Ordered Fields and the Ultrafilter Theorem

FUNDAMENTA MATHEMATICAE 159 (1999)

Ordered ﬁelds and the ultraﬁlter theorem

R. B e r r (Dortmund), F. D e l o n (Paris) and J. S c h m i d (Dortmund)

Abstract. We prove that on the basis of ZF the ultraﬁlter theorem and the theorem of Artin–Schreier are equivalent. The latter says that every formally real ﬁeld admits a total order.

Introduction. In 1900 at the international congress of mathematicians in Paris Hilbert raised his famous 23 problems. Among these was the following (the 17th): Let f ∈ R[X1,...,Xn] be a polynomial which is positive semidefinite, i.e., f(x) ≥ 0 for all x ∈ Rn. Is f a sum of squares of rational functions? Hilbert already knew that f is not necessarily a sum of squares of polynomials unless n = 1. In 1927 Artin solved Hilbert’s 17th problem affirmatively (cf. [1]). The methods he used were developed in the joint paper [2] with Schreier and are nowadays known as Artin–Schreier theory. The crucial no- tion of Artin–Schreier theory is that of a formally real field. A field K is called formally real if −1 is not a sum of squares in K. The first basic theorem proved by Artin and Schreier stated that each formally real field K admits an ordering, i.e. a subset P ⊂ K such that P + P ⊂ P , P · P ⊂ P , P ∩ −P = {0} and P ∪ −P = K. P The proof runs as follows. Let T = K2 be the set of all sums of squares of K. Then T + T ⊂ T , T · T ⊂ T , K2 ⊂ T and by the assumption −1 6∈ T , i.e. T is a preordering. An application of Zorn’s Lemma yields a preordering maximal for set-theoretic inclusion. Now an elementary computation shows that this maximal preordering is an ordering. In 1954 Tarski [6] mentioned that the axiom of choice (in the form of Zorn’s Lemma) is not necessary. The ultrafilter theorem suffices to prove the theorem of Artin–Schreier. A proof can be found in [3] (Theorem 2.2 and Example 2.3.3). As mentioned by Tarski the problem whether the

1991 Mathematics Subject Classiﬁcation: Primary 03E24; Secondary 12J15.

[231] 232 R. Berr et al. theorem of Artin–Schreier is equivalent to the ultrafilter theorem remains open. Recently, questions of similar type appeared in computer algebra. Lom- bardi–Roy [4] and Sander [5] proved independently that the existence and uniqueness of the real closure of an ordered field can be proven within ZF. In this paper we close the gap mentioned above and show that the theorem of Artin–Schreier, the ultrafilter theorem and certain other theorems from real algebra are mutually equivalent on the basis of ZF.

1. Preliminaries. It is the aim of this section to prove a couple of preliminary results which will be needed for the proof that every filter is contained in an ultrafilter provided every formally real field admits an ordering. Since these results are of quite technical nature we briefly sketch their meaning for the proof. Let S be a non-empty set and let F be a filter on S. It is our goal to show that F is contained in an ultrafilter. To this end we consider A = Q[XI | I ⊂ S] and construct in a first step a ring homomorphism ϕ : A → (K,T ) into a preordered field (K,T ) such that

F = {I ⊂ S | ϕ(XI ) ∈ T }. P In the next step we embed K into a formally real ﬁeld L with T = K ∩ L2 and show that for any ordering P ⊂ L,

U = {I ⊂ S | ϕ(XI ) ∈ P } is an ultrafilter containing F. For the construction of K one has to consider ideals in certain polynomial rings over A which are generated by elements of the form aX2 + bY 2 + c with a, b, c ∈ A. In Lemmas 1.1–1.4 we will be concerned with these ideals. The remaining statements are related to the construction of the preordering T ⊂ K and the extension field L ⊃ K. Throughout this section we are working inside ZF and all fields are as- sumed to have characteristic zero. Given a ring R we let R× denote the group of units of R. Lemma 1.1. Let R be an integral domain with K as its field of quotients. Let a, b, c ∈ R\{0} and assume that a ∈ R×. Let p, q be the ideals generated by aX2 + bY 2 + c in R[X,Y ] and K[X,Y ] respectively. Then (1) p, q are prime ideals, (2) q ∩ R[X,Y ] = p. P r o o f. We first prove (2). Let f ∈ q∩R[X,Y ]. Then f(X,Y ) = g(X,Y )· (aX2 + bY 2 + c) for some g ∈ K[X,Y ]. Let nX+2 Xn i i f(X,Y ) = fi(Y )X and g(X,Y ) = gi(Y )X i=0 i=0 Ordered fields and the ultrafilter theorem 233 for certain fi ∈ R[Y ] and gi ∈ K[Y ]. Comparing the coefficients we get agn = fn+2 ∈ R[Y ] and agn−1 = fn+1 ∈ R[Y ]. Hence gn, gn−1 ∈ R[Y ] because a is a unit in R. For 1 ≤ i ≤ n − 1 we obtain again by comparing the coefficients

2 agi−1(Y ) + (bY + c) · gi+1(Y ) = fi+1(Y ) ∈ R[Y ].

Now an easy induction yields gn, gn−1, . . . , g1, g0 ∈ R[Y ] and hence g ∈ R[X,Y ]. Therefore f ∈ p. The other inclusion is trivial. An easy computation shows that aX2 +bY 2 +c is irreducible in K[X,Y ]. Thus q is a prime ideal of K[X,Y ] and from (2) we conclude that p is a prime ideal of R[X,Y ]. This proves (1).

× Lemma 1.2. Let R be an integral domain. Let I 6= ∅ and ai, bi, ci ∈ R (i ∈ I). Let p be the ideal of R[{Xi,Yi | i ∈ I}] which is generated by the 2 2 polynomials aiXi + biYi + ci (i ∈ I). Then p is a prime ideal. P r o o f. In a ﬁrst step we assume that I is ﬁnite. We prove the assertion by induction on the number n of elements of I. We may assume that I = {1, . . . , n}. The case n = 1 follows from Lemma 1.1(1). Let

2 2 2 2 q = (a1X1 + b1Y1 + c1, . . . , an−1Xn−1 + bn−1Yn−1 + cn−1)

⊂ R[X1,Y1,...,Xn−1,Yn−1]. By the induction hypothesis q is a prime ideal. Thus the residue class ring

A = R[X1,Y1,...,Xn−1,Yn−1]/q is a domain. An easy argument yields ∼ 2 2 R[X1,Y1,...,Xn,Yn]/p = A[Xn,Yn]/(anXn + bnYn + cn). Now Lemma 1.1(1) shows that the right hand side is a domain. Thus p is a prime ideal.

In the second step we assume that I is arbitrary. Let f, g ∈ R[{Xi,Yi | i ∈ I}] and assume that fg ∈ p. Thus there are a ﬁnite set J ⊂ I and polynomials hi ∈ R[{Xi,Yi | i ∈ I}] such that X 2 2 fg = hi · (aiXi + biYi + ci). i∈J

In the polynomials f, g, hi (i ∈ J) only ﬁnitely many indeterminates occur. Enlarging J if necessary we may assume that f, g, hi ∈ R[{Xi,Yi | i ∈ J}]. Then fg ∈ q where q is the ideal of R[{Xi,Yi | i ∈ J}] generated by the 2 2 polynomials aiXi + biYi + ci (i ∈ J). By the ﬁrst step q is a prime ideal. Thus f ∈ q ⊂ p or g ∈ q ⊂ p. 234 R. Berr et al.

Lemma 1.3. Let K be a field and let T be a preordering of K. Let I 6= ∅ × −1 −1 and ai, bi, ci ∈ K (i ∈ I). Assume that for each i ∈ I either ciai or cibi is in T . Let 2 2 p = (aiXi + biYi − ci | i ∈ I) ⊂ K[{Xi,Yi | i ∈ I}] and let L be the field of quotients of the ring K[{Xi,Yi | i ∈ I}]/p. Then there is a preordering T 0 of L which extends T , i.e., T 0 ∩ K = T . P r o o f. We prove the lemma in three steps. In the first step we assume that I contains only one element. We omit all indices. Since the situation is symmetric in X and Y we may assume ca−1 ∈ T . First note that L =∼ K(Y )[X]/(X2 + ba−1Y 2 − ca−1). We denote by x ∈ L the image of X under the canonical projection K(Y )[X] → L. Now let T 0 ⊂ L be the semiring generated by T and the squares of L. We claim T = T 0 ∩ K. Obviously, T ⊂ T 0 ∩ K. So let t ∈ T 0 ∩ K. Then there are n ∈ N, ti ∈ T and g, gi, hi ∈ K[Y ], i = 1, . . . , n, with g 6= 0 and Xn Xn Xn 2 2 2 2 2 tg = (hi + xgi) · ti = (hi + x gi ) · ti + 2x · gihiti. i=1 i=1 i=1 2 −1 −1 2 P But x = ca − ba Y ∈ K(Y ) implies gihiti = 0. Therefore Xn Xn 2 2 −1 −1 2 2 tg = hi ti + (ca − ba Y ) · gi ti. i=1 i=1 −1 Now assume g(0) = 0. Since ca ∈ T we see hi(0) = 0 = gi(0) for all i. Hence we may assume that g(0) 6= 0. But then ca−1 ∈ T implies tg(0)2 ∈ T , hence t ∈ T . Therefore T 0 extends T . In the second step we assume that I is finite and proceed by induction on the number n of its elements. We may assume that I = {1, . . . , n}. The case n = 1 is exactly the first step. Let 2 2 2 2 q = (a1X1 + b1Y1 + c1, . . . , an−1Xn−1 + bn−1Yn−1 + cn−1)

⊂ K[X1,Y1,...,Xn−1,Yn−1].

By Lemma 1.2, q is a prime ideal. So A = K[X1,Y1,...,Xn−1,Yn−1]/q is a domain. Let F be its field of quotients. By the induction hypothesis T extends to a preordering T 0 of F . From the first step we see that T 0 extends 00 2 2 to a preordering T of E = Quot(F [Xn,Yn]/(anXn + bnYn + cn)). An easy argument yields ∼ 2 2 K[X1,Y1,...,Xn,Yn]/p = A[Xn,Yn]/(anXn + bnYn + cn). By Lemma 1.1(2) the right hand side embeds into E. Thus there is also an embedding L → E. Now one can easily see that T 00 induces the desired preordering on L. Ordered fields and the ultrafilter theorem 235

In the ﬁnal step we assume that I is arbitrary. Let T 0 be the semiring generated by T and the squares of L. It suﬃces to show that T 0 ∩ K = T . 0 0 The inclusion T ⊂ T ∩ K is clear. So let a ∈ T ∩ K. Then there are f0 ∈ P 2 K[{Xi,Yi | i ∈ I}] \ p, f1, . . . , fn ∈ K[{Xi,Yi | i ∈ I}] and t1, . . . , tn ∈ T such that Xn 2 af0 ≡ fi · ti mod p. i=1

Thus there is a ﬁnite set J ⊂ I and polynomials gi ∈ K[{Xi,Yi | i ∈ I}] (i ∈ J) such that Xn X 2 2 2 af0 − fi · ti = gi · (aiXi + biYi + ci). i=1 i∈J

In the polynomials f0, . . . , fn, gi (i ∈ J) only finitely many indeterminates occur. After enlarging J if necessary we may assume that f0, . . . , fn, gi ∈ K[{Xi,Yi | i ∈ J}]. 2 By Lemma 1.2 the ideal q of K[{Xi,Yi | i ∈ J}] generated by aiXi + 2 biYi − ci (i ∈ J) is prime. Let F be the field of fractions of K[{Xi,Yi | i ∈ J}]/q. By the second step T extends to a preordering T 00 of F . The above equation then yields a ∈ T 00. Since a ∈ K we obtain a ∈ T 00 ∩ K = T . This proves the lemma. Now we fix some notations. Let S be an arbitrary non-empty set. We denote by AS the ring of polynomials in the power set of S over the field Q of rational numbers. If we look at I ⊂ S as an indeterminate we write XI . So AS = Q[{XI | I ⊂ S}]. We denote by pS the ideal of AS generated by all elements of the form XI + XIc (I ⊂ S). Let BS = AS/pS. We denote by xI the residue class of the indeterminate XI .

Lemma 1.4. Let S be a non-empty set. Then pS is a prime ideal of AS and XI 6∈ pS for all I ⊂ S.

P r o o f. Let f, g ∈ AS and assume that fg ∈ pS. Then there are distinct c subsets I1,...,In ⊂ S with Ij 6= Ik for j 6= k and polynomials hi ∈ AS Pn (1 ≤ i ≤ n) such that fg = h · (X + X c ). In the polynomials i=1 i Ii Ii f, g, h1, . . . , hn only ﬁnitely many indeterminates occur. After enlarging the c c list I1,...,In we may assume f, g, h1, . . . , hn ∈ Q[XI1 ,XI1 ,...,XIn ,XIn ]. Thus fg ∈ q = (X + X c ,...,X + X c ) ⊂ Q[X ,X c ,...,X ,X c ]. I1 I1 In In I1 I1 In In Since

Q[X ,X c ,...,X ,X c ]/q ∼ Q[X ,...,X ] I1 I1 In In = I1 In we ﬁnd that q is a prime ideal. Hence f ∈ q ⊂ pS or g ∈ q ⊂ pS. Finally, a straightforward argument shows XI 6∈ pS for all I ⊂ S. 236 R. Berr et al.

Since pS is a prime ideal, BS is a domain. Its field of quotients will be denoted by KS. Now let F be a filter on S. As mentioned at the beginning of this section we will construct a preordering TF ⊂ KS with xI ∈ TF if and only if I ∈ F. To this end we need the following simple fact. Lemma 1.5. Let S be a non-empty set and let F be a filter on S. Let A1,...,An be subsets of S. Then there is a filter G on S such that (1) F ⊂ G, (2) for each i either Ai ∈ G or S \ Ai ∈ G. P r o o f. Without loss of generality we may assume n = 1. Let A ⊂ S. Only the case A 6∈ F has to be considered. But then I 6⊂ A for all I ∈ F. c Hence G1 = F ∪ {I ∩ A | I ∈ F} is a filter basis which yields the desired filter G.

If F is a ﬁlter on S we let TF denote the semiring generated by the squares of KS and the elements xI (I ∈ F). Lemma 1.6. Let F be a ﬁlter on the set S. Then

(1) TF is a preordering of KS, (2) for all I ⊂ S: I ∈ F ⇔ xI ∈ TF .

P r o o f. (1) By definition TF is additively and multiplicatively closed and contains all the squares of KS. It remains to show that −1 6∈ TF . Assume P 2 the contrary. Then there are n ∈ N, a0, . . . , an ∈ KS and t1, . . . , tn such that Xn −1 = a0 + ai · ti i=1 and each ti is a finite product of certain xI (I ∈ F). After multiplication with a common denominator we obtain Xn 2 −b = b0 + bi · ti i=1 P 2 for certain b0, . . . , bn ∈ BS, b ∈ BS \{0}. The elements b, b0, . . . , bn are the P 2 residue classes of polynomials f ∈ AS \ pS, f0, . . . , fn ∈ AS and t1, . . . , tn are residue classes of certain P1,...,Pn ∈ Q[{XI | I ⊂ S}] where each Pi is a product of certain XI with I ∈ F. Thus Xn 2 f + f0 + fi · Pi ∈ pS. i=1 In these polynomials only finitely many indeterminates occur. So there c c are J1,...,Jm ⊂ S such that J1,...,Jm,J1 ,...,Jm are pairwise distinct, c P1,...,Pn ∈ Q[XJ1 ,...,XJn ] and f, f0, . . . , fn ∈ Q[XJ1 ,...,XJm ,XJ1 ,...... ,X c ]. Now we apply the substitution homomorphism Φ : A → Jm S Ordered fields and the ultrafilter theorem 237

Q[X ,...,X ] given by X 7→ X , X c 7→ −X (1 ≤ i ≤ m), X 7→ 0 J1 Jm Ji Ji Ji Ji J c c (J 6∈ {J1,...,Jm,J1 ,...,Jm}) and obtain Xn 2 Φ(f) + Φ(f0) + Φ(fi) · Pi = 0 i=1 or Xn 2 Φ(f) = −Φ(f0) − Φ(fi) · Pi. i=1

This is a polynomial equation in the indeterminates XJ1 ,...,XJm over Q. 2 Since f0, . . . , fn are sums of squares so are Φ(f0),...,Φ(fn). Thus Φ(f)(x) m ≤ 0 for all x ∈ H = {(x1, . . . , xm) ∈ Q | x1, . . . , xm ≥ 0} and therefore

Φ(f)(x) = 0 for x ∈ H. Thus f(XJ1 ,...,XJm , −XJ1 ,..., −XJm ) = Φ(f) = c c 0. This yields the contradiction f ∈ (XJ1 + XJ1 ,...,XJm + XJm ) ⊂ pS and (1) is proved. (2) If I ∈ F then xI ∈ TF by the definition. So let xI ∈ TF . Since xI 6= 0 c we get −xI = xIc 6∈ TF because TF is a preordering. Thus I 6∈ F. Now assume that also I 6∈ F. By the proof of Lemma 1.5 we know that there is c a filter G ⊃ F on S with I ∈ G. Then xI ∈ TF ⊂ TG and by (1) we know c that TG is a preordering. But I ∈ G implies xIc = −xI ∈ TG, which gives the desired contradiction. Lemma 1.7. Let S be a non-empty set and F a filter on S. Then there is an extension field L ⊃ KS such that P 2 (1) xI ∈ L (I ∈ F), (2) each preordering T ⊃ TF extends to L.

P r o o f. Let A = KS[{YI ,ZI | I ∈ F}] and let p be the ideal of A which 2 2 is generated by the polynomials YI + ZI − xI (I ∈ F). By Lemma 1.2, p is a prime ideal. Let L be the ﬁeld of quotients of A/p. Then (1) is clear and (2) follows from Lemma 1.3.

2. The main theorem. Before we state the main theorem we recall some definitions and facts from real algebra. By a ring we always mean a commutative ring with a unit element 1 6= 0. A subset T ⊂ R is called a (quadratic) preordering if T + T ⊂ T,T · T ⊂ T,R2 ⊂ T and − 1 6∈ T. A preordering T is called total if T ∪ −T = R. Finally, an ordering P ⊂ R is a total preordering for which P ∩ −P is a prime ideal. In the case of a field this last condition reduces to P ∩ −P = {0}. Therefore the orderings of a field are exactly the total preorderings. 238 R. Berr et al.

A ring R is called real or formally real if Xn 2 ai = 0 ⇒ a1 = ... = an = 0. i=1 P R is called semi-real if −1 6∈ R2. Note that a semi-real ﬁeld is also formally real; however, for rings this is not true.

Theorem. On the basis of ZF the following statements are equivalent: (1) Each filter on a non-empty set is contained in an ultrafilter. (2) Each formally real field admits an ordering. (3) Each preordering of a field is contained in an ordering. (4) Each semi-real ring admits an ordering. (5) Each preordering of a ring is contained in a total preordering. (6) Each preordering of a ring is contained in an ordering.

P r o o f. We prove the following implications: (6) ⇒ (3) ⇓ ⇓ (4) ⇒ (2) ⇒ (1) ⇒ (5) ⇒ (6) The implications in the left square are obvious. (2)⇒(1). Let F be a ﬁlter on a non-empty set S. By Lemma 1.7 there P 2 is an extension ﬁeld L ⊃ KS such that xI ∈ L (I ∈ F) and each preordering T ⊃ TF of KS extends to L. Given a total order P ⊂ L, in general

U = {I ⊂ S | xI ∈ P } is not a ﬁlter. For example, if I,J ∈ U it may happen that I ∩ J 6∈ U. This problem leads to the following construction. Let

A = L[{YIJ ,ZIJ | (I,J) ∈ S × S}], 2 2 p = (xI YIJ + xJ ZIJ − xI∩J | (I,J) ∈ S × S) ⊂ A. By Lemma 1.2, p is a prime ideal of A. We let F denote the quotient ﬁeld of A/p. P We assume that F is not formally real. Then −1 ∈ F 2. In this repre- sentation of −1 as a sum of squares in F only ﬁnitely many indeterminates occur. Hence there are I1,...,In,J1,...,Jn ⊂ S such that E = Quot(B/q) is not formally real where

B = L[YI1J1 ,ZI1J1 ,...,YInJn ,ZInJn ], 2 2 q = (xIi YIiJi + xJi ZIiJi − xIi∩Ji | i ∈ {1, . . . , n}) ⊂ B. Ordered ﬁelds and the ultraﬁlter theorem 239

By Lemma 1.5 there is a ﬁlter G ⊃ F on S such that for each

I ∈ {I1,...,In,J1,...,Jn,I1 ∩ J1,...,In ∩ Jn} c either I ∈ G or I ∈ G. Since TF ⊂ TG the preordering TG extends to a preordering T of L. −1 Let i ∈ {1, . . . , n} and let I = Ii and J = Ji. We show that xI∩J ·xI ∈ T −1 or xI∩J · xJ ∈ T . If I ∩ J ∈ G then also I ∈ G. Thus xI∩J , xI ∈ TG and −1 therefore xI∩J · xI ∈ TG ⊂ T . If I ∩ J 6∈ G then without loss of generality c c I 6∈ G. Then (I ∩J) ,I ∈ G by the choice of G. Hence −xI∩J = x(I∩J)c ∈ TG −1 and −xI = xIc ∈ G and therefore xI∩J · xI ∈ TG ⊂ T . By Lemma 1.3, T extends to a preordering T 0 of E. Hence E is formally real and we got a contradiction. Thus F is formally real. By (2) it admits an ordering P . Let

U = {I ⊂ S | xI ∈ P }. We show that U is an ultraﬁlter that contains F. P 2 P 2 If I ∈ F then xI ∈ L ⊂ F ⊂ P . Hence I ∈ U and so F ⊂ U. Since S ∈ F we get S ∈ U. Hence xS ∈ P and therefore −xS = x∅ 6∈ P . Thus ∅ 6∈ U. 2 2 Let I,J ∈ U. Then xI , xJ ∈ P and thus xI∩J = xI yIJ + xJ zIJ ∈ P where yIJ and zIJ are the residue classes of YIJ and ZIJ respectively. So I ∩ J ∈ U. Let I ⊂ S and assume that I 6∈ U. Then xI 6∈ P and hence c xIc = −xI ∈ P . Therefore I ∈ U. Finally, we have to show that I ∈ U and I ⊂ J implies that also J ∈ U. Assume that J 6∈ U. As we have seen above this yields J c ∈ U and hence we get the contradiction I ∩ J c = ∅ ∈ U. (1)⇒(5). Let T be a preordering of the ring R. Let X be the set of all preorderings Q of R which contain T . For a ﬁnite non-empty subset A ⊂ R we let

DA = {Q ∈ X | −1 ∈ Q + aQ for each a ∈ A \ Q}. Finally, let

D = {DA | ∅ 6= A ⊂ R finite}. We first claim that D has the finite intersection property. Let A1,...,An ⊂ R be finite non-empty subsets of R. Define A = Sn Tn i=1 Ai. Then it is easy to see that DA ⊂ i=1 DAi . Hence it suffices to show that DA 6= ∅. Let Σ = {A ∩ Q| Q ∈ X}. Since A is finite Σ has a maximal element V with respect to set-theoretic inclusion. V is of the form V = A ∩ Q for some preordering Q ∈ X. We claim that Q ∈ DA. Let a ∈ A \ Q and define Q0 = Q + aQ. Then a ∈ Q0 and hence V is a proper subset of A ∩ Q0. Thus Q0 6∈ X by the maximality of V . Therefore −1 ∈ Q0 = Q + aQ. 240 R. Berr et al.

Since D has the finite intersection property it is contained in some filter FT. By (1), F is contained in some ultrafilter U. For Z ∈ U let PZ = Q∈Z Q. Obviously PZ ∈ X (Z ∈ U). Now define [ P = PZ . Z∈U We claim that P is a total preordering which contains T .

The inclusion T ⊂ P is clear. Since PY ,PZ ⊂ PY ∩Z the system (PZ )Z∈U is inductively directed. Hence P is a preordering. It remains to show that P is total. Let a ∈ R and let Z = {Q ∈ X | a ∈ Q}. If Z ∈ U then T c a ∈ Q∈Z Q = PZ ⊂ P . If Z 6∈ U then Z ∈ U because U is an ultrafilter. c Let A = {a, −a}. Then Y = Z ∩ DA ∈ U. For Q ∈ Y we get a ∈ A \ Q and thus −1 ∈ Q + aQ as Q ∈ DA. Therefore −1 6∈ Q −TaQ. Now Q ∈ DA implies −a 6∈ A \ Q and hence −a ∈ Q. Therefore −a ∈ Q∈Y Q = PY ⊂ P . (5)⇒(6). Let T be a preordering of the ring R. By (5) there is a total preordering Q which contains T . Let p = {a ∈ R | 1 + xa ∈ Q for all x ∈ R}. We first show that p is a proper prime ideal of R. Let a, b ∈ p and let x ∈ R. Then 1 + 2xa, 1 + 2xb ∈ Q and hence 2 + 2x(a + b) ∈ Q. Since Q is total, 1 + x(a + b) ∈ Q. Thus a + b ∈ p. The inclusion Rp ⊂ p is obvious by the definition. Since 1 − 2 · 1 = −1 6∈ Q we obtain 1 6∈ p and therefore p is a proper ideal of R. In order to show that p is a prime ideal let a, b ∈ R \ p. Then there are x, y ∈ R with xa − 1, yb − 1 ∈ Q. Hence xyab − 1 ∈ Q. Now assume ab ∈ p. Then 1 − 2xyab ∈ Q, which implies the contradiction −1 = 1 − 2xyab + 2(xyab − 1) ∈ Q. So ab 6∈ p. Now we show that p is convex with respect to Q, i.e., a + b ∈ p, a, b ∈ Q ⇒ a, b ∈ p. It is sufficient to show a ∈ p. Let x ∈ R. If x ∈ Q then 1 + xa ∈ Q. So assume −x ∈ Q. Then 1 + xa = 1 + x(a + b) + (−x)b ∈ Q. Hence a ∈ p. Now let P = p + Q. We claim that P is an ordering of R which contains T . Since p and Q are additively closed, so is P . Since p is an ideal, P is also multiplicatively closed. Obviously T ⊂ Q ⊂ P and therefore P ∪ −P = R because Q ∪ −Q = R. It remains to show that P ∩ −P is a prime ideal of R. We actually show P ∩ −P = p. Obviously p ⊂ P ∩ −P . Let x ∈ P ∩ −P . Then there are a, b ∈ p and p, q ∈ Q such that x = a + p and −x = b + q. Thus p + q = −(a + b) ∈ p. As p is convex with respect to Q we get p, q ∈ p and hence x = a + p ∈ p. This proves the theorem. Ordered fields and the ultrafilter theorem 241

References

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Fachbereich Mathematik CNRS–UFR de Mathématiques UniversitätDortmund UniversitéParis 7 D-44221 Dortmund, Germany 2 Place Jussieu, Case 7012 E-mail: [email protected] F-75251 Paris Cedex 05, France [email protected] E-mail: [email protected]

Received 3 March 1998; in revised form 5 May 1998