# Integer Gelfand-Kirillov dimension

Let $R$ be a (noncommutative) Noetherian affine $K$-algebra. The Gelfand-Kirillov dimension is known to be an integer for many classes of affine Noetherian algebras. I wonder, if this is true for any affine Noetherian algebra; I could not find any positive or negative result on this topic.

Most of the positive examples deal with algebras (such as somewhat commutative algebras) where the associated graded algebra $gr(R)$ is commutative (or at least a finite normalizing extension of $K$) and affine for a fixed finite filtration $\{R_n\}_{n\in\mathbb{N}_0}$. They now argue by showing the existence of a Hilbert-polynomial $HP_R(t)\in\mathbb{Q}[t]$ for the Hilbert-function $n\mapsto\dim_K(R_N)$. Hence, by definition, they get $GKdim(R)=\deg(HP_R(t))$ which must therefore be an integer as the degree of a polynomial.

But the main part of the above proof is the existence of such a Hilbert-polynomial using the theorem of Hilbert-Serre which states that the Hilbert-series $HS_R(t)=\sum_{n=0}^\infty\dim_K(R_n)t^n$ can be written as a fraction $HS_R(t)=\frac{f(t)}{\prod_{i=1}^n(1-t^{d_i})}$ of polynomials with integer coefficients where $d_i$ denotes the degrees of the appropriate homogeneous generators of the algebra $R$. This proof, however, can not be generalized to the noncommutative case...

So, are there any ideas how to deal with the noncommutative case or is there any counter-example for a Noetherian affine $K$-algebra without an integer $GKdim$?

• Maybe there's been progress since, but in the survey paper "Dimension theory of noetherian rings" (in "Infinite length modules" (Bielefeld, 1998), 129--147, Trends Math., Birkhauser, Basel, 2000), Tom Lenagan cites this as an open problem. – Jeremy Rickard May 21 '16 at 11:13