Kronecker-Product, Tensor product and isomorphism Let $F$ be a field. For an $m\times m$ matrix $A$ and an  $n \times n$ matrix B, show that the mapping $$A\otimes B\rightarrow A\odot B$$ where $A\odot B$ is the Kronecker-Product is a isomorpshims. In other words show $$M_m(F)\otimes M_n(F)\cong M_{mn}(F)$$
proof:  Note that $\phi(A,B)=A\odot B$ is a $F$-bilinear map. By the Universal Property of tensor products, the maping $\psi(A\otimes B)=A\odot B$ is $F$-algebra morphims that is well defined. All that is left to show is that $\psi$ is bijective. 
Where I am having trouble is showing one-to-one and onto. I have been trying for a while now. Mabey there is a different approach? Any hints will be appreciated.
 A: If you want to do it by hand, checking surjectivity is the easiest way. You need to convince yourself that any elementary matrix $E_{ij}$ ,$1\leq i,j\leq mn$ lies in the image, which is easy.
In fact,you can write $E_{ij}$ has the Kronecker product of two elementary matrices. To see intuitively why, separate you matrix in blocks of size $n$ (or $m$, depending your definition of Kronecker product). Then you will have zero blocks everywhere, except for one block which will be an elementary matrix. What is left to do is clear.
If you want to do it more formally, you can try to compute explicitely a formula of the type $E_{pq}\odot E_{rs}=E_{??}$, $1\leq p,q\leq m, 1\leq r,s\leq n$, and check that the indices $??$ attain any pair of integers betwenn $1$ and $mn$. 
Side note. There is a quickest way, but it relies on nontrivial results: $M_m(F)$ and $M_n(F)$ are central simple $F$-algebras. Consequently, their tensor product is also a central simple algebra (theorem) and $M_m(F)\otimes M_n(F)$ has no non trivial two-sided ideal. Hence, your map is injective (it is non trivial, so the kernel cannot be the whole tensor product)), and a dimension count shows that you get an isomorphism.
