Weyl algebras over fields of positive characteristics and their centers

Lemma 3   Let $k$ be a field of characteristic $p$. We have the following facts.

1. $\xi_i^p, \eta_j^p$ belongs to the center of $A_n(k)$.
2. More precisely, the center $Z_n(k)=Z(A_n(k))$ of the ring $A_n(k)$ is given by
   $$Z_n(k)=k[\xi_1^p,\dots,\xi_n^p,\eta_1^p,\dots,\eta_n^p].$$

3. $A_n(k)$ is a free $Z_n(k)$-module of rank $p^{2n}$.
4. Let $a=(a_1,\dots,a_n), b=(b_1,\dots b_n)$ be elements of $k^n$. Let $I_{a,b}$ be an ideal of $A_n(k)$ generated by $(\xi_1-a_1)^p,\dots, (\xi_n-a_n)^p,(\eta_1-b_1)^p,\dots, (\eta_n-b_n)^p$. Then we have
   $$A_n(k)/I_{a,b} \overset{\pi_{a,b}}{\cong } M_{p^n}(k).$$

5. If the field $k$ is algebraically closed, then any maximal (both-sided) ideal of $A_n(k)$ is of the form $I_{a,b}$ above.

(1) easy.

(2) Write

$$f= \sum_{I,J} f_{I J} \xi^I \eta^J$$

and compute $[\xi_i,f]$ and $[\eta_j,f]$ term by term.

(3)

$$A_n(k)=\bigoplus_{0\leq i_1,\dots,i_n,j_1,\dots,j_n \leq p-1} Z_n... ..._3}\dots \xi_n^{i_n} \eta_1^{i_1} \eta_2^{i_2} \eta_3^{i_3}\dots \eta_n^{i_n}.$$

(4) is a direct consequence of Lemma 1. For the reference purposes, we record here another explanation. By considering a translation in $\xi$- and $\eta$-directions, we may well assume $a=b=0$. Consider an action of $A_n(k)$ on $A=k[\xi_1,\dots,\xi_n]/(\xi_1^p,\dots,\xi_n^p)$ defined by the following formula.

$$P.f=P(\xi_1,\dots,\xi_n, \partial_1,\dots,\partial_n).f$$ (2)

We may then verify that this gives a well-defined action of $A_n(k)/I_{0,0}$ on the $p^n$-dimensional vector space $A$.

If an element $P$ in $A_n(k)/I_{0,0}$ satisfy

$$P.1=0, P.\xi=0 ,\dots, P.\xi^{p-1}=0$$

then we may easily see that $P=0$. This means the homomorphism

$$\pi_{0,0}: A_n(k)/I_{0,0} \to \operatorname{End}_{\mbox{k}\operatorname{-module}}(A)$$

is injective. A dimension argument now shows that $\pi_{0,0}$ is a bijection.

(5) Let $M$ be an ideal of $A_n(k)$. We consider a sheaf $\widetilde{(A_n(k)/M)}$ of algebras on $\operatorname{Spec}(Z_n(k))$ which corresponds to the $Z_n(k)$-module $A_n(k)/M$. Since $A_n(k)$ is a finitely generated module over a polynomial ring $Z_n(k)$, There exists a closed point $\mathfrak{m}$ of $Z_n(k)$ such that fiber $(\widetilde{(A_n(k)/M)}\vert\mathfrak{m})=A_n(k)/(M+\mathfrak{m} A_n(k))$ is nonzero. Since $M$ is maximal, this implies that $\mathfrak{m} \subset M$. The rest of the proof is easy (Use Nullstellensatz.)