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

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