## 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, ${\rm{GKdim}}(A) = {\rm{GKdim}}(B)=m. \Box$