Hello, all!
Could somebody draw a proof-sketch of next expression from tensor algebra on matrices over finite fields:
determinant of tensor product $A~ \times ~B$ of $n \times n$-matrix $A$ over finite field $GF(q)$ on $m \times m$-matrix $B$ over finite field $GF(q)$ is $\det(A)^m \cdot \det(B)^n$.


Please, give me a link or reference if it is online or in some book.
Thank you.



tensor product of matrices
Darij's first comment could be made into an answer as follows. 


Darij advised to write 


$$A \otimes B = (A \circ I\_n) \otimes (I\_m \circ B) = (A \otimes I\_m) \circ (I\_n \otimes B)$$


where the second equation follows from functoriality of the tensor product. Here both $A \otimes I\_m$ and $I\_n \otimes B$ are square matrices of size $m n \times m n$. Since the determinant from such matrices to the scalar field is a monoid homomorphism, the determinant of the last expression is 


$$\det(A \otimes I\_m) \det(I\_n \otimes B)$$ 


so we are left to determine the two determinants above. Since these are similar, we do the first. We may express an $m$-dimensional vector space $k^m$ as a direct sum of 1-dimensional vector spaces, so 


$$A \otimes I\_{k^m} = A \otimes (I\_k \oplus \ldots \oplus I\_k) = (A \otimes I\_k) \oplus \ldots \oplus (A \otimes I\_k)$$ 


because tensor products preserve direct sums. This is just $A \oplus \ldots \oplus A$. This matrix consists of $m$ blocks of $A$, so its determinant is $\det(A)^m$, and we are done. 


