Centralizers in the first Weyl algebra (3)

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

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)).


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