central localization | Abstract Algebra

Posts Tagged ‘central localization’

Algebras of GK dimension one

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
Tags: central localization, GK dimension one, Posner's theorem, Tsen's theorem

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).

von Neumann regular rings (3)

Posted: October 31, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
Tags: central localization, division ring, reduced ring, stable range one, strongly regular, von Neumann regular

We saw in part (2) that von Neumann regular rings live somewhere between semisimple and semiprimitive rings. The goal in this post is to prove a theorem of Armendariz and others which gives a necessary and sufficient condition for a ring to be both regular and reduced. This result extends Kaplansky’s result for commutative rings (see the corollary at the end of this post). We remark that a commutative von Neumann regular ring $R$ is necessarily reduced. That is because if $x^2=0$ for some $x \in R,$ then choosing $y \in R$ with $x=xyx$ we will get $x=yx^2=0.$

Definition . A von Neumann regular ring $R$ is called strongly regular if $R$ is reduced.

Theorem 1. (Armendariz, 1974) A ring $R$ with 1 is strongly regular if and only if $R_M$ is a division ring for all maximal ideals $M$ of $Z(R).$

Proof. Suppose first that $R$ is strongly regular and let $M$ be a maximal ideal of $Z(R).$ Let $0 \neq s^{-1}x \in R_M.$ So $tx \neq 0$ for all $t \in Z(R) \setminus M.$ Since $R$ is regular, there exists some $y \in R$ such that $xyx = x.$ Then $xy=e$ is an idempotent and thus $e \in Z(R)$ because in a reduced ring every idempotent is central. Since $(1-e)x=0$ we have $1-e \in M$ and hence $e \in Z(R) \setminus M.$ Thus $e^{-1}sy$ is a right inverse of $s^{-1}x.$ Similarly $f=yx \in Z(R) \setminus M$ and $f^{-1}sy$ is a left inverse of $s^{-1}x.$ Therefore $s^{-1}x$ is invertible and hence $R_M$ is a division ring. Conversely, suppose that $R_M$ is a division ring for all maximal ideals $M$ of $Z(R).$ If $R$ is not reduced, then there exists $0 \neq x \in R$ such that $x^2=0.$ Let $I=\{s \in Z(R): \ sx = 0 \}.$ Clearly $I$ is a proper ideal of $Z(R)$ and hence $I \subseteq M$ for some maximal ideal $M$ of $Z(R).$ But then $(1^{-1}x)^2=0$ in $R_M,$ which is a division ring. Thus $1^{-1}x=0,$ i.e. there exists some $s \in Z(R) \setminus M$ such that $sx = 0,$ which is absurd. To prove that $R$ is von Neumann regular, we will assume, to the contrary, that $R$ is not regular. So there exists $x \in R$ such that $xzx \neq x$ for all $z \in R.$ Let $J= \{s \in Z(R): \ xzx=sx \ \text{for some} \ z \in R \}.$ Clearly $J$ is a proper ideal of $Z(R)$ and so $J \subseteq M$ for some maximal ideal $M$ of $Z(R).$ It is also clear that if $sx = 0$ for some $s \in Z(R),$ then $s \in J$ because we may choose $z = 0.$ Thus $1^{-1}x \neq 0$ in $R_M$ and hence there exists some $y \in R$ and $t \in Z(R) \setminus M$ such that $1^{-1}x t^{-1}y = 1.$ Therefore $u(xy-t)=0$ for some $u \in Z(R) \setminus M.$ But then $x(uy)x=utx$ and so $ut \in J,$ which is nonsense. This contradiction proves that $R$ must be regular. $\Box$