Journal of Algebra 228, 406–416 (2000) doi:10.1006/jabr.1999.8232, available online at http://www.idealibrary.com on Lattice-Ordered Matrix Algebras with the Usual Lattice Order Jingjing Ma CORE Department of Mathematical Sciences, The University of Texas at El Paso, Metadata, citation and similar papers at core.ac.uk El Paso, Texas 79968 Provided by Elsevier - Publisher Connector E-mail: [email protected] Communicated by Walter Feit Received October 18, 1999 Let R be a unital lattice-ordered algebra over a totally ordered field F and Fn be the n × n n ≥ 2 matrix algebra over F. It is shown that under certain conditions C R contains a lattice-ordered subalgebra which is isomorphic to (Fn; F n. In par- ticular, let (Fn;P) be a lattice-ordered algebra over F with the positive cone P.If C a certain element is positive in Fn;P, then Fn;P is isomorphic to Fn; F n. © 2000 Academic Press Key Words: lattice-ordered algebra; matrix algebra. In the following we always assume that F is a totally ordered field and R is a unital lattice-ordered algebra (`-algebra) over F with 1 2 RC although some of the results obtained hold for a lattice-ordered ring (`-ring). Let Fn n ≥ 2 be the n × n matrix ring over F. It is well-known that Fn may be lattice-ordered by saying that a matrix in Fn is positive exactly when each of C its entries is positive, namely, the positive cone of Fn is F n. This lattice order is called the usual lattice order on Fn. Let Q be the field of rational numbers. In [6], Weinberg conjectures that C (Q n is the only lattice order of Qn (up to an isomorphism) such that Qn is an `-algebra over Q in which 1 is positive, and it was proved for n D 2. In [5], Steinberg shows that, under a certain maximum condition, Weinberg’s conjecture is true for the matrix rings over a totally ordered field. The results in this paper are motivated by their works. Given a unital `-algebra R over F with 1 2 RC. We obtain some conditions such that R is isomorphic to Fn with the usual lattice order. As a consequence, it is shown that if a certain element in an `-algebra Fn is positive, then Fn is C isomorphic to Fn; F n. 406 0021-8693/00 $35.00 Copyright © 2000 by Academic Press All rights of reproduction in any form reserved. lattice-ordered matrix algebra 407 First we collect some results about lattice-ordered groups (`-groups) and `-rings that we will use. Let G be an `-group. An element 0 <g2 G is C C called a basic element of G if a 2 G x a ≤ g is a chain. Let g1;g2 2 G . Then g1, g2 are called disjoint if g1 ^ g2 D 0. A subset S in an `-group G is called disjoint if s>0; 8s 2 S and s ^ t D 0 for any s, t 2 S with s 6D t.A subset S of G is a basis of G if (i) S is a maximal disjoint subset of G, and (ii) each element in S is basic. The following condition F was introduced by Conrad in [2], and he showed that if G satisfies F, then G has a basis [2, Theorem 5.2]: F Each 0 <a2 G is greater than at most a finite number of disjoint elements: A subset S of an `-group G is convex if whenever s; t 2 S and s ≤ g ≤ t in G for g 2 G, then g 2 S.An`-subgroup of an `-group is a subgroup and a sublattice. Let X be a nonempty subset of an `-group G. Define X? =y 2 G xy∧x=0; 8x 2 X. X? is called a polar of G, and it is a convex `-subgroup of G. The double polar X?? is denoted by X??. If X =x, then we denote x? by x?, and x?? by x??.Lets 2 G be a basic element. Then, from [2], s?? is the greatest totally ordered convex subgroup of G that contains s,sos?? is a maximal totally ordered convex subgroup of G.LetG be an `-group having a basis S =sγ x γ 2 0. Then the basis group B D BG of G is defined as the (direct) sum of ?? ?? s , γ 2 0 x B =⊕γ20s .IfG satisfies F and has no maximal totally ordered convex subgroup which is bounded from above, then G D BG ?? [2, Corollary, P. 231]. Let M =sγ x γ 2 0. Then M is the set of all maximal totally ordered convex subgroups of G. Let A be a partially ordered ring (po-ring). A module AM is called an `-module over A if M is an `-group and ACMC ⊆ MC.An`-module is called an f -module if it is isomorphic to a subdirect product of totally ordered modules. A po-ring is called directed if any two elements have a lower bound and a upper bound. Let AM be an `-module over the directed po-ring A. Then AM is an f -module if and only if x ^ y D 0 implies ax ^ y D 0, 8x; y 2 M, a 2 AC [3, Theorem 1]. An `-ring A is called an f -ring C if AA and AA are f -modules. Let A be an `-ring. The element a 2 A is called an f -element of A if b ^ c D 0 implies that ab ^ c D ba ^ c D 0 for any b; c 2 A.LetT A=a 2 A xa is an f -element of A. Then T A is a convex f -subring of A [1, p. 55]. From [4, p. 364], an `-algebra R over F is an algebra R over F which is also an f -module over F, and an `-ring. C Now let R be a unital `-algebra over F with 1 2 R . Since F R is an f - ? module, for any subset X ⊆ R, the polar X is a convex `-subspace of F R. Thus if 0 <a2 R is basic, then a?? is the greatest totally ordered convex 408 jingjing ma subspace of F R that contains a, and hence the basis group BR is a convex C `-subspace of F R.Lete 2 R be an invertible element of order n ≥ 2, that n is, e D 1, n ≥ 2. Then multiplication by e is an `-automorphism of F R, and hence if E is a totally ordered convex subspace of R, then so are eiEej, for 1 ≤ i; j ≤ n. Lemma 1. Let e 2 RC be an element of order n ≥ 2, and let E be a totally i j k l i j ordered subspace of F R.Ife Ee D e Ee for some i; j; k; l, then e xe D ekxel, 8x 2 E. Proof. Let 0 <x2 E. Since eixej 2 ekEel, there exists y 2 E such that eixej D ekyel.Ifx<y, then y D en−kCixen−lCj <en−kCiyen−lCj, and hence y<en−kCiyen−lCj <e2n−kCiye2n−lCj < ··· <enn−kCiyenn−lCj D y, which is a contradiction. Similarly, y 6<x. Thus x D y. Lemma 2 [2, Lemma 3.1]. If A and B are totally ordered convex sub- groups of an `-group, then A ⊆ B or B ⊆ A or A \ B =0. Let E be a totally ordered convex subspace of F R. As a direct conse- quence of Lemma 2 and that e is an invertible element of order n ≥ 2, we have the following: (∗) 8i; j; k; l; eiEej D ekEel or eiEej \ ekEel =0: Lemma 3. Let E be a totally ordered convex subspace of F R, and let e 2 RC be an element of order n ≥ 2.IfE \ T R2 6D 0, then we have the following. (i) For any k, Eek D E if and only if ekE D E. (ii) For any i and j,ifEei 6D Eej, then eiE 6D Eej. Proof. Let 0 <a2 E \ T R such that a2 6D 0. (i) Let Eek D E. From (∗), we have that ekE D E or ekE \ E =0. Suppose ekE \ E =0. Then eka ^ a D 0, and hence aeka ^ a D 0 since a 2 T R. By Lemma 1, we have aek D a,soa2 ^ a D 0, and hence a2 D a2 ^ a2 D 0, which is a contradiction. Thus ekE D E. Similarly if ekE D E, then Eek D E. (ii) Suppose eiE D Eej. Since Eei 6D Eej, Eei \ Eej =0. Then aei ^ aej D 0, and hence aeia ^ aej D 0. By Lemma 1, eia D aej,soa2ej ^ aej D 0, and hence a2 ^ aej D 0. Thus a2 ^ a D 0 and a2 D 0, which is a contradiction. Lemma 4. Let e 2 R be an element of order n ≥ 2, and E a subset of R. (i) If m is the smallest positive integer such that emE D E, then mn. (ii) If k is the smallest positive integer such that ekE D Eek, then kn. lattice-ordered matrix algebra 409 (iii) If m and k are in (i) and (ii) and E is a totally ordered convex subspace with E \ T R2 6D 0, then km. Proof. Let e be the cyclic group generated by e, and let PR be the set consisting of all subsets of R.