Finite dimensional representations of the Weyl algebra in characteristic $p>0$ I'm working through representation theory course notes of P. Etingof. In problem 1.26 it is asked to find all finite dimensional irreducible representations of the algebra $A=\frac{k[x,y]}{\left\langle yx-xy-1\right\rangle}$. Here $k$ is an algebraically closed field of characteristic $p$. 
I was able to show that $x^p,y^p\in Z(A)$, in fact the center is generated by these elements. More generally, I can show that $[y,p(x)]=p'(x)$ where $p$ is some polynomial in $x$. Now suppose that $V$ is an irreducible finite dimensional module over $A$. Since $k$ is algebraically closed, we have that $V=\bigoplus_{\lambda\in \text{spec}(y)}V_{\lambda}$ (Jordan normal form). Now I can show that $V_{\lambda}$ is an invariant subspace, and since $V$ is irreducible, we have that $V=V_{\lambda}$, or equivalently, $y$ has only one eigenvalue $\lambda$. Since $V$ is irreducible, we must conclude that $y$ has only one eigenvector $v$ corresponding to $\lambda$. 
I was able to show that $(y-\lambda)^n(x^{n-1}v)=0$ for $n\geq 1$ and that $\left\{v,xv,x^2v, \dots, x^{p-1}v\right\}$ are linearly independent. Moreover, we know that $$\dim(V)\equiv \text{Tr}(1) \mod p\equiv Tr(yx-xy) \mod p\equiv 0 \mod p.$$
I'm stuck on showing that $\dim(V)=p$. I already have that $\dim(V)\geq p$. How do I get the other inequality? I feel that there is something very straightforward I'm missing.
We also have that $yv=\lambda v\Rightarrow y(x^pv)=\lambda (x^pv)$ and thus $x^pv$ is a multiple of $v$, but I'm not sure whether this is useful.
Also, I'm not sure whether the author would appreciate the solutions being on the internet, so I'm willing to delete this question if necessary.
 A: Since $x^p$ and $y^p$ are in the center, they act as scalars on your simple module $V$. This means that in fact $V$ is a module over the algebra $k\langle x,y\mid yx-xy-1,x^p-\alpha,y^q-\beta\rangle$ forsome scalars $\alpha$, $\beta$ in the field.
Show that this algebra is central and simple (imitating the proofs for the Weyl algbra in characteristc zero, for example) This means that the algebra is in fact isomorphic to a matrix algebra, for the field is algebraically closed. It is easy to compute its dimension: it is $p^p$. It follows, from the easy structure theory of the modules over a matrix algebra that $\dim V=p$.
A: Is the following reasoning correct?
Define $V_{\lambda,n}:=\left\{v\in V\mid (y-\lambda)^nv=0\right\}$. Claim: $\dim(V_{\lambda,n})=n$ for all $1\leq n \leq p-1$. 
We have that $V_{\lambda,k}\subset V_{\lambda,k+1}$ for all $k$ and $x^{n-1}v\in V_{\lambda,n}$. Now suppose that $w,w'\in V_{\lambda,n}$ are independent vectors for some $n$, and $w,w'\notin V_{\lambda,r}$ with $r<n$, then $(y-\lambda)^{n-1}w$ and $(y-\lambda)^{n-1}w'$ are both eigenvectors of $y$. Hence $(y-\lambda)^{n-1}(w-\alpha w')=0$ (for some $\alpha\in k$) since there is only one eigenvector. Hence $\overline{w-\alpha w'}=0$ in $\frac{V_{\lambda,n}}{V_{\lambda,n-1}}$, which contradicts $w$ and $w'$ being independent. Thus $\dim(V_{\lambda},n)=\dim(V_{\lambda,n-1})+1$. The claim follows since $x^{n+1}v\in V_{\lambda,n}$. Thus we have that $\dim(V_{\lambda,p})=p$.
Since $(y-\lambda)^p=y^p-\lambda^p$ and $y$ can be written in Jordan normal form, we see that $y^p$ acts as $\lambda^p$ (obviously we get $\lambda^p$ on the diagonal, and the off-diagonal elements can be divided by $p$). Thus $(y-\lambda)^p=0$ and $V=V_{\lambda,p}$. Hence $\dim(V)=p$ as we needed to show.
I hope this makes sense.
A: I think I found a different method to prove that the set you describe is a basis: Since $V$  is an irreducible representation, then the submodule $A . < v, \dots , x^{p-1} v>_k = V$. That means that every element in $V$ can be written as a linear combination of elements of the form $a x^r v$. Now since $x^i y^j$ form a basis of $A$, then every element of $V$ can be written as a linear combination of elements of the form $x^i y^j x^r v$. Permuting $y^j x^r$ using the equality $yx = xy+1$ and the fact that $yv = \lambda v$, we conclude that every element of $V$ can be written as a linear combination of elements of the form $x^i v$, but now the exponent $i$ might be greater than $p-1$. But since $x^p v$ is a multiple of $v$ (this has to be proved carefully from your notes), we can reduce every exponent modulo $p$. So every element of $V$ can be written as a linear combination of elements of the form $x^iv$ with $0 \leq i \leq p-1$. Please correct if there is any mistake on this, but it looks good to me.
