PROCEEDINGSof the AMERICANMATHEMATICAL SOCIETY Volume 121, Number 4, August 1994 THE UNITABILITY OF /-PRIME LATTICE-ORDEREDRINGS WITH SQUARESPOSITIVE MA JINGJING (Communicated by Lance W. Small) Abstract. It is shown that an /-prime lattice-ordered ring with squares positive and an /-superunit can be embedded in a unital /-prime lattice-ordered ring with squares positive. In [1, p. 325] Steinberg asked if an /-prime /-ring with squares positive and an /-superunit can be embedded in a unital /-prime /-ring with squares positive. In this paper we show that the answer is yes. If R is a lattice-ordered ring (/-ring) and a e R+, then a is called an /- element of R if b A c = 0 implies ab/\c = baAc = 0. Let T = T(R) = {a e R : \a\ is an /-element of R}. Then T is a convex /-subring of R, and R is a subdirect product of totally ordered T-T bimodules [2, Lemma 1]. R is an /-ring precisely when T = R. Throughout this paper T will denote the subring of /-elements of R. An element e > 0 of an /-ring R is called a superunit if ex > x and xe > x for each x e R+ ; e is an /-superunit if it is a superunit and an /-element. R is infinitesimal if x2 < x for each x in R+ . The /-ring R is called /-prime if the product of two nonzero /-ideals is nonzero and /-semiprime if it has no nonzero nilpotent /-ideals. An /-ring R is called squares positive if a2 > 0 for each a in R. Let A be any ring and a e A . If a satisfies ab = ba = nb for some fixed integer n and all b e A, then a is said to be an «-fier of A and n is said to have an «-fier a in A. Let K = {n e Z : n has «-fiers in A} . Then K is an ideal in the ring Z of integers. The ideal K is called the modal ideal of A ; its nonnegative generator k is called the mode of A . If R is an /-ring with mode k > 0, then R has a unique k-ñer x > 0 [3, III, Lemma 2.1]. Let R be an /-ring, and let ¿ = S(R) = {a e R : \a\ > d, Md e T+} and T-1 = {a e R:\a\Ad = 0, Md e T+}. It is clear that T1- is a convex /-subgroup of R. The referee has pointed out to us that the following lemma, on which our ar- guments are based, is a special case of [4, Lemma 6.2], and P. Conrad attributes it to A. H. Clifford. Received by the editors July 20, 1992 and, in revised form, October 30, 1992. 1991 Mathematics Subject Classification. Primary 06F25; Secondary 16A86. Key words and phrases. Lattice-ordered ring, /-prime /-ring, unitability, squares positive. ©1994 American Mathematical Society 0002-9939/94 $1.00+ $.25 per page 991 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 992 MA JINGJING Lemma 1. Let R be an l-ring which contains nonzero f-elements and T a totally ordered subring of R. Then R = S U (T © T1), where the direct sum is regarded as the direct sum of l-subgroups and S n (T © TL) = 0. Proof. If a e R and a £ S, then \a\ A b < b for some b e T+. Let ax = |û|-|û|Aô and bx = b-\a\/\b. Then ax/\bx = 0, and hence (ax/\d)/\bx = 0 for each d e T+ . Since (ax A d), bx e T, which is a totally ordered ring, and bx > 0, we have ax A d = 0 for each d e T+, and hence ax e T1-. Since 0 < a+, a~ < \a\ = ax + (\a\ A b), there are a2, a3, b2, ¿»3e R such that 0 < a2, «3 < ax, 0 < b2, ¿3 < \a\ A b, a+ = a2 + b2, a~ = a%+ b-¡, and a = a+-a~ = (b2-bi) + (a2-a-i). But T and TL are the convex /-subgroups of R, so seier1. If a e S n (T © TL), then a = ax +a2, where ax e T and a2 e T1-, and hence 2\ax\ < \a\ < \ax\ + \a2\ since a e S. Thus ax = 0 and a = a2, and hence T = {0} . This contradicts T ¿ {0} , so ¿ n (T © TL) = 0 . Let R be an /-ring as in Lemma 1 and aeR. If aeT®TL, then a can be uniquely represented as the sum of elements of T and T^ , and we may write a = ÜT + aT± , where aj e T and aT± e T1- are respectively called the components of a in T and in Tx . It is clear that a V 0 = (aT V 0) + (aT± V0). Throughout this paper the following fact is used frequently. Let R be an /-ring and e an /-superunit of R. Then, for each x e R, x > 0 if and only if ex > 0 or xe > 0. More generally, if 0 < a e T is a regular element (or a non-zero-divisor), then, for each x e R, x > 0 if and only if ax > 0 or xa>0. Lemma 2. Let R be an l-ring with an f-superunit and T a totally ordered ring. Let e be an f-superunit of R. (a) If 0 < b e S, then ne + be > 0 for each neZ . (b) IfbeS and me + be > 0 for some m e Z , then b > 0. Proof, (a) Since 0 < b e S, we have -ne < \n\e < \n\e2 < be for each neZ . (b) Let me + be > 0 for some m e Z . If m < 0, then be > -me > 0, so be = \be\ = \b\e ; that is, (\b\ - b)e = 0. But e is an /-superunit, so b = \b\ > 0. If m >0,then me > -be, and hence me > (-/><?)V0 = (-bvO)e > -bvO, so -bwOeT. Since b V 0 = b + (~b V 0) and b e S, we have b V 0 e S by Lemma 1. Thus b + (-b V 0) = b V 0 > -b v 0 ; that is, b > 0. Lemma 3. Let R be an l-ring with an f-superunit and T a totally ordered ring. If the mode of R is « and the mode of T is k, then n = k and the k-fier of R equals the k-fier of T. Proof. Let e be an /-superunit of R. If x is a k-hher of T, then kd = dx = xd for each d e T, especially ke = ex = xe. Thus e(ka) = exa and (ka)e = axe for each a e R. Since e is an /-superunit of R, ka = xa = ax for each a e R, so n\k. Let y be an «-fier of R. Then na = ay = ya for each a e R. By Lemma 1 we have R = ¿U (T© TL). If y e S, then since e is an /-superunit of R and ne = ey = ye, we have ne = \ne\ = \ey\ = e\y\ > \y\ > (n + l)e . This is a contradiction, so y e T ® T-1. Let y = yT + yj*- • Then ne = ey = eyr + eyT± , and hence ne - eyj = eyT± e T n T1- = {0} , so yT± = 0 and y =yr &T. Thus k\n , and we have k = « . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use THE UNITABILITYOF /-PRIME LATTICE-ORDEREDRINGS 993 Lemma 4. Let R be an l-ring which has an f-superunit and T a totally ordered ring. If e is an f-superunit of R and neZ, b eT®TL satisfy ne + be >0, then bT± > 0 and ne + bje > 0. Proof. If ne + be > 0, then (ne + bje) + bT±e > 0. Thus ne + bje > 0 and bT±e > 0, so bT± > 0. Lemma 5. Let R be an l-ring with squares positive and a e T1- or S. Then d\a\ < a2 and \a\d < a2 for each d e T1. If R has an f-superunit, then n\a\ < a2 for each « e Z+. Proof. Since R is squares positive, 0 < (a±b)2 yields \ad+da\ < a2+d2. But R isa T-T /-bimodule, and |aí/|+|í/a| = \ad+da\ holds in any totally ordered T-T bimodule which is a homomorphic image of R since 0 < d e T ; so it also holds in R [2, Lemma 4]. If a e TL, then \a\d = \ad\ < \ad\ + \da\ = \ad + da\<a2 + d2 and \a\ Ad = 0 imply \a\d = \a\d A(a2 + d2) <(\a\d Aa2) + (\a\d A d2) = \a\d A a2 < a2. Similarly, d\a\ < a2. If a e S, then d < \a\ for each d e T+ , and hence \a\d + d\a\ = \ad\ + \da\ = \ad + da\ < a2 + d2 < a2 + d\a\. Thus \a\d < a2. Similarly, d\a\ < a2. If e is an /-superunit of R, then, as in the previous paragraph, n\a\ < ne\a\ < a2 for each « e Z+ . Theorem 1. An l-prime l-ring R with squares positive and an f-superunit can be embedded in a unit l-prime l-ring with squares positive. Proof. We first note that R is a domain [2, Theorem 1], and hence T is a totally ordered ring. Thus R = S U (T © TL) by Lemma 1. Let e be an /-superunit of R. Let R be the ring obtained by freely adjoining the integers to R, and let k be the mode of T and x the unique k-her of T.