We have seen so far that the GK dimension of a k-algebra A, where k is a field, has possible values 0, 1 and any real number \geq 2. We showed that the GK dimension of A is 0 if and only if every finitely generated k-subalgebra of A is finite dimensional as a k-vector subspace of A. In particular, a finitely generated k-algebra A has GK dimension 0 if and only if \dim_k A < \infty. Now what can we say about algebras of GK dimension one? First we show that they are not necessarily Noetherian.

Example. Let A be the k-algebra generated by x and y with the relations x^2=xyx=yxy=0. Then A is not noetherian and {\rm{GKdim}}(A)=1.

Proof. It is not noetherian because it contains the infinite direct sum of ideals \bigoplus_{n=2}^{\infty}kxy^nx. To see why the GK dimension of A is one, consider the frame V=k + kx + ky. Now, assuming that n \geq 3 and considering the relations on A, we see that the only terms which appear in V^n are

1, y, \ldots , y^n, x, xy, \ldots , xy^{n-1}, yx, y^2x , \ldots , y^{n-1}x and xy^2x, \ldots , xy^{n-2}x.

Thus \dim_k V^n = 4n-3 and hence \displaystyle {\rm{GKdim}}(A)= \lim_{n \to\infty} \log_n(4n-3)=1. \ \Box

Next theorem shows that if A is semiprime and has GK dimension 1, then A is Noetherian. In fact it will be even more than just Noetherian.

Theorem 1. If A is a finitely generated semiprime k-algebra, then {\rm{GKdim}}(A)=1 if and only if A is finitely generated as a module over some polynomial algebra in one variable k[x] \subseteq Z(A).

Proof. If A is finitely generated as a module over some polynomial algebra k[x], then

{\rm{GKdim}}(A)={\rm{GKdim}}(k[x])=1,

by this theorem and Corollary 2. Conversely, if {\rm{GKdim}}(A)=1, then by a theorem of Small, Stafford and Warfield [2], A is finitely genrated over its center Z(A) and thus {\rm{GKdim}}(Z(A))=1, by this theorem. We also have k \subseteq Z(A) \subseteq A and we know that A is both a finitely generated k-algebra and a finitely generated Z(A)-module. Thus, by Artin-Tate lemma, Z(A) is a finitely generated k-algebra. Therefore {\rm{tr.deg}}(Z(A)/k)=1, by the corollary in this post, and hence Z(A) is a finitely generated module over some polynomial algebra k[x], by the Noether normalization theorem. The result now follows because A is a finitely generated Z(A)-module. \Box

Theorem 2. Let k be an algebraically closed field and let A be a k-algebra. If A is a domain and {\rm{GKdim}}(A) \leq 1, then A is commutative.

Proof. First note that if a,b \in A, then the k-subalgebra generated by a,b has GK dimension at most one too and so we may assume that A is finitely generated. The case {\rm{GKdim}}(A)=0 easily follows because then A would be finite dimensional, and hence algebraic, over k and therefore A=k because k is algebraically closed. Now, suppose that {\rm{GKdim}}(A)=1. The algebra A is PI, by [2], and thus Q_Z(A), the central localization of A, is a finite dimensional central simple algebra by Posner’s theorem [1]. Since A is a domain, Q_Z(A) is a domain and hence Q_Z(A)=D is a finite dimensional division algebra over its center F,  which is the quotient field of Z(A). Thus {\rm{tr.deg}}(F/k)={\rm{tr.deg}}(Z(A)/k)=1, by the corollary in this post. Hence, by Tsen’s theorem [3], Q_Z(A)=F. Thus Q_Z(A), and so A itself, is commutative. \Box

Remark. Theorem 2 does not hold if k is not algebraically closed. For example, let \mathbb{H} be the division ring of real quaternions. Then, as an \mathbb{R}-algebra, \mathbb{H} is a noncommutative domain of GK dimension zero.  Similarly, if x is a central variable over \mathbb{H}, then the polynomial ring \mathbb{H}[x], as an \mathbb{R}-algebra, is a noncommutative domain of GK dimension one.

Refferences:

1. E. C. Posner, Prime rings satisfying a polynomial identity, Proc. Amer. Math. Soc. (1960) no. 2, 180-183.

2. L. W. Small, J. T. Stafford, and R. Warfield, Affine algebras of Gelfand Kirillov dimension one are PI, Math. Proc. Cambridge. Phil. Soc. (1984), 407-414.

3. C. Tsen, Divisionsalgebren uber Funktionenkorper, Nachr. Ges. Wiss. Gottingen (1933).

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