【原稿可复制编辑】段和有负平方的定向域

【原稿可复制编辑】段和有负平方的定向域 免费论文查阅: 段和有负平方的定向域 12 W. Rump，杨义川 1 Universität Stuttgart, Fachbereich Mathematik, Institut für Algebra und Zahlentheorie, Pfa,enwaldring 57 D-70550 Stuttgart, Germany 2 北京航空航天大学数学与系统科学学院数学系，数学、信息与行为教育部重点实验室 北京 100191 摘要:本文中，我们研究了段和有负平 2 方的定向域间的关系. 对于线性序域 K 和代数元 i 满足 i= ?1, 我们在用 K 的有理段参数化了 K (i) 上的一类格序群结构的理论基础上，建立了域 K (i) 上的定向序与和 K 的乘性段 V 间的对应关系. 关键词:定向域，负平方，格序，段 中图分类号: O152.6 Segments and directed partially ordered ,elds with negative squares 12RUMP Wolfgang, YANG Yichuan 1 Universität Stuttgart, Fachbereich Mathematik, Institut für Algebra und Zahlentheorie, Pfa,enwaldring 57 D-70550 Stuttgart, Germany 2 School of Mathematics and System Sciences, LMIB of the Ministry of Education, Beijing University of Aeronautics and Astronautics, 100191 Beijing, China Abstract: In this paper, we study the relationship between segments and directed orders on the ,eld with negative squares. For a linearly ordered ,eld K , we parameterize a class of 2 lattice-ordered group structures on K (i) with i= ?1 by rational segments of K . Then, we establish a correspondence between directed orderings on the ,eld K (i) and multiplicative segments V of K . Key words: directed ,eld, negative square, lattice order, segment 0 Introduction Artin and Schreier [1], Johnson [2], Fuchs [3], and many others (for instance, Szele [4], Szigeti [5], and Serre [6] etc. ) observed the non-existence of a compatible total order on rings with negative squares. On the other hand, Bourbaki [7], Birkho, and Pierce [8] etc. proved that 基金项目: Supported by the Research Fund for the Doctoral Program of Higher Education (Gran 20091102120045), t Beijing Municipal Natural Science Foundation (Gran 1102027) and NSFC (Gran 11271040) tt 作 者 简 介: RUMP Wolfgnag，male，Professor, major research direction:Algebra. Correspondence author:Yang Yichuan，male，Professor，major research direction:algebra. - 1 - 免费论文查阅: a lattice-ordered ,eld in which any square is positive must be totally ordered, and this result has been extended to skew ,elds by Yang in [9]. Schwartz [10] showed that the ,eld of algebraic numbers admits no partial order with respect to which it is a lattice-ordered ,eld. Yang [11] guranteed the existence of directed partially ordered ,elds with negative squares. Continuing in this direction, Schwartz and Yang [12] showed that almost all ,elds of characteristic 0, including the ,eld of complex numbers, can be made into a directed partially ordered ,eld. In this paper, 2 we characterize the directed orders on the ,eld K (i) with i= ?1 for a linearly ordered ,eld K by segments. We ,rst exhibit a particular class of l-group structures on K (i) and show that they can be parameterized by positive elements of K (Theorem 2). As a simple consequence, we prove that the directed ordered ,elds with negative squares constructed in [11] need not be lattice-ordered (see Remark 1). Then, we establish a correspondence between directed orderings 1 ?1 of the ,eld K (i) and segments V of K which satisfy the implication 0 < a ?/ V ? (a?a) ?/ V 2 (Theorem 3). Finally, we characterize these V as convex additive subgroups with the property ×?1 that for a ? K , either a or abelongs to V (Theorem 4). Note that the latter property holds for any valuation domain with quotient ,eld K . 1 Preliminaries A partial ly ordered group (po-group for short) G = (G, +, ?) is a (not necessarily abelian) group (G, +) (with binary operation + and identity element 0, where the inverse of a member a of G is denoted by ?a) and a partially ordered set (G, ?) in which a ? b implies that a+c ? b+c and c + a ? c b for all a, b, and c in G. An element a of G is positive if a ? 0. The positive + ?0 >0 cone and strictly positive cone of G, denoted by Gand G, is the set of all positive elements and all strictly positive elements of G respectively. For two elements a, b in a po-group, we use a ? b to denote that a and b are incomparable. A po-group is Archimedean if na ? b for all integers n implies a 0. A po-group (G, ?) is called a directed group if the partial order = +, ? is a directed order, i. e., for any a, b ? G, there exists c ? G such that c ? a and c ? b. Note that Cli,ord proved that a po-group (G, +, ?) is a directed group if and only if G can be ?0 generated by G. A directed group (G, +, ?) is called a lattice-ordered group (l-group for short) if the partial order ? is a lattice order, that is, if each pair of elements x and y in G have a unique least upper bound x ? y and a unique greatest lower bound x ? y. An l-group G is a linearly or totally ordered group if the order is a simple order: ? y or y ? in G for all x, y ? G. If (G, ?) is x x +, + an l-group and a belongs to G, then the positive part of a is a= a ? 0, the negative part of a ? ? +is a= (?a) ? 0, and the absolute value of a is |a| = a ? (?a). It is easily seen that a = a? a + ?and that |a| aa. = + A directed ring R = (R, +, ?, ?) is a ring that is partially ordered and has the following - 2 - 免费论文查阅: properties: (i) (R, +, ?) is a directed abelian group, and (ii) a ? 0 and b ? 0 imply that ab ? 0. A directed ring R is Archimedean if the directed additive group (R, ?) is Archimedean. +, Lattice-ordered rings and totally ordered rings can be de,ned similarly. A directed ,eld is a directed ring whose underlying ring is a ,eld. A partially ordered ring (R, +, ?, ?) has a negative 2 square if there is an element r ? R with r? 0. For basic theory of partially ordered algebraic systems, we refer the reader to [13]. 2 Segments and directed orders on the additive group of K (i) 2 Let (K, ?) be a linearly ordered ,eld and i= ?1. A proper subset V of K will be called a segment of K , if V satis,es (1) 0 ? V , and (2) |a| ? |b|, b ? V implies a ? V . For any segment V of K , de,ne ?1 P:= {a + bi ? K (i) | a, b ? 0; b > 0 ? (a > 0, ab?/ V )}. V Let A, B be two subsets of a additive group, we use ?A, A + B and A ? B to denote the set {?x | x ? A}, {a + b | a ? A, b ? B}, and A + (?B), respectively. Lemma 1. The Pde,ned above satis,es: V (a) P? (?P) = {0}; V V (b) P+ P? P;V V V (c) P? P= K (i). V V Proof. (a) Trivial. (b) Assume that a + bi, c + di ? P. Then a, b, c, d ? 0, without loss of generality, supposeV a+ca that b > 0, which implies a > 0 and ab ? V , so a c > 0. If d 0, then ? ? V , //+ = /b+d b a+c a c which implies ? V . There is no loss of generality in assuming that ? . It follows that/+d b d b ca a+ca+c? ? , and thus ?/ V . b b+d d b+d (c) For all a + bi ? K (i). Choose y > 0 and y > b. There exists z > 0 with z ? K \ V since V K . Choose > 0, > a, ? yz and ? a (y ? b)z since K is totally ordered. = x x x x + x x?a Then x ? a, y ? b > 0; ? z and ? z (note that a ,eld is totally ordered if and only if any y y?b x x?a ?1 square is positive, it follows that y ? 0 implies y? 0), hence , ?/ V . Finally, it is cleary y?b ()that a + bi = (x + yi) ? (x ? a) + (y ? b)i. ?0 For a partially ordered algebraic system (S, ?), we use Sto denote the positive cone. By the theorem of Cli,ord and Lemma 1 it follows that any segment determines a compatible - 3 - 免费论文查阅: partial order for which the additive group K (i) is a directed additive group. Furthermore, we have Theorem 1. V 7? Pde,nes a bijection between segments V of K and partial orders ? V on ?0 ?0 K (i) such that K ? K (i)and K (i) is a directed additive group with (1) 0 ,? bi ,? a for al l a ? K and 0 < b ? K ; (2) there exists a > 0 in K with ?a ? i. Proof. By Lemma 1 the positive cone Pde,nes a directed order on K (i) via α ? β ? V β ? α ? Pfor α, β ? K (i). Suppose that bi ? a ? K for some 0 < b ? K and a, then V a ? bi ? P, a contradiction, which implies (1) since 0 ,? bi is obvious. V = K implies that V ?0 there is an a ? K \ V , so a + i ? Pwhich implies (2). V Conversely, if (K (i), +, ?) is a directed group with the properties above. De,ne ?1 V := {ab? K | 0 ,? |a| + |b|i}. By (2) there exists 0 < a ? K with 0 ? a + i, which implies V = K . The condition 0 ,? i implies 0 ? V . Furthermore, |a| ? |b| and b ? V implies 0 ,? |b| i and a ? V since 0 ? |a| i + + is impossible by |a| + i ? |b| + i. So V is a segment. We next to show that the positive cone of K (i) is P, that is, V )(?1 0 ? a + bi ? b ? 0; b > 0 ? (a > 0, ab?/ V ).a, “?”: Assume b < 0. Then ?bi ? a, which is a contradiction and thus b ? 0. Suppose a < 0. Then b > 0, so 0 ? ?a ? bi, which contradicts (1), whence a ? 0. If b > 0, then a > 0, ?1 and ab?/ V . ?0 ?0 “?”: For b 0 and a ? 0, we get 0 ? a a bi since K ? K (i). For b > 0, we have= = + ?1 a > 0 and ab? V which implies 0 ? |a| |b|i a bi. /+ = + ?1 Finally, it is trivial that V = {ab? K | |a| + |b|i ?/ P}. V >0 We call a segment V rational if it has the form V:= {a ? K | |a| < for some α ? K . α} α Theorem 2. The order in Theorem 1 makes K (i) into an l-group if and only if V is rational. Proof. Assume that V = Vwith 0 < α ? K . To prove 0 ? (a + bi) ? K (i) for all α a + bi ? K (i). For b = 0 it is trivial. Case I. b > 0: Then 0 ? (a + bi) = c + bi with c = max{a, bα}. Case II. b < 0: Then 0 ? (a + bi) = max{0, a ? bα}. Conversely, suppose that K (i) is an l-group, but V is not rational. Let 0 ? i = a + bi. Then ?1 ?1 b ? 1, a > 0 and ab?/ V . Thus there exists α ?/ V with 0 < α < ab. However, bα + bi ? 0 bα and bα bi i ? are plain for b 1. Furthermore, if b > 1, then α ?+ = ? V which implies / b?1 bα + bi ? i. Therefore, a + bi ? (bα + bi) = a ? bα > 0, a contradiction. - 4 - 免费论文查阅: Remark 1. By Theorem 2 it follows that the directed ,elds with negative squares constructed in [11] need not be lattice-ordered: Let F be a totally ordered ,eld, K be the quotient ,eld of the polynomial ring F [], and x 2 let i be a solution of x+ 1 = 0. It is well known that F [x] is a totally ordered ring with n respect to the order: a+ ax + ? ? ? + ax> 0 if and only if a> 0 in F , and K is a 0 1 n n f (x) totally ordered ,eld with respect to the order > 0 if and only if f (x)g(x) > 0 in F [x]. g (x) Let v be the negative valuation on K de,ned by the di,erence of the degree function on K . Then the set ?0 P , = {a + bi ? K (i) | a, b ? K and if b = 0, then v(a) > v(b)} will be the positive cone of a partial order ? on K (i) for which K (i) is a directed ,eld (see Corollary 2.3 in [11]). Furthermore, it is straightforward to verify that the directed order de,ned by P satis,es the conditions of above Theorem 1. Now we show that the corresponding segment ?1 V = {ab| 0 ,? |a| + |b|i} = V= {a ? K | |a| < α} α >0 >0 for any element α ? K . Assume that there exists α ? K with V V. It is clear that = α ?0 F ? V , so v(α) > 0 and V? α ? 1 ? α. However, v(α ? 1) = v(α) and 0 ? (α ? 1) + i, a α contradiction. We also note that Remark 2.5 in [12] proved that the directed ordered in [11] is not a lattice by viewing the polynomial rings as a vector space. 3 Segments and directed orders on the ,eld K (i) ?0 ?0 Lemma 2. A segment V of K de,nes a partial order such that K ? K (i)and K (i) is a directed ,eld if and only if (i) 1 ? V , and ab?1 (ii) a, b ? V, a > 0, b > 0 implies ? V . //a+b Proof. If K (i) is a directed ,eld, that is, PP? P. Assume 1 ? V , then 1 + i ? P/V V V V 2 implies (1 + i)= 2i ? P, in contradiction with Theorem 1 (1) and thus (i) holds. Let V >0 a, b ? K \ V , then a + i, b + i ? Pand hence (a + i)(b + i) = (ab ? 1) + (a + b)i ? P, which V V ab?1 implies ? V . /ab +?0 Conversely, suppose that (i) and (ii) hold. Let a bi, c di ? P. Then a, b, c, d ? K .+ + V Case I: bd 0. Then ac ? bd ? 0. Without loss of generality, assume that d 0. If = = ?1 ad + bc > 0, then b, c > 0 which implies a > 0 and ab?/ V . So ac ? bd = ac > 0 and ac?bd ac a= = ? V ./ adbc bc b +?1 ?1 ?1 ?1 Case II: bd > 0 implies that a, c > 0 and ab, cd?/ V . Thus ab, cd> 1 since 1 ? V , a c? 1 ac?bd b d and hence ac ? bd > 0. (ii) implies that ?/ V , which implies ?/ V . a c+ adbc + b d - 5 - 免费论文查阅: Example 1. The order constructed in the proof of the main theorem in [11] can be de,ned by a segment. By the condition (i) of Lemma 2 and the de,nition of the segment we have following interesting corollary. Corollary 1. If 1 is a strong unit in K , then K (i) cannot be partial ly ordered such that K (i) is a directed ,eld for any segment V of K . Especially, if K is an l-sub,eld of the Archimedean totally ordered ,eld R, then the corol- lary above is applicable. Equivalently, the ,eld K in Theorem 3 can be considered as a non- archimedean linearly ordered one. Let us call a segment V ? K multiplicative if it satis,es 1 ?1 (iii) 0 < a ?/ V implies (a ? a) ?/ V . 2 Theorem 3. V 7? Pde,nes a bijection between multiplicative segments V and partial ly orders V >0 ? on K (i) such that K (i) is a directed ,eld with bi ,? a for al l a ? K , b ? K , and k < i for k ? K . some 2 a ?1 1 ?1 Proof. (ii) ? (iii): Set a = b, then = (a ? a) ? V ./2a 2 (iii) ? (ii): Without loss of generality, suppose that 0 < a < b. We want to show that ? ab?11 1 2 2(a ? a) ? , which is equivalent to show b(a + 1). + 1) ? a(a 2 a+b 1 ?1 (iii) ? (i): Suppose that 1 ?/ V , so (1 ? 1) ?/ V , a contradiction. 2 2 The conditions (1) and (2) in Theorem 1 are simpli,ed: If i ? 0, then i= ?1 ? 0, which is impossible. ′ 1 ?1 Remark 2. Let a= (a ? a) for a > 0. Then (iii) is equivalent to 2 ′ 0 < a ? V ? 0 < a? V.// ?1 ′ In fact, 0 < a ? V implies that a > 1 and thus a > a, which implies a> 0. Furthermore, / ′ 2 for a > 0 we have i > ?a ? i > ?asince i + a > 0 gives that (i + a)> 0 which gives 2 1 ?1 that (2a)i > 1 ? a, and thus i > ? (a ? a). 2 Corollary 2. If K (i) is an l-,eld with K as a linearly ordered sub,eld. Then i ? a for al l a ? K , or there exist a, b ? K such that a < i < b, or for al l a ? K there exists b ? K such that a < i ? b. Proof. It su,ces to prove that every directed order in Theorem 3 cannot make K (i) into ′ ′ an l-,eld. Assume that V = V. Then α ?/ V and α > 1, and thus 0 < α< α, so α? V . In α contradiction with the de,nition of a multiplicative segment. - 6 - 免费论文查阅: Remark 3. If on the ,eld C there exists a partial order such that C is an l-,eld with respect 1 a ?1 to Theorem 3, then there exists 0 > a ? R with i > a. Thus i > (a ? a) > , which2 2 aimplies i > for all n ? N by induction. However, we cannot get i ? 0 in general, which n2 implies that the order cannot be Archimedean by a simple argument. Note that all lattice orders constructed in Wilson [14] are Archimedean. Theorem 4. A segment V of K is multiplicative if and only if it satis,es the fol lowing conditions: (α) V is an additive subgroup; (β) V is convex; ?1 (γ) ?a ? K \ 0: a ? V or a? V .{} a+bProof. If V is a multiplicative segment and 0 < a < b in V , then a < < b implies 2 ?a+b a+b1 1 ? V . Assume that a + b ?/ V . Then ?((a b) ? (a b + + )) ?/ V , a contradiction. 2 2 2 Thus a + b ? V . Now let a, b ? V be arbitrary. Then |a| + |b| ?V and |a + b| ? |a| + |b| implies a + b ? V . Hence (α) is proved. (β) is clear. ?1 |a||a|?1 ?1 ?1 (γ): If = 0 and a, a?/ V . Then |a|, |a|?/ V implies a contradiction: 0 = ?/ V .?1|a|+|a| Conversely, suppose that V satis,es (α), (β) and (γ). Let 0 < a ? V . By (γ) it follows/ ?1 1 ?1 ?1 ?1 that a? V . Assume that (a ? a?1) ? V . Then a ? a? V and a (a ? a) +a? V , = +2 a contradiction. Corollary 3. Every convex valuation subdomain V = K in its quotient ,eld K (non-archimedean linearly ordered) is a multiplicative segment. Proof. The conditions (α), (β) and (γ) are clearly satis,ed. By Corollary 1 1 is not a strong unit, which implies that K is non-archimedean. Note that a multiplicative segment need not come from a real valuation ring, since there is a sea of non-archimedean linearly ordered ,elds. Example 2. Let K be the surreal number ,eld. Then K is a nonarchimedean linearly or- dered ,eld [15]. Then every convex valuation subdomain V = K in its quotient ,eld K is a multiplicative segment. Remark 4. The condition (γ) in Theorem 4 can be substituted by 1 ? V. ?1 Actually, (γ) implies 1 1? V . Conversely, assume that a ? V , then |a| ? V implies |a| > 1, = // ?1 1 1 and thus 1 > |a|= | | > 0, which implies ? V. a a - 7 - 免费论文查阅: 参考文献(References) [1] E. Artin and O. Schreier, Algebraische Konsstruction reeler Körper, Abh. Math. Sem. Hamb. Univ. 5, 1926. [2] R. E. Johnson, On ordered domains of integrity, Proc. AMS. 3 (1952), 414-416. [3] L. Fuchs, Note on ordered groups and rings, Fund. Math. 46 (1958), 167-174. [4] T. Szele, On ordered skew ,elds. Proc. AMS. 3, (1952), 410–413. [5] J. Szigeti, Linear orders on rings. Comm. Alg. 33 (2005), no. 8, 2683–2695. [6] J. P. Serre, Extensions de corps ordonnés. C. R. Acad. Sci. Paris 229 (1949), 576-577. [7] N. Bourbaki, Éléments de mathématique. XIV. 1952. [8] G. Birkho, and R. S. Pierce, Lattice-ordered rings, An. Acad. Brasil. Ci. 28 (1956), 41–69. [9] Y. C. Yang, A lattice-ordered skew-,eld is totally ordered if squares are positive, Americal Math. Monthly, vol. 113 (2006) 3, 266-267. [10] N. Schwartz, Lattice-ordered ,elds, Order, 3(1986), 179-194. [11] Y. C. Yang, On the existence of directed rings and algebras with negative squares, J. Alg., vol. 295 (2006) 2, 452-457. [12] N. Schwartz, Y. Yang, Fields with directed partial orders. J. Alg. 336, 342 –348 (2011) [13] L. Fuchs, Partially ordered algebraic systems. (1963), Pergamon Press, Oxford. [14] R. R. Wilson, Lattice orders on the real ,eld, Paci,c J. Math. 63 (1976), no. 2, 571–577. [15] J. H. Conway, On numbers and games. New York: Academic Press, 1976. - 8 -