## GK dimension; some basic facts (2)

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

Fact 4. If $A$ and $B$ are $k$-algebras and ${\rm{GKdim}}(B)=0,$ then ${\rm{GKdim}}(A \otimes_k B)={\rm{GKdim}}(A).$

Proof. Since $A \cong A \otimes_k 1 \subseteq A \otimes_k B,$ we have ${\rm{GKdim}}(A) \leq {\rm{GKdim}}(A \otimes_k B).$ Now, let $C$ be a finitely generated subalgebra of $A \otimes_k B$ with a frame $W.$ Since $\dim_k W < \infty,$ there exist finite dimensional subspaces $U, V$ of $A,B,$ respectively,  such that $1_A \in U, \ 1_B \in V$ and $W \subseteq U \otimes_k V.$ Let $A_0, B_0$ be the algebras generated by $U,V$ respectively. Now, $W^n \subseteq U^n \otimes_k V^n,$ for all $n,$ and so

$\dim_k W^n \leq (\dim_k U^n)(\dim_k V^n).$

Therefore $\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n$ and hence, taking limsup, will give us

${\rm{GKdim}}(C) \leq {\rm{GKdim}}(A_0)+ {\rm{GKdim}}(B_0)={\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A).$

Since the above holds for any finitely generated subalgebra $C$ of $A \otimes_k B,$ we have ${\rm{GKdim}}(A \otimes_k B) \leq {\rm{GKdim}}(A). \Box$

Fact 5. If $A$ is a $k$-algebra, then ${\rm{GKdim}}(M_n(A))={\rm{GKdim}}(A).$

Proof. By Fact 2, ${\rm{GKdim}}(M_n(k))=0.$ Now the result follows from $M_n(A) \cong A \otimes_k M_n(k)$ and  Fact 4. $\Box$

Fact 6. Let $A$ be a $k$-algebra and let $I$ be an ideal of $A.$ If ${\rm{l.ann}}_A(a)=(0)$ for some $a \in I,$ then ${\rm{GKdim}}(A/I) \leq {\rm{GKdim}}(A)-1.$

Proof. Let $B$ be any finitely generated subalgebra of $A$ and let $V$ be a frame of $B':=B[a]$ which contains $a.$ Let $\overline{V}=(V+I)/I.$ Clearly $\overline{V}$ is a frame of $\overline{B'}=(B'+I)/I = (B+I)/I.$ If $n$ is an integer, then, as $k$-vector spaces, $V^n=(V^n \cap I) \oplus W_n$ for some finite dimensional $k$-vector space $W_n.$ Note that

$W_n \cong V^n/(V^n \cap I) \cong (V^n+I)/I=\overline{V}^n.$

Also, since $Aa \cap W_n \subseteq I \cap V^n$ and $Aa \cap W_n \subseteq W_n,$ we have $Aa \cap W_n =(0)$ for all $n.$ Therefore, since ${\rm{l.ann}}_A(a)=(0),$ the sum $\sum_{i=0}^n W_n a^i$ is direct for all $n.$  Clearly $\sum_{i=0}^n W_na^i \subseteq V^{2n},$ for all $n,$ because both $a$ and $W_n$ are in $V^n.$ Thus

$\dim_k V^n \geq \dim_k \sum_{i=0}^n W_na^i = \sum_{i=0}^n \dim_k W_na^i=(n+1) \dim_k W_n > n \dim_k W_n$  $=n \dim_k \overline{V}^n.$

Hence ${\rm{GKdim}}(A) \geq {\rm{GKdim}}(B') \geq 1 + {\rm{GKdim}}(\overline{B'}).$ Since every finitely generated subalgebra of $A/I$ is in the form $(B+I)/I$ for some finitely generated subalgebra $B$ of $A,$ the inequality holds for any finitely generated subalgebra of $A/I.$ Thus ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(A/I). \Box$

Corollary. Let $A$ be a $k$-algebra which is also a domain. Let $B$ be a simple $k$-subalgebra of $A.$ If ${\rm{GKdim}}(A) < {\rm{GKdim}}(B) + 1,$ then $A$ is simple too.

Proof. Let $I$ be a nonzero ideal of $A.$ If $I \cap B \neq (0),$ then $I \cap B=B,$ because $B$ is simple, and so $I=A.$ Suppose now that $I \cap B=(0).$  Then the natural homomorphism $B \longrightarrow A/I$ would be injective and so

${\rm{GKdim}}(B) \leq {\rm{GKdim}}(A/I) \leq {\rm{GKdim}}(A) - 1,$

by Fact 6, which is a contradiction. $\Box$