GK dimension of polynomial rings

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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Theorem 1. Let A be a k-algebra. If x is a variable over A, then {\rm{GKdim}}(A[x])=1+{\rm{GKdim}}(A).

Proof.  Let B_0 be a finitely generated subalgebra of A[x] generated by f_1, \ldots , f_m \in A[x]. Let B be the subalgebra of A generated by the coefficients of f_i, \ i =1, \ldots , m. Then clearly B is a finitely generated subalgebra of A and B_0 \subseteq B[x]. Now, let W be a frame of B. Let V=W+kx. Then V is a generating subspace of B[x] and clearly V^n =(W+kx)^n \subseteq \bigoplus_{i=0}^nW^nx^i, for all integers n \geq 0. Hence \dim V^n \leq (n+1) \dim W^n and so

\displaystyle {\rm{GKdim}}(B_0) \leq {\rm{GKdim}}(B[x]) \leq \lim_{n\to\infty} \log_n(n+1) + {\rm{GKdim}}(B)

= 1 + {\rm{GKdim}}(B) \leq 1 + {\rm{GKdim}}(A).

Therefore {\rm{GKdim}}(A[x]) \leq 1 + {\rm{GKdim}}(A). It is also clear that V^{2n}=(W+kx)^{2n} \supseteq \bigoplus_{i=0}^n W^{2n}x^i, for all integers n \geq 0. Thus \dim V^{2n} \geq (n+1) \dim W^{2n} and so

\displaystyle {\rm{GKdim}}(A[x]) \geq {\rm{GKdim}}(B[x]) \geq \lim_{n\to\infty} \log_n(n+1) + {\rm{GKdim}}(B) = 1 + {\rm{GKdim}}(B).

Therefore {\rm{GKdim}}(A[x]) \geq 1 + {\rm{GKdim}}(A) and the result follows. \Box

Corollary 1. Let A be a k-algebra. Then {\rm{GKdim}}(A[x_1, \ldots , x_m])=m+{\rm{GKdim}}(A).

Corollary 2. {\rm{GKdim}}(k[x_1, \ldots , x_m])=m.

An immediate result of corollary 2 is that if X is an infinite set of commuting variables, then {\rm{GKdim}}(k[X])=\infty.

So, by corollary 2, for any integer m \geq 1 there exists an algebra A such that {\rm{GKdim}}(A)=m. In fact, for any real number \alpha \geq 2 there exists a k-algebra A such that {\rm{GKdim}}(A)=\alpha. It is also known that if 1 < \alpha < 2, then there is no algebra A with {\rm{GKdim}}(A)=\alpha. This result is called the Bergman’s gap theorem. We now look at the GK dimension of noncommutative polynomial algebras.

Theorem 2. If X is a set of noncommuting variables with |X| \geq 2, then {\rm{GKdim}}(k \langle X \rangle)=\infty.

Proof. Let x, y \in X and consider the k-subalgebra B=k \langle x, y \rangle of A. Choose the generating subspace V=k+kx + ky. Then it follows easily that \dim V^n = 1 + 2 + \cdots + 2^n \geq 2^n and thus

\displaystyle {\rm{GKdim}}(A) = \limsup_{n\to\infty} \log_n (\dim V^n) \geq \lim_{n\to\infty} n \log_n 2 = \infty.

Therefore {\rm{GKdim}}(A)=\infty. \Box

Corollary. Let A be a k-algebra which is a domain. If {\rm{GKdim}}(A) < \infty, then A is Ore.

Proof. If A contains a copy of k \langle x, y \rangle, where x and y are noncommuting variables, then {\rm{GKdim}}(A)=\infty, by Theorem 2. Thus A does not contain such a subalgebra and hence A is an Ore domain by the lemma in this post. \Box

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Comments
  1. […] to that the Gelfand-Kirillov dimension satisfies $mathrm{GK}dim(R[T])=mathrm{GK}dim(R)+1$ (see here) for every $K$-algebra […]

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