## GK dimension of Weyl algebras

Posted: April 10, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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We defined the $n$-th Weyl algebra $A_n(R)$ over a ring $R$ in here.  In this post we will find the GK dimension of $A_n(R)$ in terms of the GK dimension of $R.$ The result is similar to what we have already seen in commutative polynomial rings (see corollary 1 in here). We will assume that $k$ is a field and $R$ is a $k$-algebra.

Theorem. ${\rm{GKdim}}(A_1(R))=2 + {\rm{GKdim}}(R).$

Proof. Suppose first that $R$ is finitely generated and let $V$ be a frame of $R.$ Let $U=k+kx+ky.$ Since $yx = xy +1,$ we have

$\dim_k U^n = \frac{(n+1)(n+2)}{2}. \ \ \ \ \ \ \ \ \ (*)$

Let $W=U+V.$ Clearly $W$ is a frame of $A_1(R)$ and

$W^n = \sum_{i+j=n} U^i V^j,$

for all $n,$ because every element of $V$ commutes with every element of $U.$ Therefore, since $V^j \subseteq V^n$ and $U^i \subseteq U^n$ for all $i,j \leq n,$ we have $W^n \subseteq U^nV^n$ and $W^{2n} \supseteq U^nV^n.$ Thus $W^n \subseteq U^nV^n \subseteq W^{2n}$ and hence

$\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n \leq \log_n \dim_k W^{2n}.$

Therefore ${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)),$ by $(*),$ and we are done.

For the general case, let $R_0$ be any finitely generated $k$– subalgebra of $R.$ Then, by what we just proved,

$2 + {\rm{GKdim}}(R_0)={\rm{GKdim}}(A_1(R_0)) \leq {\rm{GKdim}}(A_1(R))$

and hence $2+{\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)).$ Now, let $A_0$ be a $k$-subalgebra of $A_1(R)$ generated by a finite set $\{f_1, \ldots , f_m\}.$ Let $R_0$ be the $k$-subalgebra of $R$ generated by all the coefficients of $f_1, \ldots , f_m.$ Then $A_0 \subseteq A_1(R_0)$ and so

${\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A_1(R_0))=2 + {\rm{GKdim}}(R_0) \leq 2 + {\rm{GKdim}}(R).$

Thus

${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R)$

and the proof is complete. $\Box$

Corollary. ${\rm{GKdim}}(A_n(R))=2n + {\rm{GKdim}}(R)$ for all $n.$ In particular, ${\rm{GKdim}}(A_n(k))=2n.$

Proof. It follows from the theorem and the fact that $A_n(R)=A_1(A_{n-1}(R)). \Box$

## A theorem of Borho and Kraft

Posted: April 2, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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As usual, I’ll assume that $k$ is a field. Recall that if a $k$-algebra $A$ is an Ore domain, then we can localize $A$ at $S:=A \setminus \{0\}$ and get the division algebra $Q(A):=S^{-1}A.$ The algebra $Q(A)$ is called the quotient division algebra of $A.$

Theorem (Borho and Kraft, 1976) Let $A$ be a finitely generated $k$-algebra which is a domain of finite GK dimension. Let $B$ be a $k$-subalgebra of $A$ and suppose that ${\rm{GKdim}}(A) < {\rm{GKdim}}(B) + 1.$ Let $S:=B \setminus \{0\}.$ Then $S$ is an Ore subset of $A$ and $S^{-1}A=Q(A).$ Also, $Q(A)$ is finite dimensional as a (left or right) vector space over $Q(B).$

Proof. First note that, by the corollary in this post, $A$ is an Ore domain and hence both $Q(A)$ and $Q(B)$ exist and they are division algebras. Now, suppose, to the contrary, that $S$ is not (left) Ore. Then there exist $x \in S$ and $y \in A$ such that $Sy \cap Ax = \emptyset.$ This implies that the sum $By + Byx + \ldots + Byx^m$ is direct for any integer $m.$ Let $W$ be a frame of a finitely generated subalgebra $B'$ of $B.$ Let $V=W+kx+ky$ and suppose that $A'$ is the subalgebra of $A$ which is generated by $V.$ For any positive integer $n$ we have

$V^{2n} \supseteq W^n(kx+ky)^n \supseteq W^ny + W^nyx + \ldots + W^nyx^{n-1}$

and thus $\dim_k V^{2n} \geq n \dim_k W^n$ because the sum is direct. So $\log_n \dim_k V^{2n} \geq 1 + \log_n \dim_k W^n$ and hence ${\rm{GKdim}}(A) \geq {\rm{GKdim}}(A') \geq 1 + {\rm{GKdim}}(B').$ Taking supremum of both sides over all finitely generated subalgebras $B'$ of $B$ will give us the contradiction ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B).$ A similar argument shows that $S$ is right Ore. So we have proved that $S$ is an Ore subset of $A.$ Before we show that $S^{-1}A=Q(A),$ we will prove that $Q(B)A=S^{-1}A$ is finite dimensional as a (left) vector space over $Q(B).$ So let $V$ be a frame of $A.$ For any positive ineteger $n,$ let $r(n) = \dim_{Q(B)} Q(B)V^n.$ Clearly $Q(B)V^n \subseteq Q(B)V^{n+1}$ for all $n$ and

$\bigcup_{n=0}^{\infty}Q(B)V^n =Q(B)A$

because $\bigcup_{n=0}^{\infty}V^n=A.$ So we have two possibilities: either $Q(B)V^n=Q(B)A$ for some $n$ or the sequence $\{r(n)\}$ is strictly increasing. If $Q(B)V^n = Q(B)A,$ then we are done because $V^n$ is finite dimensional over $k$ and hence $Q(B)V^n$ is finite dimensional over $Q(B).$ Now suppose that the sequence $\{r(n)\}$ is strictly increasing. Then $r(n) > n$ because $r(0)=\dim_{Q(B)}Q(B)=1.$ Fix an integer $n$ and let $e_1, \ldots , e_{r(n)}$ be a $Q(B)$-basis for $Q(B)V^n.$ Clearly we may assume that $e_i \in V^n$ for all $i.$ Let $W$ be a frame of a finitely generated subalgebra of $B.$ Then

$(V+W)^{2n} \supseteq W^nV^n \supseteq W^ne_1 + \ldots + W^ne_{r(n)},$

which gives us

$\dim_k(V+W)^{2n} \geq r(n) \dim_k W^n > n \dim_k W^n,$

because the sum $W^ne_1 + \ldots + W^ne_{r(n)}$ is direct. Therefore ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B),$ which is a contradiction. So we have proved that the second possibility is in fact impossible and hence $Q(B)A$ is finite dimensional over $Q(B).$ Finally, since, as we just proved, $\dim_{Q(B)}Q(B)A < \infty,$ the domain $Q(B)A$ is algebraic over $Q(B)$ and thus it is a division algebra. Hence $Q(B)A=Q(A)$ because $A \subseteq Q(B)A \subseteq Q(A)$ and $Q(A)$ is the smallest division algebra containing $A. \Box$

## A theorem of Smith and Zhang

Posted: May 11, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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For the definition of a PI-algebra see this post.

Lemma. Let $k$ be a field and let $A$ be a finitely generated $k$-algebra which is a domain. If $A$ is not PI, then ${\rm{GKdim}}(A) \geq 2.$

Proof. If ${\rm{GKdim}}(A)=0,$ then $A$ is finite dimensional over $k,$ and so over its center, and hence it is PI. If ${\rm{GKdim}}(A) = 1,$ then $A$ is again PI (see here). Also, there is no algebra whose GK dimension is strictly between 1 and 2, by the Bergman’s gap theorem. Thus ${\rm{GKdim}}(A) \geq 2. \ \Box$

We are now going to refine the result given in the above lemma. We will write ${\rm{GKdim}}_F$ for the GK dimension of an algebra viewed as an $F$-algebra if $F$ is not $k.$

Theorem. (Smith and Zhang, 1996) Let $k$ be a field and let $A$ be a finitely generated $k$-algebra which is a domain. If $A$ is not PI, then ${\rm{GKdim}}(A) \geq 2 + {\rm{GKdim}}(Z(A)),$ where $Z(A)$ is the center of $A.$

Proof. By the above lemma, we may assume that $2 \leq {\rm{GKdim}}(A) < \infty$ and ${\rm{GKdim}}(Z(A)) \geq 1.$ Let $Q_Z(A)$ be the central localization of $A$ and let $F$ be the center of $Q_Z(A).$ Clearly $F$ is just the quotient field of $Z(A).$ Recall, from the theorem in here, that

${\rm{GKdim}}(Q_Z(A))={\rm{GKdim}}(A) \geq 2$  and ${\rm{GKdim}}(F)={\rm{GKdim}}(Z(A)) \geq 1.$

Let $0 \leq d < {\rm{GKdim}}(F)$ and $0 \leq e < {\rm{GKdim}}_F(Q_Z(A)).$ Then there exist a finite dimensional $k$-vector subspace $V$ of $F$ which contains 1 and $\dim_k V^n \geq n^d$ for all large enough integers $n.$ Also, there exists a finite dimensional $F$-vector subspace $W \supseteq V$ of $Q_Z(A)$ which contains 1 and $\dim_F W^n \geq n^e$ for all large enough integers $n.$ Hence, for large enough integers $n$ we have

$\dim_k W^{2n} \geq \dim_k (W^nV^n) \geq (\dim_F W^n)(\dim_k V^n) \geq n^{e+d}.$

Thus ${\rm{GKdim}}(Q_Z(A)) \geq e+d.$ Since the above inequality holds for all real number $0 \leq d < {\rm{GKdim}}(F)$ and $0 \leq e < {\rm{GKdim}}_F(Q_Z(A)),$ we have

${\rm{GKdim}}(Q_Z(A)) \geq {\rm{GKdim}}_F(Q_Z(A)) + {\rm{GKdim}}(F). \ \ \ \ \ \ \ \ (*)$

Now, since $A$ is finitely generated as a $k$-algebra, $Q_Z(A)$ is a finitely generated $F$-algebra and clearly $Q_Z(A)$ is not PI because $A$ is not PI. Therefore ${\rm{GKdim}}_F(Q_Z(A)) \geq 2$ by the above lemma. We also have ${\rm{GKdim}}(F)={\rm{GKdim}}(Z(A))$ and now the result follows from $(*). \ \Box$

## Algebras of GK dimension one

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

## 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.

## GK dimension of finite extensions

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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The goal is to prove the following fundamental result: if an algebra $A$ is a finite module over some subalgebra $B,$ then ${\rm{GKdim}}(A)={\rm{GKdim}}(B).$

Lemma. Let $B$ be a subalgebra of a $k$-algebra $A.$ Suppose that, as a (left) module, $A$ is finitely generated over $B.$ Then  ${\rm{GKdim}}({\rm{End}}_B(A)) \leq {\rm{GKdim}}(B).$

Proof. So $A=\sum_{i=1}^n Ba_i$ for some $a_i \in A.$ Define $\varphi: B^n \longrightarrow A$ by $\varphi(b_1, \ldots , b_n)=\sum_{i=1}^n b_ia_i$ and let $I = \ker \varphi.$ Let

$C=\{f \in {\rm{End}}_B(B^n): \ f(I) \subseteq I \}.$

Clearly $C$ is a subalgebra of ${\rm{End}}_B(B^n) \cong M_n(B).$ Now, given $f \in C$ define $\overline{f}: A \longrightarrow A$ by  $\overline{f}(a)=\varphi f (u),$ where $u$ is any element of $B^n$ with $\varphi(u)=a.$ Note that $\overline{f}$ is well-defined because if $\varphi(v)=a$ for some other $v \in B^n,$ then $u-v \in I$ and so $f(u-v) \in I.$ Hence $0=\varphi f(u-v)=\varphi f(u) - \varphi f(v)$ and so $\varphi f(u) = \varphi f(v).$ It is easy to see that $\overline{f} \in {\rm{End}}_B(A)).$ Finally, define $\psi: C \longrightarrow {\rm{End}}_B(A))$ by $\psi(f)=\overline{f}.$ Then $\psi$ is an $k$-algebra onto homomorphism and hence

${\rm{GKdim}}({\rm{End}}_B(A)) \leq {\rm{GKdim}}(C) \leq {\rm{GKdim}}(M_n(B)) ={\rm{GKdim}}(B),$

by Fact 1 and Fact 5. $\Box$

Theorem. Let $B$ be a subalgebra of a $k$-algebra $A$ and suppose that, as a (left) module, $A$ is finitely generated over $B.$ Then ${\rm{GKdim}}(A)={\rm{GKdim}}(B).$

Proof. The algebra $A$ has a natural embedding into ${\rm{End}}_B(A)$ and so ${\rm{GKdim}}(A) \leq {\rm{GKdim}}({\rm{End}}_B(A)).$ Thus ${\rm{GKdim}}(A) \leq {\rm{GKdim}}(B),$ by the lemma. $\Box$

An important consequence of the above theorem is the following result. It shows that for finitely generated commutative algebras, GK dimension is nothing but the transcendence degree of the algebra over the base field.

Corollary. If $A$ is a finitely generated commutative $k$-algebra, then ${\rm{GKdim}}(A)={\rm{tr.deg}}(A/k).$

Proof. Let $m={\rm{tr.deg}}(A/k).$ Then $A$ contains a polynomial $k$-algebra $B=k[x_1, \ldots, x_m]$ such that $A$ is a finitely generated $B$-module, by the  Noether normalization theorem. Thus, by Corollary 2 and this theorem, ${\rm{GKdim}}(A) = {\rm{GKdim}}(B)=m. \Box$

## 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. $\Box$