## 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. $\Box$

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