The quotient of the Weyl algebra by its centre is a matrix algebra

Question

Consider the Weyl algebra, call it $W$, generated by $x_1,x_2,...x_n,\partial_1,\partial_2,...\partial_n$ over an algebraically closed field $k$ of characteristic $p>0$ subject to the usual relations: $x_ix_j=x_jx_i$, $\partial_i\partial_j=\partial_j\partial_i$ and $[\partial_i,x_j]=\delta_{ij}$ for all $i$,$j$. I want to show that the quotient of $W$ by its centre is isomorphic to some matrix algebra over $k$. I would like an explicit construction of such a matrix algebra as well as the required isomorphism. Is there any obvious way to do this? Any help, even in the $n=1$ case, will be appreciated.

Answer 1

If the center is $k[\{x_i^p\mid i=1\ldots n\}\cup\{\partial_i^p\mid i=1\ldots n\}]$, then it seems clear that $W$ is generated by the lower powers of those generators because you can always sort them out with their commutation relations and take out powers that are too high as coefficients in the center. It clearly has to be faithful because of the multiplicative identity.

I'm not completely sure how to reason $W$ is projective. It might actually be free over its center for some basis, but it's not immediately clear to me why.

If it were free of rank $R$ over its center $R$, then it is clear that $End(A_R)\cong M_n(R)$, and $A\otimes_R A\cong R^{n\times n}$, and there's probably some basic results on Hom sets that I'm forgetting that simply says it's $M_n(R)$.