Sums of values represented by a quadratic form

G. Berhuy, N. Grenier-Boley, M. G. Mahmoudi

May 22, 2011

Abstract

Let q be a quadratic form over a �eld K of characteristic di�erent from2. We investigate the properties of the smallest positive integer n suchthat �1 is a sums of n values represented by q in several situations. Werelate this invariant (which is called the q-level of K) to other invariantsof K such as the level, the u-invariant and the Pythagoras number ofK. The problem of determining the numbers which can be realized as aq-level for particular q or K is studied. We also observe that the q-levelnaturally emerges when one tries to obtain a lower bound for the indexof the subgroup of non-zero values represented by a P�ster form q. Wehighlight necessary and/or su�cient conditions for the q-level to be �nite.Throughout the paper, special emphasis is given to the case where q is aP�ster form.

2000 Mathematics Subject Classi�cation: 11E04, 11E10, 11E25, 11E39,11E81, 12D15.

Key words: Level of a �eld, quadratic form, P�ster form, Pythagorasnumber, square classes, sums of squares, formally real �elds, u-invariant,hermitian level, weakly isotropic forms, signature of a quadratic form,strong approximation property, ordering of a �eld.

1 Introduction

A celebrated theorem of E. Artin and O. Schreier states that a �eld K has anordering if and only if �1 cannot be written as a sum of squares in K. In thiscase, the �eld K is called formally real. In the situation where K is not formallyreal, one may wonder how many squares are actually needed to write �1 as asum of squares in K. This leads to the following de�nition:

De�nition 1.1. The level s(K) of a �eld K is de�ned by

s(K) = minfn j 9(x1; � � � ; xn) 2 Kn; �1 = x21 + � � �+ x2ng

if �1 is a sum of squares in K, otherwise one de�nes s(K) = +1.

A question raised by B. L. van der Waerden in the 1930s asks for the integers thatcan occur as the level of a �eld K (cf. [17]). H. Kneser obtained a partial answerto this question in 1934 by showing that the possible values for the level are 1,2, 4, 8 or the multiples of 16 (cf. [10]). In 1965, A. P�ster developed the theoryof multiplicative forms which furnished a complete answer to this question : if

1

�nite, the level of a �eld is always a power of 2 and every prescribed 2-powercan be realized as the level of a �eld, see [19]. We intend to study the followingnatural generalization of the level of a �eld.

De�nition 1.2. Let (V; q) be a quadratic form over a �eld K. We de�ne thelevel of K with respect to q (or the q-level for short) by

sq(K) = minfn j 9(v1; � � � ; vn) 2 V n such that� 1 = q(v1) + � � �+ q(vn)g;

if such an n exists and by sq(K) = +1 otherwise.

Under this setting, the (usual) level of K corresponds to the level of K withrespect to the quadratic formX2 overK. More generally, the length (see �2) of anon-zero element a 2 K coincides with the q-level of K where q is the quadraticform �aX2 over K. Studying the level of K with respect to one-dimensionalquadratic forms over K is thus nothing but investigating the length of anyelement of K.

To our best knowledge, this notion has not been explicitly de�ned - at leastin the general case of a quadratic form - but it appears implicitly in many placesand it is closely related to some invariants studied in the literature.

It is already relevant to point out that q-levels are related to some hermitianlevels studied by D. W. Lewis (see [15] and [17]). For a ring R with an identityand a non trivial involution �, recall that the hermitian level of R is de�nedas the least integer n such that �1 is a sum of n hermitian squares in R, i.e.,elements of the form �(x1)x1 + � � � + �(xn)xn where x1; � � � ; xn 2 R. Thehermitian level of (R; �) is denoted by s(R; �). If L=K (resp. Q) is a quadraticextension with L = K(

pa) (resp. a K-quaternion algebra (a; b)K) and if � is

the canonical involution of L (resp. of Q), it is easy to see that s(L;�) = sq(K)(resp. s(Q;�) = sq(K)) where q = h1;�ai (resp. q = h1;�a;�b; abi) is thenorm form of L=K (resp. of Q). Note that in both cases the hermitian level isa power of two (see [15, Prop. 1.5]) and the quadratic form q is a P�ster formover K. More generally, we observe that sq(K) is always a 2-power or in�nitewhenever q is a P�ster form (see Proposition 4.1).

Level of a noncommutative ringD, which is denoted by s(D), was introducedin [14] and [16] as the least integer n such that �1 is a sum of n squares in D. IfD is the Cli�ord algebra of a nondegenerate quadratic form q over a �eld K thenwe obviously have s(D) 6 sq(K); in particular if D is the quaternion algebra(a; b)K , generated by the elements i and j subject to the relations i2 = a 2 K�,j2 = b 2 K� and ij = �ji, then we have s(D) 6 sq(K) where q = ha; bi.

A substantial part of this paper is devoted to investigating

� the properties of sq(K), i.e., the q-level of a �eld K,� the relations of sq(K) with other invariants of K such as the (usual)level, the u-invariant, the Pythagoras number,� the calculation of sq(K) for particular q or K,� the behavior of sq(�) under �eld extensions,� for a given n, the possible values of sq(K) that are attained when q runsover all quadratic forms of dimension n over K,� for a given q, the possible values of sqK0 (K

0) when K 0=K runs over all�eld extensions of K.� criteria for the �niteness or in�niteness of sq(K) for particular q or K.

2

When q is a P�ster form there are several strong analogies between the propertiesof sq(K) and that of s(K). In particular a `Pythagoras q-number' which isrelated to sums of values represented by q is de�ned and its properties arestudied. However this does not mean that when q is a P�ster form, every resulton s(K) can be generalized to sq(K), see �4.2, Remark 4.8.

The paper is structured as follows. In the next section, we collect somede�nitions and preliminary observations about the q-level of a �eld and de�nethe q-length of a 2 K and the Pythagoras q-number of K which are respectivegeneralizations of the length of a and of the Pythagoras number of K.

Section 3 is devoted to the study of the q-level for an arbitrary quadraticform q. We �rst give upper bounds for the q-level in terms of some familiarinvariants (e.g., the usual level, the Pythagoras number and the u-invariant)before studying the behavior of the q-level and the q-length of an element withrespect to purely transcendental �eld extensions; this leads to a generalizationof a theorem due to Cassels, see Proposition 3.7. Next we make the followingsimple observations (see Proposition 3.9):

fsq(K)j dim q = 1g = f1; � � � ; p(K)g if K is non formally real,fsq(K)j dim q = 1g = f1; � � � ; p(K)g [ f+1g if K is formally real.

We then obtain some conclusions about the integers which belong to the setfsq(K)jdim q = ng; see Corollary 3.10 (2). After that, we show that

If q is a quadratic form with dim q 6 3 such that sq(K) = +1, then(1) If dim q = 1 or 2 then for any k 2 N there is a �eld extension K 0=K

such that sq(K0) = 2k.

(2) If dim q = 3 then for any k 2 N there exist �eld extensions K 0=Kand K 00=K such that sq(K

0) = 22k+23 and sq(K

00) = 22k+1+13 .

This result is proved in Corollary 3.13: for this, the key ingredient is Ho�mann'sseparation Theorem (see Theorem 3.11). Whereas the knowledge of the pos-sible q-levels over a �eld K is equivalent to the knowledge of the Pythagorasnumber of K, our result does not give an explicit way to �nd a quadratic formq with prescribed q-level in general. We make explicit calculations of q-levelsfor familiar �elds, whenever possible.

Section 4 concentrates on the particular case of P�ster forms. We �rst provethat (see Proposition 4.1 and Theorem 4.2):

If ' is a P�ster form over K, then s'(K) is a 2-power or in�nite. Moreoverfs'(K 0) : K 0=K �eld extensiong = f1; 2; � � � ; 2i; � � � ; s'(K)g.

Then, we investigate the behavior of the q-level with respect to quadratic �eldextensions: it is described in Proposition 4.7. We next show that the q-level,q-length and Pythagoras q-number share similar properties with their respectivegeneralizations (see Proposition 4.9 and Proposition 4.16).

A sharp lower bound for the cardinality of the group K�=K�2

in terms ofthe level of K was found by A. P�ster who proved that if K is a �eld whose

level is 2n then j K�

K�2 j > 2n(n+1)=2 (see [20, Satz 18 (d)]). The examples K = Fq,q � 3 mod 4 or K = Q2 show that this inequality is best possible for n � 2.Recall that an element a of K� is represented by q if there exists v 2 V suchthat q(v) = a. Denote by DK(q) the set of values represented by q. In the casewhere q is a P�ster form, DK(q) is a subgroup of K� ([11, Proposition X.2.5])

3

which contains K�2

, hence it is a natural thing to wonder if one may obtain alower bound for the cardinality of the group K�=DK(q) in terms of the q-levelof K. In fact we obtain the following result (see Theorem 4.11):

Let q be a P�ster form over a �eldK whose q-level is 2n. Then jK�=DK(q)j >2n(n+1)=2 and this lower bound is sharp.

We also draw some direct consequences of this lower bound. We then studythe behavior of the Pythagoras q-number with respect to �eld extensions L=Kof �nite degree in Proposition 4.20: in fact pq(L) is smaller than [L : K]pq(K)thus generalizing a classical result due to P�ster.

In Section 5, we investigate possible characterizations of the �niteness of theq-level. In the case of P�ster forms, a complete characterization can be given(see Proposition 5.3). In the general case, we easily observe that if sq(K) is�nite then q is not totally positive. It turns out that when dim q = 1 or 2 theconverse is always true (see Proposition 5.8) but this is not the case anymore ifdim q = 3. We also characterize all �elds K for which for every q the �nitenessof sq(K) implies that q is not totally positive (see Theorem 5.11).

2 Preliminaries

In this paper, the characteristic of the base �eld K will always supposed to bedi�erent from 2 and all quadratic forms are implicitly supposed to be nonde-generate. The notation ha1; � � � ; ani will refer to the diagonal quadratic forma1X1

2 + � � �+ anXn2. Every quadratic form q over K can be diagonalized, that

is q is isometric to a diagonal quadratic form ha1; � � � ; ani which we denote byq ' ha1; � � � ; ani. For a quadratic form q and an scalar a 2 K�, a �q denotes theform q scaled by a. We will denote by W (K) the Witt ring of K and by I(K)its fundamental ideal. Its nth power is denoted by In(K) and is additively gen-erated by the (n-fold) P�ster forms hha1; � � � ; anii := h1;�a1i � � � h1;�ani.A quadratic form � over K is a P�ster neighbor if there exists a P�ster form' and a 2 K� such that 2 dim(�) > dim(') and a � � is a subform of '. Inthis case, it is known that � is isotropic if and only if ' is isotropic if and onlyif ' is hyperbolic. For a positive integer n and a quadratic form q, we use thenotations

�n = n� h1i and �n;q = n� q:

We also denote N0 = N n f0g.The length of an element a 2 K� denoted by `(a) is the smallest integer n

such that a is a sum of n squares; if such an n does not exist, we put `(a) = +1.Note that s(K) = `(�1). Following [22, Ch. 6, p.75], we denote by �K� the setof all elements in K� which can be written as a sum of squares in K.

The Pythagoras number of K is de�ned to be

p(K) = supf`(a) j a 2 �K�g 2 N0 [ f+1g:

Recall that K is non formally real if and only if K� = �K�; in that case, p(K)is always �nite. To see this, if we put s = s(K) then the form �s+1 is isotropic,hence universal and p(K) 6 s+1. As �1 is not a sum of s�1 squares, we obtainthat p(K) 2 fs(K); s(K) + 1g. If K is formally real, p(K) can either be �niteor in�nite. D. Ho�mann has shown that each integer can in fact be realized as

4

the Pythagoras number of a certain (formally real) �eld: see [7] or [8, Theorem5.5]. For further details about the level and the Pythagoras number, the readermay also consult [22, Ch. 3, Ch. 7] or [12].

The u-invariant of K, which is denoted by u(K), is de�ned to bemax(dim q)where q ranges over all anisotropic quadratic forms over K if such a maximumexists, and we de�ne u(K) = +1 otherwise. Note that u(K) is also the mini-mal integer n for which all quadratic forms of dimension strictly greater than n(resp. greater or equal than n) are isotropic (resp. universal) over K.

Let (V; q) be a quadratic form over K. We de�ne the q-length of an elementof K and the Pythagoras q-number of K as follows.

De�nition 2.1. (1) The q-length of an element a 2 K� denoted by `q(a) isde�ned by

`q(a) = minfn j 9(v1; � � � ; vn) 2 V n; a = q(v1) + � � �+ q(vn)g

if such an n exists and by `q(a) = +1 otherwise.(2) Let �qK

� be the set of all elements x in K� for which there exists an integern such that the form �n;q represents x. The Pythagoras q-number is de�ned by

pq(K) = supf`q(a) j a 2 �qK�g;

if such an n exists and by pq(K) = +1 otherwise.

As sh1i(K) = s(K), `h1i(a) = `(a) and ph1i(K) = p(K), the q-level, theq-length of a and the Pythagoras q-number can be regarded as respective gen-eralizations of the level, the length of a and the Pythagoras number. Note alsothat sq(K) = 1 if and only if q represents �1, the number sq(K) only dependson the isometry class of q and shai(K) = `(�a).Remarks 2.2. (1) For all a 2 K�, we have pa�q(K) = pq(K), since we obviouslyhave `a�q(b) = `q(a

�1b) for all b 2 K.(2) For every positive integer n, we can easily check that `n�q(a) = d 1n`q(a)eand pn�q(a) = d 1npq(a)e .(3) If a 2 K� and L=K is a �eld extension of odd degree then `q(a) = `qL(a).

In the sequel it is convenient to introduce the following notations. If K is a�eld and n is a positive integer greater or equal than 1, we put

L(n;K) = fsq(K) j q is a quadratic form of dimension n over Kg

and we set L(K) =Sn2Nnf0g L(n;K). If q is a quadratic form over K, let

Lq(K) = fsqK0 (K 0) j K 0=K �eld extensiong:

3 General results

3.1 Upper bounds

In the following lemma, we list some properties concerning the q-level of a �eld.

5

Lemma 3.1. Let K be a �eld and q be a quadratic form over K.(1) We have 1 6 sq(K) 6 s(K) + 1.(2) We have 1 6 sq(K) 6 inffshai(K) : a 2 DK(q)g = inff`(�a) : a 2 DK(q)g.More generally, if q0 is a subform of q, then sq(K) 6 sq0(K).(3) If L=K is a �eld extension then sq(K) > sqL(L).(4) If L=K is a �eld extension whose degree is odd then sq(K) = sqL(L).

(5) For every positive integer n we have sn�q(K) =lsq(K)n

m.

Proof. To prove (1), we may assume that s(K) = n < +1. In this case, thequadratic form �n+1 is isotropic so the quadratic form �n+1;q is isotropic, henceuniversal. In particular, �n+1;q represents �1. This proves (1).

For statement (2), it su�ces to prove the second property. If �n;q0 represents�1 for a certain n, then it is also the case for �n;q, hence (2).

Finally, (3) is trivial, (4) follows from the strong version of a theorem ofSpringer (see [22, Ch. 6, 1.12]) and (5) follows from 2.2 (2).

Examples 3.2. (1) The upper bound sq(K) 6 s(K)+1 given in 3.1 (1) is sharp.If K = Q(i) and q = h2i, we easily see that s(K) = 1 and sq(K) = `(�2) = 2.(2) In the second statement of 3.1 (2), it is possible to have sq0(K) = sq(K)for a proper subform q0 of q. For instance consider the forms q0 = h�3i andq = h�3; Xi over Q(X), we have sq(Q(X)) = sq0(Q(X)) = 3.

The purpose of the following proposition is to give upper bounds for theq-level of a �eld K in terms of some classical invariants of K.

Proposition 3.3. Let K be a �eld and let q be a quadratic form over K.(1) If sq(K) < +1 then sq(K) 6 p(K) (see also Proposition 4.9 (2)).

(2) If K is not formally real, we have sq(K) 6

�u(K)

dim(q)

�6 u(K).

Proof. To prove (1), one may assume that p = p(K) < +1. Let q = ha1; � � � ; anibe a diagonalization of q. By assumption, there exists an integer m and vec-tors v1; � � � ; vm such that �1 = q(v1) + � � � + q(vm). It follows that �1 =a1�1+ � � �+an�n where �1; � � � ;�n 2 �K� are sums of at most m squares. Bythe de�nition of the Pythagoras number, each �i can be written as a sum of atmost p squares. This fact readily implies that sq(K) 6 p(K). An alternativeproof can also be obtained using Lemma 5.1 and Corollary 2.5 of [2].

(2) One may assume that u(K) < +1. Then every quadratic form q ofdimension greater or equal than u(K) is universal. It follows that if n�dim(q) >u(K) then sq(K) 6 n, hence the result.

Remark 3.4. In the previous proposition, the bound given in (1) is sharp forany non formally real �eld K as Proposition 3.9 shows. We now show that

the inequality sq(K) 6lu(K)dim(q)

mis sharp for any prescribed dimension. Let n

be a positive integer and choose m such that n < 2m. Let F be a �eld suchthat s(F ) = u(F ) = 2m (it is even possible to construct a �eld F such that

s(F ) = u(F ) = p(F ) = 2m, see [8, �5.2]). If we put r =lu(F )n

mthen

u(F )

n6 r <

u(F )

n+ 1

6

so rn > u(F ) which implies that s�n(F ) 6 r. Moreover (r� 1)n < u(F ) = s(F )so �(r�1)n+1 is anisotropic which shows that s�n(F ) = r as claimed.

Corollary 3.5. Let K be a �eld.(1) If q is a quadratic form over K, we have Lq(K) � f1; � � � ; sq(K)g.(2) If n > 1 is an integer then L(n;K) � f1; � � � ; hng [ f+1g where hn =

min(p(K);lu(K)n

m). In particular, if K is non formally real and n > u(K) then

L(n;K) = f1g.Proof. This follows from Lemma 3.1 (3) and Proposition 3.3.

3.2 Behavior under purely transcendental extensions

The following two results generalize two classical theorems concerning quadraticforms under transcendental �eld extensions in the framework of q-levels.

Proposition 3.6. Let q be a quadratic form over K and let K 0=K be a purelytranscendental �eld extension of K. Then sq(K) = sq(K

0) = sq(K((t))).

Proof. By Lemma 3.1 (3), we have sq(K) > sq(K0), hence we may assume that

sq(K0) = n < +1. The quadratic form �n;q represents �1 over K 0 and by

Cassels-P�ster's Theorem (see [11, Theorem IX.1.3]), �n;q already represents�1 over K, hence the �rst equality. The second equality is proved similarly.

A theorem due to J. W. S. Cassels asserts that the polynomial P = 1+X21 +

� � � + X2n is not a sum of n squares over L = K(X1; � � � ; Xn) if K is formally

real (see [11, Corollary IX.2.4]). Note that this is equivalent to `(P ) = n + 1over L. In the same vein we obtain the following result:

Proposition 3.7. For m > 1 and n > 0, let q = ha1; � � � ; ami be a quadratic

form over a �eld K and set L = K(X(i)j ; i = 1; � � � ; n; j = 1; � � � ;m). If �n;q is

anisotropic, then

`qL(a1 +Xi;j

aj(X(i)j )2) = n+ 1:

In particular, this is the case if sq(K) = +1.

Proof. If n = 0, we have to prove that `q(a1) = 1, which is obvious since a1 is

represented by q. Assume now that n > 1, and set a = a1 +Xi;j

aj(X(i)j )2.

Since a1 is represented by q, and thus by qL, we have `qL(a) 6 n+1. Assumethat `qL(a) < n+ 1, so that a is represented by �n;qL . Since �n;q is anisotropicover K by assumption, any subform q0 of �n;q is also anisotropic over K. Thisimplies that q0L is anisotropic over L, since L=K is purely transcendental. Con-sequently, q0M is anisotropic over M for any sub�eld M of L, and any subformq0 of �n;q. A repeated application of a theorem due to Cassels (cf. [25, Theorem

3.4, p.150]) then shows that a1+a1(X(1)1 )2 is represented by ha1i over K(X

(1)1 ).

This implies that 1 + (X(1)1 )2 is a square in K(X

(1)1 ), hence a contradiction.

Let us prove the last part of the proposition. If sq(K) = +1 but �n;q isisotropic for n > 1 then �n;q is universal, hence represents �1, so sq(K) 6 n,and we have a contradiction. Now apply the �rst part to conclude.

Corollary 3.8. With the same hypotheses as in 3.7, we have `qL(Pi;j aj(X

(i)j )2) =

n.

7

3.3 Values of the q-level

3.3.1 About L(n;K)

Proposition 3.9. If K is non formally real then L(1;K) = L(K) = f1; � � � ; p(K)g.If K is formally real then L(1;K) = L(K) = f1; � � � ; p(K)g [ f+1gProof. As L(1;K) � L(K), the direct inclusions come from Proposition 3.3 (1)in both cases. It remains to show that the sets on the right-hand sides areincluded in L(1;K).

For this, we distinguish between the cases p(K) < +1 and p(K) = +1.Note that +1 = sh1i(K) 2 L(1;K) in the formally real case.

If p = p(K) < +1, there exists a 2 �K� such that `(a) = p. Writea = x21 + � � �+ x2p where xi 2 K. For 1 6 i 6 `(a) put �i = x21 + � � �+ x2i . Then`(�i) = i = sh��ii(K), hence the result.

Suppose now that p(K) = +1 and let n be a �xed integer. By de�nition ofthe Pythagoras number, there exists a 2 �K� such that q = `(a) > n. Put a =x21+ � � �x2q where xi 2 K. If we choose b = x21+ � � �x2n then `(b) = n = sh�bi(K)and this concludes the proof.

This result has several immediate consequences.

Corollary 3.10. (1) There exists a �eld K such that for every integer n, thereexists a quadratic form q over K with sq(K) = n.(2) If n > 1 is an integer then f1; � � � ; lng � L(n;K) where ln = bp(K)=nc. Inparticular, if p(K) = +1 then L(n;K) = N0 [ f+1g for any n > 1.

Proof. (1) By Proposition 3.9, it su�ces to �nd K with p(K) = +1. Followingthe proof of [8, Theorem 5.5], putK = F (X1; X2; � � � ) with an in�nite number ofvariables Xi over a formally real �eld F . If n 2 N0, put Pn = 1+X2

1 + � � �+X2n.

Then `(Pn) = n+1 over F (X1; � � � ; Xn) (by Cassels' theorem mentioned above)and K (as K=F is purely transcendental). Thus p(K) = +1.

(2) It su�ces to prove the �rst assertion. If we �x 1 6 i 6 ln then n � i 6p(K). By Proposition 3.9, there exists a form qi = haii with sqi(K) = n� i. ByLemma 3.1 (5), the form �n;qi 2 L(n;K) has level i, hence the result.

3.3.2 About Lq(K)

We now focus on the proof of Corollary 3.13, mentioned in the Introduction, forwhich the crucial ingredient is the following result due to D. W. Ho�mann in[6, Theorem 1].

Theorem 3.11 (Ho�mann's Separation Theorem). Let q and q0 be anisotropicquadratic forms over K such that dim(q) 6 2n < dim(q0). Then q is anisotropicover K(q0), the function �eld of the projective quadric de�ned by q0 = 0.

The key fact for us is the second assertion of the following proposition. The�rst assertion is a direct consequence of Corollary 3.10 (1) but can also be provedindependently using Ho�mann's result.

Proposition 3.12. Let n be a positive integer and K be a formally real �eld.(1) There exist a �eld extension K 0=K and a quadratic form q over K 0 suchthat sq(K

0) = n.

8

(2) Let q be a quadratic form over K with sq(K) = +1. If there exists apositive integer k such that 1 + (n � 1) dim q 6 2k < 1 + n dim q then thereexists a �eld extension K 0=K such that sq(K

0) = n. In fact one can chooseK 0 = K(h1i ? �n;q).

Proof. For (1), we can suppose that n > 1. Let m and r be two integers suchthat

2r � 1

n< m 6

2r � 1

n� 1:

These two integers do exist: in fact, it su�ces to choose m and r satisfying

r >ln(n2 � n+ 1)

ln(2)which implies that

2r � 1

n� 1� 2r � 1

n=

2r � 1

n(n� 1)> 1.

For j > 1, let 'j = h1i ? �j;q where q is the anisotropic quadratic form q =�m. The quadratic forms 'n and 'n�1 are anisotropic over K. If K 0 = K('n)then, by Theorem 3.11 and by the choice of m and r, ('n�1)K0 is anisotropic,hence sq(K

0) > n. As ('n)K0 is isotropic, we obtain sq(K0) = n, hence (1). For

(2), the fact that sq(K) = +1 implies similarly that sq(K0) = n forK 0 = K( n)

where n = h1i ? �n;q.

Corollary 3.13. Let q be a quadratic form of dimension at most 3 such thatsq(K) = +1.(1) If q has dimension 1 or 2 then 2k 2 Lq(K) for any k 2 N.(2) If q has dimension 3 then 22k+2

3 , 22k+1+13 2 Lq(K) for any k 2 N.

Proof. (1) Consider the form '2k = h1i ? �2k;q. As dim q = 1 or 2, we have1 + (2k � 1) dim q 6 2k+dim q�1 < 1 + 2k dim q. By Proposition 3.12 (2), weobtain sq(K

0) = 2k where K 0 = K('2k).

(2) If n = 22k+23 , we have 1+(n�1) dim q 6 22k < 1+(n�1) dim q, therefore

the existence of K 0 follows from Proposition 3.12 (2). The proof of the secondassertion is similar and is left to the reader.

Corollary 3.14. Let K be a �eld and let n be a nonnegative integer.(1) If s(K) = +1 then there exists a �eld extension K 0=K with s(K 0) = 2n.(2) If a 2 K and `(a) = +1 over K then there exists a �eld extension K 0=Ksuch that `(a) = 2n over K 0.

3.4 Some examples and calculations

We now make explicit calculations of q-levels in many familiar �elds. Note thatsh�1i(K) = 1 for any �eld K and that sh1i(K) = +1 when K is formally real.The results 3.9, 3.5 (2) and 3.10 (2) are used without further mention.

3.4.1 Non formally real �elds

Algebraically closed �elds : in such a �eld K, L(n;K) = Lq(K) = f1g forany n and q.

Finite �elds : if K is a �nite �eld, L(K) = f1; 2g. As u(K) = 2, we haveL(n;K) = Lq(K) = f1g for n > 2, dim(q) > 2. Moreover, the quadratic formq = hai has q-level 1 if and only if �a is a square, otherwise it has q-level 2.

9

Non dyadic local �elds : in such a �eld K with residue �eld K, denote byU the group of units and choose a uniformizer �. Any quadratic form q canbe written @1(q)?�@2(q) where @1(q) = hu1; � � � ; uri, @2(q) = hur+1; � � �uni forui 2 U , i = 1; � � � ; n. The forms @1(q) and @2(q) are respectively called the �rstand second residue forms of q. By a Theorem of T. A. Springer, q is anisotropicover K if and only if @1(q) and @2(q) are anisotropic over K (see [11, PropositionVI.1.9]). This reduces the calculation of the q-levels over K to calculations ofsome levels over K.

Now, p(K) = min(s(K) + 1; 4) (see [22, Ch. 7, Examples 1.4 (1)]), henceL(K) = f1; 2g (resp. f1; 2; 3g) if jKj � 1 mod 4 (resp. jKj � 3 mod 4).

Take u 2 U with u =2 K2and put q = h�ui. Then h1;�ui is anisotropic

but h1;�u;�ui is isotropic by Springer's Theorem, hence sq(K) = 2. If jKj � 3mod 4, take q0 = h�i. As �1 is not a square in K, Springer's Theorem showsthat sq0(K) > 2, hence sq0(K) = 3.

Dyadic local �elds : we have p(K) = min(s(K) + 1; 4) and s(K) = 1; 2or 4. If K = Q2, we have s(K) = 4 (see [11, Examples XI.2.4]), henceL(Q2) = f1; 2; 3; 4g. We have sh�2i(Q2) = 2, sh2i(Q2) = 3 and sh1i(K) = 4.

The �elds Ln = K(X1; � � � ; Xn) and Mn = K((X1)) � � � ((Xn)) : if s(K) = 2m

then p(Ln) = p(Mn) = 2m + 1 ([22, Ch. 7, Proposition 1.5]), hence L(Ln) =L(Mn) = f1; � � � ; 2m + 1g. Recall that p(K) 2 f2m; 2m + 1g.

Each value in f1; � � � ; p(K)g is attained as a q-level over K and sq(K) =sq(Ln) = sq(Mn) by Proposition 3.6. If p(K) = s(K) = 2m, let q = hXni overLn. As Mn is a local �eld with residue �eld K((X1)) � � � ((Xn�1)) of level 2

m

and uniformizer Xn, we have sq(Ln) = sq(Mn) = 2m+1 by Springer's theorem.

Non formally real global �elds : if K is a number �eld, we have p(K) =min(s(K) + 1; 4) by Hasse-Minkowski principle (see [22, Examples 1.4 (2)]) ands(K) = 1; 2 or 4 ([11, Theorem XI.1.4]). If K is a function �eld, the previousexamples show that L(K) � f1; 2; 3g.

3.4.2 Formally real �elds

Real closed �elds : over such a �eld, a quadratic form q has q-level +1 ifand only if q is positive de�nite, otherwise it has q-level one.

Formally real global �elds : we have p(K) = 4 (resp. p(K) = 3) if K hasa dyadic place P such that [KP : Q] is odd (resp. otherwise) by [22, Ch. 7,Examples 1.4 (3)].

In particular, L(Q) = f1; 2; 3; 4g. By Hasse-Minkowski principle, any indef-inite quadratic form q of dimension > 5 is isotropic hence has q-level one, see[11, Ch. VI.3]. Any positive de�nite quadratic form q has an in�nite q-level. Aquadratic form q that is not positive de�nite has a �nite q-level. For examplesh�5i(Q) = 2, sh�6i(Q) = 3 and sh�7i(Q) = 4 as 6 (resp. 7) is a sum of threesquares (resp. four squares) but not a sum of two squares (resp. three squares)in Q.

The �eld R(X1; � � � ; Xn) : its Pythagoras number is 2 if n = 1, is 4 if n = 2and is in the interval [n+ 2; 2n] for n > 3 (the two last results are due to J. W.

10

S. Cassels, W. J Ellison and A. P�ster, see [3] or [11, Examples XI.5.9 (4)]).If n = 1, we thus have L(R(X)) = f1; 2g. By a Theorem due to E. Witt, we

know that any totally inde�nite quadratic form over R(X) of dimension > 3 isisotropic hence has q level one. Consider q = h�(1 +X2)i. As 1 +X2 is not asquare in R(X), q does not represent �1. But h1i ? 2 � q is totally inde�nite as1 is totally positive whereas �(1 +X2) is totally negative, hence it is isotropic.This proves that sq(R(X)) = 2.

The �eld Q(X1; � � � ; Xn) : its Pythagoras number is 5 if n = 1 (by a resultdue to Y. Pourchet, see [23]). For n > 2, we only know that p(Q(X1; X2)) 6 8,p(Q(X1; X2; X3)) 6 16 and p(Q(X1; � � � ; Xn)) 6 2n+2. The �rst two results aredue to J.-L. Colliot-Thélène and U. Jannsen in [4] and the last one is due to J.K. Arason.

If n = 1, the study done for Q together with Proposition 3.6 show thatsh�5i(Q(X)) = 2, sh�6i(Q(X)) = 3 and sh�7i(Q(X)) = 4. By a result of Y.Pourchet (see [23, Proposition 10]), for d 2 Z, the polynomial X2 + d is a sumof exactly �ve squares of Q[X] if and only if s(Q(

p�d)) = 4 which in turnis equivalent to d > 0 and d � �1(mod 8) (see [11, Remark XI.2.10]). Forq = h�(X2 + 7)i, this readily implies that sq(Q(X)) = 5.

3.4.3 A remark

Suppose that K is a non formally real �eld. In all the above examples in whichthe values of s(K), u(K) and p(K) are known, we have p(K) = min(s(K) +1; u(K)). We always have p(K) 6 min(s(K) + 1; u(K)) but there is no equalityin general.

To see this, we use the construction of �elds with prescribed even u-invariantdue to A. S. Merkurjev (see [18] or [8, Section 5]).

Theorem 3.15 (Merkurjev). Let m be an even number and E be a �eld. Thereexists a non formally real �eld F over E such that u(F ) = m and I3(F ) = 0.

Putm = 2n+2. The proof of Merkurjev's result is based upon a constructionof an in�nite tower of �elds Fi. More precisely F0 = E(X1; Y1 � � � ; Xn; Yn) andif Fi is constructed then Fi+1 is the free compositum over Fi of all function �eldsFi( ) where ranges over (1) all quadratic forms in I3(Fi) (2) all quadraticforms of dimension 2n+ 3 over Fi. Then F = [1i=0Fi is the desired �eld.

Choose a �eld E with s(E) = 4. Then there exists a �eld F over E suchthat u(F ) = 2n+ 2 and I3(F ) = 0. If we consider the 2-fold anisotropic P�sterform ' = h1; 1; 1; 1i over E, the form ' stays anisotropic over any Fi of thetower as it cannot become hyperbolic over the function �eld of a quadratic formof dimension strictly greater than 4 by [11, Theorem X.4.5], hence s(F ) = 4.Now, choose n = 2 so that min(s(F ) + 1; u(F )) = 5. Let x 2 K� and considerthe quadratic form '?h�xi: it is isotropic since it is the P�ster neighbor ofhh�1;�1; xii which is hyperbolic as I3(F ) = 0. As s(F ) = 4, this means that 'is anisotropic over F , hence every x 2 K� is a sum of at most 4 squares in K.Now, �1 is a sum of 4 squares and is not a sum of three squares which showsthat p(F ) = 4 < min(s(F ) + 1; u(F )) = 5.

11

4 The case of P�ster forms

4.1 Values of the q-level : the case of P�ster forms

Proposition 4.1. Let ' be a P�ster form over a �eld K. Then s'(K) is eitherin�nite or a power of two.

Proof. This proof is similar to P�ster's proof of the fact that the level of a �eld,when �nite, is a power of two. As pointed out in the Introduction, when 'is a one or two-fold P�ster form, the result is proved in [15, Prop. 1.5] in thecontext of hermitian levels of �elds and quaternion algebras. As sh1i(K) = s(K),one may assume that dim(') = 2n with n > 1 and that s'(K) > 1. Lets'(K) > m and let r be a positive integer such that 2r < m 6 2r+1. Byhypothesis, the quadratic form h1i ? �m�1;' is anisotropic (otherwise we wouldhave s'(K) < m). Moreover,

dim (h1i ? �m�1;') = 1 + 2n(m� 1) > 1 + 2n+r > 2n+r =1

2� 2n+r+1:

This form is a P�ster neighbor of �2r+1;' which is anisotropic. The subformh1i ? �2r+1�1;' is also anisotropic so s'(K) > 2r+1, hence the result.

Theorem 4.2. If ' is a P�ster form over K, then s'(K) is a 2-power orin�nite. Moreover L'(K) = f1; 2; � � � ; 2i; � � � ; s'(K)g.Proof. The direct inclusion follows from Proposition 4.1 together with Lemma3.1(3). To prove the converse, �rst note that the two numbers 1 and s'(K) arerespectively attained, as the '-level, over K(') and K. Let n > 0 be such that2n < s'(K). For an integer k, put 'k = �2k;'. Then 'k is a P�ster form for anyk. Put K 0 = K('n+1). As 2n < s'(K), 'n is not hyperbolic over K, Cassels-P�ster subform Theorem shows that ('n)K0 is anisotropic which implies thats'(K

0) > 2n�1. Also s'(K0) is a 2-power by the previous proposition. Moreover,

the form n = h1i ? 'n is a P�ster neighbor of 'n+1. As 'n+1 is hyperbolicover K 0, n is isotropic over K 0 so s'(K

0) = 2n, hence the result.

Corollary 4.3. Let ' be a n-fold P�ster form over K and q be a subform of 'such that dim(q) > 2n�1. Then s'(K) 6 sq(K) 6 2 s'(K).

Proof. If s'(K) = +1, we have sq(K) = +1 by Lemma 3.1 (2) so suppose thats'(K) < +1. By Proposition 4.1 we have s'(K) = 2r where r is an integerand we obtain s'(K) = 2r 6 sq(K) by Lemma 3.1 (2). If n = 1, q = ' and theresult is clear. Suppose n > 2. By the de�nition of the '-level the quadraticform h1i ? �2r;' is isotropic. This form is a P�ster neighbor of �2r+1;' which isthus hyperbolic. Finally, �2r+1;q is isotropic being a P�ster neighbor of �2r+1;'.Hence �1 is represented by �2r+1;q and thus sq(K) 6 2r+1 = 2 s'(K).

Example 4.4. The upper bound given in Corollary 4.3 is sharp. Let p 6= 2 bea prime number and K = Qp. If ' is the unique 4-dimensional anisotropic formover K and q is the pure subform of ', we have sq(K) = 2 and s'(K) = 1.

12

4.2 Behavior under quadratic extensions

We now investigate the behavior of the q-level under quadratic extensions in thecase of P�ster forms.

Lemma 4.5. Let K be a �eld, (V; ') a P�ster form over K and let L = K(pd)

be a quadratic �eld extension of K. Then we have `'(�d) 6 2s where s = s'(L).

Proof. By hypothesis, there exist 2s vectors v1; w1; � � � ; vs; ws in V such that

�1 = '(v1 1 + w1 pd) + � � �+ '(vs 1 + ws

pd): (1)

Denote by b' the bilinear form associated to '. Then for v; w 2 V we have

'(v 1+wpd) = '(v)+ d'(w)+2b'(v; w)pd. From equation (1) we obtain

the following equation

�1 = ('(v1) + � � �+ '(vs)) + d('(w1) + � � �+ d'(ws)); (2)

Thus �d('(w1) + � � �+ '(ws))2 is equal to

('(v1) + � � �+ '(vs))('(w1) + � � �+ '(ws)) + ('(w1) + � � �+ '(ws)): (3)

As �s;' is a P�ster form, it is multiplicative, hence the �rst term of the expres-sion (3) is represented by �s;'. The expression (3) can therefore be representedby the form �2s;', hence the result.

Example 4.6. Note that the bound obtained in the previous lemma is optimal.Take K = Q, d = �3 and L = Q(

pd) = Q(

p�3). Let ' = h1; 1i. As �1 =

( 1+p�32 )2+( 1�

p�32 )2, the element �1 is represented by ', hence s'(L) = 1. As

�d = 3 is represented by �2;' but it is not represented by ' we have `'(3) = 2.

Proposition 4.7. Let ' be a P�ster form over a �eld K. Let d 2 K be anelement such that `'(�d) = n. If L = K(

pd), we have s'(L) = 2k or 2k�1

where k is determined by 2k 6 n < 2k+1.

Proof. As `'(�d) = n, there exist vectors v1; � � � ; vn such that �d = '(v1) +� � �+ '(vn). We thus have

�1 = '(v1 1pd) + � � �+ '(vn 1p

d);

so s'(L) 6 n and s'(L) 6 2k by Proposition 4.1. It su�ces to prove that thecase s'(L) 6 2k�2 cannot occur. If s'(L) 6 2k�2 the previous lemma impliesthat `'(�d) 6 2k�1 < n, which is a contradiction.

Remark 4.8. In Proposition 4.7, if ' = h1i the possibility s'(K) = 2k�1 isruled out, see [25, Ch. 4, Thm. 4.3] or [12, Prop. 3.3]. In general, both values 2k

and 2k�1 can happen as we now show. For instance, by taking K = Q, d = �3and ' = h1; 1i we obtain n = `'(3) = 2 and so k = 1. In this case s'(L) = 1 =2k�1. Now take d = �1 and ' = h1i. We obtain n = `'(1) = `(1) = 1 and sok = 0. In this case we have s'(L) = s(Q(i)) = 1 = 2k.

13

4.3 Values represented by a P�ster form

Recall that if ' is a P�ster form over K, the set DK(') of non-zero valuesrepresented by ' is a subgroup ofK�. We �rst state some useful facts concernings'(K) and p'(K). The proof is adapted from [20, Satz 18].

Proposition 4.9. Let (V; ') be a P�ster form over K with s'(K) < +1.Then:(1) For every a 2 K, there exists an integer n such that a 2 DK(�n;').(2) We have p'(K) 2 fs'(K); s'(K) + 1g.(3) If t = p'(K) then DK(�t;') = K�.(4) If K is non formally real and s'(K) < p'(K) then 2 s'(K) dim(') 6 u(K).

Proof. In the proof, we will use the notations s = s'(K) and p = p'(K).(1) Note that a = (a+1

2 )2 + (�1)(a�12 )2 and that the P�ster form �s;' rep-

resents �1. As P�ster forms are multiplicative (see [11, Theorem X.2.8]), �s;'represents (�1)(a�1

2 )2, hence �s+1;' represents (a+12 )2 + (�1)(a�1

2 )2 = a.(2) comes from the fact that s = `'(�1) 6 p and from (1) as �s+1;' is

universal and (3) is a consequence of (1).(4) One may assume that u(K) < +1. As s < p, the quadratic form �s;'

is not universal (otherwise, we would have s = p). So there exists an element�a 2 K� which is not represented by this form. De�ne = h1; ai �s;'. Weclaim that is anisotropic. If it is isotropic, as �s;' is anisotropic, there existelements b; c 2 K� both represented by �s;' such that b = �ac. As �s;' ismultiplicative, it represents bc hence �a which is a contradiction. The form is thus anisotropic with dimension 2s� dim('), hence the result.

Remark 4.10. If K = Q, q = h�5i then by subsection 3.4, sq(K) = 2 andpq(K) = 4 (as `q(�35) = 4, p(Q) = 4 and DQ(�n;q) = �5Q+ if n > 4). Hencestatements (1), (2), (3) in Proposition 4.9 are not valid if ' is not a P�ster form.

Consider a �eld F with u(F ) = s(F ) = 4 (given by the construction in �3.4.3for example) and take q = h�1;�1;�1i. Then sq(F ) = 1, pq(F ) = 2 (as q doesnot represent 1 but �2;q is universal) and 2 sq(F ) dim(q) > u(F ) = 4, hencestatement (4) is also false in general.

Theorem 4.11. Let ' be a P�ster form over a �eld K whose '-level is 2n.Then jK�=DK(')j > 2n(n+1)=2.

Proof. To prove this result, we adapt P�ster's proof of the fact that jK�=K�2 j >2n(n+1)=2 whenever s(K) = 2n. In the sequel, we set Gj := DK(�j;') � K� forany integer j and write s for s'(K). We have :

� 1 = '(e1) + � � �+ '(es) (4)

Note that the conclusion is clear if n = 0 and is true if n = 1 (in this case,�1 =2 DK('), hence jK�=DK(')j > 2). One may assume that n > 2. Letj = 2i where 0 6 i < n. We claim that the elements a1 = '(e1) + � � �+ '(e2j),a2 = '(e2j+1)+ � � �+'(e4j), � � � are pairwise non congruent modulo Gj . Indeed,if we would have '(e2j+1)+� � �+'(e4j) = c('(e1)+� � �+'(e2j)) for some c 2 Gjthen we would obtain '(e1) + � � �+'(e4j) = (1+ c)('(e1) + � � �+'(e2j)) 2 G2j

which would contradict the minimality of s in (4), hence the claim.The elements a1; a2; � � � are not in Gj (otherwise, this would also contradict

the minimality in (4)), hence there are at least 1+ s2j = 1+ 2n�i�1 elements in

14

G2j=Gj (then2j elements ai together with the element 1). Now, G2j=Gj injects in

K�=DK('). If G2j=Gj has in�nite order, there is nothing to prove. Otherwise,G2j=Gj is a 2-group and what we have done above shows that [G2j : Gj ] > 2n�i.We have the following sequence of inclusions

DK(') = G1 � G2 � � � � � G2n�1 � G2n � K�:

We thus obtain jK�=DK(')j > jG2n=G1j and

jG2n=G1j = jG2n=G2n�1 j � � � � � jG2=G1j >n�1Yi=0

2n�i = 2n(n+1)

2 :

Remarks 4.12. (1) The lower bound indicated in Theorem 4.11 is attained aswe see by taking K = Qp, p 6= 2 and ' the unique 4-dimensional anisotropicform over K since jK�=DK(')j = 1 and s'(K) = 1.(2) For a �eld K of level 2n, we have already pointed out that the lower bound

j K�

K�2 j > 2n(n+1)=2 is best possible for n � 2. For higher n, however, a result

due to D. Z. Djokovi¢ and re�ned by D. B. Leep shows that j K�

K�2 j > 22n+2�n,

see [5] and [1, Remark 5.2]. In the case where n = 3 and n = 4, this boundhas in turn been re�ned by K. J. Becher: if s(K) = 8 (resp. s(K) = 16) then

j K�

K�2 j > 512 (resp. 215) (see [1, Theorem 5.3]).

Corollary 4.13. Let K be a �eld with level 2n and t = 2m > 0. Consider thesubgroup G = fa21 + � � � + a2t 6= 0 j ai 2 Kg � K�. If t > s(K) then we haveG = K�. If t 6 s(K) then we have jK�=Gj > 2k(k+1)=2 where k = n�m.

Proof. Consider the P�ster form ' = �t. We have G = DK('). If t > s(K),' is isotropic, hence G = K�. If t 6 s(K) then by Lemma 3.1 (5), we have

s'(K) =ls(K)t

m= 2n�m = 2k and the result follows from Theorem 4.11.

Corollary 4.14. Let L = K(pd) be a quadratic extension of K where d 2

K� nK�2

and ' = h1;�di. If s'(K) = 2n then jK�=NL=K(L�)j > 2n(n+1)=2.

Proof. As NL=K(L�) = DK('), the result follows from Theorem 4.11.

Remark 4.15. It is relevant to mention the following formula obtained by D.W. Lewis in [13, Cor. after Prop. 3]: if L=K is a quadratic �eld extension,

jL�=L�2jjK�=NL=K(L�)j = 1

2jK�=K�2j2:

The below result has �rst been proved by A. P�ster for the form ' = h1i in[19]. Our reformulation is taken from [1].

Proposition 4.16. Let ' be a P�ster form over a �eld K.(1) For every x; y 2 K� we have `'(xy) 6 `'(x) + `'(y)� 1.(2) For every x 2 K� we have s'(K) 6 `'(x) + `'(�x)� 1.

15

Proof. First note that (2) is a consequence of (1) as s'(K) = `'(�x2). If`'(x) = +1 or `'(y) = +1, the result is trivial. One may assume that `'(x)and `'(y) are both �nite and that `'(x) 6 `'(y) without loss of generality. Weprove the result by induction over `'(y). If `'(y) = 1 then we obtain `'(x) = 1and the result holds, because ' is multiplicative. De�ne r to be the positiveinteger such that 2r�1 < `'(x) 6 2r. One may write `'(y) = 2rk + l wherek > 0 and 1 6 l 6 2r. Two cases are to be considered: k = 0 or k > 1.

If k = 0 then `'(y) = l 6 2r and y is represented by �2r;'. As x is alsorepresented by �2r;', we deduce that xy is represented by �2r;' (because 'is multiplicative). Consequently, `'(xy) 6 2r. But 2r�1 < `'(x) 6 `'(y) so2r + 2 6 `'(x) + `'(y) which shows that `'(xy) 6 `'(x) + `'(y)� 1.

If k > 1, y is represented by the form �2rk;' ? �l;' so we can write y = y0+y00

where `'(y0) = 2rk and `'(y

00) = l. As x and y0 are respectively representedby �2r;' and �2rk;', xy

0 is represented by �2rk;', that is `'(xy0) 6 2rk. As

`'(y00) < `'(y), the induction hypothesis implies that `'(xy

00) 6 `'(x)+`'(y00)�

1 = `'(x) + l � 1. Hence `'(xy) = `'(xy0 + xy00) 6 2rk + (`'(x) + l � 1) =

`'(x) + `'(y)� 1.

Remark 4.17. The two bounds given in the previous proposition are sharp forany P�ster form ': choose x = 1, y = �1 so that `'(1) = 1 and `'(�1) = s'(K).

The �rst inequality does not hold in general. In the formally real case, choose' = h�1i and x = y = �1: then `'(�1) = 1 but `'((�1)� (�1)) = +1. In thenon formally real case, take K = Qp (p odd) and choose x = u where u is a unit

such that u =2 K2and y = � where � is a uniformizer. Choose ' = hu; �i over

K. Then `'(u) = `'(�) = 1. Now `'(u�) = 1 if and only if the quadratic formh�u;��; u�i is isotropic. As this latter form is a P�ster neighbor of hhu; �iiwhich is anisotropic, it follows that `'(u�) = 2 > `'(u) + `'(�)� 1 = 1.

4.4 Pythagoras q-number and �eld extensions

If (V; ') is a P�ster form over a �eld K then for every x; y 2 V there existsz 2 V such that '(x) �'(y) = '(z). As A. P�ster observed, it is also known thatz can be chosen in such a way that its �rst component is of the form b'(x; y)where b' is the bilinear form associated to ', see [22, Ch. 2, Cor. 2.3] or [21,Satz 1]. For the case where ' = �2r (r > 0), this implies in particular that forevery x1; � � � ; x2r ; y1; � � � ; y2r 2 K there exists an identity like

(x21 + � � �+ x22r ) � (y21 + � � �+ y22r ) = (x1y1 + �+ x2ry2r )2 + z22 + � � �+ z22r

where z2; � � � ; z2r are suitable elements of K. We obtain the following lemmaas an immediate consequence of P�ster's observation.

Lemma 4.18. Let (V; ) be a P�ster form over a �eld K. Let = h1i ? 0 where 0 is a subform of of codimension 1. Let r 2 N, m = 2r andx1; � � � ; xm; y1; � � � ; ym 2 V . Then there exist z1; � � � ; zm 2 V such that

( (x1) + � � �+ (xm)) � ( (y1) + � � �+ (ym)) = (z1) + � � �+ (zm)

and there exists z01 in the underlying vector space of 0 such that

(z1) = (b (x1; y1) + � � �+ b (xm; ym))2 + 0(z01)

where b is the bilinear form associated to .

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Proof. Apply [22, Ch. 2, Cor. 2.3] to the P�ster form ' = �2r; .

The following result was proved for the form ' = h1i in [22, Ch. 7, 1.12].

Proposition 4.19. Let (V; ') be a P�ster form over a �eld K. Suppose thats'(K) = +1. Let f(x) 2 K[x] be such that `'jK(x)

(f(x)) < +1. Thendeg(f(x)) = 2n is even and `'jK(x)

(f(x)) 6 p'(K)(n+ 1).

Proof. If p'(K) = +1 the result is trivial. Assume now that p'(K) < +1 andconsider the smallest positive integer m such that f(x) is represented by �m;'over K(x). Using the �rst representation Theorem of Cassels-P�ster (see [11,Theorem IX.1.3]) we obtain that f(x) is represented by �m;' over K[x]. Thisimplies that the degree of f(x) can not be odd, otherwise �m;' would be isotropicover K which contradicts the hypothesis s'(K) = +1. So the degree of f(x)is even and we may suppose that deg(f(x)) = 2n. We proceed by induction onn. If n = 0, then f(x) = c 2 K is a constant polynomial. As f(x) = c is arepresented by �m;' over K(x), it is represented by �m;' over K by SubstitutionPrinciple (see [22, Ch. 1, 3.1]). It follows that `K(x)(f(x)) 6 p'(K), hencethe result. Suppose now that n > 1. Take f(x) = a2nx

2n + � � � + a0 wherea2n; � � � ; a0 2 K. As f(x) is represented by �m;' over K(x) it follows that a2nis represented by �m;' over K. Thus the polynomial f(x)a2n

is also represented by

�m;' over K[x]. Let k be a positive integer such that m 6 2k. The polynomialf(x)a2n

is also represented by �2k;'. It follows that

f(x)

a2n= g1(x) + � � �+ g2k(x) (5)

where the polynomials g1(x); � � � ; g2k(x) are represented by the form ' overK[x]. Let v1(x); � � � ; v2k(x) 2 V [x] be the elements such that '(vi(x)) = gi(x)for every i = 1; � � � ; 2k. By comparing the leading coe�cients of the equation(5) we obtain a relation

1 = b1 + � � �+ b2k (6)

where b1; � � � ; b2k 2 K are represented by '. Let w1; � � � ; w2k 2 V be theelements such that '(wi) = bi for every i = 1; � � � ; 2k. By multiplying therelations of the equations (5) and (6) and applying Lemma 4.18 we obtain:

f(x)

a2n= s(x) + r(x) (7)

where s(x) = (b'(v1(x); w1)+ � � �+ b'(v2k(x); w2k))2 and r(x) is represented by

�m;' over K[x]. Note that s(x) is a monic polynomial with the same degree asf(x). It follows that r(x) is a polynomial whose degree satis�es deg(r(x)) < 2n.As r(x) is represented by �m;', the degree of r(x) should be even, other-wise �m;' would be isotropic over K which is a contradiction. We then havedeg(r(x)) 6 2n� 2. The relation (7) implies that f(x) = a2n s(x)+a2nr(x). Asa2n is represented by �m;' overK, a2nr(x) is also represented by �m;' overK[x].By the induction hypothesis we obtain `'jK[x](a2nr(x)) 6 p'(K)(n). We obvi-ously have `'jK[x](a2n s(x)) = `'(a2n) 6 p'(K). We so obtain `'jK[x]

(f(x)) 6p'(K)(n) + p'(K) = p'(K)(n+ 1).

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Proposition 4.20. (1) (P�ster) Let K be a real �eld and let L=K be a �eldextension of �nite degree. Then p(L) 6 p(K)[L : K].(2) Let L=K be a �eld extension of �nite degree. Let ' be a P�ster form overK. Suppose that s'(K) < +1. Then p'(L) 6 p'(K)[L : K].

Proof. The statement (1) is proved in [22, Ch. 7, 1.13]. It is clear that (1) isa particular case of (2) by taking ' = h1i. To prove (2), it su�ces to provethe result for the case where L = K(�) is a simple extension. Let V be theunderlying vector space of '. Suppose that [L : K] = n. If p'(K) = +1 theconclusion is trivial. Assume that p'(K) < +1. Let � 2 L be an element suchthat r := `'jL(�) < +1. In order to prove the result we have to show thatr 6 p'(K)[L : K]. Every element of the vector space V K L can be written asv01+v1�+� � �+vn�1�n�1 where vi 2 V for every i = 1; � � � ; n. There existw1; � � � ; wr 2 V L such that � = '(w1)+ � � �+'(wr). Let wj =

Pn�1i=0 vij �i

where vij 2 V and j = 1; � � � ; r. Put wj(x) =Pn�1i=0 vij xi 2 V [x] = V K

K[x]. We so have wj = wj(�) for every j = 1; � � � ; r. Consider the polynomialf(x) = '(w1(x)) + � � � + '(wr(x)). The degree of f(x) is even and satis�esdeg(f(x)) 6 2(n�1). According to Proposition 4.19, we have `'jK[x]

6 p'(K)n.By substituting x := �, we obtain `'(�) 6 p'(K)n.

5 Some results on the �niteness of the q-level

It will be convenient to use the following notation in the sequel:

~sq(K) = supfn 2 N j �n;q is anisotropicg:Note that one has ~sa:q(K) = ~sq(K) whenever a is a non-zero element of K.Recall that a quadratic form q over K is called weakly isotropic when thereexists a positive integer m such that �m;q is isotropic : in this case, the lowestsuch m is called the weak isotropy index of q and is denoted by wi(q). We havetaken this notation from [2]. We say that q is strongly anisotropic when it is notweakly isotropic and we write wi(q) = +1. We easily see that ~sq(K) is �nite ifand only if wi(q) is �nite, in which case we have ~sq(K) = wi(q)� 1.

We now compare these invariants to the q-level, when possible.

Lemma 5.1. Let q be a quadratic form over K.(1) We have wi(q) > sq(K).(2) If q represents 1 then sq(K) 2 f~sq(K); ~sq(K) + 1g.(3) If q is a P�ster form over K, we have sq(K) = wi(q)� 1 = ~sq(K).

Proof. If sq(K) = +1 then we obviously have wi(q) = +1 and the inequalityholds. Assume now that s = sq(K) < +1. Then �s�1;q is not isotropic asit would represent �1 otherwise, thus contradicting the minimality of s. Inconsequence, we also have wi(q) > sq(K) in this case, thus proving (1).

To prove (2), �rst remark that sq(K) 6 ~sq(K) + 1 by (1). One may assumethat sq(K) < +1. As h1i ? �n;q is a subform of �n+1;q, one has sq(K) > ~sq(K)and (2) follows.

When q is a P�ster form, assertion (2) is satis�ed and one may assume thatq is weakly isotropic. In this case, sq(K) is �nite, hence there exists an integern such that sq(K) = 2n by Proposition 4.1. If we would have wi(q) = 2n then�2n;q would be isotropic and its P�ster neighbor h1i ? �2n�1;q would also be

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isotropic which would imply that sq(K) 6 2n�1. This means that wi(q) = 2n+1as claimed.

Examples 5.2. (1) When wi(q) < +1 then the above result shows thatsq(K) < +1 but the converse is false as the following example shows : ifK = R, q = h�1i then wi(q) = +1 and sq(R) = 1.(2) In general, we have sq(K) 6= ~sq(K) even if q represents 1 (or is a P�sterneighbor). Take K to be a local �eld with jKj � 3 mod 4 and let � be anuniformizer of K. If q = h1; �; �i then sq(K) = 2 and ~sq(K) = 1.

In the following result, if q is a P�ster form, we aim at characterizing thein�niteness of the q-level in terms of usual notions in quadratic form theory.Denote �qK = �qK

� [ f0g.Proposition 5.3. Let q be a n-fold P�ster form over a formally real �eld K.Then the following are equivalent :(1) sq(K) = +1.(2) �qK is a preordering of K.(3) There exists an ordering P of K containing the preordering �qK.(4) q is strongly anisotropic.(5) q is not torsion.(6) There exists an ordering P of K for which sgnP (q) = dim(q).

Proof. First note that we easily have (3) ) (2) ) (1) (as a preordering doesnot contain �1). Note also that (5) () (6) by P�ster's local global principleand the fact that the signature of an n-fold P�ster form q at any ordering is 0or dim(q).

Suppose now that (1) holds. Then �1 =2 �qK, �qK + �qK � �qK and�qK � �qK � �qK as q is multiplicative. As q represents 1, we also have�K � �qK which implies (2).

If (2) holds then there exists a maximal preordering P containing �qK byZorn's lemma. A well-known argument then shows that P is in fact an orderingof K (that is P [ �P = K and P \ �P = f0g) and (3) follows.

Now, we obviously have (1) () (4) by the previous lemma. If q is torsionthen there exists an integer l for which �2l;q is hyperbolic, hence sq(K) 6 2l and(1)) (5) holds. If q is not torsion, suppose that there exists an integer m suchthat �m;q is isotropic. Take k such that 2k > m. Then h1i ? �2k;q is isotropicand is a P�ster neighbor of �2k+1;q which is thus hyperbolic and q is torsion,contradiction. This shows that (5)) (4) and concludes the proof.

Remarks 5.4. (1) When q is a P�ster form, Proposition 4.9 (1) shows that�qK = K if and only if sq(K) < +1.(2) The equivalences (1) () (2) () (3) in the above proposition generalizesArtin-Schreier's Theorem which says that s(K) is in�nite if and only if �K is apreordering if and only if K has an ordering.

Proposition 5.5. Let q be a quadratic form over a formally real �eld K. Ifthere exists an ordering P of K such that sgnP (q) = dim(q), then sq(K) isin�nite.

Proof. In fact the condition sgnP (q) = dim(q) implies that q is positive de�nitewith respect to P , thus for every positive integer n, the form �n;q is also positivede�nite and it does not represent �1.

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Remark 5.6. If K = R((X))((Y )) and q = hX;Y;�XY i then using a theoremdue to T. A. Springer (see [25, Ch. 6, 2.6]), sq(K) is in�nite. On the other handfor any ordering P of K, the signature is never 3, so the converse of 5.5 does nothold in general. Nonetheless, in the case where dim q = 1 or 2, or over number�elds, the converse is true as the following propositions show.

Proposition 5.7. Let K be a number �eld and let q be a quadratic form overK. Then sq(K) is in�nite if and only if there exists an ordering P of K suchthat sgnP (q) = dim(q).

Proof. Regarding to Proposition 5.5, it is enough to show that if sq(K) is in�nite,then there exists an ordering P of K such that sgnP (q) = dim(q). If it is notthe case, then for every ordering P of K we have sgnP (q) < dim(q). Put' := h1i ? �4;q, we have sgnP (') < 1 + 4dim(q). Furthermore, as ' isnot negative de�nite, we have sgnP (') > �dim('). The form ' is therefore atotally inde�nite form of dimension > 5 and it is isotropic by Hasse-Minkowskitheorem. Thus sq(K) 6 4, which yields a contradiction.

Proposition 5.8. (1) Consider the one-dimensional quadratic form q = haiover a �eld K. Then sq(K) is in�nite if and only if there exists an ordering Pof K such that sgnP (hai) = 1.(2) Consider the two-dimensional quadratic form q = ha; bi over a �eld K.Then sq(K) is in�nite if and only if there exists an ordering P of K such thatsgnP (q) = 2.

Proof. (1) In fact sq(K) is in�nite if and only if �a is not a sum of squares ifand only if �a is not totally positive.

(2) If the conclusion does not hold then for every ordering P of K we havesgnP (q) = 0 or sgnP (q) = �2, which means that the elements a and b are eitherboth negative or one of them is positive and the other one is negative. In bothcases the signature of the form h1; a; b; abi with respect to P is zero. P�ster'slocal-global principle so implies that there exists a positive integer n such thatn�h1; a; b; abi is hyperbolic. We then obtain n�ha; bi ' n�h�1;�abi. The formn�ha; bi therefore represents �1, so sq(K) 6 n which yields a contradiction.

Corollary 5.9. Let q be a quadratic form over a �eld K such that ' = h1i ? qis weakly isotropic, then sq(K) <1.

Proof. As ' is weakly isotropic, there exists a positive integer n such that �n;' isisotropic. There exists so a element a 2 K� which is simultaneously representedby �n and �n;�q. The form �n;q represents therefore the totally negative element�a. Lemma 3.1 (3) implies that sn�q(K) 6 sh�ai(K). Proposition 5.8 (1),implies that sh�ai(K) <1. Lemma 3.1 (5) concludes the proof.

Remark 5.10. Recall that a �eld K satis�es the Strong Approximation Prop-erty (SAP) if for any disjoint closed subsets A and B of the space orderingof K (endowed with the Harrison topology), there exists a 2 K� such thata is positive (resp. negative) with respect to every ordering in A (resp. B).This notion was introduced by M. Knebusch, A. Rosenberg and R. Ware [9]. Acharacterization of the �elds for which every totally inde�nite form is weaklyisotropic was given by A. Prestel [24]. It turned out that these are exactly theSAP-�elds or equivalently the �elds for which every quadratic form of the shape

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h1; a; b;�abi is weakly isotropic. Formally real number �elds and more gener-ally every formally real algebraic extension of Q, every formally real algebraicextension of R(X) and Q((t)) are examples of SAP-�elds.

In the following result we characterize the �elds for which the converse ofProposition 5.5 is true.

Theorem 5.11. Let K be a formally real �eld. The �elds K for which forevery quadratic form q the relation sq(K) =1 would imply the existence of anordering P of K such that sgnP (q) = dim(q) are exactly the �elds for which thestrong approximation property holds.

Proof. First suppose that K is a SAP-�eld. Let q be a quadratic form overK with sq(K) = 1. If there is no ordering P of K such that sgnP (q) =dim(q), then q is not totally positive. It follows that the form ' = h1i ? qis totally inde�nite. By Prestel's theorem [24, Satz. 3.1], ' is weakly isotropicand Corollary 5.9 yields a contradiction.

Conversely suppose that for every quadratic form q the relation sq(K) =1 implies that there exists an ordering P of K such that sgnP (q) = dim(q).In order to prove that K is a SAP-�eld, it is enough to show that the formh1; a; b;�abi is weakly isotropic. Consider the form q = ha; b;�abi. For anyordering P of K, the elements a, b and �ab can not be simultaneously positivewith respect to P . It follows that there is no ordering P for which sgnP (q) =dim(q), thus sq(K) < 1. There exists so a positive integer n such that h1i ?�n;q is isotropic. It follows that n� h1; a; b;�abi is isotropic as well.Lemma 5.12. Let K be a formally real �eld and let q be an n-dimensionalquadratic form over K. Then the following statement are equivalent:(1) For any ordering P of K on has sgnP (q) 6 2� dim(q).(2) The form := n(n� 2)�h1i ? (2n� 2)� q ? q q is weakly hyperbolic.

Proof. It is easy to verify that sgnP ( ) = 0 if and only if sgnP (q) 6 2�dim(q).The conclusion then follows from P�ster's local-global principle.

Proposition 5.13. Let K be a formally real �eld and let q be a quadratic formof dimension n > 2 over K. If for every ordering P of K on has sgnP (q) 62� dim(q) then sq(K) <1.

Proof. Put '1 := (2n� 2)� q and '2 := n(n� 2)� h1i ? q q. According to5.12, the form '1 ? '2 is weakly hyperbolic. There exists so a positive integerm such that m � ('1 ? '2) is hyperbolic. As dim('1) = dim('2), we obtain�m;'1

' �m;�'2, therefore m(2n � 2) � q ' m � h�1; � � � i, this implies that

sq(K) 6 m(2n� 2).

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Institut Fourier, 100 rue des Maths, BP 74, 38402 Saint Martin

d'Héres cedex, France

E-mail-address: berhuy@fourier.ujf-grenoble.fr

Université de Rouen at IUFM de Rouen, 2 rue du Tronquet, BP 18,

76131 Mont-Saint-Aignan Cedex, France and Laboratoire de Didac-

tique André Revuz.

E-mail address: nicolas.grenier-boley@univ-rouen.fr

Department of Mathematical Sciences, Sharif University of Tech-

nology, P. O. Box 11155-9415, Tehran, Iran.

E-mail address: mmahmoudi@sharif.ir

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