $R$-linear maps between $U$ and $V$ is not an $R$-module in any way. One is tempted to define a left action of $R$ on the hom space by postcomposition by $(\lambda, \varphi) \mapsto 1_\lambda \circ \varphi$, where $1_\lambda \colon V \to V$ is the map which performs scalar multiplication by $\lambda$. However, in the noncommutative world the map $1_\lambda$ is no longer $R$-linear, and neither is the composition $1_\lambda \circ \varphi$.

This scaling problem is showing that $\operatorname{Hom}_R(U, V)$ is only an abelian group for noncommutative $R$, it is not an $R$-module, and so taking tensor products is not well-defined. (However, it is a module over the centre of $R$, so for example if $R$ is the quaternions then $\operatorname{Hom}_R(U, V)$ is a real vector space, and a tensor product could be taken over $\mathbb{R}$. I haven't worked out what operation this corresponds to on matrices).