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I was doing some reading on ring theory, and there's something mentioned in a footnote I'd like to clarify.

Suppose you have some vector space $V$ of even finite dimension, and some nondegenerate, alternating bilinear form $g$. I'll use $W_g(V)$ to denote the corresponding Weyl algebra.

Now if the underlying field $F$ is such that $\mathrm{char}(F)=0$, then why does the center of the Weyl algebra coincide with $F$?

I'd appreciate any proof or reference to a proof I could read, as I could not find or come up with one. Cheers!

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    $\begingroup$ This is proved in pretty much any text which deals with the Weyl algebra! $\endgroup$ Apr 9, 2012 at 5:53
  • $\begingroup$ For example, Miličić's notes on $\mathcal D$-modules have a proof of this; see proposition 5.12 www.math.utah.edu/~milicic/Eprints/dmodules.pdf $\endgroup$ Apr 9, 2012 at 6:22

1 Answer 1

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I'll do the case where $n=1$, with with a little extra generality: let $F$ be an algebra over a field of characteristic zero, and let $A=F[x,\partial]$ be the Weyl algebra over the ring $F$.

If $a\in A$, there exists unique $k\geq-0$ and polynomials $f_0,\dots,f_k\in F[T]$ such that $a=\sum_{i=0}^kf_i(x)\partial^i$. Suppose that $a$ is central.

One can compute easily that $[\partial,f(x)]=f'(x)]$ for each polynomial $f\in F[x]$, so that $$[\partial, a]=\sum_{i=0}f_i'(x)\partial^i.$$ Since $a$ is central, this is zero, so we must have $f_0'(x)=\dots=f_k'(x)=0$, and therefore the $f_i$ are constant polynomials. It follows that $a=g(\partial)$ for some polynomial $g\in F[T]$. Now another easy computation shows that $$[x,a]=g'(\partial)]$$ and this again is zero because $a$ is central. This means that $g$ is constant, so $a$ is in $F$. Since $a$ is central in $A$, it is central in $F$.

It follows that the center of $A$ is the center of $F$.

Now, we can prove that the center of the usual Weyl algebra $A_n=k[x_1,\dots,x_n,\partial_1,\dots,\partial_n]$ over a field $k$ of characteristic zero has center $k$ by induction. Indeed, there is an obvious isomorphism $$A_n=A_{n-1}[x_n,\partial_n]$$ so that the above reasoning shows that the center of $A_n$ is the center of $A_{n-1}$. Since the center of $A_0$ is $k$, this proves what we want.

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  • $\begingroup$ Thank you Mariano. Just to clarify, is the $F$ in your answer different from the $F$ as a field in my question? I'm slightly confused because the field $k$ of characteristic $0$ is introduced at the end of this answer, and then $F$ is mentioned as the center of $A_0$. $\endgroup$ Apr 12, 2012 at 22:59
  • $\begingroup$ The last mention of $F$ should have been $k$. My $F$ is, all the way up to the now any algebra, as described in the first line. $\endgroup$ Apr 12, 2012 at 23:22
  • $\begingroup$ Thanks for clarifying. I suspected this, but didn't want to make any stupid assumptions. $\endgroup$ Apr 12, 2012 at 23:29
  • $\begingroup$ Amazing answer as always Mariano! :) $\endgroup$
    – ABIM
    Apr 13, 2014 at 18:23
  • $\begingroup$ Just a note: we actually have $[x,a]=-g'(\partial)$. $\endgroup$
    – Tomo
    Nov 16, 2015 at 23:50

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