A theorem of Smith and Zhang

Posted: May 11, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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For the definition of a PI-algebra see this post.

Lemma. Let k be a field and let A be a finitely generated k-algebra which is a domain. If A is not PI, then {\rm{GKdim}}(A) \geq 2.

Proof. If {\rm{GKdim}}(A)=0, then A is finite dimensional over k, and so over its center, and hence it is PI. If {\rm{GKdim}}(A) = 1, then A is again PI (see here). Also, there is no algebra whose GK dimension is strictly between 1 and 2, by the Bergman’s gap theorem. Thus {\rm{GKdim}}(A) \geq 2. \ \Box

We are now going to refine the result given in the above lemma. We will write {\rm{GKdim}}_F for the GK dimension of an algebra viewed as an F-algebra if F is not k.

Theorem. (Smith and Zhang, 1996) Let k be a field and let A be a finitely generated k-algebra which is a domain. If A is not PI, then {\rm{GKdim}}(A) \geq 2 + {\rm{GKdim}}(Z(A)), where Z(A) is the center of A.

Proof. By the above lemma, we may assume that 2 \leq {\rm{GKdim}}(A) < \infty and {\rm{GKdim}}(Z(A)) \geq 1. Let Q_Z(A) be the central localization of A and let F be the center of Q_Z(A). Clearly F is just the quotient field of Z(A). Recall, from the theorem in here, that

{\rm{GKdim}}(Q_Z(A))={\rm{GKdim}}(A) \geq 2  and {\rm{GKdim}}(F)={\rm{GKdim}}(Z(A)) \geq 1.

Let 0 \leq d < {\rm{GKdim}}(F) and 0 \leq e < {\rm{GKdim}}_F(Q_Z(A)). Then there exist a finite dimensional k-vector subspace V of F which contains 1 and \dim_k V^n \geq n^d for all large enough integers n. Also, there exists a finite dimensional F-vector subspace W \supseteq V of Q_Z(A) which contains 1 and \dim_F W^n \geq n^e for all large enough integers n. Hence, for large enough integers n we have

\dim_k W^{2n} \geq \dim_k (W^nV^n) \geq (\dim_F W^n)(\dim_k V^n) \geq n^{e+d}.

Thus {\rm{GKdim}}(Q_Z(A)) \geq e+d. Since the above inequality holds for all real number 0 \leq d < {\rm{GKdim}}(F) and 0 \leq e < {\rm{GKdim}}_F(Q_Z(A)), we have

{\rm{GKdim}}(Q_Z(A)) \geq {\rm{GKdim}}_F(Q_Z(A)) + {\rm{GKdim}}(F). \ \ \ \ \ \ \ \ (*)

Now, since A is finitely generated as a k-algebra, Q_Z(A) is a finitely generated F-algebra and clearly Q_Z(A) is not PI because A is not PI. Therefore {\rm{GKdim}}_F(Q_Z(A)) \geq 2 by the above lemma. We also have {\rm{GKdim}}(F)={\rm{GKdim}}(Z(A)) and now the result follows from (*). \ \Box

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