GK dimension of Weyl algebras

Posted: April 10, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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We defined the n-th Weyl algebra A_n(R) over a ring R in here.  In this post we will find the GK dimension of A_n(R) in terms of the GK dimension of R. The result is similar to what we have already seen in commutative polynomial rings (see corollary 1 in here). We will assume that k is a field and R is a k-algebra.

Theorem. {\rm{GKdim}}(A_1(R))=2 + {\rm{GKdim}}(R).

Proof. Suppose first that R is finitely generated and let V be a frame of R. Let U=k+kx+ky. Since yx = xy +1, we have

\dim_k U^n = \frac{(n+1)(n+2)}{2}. \ \ \ \ \ \ \ \ \ (*)

Let W=U+V. Clearly W is a frame of A_1(R) and

W^n = \sum_{i+j=n} U^i V^j,

for all n, because every element of V commutes with every element of U. Therefore, since V^j \subseteq V^n and U^i \subseteq U^n for all i,j \leq n, we have W^n \subseteq U^nV^n and W^{2n} \supseteq U^nV^n. Thus W^n \subseteq U^nV^n \subseteq W^{2n} and hence

\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n \leq \log_n \dim_k W^{2n}.

Therefore {\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)), by (*), and we are done.

For the general case, let R_0 be any finitely generated k– subalgebra of R. Then, by what we just proved,

2 + {\rm{GKdim}}(R_0)={\rm{GKdim}}(A_1(R_0)) \leq {\rm{GKdim}}(A_1(R))

and hence 2+{\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)). Now, let A_0 be a k-subalgebra of A_1(R) generated by a finite set \{f_1, \ldots , f_m\}. Let R_0 be the k-subalgebra of R generated by all the coefficients of f_1, \ldots , f_m. Then A_0 \subseteq A_1(R_0) and so

{\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A_1(R_0))=2 + {\rm{GKdim}}(R_0) \leq 2 + {\rm{GKdim}}(R).

Thus

{\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R)

and the proof is complete. \Box

Corollary. {\rm{GKdim}}(A_n(R))=2n + {\rm{GKdim}}(R) for all n. In particular, {\rm{GKdim}}(A_n(k))=2n.

Proof. It follows from the theorem and the fact that A_n(R)=A_1(A_{n-1}(R)). \Box

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