GK dimension and localization

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
Tags: , ,

Theorem. Let $A$ be a $k$-algebra and suppose that $S$ is a regular submonoid of $A$ contained in the center of $A.$ Then ${\rm{GKdim}}(S^{-1}A) = {\rm{GKdim}}(A).$

Proof. Let $T$ be a finitely generated $k$-subalgebra of $S^{-1}A$ and suppose that $W=\{w_1=1, \ldots , w_m\}$ is a frame of $T.$ Choose $s \in S$ and $a_1, \ldots, a_m \in A$ such that $w_i=s^{-1}a_i$ for all $i.$ Let $B$ be the $k$-subalgebra of $A$ generated by $a_i$ and let $V$ be the $k$-subspace generated by $1$ and $a_i.$ Now, since $S$ is in the center of $A,$ we have $s^nW^n \subseteq V^n.$ Thus $\dim W^n = \dim s^nW^n \leq \dim V^n.$ Therefore

${\rm{GKdim}}(T) \leq {\rm{GKdim}}(B) \leq {\rm{GKdim}}(A),$

for every finitely generated $k$-subalgebra of $T$ of $S^{-1}A,$ and so ${\rm{GKdim}}(S^{-1}A) \leq {\rm{GKdim}}(A).$ On the other hand, $A \subseteq S^{-1}A,$ because $S$ is regular, and thus ${\rm{GKdim}}(A) \leq {\rm{GKdim}}(S^{-1}A). \Box$

Using the above result we can now evaluate the GK dimension of a Laurent polynomial ring.

Corollary. Let $A$ be a $k$-algebra and let $x$ be a variable over $A.$ Then ${\rm{GKdim}}(A[x,x^{-1}])=1+ {\rm{GKdim}}(A).$

Proof. Since $A[x,x^{-1}]$ is the localization of $A[x]$ at the central regular submonoid $S=\{1,x,x^2, \ldots \},$ we have ${\rm{GKdim}}(A[x,x^{-1}])={\rm{GKdim}}(A[x]).$ The result now follows from Theorem 1. $\Box$

Let $A$ be a $k$-algebra. We showed that ${\rm{GKdim}}(A)=0$ if and only if $A$ is locally finite. We also saw that there is no algebra of GK dimension strictly between 0 and one. Now, what can we say about the case ${\rm{GKdim}}(A)=1$? There is a partial answer to this question and we will see it in the next post.