With Squares Positive and an /-Superunit Can Be Embedded in a Unital /-Prime Lattice-Ordered Ring with Squares Positive

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.

991

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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 > (-/> -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 = « .

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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\ 0 for all c e R+ , which is equivalent to ne + ae > 0. Since nc + ac >0 if and only if (« + mk)c + (a - mx)c > 0 for each meZ , (R')+ is well defined. It is obvious that (R')+ has the following properties: (i) (Ö7Ö)e(R')+, (ii) (R')+n[-(R')+] = {0), (iii) (R')+ + (R')+ ç (R')+ , (iv) (R')+(R')+ç(R')+. Hence, the ring R' is a partially ordered ring, and a' < b' if and only if (b' - a') e (R')+ for all a', b' e R'. It is clear that, for each x' e R', x' > 0 if and only if x'(0,e) >0. We show below that R' is a lattice-ordered ring under the partial order defined above. In fact, if a e S, then ,_v,.n /(ïïTTvÔ) ifav0e¿, («, a) VO = { _ 1(0, avo) if avoe r©rx. If a e T © TL, then since T is a totally ordered ring and ne + are e T,

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ne + aTe > 0 or ne + aTe < 0. Thus

,_Xwn f (n,aT + aT±. v 0) if «e + are > 0, l (0,ar± V0) if «e + ar^<0. We show the above results according to the following four cases. (a) a e S and avOeS. By Lemma 2(a) we have («, a V 0) > («To), 0. If b' = (m, b) e R' and />' > (TTTä),0, then me+be > 0 and (m-n)e+(b-a)e > 0. If b e S, then è > 0 by Lemma 2(b), and hence (m - n)e + be > 0 by Lemma 2(a). Thus (m - n)e + be > ae V 0 = (a V 0)e ; that is, b' > («, avO). If 6 € r © T-1, then ôj-i > 0 by Lemma 4. Since (6 - a) e S, b-a>0 by Lemma 2(b), and hence bT± > a , so bT± > a V 0. Since èj-j. > a V 0, èr± is in S, a contradiction. (b) a e S and aW 0 e T @T^ . Since a = (a V 0) - (-a V 0), we have (-a V 0) e ¿. It follows from (a) that (-«, -a) V 0 = (-«, -avO), so («To) V 0 = (0,flV0). _ (c) aeT @TL and «e + a^e > 0. It is evident that (n, ar + (aT± V 0)) > (rTTa), 0. Let V = (m, b) e R' and V > (WTa), 0. Then me + be > 0 and (m - n)e + (b - a)e > 0. If b e S, then b > 0, and hence 0 < b - aT e S. Since (m - n)e + (b - ar)e > 0 by Lemma 2, we have (m - n)e + (b - ar)e > aT±e V 0 = (aT± V 0)e; that is, V > (n,aT + (aT±. V0)). If b e T © T1-, then bT± > aT± V 0 and (m - n)e + (br - aj)e > 0 by Lemma 4. Thus (m - n)e + (b - ar - aT± V 0)e > 0. Again we have V > (n, ar + (aT± V 0)). (d) a e T© T1 and ne + aTe < 0. It is clear that (0, aT± V 0) > (TTTä), 0. If V = (m, b) e R' and b' > (n7~â), 0, then (me + be)>0 and (m - n)e + (b - a)e > 0. If b e S, then since b > 0 and b > a, 0 < (b - aT± v 0) e S, and hence me + (b- aT±.V 0)e > 0 by Lemma 2(a); that is, b' > (0, aT± V 0). If b e T © rx , then bT±. > aT± V 0 and me + bTe > 0 by Lemma 4. Hence me + (b- aT±. V 0)e > 0. Thus V > (0, aT± v 0). So far we have shown that R' is an /-ring. Finally, we show that R' is squares positive. _ Let a' = (WTa) e R', ae R. Then (a')2 = (n2, 2na + a2). If a e S, by Lemma 5 we have 2« 0, and hence (a')2 > 0. If a e T © TL and a = ÜT + aT± , then 2«a + a2 = 2naj + 2naT± + a\ + araT± + aT± ar + (aT± )2 , and hence (a')2 = («, ar)2 + (0, 2naT± + araT± + aTd.ar + (aT±)2). Since T is a totally ordered ring, (n, ar)2 > 0, and, from Lemma 5 and aT± e TL , 2naT± + aTaT± + aT±ar + (a^)2 > 0, so (a')2 > 0; that is, R' is squares positive. _ Let y be the map given by a -> (0, a) for every a e R. It is evident that \p is a ring monomorphism from R into R' and y/(a V 0) = (0, a V 0) = (0, a) V 0 = ip(a) V 0 for each a e R; that is, \p is an /-ring monomorphism. The proof of Theorem 1 is now complete.

Corollary. An l-semiprime l-ring with squares positive and an f-superunit can be embedded in a unital l-semiprime l-ring with squares positive. Proof. Since an /-semiprime /-ring with squares positive and an /-superunit is a subdirect product of its /-prime homomorphic images which are squares positive and contain /-superunits, the proof is completed by Theorem 1.

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Lemma 6. Let R be an l-prime l-ring with squares positive and an f-superunit, and let R' be as in Theorem 1. If e is an f-superunit of R, then (0, e) is an f-superunit of R'. Proof. Since e is an /-superunit of R, (1,0) < (0,e), and hence (0,e) is a superunit of R'. We show (0, e) is an /-element of R'. Suppose that (ñ~7~x)A (m, y) = 0. First we show that (ne + xe) A (me + ye) = 0 in R. Note that x and y are not both in ¿ ; otherwise, we have (1,0) < (W7x), (m, y) by Lemma 2. If x e S and y e T © T1-, then (m, yT) = 0 and (ñ~7~x)A (0, yT± ) = 0, so me + yje = 0 and x A yT± = 0 since 0 < (0, x) < 2(W7x). Thus (ne + xe) A (me + ye) = (ne + xe) A yT±e = 0. If x, ye T©rx , then («, xT) = 0 or (m,yT) = 0 and (0,xT±) A(0,yT±) = 0. Thus (ne+xe) A(me+ye) < [(ne+XTe)A(me+yre)] + (xT±eAyT±e) = 0, and we get (ne + xe) A (me + ye) = 0. Similarly, we also have (ne + ex) A (me + ey) = 0. These yield («7*)(0, e) A (m, y)(0, e) = (0, ne + xe) A (0, me + ye) = 0 and (0, e)(W7~x)A (0, e)(m, y) = (0, ne + ex) A(0, me + ey) = 0. So (0, e) is an /-element of R' since (T7Ö)< (0~7ë). Let R be an /-prime /-ring with squares positive and an /-superunit, and let R' be as in Theorem 1. It is easy to verify T(R') = {(WTa) : a e T} by Lemma 6. Thus R' = S(R') u [T(R7)_+T(R')X] by Lemma 1, where S(R') = {(WTa) :aeS} and T(R')1 = {(0, a) : a e Tx}. Theorem 2. Let R be an l-prime l-ring with squares positive and an f-superunit, and let R' be as in Theorem 1. If R can be embedded as an l-subring in a unital l-ring B and some f-superunit of R is an f-superunit of B, then there are an l-subring Rx of B containing R andan ¡-isomorphism ip from l-ring R' to l-ring Rx, which satisfies y/(a) = a for each a e R. Proof. Let 1R be the identity element of B and Rx thesubringof B generated by Iß and R. Let ip be the map from R' into Rx given by («7~ß)-* nlß+a, and let e be an /-superunit of R which is also an /-superunit of B. If (WTa) = (m, b), then (n-m)e+(a-b)e = 0, and hence [(n-m)lß+(a-b)]e = 0 (in B). Since e is an /-superunit of B, we have (n-m)ÍB+(a-b) = 0, and hence y/((W7~ä))= y ((m, b)); that is, y/ is well defined. It is evident that ip is an isomorphism of rings and y/((0, a)) = a . If a e R and (nlR + a) VO = b (in B), then be = (nlR + a)e V 0 = (ne + ae) V 0 since e is an /-element of B. Let (WTa) V0= («i, ax) (in R'). Since (0, e) is an /-element of R' by Lemma 6, we get (tT7~ä)(0,e) V 0 = («i, ax)(0, e) and (0, ne + ae) V 0 = (0, nxe + axe), and hence (ne + ae) V 0 = nxe + axe . Since R is an /-subring of B, be = nxe + axe = («ils + ax)e, and hence b = nxlR + ax e Rx since e is an /-superunit of B ; that is, Rx is an /-subring of B . Finally, it is easy to verify that y/ is also an /-isomorphism from R' to Rx . Let R be an /-prime /-ring with squares positive and an /-superunit. The- orem 2 shows that the R' in Theorem 1 is the unique smallest unital /-ring which contains R and has an /-superunit that is an /-superunit of R. In general, no such uniqueness holds [1, p. 331]. Now we determine the conditions which ensure that an /-prime /-ring with squares positive which has nonzero /-elements can be embedded as a convex /-subring in a unital /-prime /-ring with squares positive. If R is an /-prime

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/-ring with squares positive which has nonzero /-elements, then R is a domain [5, Theorem 1], and hence T is a totally ordered ring. Thus R = ¿U (T© rx) by Lemma 1. In [6, Remark, p. 367] Steinberg defined an /-ring R to be supertessimal if n\x\ < \x2\ for each n e Z+ and x e R. Theorem 3. Let R be an l-prime l-ring with squares positive which has nonzero f-elements, and suppose R has no identity element. R can be embedded as a convex l-subring in a unital l-prime l-ring with squares positive if and only if R is supertessimal; or S = 0, T is infinitesimal, and n\a\ < a2 for each n e Z+ and aeTL. Proof. Let B be a unital /-prime /-ring with squares positive and R a convex /-subring of B. Since R is a convex /-subring of B, rx ç T(B)1-, and, also, since R does not have an identity element, R ç T(B) + T(B)-L by Lemma 1. If ¿ ¿ 0 , then, for any a e S, we have \a\ e T(B)L . This yields S ç T(B)X , and hence T ç T(B)± , so R ç T(B)X by Lemma 1. From Lemma 5 we get R to be supertessimal. If ¿ = 0, but T is not infinitesimal, then there is d e T which satisfies 0 < d < d2 . Let d = dx+d2, where 0 < dx e T(B) and 0 < d2 e T(B)1-. Since R is a convex /-subring of B, dx, d2e R, and hence dx, d2 e T. But T is a totally ordered ring and dx A d2 = 0, so dx — 0 or d2 = 0. If d2 = 0, then d e T(B), and hence d > 1B (identity element of B) since d is comparable with 1R and d < d2. This contradicts the fact that R has no identity element, so dx = 0 and d e T(B)L . This yields T ç T(B)1-, and hence R ç T(B)1. Thus, again, we get R to be supertessimal by Lemma 5. Conversely, first suppose R is supertessimal. Let Z © R be as an /- group with multiplication defined by (n, a)(m, b) = (nm, nb + ma + ab). Then Z ©R is a unital /-ring with squares positive [6, p. 367]. If («, a)(m, b) = 0, then nm = 0 and nb + ma + ab = 0. We may assume « = 0. If a ^ 0, then, since R is domain and ma + ab = 0, we have mb + b2 = 0, and hence (|w| + l)\b\ < b2 = -mb = \mb\ = \m\\b\ since R is supertessimal. This yields b = m = 0, and hence (m, b) = 0 ; that is, Z @ R is a domain. Let y/ be the map given by a —>(0, a). It is evident that ip is an /-monomorphism and xp(R) is a convex /-subring of Z ® R. Finally, let R satisfy the following conditions: ¿ = 0, T is infinitesimal, and n\a\ < a2 for each a e TL and n e Z+. From Lemma 1 we have R = T®T±. Let T' be the ring obtained by freely adjoining the Z to T with partial order defined by («, a) > 0 if and only if « > 0 or « = 0 and a > 0. Then V is a totally ordered domain ring with identity (1,0) [7, 1.4]. Let R" = V© Tx as an /-group. If a" = ((n,ax), a2) and b" = ((m,bx), ¿2) £ R" , where ax, bx e T and 02, b2eTL , we define the multiplication in R" as a"Z>"= ((«, ax)(m, bx) + (0, (a2b2)T), nb2 + axb2 + ma2 + a2bx + (a2b2)T±). It is easy to verify R" to be a ring with identity ((1, 0), 0). Since T has mode 0 and R = T © rx, R has mode 0, so R" is a domain. Also, the product of positive elements of R" is clearly positive by the definition of the multiplication in R" ; that is, R" is an /-ring. Let a" = ((«, ax), a2) e R", where ax e T and a2 e Tx . Then (a")2 = ((«, ûi)2 + (0, (a\)T), 2na2 + axa2 + a2ax + (oI)tl) . From the hypothesis and

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Lemma 5, we get 2«a2 + #1^2 + ti2ax + a\ > 0, and hence 2«a2 + axa2 + a2ax + (a2)T± > 0, (a¡)T > 0. Thus (a")2 > 0. So far we have shown that R" is a unital /-prime /-ring with squares positive. Let y/ be the map from R into R" given by y/(a) = ((0,ar),ar±) for each a e R. It is clear that y/ is an /-monomorphism. If ((«, ax), a2) and ((0, bx),b2) e R" satisfy 0 < ((n,ax),a2) < ((0, bx), b2), then « = 0 by the definition of the partial order of R" , and hence y/(R) is a convex /-subring of R". The proof of Theorem 3 is now completed.

ACKNOWLEDGMENT The author thanks the referee for simplifying the proofs of Theorems 1 and 3.

References

1. S. A. Steinberg, On the unitability of a class of partially ordered rings that have squares positive,J. Algebra100 (1986), 325-343. 2. _, Unital l-prime lattice-ordered rings with polynomial constraints are domains, Trans. Amer. Math. Soc. 276 (1983), 145-164. 3. D. G. Johnson, A structure theory for a class of lattice-ordered rings, Acta. Math. 104 ( 1960), 163-215. 4. P. Conrad, Some structure theorems for lattice-ordered groups, Trans. Amer. Math. Soc. 99 (1961), 212-240. 5. Ma Jingjing, On lattice-ordered rings with polynomial constraints, J. Math. Res. Exposition 11 (1991),325-330. (Chinese) 6. S. A. Steinberg, On lattice-ordered rings in which the square of every element is positive, J. Austral. Math. Soc. 22 (1976), 362-370. 7. M. Henriksen and J. Isbell, Lattice-ordered rings and function rings, Pacific J. Math. 12 (1962), 533-565.

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