The Witt group
and the problem of L¨uroth
Manuel Ojanguren
1990 ii
TABLE OF CONTENTS
Introduction 7 Acknowledgments 10 1. Unirational fields 11 2. The Witt group of a commutative field 23 3. The Witt group and the theorem of Tsen 33 4. The spectrum of the Witt ring 44 5. Also sprach Brahmagupta 54 6. Dedekind domains 64 7. The unramified Witt group 79 8. Dulcis in fundo 85 References 92 Index 96 iii
INTRODUCTION
by Inta Bertuccioni
When I discovered, purely by accident, that Manuel was writing up the notes of a course he had given in Pisa with the intention of publishing them, I first tried to dissuade him. I knew that he was going to waste his time fighting with word processing programs, laser printers and macros, instead of seriously pondering on the mathematical content and the pedagogical aspects of his brain child. But he was more stubborn than I had anticipated. So, with a justified apprehension for the results of his misplaced tenacity — pigheadedness might be a more apt term — I suggested, in the most delicate and diplomatic manner, that he use some not too well known language: “Write them in Basque or, even better, in Nicobarese” I told him. “If your fluency in Nicobarese is not quite up to the mark, well, then write them in Italian. But English! Half of the planet speaks it, and rather better than you and me! Why unnecessarily make yourself conspicuous?” Again I failed. He mumbled some excuse, putting the blame on the linguistic prejudices of the editors, and then suggested that instead of pestering him I write the introduction. Reluctantly, I decided to humour him.
An algebraic variety defined over a field k is rational if it is birationally isomorphic to a projective space over k or, equivalently, if its field of rational functions is a purely transcendental extension of k. A variety is unirational if it is dominated by a projective space or, equivalently, if its field of rational functions is contained in a purely transcendental extension of k. For a given class C of varieties we may ask if its unirational members are rational: this is L¨uroth’s problem for C. In 1876 L¨uroth solved this problem for the class of all curves. He proved [26] that unirational curves are rational. Although he only considered complex curves, his proof can be extended to arbitrary fields. iv
Two decades later, when the italian school of geometry was embarking on the study of algebraic surfaces (1893 is also the year of Enriques’s doctoral dis- sertation), Castelnuovo proved [8] that unirational complex surfaces are rational. In this case the assumption that k is the complex field is essential: Zariski [42] gave examples of unirational non-rational surfaces over any algebraically closed field of positive characteristic and Segre [38] did the same with real cubic sur- faces. The study of the three-dimensional case started with Fano’s claim [15] that the generic hypersurface of degree 4 in P4 and the generic intersection of a 5 smooth cubic with a smooth quadric in P are not rational. Prompted by this claim, Enriques proved that the latter is unirational [14], thus giving a negative solution to L¨uroth’s problem in dimension 3. Although the truth of these results was continued to be believed, as time passed Fano’s proofs were viewed with growing suspicion. The gaps that mar them are described in Roth’s monograph [35] and subjected to close scrutiny in [21]. Only as late as in 1970 three dif- ferent approaches to L¨uroth’s problem led to rigorous proofs of the existence of non-rational unirational threefolds. Iskovskih and Manin, in their joint paper [21], completed and extended part of the work of Fano and proved that the group of birational automorphisms of a 4 smooth quartic in P is finite. This result clearly confirms Fano’s claim that such a quartic is not rational. On the other hand B. Segre [37] had already discovered examples of smooth quartics which are unirational. (It does not seem to be known whether every smooth quartic threefold is unirational). Clemens and Griffiths [10], after a deep study of the so called “intermediate Jacobian” — a principally polarized abelian variety associated to any smooth threefold with Hodge number h3,0 = 0 — proved that no smooth cubic hyper- 4 surface of P can be rational because its intermediate Jacobian is different from that of a rational threefold. Finally, Artin and Mumford [1] constructed a unirational conic bundle X v
2 over P and proved its irrationality by showing that its third cohomology group H3(X, Z) contains non-trivial torsion elements.
These notes give an elementary, selfcontained construction of a unirational non-rational threefold. The method used to this effect is an algebraic version of the method of Artin and Mumford. It has been briefly described in the appendix to [11], of which these notes constitute an expanded version. In short, varieties are replaced by their function fields and the torsion of H3(X, Z) by the unramified Witt group of C(X). Most of the material contained here is treated more systematically in several textbooks, a fact of which the author seems to be fully aware: the lectures from which these notes originated, he told me, were not meant to expose in full the theory of bilinear spaces, but only to give the indispensable background for a couple of computations involving the Witt group of a field. This brings us to the topic of prerequisites. It is a subject on which authors frequently lie, claiming that whoever can count up to ten and recognize a few greek letters will immensely profit from the purchase of their books. A little more than that is required here, but very little indeed: the most elementary results in commutative algebra, some familiarity with Dedekind domains and a smattering of algebro-geometric jargon.
Finally, Manuel urges me to stress his indebtedness to Jean-Louis Colliot- Th´el`ene: the results of the last section, which are the only original ones, are entirely due to their collaboration.
Pisa, Spring 1989 vi
ACKNOWLEDGMENTS
I thank Inta for the quotation of Brahmagupta (well, this is what she claims, for all I know it could be the recipe of bhelpoori) and Ghazi, Max, Parimala, Sridharan and Sujatha for several suggestions and (benevolent) comments. I thank Lorenzo who, after having endured my lectures in Pisa, went through the first draft of the manuscript and corrected all the mistakes, mathematical and otherwise. (But, as the reader will notice, I surreptitiously inserted a few new ones.) I thank Margherita for inviting me to spend a pleasent semester in Pisa and for taking care, among thousend other things, of the publication of these notes.
M.O. 1 1. Unirational fields
A li principianti parera` forse ch’io sia stato troppo breve, et massime ne gl’ultimi quarantadue Capitoli. Et a li intendenti troppo lungo. Rafael Bombelli
Let K be an extension of a fixed ground field k. We say that K is rational over k if it is of the form k(t1,...,tn) where t1,...,tn are algebraically independent over k. We say that K is unirational over k if it is contained in some rational extension of k:
(1) k K k(t ,...,t ). ⊆ ⊆ 1 n
In this case n must be at least equal to the transcendence degree d of K over k. We show that, in fact, n can always be chosen equal to d.
Proposition 1.1. Any unirational extension K of k is contained in a rational extension k(t1,...,td), where d is the transcendence degree of K over k.
To prove Proposition 1.1 we need the following general fact, due essentially to E. Noether [31].
Proposition 1.2. Let L = k(x1,...,xn) be any finitely generated extension of k. Any field K containing k and contained in L is finitely generated over k.
Proof . Let t ,...,t be a transcendence basis of L over K. Each x satisfies { 1 d} i an algebraic equation
a xNi + + a =0, i0 i ··· iNi 2
with coefficients aij in K[t1,...,td]. Let K0 be the field obtained by adjoining to
k all the coefficients of all the polynomials aij; thus each aij lies in K0[t1,...,td].
The field L is an algebraic extension of K0(t1,...,td) generated by x1,...,xn, and
therefore its dimension over K0(t1,...,td) is finite. This implies that K(t1,...,td)
is also finite-dimensional over K0(t1,...,td). Since t1,...,td are algebraically
independent over K, the dimension of K over K0 coincides with the dimension of
K(t1,...,td) over K0(t1,...,td) and is, in particular, finite. By construction, K0 is a finitely generated extension of k, hence K is a finitely genereted extension of k as well. Proof of Proposition 1.1. We know from Proposition 1.2 that K is generated over k by a finite set x ,...,x . Let A be the affine k-algebra k[x ,...,x ]. { 1 m} 1 m By Emmy Noether’s normalization lemma [29, 14.G, Corollary 1], [31], we can
choose the generators of A such that x1,...,xd are algebraically independent over
k and xd+1,...,xm are integral over B = k[x1,...,xd]. It suffices to prove that if n>d there exists a monomorphism of k-algebras A֒ k(t1,...,tn 1). Each xi → − is a rational function in t1,...,tn. Let f(t1,...,tn) be a common denominator
of x1,...,xm. The transcendence degree of k(t1,...,tn) over k(x1,...,xd) is n d, which we assume to be positive. Renumbering, if necessary, t ,...,t , − 1 n we may assume that td+1,...,tn are algebraically independent over k(x1,...,xd)
and that t1,...,td are algebraic over k(x1,...,xd,td+1,...,tn). This means that there exist d non-vanishing polynomials
F (x ,...,x ,t ,...,t , T ) k[x ,...,x ,t ,...,t , T ] i 1 d d+1 n ∈ 1 d d+1 n such that for each i between 1 and d,
Fi(x1,...,xd,td+1,...,tn,ti)=0.
We observe that, given any polynomial F (t1,...,tn) not identically zero, the ν polynomial F (t1,...,tn 1,t1) is also different from zero, provided ν is sufficiently − large: it suffices to choose ν larger than the total degree of F in t1 and tn. 3
Put C = k[t1,...,tn, 1/f], where f is the common denominator chosen above. We may write
p q F1 = apq(x1,...,td+1,...,tn 1)tnT − p,q X and, for 2 i d, ≤ ≤
j Fi = bi,j(x1,...,xd,td+1,...,tn)T . i,j X
The coefficients ap,q and bi,j are rational functions of the form
g(t1,...,tn) r . (f(t1,...,tn))
By the observation made above we can choose an integer ν so large that
ν f(t1,...,tn 1,t1) =0 − 6 and for each numerator g,
ν g(t1,...,tn 1,t1) =0. − 6
We may further impose that for any integers p, q, p′, q′ between 0 and n,if(p, q) = 6 (p′, q′) then νp + q = νp′ + q′. 6 Let C be the reduction of C modulo t tν and x ,..., x the images of x ,...,x n − 1 1 d 1 d in C. The polynomials
νp+q F1′ = ap,q(x1,..., xd,td+1,...,tn 1)T − p,q X and, for 2 i d, ≤ ≤
ν j Fi′ = bi,j(x1,..., xd,td+1,...,tn 1,t1)T − i,j X 4
are all different from zero. Since the first one vanishes for T = t1, t1 is alge-
braic over k(x1,..., xd,td+1,...,tn 1). Similarly, for 2 i d, Fi vanishes − ≤ ≤ for T = ti, hence t2,...,td are algebraic over k(x1,..., xd,td+1,...,tn 1). Thus − x1,..., xd,td+1,...,tn 1 are algebraically independent over k, otherwise the tran- − scendence degree of k(t1,...,tn 1) would be less than n 1. In particular, − − x1,..., xd are algebraically independent over k and hence the homomorphism -B C induced by B ֒ C is injective. We claim that A C is in −→ → −→ jective too. In fact, its kernel is a prime ideal p of A, intersecting B in (0). Since the domain A is integral over B, by the theorem of Cohen–Seidenberg E, Theorem 5] p must be (0). The injection A ֒ C induces an injection.5 ,29] → .k(x1,...,xm) ֒ k(t1,...,tn 1) and thus the proposition is proved → −
We now prove L¨uroth’s theorem [26].
Theorem 1.3. Let k be a field and K a unirational extension of k. If the transcendence degree of K over k is 1, K is rational.
Proof. In view of Proposition 1.1 we may assume that K is contained in k(t), where t is transcendental over k. Observe that if the theorem is correct, then K = k(α(t)), where α(t) is some non-constant rational function. Writing α(t) as a reduced fraction
i i α(t) = ϕ(t)/ψ(t) with ϕ(t) = ait and ψ(t) = bit , X X we see that t is a zero of the polynomial
F (X) = ϕ(X) α(t)ψ(X) −
of K[X]. Since ϕ(t) and ψ(t) are coprime, F is irreducible over k(α) = K and is, therefore, the minimal polynomial of t over K. The coefficients of F are of the form a α(t)b , hence K = k(α) = k(a α(t)b ) unless b =0: K is generated i − i i − i i 5 by any nonzero coefficient of F ! To prove the theorem we show that, conversely, if n n 1 f = X + a X − + + a 1 ··· n is the minimal polynomial of t over K, then K = k(ai) for any non-constant ai.
First of all, at least one of the ai must be non-constant, otherwise t would be algebraic over k. Choose therefore m such that am = ϕ(t)/ψ(t), where ϕ and ψ are coprime elements of k[t] and at least one of them has positive degree. Then t annihilates the polynomial
F (X) = ϕ(X) a ψ(X), − m whose coefficients are in k(a ) K; hence f divides F in K[X] and we may m ⊆ write
(2) ϕ(X) a ψ(X) = f(X)g(X), − m where g(X) is in K[X]. It will suffice to show that g is a constant, because if such is the case, the degree of k(t) over K is the same as the degree of k(t) over k(am) and thus K = k(am). Multiplying (2) by a common denominator of the coefficients of f and g we can rewrite it as
c(t) ψ(t)ϕ(X) ϕ(t)ψ(X) = f (X)g (X), − 1 1 where f1 and g1 lie in k[t,X] and are multiples, respectively, of f and g by a polynomial of k[t]. Since k[t,X] is factorial each prime factor of c(t) must divide f1 or g1 and, dividing it out, we may assume that c(t)=1. If
f = γ (t)Xn + + γ (t), 1 0 ··· n its degree in t cannot be less than the degree of ϕ and ψ because γm/γ0 = am.
Hence g1 is constant in t. Let ξ be a root of ϕ (in some algebraic closure of k). 6
Since ϕ and ψ are coprime, ψ(ξ) = 0, hence g divides ϕ(X). Similarly it divides 6 1 ψ(X) and therefore it is a constant. This shows that ϕ(X) a ψ(X) is indeed − m the minimal polynomial of t over K.
The original version of L¨uroth’s theorem was phrased in geometrical terms and dealt with the parametrization of complex curves. Let be an irreducible C curve in Cnand suppose that it can be parametrized by n rational functions ϕ ,...,ϕ in C(t). This means that for every ζ C where these functions are 1 n ∈ defined, the point
ϕ(ζ) = ϕ1(ζ),...,ϕn(ζ)
belongs to . In general the point ϕ(ζ) will be the image of more than just C one complex number. L¨uroth proved that if can be parametrized at all, the C parametrization can be chosen to be bijective, except for a finite number of values of t. We shall deduce this from Theorem 1.3. Let p be the prime ideal of all polynomials vanishing on , A the affine algebra C C[X1,...,Xn]/p and K the field of fractions of A. The C –algebra homomorphism
C[X ,...,X ] C(t) 1 n −→ that maps each X to ϕ (t) induces, by assumption, a homomorphism ϕ : A i i −→ C(t). At least one among the ϕ1,...,ϕn is not constant, hence the image of A has transcendence degree 1 over C. Since A is integral and one-dimensional, ϕ is injective and induces a monomorphism of K into C(t). By Theorem 1.3
ϕ(K) = C(ϕ1,...,ϕn) = C(τ)
for some τ C(t). We can express each ϕ as ϕ (t) = ψ (τ(t)), where ψ is a ra- ∈ i i i i tional function in one indeterminate. On the other hand, since τ C(ϕ ,...,ϕ ), ∈ 1 n we can also write F (ϕ ,...,ϕ ) τ = 1 n , G(ϕ1,...,ϕn) 7 where F and G are polynomials in n indeterminates and G(ϕ ,...,ϕ ) = 0. 1 n 6 Replacing each ϕi by ψi(τ) and observing that τ is transcendental over C we get the identity F (ψ (τ),...ψ (τ)) (3) τ = 1 n . G(ψ1(τ),...ψn(τ)) We claim that ψ =(ψ ,...,ψ ) is a parametrization of and that it is bijective 1 n C up to a finite number of exceptional values of t. First of all, if η = τ(ζ) where ζ is a complex number for which τ, ϕ ,...,ϕ are defined, ψ(η) = ϕ(ζ) belongs to . 1 n C Since is closed, ψ(η) whenever ψ(η) is defined. If G(ψ (η),...,ψ (η)) = 0, C ∈ C 1 n 6 η is uniquely determined by (3), hence ψ is bijective on the set of all η C for ∈ which ψ is defined and G(ψ(η)) = 0. 6
The proof of L¨uroth’s theorem actually furnishes the new parameter τ(t). Let us look at the example given in L¨uroth’s paper. Let be the curve in C2 parametrized by C (t2 + 1)2 t(t2 + 1) ϕ = and ϕ = . 1 t4 +3t2 +1 2 t4 +3t2 +1 Then t is a root of the polynomial
X2ϕ Xϕ + ϕ C(ϕ , ϕ )[X]. 2 − 1 2 ∈ 1 2 Clearly K = C(ϕ , ϕ ) = C(t) because the involution mapping t to 1/t fixes 1 2 6 ϕ1 and ϕ2, hence the above polynomial is irreducible over K and the minimal polynomial of t over K is ϕ f = X2 1 X +1. − ϕ2 According to the proof of L¨uroth’s theorem, K is generated over C by the non- constant coefficient τ = ϕ /ϕ of f: indeed, − 1 2 τ 2 τ ϕ = and ϕ = . 1 τ 2 +1 2 −τ 2 +1 8
We now consider extensions of transcendence degree two over k. We start with a simple example of a unirational non-rational extension of the field R of real numbers.
Consider the surface in R3 defined by the equation S
Y 2 + Z2 = X3 3X. −
p intersection with the plane Z=21/2
umbrella
It can be obtained by rotating (very rapidly) the plane elliptic curve
Y 2 = X3 3X −
around its symmetry axis. The plane Z = √2 cuts along the cubic S C Z = √2 ( Y 2 +(X + 1)2 =(X + 1)2(X 1) −
of which p = ( 1, 0, √2) is an isolated real double point. Cutting with the − C pencil of “horizontal” lines through p
Y = s(X + 1) ( Z = √2 9 and computing the coordinates of the intersection point q(s) = p we find a 6 parametrization of : C X = 2 + s2 2 Y = s(3 + s ) Z = √2. To get a parametrization of it suffices to transform Y and Z by a family of S rotations like 1 t2 2t − 2 2 1+t 1+t . 2t 1 t2 2 − 2 ! − 1+t 1+t This yields
X = ϕ(s,t)=2+ s2 s(3 + s2)(1 t2) 2√2t Y = ψ(s,t) = − − (4) 2 1 + t 2st(3 + s2) + √2(1 t2) Z = χ(s,t) = − . 1 + t2 Let A = R[X,Y,Z]/(Y 2 + Z2 +3X X3) = R[x,y,z] be the affine algebra of , − S where x is the class of X, and so on. Let K be the field of fractions of A. The formulae (4) define a map Φ : A R(s,t). It is clear that the image of A has −→ transcendence degree 2 over R. Since A is a two-dimensional integral domain, Φ is injective and induces an inclusion of K into R(s,t). From
y s(3 + s2) = t z + √2 − − we get R(s,t) = R(s,y,z) and since s is quadratic over R(x) the degree of R(s,t) over R(x,y,z) = K is at most 2. It would be easy to prove that it is in fact equal to 2 by checking that s = s and t = √2t s(3 + s2) / √2 + st(3 + s2) define − − an involution of R(s,t) that fixes K. But, as Father Brown once said, we can do more than just prove it: we can see it. If we look at (R) (the set of R-rational S points of ) from a point situated far left on the X-axis, we see that every real S 10
point p =(ξ,η,ζ) with ξ > 2 is obtained from two different rotations of the plane Z = √2:
p
umbrella
This, of course, does not prove that K is not rational; it merely proves that it does not coincide with R(s,t). The irrationality of K results from a general theorem on real surfaces.
Theorem 1.4. Let and ′ be two real irreducible projective surfaces. Assume S S that every real point of and ′ is smooth. If their fields of rational functions S S R( ) and R( ′) are R-isomorphic, (R) and ′(R) have the same number of S S S S connected components.
Theorem 1.4 implies that the field K constructed above is not rational. In 3 fact the projective closure of is the surface of P defined by S S
(Y 2 + Z2)T = X3 3XT 2. −
Using the Jacobian criterion [29] it is easily seen that the singular locus of does S not contain any real point. If K = R( ) were R-isomorphic to R(s,t) = R(P2), 2 S (R) would have, like P (R), only one connected component instead of two. S 11
Proof of Theorem 1.4. Let Φ : R( ′) R( ) be an R-isomorphism. Let ′ S −→ S U and be affine open dense subsets of ′ and respectively, A′ R( ′) and V S S ⊂ S B R( ) their affine algebras. Since Φ(A′) is finitely generated and R( ) is the ⊂ S S field of fractions of B, we may choose an s B such that Φ(A′) B[1/s]. Then, ∈ ⊂ if = SpecB[1/s] = p s(p) =0 , Φ induces a morphism of affine varieties U { ∈S| 6 } f : ′ ′. Let be the open set of regular points of . It is well U −→ U ⊂ S Sreg S known [20, Th. 2.17] that f can be extended to an open set such that U0 U0 ∩Sreg is of codimension 2 in . In particular, since all real points of are smooth, Sreg S the complement of (R) in (R) is a finite set F . We thus have a continuous U0 S map f : (R) F ′(R). S \ −→ S The number of connected components of (R) F is the same as that of (R) S \ S because every real smooth point has an open real neighbourhood homeomorphic to a disc. Hence the number of connected components of f( (R) F ) is at most S \ equal to that of (R). We now show that f( (R) F ) intersects every component S S \ of ′(R). This will imply that (R) has at least as many components as ′(R), S S S hence equally many by symmetry.
We choose a suitable open set ′ ′ and a morphism g : ′ which U ⊂ S U −→ S 1 1 induces Φ− . Then the open subscheme ′ = g− ( ) = f( ) ′ of ′ is W U0 U0 ∩ U S 1 isomorphic to = f − ( ′). Since f( )(R)) contains ′(R) = f( (R)), it W U0 ∩ U U0 W W suffices to show that ′(R) intersects each connected component of ′(R). More W S generally, we prove the following result.
Proposition 1.5. Let be a real projective variety, a (non-empty) open S W subscheme of and p a smooth real point of . Then (R) intersects any open S S W disc around p.
Proof. Replacing by a suitable affine neighbourhood of p we may assume [30, S Ch. 3, 7, Theorem 2] that is a hypersurface in An+1, defined by an equation § S
N N 1 f = Z + a (X ,...,X )Z − + + a (X ,...,X )=0. 1 1 n ··· N 1 n 12
We may further assume that p = (0,..., 0) and that f(0) = 0 but ∂f (0) = 0. ∂Z 6 The open set will then contain a principal open set defined by g = 0 for some W 6 polynomial (X ,...,X ,Z) that does not vanish identically on . 1 n S We only have to show that for any ǫ > 0 there are points (ξ ,...,ξ ,ζ) (R) 1 n ∈ S such that
ξ2 + ,ξ2 + ζ2 < ǫ2 and g(ξ ,...,ξ ,ζ) =0. 1 ··· n 1 n 6 The affine algebra A = R[X ,...,X ,Z]/(f) of is a free module of rank N over 1 n S B = R[X1,...,Xn]. Let g be the image of g in A. By Dini’s theorem * there exists a (euclidean) neighbourhood W (R) of p which projects onto a disc 0 ⊂ S around (0,..., 0) in Rn, of radius less than ǫ. Suppose that g = 0 and consider |W0 the norm map : A B. The image (g) of g, being of the form gh for some N −→ N h A, vanishes identically on (the projection of) W ; hence (g) = 0. Since A ∈ 0 N is a domain this implies g = 0, against the assumption.
Cubic surfaces show a marked tendency to be unirational. The unirationality of the field K = R( ) constructed above is a very special case of a general theorem S [28, Ch. 2, Theorem 12.11].
It is worth mentioning that L¨uroth’s theorem extends to fields K of tran- scendence degree 2, provided the base field k is algebraically closed and of char- acteristic 0. This is a deep theorem, discovered and proved by Castelnuovo in 1893 [8].
* It takes no time, in Pisa, to learn the correct name of the implicit function theorem. 13 2. The Witt group of a commutative ring
In der Mathematik gibt es viele Lehrsatze,¨ welche sich nur dadurch beweisen lassen, daß man von vorne anfangt.¨ Professor Galletti
Let A be a commutative ring, M an A-module and M ∗ = HomA(M,A) its dual. For any f Hom (M,N) we define its dual f ∗ Hom (N ∗,M ∗) by ∈ A ∈ A f ∗(α) = α f for any α N ∗. Let ◦ ∈
i : M M ∗∗ −→ be the canonical homomorphism of M into its double dual, defined by
i(x)(f) = f(x) forany x M and any f M ∗. ∈ ∈ For any A-bilinear form ϕ : M M A the adjoint of ϕ is the map ϕ : M × −→ −→ M ∗ defined by ϕ(x) (y) = ϕ(x,y). Conversely, any f : M M ∗ yields a −→ e bilinear form defined by ϕ(x,y) = f(x) (y) of which f is the adjoint. Giving ϕ e is clearly the same as giving ϕ. The identity (ϕ∗ i)(x) (y) = ϕ(y) (x) = ϕ(y,x) ◦ shows that ϕ is symmetric if and only if its adjoint is self-dual : ϕ∗ i = ϕ. e e e ◦ A symmetric bilinear space (sbs or simply “space”) over A isa pair (P, ϕ) con- e sisting of a finitely generated projective A-module P and a unimodular symmetric bilinear form ϕ : P P A. Unimodular means that ϕ is an isomorphism and × −→ is the same as requiring that every A-linear map f : P A be of the form e −→ f(x) = ϕ(y,x) for some fixed y P . From now on any projective module over A ∈ will be tacitly assumed to be of finite rank and will be identified with its double 14
dual via the canonical isomorphism i, which will be omitted from the formulas. Thus a sbs over A can also be described as a pair (P,f) where f is an isomorphism
of P onto P ∗ that coincides with its dual: f = f ∗. When the explicit mention of ϕ is not necessary we shall denote (P, ϕ) by (P, ) and ϕ(x,y) by x y. The scalar · · x x is called the norm of x and an element a of A is said to be represented by · (P, ) if it is the norm of some x in P . · If P has a basis e ,....e over A, a symmetric bilinear form ϕ on P is { 1 n} determined by the symmetric n n matrixϕ ˆ with entries ϕ(e , e ) and ϕ is × i j unimodular if and only if detϕ ˆ is invertible in A. Ifϕ ˆ is a diagonal matrix
diag (a1,...,an), the space (P, ϕ) is denoted by
If (P,f) is a sbs over A and B is any A-algebra,
(P,f) B =(P B,f id ) ⊗A ⊗A ⊗A B
is a sbs over B : the space obtained from (P,f) by extending the scalars to B. In particular, for any prime ideal p of A, we can define the localization of (P, ) at p · as (P, ) Ap. We say that a sbs over B is extended from A if it can be obtained · ⊗A from a sbs over A by scalar extension. A symmetric bilinear form ϕ on P yields
a sbs (P, ϕ) if and only if, for every maximal ideal m of A the pair (Pm, ϕm) is a
sbs over Am.
For any projective A-module Q we define a sbs over A, called the hyperbolic space over Q as
H(Q)=(Q Q∗, χ), where χ(x f,y g) = f(y) + g(x). ⊕ ⊕ ⊕
We shall occasionally describe homomorphisms
f : M N M ′ N ′ ⊕ −→ ⊕ 15
α β between direct sums by a matrix γ δ , whose entries are the linear maps deter- mined by f(x,y) = α(x) + β(y),γ(x) + δ(y) .
With this notation, identifying (Q Q∗)∗ with Q∗ Q, we get ⊕ ⊕ 0 1 H(Q) = Q Q∗, . ⊕ 1 0 The orthogonal sum of two sbs is (P, ϕ) (Q,ψ)=(P Q, ϕ ψ), ⊥ ⊕ ⊥ where (ϕ ψ)(x y, w z) = ϕ(x, w) + ψ(y, z). ⊥ ⊕ ⊕ An isometry σ :(P, ϕ) (Q,ψ) is an A-isomorphism σ : P Q for −→ −→ which the following diagram commmutes: σ P Q −−−−→ ϕ ψ σ∗ P∗ Q∗ e ←−−−− e This condition is equivalent toy the requirementy that x y = σ(x) σ(y) for any · · x and y in P . In particular, if P = Q,(P, ϕ) and (P,ψ) are isometric if and
only if there exists an automorphism σ of P such that σ∗ψσ = ϕ. The group of isometries of (P, ) into itself is the orthogonal group of (P, ) We say that a sbs is · e· e hyperbolic if it is isometric to some H(P ). Given a sbs (P, ϕ) and a submodule N P we define the orthogonal of N as ⊂ the submodule N ⊥ of all x P such that x y = 0 for every y N. If we denote ∈ · ∈ by i the inclusion of N in P , N ⊥ can also be described by the exact sequence
i∗ϕ (5) 0 N ⊥ P N ∗. −→ −→ −−−−→ e Example 2.1. If T is a rank 2 projective A-module, its endomorphism algebra
E = EndAT carries a structure of sbs defined by
f g = (Trf)(Trg) Tr(fg) , · − 16
where Tr denotes the map “trace” from E to A. If A is local, T A2, E M (A) ≃ ≃ 2 and ab a′ b′ = ad′ + da′ bc′ cb′ , cd · c d − − ′ ′ hence in this case (E, ) is the space · 2 0 1 2 0 1 2 M2(A), = A , 1 0 A , 1− 0 H(A ). · ⊥ − ≃ Thus, for any A,(E, ) is a sbs over A. The normof f E is twice the determinant · ∈ of f. The orthogonal of the submodule generated by idT is the submodule of endomorphisms of trace zero.
Proposition 2.2. Let (P, ϕ) be a sbs over A and Q a submodule of P . Assume
that (Q, ϕ ) is a sbs. Then (Q⊥, ϕ ⊥ ) is a sbs and |Q |Q (P, ϕ)=(Q, ϕ ) (Q⊥, ϕ ⊥ ). |Q ⊥ |Q Proof. Consider the diagram i 0 Q⊥ P Q −→ −→ ϕ ←− i∗ P∗ Q∗ , ey −→ y where i is the inclusion of Q in P . By assumption i∗ϕi is an isomorphism, hence
i∗ϕ has a section and Q⊥ is a direct summand of P . Clearly P = Q + Q⊥ and e Q Q⊥ = (0) because, by assumption, (Q, ϕ Q) is a sbs. Hence P = Q Q⊥ , e∩ | ⊕ which immediately implies the assertion.
A sbs over A is said to be diagonalizable if it is isometric to a diagonal space
Note that if u1,...,un are invertible in A the spaces
2 2
Corollary 2.3. Let K be a field. Any anisotropic space over K is diagonalizable.
Proof. If v V is a nonzero vector, (Kv, ) is a sbs. By the proposition above, ∈ ·
(V, )=(Kv, ) (Kv)⊥, , · · ⊥ · and we conclude by induction on the dimension of V .
Corollary 2.4. Let K be a field of characteristic different from 2. Any sbs over K is diagonalizable.
Proof. Let v be a nonzero vector of V . If v v = 0 we proceed as above. If v v =0 · 6 · there exists a linear function on V mapping v to 1. This linear function is given by the scalar product with a fixed w V . Putting ∈ e = 1 (v + w) 1 (w w)v and e = 1 (v w) + 1 (w w)v 1 2 − 4 · 2 2 − 4 · we see that (Ke + Ke , ) is isometric to <1, 1>. Hence 1 2 · −
(V, )=(Ke + Ke , ) (V ′, ), · 1 2 · ⊥ · where V ′ =(Ke1 + Ke2)⊥. We conclude, as in Corollary 2.3, by induction on the dimension of V .
Proposition 2.5. Let (P, ) be a sbs over A and N a direct factor of P . Then ·
1 rank N + rank N ⊥ = rank P ,
2 N = N ⊥⊥ .
Proof. If N is a direct factor of P the restriction map P ∗ N ∗ is surjective −→ and therefore (5) becomes a split exact sequence
(6) 0 N ⊥ P N ∗ 0 . −→ −→ −→ −→
This shows that rank N ⊥ + rank N ∗ = rank P , but rank N = rank N ∗, whence
1 .
18
The sequence (6) shows that N ⊥ is also a direct factor of P and therefore, by
1 , rank N ⊥ + rank N ⊥⊥ = rank P , which implies that rank N ⊥⊥ = rank N.
Obviously N N ⊥⊥, hence N, being a direct factor of P , is a direct factor of ⊆ N ⊥⊥ and, having the same rank, coincides with it.
A sublagrangian of a sbs (P, ) is a direct factor L of P such that L L⊥. · ⊆ A lagrangian of (P, ) is a direct factor L of P that coincides with its orthogonal: · L = L⊥. For example, Q and Q∗ are both lagrangians of H(Q). A sbs is metabolic if it contains a lagrangian. In particular, hyperbolic spaces are metabolic. Note that a metabolic space has even rank because, by the proposition above, rank P =
rank L + rank L⊥ = 2rank L. A sublagrangian is a lagrangian if and only if its rank is half the rank of P .
We say that a sbs (P, ϕ) is even if ϕ can be written as ϕ = ψ + ψ∗ for
some A-linear homomorphism ψ : P P ∗. (We do not require that ψ be an −→ e e isomorphism). Clearly if 2 is invertible in A any sbs over A is even because we can 1 take ψ = 2 ϕ. If 2 is not invertible in A metabolic spaces need not be hyperbolic. For instance, the space S =<1, 1> is metabolic because its diagonal submodule e − L = (a,a) a A { | ∈ } is a lagrangian, but if 2A = A, S cannot be hyperbolic. In fact, for any Q, every 6 x H(Q) has norm x x 2A, whereas S contains elements of norm 1; hence S ∈ · ∈ and H(Q) cannot be isometric.
Proposition 2.6. Any even metabolic space is hyperbolic. In particular, if 2 is invertible in A every metabolic space is hyperbolic.
Proof. Let L be a lagrangian of (P, ϕ). By definition, L is a direct summand of P . We may therefore write a b P = L M and ϕ = , ⊕ b∗ c e 19 where
a Hom (L,L∗) , b Hom (M,L∗) and c Hom (M,M ∗). ∈ A ∈ A ∈ A
Since L = L⊥, a vanishes and b is an isomorphism. On the other hand, by assumption, c = d + d∗ for some d Hom (M,M ∗). The homomorphism σ : ∈ A L L∗ L M given by ⊕ −→ ⊕
1 1 1 b − db σ = ∗ − 0 − b 1 − 0 1 satisfies σ∗ϕσ = 1 0 and is therefore an isometry of H(L) onto (P, ϕ). The sete of isometry classes of symmetric bilinear spaces over A gives rise, with respect to the orthogonal sum as operation, to a commutative monoid ⊥ S(A).The set of isometry classes of metabolic spaces is a submonoid N(A) ⊂ S(A). We denote by W(A) the quotient S(A)/N(A). Officially it is a monoid but it will soon turn into a group (the Witt group of A) and a little bit later into a ring (the Witt ring of A).
In what follows, the same symbol will frequently be used to denote a sbs, its isometry class and, pushing the confusion to its logical conclusion, its class in the Witt group.
It is clear that any ring homomorphism A B induces, via the extension −→ of scalars from A to B, a homomorphism W(A) W(B) and that W behaves −→ functorially.
We shall write (P, ) (Q, ) and say that (P, ) and (Q, ) are Witt equivalent · ∼ · · · when (P, ) and (Q, ) represent the same element of W(A). By the very definition · · of W(A) this means that there exist two metabolic spaces (M, ) and (N, ) such · · that (P, ) (M, ) and (Q, ) (N, ) are isometric. · ⊥ · · ⊥ · 20
Proposition 2.7. Let (P, ϕ) be a sbs over A and L a sublagrangian of (P, ϕ).
1 ϕ induces a structure of sbs on L⊥/L.
2 (P, ϕ) (L⊥/L,ϕ) is metabolic. − ⊥
Proof. Since L L⊥, for any x,y L⊥ the scalar ϕ(x,y) only depends on the ⊆ ∈ classes of x and y modulo L, hence (L⊥/L,ϕ) is well-defined. We must show that
the adjoint of ϕ is an isomorphism of L⊥/L onto its dual. Injectivity is clear
(but superfluous, being a consequence of surjectivity) because if ϕ(x,L⊥)=0
for some x L⊥, then x L⊥⊥ = L. To show surjectivity consider any linear ∈ ∈ map f : L⊥/L A. Since L⊥ is a direct factor of P (as we observed in the −→ proof of Proposition 2.5) f extends to a g : P A, which must be of the form −→ g(x) = ϕ(u,x) for some fixed u P . Since f(L)=0, u must lie in L⊥ and thus ∈ the class of u in L⊥/L induces f. To prove the second assertion denote by x the class of any x in P/L and define the “diagonal” map
∆: L⊥ P L⊥/L −→ ⊕ by ∆(x)=(x, x). The sequence ∆ π 0 L⊥ P L⊥/L P/L 0 , −→ −−−−→ ⊕ −−−−→ −→ where π(x, y) = x y, is easily seen to be exact. Since L is a direct summand − of P the quotientP/L is projective and hence π has a section. This shows that
∆(L⊥) is a direct factor of P L⊥/L. On the other hand, if (a, b) P L⊥/L is ⊕ ∈ ⊕ orthogonal (with respect to the bilinear form ϕ ϕ) to every (x, x) ∆(L⊥), − ⊥ ∈ ϕ(a,x) ϕ(b,x)= 0 for any x L⊥, hence a b L⊥⊥ = L and (a, b)=(a, a) = − ∈ − ∈ ∆(a). This shows that ∆(L⊥) is a lagrangian of (P, ϕ) (L⊥/L,ϕ). − ⊥ Corollary 2.8. The orthogonal sum (P, ϕ) (P, ϕ) is metabolic. In particular ⊥ − W(A) is an abelian group in which the class of (P, ϕ) is the opposite of the class − of (P, ϕ). 21
Corollary 2.9. If L is a sublagrangian of (P, ϕ) we have (P, ϕ) (L⊥/L,ϕ). ∼ As we mentioned earlier, W(A) has the structure of a commutative ring: define a multiplication on S(A) by ⊗ (P, ϕ) (Q,ψ)=(P Q, ϕ ψ), ⊗ ⊗ ⊗ where ϕ ψ is defined by ⊗
(ϕ ψ)(x y,x′ y′) = ϕ(x,x′)ψ(y,y′). ⊗ ⊗ ⊗ If (Q,ψ) is metabolic with a lagrangian L,(P, ϕ) (Q,ψ) is also metabolic, with ⊗ a lagrangian P L. Hence induces a multiplication on W(A) and it is easily ⊗ ⊗ checked that the additive goup W(A) becomes a commutative ring with <1> as unit element.
We have already observed that a sbs (P, ) represents 0 in W(A) if and only · if there exist metabolic spaces (M, ) and (N, ) such that (P, ) (M, ) (N, ). · · · ⊥ · ≃ · This does not imply, in general, that (P, ) is itself metabolic, although this is · indeed the case if A is a field or a Dedekind domain, as we shall see later on.
Example 2.10. Let T be a projective module of rank 2 over A, E = EndAT its algebra of endomorphisms and (E, ) the sbs defined in Example 2.1. The · A-module E can be identified with T T ∗ via the isomorphism that maps t f ⊗ ⊗ to the endomorphism tf : T T defined by (tf)(x) = f(x)t. Thus there exists −→ a canonical isomorphism
2 2 2 2 2 Φ : End T T T ∗ (T T ∗) = T T ∗ T T ∗ . A ⊕ ⊕ −→ ⊕ ⊗ ⊕ ⊕ V V V 4 V V For any projective module P of rank 4 such that P = A (in particular for T T ∗) 2 ⊕ P has a natural structure of sbs defined by (pV p′) (q q′) = p p′ q q′. It ∧ · ∧ ∧ ∧ ∧ turnsV out that Φ is actually an isometry between
2 2 (E, ) H T and (T T ∗), . · ⊥ ⊕ · V V 22
2 4 3 If T T ∗ = A , the space (T T ∗), is just H(A ) and hence (E, ) 0. But ⊗ ⊕ · · ∼ in general, as we are going V to see, (E, ) is not metabolic. · Let A be the affine coordinate ring of the real sphere S2,
A = R[X,Y,Z]/(X2 + Y 2 + Z2 1) = R[x,y,z] , −
and T the projective A-module corresponding to the tangent bundle of S2: T = ker π, where π : A3 A is the map sending (a,b,c) to ax + by + cz,[39, p.269]. −→ We claim that in this case (E, ) is not metabolic. In fact, by Proposition 2.6, if · it were metabolic it would be hyperbolic, isometric to some H(Q), Q a projective module. Since E contains an element of norm 1 (for instance 1 id ), H(Q) would √2 T also contain one, say (u, v) Q Q∗. But then ( u, v) would have norm 1 and ∈ ⊕ − − its inverse image in E would be an endomorphism of T of determinant 1. Since − the tangent bundle to S2 is topologically indecomposable this would contradict the following proposition.
Proposition 2.11. Let be an indecomposable real vector bundle of rank 2 T over a compact space S. No endomorphism of has determinant 1. T − Proof. Suppose f End has determinant 1. Since the characteristic polyno- ∈ T − mial of f is t2 Tr(f)t + det(f) = t2 Tr(f)t 1 , − − − f has, on every fibre, one positive and one negative eigenvalue. Every fibre splits in a unique way as a direct sum of two one-dimensional eigenspaces and, this splitting being continuous on S, splits into the direct sum of two line bundles. T 23 3. The Witt group and the theorem of Tsen
Tout a et´ e´ dit cent fois Et beaucoup mieux que par moi. Boris Vian
In this section we give some examples of Ci-fields and compute the Witt group of a C1-field.In what follows K denotes a field, possibly of characteristic 2.
For any positive integer n and any space (V, ) over K we denote by n(V, ) · · the orthogonal sum of n copies of (V, ) and by (V, )n the tensor product over K · · of n copies of (V, ). To simplify the notation we occasionaly write ϕ instead of · (V, ϕ) and drop the when we write the tensor product of two spaces. ⊗ Proposition 3.1. Every element of W(K) is represented by an anisotropic space.
Proof. Let L be a sublagrangian of maximal dimension of a given space (V, ). · By Corollary 2.9, (V, ) (L⊥/L, ) and the latter is anisotropic. · ∼ · Proposition 3.2. Let L and M be sublagrangians of (V, ϕ). Then
(S, )=(L⊥/L,ϕ) (M ⊥/M, ϕ) · ⊥ − is metabolic.
Proof. As in the proof of Proposition 2.7 we see that the image of the “diagonal” map
∆:(L⊥ M ⊥)/(L M) L⊥/L M ⊥/M ∩ ∩ −→ ⊕ is a lagrangian of (S, ). The fact that K is a field is used here only to make sure · that the image of ∆ is a direct factor of S. 24
Corollary 3.3. If (V, ϕ) 0 then (V, ϕ) is metabolic. ∼ Proof. By assumption there exists a metabolic space (W, ψ) such that
(S, χ)=(V, ϕ) (W, ψ) ⊥
is metabolic. Let L be a lagrangian of (S, χ) and M a lagrangian of (W, ψ). Clearly M is a sublagrangian of (S, χ), hence, by the proposition above,
(L⊥/L, χ) (M ⊥/M, χ) − ⊥
is metabolic. (The exponent denotes the orthogonal complement in S). But ⊥ L⊥/L =0 and (M ⊥/M, χ)=(V, ϕ).
Corollary 3.4. Any two anisotropic representatives of an element of W(K) are isometric.
Proof. Suppose that (V, ϕ) (W, ψ) and that both are anisotropic. By Corol- ∼ lary 3.3, (S, χ)=(V, ϕ) (W, ψ) is metabolic and has therefore a lagrangian ⊥ − L. Since every element of L is isotropic, V L = W L = (0). Hence the two ∩ ∩ orthogonal projections of S onto V and W induce injections f : L V and −→ g : L W . On the other hand 2 dim L = dim S = dim V + dim W implies −→ dim L = dim V = dim W , and therefore f and g are isomorphisms. It is easy to 1 check that Φ = gf − : V W is an isometry. −→ When K is not of characteristic 2 the above corollaries are consequences of the following cancellation theorem of Witt [41].
Theorem 3.5. Let K be a field of characteristic different from 2 and (V , ), 1 · (V , ), (W, ) spaces over K. If (V , ) (W, ) and (V , ) (W, ) are isometric, 2 · · 1 · ⊥ · 2 · ⊥ · (V , ) and (V , ) are isometric too. 1 · 2 · Proof. Since K is not of characteristic 2, by Corollary2.4 (W, ) is an orthogo- · nal sum of one-dimensional spaces. By induction on dim W it suffices to prove 25 the theorem for a one-dimensional W . Put (V, )=(V , ) (W, ) and let σ be · 2 · ⊥ · an isometry of (V , ) (W, ) onto (V, ). It suffices to find an orthogonal trans- 1 · ⊥ · · formation τ of (V, ) such that τσ(W ) = W , because τσ(V ) must then be the · 1 orthogonal complement of τσ(W ) in V , which is nothing but V2. To construct τ we introduce a special kind of orthogonal transformations: the reflections. Let (V, ) be any space over K and v V an anisotropic vector. Put, · ∈ for any x V , ∈ x v τ (x) = x 2 · v . v − v v · It can be easily checked that τ is an isometry, that τ (v) = v and that the v v − restriction of τv to the hyperplane (Kv)⊥ is the identity.
Let now w be a generator of W and w′ = σ(w) . Note that w w = w′ w′. If · · v = w w′ is anisotropic the reflection τ will exchange w and w′ and we can take − v τ = τ . If w w′ is isotropic, w w + w′ w′ =2w w′, hence (w + w′) (w + w′) = v − · · · · 2(w w + w w′)=4w w = 0. The vector v′ = w + w′ is anisotropic and we can · · · 6 take τ = τ ′ . − v Corollary 3.6. Let K be a field of characteristic different from 2. If two spaces over K are equivalent in the Witt group and have the same dimension they are isometric.
Proof. If (V , ) and (V , ) are equivalent there exist hyperbolic spaces H(Km) 1 · 2 · and H(Kn) such that
(V , ) H(Km) (V , ) H(Kn) . 1 · ⊥ ≃ 2 · ⊥ If V and V have the same dimension we can apply Theorem 3.5 with (W, ) = 1 2 · H(Km)=H(Kn).
It is equally easy to deduce from Witt’s cancellation theorem its two corol- laries 3.3 and 3.4 in characteristic different from 2. 26
Note that Witt’s theorem is false in characteristic 2. For any ring A the free spaces of rank 3 defined by the symmetric matrices
1 0 0 0 1 0 α = 0 1 0 and β = 1 0 0 0− 0 1 0 0 1 are isometric because if α describes a scalar product with respect to a basis
(e1, e2, e3), β describes the same scalar product with respect to the basis
(e + e , e e , e + e + e ). 1 2 − 2 − 3 1 2 3
But we have seen that if 2 is not invertible in A,
1 0 0 1 and 0 1 1 0 − define non-isometric spaces.
We observe that the reflections τv introduced above generate the orthogonal group of (V, ). More precisely , according to a theorem proved by Cartan for · K = R or C [6, Ch.1, 10] and generalized by Dieudonn´e[12, Ch.3, 10] to § § arbitrary fields, if (V, ) is a sbs of dimension n over a field K of characteristic · different from 2 every isometry σ O(V, ) can be expressed as the product of at ∈ · most n symmetries (see also [23, Ch.1, Appendix]).
Any element ξ W(K) may be represented by several spaces (V, ) of dif- ∈ · ferent dimensions but since metabolic spaces are even-dimensional the parity of dim V only depends on ξ and is called the rank of ξ. Mapping every ξ W(K) ∈ to its rank yields a ring homomorphism ρ : W(K) Z/2Z. Its kernel I(K) is −→ the ideal of even rank spaces. Note that I(K) is additively generated by spaces of the type
Given a space (V, ϕ) and a basis (e , , e ) of V , we may describe ϕ by the 1 ··· n symmetric matrixϕ ˆ =(e e ). Changing the basis will change the determinant i · j ofϕ ˆ by a square, hence the class of detϕ ˆ in K/˙ K˙ 2 only depends on the isometry class of (V, ϕ). We define the discriminant of (V, ϕ) as
n n− ( 1) 2 d(V, ϕ) =class of ( 1) 2 detϕ ˆ in K/˙ K˙ . −
n(n−1) The factor ( 1) 2 owes its presence to the desire of making the discriminant − of metabolic spaces equal to (the class of) 1. Indeed, the matrix of a metabolic space of dimension 2n is, with respect to a suitable basis,
0 1 , 1 c
n(n−1) where c M (K), and its determinant is precisely ( 1) 2 . ∈ n − If (V, ) and (W, ) are two spaces over K of dimension m and n respectively, · · we have d (V, ) (W, ) =( 1)mnd(V, ) d(W, ); · ⊥ · − · · hence, if one of them is even-dimensional,
d (V, ) (W, ) = d(V, ) d(W, ), · ⊥ · · · and if (W, ) is metabolic, ·
d (V, ) (W, ) = d(V, ). · ⊥ · · All this shows that d induces a map (the discriminant)
d:W(K) K/˙ K˙ 2 −→ 28
and that d:I(K) K/˙ K˙ 2 −→ is a group homomorphism.
Proposition 3.7. d induces a group isomorphism
δ : I(K)/I(K)2 K/˙ K˙ 2. −→ Proof. Since d(< 1, a >< 1, b >) = 1, δ is well defined. Put, for any a K˙ , ∈ ǫ(a) = <1, a>. Then ǫ induces a map from K/˙ K˙ 2 to I(K)/I(K)2 which is a − set-theoretical inverse of δ. But
ǫ(a) + ǫ(b) ǫ(ab) =<1, a> + <1, b> <1, ab> − − − − − =<1, a><1, b> I(K)2, − − ∈ hence ǫ is a homomorphism and δ is invertible.
In order to compute the Witt group of some fields we introduce, following
Lang [24], the notion of a Ci-field. A field K is said to be Ci if it satisfies the following condition: every homogeneous non-constant form F (X ,...,X ) 1 n ∈ i K[X1,...,Xn] of degree d has a non-trivial zero provided n>d . A field is
C0 if and only if it is algebraically closed. C1-fields are also said to be quasi algebraically closed. The next theorem was conjectured by E. Artin and proved by Chevalley [9 (“M.” for “Monsieur”, not for “Michael”!)].
Theorem 3.8. Finite fields are C1.
e Proof. Let K be a field with q = p elements, p a prime. Let F (X1,...,Xn)bea homogeneous polynomial of degree d in n indeterminates and assume that n>d. We want to show the existence of a ξ = (ξ ,...,ξ ) Kn which is = (0,..., 0) 1 n ∈ 6 and satisfies F (ξ) = 0. Since K˙ is a group of order q 1 we have −
q 1 1 if F (ξ) =0, F (ξ) − = 6 0 if F (ξ)=0, 29
q 1 n hence ξ Kn F (ξ) − is the reduction modulo p of the number N of ξ K for ∈ ∈ which F (ξ) = 0. The polynomial P 6
q 1 i1 in F (X ,...,X ) − = a X X 1 n i1,...,in 1 ··· n X is homogeneous of degree d(q 1), hence in each monomial Xi1 Xin − 1 ··· n (7) i + + i = d(q 1) q 1 i1 in To evaluate F (ξ) − it suffices to evaluate each ξ ξ . If one of the ξ ξ 1 ··· n exponents isP zero the sum decomposes into q equal termsP and is therefore zero. Suppose that each one of the exponents is positive. Then, in view of (7), at least one of them, say i , is less than q 1. Let us write 1 − ξi1 ξin = ξi1 ξi2 ξin , 1 ··· n 1 2 ··· n ξ ξ ξ′ X X1 X where ξ =(ξ1,ξ′). If ζ generates K˙ , q 2 − i1 i1 i1j ξ1 = ξ1 = ζ , ξ1 K ξ K˙ j=0 X∈ X1∈ X and since ζi1 = 1, the last sum is 6 i q 1 (ζ 1 ) − 1 − =0. ζi1 1 − This shows that N is a multiple of p. But then the number Z of zeros of F is also a multiple of p because N + Z = qn. On the other hand, since F (0,..., 0) = 0, Z = 0, hence there exist at least p 1 non-trivial zeros of F in Kn. 6 − A large class of Ci-fields is described by the following result of Lang [24]. Theorem 3.9. If k is algebraically closed every algebraic extension K of the rational function field k(t1,...,tm) is a Cm-field. For the proof we need the following theorem. 30 Theorem 3.10. If K is Ci any algebraic extension of K is also Ci. Proof. Let L be algebraic over K and F L[X ,...,X ]. To show that F has ∈ 1 n n n a non-trivial zero in L it suffices to show that F has a non-trivial zero in L0 , where L0 is a subfield of L containing K and all the coefficients of F . Hence to prove the theorem we may assume that L is a finite extension of K of degree, say, m. Suppose that the theorem is false and that for some n>di there exists a ho- mogeneous polynomial F (X ,...,X ) L[X ,...,X ] of degree d without zeros 1 n ∈ 1 n in Ln 0 . Let G(X ,...,X ) K[X ,...,X ] be the norm of F . G is the \{ } 1 n ∈ 1 n determinant of the K[X1,...,Xn]-linear map induced on the free K[X1,...,Xn]- module L[X1,...,Xn] by the multiplication by F . Clearly G is homogeneous of degree dm. Since for any ξ Kn 0 F (ξ) = 0, the norm G(ξ) of F (ξ) is also ∈ \{ } 6 different from zero. We now define a sequence of forms of degrees d,d2,d3,... in n,n2,n3,... indeterminates by F1(X1,...,Xn) = F (X1,...,Xn), F2(X1,...,Xn2 ) = F1(F (X1,...,Xn),...,F (Xn2 n+1,...,Xn2 )), − and in general Fν(X1,...,Xnν ) = Fν 1(F (X1,...,Xn),...,F (Xnν n+1,...,Xnν )). − − We claim that the only zero of Fν is the trivial one. This is true by assumption for ν =1. If it is true for ν 1 then F (ξ ,...,ξ ν ) = 0 implies − ν 1 n F (ξ1,...,ξn) = ... = F (ξnν n+1,...,ξnν )=0 − and hence ξ1 = ... = ξnν = 0. Consider now, for each ν, the norm Gν of Fν . As we have already observed, it is a homogeneous polynomial of degree dνm in nν indeterminates, without non-trivial zeros. But since n>di, taking ν sufficiently large we get nν > (dνm)i and for 31 such a ν, Gν does have a non-trivial zero because K is Ci. Hence F itself must have a non-trivial zero. In view of Theorem 3.10 it suffices to prove Theorem 3.9 in the case K = k(t1,...,tm). This we do by induction on m, but to make the induction step with more nimbleness it is convenient to state the result in a more general form. Theorem 3.11. Let k be an algebraically closed field and K = k(t1,...,tm) the field of rational functions in m indeterminates over k. Let F1,...,Fr be r homogeneous non-constant polynomials in X1,...,Xn, of degree at most d. If m n n>rd , the polynomials F1,...,Fr have a common non-trivial zero in K . Proof. If m = 0, K coincides with k and the theorem is a consequence of Hilbert’s Nullstellensatz. In fact, assuming that (0,..., 0) is the unique common zero of F ,...,F , by the Nullstellensatz [19, 3] there exists an exponent N 1 such 1 r § ≥ N that each Xi belongs to the ideal a of B = k[X1,...,Xn] generated by F1,...,Fr. Let A = k[F1,...,Fr] be the subalgebra of B generated by F1,...,Fr, M the set of monomials Xi1 Xin for which 0 i ,...,i < N and C the A-submodule 1 ··· n ≤ 1 n of B generated by M. We claim that B = C. Clearly every f B of degree ∈ less than N is in C. Take, if possible, an f B C of minimal degree. We may ∈ \ assume that f is a single monomial Xj1 Xjn . At least one of the exponents, 1 ··· n say j , must be N. Since XN a we may write XN = G F . The F ,...,F ν ≥ ν ∈ ν i i 1 r being homogeneous, we may choose each G homogeneous of degree N degF . i P − i Then r j1 jn j1 jν N jn X X = X X − X G F . 1 ··· n 1 ··· ν ··· n i i i=1 X j1 jν N jn Since degF > 0, the degree of g = X X − X G is less than that of i i 1 ··· ν ··· n i f and therefore g C, whence f C. i ∈ ∈ Since B is a finite module over A, its field of fractions k(X1,...,Xn) is a 32 finite extension of k(F1,...,Fr). In particular these two fields have the same transcendence degree over k, which is impossible unless r n. Thus, if n>r, ≥ F1,...,Fr must have a non-trivial common zero. Assume now that m 1 and that K′ = k(t1,...,tm 1) is a Cm 1-field. Multi- ≥ − − plying F1,...,Fr by a suitable element of K˙ we may assume that each Fi is a polynomial in t , of degree h for some integer h. To find a point in Kn 0 m ≤ \{ } at which F1,...,Fr vanish we make the substitution N ν Xj = Xjνtm, ν=0 X where N is an integer to be chosen later. Expanding the Fi’s as polynomials in tm we get Nd+h ν Fi(X1,...,Xn) = Giν (X10,...,XnN )tm, ν=0 X where G K′[X ,...,X ] are non-constant forms of degree d. iν ∈ 10 nN ≤ It will now suffice to show the existence of a common non-trivial zero of the Giν ’s, with coordinates in K′. The number of the Giν ’s is r(Nd+h+1) and the number of variables Xjν is n(N + 1). Since K′ is a Cm 1-field it will be enough to choose − N such that m 1 n(N + 1) >r(Nd + h + 1)d − . This inequality is certainly satisfied if N is large enough because by assumption n>rdm. As a special case of Theorem 3.9 we record Tsen’s theorem [40, Satz 8]. Corollary 3.12. If k is algebraically closed every algebraic extension of k(t) is quasi algebraically closed. A special case of Tsen’s theorem is the following result, proved in geometrical disguise by M. Noether [32] for K = C(t) and by Enriques [13] in general. 33 Corollary 3.13. Let K be the rational function field of a complex curve. Every conic over K has a K-rational point. 2 Proof. The equation of a conic in PK is a form of degree 2 in 3 indeterminates and 3 is larger than 21. We now determine the Witt group of a C1-field. Theorem 3.14. Let K be a quasi algebraically closed field. There is an exact sequence ρ 0 K/˙ K˙ 2 W(K) Z/2Z 0 −→ −→ −→ −→ which splits if and only if 1 is a square in K. − Proof. We know that I(K) is the kernel of ρ. Let us prove that, under the assumption of the theorem, the discriminant homomorphism δ : I(K) K/˙ K˙ 2 −→ is an isomorphism. 1 Since K is C1 every quadratic form in n > 2 = 2 variables has a non-trivial zero, hence any even-dimensional anisotropic space (V, ) over K must be of dimension · 2. Since anisotropic spaces are diagonalizable (Corollary 2.3), we may assume that (V, ) = If 1 is a square in K, the class of < 1 > is of order 2 in W(K) and yields a − splitting of ρ. Conversely, if the sequence splits there is an element ξ of order 2 and of odd rank in W(K). An anisotropic representative of ξ can only be of dimension 1, hence there is an a K˙ such that is metabolic. This ∈ ⊥ means that aX2 + aY 2 has a non-trivial zero (x,y) in K and hence 1=(x/y)2. − Corollary 3.15. If K is finite field with q elements we have 34 Z/2Z Z/2Z if q 1 (mod 4), ⊕ ≡ W(K) = Z/4Z if q 3 (mod 4), ≡ Z/2Z if q 0 (mod 2). ≡ 35 4. The spectrum of the Witt ring Practice juggling and you practice simplifying complexity. Dave Finnigan This section contains a short exposition of the most elementary properties of the Witt ring of a field, along the lines of the elegant paper of Lewis [25]. As the reader will notice, as far as the main result of the last section is concerned, most of the content of this chapter is totally irrelevant (and, as Perry Mason would add, incompetent and immaterial). A field K is said to be formally real if 1 is not a sum of squares in K. − For instance, any subfield of R is formally real. (Some fields may look rather imaginary and still be formally real, like Q 21/3(1 + i31/2) .) Equivalently, we 2 2 may say that K is formally real if, for any a1,...,an in K, a1 + + an = 0 ··· implies a = = a = 0: in other words, if < 1,..., 1 > is anisotropic. A 1 ··· n formally real field must be of characteristic zero because 12 + +12 = 0. ··· 6 Let Zm be the kernel of the canonical map Z W(K). If K is formally real, −→ m = 0 because if < 1 > ... < 1 > were 0 in W(K) it would, in particular, ⊥ ⊥ ∼ be isotropic. If K is a finite field, according to Corollary 3.15, m is 2 or 4. We shall prove that m is either 0 or a power of 2, but for the moment let it be any non-negative integer. Proposition 4.1. Define, for any positive integer n, a polynomial F Z[X] by n ∈ (X2 12)(X2 32) (X2 n2) if n is odd, Fn = − − ··· − X(X2 22)(X2 42) (X2 n2) if n is even. − − ··· − 36 Any ξ W(K) represented by an n-dimensional space (V, ) satisfies F (ξ)=0. ∈ · n Proof. If V is of dimension 1 the assertion is obvious. Assuming therefore that n 2 we proceed by induction. ≥ For n 3, Fn 2 divides Fn, hence we may replace (V, ) by an equivalent ≥ − · anisotropic space and assume that it is diagonal: (V, ) =. The · 1 ··· n assertion F (ξ) = 0 is equivalent to F ( ξ)= 0 for any a K˙ . Scaling (V, ) n n ∈ · by a we may assume that (V, ) =< 1 > , so that ξ = 1 + η, 1 · ⊥ 2 ··· n where η is represented by an (n 1)-dimensional space Fn 1 and Fn 1(η) = 0. − − − Inspecting the roots of F (among other possibilities) we see that for any n 2, n ≥ Fn(X)=(X + n)Fn 1(X 1) − − Thus Fn(ξ)=(ξ + n)Fn 1(η)=0. − Corollary 4.2. The ring W(K) is an integral extension of Z/mZ. Its (Krull–) dimension is 1 if m =0 and 0 if m =0. 6 Proposition 4.3. I(K) is the unique prime ideal of W(K) that contains 2 <1>. Proof. Let p SpecW(K). If <1> + <1> p, for any a K˙ ∈ ∈ ∈ and 0=( <1>)( + <1>) ( + <1>)2 (mod p), − ≡ hence < 1, a > p for any a K˙ . This proves that I(K) p. Since I(K) is a ∈ ∈ ⊆ maximal ideal of W(K) it coincides with p. We recall that an ordering of K is the assignment of a subset P of K, called the set of positive elements of K, which satisfies the following properties: 37 1 P P = K˙ ∪ − 2 P P = ∩ − ∅ 3 P P P · ⊆ 4 P + P P ⊆ If K contains at least one such set we say that K is orderable. Given such a P we can define on K a structure of (totally) ordered set by a By 1 , for any a K˙ , a P or a P, and in both cases a2 = aa =( a)( a) ∈ ∈ − ∈ − − ∈ P by 3 . Hence, for arbitrary a ,...,a not all 0, a2 + + a2 P. In particular 1 n 1 ··· n ∈ a2 + + a2 = 0: every orderable field is formally real. 1 ··· n 6 It is easy to see that the only orderings of Q and R are the usual ones. There are fields with any prescribed number of orderings. For instance if f Q[X] ∈ is irreducible and has exactly n real zeros, the field K = Q[X]/(f) has exactly n orderings, obtained from the n different imbeddings of K into R. This can be shown using standard results on ordered fields [22, Ch. 8]. Some fields have infinitely many orderings. For instance, given an ordered field K and any a K we can define an ordering on the rational function field K(t) ∈ for which t is “infinitesimally larger” than a. We write every f K˙(t) as (t ∈ − a)mg with m Z and g neither zero nor infinity at t = a, and then declare ∈ that f is positive if and only if g(a) is positive in K. Similarly we can make t “infinitesimally smaller” than a, or larger than any a K, or smaller than any ∈ a K. ∈ Proposition 4.4. Every prime ideal p = I(K) gives rise to an ordering P of K 6 defined by P = P(p) = a K˙ <1> p { ∈ | − ∈ } 38 Proof. We must check the four properties of an ordering. Since, for any a K˙ , ∈ (<1> )(<1> + )=0 p, − ∈ if a / P then a P, hence P satisfies 1 . ∈ − ∈ If a and a were both in P we would have − (<1> )+(<1> + ) =<1> + <1> p − ∈ and this would imply, by Proposition 4.3, that p = I(K), against the assumption. Hence P satisfies 2 . Suppose that a and b are in P. Then hence which proves 3 . Further, a + b = 0, otherwise we would have a P and a P, 6 ∈ − ∈ contradicting 2 , which is already proved. Since for some c K˙ . From ∈ and the preceding equation we get <1> + <1> <1> Corollary 4.5. If K does not admit any ordering, W(K) is a local ring of di- mension zero, I(K) is its unique prime ideal and 2nW(K)=0 for some natural number n. In particular, a formally real field is always orderable. Proof. By the proposition above, I(K) is the unique prime ideal of W(K), hence W(K) is local and zero-dimensional. Its nilradical is I(K) and 2 < 1 > I(K), ∈ hence 2n <1>= 0 for some n. A signature of K is a (unitary) ring homomorphism σ : W(K) Z. −→ Given an ordering P of K we can associate to P a signature σP in the following way. Choose for any ξ W(K) a representative (V, ). Since K is of characteristic ∈ · zero, (V, ) . Define σP(V, ) = p q, where p is the number of · ≃ 1 n · − a ’s in P and q the number of those in P. The integers p and q do not depend i − on the chosen diagonalization. If (V, ) and p′ (q′) is the number · ≃ 1 n of positive (negative) bi’s, then V contains a subspace W ′ of dimension p′ whose nonzero vectors have all positive norm, and a subspace W of dimension q whose nonzero vectors have all negative norm. If p′ p then q q′ but since W W ′ = 0, ≥ ≥ ∩ p′ + q n = p + q = p′ + q′, hence p′ = p and q′ = q. Clearly, modifying (V, ) ≤ · by a hyperbolic space does not alter p q, hence σP(V, ) only depends on ξ. It − · is easily checked that σP is a surjective ring homomorphism of W(K) onto Z. The kernel of any signature σ is a prime ideal p of W(K) and since the composition σ Z W(K) Z −→ −→ 40 is the identity of Z, Z injects into W(K) (which we already knew) and p Z = 0. ∩ Proposition 4.6. Let K be a formally real field. Mapping σ to ker σ, p to P(p) and P to σP yields bijections F, G and H between the set Σ of signatures, the set P of prime ideals lying above (0) Z and the set O of orderings . ⊂ Proof. A signature σ is determined by its kernel. In fact, given ker σ with (ker σ) Z = (0), for any ξ W(K) there exists a unique integer n such that ∩ ∈ ξ n <1> ker σ, and σ(ξ) is then equal to this n. This shows that F :Σ P − ∈ −→ is injective. A prime p such that p Z = (0) defines, as we have seen, an ordering P(p). ∩ Let σ be the corresponding signature. A class a ,...,a P(p) and a ,...,a P(p). 1 m ∈ m+1 n ∈ − This means that and ... <1> (mod p), m+1 ≡ ≡ n ≡ − hence 0 (mod p); 1 n ≡ in other words, ker σ p. Since ker σ and p are both prime ideals intersecting Z ⊆ in (0) and since W(K) is integral over Z, by the theorem of Cohen–Seidenberg [29, 5.E, Theorem 5] ker σ = p. This shows that the composite map F H G ◦ ◦ is the identity of P; in particular F is surjective (hence an isomorphism), H is surjective and G is injective. 41 Given two different orderings of K, P and P′, for any element a P P′ we have ∈ \ σP(< a >) = 1 and σP′ (< a >) = 1. This shows that σP = σP′ and proves the − 6 injectivity of H. Thus H, and hence G, are isomorphisms. Proposition 4.7. let K be a formally real field. The set of prime ideals of W(K) consists of all p P, the maximal ideal I(K) and the maximal ideals p + W(K)p ∈ where p P and p is an odd rational prime. Every p P is contained in I(K). ∈ ∈ The set of minimal prime ideals of W(K) is precisely P. Proof. A prime ideal q different from I(K) defines an ordering of K and hence a signature σ for which, by the proof of proposition 4.6, ker σ q. If ker σ = q, ⊆ 6 q is maximal and since W(K)/ker σ Z we must have W(K)/q Z/pZ for ≃ ≃ some rational prime p. Hence q = ker σ + pW(K). Since q = I(K), in view of 6 Proposition 4.3 the prime p must be odd. By Proposition 4.6 every p P is the kernel of a signature and only even- ∈ dimensional spaces can have signature 0, hence p I(K). ⊂ Finally, by the theorem of Cohen–Seidenberg the minimal prime ideals of W(K) are precisely those lying above (0). Proposition 4.8. Let K be a formally real field. The torsion of W(K) is its nilradical . Proof. Let ξ be an element of W(K) and m a positive integer such that mξ = 0. Since no p P contains m, ξ belongs to every p and is nilpotent by [29, 1.E]. ∈ Conversely, assume that ξm = 0 for some positive integer m. Clearly ξ I(K). ∈ By Proposition 4.1 there exists an even n such that F (ξ) = ξ(ξ2 22) (ξ2 n2)=0. n − ··· − Expanding the product we get (8) 22 n2ξ = ξ3G(ξ), ··· 42 where G is some polynomial with integral coefficients. From (8) we get, putting N =22 n2, ··· N 2ξ = N Nξ = Nξ3G(ξ) = Nξ ξ2G(ξ) = ξ5G(ξ)2 , · · N 3ξ = N N 2ξ = N ξ5G(ξ)2 = Nξ ξ4G(ξ)2 = ξ7G(ξ)3,... · · · which shows that ξ is annihilated by some power of N. The next theorem is a well-known result of Pfister [34, Satz 22]. Theorem 4.9. Let K be an arbitrary field. Every torsion element of W(K) is annihilated by a power of 2. Before going into the proof we must juggle a little with Pfister forms. For any a =(a ,...,a ) Kn we put 1 n ∈ ψ = and π =<1,a > <1,a >. a 1 n a 1 ⊗···⊗ n The symbol ǫ will always denote a vector (ǫ ,...,ǫ ) with ǫ = 1 for every i. 1 n i ± For a =(a1,...,an) and b =(b1,...,bn) we put ab =(a1b1,...,anbn). Lemma 4.10. For any K and any n we have a π2 = 2nπ , a a b ψaπaǫ = ψǫπaǫ, c 2nψ = ψ π . a ǫ ǫ aǫ Proof. If n =P 1, a and b are obvious and c is easily checked. We prove them by induction on n, putting a =(a1,...,an)=(a′,an) and ǫ =(ǫ1,...,ǫn)=(ǫ′, ǫn), so that ψ = ψ ′ and π = π ′ <1,a > a a ⊥ n a a ⊗ n 43 and similarly for ǫ. 2 2 2 n 1 n a : π = π ′ <1,a > =2 − π ′ (2 <1,a >)=2 π . a a n a n a b : ψaπaǫ = ψa′ πaǫ+ = ψa′ πa′ǫ′ <1,anǫn > +πa′ǫ′ = ψǫ′ πa′ǫ′ <1,anǫn > + <ǫn >πa′ǫ′ <1,anǫn > = ψǫπaǫ. c : ψ π = ′ (ψ ′ + <ǫ >)(π ′ ′ <1,a ǫ >) ǫ ǫ aǫ ǫn ǫ ǫ n a ǫ n n ′ ′ ′ ′ ′ P = Pǫ′ (Pψǫ + <1>)(πa ǫ + + ′ (ψ ′ <1>)(π ′ ′ π ′ ′ ) P ǫ ǫ − a ǫ − n a ǫ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ = Pǫ′ ψǫ πa ǫ + ǫ′ πa ǫ + + ′ ψ ′ π ′ ′ ′ π ′ ′ ′ ψ ′ π ′ ′ + ′ π ′ ′ P ǫ ǫ a ǫ −P ǫ a ǫ − n P ǫ ǫ a ǫ n P ǫ a ǫ n ′ ′ ′ = 2 Pψa +2 n ′ ′ ′ n ′ n n 2 ψa +2 The theorem is true for Pfister forms. In fact, if πa is torsion, by Proposition 4.8 it is nilpotent and hence by a it is 2-torsion. Suppose now that ψa is torsion. Then, for every ǫ, ψaπaǫ is also torsion. By b ψ π = ψ π = mπ , where m = ǫ . Hence, if m = 0, π is torsion and a aǫ ǫ aǫ aǫ i 6 aǫ therefore 2-torsion. In any case ψ π is 2-torsion and c shows that ψ itself is a aǫP a 2-torsion. Corollary 4.11. An odd-dimensional space cannot be a zero-divisor in W(K). 44 Proof. Let ξ W(K) I(K) and assume that ξη = 0 for some η W(K). From ∈ \ ∈ (ξ2 12)(ξ2 32) (ξ2 n2)=0, − − ··· − multiplying by η we get 12 32 n2η =0, · ··· hence η = 0. Corollary 4.12. Spec W(K) is connected. Proof. If e W(K) is an idempotent e(e 1) = 0, but either e or e 1 must be ∈ − − of odd rank, hence e =0 or e 1=0. − The next proposition will be used in the last section. Proposition 4.13. Let K be an ordered field and a K˙ K˙ 2 a positive element ∈ \ of K. The ordering of K can be extended to L = K(√a). Proof. Instead of giving a direct proof we use Proposition 4.6. Let σ : W(K) Z −→ be the signature defined by the ordering of K and p its kernel. By Proposition 5.10 (whose proof is independent of the results in this section)*, the kernel of the natural map α : W(K) W(L) −→ is the ideal generated by < 1, a >. Since a is positive σ(< 1, a >) = 0, hence − − ker α p. This already shows that Z injects into W(L) because ⊆ Z ker α Z p =0. ∩ ⊆ ∩ By Corollary 4.2, W(L) is integral over Z, hence a fortiori over α(W(K)). By the theorem of Cohen–Seidenberg W(L) contains a minimal prime ideal q lying above the minimal prime α(p) of α(W(K)). It is clear that the ordering defined by q on L is an extension of the ordering of K. * This is a bit unfair, I agree. 45 5. Also sprach Brahmagupta m´ l\EŠD˜čvgA‚Ń` Zkg` nAEdčy` tEvEhnAÎ. aAŊvDog` Zkg` Z,shA˚(yGAt˜nkˆ tm˚(ym^ ; vĚvD{È\þTm\þ"˜p,"˜pvDt` Sy,. þ"˜pfoDkãt˜m´ l˜þ"˜pk˜!p˜; In this section K is a field of characteristic different from 2. We will denote symmetric bilinear spaces (V, ϕ) simply by ϕ and, for any v V we will write ∈ ϕ(v) instead of ϕ(v, v). In [5, Ch.18] Brahmagupta proves, in the slightly different formulation that you see quoted above, the following lemma. Lemma 5.1. If c K˙ is represented by < 1, a > the space < 1, a > is isometric ∈ to 2 2 s t Proof. If c = s + at the matrix at− s is regular and s t 1 0 s at c 0 = . at− s 0 a t s 0 ca − This, of course, is nothing but the identity (s2 + at2)(X2 + aY 2)=(sX + atY )2 + a(tX sY )2. − A space ϕ is said to be multiplicative if either ϕ is anisotropic and If ϕ is multiplicative the set Dϕ(K) of all nonzero elements of K represented by ϕ is a subgroup of K˙ . Pfister [33] generalized Brahmagupta’s lemma as follows. Theorem 5.2. Pfister forms are multiplicative. Proof. Let π =<1,a > <1,a > be a Pfister form. We proceed by induction a 1 ··· n on n. The case n = 0 (that is πa =< 1 >) is trivial and the case n = 1 has been dealt with in the seventh century. By induction, it suffices to prove that if ϕ is a multiplicative space, ϕ <1, a> is multiplicative too. We distinguish three cases. i) ϕ is hyperbolic. Then ϕ <1, a> is hyperbolic too. ii) ϕ is anisotropic but ϕ <1, a> is isotropic. In this case there are two vectors u and v not both zero and such that ϕ(u) + aϕ(v) = 0. Since ϕ is anisotropic ϕ(u) and ϕ(v) are both = 0, therefore 6 ϕ ϕ(u)ϕ , aϕ aϕ(v)ϕ = ϕ(u)ϕ ≃ ≃ − and ϕ aϕ ϕ(u)ϕ ( ϕ(u)ϕ) ⊥ ≃ ⊥ − is hyperbolic. iii) ϕ and ϕ < 1, a > are both anisotropic. Let c = ϕ(u) + aϕ(v) K˙ be an ∈ element represented by ϕ <1, a>= ϕ aϕ. If ϕ(v)=0, ⊥ c(ϕ aϕ) = ϕ(u)ϕ aϕ(u)ϕ ϕ aϕ. ⊥ ⊥ ≃ ⊥ The case ϕ(u) = 0 is similar. Suppose that r = ϕ(u) and s = ϕ(v) are both = 0. 6 Since ϕ rϕ s ϕ , ≃ ≃ r c(ϕ aϕ)=(1+ as )(ϕ as ϕ)=(1+ as )ϕ <1, as >. ⊥ r ⊥ r r r 47 By Lemma 5.1 (1 + as ) <1, as > <1, as >, r r ≃ r hence c(ϕ aϕ) ϕ as ϕ ϕ aϕ. ⊥ ≃ ⊥ r ≃ ⊥ Corollary 5.3. For any field K and any integer n the set of nonzero elements that can be represented as sums of 2n squares is a subgroup of K˙ . n n Proof. The Pfister form ϕ =< 1, 1 > = 2 < 1 > is multiplicative, hence Dϕ(K) is a group. We say that the level of a field K is l if 1 can be represented as the sum of − l but not less squares. Corollary 5.4. The level of a field is a power of 2. Proof. Let K be of level l and choose n such that 2n l < 2n+1. Then < ≤ 1, 1>n+1=2n <1> 2n <1> is isotropic and hence, by theorem 5.2, hyperbolic. ⊥ By Corollary 3.6, 2n <1> 2n <1>, hence 2n <1> <1> is isotropic, which ≃ − ⊥ means that 1 is a sum of 2n squares. − We now study symmetric bilinear spaces over rational function fields. A key result is theorem 5.6, proved by Cassels [7] for the diagonal space < 1,..., 1 > and generalized by Pfister [33] to arbitrary spaces. For further reference we first record the following obvious facts. Proposition 5.5. Let t be transcendental over K. If a space ϕ over K is anisotropic its extension ϕ K(t) to K(t) is anisotropic too. In particular the ⊗ canonical homomorphism W(K) W K(t) −→ 48 is injective. Theorem 5.6. Let ϕ be a sbs over K and f K[X] a polynomial in one ∈ indeterminate. If ϕ represents f over K(X) then it represents it over K[X]. Proof. An isotropic space ϕ represents everything because it contains an orthog- onal summand <1, 1> and − 1 + f 2 1 f 2 f = − . 2 − 2 We may therefore assume that ϕ is an anisotropic space of dimension n. Consider the (n + 1)-dimensional space over K(X) Φ = fi = f0qi + ri, where q ,r K[X] and r is either zero or a polynomial of degree less than that i i ∈ i of f0. Further we put q0 = 1 and r0 = 0 so that the equations above are valid for 0 i n. Let q = (q ,...,q ) and r = (r ,...,r ). If Φ(q) = 0 we have ≤ ≤ 0 n 0 n f = ϕ(q ,...,q ) , which is what we wanted. Assuming that Φ(q) = 0, we shall 1 n 6 find a new point [s] K(X) with s =(s ,...,s ) K[X]n+1 and s =0 of ∈ Q 0 n ∈ 0 6 degree less than the degree of f0. Iterating the construction of s will ultimately yield an s with s K˙ and hence a representation of f over K[X]. 0 ∈ The new point [s] is simply the “other” intersection point of with the line Q through [p] and [q]. Solving Φ(p + λq) = 0 we get, apart from λ = 0, λ = 2Φ(p, q)/Φ(q). We put − s = Φ(q)p 2Φ(p, q)q. − 49 Clearly s K[X]n+1 and [s] . The first coordinate of s is the polynomial ∈ ∈ Q Φ(f0q) Φ(p, f0q) Φ(p r) 2Φ(p, p r) Φ(r) s0 = f0Φ(q) 2Φ(p, q) = 2 = − − − = . − f0 − f0 f0 f0 Since r = 0 (otherwise Φ(q)=(1/f )2Φ(p) =0), Φ(r) = ϕ(r) = 0, because ϕ, by 6 0 6 Proposition 5.5, remains anisotropic over K(X). The degree of Φ(r) is at most twice the maximum degree of the r ’s, that is at most 2degf 2, hence i 0 − degs = deg Φ(r) degf < degf . 0 − 0 0 It is worth mentioning that by the same method one can prove that any integer which is the sum of three rational squares is the sum of three integral squares. A space ψ is said to be contained in a space ϕ or a partial form of ϕ if ϕ ψ ϕ ≃ ⊥ 0 for some ϕ0. The next theorem is known as Pfister’s Teilformensatz. Theorem 5.7. Let ϕ = a ψ is a partial form of ϕ. b for every extension L of K, Dψ(L) Dϕ(L). ⊆ c D K(X ,...,X ) contains b X2 + + b X2 . ϕ 1 m 1 1 ··· m m We first prove two lemmas. 50 Lemma 5.8. Let ϕ = ψ be an anisotropic space over K. If ψ represents 1 ⊥ 2 a1X + b over K(X), ψ represents b over K. 2 Proof. By the theorem of Cassels–Pfister ϕ represents a1X + b over K[X]. If ψ = we may write (9) a X2 + b = a f 2 + + a f 2 , with f ,...,f K[X]. 1 1 1 ··· n n 1 n ∈ Since ϕ is anisotropic the coefficients of the dominant terms in X on the right hand side do not cancel, hence each fi is linear or constant: fi = piX + qi. We solve X2 = f 2 by X = q /(1 p ) or, if p happens to be 1, X = q /(1 + p ). 1 1 − 1 1 − 1 1 Then the first terms on each side of equation (9) cancel each other and we get a representation of b by ψ. Lemma 5.9. Let F be a polynomial in K[X ,...,X ] and c = (c ,...,c ) 1 m 1 m ∈ Km such that F (c) =0. If ϕ represents F over K(X ,...,X ) it also represents 6 1 m F (c) over K. Proof. Note that the theorem of Cassels–Pfister does not extend to polyno- mials in more than one indeterminate: in general ϕ will not represent F over K[X1,...,Xm]. But we can deal with one indeterminate at a time: ϕ repre- sents F over K(X1,...,Xm 1)[Xm], hence it represents F (X1,...,Xm 1, cm) − − over K(X1,...,Xm 1), and so on. − We now prove the Teilformensatz. Obviously a implies b and b implies c , so we only have to show that c implies a . If ϕ represents F = b X2 + + b X2 over K(X ,...,X ), by Lemma 5.9 1 1 ··· m m 1 m it represents F (0,..., 0, 1) = b over K. We can then write ϕ = ϕ . m m ⊥ 1 Applying Lemma 5.8 with Xm instead of X and K(X1,...,Xm 1) instead of K − 51 2 2 we see that ϕ1 represents b1X1 + + bm 1Xm 1 over K(X1,...,Xm 1) and we ··· − − − finish the proof by induction on m. In passing we have proved that c implies m dim ϕ. ≤ We now study the kernel of W(K) W(L) for certain extensions K L. −→ ⊂ Proposition 5.10. Let L = K(√a) be a quadratic extension of K. The kernel of W(K) W(L) −→ is the ideal generated by <1, a>. − We prove the following, more precise result, of which Proposition 5.10 is an immediate corollary. Proposition 5.11. Let L = K(√a) be a quadratic extension of K and (V, ϕ) an anisotropic space over K. If ϕ L is isotropic ϕ contains an orthogonal summand ⊗ of the form c <1, a>. − Proof. Let u+√av =0(u, v V ) be an isotropy of (V, ϕ) L. From ϕ(u+√av) = 6 ∈ ⊗ 0 we get ϕ(u) = aϕ(v) and ϕ(u, v) = 0. Since ϕ is anisotropic over K, u and − v are both different from zero and, being orthogonal to each other, span a 2- dimensional space isometric to < ϕ(u), ϕ(v) >= ϕ(v) < 1, a >. By Proposition − 2.2, Ku + Kv is an orthogonal summand of (V, ϕ). Let now (Kn, ϕ) be a space of dimension n 3. The equation ϕ(u)=0 ≥ defines an irreducible quadratic cone in An and an irreducible smooth quadric Cϕ K n 1 in P − . We denote by K(ϕ) and K( ) their fields of rational functions. If Qϕ K Qϕ ϕ = K[X ,...,X ]/ a X2 + + a X2 1 n 1 1 ··· n n 52 and K( ) the field of fractions of Qϕ 2 2 K[X1,...,Xn 1]/ a1X1 + + an 1Xn 1 + an . − ··· − − Note that K(ϕ) = K( )(t) and that K(ϕ) and K( ) do not change if ϕ is Qϕ Qϕ replaced by some scalar multiple ϕ. Hence, in dealing with K(ϕ) we may always assume that ϕ represents 1. In this case ϕ =<1,a2,...,an > and K(ϕ) = K(X ,...,X ) a X2 + + a X2 . 2 n − 2 2 ··· n n q Proposition 5.12. K(ϕ) and K( ) are rational over K if and only if ϕ is Qϕ isotropic. Proof. If ϕ is isotropic has a rational point and is K-birationally isomorphic to Qϕ n 2 P − ; hence K( ) is rational and K(ϕ) = K( )(t) is rational too. Conversely, K Qϕ Qϕ suppose that K(ϕ) = K(t1,...,tn). Since ϕ is (tautologically) isotropic over K(ϕ), it is isotropic over K by Proposition 5.5. Theorem 5.13. Let ϕ be a space of dimension at least 3 over K. Let ψ be an anisotropic space over K. If ψ 0 in W K(ϕ) then, for every c D (K), cϕ is ∼ ∈ ψ contained in ψ. If ϕ is a Pfister form there exists a space ρ such that ψ = ϕ ρ. ⊗ Corollary 5.14. For any Pfister form ϕ of rank at least 4 the kernel of W(K) W K( ) −→ Qϕ is the ideal generated by the class of ϕ. Proof of Theorem 5.13. As mentioned before, we may assume that ϕ represents 1: ϕ =<1> ϕ′ =<1,a ,...,a >. ⊥ 2 n 53 Then K(ϕ) = K(X ,...,X ) f , where f = a X2 + + a X2. 2 n − 2 2 ··· n n p Since ψ 0 over K(ϕ), by Proposition 5.11 ∼ ψ =<1,f>ψ′ where ψ′ is some space over K(X2,...,Xn); hence 2 2 (X + f)ψ (X + f) <1,f>ψ′, 1 ≃ 1 and from Brahmagupta’s lemma we get an isometry (X2 + f)ψ ψ 1 ≃ over K(X ,...,X ). If c K˙ is an element represented by ψ over K, ψ will 1 n ∈ represent c(X2 + a X2 + + a X2) over K(X ,...,X ). By Theorem 5.7 1 2 2 ··· n n 1 n ψ = cϕ ψ . ⊥ 1 This proves the first part of the theorem. Suppose now that ϕ is a Pfister form. In this case cϕ becomes hyperbolic over K(ϕ), hence ψ 0 over K(ϕ). We can therefore repeat the argument with ψ 1 ∼ 1 instead of ψ and write ψ = c ϕ ψ and so on, until we finally get 1 2 ⊥ 2 ψ = The next result is a simple but useful generalization of the above corollary. We say that ϕ is a neighbour of a Pfister form τ if a suitable scalar multiple of ϕ is contained in τ. 54 Theorem 5.15. Let ϕ be a neighbour of a Pfister form τ over K and assume that dimϕ > 1 dimτ 2. The kernel of 2 ≥ W(K) W K( ) −→ Qϕ is the ideal generated by the class of τ. Proof. Consider the commutative diagram α W(K) W K(ϕ) −−−−→ γ δ β W K(τ) W K(ϕ)(τ) y −−−−→ y of canonical maps. The class of τ maps to zero in W K(τ) because τ becomes isotropic over K(τ) 1 and hence, being a Pfister form, hyperbolic. Since dim ϕ > 2 dim τ, ϕ becomes isotropic over K(τ) and therefore (Proposition 5.12) K(τ)(ϕ) = K(ϕ)(τ) is ra- tional over K(τ). This implies that β is injective. Similarly, since τ is isotropic over K(ϕ), δ is injective as well, hence ker α = ker γ. The result now follows from Corollary 5.14. 55 6. Dedekind domains Ithaga ryene renogyaga ngingo. Kikuyu proverb Let A be a Dedekind domain and K its field of fractions. For every maximal ideal p of A we denote by k(p) its residue field A/p. Liberally plagiarizing [2] we are going to describe the relations between W(A), W(K) and all the W(k(p)). For any projective A-module P we denote by PK the vector space P K. ⊗A Proposition 6.1. Let (P, ) be a sbs over A. If W is a sublagrangian of (PK, ) · · the A-module L = P W is a sublagrangian of (P, ). ∩ · Proof. Clearly L L⊥. Since the natural inclusion of P into PK induces an ⊆ injection P/L PK/W, −→ P/L is torsionfree and therefore projective over A. The exact sequence 0 L P P/L 0 −→ −→ −→ −→ splits, hence L is a direct factor of P . Corollary 6.2. The canonical homomorphism W(A) W(K) is injective. −→ Proof. We have seen that if (PK, ) 0 in W(K),(PK, ) is metabolic. By the · ∼ · above proposition P has a sublagrangian L whose rank is half the rank of P and is, therefore, a lagrangian. 56 To study the cokernel of W(A) W(K) we introduce an analogue of the −→ Witt group for A-modules of finite length. For any A-module M of finite length we put M = HomA(M,K/A). We denote by l(M) the length of M: it is, by definition, the length of a filtration c M = M M M =0 0 ⊃ 1 ⊃···⊃ l with simple quotients. We record some elementary facts. Proposition 6.3. 1 For any A-module M of finite length 1 HomA(M,K/A) = ExtA(M,A). 2 Hom ( , K/A) is an exact functor from the category of A-modules of finite A − length into itself. 3 M and M have the same length. 4 The evaluation pairing M M K/A defines a natural isomorphism c × −→ jc: M ∼ M. −→ cc Proof. The four statements can easily be proved by reduction to the local case, but the following, more complicated arguments work almost as well. 1 : The short exact sequence 0 A K K/A 0 −→ −→ −→ −→ yields the exact sequence (10) Hom (M,K) Hom (M,K/A) Ext1 (M,A) Ext1 (M,K) . A −→ A −→ A −→ A 57 For any i, any a A and any two A-modules S, T the multiplication by a on S ∈ i or T induces the same map on ExtA(S, T ). Since M is annihilated by a nonzero element a A, ∈ i i i ExtA(M,K) = ExtA(M,aK)= ExtA(aM,K)=0. In particular 1 HomA(M,K) = ExtA(M,K)=0 and (10) shows that Hom (M,K/A) Ext1 (M,A). A ≃ A 2 : Let 0 M ′ M M ′′ 0 −→ −→ −→ −→ be an exact sequence of A-modules of finite length. The sequence 0 Hom (M ′′, K/A) Hom (M,K/A) → A → A → 1 Hom (M ′, K/A) Ext (M ′′, K/A) → A → A 1 2 is exact. As in the proof of 1 , Ext (M ′′, K/A) = Ext (M ′′,A) and this last A A group is trivial because A has homological dimension 1. Hence the sequence 0 Hom (M ′′, K/A) Hom (M,K/A) Hom (M ′, K/A) 0 −→ A −→ A −→ A −→ is exact. 3 : Let p be a maximal ideal of A and p p a local parameter: pAp = pAp. ∈ Using p we can define a (non-canonical) isomorphism i : A/p ∼ Hom (A/p, K/A) p −→ A 58 which maps 1 to f : A/p K/A defined by f(x+p) = x +A. Thus the assertion −→ p is true for modules of length 1. Using 2 we finish by induction on l(M). 4 : The evaluation pairing M M K/A defines a homomorphism × −→ j :cM M. −→ c If N = kerj, by definition N M K/Ac is zero. Since, by 2 , the restriction × −→ map M N is surjective, N N K/A is also zero. This means that N =0 −→ ×c −→ and hence, by 3 , N = 0. Since l(M) = l(M), j must be an isomorphism. c b b b c A space of finite length (sfl) over A isa pairc (M,f) where M is an A-module of finite length and f : M M K/A is a non-degenerate symmetric bilinear form. × −→ Giving f is the same as giving an isomorphism (the adjoint of f) f ♯ : M M −→ which satisfies f ♯ = fˆ♯ j. ◦ c -Let A ֒ B be an extension of Dedekind domains and K ֒ L the corre → → sponding extension of their fields of fractions. A sfl (M,f) over A gives rise, by scalar extension, to a sfl (M B,f 1 ) over B. To prove this observe that ⊗A ⊗A B B, being a torsionfree A-module, is flat over A (the only reference I know by heart is [4, Ch. 1, 2, n. 4, Proposition 3 (ii) and Ch. 2, 3, n. 4, Corollaire de la § § Proposition 15]) and hence (for instance by [3, Ch. 10, 6, n. 8, Proposition 10 § (b)]) for any i the groups Exti (M,A) B and Exti (M B,B) A ⊗A B ⊗A are naturally isomorphic. In particular, taking i = 1 and using Proposition 6.3 1 , we get a natural isomorphism Hom (M,K/A) B Hom (M B,L/B). A ⊗A ≃ B ⊗A Thus tensoring f ♯ : M M by B yields an isomorphism of M B with its −→ ⊗A B-dual. The rest of the proof is clear. c 59 In what follows we will only use the above observation in the case B = Ap, p a maximal ideal of A. Note that, given an A-module of finite length M and a symmetric bilinear form f : M M K/A, to check that (M,f) is a sfl it suffices to verify that × −→ the adjoint f ♯ of f is a monomorphism: for M and M have the same length, hence a monomorphism of one into the other is an isomorphism. c Isometries and orthogonal sums are defined in the obvious way. For any submodule N M we put ⊆ N ⊥ = x M f(x, N) = (0) . { ∈ | } As in the case of sbs over fields, we have an exact sequence ˆi 0 N ⊥ M N 0. −→ −→ −→ −→ Exactness on the right is due to the fact that, byb Proposition 6.3 2 , the restriction map M N is surjective. −→ Proposition 6.4. Let (M, ) be a sfl over A. For any submodule N M, c b · ⊆ l(N) + l(N ⊥) = l(M) and N = N ⊥⊥. Proof. It is an immediate consequence of the above exact sequence, using l(N) = l(N). b A sublagrangian of (M, ) is a totally isotropic submodule L of M: L L⊥. A · ⊆ lagrangian of (M, ) is a submodule of M that coincides with its own orthogonal: · L = L⊥. Asfl(M, ) is said to be metabolic if it has a lagrangian. · Let T(A) be the monoid of the isometry classes of sfls over A with the or- thogonal sum as operation. LetL(A) be the submonoid of T(A) consisting of metabolic spaces. We put W(K,A) = T(A)/L(A) 60 and call this quotient the Witt group of torsion A-modules: in fact it will soon, quite unexpectedly, turn from a monoid into a group. Proposition 6.5. Let (M,f) be a sfl over A and I, J M two sublagrangians. ⊆ Then f induces on I⊥/I and on J ⊥/J a structure of sfl and (I⊥/I,f) (J ⊥/J, f) ⊥ − is metabolic. Proof. Let α : I⊥ I⊥/I and β : J ⊥ J ⊥/J −→ −→ be the canonical projections. Since f(I,I⊥)=0, f induces a bilinear map f : I⊥/I I⊥/I K/A. × −→ To show that it is non-degenerate it suffices, as we mentioned before, to verify that its adjoint is injective. Suppose that, for some x I⊥, ∈ f(α(x),α(y)) = f(x,y) = 0 for every y I⊥. ∈ Then x I⊥⊥ = I, hence α(x)=0. ∈ Let now I⊥ J ⊥ ∆: ∩ I⊥/I J ⊥/J I J −→ ⊕ ∩ be the homomorphism defined by ∆(x) = α(x), β(x) . It is obviously injective and we claim that its image L is a lagrangian of (I⊥/I,f) (J ⊥/J, f). ⊥ − It is clear that L L⊥. Suppose that α(y), β(z) L⊥. Then f(x,y) f(x, z) = ⊆ ∈ − 0 for every x I⊥ J ⊥ =(I + J)⊥. This means that y z (I + J)⊥⊥ = I + J. ∈ ∩ − ∈ 61 Writing y z = s + t with s I,t J and putting x = y s = z + t we get − ∈ ∈ − α(y), β(z) = ∆(w) L. Hence L is a lagrangian. ∈ Corollary 6.6. Let I be a sublagrangian of (M, ). Then (M, ) (I⊥/I, ) in · · ∼ · W(K,A). Corollary 6.7. W(K,A) is a group. The inverse of (M,f) is (M, f). − Corollary 6.8. A sfl (M, ) which represents 0 in W(K,A) is metabolic. · Proof. If (M, ) 0 there exists a metabolic space (N, ) such that · ∼ · (S, )=(M, ) (N, ) · · ⊥ · is metabolic. Let I be a lagrangian of (N, ) and (J, ) a lagrangian of (S, ). · · · Clearly I is a sublagrangian of (S, ) and its orthogonal in (S, ) is I M. By · · ⊕ Proposition 6.5 the space (I M)/I, (J ⊥/J, )=(M, ) ⊕ · ⊥ · · is metabolic. Let V be an n-dimensional vector space over K. A lattice in V (more precisely an A-lattice) is a finitely generated A-submodule P of V that spans V over K. Since A is a Dedekind domain and V a torsion-free A-module, P is a projective A-module of rank n. If (V, ϕ) is a sbs over K for any lattice P of V we define its dual by P ♯ = v V ϕ(v,P ) A . { ∈ | ⊆ } The term “dual” is totally appropriate because the adjoint of ϕ identifies P ♯ with ♯ its official dual P ∗ = Hom (P,A). In fact, for any v P , ϕ(v, ) defines a A ∈ − linear map P A and conversely, any linear map α : P A extends to a −→ −→ 62 map α : PK = V K given by α (x) = ϕ(v,x) for some fixed v V . Since K −→ K ∈ ♯ ♯♯ α (P ) A, v must be in P . From P = P ∗∗ we deduce that P = P . K ⊆ Proposition 6.9. Let P be a lattice in (V, ϕ). If P P ♯, ϕ induces on P ♯/P a ⊆ structure of sfl over A. Proof. The A-module P ♯/P is finitely generated and torsion, hence of finite length. Since ϕ(P,P ♯) = A, for any x,y P ♯ the class ϕ(x,y) of ϕ(x,y) in K/A ∈ only depends on the classes x, y of x and y in P ♯/P . Since ϕ(x, y) = 0 for all y P ♯/P is equivalent to ϕ(x,P ♯) A, x P ♯♯ = P and therefore x = 0. This ∈ ⊆ ∈ shows that ϕ♯ is injective, which implies that (P ♯/P, ϕ) is a sfl over A. Proposition 6.10. Let P and Q be two lattices of (V, ) both contained in their · duals. Then (P ♯/P, ) (Q♯/Q, ) in W(K,A). · ∼ · Proof. The submodule R = P Q of V is also a lattice and ∩ R P P ♯ R♯. ⊆ ⊆ ⊆ The quotient P/R is a sublagrangian of (R♯/R, ) because · ♯ P/R P /R =(P/R)⊥, ⊆ hence by Corollary 6.6 (R♯/R,.) (P ♯/R) (P/R), =(P ♯/P, ), ∼ · · and similarly (R♯/R, ) (Q♯/Q, ). · ∼ · The last proposition shows that we can associate to any space (V, ) over K · a well-defined element ∂(V, ) of W(K,A): the class of (P ♯/P, ), where P is an · · 63 arbitrary lattice of V contained in its dual (it is easy to see that such lattices do exist). This map ∂ defines a homomorphism ∂ : S(K) W(K,A) −→ from the monoid of the isometry classes of sbs over K to the Witt group of torsion modules over A. For a space (PK, ) obtained from a sbs (P, ) over A by scalar · · extension, ∂(PK, ) = 0 because P = P ♯. Now, any metabolic space over K is · isometric to a space (K2n, ϕ), where 0 1n ϕˆ = with ψ Mn(K). 1n ψ ∈ For any a K˙ , ∈ a 1 0 0 1 a 1 0 0 1 − n − = n , 0 a 1 ψ 0 a 1 a2ψ n n hence, choosing a suitable a, we may assume that ψ M (A). This shows that ∈ n every metabolic space over K is extended from A, thus ∂(N(K))= 0. We can fit ∂ into an exact sequence ǫ ∂ (11) 0 W(A) W(K) W(K,A) . −→ −→ −→ Proposition 6.11. The above sequence is exact. Proof. We know that W(A) W(K) is injective and that ∂ǫ = 0. Suppose that −→ ∂(V, ) = 0 and let P be a lattice of V such that P P ♯. In view of Corollary · ⊆ ♯ ♯ ♯ 6.8, (P /P, ) has a lagrangian L = Q/P P /P . From L⊥ = L we get Q = Q, · ⊆ hence the restriction of the bilinear form on V defines on Q a structure of sbs over A such that (Q, ) K = (V, ). This shows that (V, ) (and not only its · ⊗A · · class, which would suffice) is extended from A. Any A-module of finite length M can be decomposed as direct sum M = pMp, where p runs over the maximal ideals of A and Mp = 0 for all but a finite ⊕ 64 number of them. Clearly if (M, ) is a sfl over A the above decomposition of M · yields an orthogonal decomposition (Mp, ), p| · where (Mp, ) can be identified with the sfl over Ap obtained by localization at p. · From this, observing that the localization of a metabolic space is metabolic, we obtain W(K,A) = W(K,Ap). p M Thus, for each p we have a homomorphism ∂p : W(K) W(K,Ap) −→ and ∂ = p∂p. ⊕ Proposition 6.12. Every choice of a local parameter p at p yields an isomor- phism i : W(k(p)) ∼ W(K,Ap). p −→ Proof. Assume that A is a dicrete valuation ring and p its maximal ideal. Let 1 p be a generator of p. Mapping the class of a in k(p) = A/p to the class of p a in K/A we may identify k(p) with a submodule of K/A and thus view bilinear forms with values in k(p) as bilinear forms with values in K/A. In this way we obtain a homomorphism i : W(k(p)) W(K,Ap). p −→ The injectivity of i follows from Corollary 6.8: a space (V, ) over k(p) that p · represents 0 in W(K,A) contains a lagrangian L and therefore represents 0 in W(k(p)) as well. To show that i is surjective consider an arbitrary sfl (M, ). Replacing it by p · (I⊥/I, ) where I is a maximal sublagrangian (one of maximal length) we may · 65 assume that (M, ) is anisotropic. Let pm be the lowest power of p that annihilates · M. Were m 2, then ≥ m 1 m 1 2m 2 m p − M p − M = p − M p M =0 · ⊆ m 1 and p − M would be a non-trivial sublagrangian. Hence m = 1 and M is a vector space over k(p). The scalar products x y of elements of M are in the · 1 submodule of K/A annihilated by p, which is precisely ( p A)/A, hence the class of (M, ) is in the image of i . · p Proposition 6.12 allows us to write (11) as ǫ p∂p (12) 0 W(A) W(K) ⊕ W k(p) , −−−−→ −−−−→ −−−−−−→ p M 1 where ∂ is the composite map i− ∂p. An explicit description of ∂ will prove p p ◦ p useful. Let In general the determination of coker ∂ is rather difficult. In one case, though, we can show that ∂ is surjective. Proposition 6.13. If A is a Euclidian domain ∂ is surjective. Proof. Recall that A is said to be Euclidian if there exists a map : A 0 N k k \{ } −→ 66 with the following two properties: 1 for any a,b A,b = 0, there exist q,r A such that ∈ 6 ∈ a = bq + r and either r =0 or r < b . k k k k 2 for any a,b A 0 ∈ \{ } ab a . k k≥k k (If a domain A admits a function satisfying 1 then it also admits a function k k1 satisfying 1 and 2 [36, Proposition 4].) k k2 We have to show that ∂ ′ = ∂ is surjective, where p runs over a set of non- ⊕p p associated prime elements of A. Assume that, for some prime q, the image of ∂ ′ contains W(A/Ap) whenever p < q . Let < a > be a one-dimensional space k k k k over A/Aq. Dividing a by q as in 1 and replacing it by its remainder we may assume that a < q . If p divides a, by 2 we have p a < q . As k k k k k k≤k k k k we mentioned before and for any p not associated to q either p > q , in which case ∂ Corollary 6.14. Let k be a field. The sequence p∂p 0 W(k[t]) W(k(t)) ⊕ W(k[t]/(p)) 0, −−−−−−→ −−−−−−→ −−−−−−→ −−−−−−→ p M where p runs over all the monic irreducible polynomials of k[t], is exact. We now show that, if the characteristic of k is not 2, W(k[t]) = W(k). In fact, we can show the following, more precise result. 67 Theorem 6.15. Let k be a field of characteristic different from 2. Any sbs over k[t] is extended from k. This result, first proved by Harder [22] using algebro-geometric methods, is an easy consequence of Theorem 6.16 below, for the formulation of which we need some terminology. For any f K = k(t) let f be the degree of f in t. Given ∈ k k a sbs (V, ) over K and a basis e ,...,e of V , we put D(e ,...,e ) = det(e e ). · 1 n 1 n i · j Every lattice P of V is a free k[t]-module. If (e1,...,en) and (e′1,...,e′n) are bases of P over k[t] the determinants D(e1,...,en) and D(e′1,...,e′n) only differ 2 by a constant square factor, hence the class D(P ) of D(e1,...,en) in K/˙ k˙ only depends on P : it is the “discriminant” of P . We denote by P the degree of k k D(P ): it is the “volume” of V/P . Theorem 6.16. Let (V, ) be an anisotropic n-dimensional sbs over k(t) and P · a k[t]-lattice in V . There exists a nonzero vector x P such that ∈ P x x k k. k · k ≤ n Proof. The assertion is obvious if n = 1. We prove it by induction on n 2. ≥ Let x P 0 be an element for which x x is minimal. Let V ′ =(Kx)⊥ and ∈ \{ } k · k π : V V ′ the orthogonal projection of V onto V ′. The A-module P ′ = π(P ) −→ is a lattice of V ′ and hence, by induction assumption, it contains a y = 0 such 6 that P ′ y y k k. k · k ≤ n 1 − This y is the projection of some z = y + λx P , where λ K. Since x P we ∈ ∈ ∈ may modify λ by an arbitrary element of k[t], in particular we may subtract from λ its integral part and assume that λ 1. k k ≤ − Let us now compute P . Since x x is minimal, x must be primitive , hence k k k · k there is a basis e1,...,en of P in which en = x. The projections π(e1),...,π(en 1) − 68 then form a basis of P ′. For 1 i n 1 we may write ≤ ≤ − ei = π(ei) + λien with λ K. The matrix that transforms the K-basis (e ,...,e ) of V into the i ∈ 1 n basis (π(e1),...,π(en 1), en) is upper triangular (unless it is lower triangular, − which amounts to the same) and has therefore determinant 1. Hence D(e1,...,en)=D(π(e1),...,π(en 1), en)=D(π(e1),...,π(en 1))(en en) − − · and P = P ′ + x x . k k k k k · k On the other hand y y = z z λ2(x x) , k · k k · − · k but x x z z (because x x is minimal) and λ2 < 0, hence k · k≤k · k k · k k k z z λ2(x x) = z z k · − · k k · k and P ′ P x x x x z z = y y k k = k k−k · k , k · k≤k · k k · k ≤ n 1 n 1 − − which implies P x x k k. k · k ≤ n Corollary 6.17. If (P, ) is an anisotropic sbs over k[t] there exists an x P · ∈ such that x x k˙ . · ∈ Proof. P is a lattice in (PK, ) with P = 0. · k k 69 We now deduce Theorem 6.15 from the above corollary. Let (P, ) be a sbs · over A = k[t]. If (PK, ) is isotropic, P contains a primitive vector x of norm · zero. Since x generates a free summand of rank 1 of P , there exists an y P ∈ such that x y = 1. Then Ax + Ay is a metabolic summand of P and we may · write (P, )=(P ′, ) (Ax + Ay, ). · · ⊥ · Since 2 is invertible in A metabolic spaces are hyperbolic and (Ax + Ay, ) H(A)=H(k) A · ≃ ⊗k is extended from k. By induction on the rank of P we may assume that (P ′, ) is · extended, hence so is (P, ). · If (PK, ) is anisotropic, by Corollary 6.17 P contains an x such that (Ax, )is a · · sbs over A, hence (P, )=(P ′, ) (Ax, ). · · ⊥ · Since (Ax, ) p∂p (13) 0 W(k) W(k(t)) ⊕ W(k[t]/(p)) 0 , −−−−−−→ −−−−−−→ −−−−−−→ −−−−−−→ p M where p runs over all the monic irreducible polynomials of k[t], is exact. Proof. Immediate from Corollary 6.14 and Theorem 6.15. Note that Theorem 6.15 is false in characteristic 2. For instance the space 0 1 k[t]2, 1 t ! is not extended from k if k has characteristic 2. 70 It is a pleasant and instructive exercise to prove a number-theoretic analogue of Theorem 6.16. Denote by a the usual absolute value of a Z. Hermite, in k k ∈ 1845 [18], proved that every lattice P in an anisotropic n-dimensional space over Q contains a nonzero x such that n−1 2 4 1 x x P n . k · k ≤ 3 k k The proof of Theorem 6.16 given here is due to Gerstein [16,17]. 71 7. The unramified Witt group Sept indique une premiere` recette. Jean•Pierre Brisset We fix a ground field k (for instance C) of characteristic different from 2, and define, for any extension K of k, a subgroup Wnr(K) of W(K), called the unramified Witt group of K. Let be the set of discrete rank 1 valuations of K that are trivial on k. VK To any v we associate its discrete valuation ring ∈VK O = a K v(a) 0 . v { ∈ | ≥ } It is a principal local domain with maximal ideal P = a K v(a) > 0 , v { ∈ | } whose field of fractions is K. Conversely, given a local principal domain A with maximal ideal P we can define a discrete valuation of rank one on its field of fractions K by v(f) = max n N f Pn for f A 0 and v(f/g) = { ∈ | ∈ } ∈ \{ } v(f) v(g) for any f/g K˙ . − ∈ We denote by k(v) the residue field Ov/Pv of Ov. By the results of section 6 the choice of a generator πv of Pv allows us to define a homomorphism ∂ : W(K) W(k(v)). v −→ The unramified Witt group of K (more precisely, of K/k, but we keep k fixed and innominate) is 72 Wnr(K) = ker ∂v. v K ∈V\ To be sure, ∂v depends on the choice of πv, but ker ∂v does not, hence the left hand side is well-defined. An element ξ W(K) is said to be unramified (over k, but ...) if it belongs ∈ to Wnr(K). More generally we define the ramification of ξ as Ram ξ = v ∂ (ξ) =0 , { ∈VK | v 6 } and we say that ξ is unramified at v if ∂v(v) = 0. By the exactness of the sequence (11), ξ is unramified at v if and only if it lies in the image of W(O ) W(K); v −→ in other words, if and only if it is extended from Ov. Proposition 7.1. Let σ : K L be a k-homomorphism, ξ W(K) and ξ −→ ∈ L the image of ξ in W(L). If w Ram ξ , its restriction v = w to K belongs to ∈ L |K Ram ξ. In particular, if ξ is unramified ξL is unramified too and hence σ induces a homomorphism W (K) W (L). nr −→ nr Proof. If w restricts to the “trivial” valuation of K, Ow contains K, hence ξL is extended from O and is therefore unramified at w. If w restricts to a v and w ∈VK ξ is unramified at v, ξ is extended from O , hence ξ is extended from O O v L w ⊃ v and is therefore unramified at w. Proposition 7.2. For any extension K of k Wnr(K(t))=Wnr(K). Proof. Let p K[t] be an irreducible polynomial and v the corresponding p-adic ∈ valuation. Clearly v , hence ∈VK(t) 73 W (K(t)) ker ∂ : W K(t) W K[t]/(p) nr ⊆ v −→ for any such p. By Corollary 6.18 this implies that W K(t) W(K). nr ⊆ Let now v be a valuation in . The localization of O [t] at its prime ideal VK v generated by p is the discrete valuation ring corresponding to an extension w of v to K(t) . Since k(w) = k(v)(t), the canonical isomorphism W(k(v)) W(k(w)) −→ is injective. The commutative diagram ∂ W(K) v W k(v) −−−−→ ∂ W K(t) w W k(w) y −−−−→ y shows that any ξ W(K ) which is ramified at v is also ramified at w. Hence ∈ W (K(t)) W (K). nr ⊆ nr Corollary 7.3. For every n Wnr(k(t1,...,tn)) = W(k). We define, for every m, Im (K) = I(K)m W (K) and record, for further nr ∩ nr reference, the following fact: Proposition 7.4. For every m > 0 m Inr(C(t1,...,tn))=0. m Proof. Immediate from Corollary 7.3 and the fact that Inr(C)=0. 74 We now recall the notion of the centre of a valuation. Let be a projective V n variety over k: by this we mean an irreducible reduced closed subscheme of Pk . Let K = k( ) be its field of rational functions and, for any (not necessarily closed) V point p , let be the local ring of all f K that are regular at p and ∈ V Op ∈ Mp its maximal ideal. Given v we denote by C the set of all points p for ∈VK v ∈V which O and P . Op ⊆ v Mp ⊇ v ∩ Op By definition the centre of v on is C . V v Proposition 7.5. The centre C of any v is an non-empty closed irre- v ∈ VK ducible subset of . If ζ is the generic point of C the inclusion ֒ O induces V v Oζ → v .an inclusion k(ζ) ֒ k(v) of their respective residue fields → Proof. Let = Proj(k[X ,...,X ]/a) = Proj(k[x ,...,x ]), V 0 n 0 n where a is the homogeneous prime ideal defining and x ,...,x are the classes of V 0 n X0,...,Xn modulo a. We may assume (after a change of coordinates, if necessary) that no Xi lie in a. Then for any pair (i, j) of indices v(xi/xj) is finite and we may assume that it is minimal for, say, x0/xi. Since v(x /x ) = v(x /x ) v(x /x ) 0, j 0 j i − 0 i ≥ v is non-negative on A = k[x /x ,...,x /x ] and we have A O . The prime 0 1 0 n 0 0 ⊂ v ideal p = P A of A defines a closed subset of and it is clear that v ∩ 0 0 W0 V0 = C . In particular C = and C is closed and irreducible. W0 v ∩V0 v 6 ∅ v ∩V0 For any affine open subset = SpecB of either B O , in which case O U V 6⊆ v Op 6⊆ v for all p and C = , or, by the same argument used for , C is an ∈ U v ∩ U ∅ V0 U ∩ v irreducible closed subset of . This shows that C is indeed closed and irreducible. U v 75 The second assertion is clear because the generic point of C is p = A P v 0 ∩ v ∈ SpecA0. To avoid misunderstandings about the nature of Cv we look at some exam- 2 ples, taking = PC = ProjC[X ,X ,X ]. We put x = X /X and y = X /X , V 0 1 2 1 0 2 0 so that K = C(x,y). 1 Let f C[x,y] be an irreducible polynomial. Every g K˙ can be written ∈ ∈ m as f (h1/h2), where h1 and h2 are polynomials prime to f. Putting v(g) = m we define a valuation v whose centre is the projective closure of the affine ∈ VK curve f = 0. 2 Let m be the maximal ideal of C[x,y] generated by x and y and let v be the valuation of C[x,y] defined by v(f) = r if f mr mr+1. This valuation extends ∈ \ to a discrete rank one valuation of K in an obvious way. Since C[x,y] O and ⊂ v 2 m = P C[x,y], the centre of v is the point [1, 0, 0] PC. Blowing it up we v ∩ ∈ obtain a surface ′ with the same field K of rational functions and such that the V centre of v on it is the line that contracts to [1, 0, 0]. The residue field of v is isomorphic to C(x/y). 3 Let ∞ tn ǫ(t) = C[[t]] n! ∈ n=1 X and w the t-adic valuation of C[[t]]. Mapping x to t and y to ǫ(t) defines a homomorphism of C-algebras σ : C[x,y] C[[t]]. −→ Since t and ǫ(t) are algebraically independent over C, σ is injective. The restric- 2 tion of w to C[x,y] defines a v whose centre on PC is again [1, 0, 0] and ∈ VK whose residue field is C. The first example can be generalized to any normal projective variety of V dimension d. Let be an irreducible subvariety of of codimension 1 and ζ its W V 76 generic point. Since is normal, is a discrete valuation ring. The valuation V Oζ w associated to has centre and its residue field is k( ). In particular k(w) Oζ W W has transcendence degree d 1 over k. − The valuation v costructed in the third example is of a basically different kind because k(v) is of transcendence degree zero over k. The following remark will be used later on. Proposition 7.6. Let be a normal projective variety, K its field of rational V functions and v . Suppose that the centre of v on is an irreducible ∈ VK V subvariety of codimension 1. Then v coincides with the valuation w associated W to . W Proof. Let π be a generator of P . By the very definition of centre, O O , w w ⊆ v hence v(f) 0 whenever w(f) 0 and v(f)= 0if w(f) = 0. Every element ≥ ≥ g K˙ can be written as πmf with w(f) = 0, hence v(g) = mv(π) 0 and since, ∈ ≥ by definition, v : K˙ Z is surjective, we must have v(π) = 1 , which implies −→ v = w. 77 8. Dulcis in fundo Toda la gracia de un soneto estriba en el ultimo´ verso; los demas,´ con que rimen basta. Noel Claraso´ Here comes, at last, the example promised in the introduction. We consider 2 the following configuration in PC L L L 3 L 1 2 4 R p4 q1 p3 q2 p p 2 1 q 3 q 4 S M 2 M 3 M1 M4 and assume for simplicity that R and S are the coordinate axes x = 0 and y =0 (as usual, x = X1/X0 and y = X2/X0). The other lines Li, Mj will be defined by affine equations li = 0, mj = 0. We put g1 = l1l2m1m2 , g2 = l3l4m3m4 and f = xy. 78 Let K be the field C(x,y) and L the rational function field of the conic X2 fX2 g g X2 =0 0 − 1 − 1 2 2 in P2 . In the terminology of section 5, L is K( ) where K Qϕ ϕ =<1, f, g g > − − 1 2 is a neighbour of the Pfister form < Note that we adopt the most fashionable notation and define <> as <1, a > <1, a >. 1 n − 1 ··· − n We claim that L is unirational over C but not rational. The first assertion is nearly obvious because C L L f = K f ( ) ⊆ ⊆ Qϕ p p and since ϕ is isotropic over K √f K f ( ) K f (t) = C x,y, √xy,t = C x, √xy,t C(t ,t ,t ). Qϕ ≃ ≃ 1 2 3 p p 2 To show that L is not rational we exhibit a non-trivial element in Inr(L). To accomplish this arduous task we consider the Pfister forms α1 = < over K and try to determine their ramification. A general observation will be helpful. 79 Proposition 8.1. Let F be an extension of the ground field k, v and ∈ VF < ∂ < Proof. Follows immediately from the explicit description of ∂p given in section 6. 2 We now examine the centres on PC of the valuations v in Ram α1. By Propo- sition 8.1, unless v(f) or v(g1) is odd, ∂v α1 = 0. Assume first that Cv is a curve. We have seen (Proposition 7.6) that in this case v is the valuation associated to Cv. If Cv is not one of the lines R, S, L1, L2, M1, M2 we have v(f) = v(g1)=0 and, by Proposition 8.1, ∂v α1 = 0. Suppose now that Cv = R. In this case v(g1) = 0 and v(f) = 1. The residue field of v is 1 k(v) = k(R) = k(PC) C(t), ≃ and the restriction g1 = l1l2m1m2 of g1 to R is a rational function with two double zeros at p1 and p2 and one quadruple pole at infinity. Hence g1 is a square in k(v) and by Proposition 8.1, ∂v α1 = 0. Similarly, ∂v α1 = 0 if the centre of v is S. Let us now suppose for instance that the centre of the valuation v is L1. In this case v(f) = 0 and v(g1) = 1. The restriction of f to k(v) = k(L1) is a rational function f with one simple zero at p1, one simple zero at q1 and one double pole at infinity. Clearly f cannot be a square, hence ∂ α = 0 by Proposition 8.1. v 1 6 Similarly one sees that α1 is ramified at the valuations with centre L2, M1 and M2. 2 We now turn our attention to valuations v whose centre is a closed point p of PC. 80 Lemma 8.2. Let < ∂v α1 =0. Proof. Assume that v(a) is even. Modifying a by a suitable square we may assume that v(a) = 0. Then by Proposition 8.1 we only have to check that a is a square in k(v). This is indeed the case because by Proposition 7.5 the homomorphism k(v) factors through k(p) = C. Op −→ The above lemma restricts the possible positions of p to the points at in- finity (that is on the line X0 = 0) and p1, p2, q1 or q2. Assume that p lies on X0 = 0. If for instance p is not the point at infinity of R or of S, writing 2 f = (X1/X2)(X2/X0) we see that v(f) is even because X1/X2 is regular and =0 at p. If p is the point at infinity of R or of S we may reason in the same way 6 with g1 instead of f. Summing up, we have proved two facts: a Ram α = . 1 6 ∅ b if v Ram α1, its centre is one of the following closed sets: ∈ L1, L2, M1, M2, p1, p2, q1, q2. Similarly we can prove that c Ram α = . 2 6 ∅ d if v Ram α2, its centre is one of the following closed sets: ∈ L3, L4, M3, M4, p3, p4, q3, q4. From all this we retain that 1 α = 0, α = 0 and α = α in W(K). 1 6 2 6 1 6 2 2 (Ram α ) (Ram α ) = . 1 ∩ 2 ∅ By Proposition 5.15 the kernel of W(K) W(L) is the ideal W(K)α generated −→ 3 by α = < Thus W(K)α = 0,α : the kernel of W(K) W(L) contains at most one non- { } −→ trivial element. We have seen above that α1 and α2, having different ramifications, are different. On the other hand α α = <1, f, g ,fg > <1, f, g ,fg > 1 − 2 − − 1 1 − − − 2 2 = < g ><1, f, g g ,fg g > − 1 − − 1 2 1 2 ker W(K) W(L) , ∈ −→ hence ker W(K) W(L) contains exactly one non-trivial element, namely −→ α1 α2; The images of α1 and α2 in W(L) are therefore the same element β. − This β cannot be zero, otherwise ker W(K) W(L) would contain the two −→ distinct and non-trivial elements α1 and α2. By Proposition 7.1 the ramification of β restricts to that of α and also to that of α , hence by 2 , Ram β = . We 1 2 ∅ have shown that 0 = β I2 (L): by Proposition 7.4, L is not rational. 6 ∈ nr Variations on this example, including the original construction of Artin and Mumford [1] can be found in [11]. We now return to the cubic real surface of section 1 and show that the S unramified Witt group of its function field K = R( ) is larger than W(R)) = Z. S Recall that K = R(x,y,z) where x, y, and z satisfy the relation y2 + z2 = x3 3x. − Let α1, α2 and α3 be the classes in W(R(x)) of << 1, x>> , << 1,x2 3>> and << 1,x3 3x>>. − − − − − The field K is nothing but R(x)( ), where Qϕ ϕ = < 1, 1,x3 3x> − − − 82 is a neighbour of the Pfister form α = << 1,x3 3x>>. By Proposition 5.15 the − − kernel of W R(x) W(K) −→ is generated by the class of α. Since α α = << 1, x>> << 1,x2 3>> = < x><< 1,x3 3x>> 1 − 2 − − − − − − − = < x> α, − α1 and α2 have the same image β in W(K). We claim that β is unramified. As in the example of the complex threefold we first examine the ramification of α1 and α . Here the situation is much simpler because the valuations in R correspond 2 V (x) 1 bijectively to their centres, which are the closed points of PR. Denoting by vp the valuation with centre 1 p PR = Proj R[X ,X ], ∈ 0 1 we have vp(x) = 0 unless p = [1, 0] or p = [0, 1], hence, by Proposition 8.1, Ram α1 contains at most the valuations corresponding to these two points. Similarly if p is different from [1, √3], [1, √3] and [0, 1], then v (x2 3) = 0. On the other − p − hand, for p = [0, 1] = (the point at infinity), v (x2 3) = 2 is even and ∞ ∞ − − hence, again by Proposition 8.1, Ram α2 contains at most the valuations centred at [1, √3] and [1, √3]. This shows that − Ram α Ram α = , 1 ∩ 2 ∅ from which we conclude, in view of Proposition 7.1, that β is not ramified. We now show that β is not in the image of W(R). As explained in section 4 we can order R(x) making x “infinitesimally larger” than 1. With respect to this − ordering the function x3 3x is positive, hence the ordering can be extended to − K = R x, x3 3x . 0 − p 83 Adjoining w = √x3 3x to K we get − K′ = R(x,y,z,w) = K0(u, v) where u = y/w and v = z/w satisfy the equation of a circle u2 + v2 =1. Hence K′ is a rational function field over K0. As we have remarked in section 4 the ordering of K0 can be extended to K′ and yields by restriction an ordering P of K for which x is infinitesimally near to 1. In particular x is negative. 1 − Similarly, if we order R(x) in such a way that x > 2 we can define an ordering P2 of K for which x is positive. It is clear now that β is not the image of a γ W(R): with respect to P the ∈ 1 signature of β =< 1, 1, x, x > is 4 and with respect to P it is zero, whereas − − 2 the signature of the image of any γ W(R) is constant. ∈ At this point the reader may suspect a connection between real components and signatures. If he does, he should consult Mah´e’s paper [27]. 84 REFERENCES 1. M. 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Aguilar, 1951. 88 INDEX adjoint 13,57 orthogonalgroup 15 anisotropic 16 orthogonalof 15 bhelpoori vi orthogonalsum 15 centre 72 partialform 48 Ci-field 28 Pfister form 26 contained 48 positive 35 diagonalizable 16 quasialgebraicallyclosed 28 discriminant 27 ramification 70 Euclidean 64 rank 26 even 18 rational 1 extended from 14 reflection 25 extending the scalars 14 sbs 13 formallyreal 34 sfl 57 hyperbolic 15 signature 38 hyperbolicspace 14 spaceoffinitelength 57 ideal of even rank spaces 26 sublagrangian 18, 58 isometry 15 symmetricbilinearspace 13 isotropy 16 unimodular 13 lagrangian 18,58 unirational 1 lattice 60 unirationalextension 1 metabolic 18,58 unramified 70 multiplicative 44 unramifiedWittgroup 69 neighbour 52 Witt equivalent 19 norm14 Wittgroup19 orderable 36 Wittgroupoftorsionmodules 58 ordering 35 Wittring 19