Ordered fields and the ultrafilter theorem

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1 F U N D A M E N T A MATHEMATICAE 59 (999) Ordered fields and the ultrafilter theorem by R. B e r r (Dortmund), F. D e l o n (Paris) and J. S h m i d (Dortmund) Abstrat. We prove that on the basis of ZF the ultrafilter theorem and the theorem of Artin Shreier are equivalent. The latter says that every formally real field admits a total order. Introdution. In 900 at the international ongress of mathematiians in Paris Hilbert raised his famous 23 problems. Among these was the following (the 7th): Let f R[X,..., X n ] be a polynomial whih is positive semidefinite, i.e., f(x) 0 for all x R n. Is f a sum of squares of rational funtions? Hilbert already knew that f is not neessarily a sum of squares of polynomials unless n =. In 927 Artin solved Hilbert s 7th problem affirmatively (f. []). The methods he used were developed in the joint paper [2] with Shreier and are nowadays known as Artin Shreier theory. The ruial notion of Artin Shreier theory is that of a formally real field. A field K is alled formally real if is not a sum of squares in K. The first basi theorem proved by Artin and Shreier stated that eah formally real field K admits an ordering, i.e. a subset P K suh that P + P P, P P P, P P = {0} and P P = K. The proof runs as follows. Let T = K 2 be the set of all sums of squares of K. Then T + T T, T T T, K 2 T and by the assumption T, i.e. T is a preordering. An appliation of Zorn s Lemma yields a preordering maximal for set-theoreti inlusion. Now an elementary omputation shows that this maximal preordering is an ordering. In 954 Tarski [6] mentioned that the axiom of hoie (in the form of Zorn s Lemma) is not neessary. The ultrafilter theorem suffies to prove the theorem of Artin Shreier. A proof an be found in [3] (Theorem 2.2 and Example 2.3.3). As mentioned by Tarski the problem whether the 99 Mathematis Subjet Classifiation: Primary 03E24; Seondary 2J5. [23]

2 232 R. Berr et al. theorem of Artin Shreier is equivalent to the ultrafilter theorem remains open. Reently, questions of similar type appeared in omputer algebra. Lombardi Roy [4] and Sander [5] proved independently that the existene and uniqueness of the real losure of an ordered field an be proven within ZF. In this paper we lose the gap mentioned above and show that the theorem of Artin Shreier, the ultrafilter theorem and ertain other theorems from real algebra are mutually equivalent on the basis of ZF.. Preliminaries. It is the aim of this setion to prove a ouple of preliminary results whih will be needed for the proof that every filter is ontained in an ultrafilter provided every formally real field admits an ordering. Sine these results are of quite tehnial nature we briefly sketh 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 ontained in an ultrafilter. To this end we onsider A = Q[X I I S] and onstrut in a first step a ring homomorphism ϕ : A (K, T ) into a preordered field (K, T ) suh that F = {I S ϕ(x I ) T }. In the next step we embed K into a formally real field L with T = K L 2 and show that for any ordering P L, U = {I S ϕ(x I ) P } is an ultrafilter ontaining F. For the onstrution of K one has to onsider ideals in ertain polynomial rings over A whih are generated by elements of the form ax 2 + by 2 + with a, b, A. In Lemmas..4 we will be onerned with these ideals. The remaining statements are related to the onstrution of the preordering T K and the extension field L K. Throughout this setion we are working inside ZF and all fields are assumed to have harateristi zero. Given a ring R we let R denote the group of units of R. Lemma.. Let R be an integral domain with K as its field of quotients. Let a, b, R \{0} and assume that a R. Let p, q be the ideals generated by ax 2 + by 2 + in R[X, Y ] and K[X, Y ] respetively. Then () 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 ) (ax 2 + by 2 + ) for some g K[X, Y ]. Let n+2 f(x, Y ) = f i (Y )X i and g(x, Y ) = g i (Y )X i i=0 i=0

3 Ordered fields and the ultrafilter theorem 233 for ertain f i R[Y ] and g i K[Y ]. Comparing the oeffiients we get ag n = f n+2 R[Y ] and ag n = f n+ R[Y ]. Hene g n, g n R[Y ] beause a is a unit in R. For i n we obtain again by omparing the oeffiients ag i (Y ) + (by 2 + ) g i+ (Y ) = f i+ (Y ) R[Y ]. Now an easy indution yields g n, g n,..., g, g 0 R[Y ] and hene g R[X, Y ]. Therefore f p. The other inlusion is trivial. An easy omputation shows that ax 2 +by 2 + is irreduible in K[X, Y ]. Thus q is a prime ideal of K[X, Y ] and from (2) we onlude that p is a prime ideal of R[X, Y ]. This proves (). Lemma.2. Let R be an integral domain. Let I and a i, b i, i R (i I). Let p be the ideal of R[{X i, Y i i I}] whih is generated by the polynomials a i Xi 2 + b iyi 2 + i (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 indution on the number n of elements of I. We may assume that I = {,..., n}. The ase n = follows from Lemma.(). Let q = (a X 2 + b Y 2 +,..., a n X 2 n + b n Y 2 n + n ) R[X, Y,..., X n, Y n ]. By the indution hypothesis q is a prime ideal. Thus the residue lass ring A = R[X, Y,..., X n, Y n ]/q is a domain. An easy argument yields R[X, Y,..., X n, Y n ]/p = A[X n, Y n ]/(a n X 2 n + b n Y 2 n + n ). Now Lemma.() shows that the right hand side is a domain. Thus p is a prime ideal. In the seond step we assume that I is arbitrary. Let f, g R[{X i, Y i i I}] and assume that fg p. Thus there are a finite set J I and polynomials h i R[{X i, Y i i I}] suh that fg = i J h i (a i X 2 i + b i Y 2 i + i ). In the polynomials f, g, h i (i J) only finitely many indeterminates our. Enlarging J if neessary we may assume that f, g, h i R[{X i, Y i i J}]. Then fg q where q is the ideal of R[{X i, Y i i J}] generated by the polynomials a i Xi 2 + b iyi 2 + i (i J). By the first step q is a prime ideal. Thus f q p or g q p.

4 234 R. Berr et al. Lemma.3. Let K be a field and let T be a preordering of K. Let I and a i, b i, i K (i I). Assume that for eah i I either i a i or i b i is in T. Let p = (a i X 2 i + b i Y 2 i i i I) K[{X i, Y i i I}] and let L be the field of quotients of the ring K[{X i, Y i i I}]/p. Then there is a preordering T of L whih extends T, i.e., T K = T. P r o o f. We prove the lemma in three steps. In the first step we assume that I ontains only one element. We omit all indies. Sine the situation is symmetri in X and Y we may assume a T. First note that L = K(Y )[X]/(X 2 + ba Y 2 a ). We denote by x L the image of X under the anonial projetion K(Y )[X] L. Now let T L be the semiring generated by T and the squares of L. We laim T = T K. Obviously, T T K. So let t T K. Then there are n N, t i T and g, g i, h i K[Y ], i =,..., n, with g 0 and tg 2 = (h i + xg i ) 2 t i = (h 2 i + x 2 gi 2 ) t i + 2x g i h i t i. But x 2 = a ba Y 2 K(Y ) implies g i h i t i = 0. Therefore tg 2 = h 2 i t i + (a ba Y 2 ) gi 2 t i. Now assume g(0) = 0. Sine a T we see h i (0) = 0 = g i (0) for all i. Hene we may assume that g(0) 0. But then a T implies tg(0) 2 T, hene t T. Therefore T extends T. In the seond step we assume that I is finite and proeed by indution on the number n of its elements. We may assume that I = {,..., n}. The ase n = is exatly the first step. Let q = (a X 2 + b Y 2 +,..., a n X 2 n + b n Y 2 n + n ) K[X, Y,..., X n, Y n ]. By Lemma.2, q is a prime ideal. So A = K[X, Y,..., X n, Y n ]/q is a domain. Let F be its field of quotients. By the indution hypothesis T extends to a preordering T of F. From the first step we see that T extends to a preordering T of E = Quot(F [X n, Y n ]/(a n X 2 n + b n Y 2 n + n )). An easy argument yields K[X, Y,..., X n, Y n ]/p = A[X n, Y n ]/(a n X 2 n + b n Y 2 n + n ). By Lemma.(2) the right hand side embeds into E. Thus there is also an embedding L E. Now one an easily see that T indues the desired preordering on L.

5 Ordered fields and the ultrafilter theorem 235 In the final step we assume that I is arbitrary. Let T be the semiring generated by T and the squares of L. It suffies to show that T K = T. The inlusion T T K is lear. So let a T K. Then there are f 0 K[{X i, Y i i I}] \ p, f,..., f n K[{X i, Y i i I}] 2 and t,..., t n T suh that af0 2 f i t i mod p. Thus there is a finite set J I and polynomials g i K[{X i, Y i i I}] (i J) suh that af0 2 f i t i = g i (a i Xi 2 + b i Yi 2 + i ). i J In the polynomials f 0,..., f n, g i (i J) only finitely many indeterminates our. After enlarging J if neessary we may assume that f 0,..., f n, g i K[{X i, Y i i J}]. By Lemma.2 the ideal q of K[{X i, Y i i J}] generated by a i Xi 2 + b i Yi 2 i (i J) is prime. Let F be the field of frations of K[{X i, Y i i J}]/q. By the seond step T extends to a preordering T of F. The above equation then yields a T. Sine a K we obtain a T K = T. This proves the lemma. Now we fix some notations. Let S be an arbitrary non-empty set. We denote by A S 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 X I. So A S = Q[{X I I S}]. We denote by p S the ideal of A S generated by all elements of the form X I + X I (I S). Let B S = A S /p S. We denote by x I the residue lass of the indeterminate X I. Lemma.4. Let S be a non-empty set. Then p S is a prime ideal of A S and X I p S for all I S. P r o o f. Let f, g A S and assume that fg p S. Then there are distint subsets I,..., I n S with I j Ik for j k and polynomials h i A S ( i n) suh that fg = n h i (X Ii + X I i ). In the polynomials f, g, h,..., h n only finitely many indeterminates our. After enlarging the list I,..., I n we may assume f, g, h,..., h n Q[X I, X I,..., X In, X I n ]. Thus fg q = (X I + X I,..., X In + X I n ) Q[X I, X I,..., X In, X I n ]. Sine Q[X I, X I,..., X In, X I n ]/q = Q[X I,..., X In ] we find that q is a prime ideal. Hene f q p S or g q p S. Finally, a straightforward argument shows X I p S for all I S.

6 236 R. Berr et al. Sine p S is a prime ideal, B S is a domain. Its field of quotients will be denoted by K S. Now let F be a filter on S. As mentioned at the beginning of this setion we will onstrut a preordering T F K S with x I T F if and only if I F. To this end we need the following simple fat. Lemma.5. Let S be a non-empty set and let F be a filter on S. Let A,..., A n be subsets of S. Then there is a filter G on S suh that () F G, (2) for eah i either A i G or S \ A i G. P r o o f. Without loss of generality we may assume n =. Let A S. Only the ase A F has to be onsidered. But then I A for all I F. Hene G = F {I A I F} is a filter basis whih yields the desired filter G. If F is a filter on S we let T F denote the semiring generated by the squares of K S and the elements x I (I F). Lemma.6. Let F be a filter on the set S. Then () T F is a preordering of K S, (2) for all I S: I F x I T F. P r o o f. () By definition T F is additively and multipliatively losed and ontains all the squares of K S. It remains to show that T F. Assume the ontrary. Then there are n N, a 0,..., a n KS 2 and t,..., t n suh that = a 0 + a i t i and eah t i is a finite produt of ertain x I (I F). After multipliation with a ommon denominator we obtain b 2 = b 0 + b i t i for ertain b 0,..., b n BS 2, b B S \{0}. The elements b, b 0,..., b n are the residue lasses of polynomials f A S \ p S, f 0,..., f n A 2 S and t,..., t n are residue lasses of ertain P,..., P n Q[{X I I S}] where eah P i is a produt of ertain X I with I F. Thus f 2 + f 0 + f i P i p S. In these polynomials only finitely many indeterminates our. So there are J,..., J m S suh that J,..., J m, J,..., J m are pairwise distint, P,..., P n Q[X J,..., X Jn ] and f, f 0,..., f n Q[X J,..., X Jm, X J,......, X J m ]. Now we apply the substitution homomorphism Φ : A S

7 Ordered fields and the ultrafilter theorem 237 Q[X J,..., X Jm ] given by X Ji X Ji, X J i X Ji ( i m), X J 0 (J {J,..., J m, J,..., Jm}) and obtain Φ(f) 2 + Φ(f 0 ) + Φ(f i ) P i = 0 or Φ(f) 2 = Φ(f 0 ) Φ(f i ) P i. This is a polynomial equation in the indeterminates X J,..., X Jm over Q. Sine f 0,..., f n are sums of squares so are Φ(f 0 ),..., Φ(f n ). Thus Φ(f)(x) 2 0 for all x H = {(x,..., x m ) Q m x,..., x m 0} and therefore Φ(f)(x) = 0 for x H. Thus f(x J,..., X Jm, X J,..., X Jm ) = Φ(f) = 0. This yields the ontradition f (X J + X J,..., X Jm + X J m ) p S and () is proved. (2) If I F then x I T F by the definition. So let x I T F. Sine x I 0 we get x I = x I T F beause T F is a preordering. Thus I F. Now assume that also I F. By the proof of Lemma.5 we know that there is a filter G F on S with I G. Then x I T F T G and by () we know that T G is a preordering. But I G implies x I = x I T G, whih gives the desired ontradition. Lemma.7. Let S be a non-empty set and F a filter on S. Then there is an extension field L K S suh that () x I L 2 (I F), (2) eah preordering T T F extends to L. P r o o f. Let A = K S [{Y I, Z I I F}] and let p be the ideal of A whih is generated by the polynomials YI 2 + Z2 I x I (I F). By Lemma.2, p is a prime ideal. Let L be the field of quotients of A/p. Then () is lear and (2) follows from Lemma The main theorem. Before we state the main theorem we reall some definitions and fats from real algebra. By a ring we always mean a ommutative ring with a unit element 0. A subset T R is alled a (quadrati) preordering if T + T T, T T T, R 2 T and T. A preordering T is alled total if T T = R. Finally, an ordering P R is a total preordering for whih P P is a prime ideal. In the ase of a field this last ondition redues to P P = {0}. Therefore the orderings of a field are exatly the total preorderings.

8 238 R. Berr et al. A ring R is alled real or formally real if a 2 i = 0 a =... = a n = 0. R is alled semi-real if R 2. 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: () Eah filter on a non-empty set is ontained in an ultrafilter. (2) Eah formally real field admits an ordering. (3) Eah preordering of a field is ontained in an ordering. (4) Eah semi-real ring admits an ordering. (5) Eah preordering of a ring is ontained in a total preordering. (6) Eah preordering of a ring is ontained in an ordering. P r o o f. We prove the following impliations: (6) (3) (4) (2) () (5) (6) The impliations in the left square are obvious. (2) (). Let F be a filter on a non-empty set S. By Lemma.7 there is an extension field L K S suh that x I L 2 (I F) and eah preordering T T F of K S extends to L. Given a total order P L, in general U = {I S x I P } is not a filter. For example, if I, J U it may happen that I J U. This problem leads to the following onstrution. Let A = L[{Y IJ, Z IJ (I, J) S S}], p = (x I Y 2 IJ + x J Z 2 IJ x I J (I, J) S S) A. By Lemma.2, p is a prime ideal of A. We let F denote the quotient field of A/p. We assume that F is not formally real. Then F 2. In this representation of as a sum of squares in F only finitely many indeterminates our. Hene there are I,..., I n, J,..., J n S suh that E = Quot(B/q) is not formally real where B = L[Y I J, Z I J,..., Y In J n, Z In J n ], q = (x Ii Y 2 I i J i + x Ji Z 2 I i J i x Ii J i i {,..., n}) B.

9 Ordered fields and the ultrafilter theorem 239 By Lemma.5 there is a filter G F on S suh that for eah I {I,..., I n, J,..., J n, I J,..., I n J n } either I G or I G. Sine T F T G the preordering T G extends to a preordering T of L. Let i {,..., n} and let I = I i and J = J i. We show that x I J x I T or x I J x J T. If I J G then also I G. Thus x I J, x I T G and therefore x I J x I T G T. If I J G then without loss of generality I G. Then (I J), I G by the hoie of G. Hene x I J = x (I J) T G and x I = x I G and therefore x I J x I T G T. By Lemma.3, T extends to a preordering T of E. Hene E is formally real and we got a ontradition. Thus F is formally real. By (2) it admits an ordering P. Let U = {I S x I P }. We show that U is an ultrafilter that ontains F. If I F then x I L 2 F 2 P. Hene I U and so F U. Sine S F we get S U. Hene x S P and therefore x S = x P. Thus U. Let I, J U. Then x I, x J P and thus x I J = x I yij 2 + x JzIJ 2 P where y IJ and z IJ are the residue lasses of Y IJ and Z IJ respetively. So I J U. Let I S and assume that I U. Then x I P and hene x I = x I P. Therefore I U. Finally, we have to show that I U and I J implies that also J U. Assume that J U. As we have seen above this yields J U and hene we get the ontradition I J = U. () (5). Let T be a preordering of the ring R. Let X be the set of all preorderings Q of R whih ontain T. For a finite non-empty subset A R we let Finally, let D A = {Q X Q + aq for eah a A \ Q}. D = {D A = A R finite}. We first laim that D has the finite intersetion property. Let A,..., A n R be finite non-empty subsets of R. Define A = n A i. Then it is easy to see that D A n D A i. Hene it suffies to show that D A. Let Σ = {A Q Q X}. Sine A is finite Σ has a maximal element V with respet to set-theoreti inlusion. V is of the form V = A Q for some preordering Q X. We laim that Q D A. Let a A \ Q and define Q = Q + aq. Then a Q and hene V is a proper subset of A Q. Thus Q X by the maximality of V. Therefore Q = Q + aq.

10 240 R. Berr et al. Sine D has the finite intersetion property it is ontained in some filter F. By (), F is ontained in some ultrafilter U. For Z U let P Z = Q Z Q. Obviously P Z X (Z U). Now define P = P Z. Z U We laim that P is a total preordering whih ontains T. The inlusion T P is lear. Sine P Y, P Z P Y Z the system (P Z ) Z U is indutively direted. Hene 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 a Q Z Q = P Z P. If Z U then Z U beause U is an ultrafilter. Let A = {a, a}. Then Y = Z D A U. For Q Y we get a A \ Q and thus Q + aq as Q D A. Therefore Q aq. Now Q D A implies a A \ Q and hene a Q. Therefore a Q Y Q = P Y P. (5) (6). Let T be a preordering of the ring R. By (5) there is a total preordering Q whih ontains T. Let p = {a R + 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 + 2xa, + 2xb Q and hene 2 + 2x(a + b) Q. Sine Q is total, + x(a + b) Q. Thus a + b p. The inlusion Rp p is obvious by the definition. Sine 2 = Q we obtain 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, yb Q. Hene xyab Q. Now assume ab p. Then 2xyab Q, whih implies the ontradition = 2xyab + 2(xyab ) Q. So ab p. Now we show that p is onvex with respet to Q, i.e., a + b p, a, b Q a, b p. It is suffiient to show a p. Let x R. If x Q then + xa Q. So assume x Q. Then + xa = + x(a + b) + ( x)b Q. Hene a p. Now let P = p + Q. We laim that P is an ordering of R whih ontains T. Sine p and Q are additively losed, so is P. Sine p is an ideal, P is also multipliatively losed. Obviously T Q P and therefore P P = R beause Q Q = R. It remains to show that P P is a prime ideal of R. We atually show P P = p. Obviously p P P. Let x P P. Then there are a, b p and p, q Q suh that x = a + p and x = b + q. Thus p + q = (a + b) p. As p is onvex with respet to Q we get p, q p and hene x = a + p p. This proves the theorem.

11 Ordered fields and the ultrafilter theorem 24 Referenes [] E. Artin, Über die Zerlegung definiter Funktionen in Quadrate, Abh. Math. Sem. Univ. Hamburg 5 (927), [2] E. Artin und O. Shreier, Algebraishe Konstruktion reeller Körper, ibid., [3] T. Jeh, The Axiom of Choie, North-Holland, 973. [4] H. Lombardi and M.-F. Roy, Construtive elementary theory of ordered fields, in: Effetive Methods in Algebrai Geometry, Progr. Math. 94, Birkhäuser, 99, [5] T. Sander, Existene and uniqueness of the real losure of an ordered field without Zorn s Lemma, J. Pure Appl. Algebra 73 (99), [6] A. Tarski, Prime ideal theorems for set algebras and ordering priniples, preliminary report, Bull. Amer. Math. So. 60 (954), 39. Fahbereih Mathematik Universität Dortmund D-4422 Dortmund, Germany berr@math.uni-dortmund.de shmid@math.uni-dortmund.de CNRS UFR de Mathématiques Université Paris 7 2 Plae Jussieu, Case 702 F-7525 Paris Cedex 05, Frane delon@logique.jussieu.fr Reeived 3 Marh 998; in revised form 5 May 998