## 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 $A. \Box$