Centralizers in the first Weyl algebra (2)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
Tags: , ,

Theorem 1. (Amitsur, 1957) Let k be a field of characteristic zero and let f \in A_1(k) with \deg f =n \geq 1. Then C(f;A_1(k)) is a free k[f]-module of rank d, where d is a divisor of \deg f.

Proof. Suppose that S is the set of all integers m \geq 0 for which there esists some g \in C(f;A_1(k)) such that \deg g = m. Clearly S is a submonoid of \mathbb{Z}.  For any m \in S let \overline{m} be the image of m in \mathbb{Z}/n\mathbb{Z} and put \overline{S}=\{\overline{m}: \ m \in S \}. Since \overline{S} is a submonoid of a finite cyclic group, it is a cyclic subgroup and hence d=|\overline{S}| divides |\mathbb{Z}/n\mathbb{Z}|=n. Let S=\{\overline{m_i}: \ 1 \leq i \leq d \}, where m_1=0 and, in general, each m_i is chosen to be non-negative and the smallest member of its class \overline{m_i}. That means if m \equiv m_i \mod n and m \geq 0, then m \geq m_i. For any 1 \leq i \leq d, let g_i \in C(f;A_1(k)) with \deg g_i=m_i. So we can choose g_1 to be any element of degree zero in C(f;A_1(k)). We choose g_1=1. To complete the proof of the theorem, we are going to show that, as a k[f]-module, g_1, \ldots , g_d generate C(f;A_1(k)) and g_1, \ldots , g_d are linearly independent over k[f]. We first show that C(f;A_1(k))=\sum_{i=1}^d g_ik[f]. Clearly \sum_{i=1}^d g_ik[f] \subseteq C(f;A_1(k) because f, g_i \in C(f;A_1(k)), for all 1 \leq i \leq d. Now let g \in C(f;A_1(k) and suppose that \deg g = \ell. If \ell = 0, then \deg g = \deg 1 and hence, by Lemma 1, g \in k \subseteq g_1k[f] \subseteq \sum_{i=1}^d g_ik[f]. If \ell \geq 1, then \overline{\ell}=\overline{m_j}, for some j. We also have \ell \geq m_j by minimality of m_j. Thus \ell=m_j+nu for some integer u \geq 0. Therefore \deg g = \ell = m_j+nu=\deg g_jf^u. Now both g and g_jf^u are obviously in C(f;A_1(k)). So if s and t are the leading coefficeints of g and g_jf^u, respectively, then by Lemma 1, s=at for some a \in k. Therefore \deg(g - ag_if^u) \leq \ell - 1 and, since g - ag_if^u \in C(f;A_1(k)), we can apply induction on \deg g to get g - ag_jf^u \in \sum_{i=1}^d g_i k[f]. Thus g \in \sum_{i=1}^d g_i k[f]. It remains to show that g_1, \ldots , g_d are linearly independent over k[f]. Suppose, to the contrary, that

g_1 \mu_1(f) + \ldots + g_d \mu_d(f)=0, \ \ \ \ \ \ \ \ (*)

for some \mu_i(f) \in k[f] and not all \mu_i(f) are zero. Note that if i \neq j and \mu_i(f) \mu_j(f) \neq 0, then \deg (g_i \mu_i(f)) \equiv m_i \mod n and \deg (g_j \mu_j(f)) \equiv m_j \mod n. Since i \neq j, we have m_i \not\equiv m_j \mod n and hence  \deg (g_i \mu_i(f)) \neq \deg (g_j \mu_j(f)). Thus the left hand side of (*) is a polynomial of degree \max \{\deg(g_i \mu_i(f)): \ g_i \mu_i(f) \neq 0 \} and so it cannot be equal to zero. \Box

In part (3) we will prove that C(f;A_1(k)) is commutative.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s