## 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 $(*). \ \Box$