# Weyl algebra as an Azumaya algebra over its centre

Assume that $$k$$ is an algebraically closed field of positive characteristic $$p$$. On page 3 (page 6 of the PDF file) of Bezrukavnikov, Mirković, and Rumynin - Localisation of Modules for a semisimple Lie Algebra in prime characteristic, we have the following sentence:

The sheaf $$D_X$$ of crystalline differential operators on a smooth variety $$X$$ over $$k$$ has a non-trivial center, canonically identified with the sheaf of functions on the Frobenius twist $$T^∗X^{(1)}$$ of the cotangent bundle. Moreover $$D_X$$ is an Azumaya algebra over $$T^∗X^{(1)}$$.

Instead of going through the general proof given, I only want to understand, in as simple a manner as possible, the situation when $$X$$ is the affine $$n$$-space over $$k$$. In this case, $$D_X$$ is simply the Weyl algebra and the Azumaya property, if I understand correctly, means that the quotient of the Weyl algebra by its centre is isomorphic to some matrix algebra over $$k$$. Is there a way to construct such a matrix algebra and a corresponding isomorphism to the quotient explicitly? Any help, even in the case of the affine line, would be highly appreciated.

P.S. If my understanding is incorrect, could you please point out the flaw(s) and how the question could be turned into something reasonable?

Major Edit It has been pointed out that my understanding of the Azumaya property is incorrect. But my question remains the same: is there a direct way to prove the claim of the paper in the case when $$X$$ is the affine $$n$$-space (or even the affine line) over $$k$$.

• Your definition of Azumaya algebra is incorrect — see Azumaya algebra.
– abx
Aug 11, 2020 at 16:36
• Thank you very much for the quick reply. Even if that is the case, will the quotient of the Weyl algebra by its centre be isomorphic to a matrix algebra? Aug 11, 2020 at 16:39
• The trivial Azumaya algebra (say, over a field) is a matrix algebra, so the quotient by its center is not a matrix algebra.
– abx
Aug 11, 2020 at 16:41
• @abx If I consider $n=1$ and consider $k[x]/x^p$ and let $x$ and $\partial$ define the standard actions on $k[x]/x^p$. Then since the centre of the Weyl algebra is just $k[x^p, \partial^p]$, each element of the quotient will define an endomorphism of $k[x]/x^p$. Will/can this be an isomorphism? Aug 11, 2020 at 16:50
• No. Even briefer version of my previous argument: $k[x, \partial]/k[x^p, \partial^p]$, or even just $k[x, \partial]/k$, is Abelian, so can't be $\operatorname{End}_k(k[x]/(x^p))$. Aug 11, 2020 at 16:59

If you are only interested in the $$n=1$$ case $$A=k \langle X,Y \mathrel| XY-YX=1 \rangle$$, then give $$X$$ degree $$1$$ and $$Y$$ degree $$-1$$, so that $$A$$ becomes a $$\mathbb{Z}$$-graded ring. It is even strongly graded, meaning $$A_1\cdot A_{-1} = A_0$$, so all graded info is determined by the zero-part $$A_0=k[XY]$$. In particular, if you divide out a graded maximal ideal you get a strongly graded ring with part of degree $$0$$ a field and finite over its center, and these must be central simple algebras (in this case, just $$p \times p$$ matrices) proving that $$A$$ is what is called a graded Azumaya algebra. Now, for the fun part, as there exist homogeneous central identities, a graded Azumaya algebra is a genuine Azumaya algebra. Done.