# On Weyl algebras and their centers over fields of positive characteristic.

I was trying to read some notes leading up the Dixmier conjecture, but I was hoping to see clarify a lemma.

Suppose you have the Weyl algebra $A_n$ over a field $k$ of positive characteristic $p$. Why is the center of the Weyl algebra isomorphic to the algebra of polynomials in $2n$ variables over $k$?

A cursory remark says this follows by writing $f=\sum_{I,J}f_{IJ}x^Iy^J$ and computing $[x_i,f]$ and $[y_j,f]$ term by term.

I must be dense, but I don't see what this is getting at. Can anyone please flesh out this remark to make it more transparent why the result is true? Thank you.

Edit: The thing in question is statement (2) of Lemma 3 in these notes. I'm just hoping to see a more detailed proof than the one given.

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Let us show by induction that in $A_1$ $[x,y^n]=n y^{n-1}$:

• $[x,y]=xy-yx=1$
• $[x,y^n]=xy^n-y^n x= xy^{n-1}y-y^{n-1}xy+y^{n-1}xy-y^{n-1} yx=[x,y^{n-1}]y+y^{n-1}[x,y]=n y^{n-1}$

(this also can be shown by considering $[x,-]$ as a derivation and noticing that $[x,y]=1$)

Now, since $[x_i,x_j]=0$ for all $i,j$ and $[x_i,y_j]=0$ if $i\neq j$, it follows that $[x_i,x^Iy^{J_1}(y_i)^ry^{J_2}]=rx^Iy^{J_1}(y_i)^{r-1}y^{J_2}$.

Applying these formulas (and similar ones with $y_j$) to the sum in the remark and using that the monomials are linearly independent the result should follow.

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Thank you Quimey :). I guess my problem is, I don't understand why this computation shows that the center is isomorphic to $k[x^p_1,\dots,x^p_n,y^p_1,\dots,y^p_n]$ as claimed. Do you know why that is? – Tiffany Hwang Apr 11 '12 at 3:18
The above formulas (together with a similar identity for $[y, x^n]$) show that all $p$th powers are central. To see that there's nothing else, if some $y^n$ were central, then $0 = [x, y^n] = n y^{n-1}$, so $n=0$ in the field and therefore $n$ is a multiple of $p$. Same with powers of $x$. This generalizes to arbitrary polynomials because the derivations $[x, -]$ and $[y, -]$ act separately on each power of $y$, resp., $x$. – Ted Apr 11 '12 at 15:54