## Centralizers in the first Weyl algebra (3)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
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We are now going to prove that $C(f;A_1(k))$ is a commutative $k$-algebra for all $f \in A_1(k)$ with $\deg f \geq 1.$ Recall that by $\deg f$ we mean the largest power of $y$ in $f.$ First a simple lemma.

Lemma 2. Let $k$ be a field of characteristic zero and let $f \in A_1(k)$ with $\deg f \geq 1.$ If $m \geq 0$ is an inetger, then the set $V_m$ consisting of all elements of $C(f;A_1(k))$ of degree at most $m$ is a finite dimensional $k$-vector space.

Proof. It is clear that $V_m$ is a $k$-vector space. The proof of finite dimensionality of $V_m$ is by induction over $m.$ If $m=0,$ then $V_m=k$ and there is nothing to prove. So suppose that $m \geq 1$ and fix an element $g \in V_m$ with $\deg g=m.$ Of course, if there is no such $g,$ then $V_m=V_{m-1}$ and we are done by induction. Now, let $h \in V_m.$ If $\deg h < m,$ then $h \in V_{m-1}$ and if $\deg h = m,$ then there exists some $a \in k$ such that $h - ag \in V_{m-1},$ by Lemma 1. Thus $V_m = kg + V_{m-1}$ and hence $\dim_k V_m = \deg_k V_{m-1} + 1$ and we are done again by induction. $\Box$

Theorem 2. (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 commutative.

Proof. Let $S$ and $\overline{S}$ be as defined in the proof of Theorem 1. As we mentioned in there, $\overline{S}$ is a cyclic subgroup of $\mathbb{Z}/n\mathbb{Z}$ of order $d,$ for some divisor $d$ of $n.$ Let $\overline{m}, \ m > 0,$ be a generator of $\overline{S}$ and choose $g \in C(f;A_1(k))$ such that $\deg g = m.$ Now let

$B = k[f] + gk[f] + \ldots + g^{d-1}k[f].$

Clearly $B \subseteq C(f;A_1(k)).$ Let $T = \{mi+nj: \ 0 \leq i \leq d-1, \ j \in \mathbb{Z}, j \geq 0 \}.$ So basically $T$ is the set of all non-negative integers which appear as the degree of some element of $B.$ Let $p \in S.$ Then $p \equiv mi \mod n,$ for some inetger $0 \leq i \leq d-1$ because $\overline{m}$ is a generator of $\overline{S}.$ Hence $p = mi + nj,$ for some integer $j.$ If $j \geq 0,$ then $p \in T$ and if $j < 0,$ then $0 \leq p \leq mi \leq m(d-1).$ Thus if $h \in C(f;A_1(k))$ and $\deg h > m(d-1),$ then $\deg h \in T.$ Let $V$ be the set of all elements of $C(f;A_1(k))$ of degree at most $m(d-1).$ By Lemma 2, $V$ is $k$-vector space and $\dim_k V = v < \infty.$ The claim is that

$C(f;A_1(k))=B + V. \ \ \ \ \ \ \ \ (*)$

Clearly $B+V \subseteq C(f;A_1(k))$ because both $B$ and $V$ are in $C(f;A_1(k)).$ To prove $C(f;A_1(k)) \subseteq B+V,$ let $h \in C(f;A_1(k)).$ We use inducton over $\deg h.$ If $\deg h=0,$ then $\deg h = \deg 1$ and hence $h \in k,$ by Lemma 1. If $\deg h \leq m(d-1),$ then $h \in V$ and we are done. Otherwise, $\deg h \in T$ and hence there exists some $h_1 \in B$ such that $\deg h = \deg h_1.$ Thus, by Lemma 1, there exists some $a_1 \in k$ such that $\deg(h - a_1h_1) < \deg h.$ Therefore by induction $h-a_1h_1 \in B+V$ and hence $h \in B+V$ because $a_1h_1 \in B.$ This completes the proof of $(*).$

Now let $h \in C(f;A_1(k))$ and let $0 \leq i \leq v = \dim_k V.$ Clearly $f^i h \in C(f;A_1(k))$ and hence

$f^ih - h_i \in B, \ \ \ \ \ \ \ \ (**)$

for some $h_i \in V.$ Since $\dim_k V=v,$ the elements $h_0, \ldots , h_v$ are $k$-linearly dependent and so $\sum_{i=0}^v a_ih_i=0$ for some $a_i \in k$ which are not all zero. It now follows from $(**)$ that $(\mu(f)h \in B,$ where $0 \neq \mu(f)=\sum_{i=0}^v a_if^i \in k[f].$ So we have proved that for every $h \in C(f;A_1(k))$ there exists some $0 \neq \mu(f) \in k[f]$ such that $\mu(f)h \in B.$ Let $h_1, h_2 \in C(f;A_1(k))$ and let $0 \neq \mu_1(f), \mu_2(f) \in k[f]$ be such that $\mu_1(f)h_1 \in B$ and $\mu_2(f)h_2 \in B.$ Then, since $B$ is clearly commutative, we have $\mu_1(f)h_1 \mu_2(f)h_2 = \mu_2(f)h_2 \mu_1(f)h_1.$ Therefore, since $k[f]$ is commutative and $h_1$ and $h_2$ commute with $f,$ we have $\mu_1(f) \mu_2(f)h_1h_2=\mu_1(f) \mu_2(f)h_2h_1.$ Thus, since $A_1(k)$ is a domain and $\mu_1(f), \mu_2(f) \neq 0,$ we have $h_1h_2=h_2h_1.$ Hence $C(f;A_1(k))$ is commutative. $\Box$

In part (4), which will be the last part, we will find the field of fractions of $C(f;A_1(k)).$