## Centralizers in the first Weyl algebra (2)

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