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.

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