Centralizers in the first Weyl algebra (4)

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

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

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