# Nilpotent operator of the Weyl algebra

For a research project I'm currently working on, I came across the following problem:

Let $A=$ $^{k <x,y> }\Big/_{(yx-xy-1)}$ be the Weyl Algebra over a field $k$ of characteristic $p$, where $p$ is an odd prime. For an element $q(x,y) \in A$, define the operator $ad_q(h):=[q,h]=qh-hq$. My question is:

For which $q$ is $ad_q$ a nilpotent operator?

If $q \implies ad_q$ nilpotent, I say that $q$ is good. Here's what I have so far:

• $q(x,y)=r(x)+s(y)$ is good if $deg(r)=0,1$ or $deg(s)=0,1$.

• $q(x,y)=p(x^p)+q(y)$ is good

• $q(x,y)=xp(x^p)+q(y)$ is good.

All of these observations are straightforward to prove (I will give details if necessary).

It might help to know that the center of $A$ is $Z(A)=k[x^p,y^p]$ and that $ad_q$ is a derivation (i.e., it satisfies the Leibniz Rule). Also, if I'm not mistaken, the following formula is valid: $ad_q(h)^{n}=\sum_{k=0}^{n} (-1)^k {n \choose k} q^k h q^{n-k}$ $\forall h \in A$. Hence, if $q$ is good, there exists a $p^j$ such that $0=ad_q(h)^{p^j}=hq^{p^j}-q^{p^j}h=[h,q^{p^j}]$, so $q^{p^j} \in Z(A) = k[x^p,y^p]$.

I would be satisfied if someone had any hint or idea on how to attack this problem. Thank you!

PS: I don't usually post on MO, so I apologize if I made any mistakes or if this question is not appropriate for this website.