GK dimension and localization

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

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