## Centralizers in the first Weyl algebra (4)

Posted: April 16, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
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So we have proved that if $k$ is a field of characteristic zero and $f \in A_1(k)$ with $\deg f \geq 1,$ then $C(f;A_1(k))$ is a commutative domain and it is a free module of finite rank over $k[f].$ What can be said about the field of fractions $Q$ of $C(f;A_1(k))$? The next theorem shows that $Q$ has a very simple form.

Theorem 2. (Amitsur, 1957) Let $k$ be a field of characteristic zero and let $f \in A_1(k)$ with $\deg f \geq 1.$ Let $Q$ and $k(f)$ be the field of fractions of $C(f;A_1(k))$ and $k[f]$ respectively. Then $Q$ is an algebraic extension of $k(f)$ and $Q=k(f)[g],$ for some $g \in C(f;A_1(k)).$

Proof. First note that, by Theorem 1, $C(f;A_1(k))$ is a commutative domain and hence its field of fractions exists. Now let $g, d$ and $B$ be as defined in the proof of Theorem 1. We proved in that theorem that for every $h \in C(f;A_1(k))$ there exists some $0 \neq \mu(f) \in k[f]$ such that

$\mu(f)h \in B=k[f]+gk[f] + \ldots + g^{d-1}k[f]. \ \ \ \ \ \ \ \ (*).$

If in $(*)$ we choose $h=g^d,$ then we will get $g^d \in k(f) + gk(f) + \ldots + g^{d-1}k(f).$ So $g$ is algebraic over $k(f)$ and thus $k(f)[g]$ is a subfield of $Q.$ Also $(*)$ shows that $h \in k(f)[g],$ for all $h \in C(f;A_1(k))$ and thus $C(f;A_1(k)) \subseteq k(f)[g].$ Therefore $C(f;A_1(k)) \subseteq k(f)[g] \subseteq Q$ and hence $Q=k(f)[g]. \ \Box$