## Centralizers in the first Weyl algebra (1)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
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In this post I am going to look at the centralizer of non-central elements in the first Weyl algebra over some field $k$ of characteristic zero. Recall that the first Weyl algebra $A_1(k)$ is defined to be the $k$-algebra generated by $x$ and $y$ with the relation $yx = xy+1.$ It then follows easily that $yr = ry + r',$ for every $r \in k[x],$ where $r' = \frac{dr}{dx}.$ It is easily seen that the center of $A_1(k)$ is $k.$ Also, every non-zero element of $A_1(k)$ can be uniquely written in the form $\sum_{i=0}^n r_i y^i,$ where $r_i \in k[x]$ and $r_n \neq 0.$ We call $n$ the degree of $f.$ It is easy to see that $A_1(k)$ is a domain. For every $f \in A_1(k),$ we will denote by $C(f;A_1(k))$ the centralizer of $f$ in $A_1(k).$ The goal is to show that if $f \notin k,$ then $C(f;A_1(k))$ is a commutative algebra and also a free $k[f]$-module of finite rank. This result is due to Amitsur.

Remark 1. If $r \in k[x],$ then $y^nr = \sum_{i=0}^n \binom{n}{i}r^{(i)}y^{n-i},$ where $r^{(i)}$ means the $i$-th derivative of $r$ with respect to $x.$ This follows easily by induction and the fact that $yr=ry+r'.$

Remark 2. If $f \in k[x],$ then $C(f;A_1(k))=k[x].$ This is easy to see: clearly $k[x] \subseteq C(f;A_1(k)).$ Conversely, if $g = r_ny^n + r_{n-1}y^{n-1} + \ldots + r_0, \ r_i \in k[x], \ r_n \neq 0$ commutes with $f$ and $n \geq 1,$ then comparing the coefficients $y^{n-1}$ in both sides of $fg=gf$ will give us $nr_nf'=0,$ which is a contradiction. Thus $n=0$ and so $g \in k[x].$

So, by the above remark, we only need to find the centralizer of an element of $A_1(k)$ in the form $f = \sum_{i=0}^n r_iy^i, \ r_i \in k[x], \ n = \deg f \geq 1.$

Lemma 1. Let $k$ be a field of characteristic zero and let $f = r_n y^n + \ldots + r_0 \in A_1(k), \ n \geq 1, \ r_n \neq 0.$ Let $g=s_my^m + \ldots + s_0, \ s_m \neq 0,$ and $h=t_my^m + \ldots + t_0, \ t_m \neq 0,$ be two elements of $C(f;A_1(k)).$ Then $s_m=a t_m,$ for some $a \in k.$

Proof.  Since $yr=ry+r',$ for any $r \in R=k[x],$ induction on $\ell$ shows that $y^{\ell}r = ry^{\ell} + \ell r'y^{\ell - 1} + \ldots,$ for any integer $\ell \geq 1.$ Therefore the coefficient of $y^{n+m-1}$ in $fg$ and $gf$ are $nr_ns_m' + r_ns_{m-1}+r_{n-1}s_m$ and $ms_mr_n' + s_mr_{n-1} + s_{m-1}r_n,$ respectively. Thus, since $fg=gf,$ we must have

$nr_ns_m' + r_ns_{m-1}+r_{n-1}s_m = ms_mr_n' + s_mr_{n-1} + s_{m-1}r_n.$

Hence, since $R$ is commutative, we will have

$nr_ns_m'=mr_n's_m. \ \ \ \ \ \ \ \ (1)$

A similar arguemnt shows that $fh=hf$ implies that

$nr_nt_m'=mr_n't_m. \ \ \ \ \ \ \ \ (2)$

Now, multiplying both sides of (1) by $t_m$ and both sides of (2) by $s_m$ and then subtracting the resulting identities will give us $nr_n(t_ms_m'-s_mt_m')=0.$ Thus

$t_ms_m'-s_mt_m'=0, \ \ \ \ \ \ \ \ (3)$

because $R$ is a domain, $r_n \neq 0$ and the characteristic of $k$ is zero. Now look at $R$ as a subalgebra of the field of rational functions $k(x).$ Then, since $s_m \neq 0,$ by (3) we have $(s_m/t_m)'=0$ and hence $s_m/t_m \in k,$ i.e. $s_m=at_m,$ for some $a \in k. \ \Box$

In part (2) we will prove that $C(f;A_1(k))$ is a free $k[f]$-module of finite rank.