## Algebras of GK dimension one

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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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
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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. $\Box$

Corollary. (Kaplansky) A commutative ring $R$ is regular if and only if $R_M$ is a field for all maximal ideals $M$ of $R. \ \Box$

At the end let me mention a nice property of strongly regular rings.

Theorem 2. (Pere Ara, 1996) If $R$ is strongly regular and $Ra+Rb=R,$ for some $a, b \in R,$ then $a+rb$ is a unit for some $r \in R.$