Integral Spinor Norm Groups Over Dyadic Local Fields

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Journal of Number Theory 102 (2003) 125–182 http://www.elsevier.com/locate/jnt

Integral spinor norm groups over dyadic local ﬁelds

Constantin N. Beli Institute of Mathematics of the Romanian Academy, P.O. Box 1-764, RO-70700 Bucharest, Romania

Received30 September 2002; revised3 December 2002

Communicatedby J.S. Hsia

Abstract

The spinor norms of integral rotations of an arbitrary quadratic lattice over an arbitrary dyadic local ﬁeld are determined. The results are given in terms of BONGs, short for ‘‘bases of norm generators’’. This approach provides a new way to describe lattices over dyadic local ﬁelds. r 2003 Elsevier Science (USA). All rights reserved.

MSC: 11E08

Keywords: Quadratic forms; Dyadic ﬁelds; Spinor norms; BONGs

Introduction

In the theory of quadratic lattices over global fields, where the local–global principle does not exist, the genus of a lattice usually contains many classes. The spinor genera are an intermediate step between classes and genera. (In the indefinite case when the rank is at least three the notions of spinor genera andclasses coincide.) In the theory of spinor genera in particular for the purpose of calculating the number of spinor genera in a genus, a key element is the knowledge of spinor norms of integral rotations associatedto the localizations of the given lattice at all primes. Until now these groups have been computedfor the case when the local fieldis non- dyadic by Kneser [K] or 2-adic by Earnest and Hsia [EH2]. More recently Xu [X3]

Correspondence address: Institute of Mathematics ‘‘Simion Stoilow’’ of the Romanian Academy, 21, Calea Grivitei Street, RO-781011 Bucharest, Sector 1, Romania. Fax: +40-21-212-51-26. E-mail address: [email protected].

126 C.N. Beli / Journal of Number Theory 102 (2003) 125–182 solved the case of dyadic fields where the ramification index e :¼ ord2 is 2. In this paper we completely solve this problem for arbitrary dyadic fields. See Theorems 1 (Section 5) and2 (Section 7) below. Throughout this paper F will be a dyadic local field, O ¼ OF the ring of integers, p the maximal ideal, OÂ the group of units in O; e :¼ ord2 ; p a fixedprime element in F; Dð:Þ the quadratic defect function and D ¼ 1 À 4r a fixedunit with DðDÞ¼4O: Also we denote for any two lattices NDM by XðM=NÞ :¼fsAOþðFMÞjNDsMg the set of relative integral rotations. All unexplainednotations are from [OM]. In order to simplify the notation we put yðLÞ :¼ yðOþðLÞÞ and yðM=NÞ :¼ yðXðM=NÞÞ: If V is a quadratic space we denote LðVÞ the set of all lattices on V: We make the convention that if V is a quadratic space then det V is regarded as an element of 2 F’=F’2 while if L is a lattice then det L is regarded as an element of F’=OÂ : (Hence if aAF’ we have det L ¼ a iff det FL ¼ a andvol ðLÞ¼aO:) One of the main difficulties in studying lattices over dyadic local fields is the fact that, unlike in the non-dyadic case, lattices no longer have orthogonal bases. Instead they split into unary or binary improper modular lattices. The 2-adic case is somewhat easier because the improper binary unimodular lattices are totally improper andthere are only two possibilities; namely, Að0; 0Þ and Að2; 2rÞ: When e41 things become more complicated. In order to overcome this we introduce in Section 2 the notion of a BONG (short for ‘‘basis of norm generators’’) of a lattice which is a generalization of the usual concept of an orthogonal basis. In many ways BONGs, especially the so-called‘‘goodBONGs’’ (introducedinSection 4) behave like orthogonal bases. This allows us to study a general lattice almost as if it is diagonalizable and thereby we simplify considerably both the reasonings and the statements which are often very complicated when dealing with lattices over dyadic fields. On the other hand, the use of BONGs is not intended to replace the usual Jordan splittings. In fact, our proofs incorporate both concepts and sometimes we switch back andforth between them. We shall use the notation L ¼ !x1; y; xng to denote the fact that x1; y; xn is a BONG for L: The approach we use for calculating yðLÞ is quite classical, namely finding generators for OþðLÞ: O’Meara andPollak [OP1,OP2] provedthat in the special cases when L is either modular or when F is 2-adic (i.e., e ¼ ord2 ¼ 1) the group OþðLÞ is the group XðLÞ generatedby SþðLÞ andthe Eichler transformations. Since the Eichler transformations always have trivial spinor norms they do not contribute to yðLÞ: Hence, we still have yðLÞ¼yðSþðLÞÞ: This fact allowedthe calculation of yðLÞ in the case when L is modular by Hsia [H] andin the general 2-adic case by Earnest andHsia [EH2]. Unfortunately, when e41 it is not known whether OþðLÞ¼XðLÞ andthe generation problem of OþðLÞ it is still unsolved. As already observed in [EH1] that since one is only interestedin yðLÞ one may assume that yðLÞaF’: This restriction on L simplifies the problem as it severely limits the possible structures of the Jordan splittings of L: Xu [X2] gave a set of complicatednecessary conditions for a lattice L to satisfy in order that yðLÞaF’: He also showedthat these conditionsimply that ARTICLE IN PRESS

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OþðLÞ¼XðLÞ; thereby bringing us back to calculating yðSþðLÞÞ: Nevertheless, the explicit calculation of yðSþðLÞÞ is still a very technically formidable challenge. One of the benefits of using BONGs is that we can give necessary conditions such that yðLÞaF’: These conditions, called ‘‘property B’’ (see Section 4), are similar to those given in [X2] in terms of Jordan decompositions. But, they are expressible in a more compact form andit is not necessary to distinguish between the unary andthe binary components. Another advantage of using BONGs is that the formulas for yðLÞ given for the binary lattices in [H,X,X1] can be unifiedso that the modularand non-modular cases need not be separated. Our main result is the theorem given in Section 5. Its proof is split into two parts; namely, first we prove yðLÞ+G in Section 5 andthen yðLÞDG in Section 6. Here G is the group predicted for yðLÞ: The first part is easier andinvolves some techniques of multiplying the spinor norm groups of some binary lattices which are developed in Section 3. As for the secondpart we employ a technique from [X2] except that the Jordan decomposition used there is replaced by a BONG here. More specifically, given a sAOðLÞ; L ¼ L1>y>Lt; the necessary conditions in [X2] are usedto 0 0 express s as a product of symmetries of L anda s ; s AOðL2>y>LtÞ: This reduces the case to a lattice with fewer Jordan components. In our case, if we express L ¼ !x1; y; xng then property B allows us to have a similar expression for s with 0 þ s AO ð!x2; y; xngÞ: Next, we prove that the spinor norm of the above product of symmetries of L belongs to G: In fact, these symmetries belong to the orthogonal groups of some binary lattices of the form !x1; yg which have their spinor norms lying in G: Although our main result is given in terms of BONGs it can be readily translated to the traditional language of Jordan decompositions. However, the formulas will become somewhat less compact. It seems that the calculation of yðLÞ can also be done without the use of BONGs by using the methods of [X2] andthe formulas for the binary case obtainedin [H,X,X1]. However, this wouldrequire considerably more effort. Finally Section 7 is devoted to lattices not having ‘‘property A’’ (see Section 4). In this case yðLÞ can only be F’ or OÂF’2 andwe settle their distinctions. The use of BONGs has some further applications. In a future paper we will compute the relative spinor norm group yðM=NÞ in terms of goodBONGs of M and N: Also this methodprovidessome new ways to approach the problem of representation of lattices over dyadic local fields.

1. Duality lemmas

’ ’2 Deﬁnition 1. We deﬁne d : F=F -N,fNg to be the order of the ‘‘relative quadratic defect’’ by dðaÞ¼ord aÀ1DðaÞ:

If a ¼ pRe with eAOÂ then dðaÞ¼0ifR is odd and dðaÞ¼dðeÞ¼ord DðeÞ if R is even. Hence dðF’Þ¼f0; 1; 3; y; 2e À 1; 2e; Ng: ARTICLE IN PRESS

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Since ð1 þ pkÞF’2 ¼faAF’ j dðaÞXkg for any positive integer k we make the a ’2 ’ convention that ð1 þ p ÞF ¼faAFjdðaÞXag for any aAR,fNg: In particular N ’2 ’2 a ’2 ’ ð1 þ p ÞF ¼ F and ð1 þ p ÞF ¼ F for ap0:

Lemma 1.1. d satisﬁes the domination principle: dðabÞXminfdðaÞ; dðbÞg:

Proof. If one of a; b has odd order then the above inequality is trivial since the right side is 0. Otherwise, by multiplying with squares we may assume that a; bAOÂ andmoreover a ¼ 1 þ a; b ¼ 1 þ b where DðaÞ¼aO and DðbÞ¼bO: We have ab ¼ 1 þ a þ b þ ab so dðabÞXordða þ b þ abÞXminford a; ord bg¼ minfdðaÞ; dðbÞg: &

’ ’2 ’ ’2- 7 Let ð ; Þp be the Hasse symbol. It is known that ð ; Þp : F=F Â F=F f 1g is a ’ ’2 non-degenerate symmetric bilinear form. For any subgroup HpF=F we denote by H> its orthogonal complement. pffiffiffi A ’ A ’ If a F we put NðaÞ :¼ NðFð aÞ=FÞ: Note that NðaÞ¼fx Fjða; xÞp ¼ 1g¼ ’2 > ’ ’2 ð/aSF Þ : Hence if HpF=F the inclusion HDNðaÞ is equivalent by dualization to H>+/aSF’2; i.e. to aAH>:

Lemma 1.2.

> > (i) We have ðð1 þ p0ÞF’2Þ ¼ð1 þ pNÞF’2; ðð1 þ p1ÞF’2Þ ¼ð1 þ p2eÞF’2 and for any a ’2 > 2e 2 a ’2 odd integer 1oao2e þ 1 we have ðð1 þ p ÞF Þ ¼ð1 þ p þ À ÞF : ’ b ’2 > a ’2 (ii) If aAdðFÞ and bAR,fNg then ðð1 þ p ÞF Þ +ð1 þ p ÞF iff ðð1 þ > paÞF’2Þ +ð1 þ pbÞF’2 iff a þ b42e: > (iii) If aAF’ then ð1 þ pkÞF’2DNðaÞ iff ð1 þ pkÞF’2Dðð1 þ pdðaÞÞF’2Þ iff dðaÞþk42e:

Proof. (i) The ﬁrst two equalities are obvious since ð1 þ p0ÞF’2 ¼ F’; ð1 þ p1ÞF’2 ¼ OÂF’2; ð1 þ p2eÞF’2 ¼ /DSF’2 and ð1 þ pNÞF’2 ¼ F’2: > For ðð1 þ paÞF’2Þ Dð1 þ p2eþ2ÀaÞF’2 we use Lemma 3 in [H]: Let Zeð1 þ 2eþ2Àa ’2 e a ’2 > p ÞF : If ord Z is odd then ðZ; DÞp ¼À1soZ ðð1 þ p ÞF Þ : If ord Z is even we may suppose that ZAOÂ with dðZÞ¼bo2e þ 2 À a: Since b is odd we must have A Â bp2e À a: By Lemma 3 in [H] there is x O with dðxÞ¼2e À b s.t. ðZ; xÞp ¼À1: DÀZ (Note: If Z ¼ 1 þ d with ord d ¼ b then x may be chosen to be 1ÀZ:) But dðxÞ¼ > 2e À bX2e Àð2e À aÞ¼a so xAð1 þ paÞF’2: Hence Zeðð1 þ paÞF’2Þ : For the reverse inclusion if ZAð1 þ p2eþ2ÀaÞF’2 let xAð1 þ paÞF’2 be arbitrary. A Â We have to prove that ðZ; xÞp ¼ 1: We may assume Z; x O ; Z ¼ 1 þ a and x ¼ 1 þ b with DðZÞ¼aO and DðxÞ¼bO: Hence ord aX2e þ 2 À a andord bXa which implies ord abX2e þ 2: Hence 1 À ab ¼ 1 þ a À að1 þ bÞ is a square so

ð1 þ a; Àað1 þ bÞÞp ¼ 1: We also have that ð1 þ a; ÀaÞp ¼ 1soðZ; xÞp ¼ð1 þ a; 1 þ bÞp ¼ 1: ARTICLE IN PRESS

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(ii) is an obvious consequence of (i) since ð1 þ pxÞF’2 ¼ð1 þ p0ÞF’2 ¼ F’ whenever x ’2 b ’2 xp0 and ð1 þ p ÞF ¼ð1 þ p ÞF whenever aoxpb where a; b are two consecutive elements in dðF’Þ¼f0; 1; 3; y; 2e À 1; 2e; Ng: > > (iii) We have ð1 þ pkÞF’2DNðaÞ iff aAðð1 þ pkÞF’2Þ : Now since ðð1 þ pkÞF’2Þ is > of the form ð1 þ pd ÞF’2 for some dAdðF’Þ we have aAðð1 þ pkÞF’2Þ ¼ð1 þ pd ÞF’2 iff > dðaÞXd; i.e. iff ð1 þ pdðaÞÞF’2Dð1 þ pd ÞF’2 ¼ðð1 þ pkÞF’2Þ : By (ii) this is equivalent to dðaÞþk42e: &

’ ’ ’2 Lemma 1.3. Let a; bAF and let HpF=F : We have:

(i) NðaÞ-HDNðbÞ iff either HDNðbÞ or HDNðabÞ: (ii) NðaÞ-H ¼ NðbÞ-H iff HDNðabÞ:

> > Proof. We have NðaÞ¼ð/aSF’2Þ andN ðbÞ¼ð/bSF’2Þ : Hence our inclusion ’2 > ’2 > can be written as ð/aSF Þ -HDð/bSF Þ : By dualization this is equivalent > ’2 > > > to /aSH +/bSF : But since /aSH ¼ H ,aH our inclusion is equivalent to bAH> or bAaH>; i.e. to b or abAH>: But bAH>3HDNðbÞ and abAH>3HDNðabÞ hence the conclusion. For the secondclaim we have N ðaÞ-H ¼ NðbÞ-H3NðaÞ-HDNðbÞ and NðbÞ-HDNðaÞ which by (i) are equivalent to HDNðbÞ or HDNðabÞ and, resp. HDNðaÞ or HDNðabÞ: If HDNðabÞ then both conditions are satisﬁed so we have the ‘‘if’’ part. If HD/NðabÞ then we must have HDNðbÞ and HDNðaÞ: But NðaÞ-NðbÞDNðabÞ so HDNðabÞ so we also have the ‘‘only if’’ part. &

2. Bases of norm generators (BONGs)

Deﬁnition 2. An element xAL is calleda norm generator if nL ¼ QðxÞO: We call a basis of norm generators (BONG) for L a sequence x1; y; xn of vectors in V s.t. x1 is a norm generator for L and x2; y; xn is a base of norm generators for pr > L: (This is x1 a recursive deﬁnition.)

Lemma 2.1. If x1; y; xn is a BONG for L then det L ¼ Qðx1ÞyQðxnÞ:

Proof. We use induction by rank L: If our statement is true for n À 1 then det pr > L ¼ Qðx2ÞyQðxnÞ so we reduced to proving that for any norm generator x x1 of L we have det L ¼ QðxÞ det prx> L: This actually happens for any xAL primitive andanisotropic. If L ¼ Ox þ Ox2 þ ? þ Oxn then we write xi ¼ yi þ aix for 2pipn where aiAF and Bðx; yiÞ¼0: We have prx> L ¼ Oy2 þ ? þ Oyn so det L ¼ dðx; x2; y; xnÞ¼dðx; y2 þ a2x; y; yn þ anxÞ¼dðx; y2; y; ynÞ¼QðxÞdðy2; y; ynÞ¼ QðxÞ det prx> L: (Note: Here dðx1; y; xnÞ¼detðBðxi; xjÞÞ; see [OM, p. 87].) & ARTICLE IN PRESS

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Lemma 2.2. Let M be a lattice and xAM a norm generator. Then if N is another lattice s.t. nNDnM and prx> NDprx> M we have NDM:

D D1 Proof. WLOG we may assume that QðxÞ¼1sonN nM ¼ O and sM; sN 2 O: Any element of M can be written as ax þ y with aAF and yAprx> M: We have A D1 A1 a ¼ Bðax þ y; xÞ sM 2 O: So a 2 O: D1 A A We want to prove that BðM; NÞ 2 O: Let ax þ y M and bx þ z N with yAprx> M and zAprx> N: We have to prove that ab þ Bðy; zÞ¼Bðax þ y; bx þ 1 1 A A > D > A A A zÞ 2 O: Since z prx N prx M there is g F s.t. gx þ z M: We have a; g 2 O: We A D1 also have that ag þ Bðy; zÞ¼Bðax þ y; gx þ zÞ sM 2 O: Now b2 þ QðzÞ¼Qðbx þ zÞAnNDO and g2 þ QðzÞ¼Qðgx þ zÞAnM ¼ O: By subtraction we get that b2 À g2AO: It follows that either b À g or b þ g belongs A1 A A to O: But g 2 O so 2g O so even in the case when b þ g O we still have b À g ¼ A A A1 A1 b þ g À 2g O: Hence anyway b À g O: We also have a 2 O so aðb À gÞ 2 O: By A1 A1 D1 adding it to ag þ Bðy; zÞ 2 O we get ab þ Bðy; zÞ 2 O: So BðM; NÞ 2 O: D D 1 We have O ¼ nM nðM þ NÞ¼nM þ nN þ 2BðM; NÞ O þ O þ 2 Á 2 O ¼ O: Hence nðM þ NÞ¼O and x is also a norm generator for M þ N: We have det M ¼ QðxÞ det prx> M ¼ det prx> M anddet ðM þ NÞ¼QðxÞ det prx> ðM þ NÞ¼ det prx> ðM þ NÞ: But prx> ðM þ NÞ¼prx> M þ prx> N ¼ prx> M: So detðM þ NÞ¼ det M: This implies that the lattices MDM þ N have the same volume so M þ N ¼ M so NDM: &

2.3. Let now NDM be arbitrary lattices over V andlet V ¼ U>W where U V - and W are non-degenerate. Let pr ¼ prW : V W be the projection map andlet M ¼ prM and N ¼ prN: Let sAXðM=NÞ s.t. sjU ¼ 1U : We have that 0 0 0 s ¼ 1U >s where s AOðWÞ: Since s Á pr ¼ pr Á s (easy to verify ﬁrst for elements in U then for elements in W) we have s0M ¼ s0ðprMÞ¼prðsMÞ+prN ¼ N so s0AXðM=NÞ:

Corollary 2.4. With the above notation in the case when U ¼ Fx where xAM is a norm generator the converse is also true, i.e. if M ¼ prx> M; N ¼ prx> N and 0 0 s AXðM =N Þ then s :¼ 1Fx>s AXðM=NÞ:

0 0 0 Proof. We have prx> Á s ¼ s Á prx> so prx> ðsMÞ¼s ðprx> MÞ¼s M +N ¼ prx> N: We also have nNDnM ¼ nsM: Since nNDnsM and prx> NDprx> sM we have by 2.2 that NDsM so sAXðM=NÞ: &

0 0 2.5. As a consequence of 2.3 and2.4 we have that the mapping s /1Fx>s is an embedding of XðM=NÞ into XðM=NÞ: In particular yðM=NÞDyðM=NÞ: When we take M ¼ N we get that OðMÞDOðMÞ:

Corollary 2.6. A lattice is uniquely determined by a BONG. ARTICLE IN PRESS

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0 Proof. Let M; M be two lattices with the same BONG x1; y; xn: We use induction 0 0 over n: We have nM ¼ nM ¼ Qðx1ÞO: Also pr > M and pr > M have the same x1 x1 BONG namely x2; y; xn so by the induction step they must be equal. Now 0 0 0 0 nM DnM and pr > M Dpr > M: By 2.2 we have M DM: Similarly, MDM so x1 x1 M0 ¼ M: &

Notation. If x1; y; xn is a BONG for M we will write M ¼ !x1; y; xng: (We can do this since x1; y; xn determines uniquely M:)IfQðxiÞ¼ai we will also say that MD!a1; y; ang adapted to the BONG x1; y; xn: Let us note that if x1; y; xn is a BONG for M then e1x1; y; enxn is also a BONG Â for M if eiAO : Hence the equality M ¼ !x1; y; xng remains true if we multiply the xi’s by units. It follows that MD!a1; y; ang remains true if we multiply the ai’s by squares of units. Note that not every orthogonal basis of a quadratic space V is a BONG for a certain lattice M over V so !x1; y; xng does not always make sense. In the following lemma we give some sufﬁcient conditions for an orthogonal basis of a quadratic space to be a BONG of a lattice.

Lemma 2.7. Let L ¼ !x1; y; xng be a lattice on a quadratic space V: We write V ¼ Uk>Vk for 0pkpn where Uk ¼ Fx1>y>Fxk and Vk ¼ Fxkþ1>y>Fxn: Then:

(i) The lattice !xkþ1; y; xng exists for any 0pkpn and we have !x ; y; x g prV L and O !x ; y; x g DO L : kþ1 n ¼ Vk ð kþ1 n Þ ð Þ (ii) If !ykþ1; y; ymg exists and !ykþ1; y; ymgD!xkþ1; y; xng then !x1; y; xk; ykþ1; y; ymg also exists and we have !x1; y; xk; ykþ1; y; y g xAL prV xA!y ; y; y g DL: m ¼f j Vk kþ1 m g (iii) !x1; y; xkg exists and !x1; y; xkg ¼ Uk-LDL: (iv) If !xkþ1; y; xng ¼ !ykþ1; y; yng then L ¼ !x1; y; xk; ykþ1; y; yng:

Proof. (i) From the recursive deﬁnition of a BONG we have inductively that !x ; y; x g exists and prViÀ1 !x ; y; x g !x ; y; x g (x> in V is V ). i n Vi i n ¼ iþ1 n i iÀ1 i Since prVkÀ1 ? prV0 prV (we have V V) we have !x ; y; x g prV L as Vk Á Á V1 ¼ Vk 0 ¼ kþ1 n ¼ Vk claimed. We also have that xi is a norm generator of !xi; y; xng and pr > !xi; y; xng ¼ !xiþ1; y; xng so from 2.5 we have Oð!xiþ1; y; xngÞD xi Oð!xi; y; xngÞ: Hence Oð!xkþ1; y; xngÞD?DOð!x2; y; xngÞDOðLÞ so we have the secondclaim of (i). (ii) By (i) we have that f : prV : L-!x ; y; x g is well deﬁned and is ¼ Vk kþ1 n surjective. Since f À1 !y ; y; y g xAL prV xA!y ; y; y g we actu- ð kþ1 m Þ¼f j Vk kþ1 m g À1 ally have to prove that f ð!ykþ1; y; ymgÞ¼!x1; y; xk; ykþ1; y; ymg: Case k ¼ 0 is trivial. For k ¼ 1 let K :¼ !y2; y; ymg: We want to prove that À1 À1 f ðKÞ¼!x1; y2; y; ymg: We have f ðx1Þ¼0sox1Af ðKÞDL: Since x1 is a norm generator for L it will also be a norm generator for f À1ðKÞ: Now f is surjective ARTICLE IN PRESS

132 C.N. Beli / Journal of Number Theory 102 (2003) 125–182

À1 À1 so pr > ð f ðKÞÞ ¼ f ð f ðKÞÞ ¼ K: But y2; y; ym is a BONG for K so x1; y2; y; ym x1 is a BONG for f À1ðKÞ: For k41 we use induction on k: We write f hg where g prV ¼ ¼ V1 ¼ V1 pr > : L-!x2; y; xng and h ¼ pr : !x2; y; xng-!xkþ1; y; xng: Now x1 Vk from the induction step k À 1 appliedto !x2; y; xng and !ykþ1; y; ymgD À1 !xkþ1; y; xng we have h ð!ykþ1; y; ymgÞ¼!x2; y; xk; ykþ1; y; ymg: À1 From the case k ¼ 1 appliedto L and !x2; y; xk; ykþ1; y; ymg ¼ h ð!ykþ1; À1 À1 y; ymgÞD!x2; y; xng we have g ðh ð!ykþ1; y; yngÞÞ ¼ !x1; y; xk; ykþ1; À1 À1 À1 À1 y; yng: But g h ¼ f so f ð!ykþ1; y; yngÞ¼!x1; y; xk; ykþ1; y; yng: (iii) and(iv) are just ‘‘degenerate’’ cases ( !ykþ1; y; ymg ¼ 0; resp. !ykþ1; y; ymg ¼ !xkþ1; y; xmg) of (ii). &

Corollary 2.8. If !x1; y; xng exists then !xi; y; xjg also exists for any 1pipjpn:

Proof. From 2.7(i) we have that !xi; y; xng exists which by 2.7(iii) implies that !xi; y; xjg exists. &

3. Binary lattices

In this section we study binary lattices in terms of BONGs. We introduce the invariant aðLÞ which fully characterizes the class of a binary lattice L up to scaling. By using the results on binary lattices from [H,X,X1] we will write yðLÞ in terms of aðLÞ: We will also give formulas for products between spinor norm groups associated to different binary lattices as well as obtain other results needed in the later sections. Let L be a binary lattice. If L is not modular then it is isomorphic after scaling to /1; eprS for some eAOÂ and r40: In general if LD/a; bS with ord aoord b then D / rS r b r L a 1; ep where ep ¼ a: The number a ¼ ep is uniquely determined up to a r b À2 À2 square of a unit. We have ep ¼ a ¼ðabÞa ¼ det La : We also have that r ¼ ord pre ¼ ordðdet LaÀ2Þ¼ordvol ðLÞÀ2 ord nL: These can naturally be extended to modular lattices so we get the following deﬁnition:

Definition 3. If L is a binary lattice (modular or not) then we define the number 2 a ¼ aðLÞAF’=OÂ by a :¼ det LaÀ2 where aAF’ with nL ¼ aO: We denote by A ¼ AF to be the set of all possible values of aðLÞ: We also define the number R ¼ RðLÞ by R :¼ ordvol ðLÞÀ2 ord nL: (Obviously RðLÞ¼ord aðLÞ:)

3.1. Since aðLÞ differs from det L by a square we have that det FL ¼ aðLÞ: Also if LD!a; bg adapted to a certain BONG then from 1.3 we have ab ¼ det L so b ¼ det LaÀ1 ¼ a det LaÀ2 ¼ aaðLÞ: Hence LD!a; aaðLÞg ¼ a!1; aðLÞg: We ARTICLE IN PRESS

C.N. Beli / Journal of Number Theory 102 (2003) 125–182 133 also have that ord b ¼ ord aaðLÞ¼ord a þ RðLÞ: Note that aðLÞ and RðLÞ are invariant under scaling and isomorphisms.

Lemma 3.2. Let V be a binary quadratic space. We have:

(i) If L; L0 are two lattices on V that have a common norm generator then LDL03RðLÞXRðL0Þ: Moreover if LDL0 then OðLÞDOðL0Þ: (ii) A lattice L on V is uniquely determined by a norm generator and RðLÞ:

Proof. Let x be the common norm generator of L and L0: We have V ¼ Fx>Fy for some yAV: Since x is a norm generator in L and L0 we have L ¼ !x; pkyg and 0 k0 0 0 k 0 k0 L ¼ !x; p yg for some k; k AZ: Now if LDL then p y ¼ prx> LDprx> L ¼ p y so kXk0: Conversely, if kXk0 then !pkygD!pk0 yg so by 2.7(ii) we have L ¼ !x; pkygD!x; pk0 yg ¼ L0: Hence LDL03kXk0: But RðLÞ¼ord QðpkyÞÀ ord QðxÞ¼2k þ ord QðyÞÀord QðxÞ andsimilarly RðL0Þ¼2k0 þ ord QðyÞÀ ord QðxÞ: Hence kXk03RðLÞXRðL0Þ andwe have the ﬁrst part of (i). (ii) If RðLÞ¼RðL0Þ then RðLÞXRðL0Þ which by the ﬁrst part of (i) implies LDL0: Similarly L0DL so L ¼ L0: For the secondclaim of (i), let sAOðLÞ: We have xAL so sxALDL0: We also have xAL0 so sxAsL0: Now QðsxÞO ¼ QðxÞO ¼ nL0 ¼ nsL0 so sx is a common norm generator of L0 and sL0: But we also have RðL0Þ¼RðsL0Þ so from (ii) we must have L0 ¼ sL0 so sAOðL0Þ: &

Lemma 3.3.

(i) If L ¼ !x1; x2g then L is modular iff RðLÞp0 iff ord Qðx1ÞXord Qðx2Þ: r1 r2 (ii) If LD/p e1; p e2S with r1or2 relative to a basis fx1; x2g then we have L ¼ r r !x1; x2gD!p 1 e1; p 2 e2g: We also have RðLÞ¼r2 À r1 and aðLÞ¼ r Àr p 2 1 e1e2: r Â u (iii) If L is p -modular with det FL ¼ e where eAO and nL ¼ p and x1AL is a norm u u 2rÀu generator s.t. Qðx1Þ¼p e1 then LD!p e1; p ee1g relative to a certain 2rÀ2u BONG x1; x2: We also have RðLÞ¼2r À 2u and aðLÞ¼p e:

r r Proof. (ii) Obviously x1; x2 is a BONG for L so L ¼ !x1; x2gD!p 1 e1; p 2 e2g: r1 r2 r2Àr1 ’ ’2 By 3.1 we have aðLÞ¼p e1=ðp e2Þ¼p e1e2 (in F=F ) and RðLÞ¼ord aðLÞ¼ r2 À r1: (iii) If L is pr-modular and det FL ¼ e we also have vol L ¼ p2r so det L ¼ p2re: Since nL ¼ pu ¼ puO we have aðLÞ¼p2reðpuÞÀ2 ¼ p2rÀ2ue and RðLÞ¼ord aðLÞ¼ u 2r À 2u: Since x1 is a norm generator with Qðx1Þ¼p e1 we have from 3.1 that u u u 2rÀu LD!p e1; p e1aðLÞg ¼ !p e1; p ee1g: (i) If L is in case (ii), i.e. non-modular we have that RðLÞ¼r2 À r140: If L is in case (iii), i.e. modular we have RðLÞ¼2r À 2up0: (We have pu ¼ nLDsL ¼ pr so ARTICLE IN PRESS

134 C.N. Beli / Journal of Number Theory 102 (2003) 125–182 rpu:) Since ord Qðx2Þ¼ord Qðx1ÞþRðLÞ we have RðLÞp0 iff ord Qðx1ÞX ord Qðx2Þ: &

R R Corollary 3.4. If LD!p 1 e1; p 2 e2g then:

R2ÀR1 (i) aðLÞ¼p e1e2 and RðLÞ¼R2 À R1: R1 R2 (ii) If R1pR2 then L is diagonalizable and we have LD/p e1; p e2S: (iii) If R1XR2 then R1 R2 ðmod2 Þ; L is modular and we have sL ¼ pðR1þR2Þ=2; nL ¼ pR1 and nL# ¼ pÀR2 :

R R Proof. Let us suppose that LD!p 1 e1; p 2 e2g relative to the BONG x1; x2: (i) Follows from 3.1. 0 (ii) We consider the lattice L ¼ Ox1>Ox2: Since ord Qðx1Þ¼R1pR2 ¼ 0 ord Qðx2Þ we have from 3.3(ii) that L ¼ !x1; x2g ¼ L: Hence the BONG x1; x2 0 R1 R2 of L is actually an usual basis. We have L ¼ L D/p e1; p e2S relative to the basis x1; x2: (iii) Since ord Qðx1Þ¼R1XR2 ¼ ord Qðx2Þ we have from 3.3(ii) that L is r u modular. If sL ¼ p and nL ¼ p then from 3.3(i) we have R1 ¼ u and R2 ¼ ðR þR Þ=2 2r À u: It follows that R1 R2 ðmod2 Þ and r ¼ðR1 þ R2Þ=2: Hence sL ¼ p 1 2 À2 and nL ¼ pR1 : We also have nL# ¼ nLðsLÞ ¼ pÀR2 : &

Lemma 3.5. A is the set of all aAF’ for which either ÀaeF’2 and DðÀaÞDO or ÀaAF’2 X À1 1 and ord a À 2e: For short A ¼ÀD ðOÞ-4 O:

Proof. We know that for any eAOÂ and R40 we have that að/1; epRSÞ¼epR so aðLÞ can be any number of the form epR with R40: If we want non-positive values of R ¼ RðLÞ we needto check the modularlattices. If V is a binary quadratic space with det V ¼ eAOÂ then we may suppose that Àe ¼ 1 þ d where DðÀeÞ¼dO ¼ pd : If LALðVÞ is pr-modular with nL ¼ pu we can D r uÀr À1 rÀu r ÀR=2 À1 R=2 express L p Aðe1p ; Àe1 p dÞ¼p Aðe1p ; Àe1 p dÞ adapted to a certain basis fx; yg (see [OM, 93:17]). The condition that xAL is a norm generator ÀR=2 is equivalent to the fact that e1p is a norm generator for the unimodular ÀR=2 À1 R=2 ÀR=2 lattice Aðe1p ; Àe1 p dÞ; i.e. with ÀR=2 ¼ ord e1p pe and ÀR=2 ¼ ÀR=2 À1 R=2 ord e1p pord e1 p d ¼ R=2 þ d: The two inequalities are equivalent to RX À2e and RX À d: If Àe is not a square then dp2e so we only have the condition that RX À d: If Àe is a square then d ¼ dðÀeÞ¼N so we only have the condition RX À 2e: In conclusion if V is a binary quadratic space the values of the map R : LðVÞ-Z are all positive odd numbers if det V ¼ ep with eAOÂ andall even integers X À d if det V ¼ e with dðÀeÞ¼doN: If V is hyperbolic, i.e. det V ¼ e where Àe is a square then R ¼ RðLÞ has to be even and X À 2e: ARTICLE IN PRESS

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Hence for any lattice L we have that aðLÞ is of the form aðLÞ¼epR where eAOÂ with dðÀeÞ¼d and R ¼ RðLÞ are in one of the following situations: (i) R40 is odd, (ii) Àe is not a square, R is even and RX À d; (iii) Àe is a square, R is even and RX À 2e:

Now in the case when R is odd we have R403RX03DðÀpReÞ¼pRDO: In case (ii) we have that DðÀpReÞ¼pRDðÀeÞ¼pRþd so the condition R þ dX0 means DðÀpReÞDO: Finally if Àe is a square then the condition that R is even is equivalent R A ’2 X 3 R A1 to Àp e F while R À 2e À p e 4 O: Hence (i)–(iii) are equivalent to saying that ÀaðLÞ¼ÀepR is either a non-square with DðÀaðLÞÞDO or a square belonging 1 & to 4 O:

Note: As seen above we always have RX À 2e: The only situations when R ¼ 2r À 2u ¼À2e are when u ¼ r þ e; i.e. nL ¼ pu ¼ 2pr ¼ 2sL which means LDprAð0; 0Þ or LDprAð2; 2rÞ: Let us note that det Að0; 0Þ¼À1 anddet Að2; 2rÞ¼ À2 1 ÀD: Since nAð0; 0Þ¼nAð2; 2rÞ¼2O we have aðAð0; 0ÞÞ ¼ À1 Á 2 ¼À4 and À2 D aðAð2; 2rÞÞ ¼ ÀD Á 2 ¼À4:

We also note that if R is even the inequality RX À d is usually strict. The only case when R ¼Àd is when d is even so d ¼ 2e and R ¼À2e: Since d ¼ dðÀeÞ¼2e we A Â2 R À2e A À2e Â2 D Â2 have e À DO so p e ¼ p e À p DO ¼À4 O which corresponds to the case of lattices of the form LDprAð2; 2rÞ: Summary of properties of A: If L is a binary lattice with aðLÞ¼pRe then: RX À 2e with equality iff L is totally improper (i.e. iff nL ¼ 2sL). Moreover we 1 D r D D r have aðLÞ¼À4 if L p Að0; 0Þ and aðLÞ¼À4 if L p Að2; 2rÞ: X R A D Â2 D r R À dðÀeÞ with equality iff p e À 4 O ; i.e. iff L p Að2; 2rÞ for some r: R40ifR is odd. If Ro0 then L is improper modular. If R ¼ 0 then L is proper modular. If R40 then L is not modular.

Lemma 3.6. Let V be a binary quadratic space and let x1; x2AV: Then !x1; x2g exists iff Qðx2Þ=Qðx1ÞAA:

Proof. If L ¼ !x1; x2g then Qðx2Þ=Qðx1Þ¼aðLÞAA so we have the ‘‘only if’’ 0 part. If a ¼ Qðx2Þ=Qðx1ÞAA then there is a binary lattice L over a quadratic space 0 0 0 V s.t. aðL Þ¼a: By scaling we may assume that L DQðx1Þ!1; ag ¼ 0 0 0 0 !Qðx1Þ; Qðx2Þg relative to a BONG x1 ; x2 : Since Qðx1 Þ¼Qðx1Þ and Qðx2 Þ¼ 0 0 0 0 0 Qðx2Þ there is an isometry s : V -V s.t. sx1 ¼ x1 and sx2 ¼ x2: Since x1 ; x2 is a 0 0 BONG for L we have that x1; x2 is a BONG for L ¼ sL so we have the ‘‘if’’ part. & ARTICLE IN PRESS

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2 Definition 4. We define the function G : F’=OÂ -SgpðF’=F’2Þ as follows: e 1 Â ’2 R A 1 Â2 GðaÞ¼NðÀaÞ for a A and GðÀ4Þ¼O F : For a ¼ p e A ÀfÀ4 O g with dðÀaÞ¼d define:

(I) GðaÞ¼/aSF’2 when R44e (i) (II) When 2eoRp4e: ( R d 2e ’2 /aSð1 þ p þ À ÞF -NðÀaÞ if dp2e À R=2 ðiiÞ GðaÞ¼ /aSð1 þ pR=2ÞF’2 if Rd42e À R=2 ðiiiÞ

(III) When Rp2e: 8 <> NðÀaÞ if Rdpe À R=2 ðivÞ R=2 d e ’2 GðaÞ¼ ð1 þ p þ À ÞF -NðÀaÞ if e À R=2odp3e=2 À R=4 ðvÞ :> ð1 þ peÀ½e=2ÀR=4ÞF’2 if d43e=2 À R=4 ðviÞ

Remark. If R is odd then d ¼ 0soif2eoRo4e then (ii) gives GðpReÞ¼/peSð1 þ R 2e ’2 R R p À ÞF -NðÀpeÞ while if Ro2e then (iv) gives Gðp eÞ¼NðÀpeÞ: In fact Gðp eÞ¼ NðÀpeÞ also holds R ¼ 2e þ 1: Indeed formula (ii) gives Gðp2eþ1eÞ¼/peSð1 þ 1 ’2 1 ’2 Â ’2 ’ 2e 1 p ÞF -NðÀpeÞ: But /peSð1 þ p ÞF ¼ /peSO F ¼ F which implies Gðp þ eÞ¼ NðÀpeÞ:

Lemma 3.7. For any binary lattice L we have yðLÞ¼GðaðLÞÞ:

Proof. For R40 this is just Theorems 2.1, 2.2 and2.3 in [X1]. These formulas turn out to be true also for Rp0; i.e. for the case L modular. Case Rp0 when dpe À R=2 is proposition B in [H], when d43e=2 À R=4 it is C andD in [H] andwhen e À R=2odp3e=2 À R=4itisin[X]. (The results for Rp0in[H] and [X] are in a different form. Note that in [H] and [X] we have r ¼ ord sL ¼ 0 and u ¼ ord nL ¼ ord a ¼ n so R ¼ 2r À 2u ¼À2n:) The only exception to these formulas is the case when L is hyperbolic, i.e. aðLÞ¼ 1 0 ’2 ’ À4: Formula (III) wouldgive yðLÞ¼ð1 þ p ÞF ¼ F but it is known that Â ’2 1 Â ’2 & yðAð0; 0ÞÞ ¼ O F : This explains the exception GðÀ4Þ¼O F :

D! g b ! g b Note: For L a; b we have aðLÞ¼a so yð a; b Þ¼GðaÞ:

Deﬁnition 5. If a ¼ pRe with dðÀaÞ¼d we deﬁne G0ðaÞ as follows:

(I) G0ðaÞ¼F’2 when R44e (i) ARTICLE IN PRESS

C.N. Beli / Journal of Number Theory 102 (2003) 125–182 137

(II) When 2eoRp4e: ( R d 2e ’2 ð1 þ p þ À ÞF -NðÀaÞ if dp2e À R=2 ðiiÞ; G0ðaÞ¼ ð1 þ pR=2ÞF’2 if d42e À R=2 ðiiiÞ:

Obviously, the formula for G0ðaÞ is the same as the formula for GðaÞ in cases (I) and(II) but with the factor /aS ignored. Hence we have GðaÞ¼/aSG0ðaÞ: R R 2e ’2 Also note that if 2eoRo4e is odd then (ii) gives G0ðp eÞ¼ð1 þ p À ÞF - NðÀpeÞ:

0 Lemma 3.8. If pReAA and R0XR and R0 R ðmod2 Þ then pR eAA and GðpR0 eÞDGðpReÞ:

Proof. We couldgive a computational proof using formulas (I)–(III) from Deﬁnition 4. However there is a more conceptual proof using 3.2. Let R0 ¼ R R þ 2k where kX0 is an integer. Let L ¼ !x1; x2g a lattice with aðLÞ¼p e: Since k 0 k !p x2gD!x2g we have from 2.7(ii) that there is a lattice L ¼ !x1; p x2g and 0 0 L DL: Since also L and L have the same norm generator x1 we have from 3.2(i) that 0 0 0 k 2k OðL ÞDOðLÞ so yðL ÞDyðLÞ: But aðL Þ¼Qðp x2Þ=Qðx1Þ¼p Qðx2Þ=Qðx1Þ¼ p2kaðLÞ¼pRþ2ke ¼ pR0 e: Hence GðpR0 eÞ¼yðL0ÞDyðLÞ¼GðpReÞ: &

3.9. Let us see how a binary lattice L with aðLÞ¼pRe looks like if we suppose that there is a norm generator xAL with QðxÞ¼1: For R40 we have already seen that L corresponds to a form of the type X 2 þ epRY 2: For Rp0 L must be pr-modular for certain r: We have u ¼ ord nL ¼ ord QðxÞ¼0so R ¼ 2r À 2u ¼ 2r: Since x is a norm generator and QðxÞ¼1 we have D r Àr r R=2 ÀR=2 R=2 1 pR=2 L p Aðp ; Àp dÞ¼p Aðp ; Àp dÞ¼ pR=2 ÀpRd relative to a basis x; y: (As usual we assume that Àe ¼ 1 þ d and DðÀeÞ¼dO:)SoL corresponds to the 2 R=2 R 2 form X þ 2p XY À dp Y : Let us note that this formula applies also in the case 1 0 when R40 is even since if LD R adapted to the basis fx; yg then 0 ðÀ1ÀdÞp D 1 pR=2 R=2 L pR=2 ÀpRd adapted to the basis fx; y þ p xg: Hence it is not necessary to treat the two cases separately.

Corollary 3.10. If LD!1; pReg then:

(a) If R40 then QðLÞDO2 þ pR: If Rp2e is even then: (b) If dXe À R=2 then QðLÞDO2 þ pR=2þe: (c) If dpe À R=2 then QðLÞDO2 þ pRþd : ARTICLE IN PRESS

138 C.N. Beli / Journal of Number Theory 102 (2003) 125–182

Proof. If R40 then LDo1; pRe adapted to a basis fx; yg: For v ¼ ax þ byAL with 2 R 2 2 2 R 2 R 2 a; bAO we have QðvÞ¼a þ p eb : Since a AOand p eb Ap we have QðvÞAO þ R D 1 pR=2 p which implies (a). If Rp2e is even then L pR=2 ÀpRd adapted to a basis fx; yg: Let v ¼ ax þ byAL with a; bAO: We have QðvÞ¼a2 þ 2pR=2ab À dpRb2: But a2AO2; 2pR=2abApR=2þe and dpRb2ApRþd : It follows that QðLÞDO2 þ pa where a ¼ minfR=2 þ e; R þ dg; i.e. we have (b) and(c). &

2 Definition 6. We define the function g : A-SgpðOÂ=OÂ Þ as follows: For a ¼ pReAA with dðÀaÞ¼d define:

2 (I) gðaÞ¼OÂ when R42e (i) (II) When Rp2e: ( 2 ð1 þ pRþd ÞOÂ -NðÀaÞ if Rdpe À R=2 ðiiÞ gðaÞ¼ 2 ð1 þ pR=2þeÞOÂ if Rd4e À R=2 ðiiiÞ

Note that for Ro2e odd we have d ¼ 0 so (ii) gives gðÀpReÞ¼ð1 þ 2 pRÞOÂ -NðÀpeÞ:

Lemma 3.11. If L is a binary lattice and xAL is a norm generator then fQðvÞ=QðxÞjvAL is a norm generatorg¼gðaðLÞÞ:

Proof. Let L be a lattice with aðLÞ¼a ¼ pRe andlet xAL be a norm generator. By scaling we can assume that QðxÞ¼1soLD!1; ag: In order for an element vAL to be a norm generator we needto have QðvÞAOÂ: Hence fQðvÞ=QðxÞjvAL is a norm generatorg¼QðLÞ-OÂ so we have to prove that QðLÞ-OÂ ¼ gðaÞ: First, we consider the case R ¼ RðLÞ¼À2e; i.e. when LDprAð0; 0Þ or D r D1 D1 L p Að2; 2rÞ: Since nL ¼ O we have L 2 Að0; 0Þ; resp. L 2 Að2; 2rÞ: In the ﬁrst 1 À2e Â Â2 N case we have aðLÞ¼À4 ¼Àp (in O =O )andd ¼ 4e ÀðÀ2eÞ=2so 2 2 gðÀpÀ2eÞ¼ð1 þ pðÀ2eÞ=2þeÞOÂ ¼ð1 þ p0ÞOÂ ¼ OÂ while in the secondcase we D À2e À2e have aðLÞ¼À4 ¼ÀDp and d ¼ 2epe ÀðÀ2eÞ=2sogðÀDp Þ¼ð1 þ À2eþ2e Â2 Â 1 D1 p ÞO -NðDÞ¼O : But for both L ¼ 2 Að0; 0Þ and L 2 Að2; 2rÞ we have QðLÞ*OÂ so we are done. From now on we will assume that L is not of the form 1 1 2 Að0; 0Þ or 2 Að2; 2rÞ: This will imply that R4 À 2e so R=2 þ e40 andalso R4 À d so R þ d40: From 3.10 we have QðLÞDO2 þ pa for certain integer a: This implies that 2 QðLÞ-OÂDð1 þ paÞOÂ : Also QðLÞDQðFLÞ¼Qð½1; aÞ ¼ NðÀaÞ so in fact 2 QðLÞ-OÂDð1 þ paÞOÂ -NðÀaÞ: We use now the cases (a)–(c) of 3.10 in order to prove the D part of the lemma. ARTICLE IN PRESS

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2 2 If R42e we use (a) so a ¼ R andwe get QðLÞ-OÂDð1 þ pRÞOÂ ¼ OÂ ¼ gðaÞ: For 0oRo2e odd we use again (a) and we have QðLÞ-OÂDð1 þ 2 pRÞOÂ -NðÀaÞ¼gðaÞ: If Ro2e is even and d4e À R=2 we use (b) so a ¼ 2 R=2 þ e andwe get QðLÞ-OÂDð1 þ pR=2þeÞOÂ ¼ gðaÞ andif dpe À R=2 we use (c) 2 andso a ¼ R þ d so we get QðLÞ-OÂDð1 þ pRþd ÞOÂ -NðÀaÞ¼gðaÞ: 2 We prove now the reverse inclusion. For R42e it is trivial (OÂ DQðLÞ is obvious). If Ro2e is odd let ZAgðaÞ; i.e. ZAOÂ with dðZÞXR and ZANðÀpeÞ: We needto prove that ZAQðLÞ: Now let QðLÞD/1; pReS adapted to a basis fx; yg: Since ZANðÀpeÞ¼QðFLÞ there is v ¼ ax þ byAFL with a; bAF s.t. QðvÞ¼a2 þ pReb2 ¼ Z: We have ordða2 þ pReb2Þ¼ord Z ¼ 0: But ord a2 is even andord pReb2 is odd. It follows that ord a2 ¼ 0; i.e. aAOÂ andord pReb240: If beO then 0oord pReb2oRo2e is odd so ord DðZÞ¼ord Dða2 þ pReb2Þ¼ord pReb2oR: Contradiction. So bAO and v ¼ ax þ byAL: For the case when Rp2e is even we use the basis fx; yg adapted to which D 1 pR=2 A L pR=2 ÀpRd with the notation of 3.10. If a; b F and v ¼ ax þ by then QðvÞ¼ a2 þ 2pR=2ab À pRdb2: 2 If dpe À R=2 let ZAgðaÞ¼ð1 þ pRþd ÞOÂ -NðÀeÞ: We may assume Z ¼ 1 þ a where ord a ¼ dðZÞXR þ d: Now d must be odd since if d ¼ 2epe À R=2 then Rp À 2e which contradicts our assumption that R4 À 2e: We have ZANðÀeÞ¼ QðFLÞ so there is v ¼ ax þ byAFL s.t. Z ¼ QðvÞ¼a2 þ 2pR=2ab À pRdb2: Since do2e we have ord a2 þ ord pRdb2 ¼ 2 ord a þ 2 ord b þ R þ do2 ord a þ 2 ord b þ R þ 2e ¼ 2 ord2 abpR=2: Hence ord2 abpR=24minford a2; 2abpR=2; ord pRdb2g: We also have that ord a2 is even andord pRdb2 is odd so they are different. It follows that 0 ¼ ord QðvÞ¼minford a2; 2abpR=2; ord pRdb2g¼ord a2: Hence aAOÂ: Now we only have to prove that bAO: If not then ord bo0 so 2 ord boord b: We also have dpe À R=2soR þ dpR=2 þ e: We add and we get ord pRdb2 ¼ 2 ord b þ R þ doord b þ R=2 þ e ¼ ord2 abpR=2d: Hence ordð2pR=2ab À pRdb2Þ¼ord pRdb2 ¼ 2 ord b þ R þ d: Since 2 ord b þ R þ doR þ dpR=2 þ ep2e is odd we have R þ dpord DðZÞ¼ord Dða2 þ 2pR=2ab À pRdb2Þ¼2 ord b þ R þ doR þ d: Contradic- tion. So bAO and vAL: 2 If d4e À R=2 let ZAgðaÞ¼ð1 þ pR=2þeÞOÂ with dðZÞXR=2 þ e: We may assume that Z ¼ 1 þ a where ord a ¼ dðZÞXR=2 þ e: The inequality d4e À R=2 implies R þ d4R=2 þ e: We are looking for vAL of the form v ¼ x þ by with bAO s.t. QðvÞ¼Z: Now b has to satisfy the equation: 1 þ 2pR=2b À pRdb2 ¼ Z ¼ 1 þ a i.e. pRdb2 À 2pR=2b þ a ¼ 0: We use the fact that F is henselian. We have ord pRd ¼ R þ d andord2 pR=2 ¼ R=2 þ e: Since R þ d þ ord aXR þ d þ R=2 þ e42ðR=2 þ eÞ the Newton polygon associatedto pRdb2 À 2pR=2b þ a ¼ 0 has two segments of length 1 andwith slopes R=2 þ e ÀðR þ dÞ¼e À R=2 À do0 andord a ÀðR=2 þ R 2 R=2 eÞX0: Hence the equation p db À 2p b þ a ¼ 0 has two roots b1; b2AF of orders e À R=2 À d; resp. ord a ÀðR=2 þ eÞ: Since ord b2 ¼ ord a ÀðR=2 þ eÞX0 we have b2AO so we may chose b ¼ b2 andwe are done. & ARTICLE IN PRESS

140 C.N. Beli / Journal of Number Theory 102 (2003) 125–182

Remark. As seen in the proof of 3.11 for R40 we can use 3.10(a) andtake a ¼ R so 2 2 QðLÞ-OÂDð1 þ pRÞOÂ : Hence gðpReÞDð1 þ pRÞOÂ whenever R40: (This can also be checked directly from deﬁnition.)

3.12. As a consequence if LD!a; bg then there is a norm generator xAL with A b QðxÞ¼a: Lemma 3.11 implies that for any Z gðaðLÞÞ ¼ gðaÞ there is another norm 0A 0 D ! g ! bg generator x L with Qðx Þ¼Za: It follows that L Za 1; aðLÞ ¼ Za 1; a ¼ ! g ! gD! g A b Za; Zb : For short a; b Za; Zb for any Z gðaÞ:

Lemma 3.13. Let a ¼ pReAA with dðÀaÞ¼d: We have:

(i) If R42e then G0ðpReÞ¼gðpRÀ2eeÞF’2 and G0ðpReÞDð1 þ pRÀ2eÞF’2: R 2 e=2 R=4 ’2 (ii) If Rp2e is even and d4e À R=2 then Gðp eÞ¼gðpÀ ½ À eÞF :

2 Proof. (i) If R44e then G0ðpReÞ¼F’2 and R À 2e42e so gðpRÀ2eeÞ¼OÂ which implies gðpRÀ2eeÞF’2 ¼ F’2 ¼ G0ðpReÞ: If 2eoRp4e then 0oR À 2ep2e so pRÀ2eeAA andit is in case (II) of Definition 6. R R d 2e ’2 If dp2e À R=2 then G0ðp eÞ¼ð1 þ p þ À ÞF -NðÀeÞ: We also have dp2e À 2 R=2 ¼ e ÀðR À 2eÞ=2 which implies that gðpRÀ2eeÞ¼ð1 þ pRÀ2eþd ÞOÂ -NðÀeÞ so R 2e ’2 R d 2e ’2 R gðp À eÞF ¼ð1 þ p þ À ÞF -NðÀeÞ¼G0ðp eÞ: If d42e À R=2 then G0ðpReÞ¼ð1 þ pR=2ÞF’2: We also have d42e À R=2 ¼ e À 2 2 ðR À 2eÞ=2sogðpRÀ2eeÞ¼ð1 þ pðRÀ2eÞ=2þeÞOÂ ¼ð1 þ pR=2ÞOÂ : This implies that gðpRÀ2eeÞF’2 ¼ð1 þ pR=2ÞF’2 ¼ G0ðpReÞ: Now R42e so R À 2e40 so by Remark to 3.11 we have gðpRÀ2eeÞDð1 þ 2 pRÀ2eÞOÂ which implies that G0ðpReÞ¼gðpRÀ2eeÞF’2Dð1 þ pRÀ2eÞF’2: X 1 (ii) First, we note that d4e À R=2 2½2ðe À R=2Þ ¼ 2½e=2 À R=4 so À2½e=2 À R=44 À d: Also À2epRp2e which implies that 0 ¼½e=2 Àð2eÞ=4p½e=2 À R=4p½e=2 ÀðÀ2eÞ=4¼e so À2ep À 2½e=2 À R=4p0: Now À2½e=2 À R=4 is even, X À d and X À 2e so pÀ2½e=2ÀR=4eAA: Also since À2½e=2 À R=4p0o2e we have that pÀ2½e=2ÀR=4e is in case (II) of Definition 6. We have e ÀðÀ2½e=2 À R=4Þ=2 ¼ e þ½e=2 À R=4¼½3e=2 À R=4: It follows that d43e=2 À R=43d4½3e=2 À R=43d4e ÀðÀ2½e=2 À R=4Þ=2: Hence if pRe is in case (v), resp. (vi), of Definition 4 then pÀ2½e=2ÀR=4e is in case (ii), resp. (iii), of Definition 6. Let us consider the two cases. If d43e=2 À R=4 then GðpReÞ¼ð1 þ peÀ½e=2ÀR=4ÞF’2 and gðpÀ2½e=2ÀR=4eÞ¼ð1 þ 2 2 pðÀ2½e=2ÀR=4Þ=2þeÞOÂ ¼ð1 þ peÀ½e=2ÀR=4ÞOÂ which implies that gðpÀ2½e=2ÀR=4eÞF’2 ¼ ð1 þ peÀ½e=2ÀR=4ÞF’2 ¼ GðpReÞ: R R=2 d e ’2 If e À R=2odp3e=2 À R=4 then Gðp eÞ¼ð1 þ p þ À ÞF -NðÀeÞ and 2 gðpÀ2½e=2ÀR=4eÞ¼ð1 þ pÀ2½e=2ÀR=4þd ÞOÂ -NðÀeÞ which implies that gðpÀ2½e=2ÀR=4eÞ ’2 2 e=2 R=4 d ’2 F ¼ð1 þ pÀ ½ À þ ÞF -NðÀeÞ: ARTICLE IN PRESS

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1 Now 2½e=2 À R=4¼2½2ðe À R=2Þ so if e À R=2 is even then 2½e=2 À R=4¼ e À R=2 while if e À R=2 is odd then 2½e=2 À R=4¼e À R=2 À 1: In the ﬁrst case we have À2½e=2 À R=4þd ¼Àðe À R=2Þþd ¼ R=2 þ d À e which implies 2 e=2 R=4 ’2 R=2 d e ’2 R that gðpÀ ½ À eÞF ¼ð1 þ p þ À ÞF -NðÀeÞ¼Gðp eÞ: In the secondcase we have À2½e=2 À R=4þd ¼Àðe À R=2 À 1Þþd ¼ R=2 þ d À e þ 1 which 2 e=2 R=4 ’2 R=2 d e 1 ’2 implies that gðpÀ ½ À eÞF ¼ð1 þ p þ À þ ÞF -NðÀeÞ: We have three cases: (1) d is odd. Since R=2 À e is also odd we have that R=2 þ d À e is even. But we also have R=2 þ d À epð2eÞ=2 þ d À e ¼ do2e which implies that ð1 þ R=2 d e ’2 R=2 d e 1 ’2 k ’2 p þ À ÞF ¼ð1 þ p þ À þ ÞF : (If 0oko2e is even then ð1 þ p ÞF ¼ k 1 ’2 2 e=2 R=4 ’2 R=2 d e 1 ’2 ð1 þ p þ ÞF :) It follows that gðpÀ ½ À eÞF ¼ð1 þ p þ À þ ÞF -NðÀeÞ¼ð1 þ R=2 d e ’2 R p þ À ÞF -NðÀeÞ¼Gðp eÞ: (2) d ¼ N: We have R=2 þ d À e ¼ R=2 þ d À e þ 1 ¼ N andagain we are done. (3) d ¼ 2e: Let us note that we cannot have R ¼À2e since otherwise the relation d4e À R=2 would be contradicted. Hence R4 À 2e which implies that 3e=2 À R=4o3e=2 ÀðÀ2eÞ=4 ¼ 2e ¼ d which contradicts our assumption that dp3e=2 À R=4: Hence the case d ¼ 2e does not happen. (Actually case 2, d ¼ N does not happen either.) &

Lemma 3.14. If Rp2e and pRe; pRZAA then gðpReÞgðpRZÞ¼ð1 þ 2 pRþdðeZÞÞOÂ gðpReÞ:

Proof. We denote a ¼ pRe; b ¼ pRZ; g ¼ gðaÞgðbÞ and d ¼ dðeZÞ: Hence we have to 2 prove that g ¼ð1 þ pRþd ÞOÂ gðaÞ: We have d ¼ dðabÞXminfdðÀaÞ; dðÀbÞg with equality if dðÀaÞadðÀbÞ: We consider three cases: (1) dðÀaÞ; dðÀbÞ4e À R=2: This implies d4e À R=2: We have gðaÞ¼gðbÞ¼ 2 2 ð1 þ pR=2þeÞOÂ so g ¼ð1 þ pR=2þeÞOÂ : Hence we only have to prove that 2 ð1 þ pRþd ÞOÂ2Dð1 þ pR=2þeÞOÂ : But this is follows from R þ d4R þ e À R=2 ¼ R=2 þ e: (2) minfdðÀaÞ; dðÀbÞgpe À R=2anddðÀaÞadðÀbÞ: In this case we claim that 2 2 g ¼ð1 þ pRþd ÞOÂ which makes the equality g ¼ð1 þ pRþd ÞOÂ gðaÞ trivial. WLOG we may assume that dðÀaÞodðÀbÞ so d ¼ dðÀaÞpe À R=2: Hence gðaÞ¼ð1 þ 2 pRþd ÞOÂ -NðÀaÞ: There are two cases: 2 (a) If dðÀbÞ4e À R=2 then gðbÞ¼ð1 þ pR=2þeÞOÂ : Since R þ dpR þ e À R=2 ¼ 2 R=2 þ e we have gðbÞDð1 þ pRþd ÞOÂ : We also have gðaÞ¼ð1 þ 2 2 pRþd ÞOÂ -NðÀaÞCð1 þ pRþd ÞOÂ with index 2. Hence g can only be gðaÞ or ð1 þ 2 pRþd ÞOÂ : In order to have g ¼ gðaÞ we needthat ð1 þ pR=2þeÞOÂ2 ¼ gðbÞDgðaÞ¼ 2 2 ð1 þ pRþd ÞOÂ -NðÀaÞ; i.e. ð1 þ pR=2þeÞOÂ DNðÀaÞ: By 1.2(iii) this is equivalent to dðÀaÞþR=2 þ e42e: But this is impossible since dðÀaÞpe À R=2: Hence g ¼ 2 ð1 þ pRþd ÞOÂ : ARTICLE IN PRESS

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2 (b) If dðÀbÞpe À R=2 then gðbÞ¼ð1 þ pRþdðÀbÞÞOÂ -NðÀbÞ: Since R þ doR þ 2 e À R=2 ¼ R=2 þ e we have gðbÞDð1 þ pRþd ÞOÂ : By a similar reasoning as in case 2 (a) we have g ¼ gðaÞ¼ð1 þ pRþd ÞOÂ -NðÀaÞ if gðbÞDNðÀaÞ and g ¼ð1 þ 2 2 pRþd ÞOÂ otherwise. If ð1 þ pRþdðÀbÞÞOÂ -NðÀbÞ¼gðbÞDNðÀaÞ then by 1.3(i) 2 we have that ð1 þ pRþdðÀbÞÞOÂ is included in either NðÀaÞ or NðabÞ which by 1.2(iii) implies that either dðÀaÞþR þ dðÀbÞ42e or dðabÞþR þ dðÀbÞ42e: But this is 2 impossible since dðÀaÞ¼dðabÞodðÀbÞpe À R=2: So g ¼ð1 þ pRþd ÞOÂ : (3) dðÀaÞ¼dðÀbÞpe À R=2: This implies dXdðÀaÞ: We have gðaÞ¼ 2 2 ð1 þ pRþdðÀaÞÞOÂ -NðÀaÞ and gðbÞ¼ð1 þ pRþdðÀaÞÞOÂ -NðÀbÞ: Hence gðaÞ; gðbÞ 2 are subgroups of ð1 þ pRþdðÀaÞÞOÂ of index 2. It follows that g ¼ gðaÞ if gðaÞ¼gðbÞ 2 2 and g ¼ð1 þ pRþdðÀaÞÞOÂ otherwise. But the condition that ð1 þ pRþdðÀaÞÞOÂ - 2 NðÀaÞ¼gðaÞ¼gðbÞ¼ð1 þ pRþdðÀaÞÞOÂ -NðÀbÞ is equivalent by 1.3(ii) with ð1 þ 2 pRþdðÀaÞÞOÂ DNðabÞ which by 1.3(ii) is equivalent to dðabÞþR þ dðÀaÞ42e; i.e. to d þ dðÀaÞ42e À R: 2 2 On the other hand, we have dXdðÀaÞ so ð1 þ pRþd ÞOÂ Dð1 þ pRþdðÀaÞÞOÂ : Since 2 2 also gðaÞ¼ð1 þ pRþdðÀaÞÞOÂ -NðÀaÞDð1 þ pRþdðÀaÞÞOÂ with index 2 we 2 have by a similar reasoning as in case (2) that ð1 þ pRþd ÞOÂ gðaÞ equals gðaÞ if 2 2 ð1 þ pRþd ÞOÂ DNðÀaÞ andit is ð1 þ pRþdðÀaÞÞOÂ otherwise. But by Lemma 1.2(iii) 2 ð1 þ pRþd ÞOÂ DNðÀaÞ iff dðÀaÞþR þ d42e i.e. iff d þ dðÀaÞ42e À R: 2 In conclusion g ¼ð1 þ pRþd ÞOÂ gðaÞ¼gðaÞ if d þ dðÀaÞ42e À R and g ¼ð1 þ 2 2 pRþd ÞOÂ gðaÞ¼ð1 þ pRþdðÀaÞÞOÂ otherwise. &

Corollary 3.15. Let pRe; pRZAA with e; ZAOÂ: We have:

R R R d eZ 2e ’2 R (i) If 2eoRp4e then G0ðp eÞG0ðp ZÞ¼ð1 þ p þ ð ÞÀ ÞF G0ðp eÞ and GðpReÞGðpRZÞ¼/eZSð1 þ pRþdðeZÞÀ2eÞF’2GðpReÞ: (ii) If Rp2e is even and dðÀeÞ; dðÀZÞ4e À R=2 then GðpReÞGðpRZÞ¼ð1 þ pR=2þdðeZÞÀeÞF’2GðpReÞ:

Proof. (i) From 3.13(i) we have G0ðpReÞ¼gðpRÀ2eeÞF’2 and G0ðpRZÞ¼gðpRÀ2eZÞF’2 R R R 2e R 2e ’2 so G0ðp eÞG0ðp ZÞ¼gðp À eÞgðp À ZÞF : Since R À 2ep2e we have from 3.14 that 2 gðpRÀ2eeÞgðpRÀ2eZÞ¼ð1 þ pRÀ2eþdðeZÞÞOÂ gðpRÀ2eeÞ: It follows that G0ðpReÞG0ðpRZÞ¼ 2 ð1 þ pRÀ2eþdðeZÞÞOÂ gðpRÀ2eeÞF’2 ¼ð1 þ pRÀ2eþdðeZÞÞF’2gðpRÀ2eeÞF’2 ¼ð1 þ pRþdðeZÞÀ2eÞ F’2G0ðpReÞ: For the other claim of (i) we note that GðpReÞGðpRZÞ¼/pRe; pRZS G0ðpReÞG0ðpRZÞ and /eZSð1 þ pRþdðeZÞÀ2eÞF’2GðpReÞ¼/eZ; pReSð1 þ pRþdðeZÞÀ2eÞ F’2G0ðpReÞ: Since G0ðpReÞG0ðpRZÞ¼ð1 þ pRþdðeZÞÀ2eÞF’2G0ðpReÞ and /pRe; pRZSF’2 ¼ /eZ; pReSF’2 we get GðpReÞGðpRZÞ¼/eZSð1 þ pRþdðeZÞÀ2eÞF’2GðpReÞ: ARTICLE IN PRESS

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(ii) We have GðpReÞ¼gðpÀ2½e=2ÀR=4eÞF’2 and GðpRZÞ¼gðpÀ2½e=2ÀR=4ZÞF’2 from 3.13(ii). Since À2½e=2 À R=4p0o2e we can apply 3.14 andwe 2 get gðpÀ2½e=2ÀR=4eÞgðpÀ2½e=2ÀR=4ZÞ¼ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞOÂ gðpÀ2½e=2ÀR=4eÞ so 2 GðpReÞGðpRZÞ¼ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞOÂ gðpÀ2½e=2ÀR=4eÞF’2 ¼ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞ F’2gðpÀ2½e=2ÀR=4eÞF’2 ¼ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞF’2GðpReÞ: But, as seen in the proof of Lemma 3.13(ii) we have ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞF’2 ¼ð1 þ pR=2þdðeZÞÀeÞF’2 with the exception of the case when dðeZÞ¼2e and e À R=2 is odd. Hence in this case GðpReÞGðpRZÞ¼ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞF’2GðpReÞ¼ð1 þ pR=2þdðeZÞÀeÞF’2GðpReÞ as claimed. If dðeZÞ¼2e and e À R=2 is odd then we have ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞF’2 ¼ð1 þ pR=2þdðeZÞÀeþ1ÞF’2 ¼ð1 þ pR=2þeþ1ÞF’2: It follows that GðpReÞGðpRZÞ¼ð1 þ pÀ2½e=2ÀR=4þdðeZÞÞF’2GðpReÞ¼ð1 þ pR=2þeþ1ÞF’2GðpReÞ andwe want to prove that GðpReÞGðpRZÞ¼ð1 þ pR=2þdðeZÞÀeÞF’2GðpReÞ¼ð1 þ pR=2þeÞF’2GðpReÞ: But we have 2 dðÀeÞ4e À R=2sogðpReÞ¼ð1 þ pR=2þeÞOÂ : Now gðpReÞDGðpReÞ (see 3.16) 2 so GðpReÞ+ð1 þ pR=2þeÞOÂ which implies that GðpReÞ+ð1 þ pR=2þeÞF’2+ð1 þ pR=2þeþ1ÞF’2: It follows that ð1 þ pR=2þeÞF’2GðpReÞ¼ð1 þ pR=2þeþ1ÞF’2GðpReÞ¼ GðpReÞ so we are done. &

3.16. Let us note that we always have gðaÞDGðaÞ and aAGðaÞ: Indeed if LD!1; ag then there is a norm generator xAL with QðxÞ¼1: By 3.11 for any 0 0 ZAgðaÞ there is a norm generator x AL with Qðx Þ¼Z: Now tx; tx0 AOðLÞ: Hence þ Z ¼ yðtxtx0 ÞAyðLÞ¼GðaÞ: We also have À1FLAO ðLÞ so a ¼ det FL ¼ yðÀ1FLÞAyðLÞ:

Lemma 3.17. Let a ¼ pReAA and let d ¼ dðÀeÞ: The following are equivalent: (1) There is a binary lattice L with aðLÞ¼a and two norm generators x; x0AL s.t. QðxÞ¼Qðx0Þ and L ¼ Ox þ Ox0: (2) For any binary lattice L with aðLÞ¼a and any norm generator xAL there is x0AL with Qðx0Þ¼QðxÞ s.t. L ¼ Ox þ Ox0: (3) a ¼ pRe is in one of the following situations: (i) R42e: (ii) R ¼ 2e À 2d: (iii) R e 1 Â2 2e À 2doRo2e is even and R=2 þ e is even and p e À 4 O : (iv) R A 1 Â2 R ¼ 2eorp e À 4 O and the residual ﬁeld O=p has more than two elements.

Proof. ð2Þ-ð1Þ is obvious. Let us do ð1Þ-ð2Þ: Let L be a binary lattice with aðLÞ¼ a andlet xAL be a norm generator. From (1) we have that there is a binary lattice K with aðKÞ¼a andtwo norm generators y; y0AK s.t. QðyÞ¼Qðy0Þ and K ¼ Oy þ Oy0: We have LD!QðxÞ; aQðxÞg; KD!QðyÞ; aQðyÞg adapted to some BONGs x; x00; resp. y; y00: By scaling we may assume that QðyÞ¼QðxÞ: It follows ARTICLE IN PRESS

144 C.N. Beli / Journal of Number Theory 102 (2003) 125–182 that there is an isometry s : K-L given by y/x; y00/x00: Since L ¼ sK ¼ Osy þ Osy0 ¼ Ox þ Osy0 we can take x0 ¼ sy0 andwe are done. ð2Þ-ð3Þ: Let us suppose that aAA satisﬁes (2) but it is not in any of cases (i)–(iv). There are several cases: (a) 0oRo2e is odd. In this case we can take L ¼ Ox>Oy with QðxÞ¼1 and QðyÞ¼pRe: Let us assume that there is x0AL with Qðx0Þ¼QðxÞ¼1 s.t. L ¼ Ox þ Ox0: We have x0 ¼ ax þ by with a; bAO: The condition Ox þ Oy ¼ L ¼ Ox þ Ox0 implies that bAOÂ: We have a2 þ pReb2 ¼ Qðx0Þ¼1so1À pReb2 is a square. But this is impossible since ord pReb2 ¼ R which is an odd number o2e: (b) R 2e is even but either R 2e À 2d or 2e À 2d R and R=2 þ e is odd. We o À o o D 1 pR=2 take a lattice L pR=2 ÀdpR relative to a basis x; y where as usual Àe ¼ 1 þ d with dO ¼ pd : We have QðxÞ¼1 andwe suppose that there is x0AL s.t. Qðx0Þ¼QðxÞ¼1 and L ¼ Ox þ Ox0: As before we write x0 ¼ ax þ by andthe condition Ox þ Oy ¼ L ¼ Ox þ Ox0 implies bAOÂ: Let us note that we cannot have R ¼À2e: (If R ¼À2e R A 1 Â2 D Â2 N then p e À 4 O or À4O so d ¼ or 2e which makes Ro2e À 2d impossible. Also R=2 þ e ¼ðÀ2eÞ=2 þ e ¼ 0 is even.) This implies that the inequality RX À d is strict. Hence R=2 þ e40andR þ d40: We have 1 ¼ Qðx0Þ¼a2 þ 2pR=2ab À dpRb2: Since ord2 pR=2abXord2 pR=2 ¼ R=2 þ e40 andord dpRb2 ¼ R þ d40 we must have ord a2 ¼ 0soaAOÂ: Since a; bAOÂ we have ord2 pR=2ab ¼ R=2 þ e and ord dpRb2 ¼ R þ d: If Ro2e À 2d then doe À R=2: It implies R þ doR=2 þ e so ordð2pR=2ab À dpRb2Þ¼R þ d: But R þ d is odd (doe À R=2oe ÀðÀ2eÞ=2 ¼ 2e so d is odd) and R þ do2e À 2d þ d ¼ 2e À do2e: Since aAOÂ we get dð1Þ¼dða2 þ 2pR=2ab À dpRb2Þ¼R þ d: Contradiction. If 2e À 2doRo2e and R=2 þ e is odd then the inequality 2e À 2doR implies d4e À R=2soR þ d4R=2 þ e: It follows that ordð2pR=2ab À dpRb2Þ¼R=2 þ e: But R=2 þ e is odd and R=2 þ eoð2eÞ=2 þ e ¼ 2e: Since aAOÂ this implies dð1Þ¼ dða2 þ 2pR=2ab À dpRb2Þ¼R=2 þ e: Contradiction. R A 1 Â2 (c) p e À 4 O and O=p has two elements. This corresponds to the case of hyperbolic planes. We take LDAð0; 0Þ relative to a basis u; v: We take x ¼ u þ v with QðxÞ¼2: Any other element x0AL with Qðx0Þ¼2 is of the form x0 ¼ au þ aÀ1v Â 0 for an À aAOÁ : The condition Ox þ Ox ¼ L ¼ Ou þ Ov is equivalent to Â{ 1 1 À1 2 2A Â O det a aÀ1 ¼ a À a ¼ð1 À a Þ=a; i.e. to 1 À a O : But this is impossible since O=p has only two elements so any unit is 1 ðmod pÞ so 1 À a2 1 À 12 ¼ 0 ðmod pÞ: (d) R ¼ 2e and O=p has two elements. Let D ¼ 1 À 4r be as usual. We take L ¼ 2 Ox>Oy with QðxÞ¼1andQðyÞ¼4e: (We have aðLÞ¼pRe since 4e ¼ p2ee in F’=OÂ :) If x0 ¼ ax þ bx is another norm generator with Qðx0Þ¼1thenOx þ Ox0 ¼ L ¼ Ox þ Oy implies bAOÂ: We have a2 þ 4eb2 ¼ Qðx0Þ¼1so1À 4eb2 is a square. But eb2; rAOÂ and O=p has two elements we must have eb2 r 1 ðmod pÞ: Hence 1 À 4eb2 1 À 4r ¼ D ðmod4 pÞ: This implies that a2 ¼ 1 À 4eb2ADF’2: Contradiction. ARTICLE IN PRESS

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ð3Þ-ð1Þ: We prove first case (iv). If R ¼ 2e we write L ¼ Ox þ Oy where QðxÞ¼1 2 and QðyÞ¼4e: We have aðLÞ¼p2ee since 4e ¼ p2ee in F’=OÂ : We needto find x0AL with Qðx0Þ¼QðxÞ¼1 s.t. L ¼ Ox þ Ox0: Since O=p has more than two elements there is a unit ZAOÂ s.t. Zc1 ðmod pÞ: Since Zc0; 1 ðmod pÞ we have that Z2 þ Zc0 ðmod pÞ so Z2 þ ZAOÂ: Let now aAOÂ s.t. a2e Z2 þ Z ðmod pÞ: We have 1 þ 2 4a2e 1 þ 4ðZ2 þ ZÞ¼ð1 þ 2ZÞ2 ðmod4 pÞ: It follows that 1 þ 4a2eAOÂ : Let bAOÂ s.t. b2 ¼ 1 þ 4a2e: Let x0 ¼ bÀ1ðx þ ayÞ: We have Qðx0Þ¼bÀ2ð1 þ 4a2eÞ¼1 and Ox þ Ox0 ¼ Ox þ Oðx þ ayÞ¼Ox þ Oy ¼ L (because a; bAOÂ) so we are done. R A 1 Â2 0A If p e À 4 O then we needto finda hyperbolic plane L and x; x L norm generators s.t. QðxÞ¼Qðx0Þ and Ox þ Ox0 ¼ L: Like in case (c) of ð2Þ-ð3Þ we take L ¼ Ou þ Ov with QðuÞ¼QðvÞ¼0 and Bðu; vÞ¼1: We take x ¼ u þ v and x0 ¼ au þ aÀ1v with aAOÂ: We have QðxÞ¼Qðx0Þ¼2 andthe condition L ¼ Ox þ Ox0 will be again equivalent to 1 À a2AOÂ; i.e. to 1 À a2c0 ðmod pÞ or ac1 ðmod pÞ: But it is possible to choose such aAOÂ since O=p has more than two elements. D 1 pR=2 Let us now consider the case (ii). We take L pR=2 ÀdpR with the usual notations relative to the basis x; y: We needto find x0AL with Qðx0Þ¼QðxÞ¼1 s.t. Ox þ Ox0 ¼ L: We will take x0 ¼ x þ ay with aAF’: The condition Ox þ Oðx þ ayÞ¼L ¼ Ox þ Oy is equivalent to aAOÂ: Hence we needto find aAOÂ s.t. 1 ¼ Qðx þ ayÞ¼ 1 þ 2pR=2a À dpRa2; i.e. dpRa2 À 2pR=2a ¼ 0: Now a ¼ 2pÀR=2dÀ1 is a solution of the equation dpRa2 À 2pR=2a ¼ 0: But ord2 pÀR=2dÀ1 ¼ e À R=2 À d ¼ e Àð2e À 2dÞ=2 À d ¼ 0soa ¼ 2pÀR=2dÀ1AOÂ has the requiredproperties.

For cases (i) and(iii) we prove the following:

Claim. If pReAA s.t. gðpReÞ¼gðpRþ2eÞ andthere is a binary lattice L with aðLÞ¼ pRe that can be spannedby two norm generators then (1) is true.

Proof. Let L be as above andlet x1; xAL be norm generators s.t. L ¼ Ox1 þ Ox: Let pr > x ¼ x2: We have pr > ðOx1 þ OxÞ¼Ox2 so L ¼ !x1; x2g: We consider the x1 x1 0 0 0 sub-lattice L ¼ Ox1 þ px: Since pr > L ¼ px2 we have L ¼ !x1; px2g: Now x1 0 0 À1 2 À1 2 Rþ2 aðL Þ¼det L Qðx1Þ ¼ p det LQðx1Þ ¼ p aðLÞ¼p e: We have QðxÞ=Qðx1ÞA R Rþ2 0 0 gðp eÞ¼gðp eÞ¼gðaðL ÞÞ and x1AL is a norm generator so by Lemma 3.11 0 0 0 0 there is x AL s.t. Qðx Þ=Qðx1Þ¼QðxÞ=Qðx1Þ so QðxÞ¼Qðx Þ: We also 0 0 0 have x AOx1 þ px so x ¼ ax1 þ bx with aAO; bAp: Since x ¼ ax1 þ bxAL is a norm generator ðQðx0Þ¼QðxÞÞ it has to be primitive. Since bAp we must have Â 0 aAO : It follows that Ox þ Ox ¼ Ox þ Oðax1 þ bxÞ¼Ox þ Ox1 ¼ L so we are done. Let us prove now that in both cases (i) and(iii) pRe satisﬁes the conditions of our 2 claim. If R42e (case (i)) then gðpReÞ¼gðpRþ2eÞ¼OÂ : If R is as in case (iii) then ARTICLE IN PRESS

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Ro2e so R þ 2p2e: We also have that 2e À 2doR which implies that d4e À 2 R=24e ÀðR þ 2Þ=2: It follows that gðpReÞ¼ð1 þ pR=2þeÞOÂ and gðpRþ2eÞ¼ð1 þ 2 2 pðRþ2Þ=2þeÞOÂ ¼ð1 þ pR=2þeþ1ÞOÂ : But R=2 þ e is even and R=2 þ eoð2eÞ=2 þ e ¼ 2 2 2e ðRo2eÞ so ð1 þ pR=2þeÞOÂ ¼ð1 þ pR=2þeþ1ÞOÂ : Hence again gðpReÞ¼gðpRþ2eÞ: For the other condition of the claim in case (i) we take L ¼ Ox>Oy with QðxÞ¼1 and QðyÞ¼pRe: We have that Qðx þ yÞ¼1 þ pReAOÂ so x þ y is a norm generator for L: But x; x þ y span L ¼ Ox þ Oy so we are done. In case (iii) let us note that R A D Â2 R4 À 2e: Indeed if R ¼À2e then we must have p e À 4 O so d ¼ 2e: (Case R A 1 Â2 p e À 4 O is excluded from (iii).) But this implies À2e ¼ R42e À 2d ¼ 2e À 2ð2eÞ¼À2e: Contradiction. So R4 À 2e which, as before, implies R=2 þ e40 and D 1 pR=2 R þ d40: We take L pR=2 ÀdpR with the usual notation relative to a basis x; y: Now Qðx þ yÞ¼1 þ 2pR=2 À dpR: Since ord2 pR=2 ¼ R=2 þ e40 andord dpR ¼ R þ d40 we have Qðx þ yÞAOÂ so x þ y is a norm generator. But x; x þ y span L so again we are done. (Note: One can actually prove that the only case where a binary lattice L cannot be spannedby two norm generators is when L is a hyperbolic plane and O=p has just two elements.) &

Lemma 3.18. Let L be a binary lattice with aðLÞ¼pRe and let x; x0AL be two norm generators s.t. QðxÞ¼Qðx0Þ and x À x0 is anisotropic. We have:

0 R (i) txÀx0 AOðLÞ and QðxÞQðx À x ÞAGðp eÞ: (ii) If R42e then ord Qðx7x0ÞXord QðxÞþ2e and for exactly one choice of the sign 7 we have ord Qðx7x0Þ¼ord QðxÞþ2e: (iii) If R42e and ord Qðx À x0Þ¼ord QðxÞþ2e then QðxÞQðx À x0ÞAG0ðpReÞ:

0 0 0 Proof. (i) Let L ¼ txÀx0 L: Since xAL is a norm generator x ¼ txÀx0 xAtxÀx0 L ¼ L is also a norm generator. Since x0 is a common norm generator of L and L0 and 0 0 0 RðLÞ¼RðL Þ (because LDL ) we have L ¼ L ¼ txÀx0 L by 3.2(ii) so txÀx0 AOðLÞ: þ Since xAL is a norm generator we also have txAOðLÞ so txtxÀx0 AO ðLÞ: It follows 0 R that QðxÞQðx À x Þ¼yðtxtxÀx0 ÞAyðLÞ¼Gðp eÞ: (ii) By scaling we can suppose WLOG that QðxÞ¼Qðx0Þ¼1: We have R ¼ RðLÞ40soL is not modular. Since xAL is a norm generator we have L ¼ !x; yg ¼ Ox>Oy for some yAL: We have QðyÞ¼QðyÞ=QðxÞ¼aðLÞ¼pRe: Now x0AL ¼ Ox>Oy so x0 ¼ ax þ by where a; bAO: We have Bðx; x0Þ¼Bðx; ax þ byÞ¼ a so Qðx7x0Þ¼QðxÞþQðx0Þ72Bðx; x0Þ¼2ð17aÞ: We have 1 ¼ Qðx0Þ¼Qðax þ byÞ¼a2 þ b2pRe so 1 À a2 ¼ b2pRe: It follows that ordð1 À aÞþordð1 þ aÞ¼ ordð1 À a2ÞXR42e so for at least one of the choices of 7 we have ordð17aÞ4e: Let us suppose that ordð1 À aÞ4e ¼ ord2 : It follows that ordð1 þ aÞ¼ordð2 À ð1 À aÞÞ ¼ ord2 ¼ e: Since ord Qðx7x0Þ¼ord2 ð17aÞ¼ordð17aÞþe we get ord Qðx À x0Þ42e andord Qðx þ x0Þ¼2e: If ordð1 þ aÞ4e we get similarly that ordð1 À aÞ¼e so ord Qðx þ x0Þ42e andord Qðx À x0Þ¼2e: ARTICLE IN PRESS

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(iii) With the same notation as for (ii) we have that ordð1 À aÞþe ¼ ord Qðx À x0Þ¼2e so ordð1 À aÞ¼e: It follows that x À x0 ¼ð1 À aÞx À byAK where K ¼ pex>Oy: Since QðpexÞ¼p2e and QðyÞ¼pRe and2 eoR we have from 3.3(ii) that K ¼ !pex; yg and aðKÞ¼pRÀ2ee: Now ord Qðx À x0Þ¼2e ¼ ord nK so x À x0 is a norm generator in K: Since pex is another norm generator of K we have that pÀ2eQðx À x0Þ¼Qðx À x0Þ=QðpexÞAgðaðKÞÞ ¼ gðpRÀ2eeÞ: It follows that QðxÞQðx À x0Þ¼Qðx À x0ÞAgðpRÀ2eeÞF’2 ¼ G0ðpReÞ: &

Lemma 3.19. If x; y are two vectors and R an integer s.t. ord QðxÞ; ord QðyÞ4R and ord Qðx þ yÞ¼R then Ox þ OyDpRÀeAð0; 0Þ:

1 Proof. By scaling we may assume WLOG that R ¼ e: We have Bðx; yÞ¼2ðQðx þ yÞÀQðxÞÀQðyÞÞ: Since ord QðxÞ; ord QðyÞ4e andord Qðx þ yÞ¼e we have ord Bðx; yÞ¼0: It follows that Ox þ Oy is unimodular with nðOx þ OyÞ¼2O: (We have QðxÞ; QðyÞA2O:) Hence Ox þ OyDAð0; 0Þ or Að2; 2rÞ: But since x is primitive in Ox þ Oy andord QðxÞ4e we cannot have Ox þ OyDAð2; 2rÞ: Hence Ox þ OyDAð0; 0Þ as claimed. &

4. More on BONGs

Xu [X2] introduced the so-called ‘‘minimal norm splittings’’ and he proved that every lattice has a minimal norm splitting. A minimal norm splitting is a Jordan splitting L ¼ L1>y>Lt having the property that for any other Jordan splitting 0 0 0 0 L ¼ L1 >y>Lt we have nLiDnLi andif Li is a hyperbolic lattice then so is Li: As a corollary to the existence of minimal norm splittings he proves that any lattice admits a splitting L ¼ L0>H where H is a maximal hyperbolic lattice that can be 0 0 0 0 split by L: If L ¼ L0 >H is another splitting of this kindthen L0DL0 andH DH : (Theorem 1.1 andCorollary 1.1.) If L ¼ L1>y>Lt is a minimal norm splitting with sL1*?*sLt then for any 1piojpt we have that either 0oord nLj À ord nLio2ðord sLj À ord sLiÞ or ord nLi ord nLj þ 1 ðmod2 Þ or both. [X2, Lemma 1.1].IfL does not split any hyperbolic planes and L ¼ L1>y>Lt is a minimal norm splitting andthere are ioj s.t. ord nLi ord nLj þ 1 ðmod2 Þ and0 oord nLj À ord nLio2ðord sLj À ord sLiÞ fails or if any of the Li’s has rank X3 then yðLÞ¼F’: ’ Indeed if rank LiX3 and Li does not split any hyperbolic plane then yðLiÞ¼F by [H] Proposition A and [OM, 93:18] so yðLÞ¼F’: If all ranks of Li’s are p2 we denote ri ¼ ord sLi and ui ¼ ord nLi: If for some ioj we have uicuj ðmod2 Þ and0 ouj À uio2ðrj À riÞ fails then one of Li; Lj is binary ’ andwe have uj þ ui À 2rip2e which by [X2] Lemma 2.2(2) implies yðLÞ¼F: (See [HSX], Proof of 2.1.) This leads to the following deﬁnitions: ARTICLE IN PRESS

148 C.N. Beli / Journal of Number Theory 102 (2003) 125–182

Deﬁnition 7. We say that a lattice L has property A if it has a Jordan splitting L ¼ L1>y>Lt s.t. rank Lip2 andfor any ioj we have

0oord nLj À ord nLio2ðord sLj À ord sLiÞ: ð1Þ

As we have just notedabove any lattice L which is not split by a hyperbolic plane andfor which yðLÞaF’ has property A. Condition (1) is equivalent to the condition sLi that nð>jaiLjÞ CnLi for any 1pipt andalso equivalent to the inclusions: *?* xC?C x nL1 nLt and nL1 nLt : They imply [X2, Proposition 1.1] that the norms of the Jordan components are invariants.

Deﬁnition 8. A splitting L ¼ L1>y>Lt is calleda maximal norm splitting if each Li sLi is either unary or modular binary with sL1+?+sLt and nLi ¼ nL : (Note that a maximal norm splitting is not in general a Jordan splitting. However if L has property A then any Jordan splitting is also a maximal norm splitting).

sL Remark. The condition nLi ¼ nL i is equivalent to 0pord nLj À ord nLip 2ðord sLj À ord sLiÞ for any 1piojpn andalso equivalent to the inclusions: +?+ xD?D x nL1 nLt and nL1 nLt :

Lemma 4.1. (i) If L ¼ L1>y>Lt is a maximal norm splitting then a BONG for L can be obtained by putting together any BONGs of L1; y; Lt: (ii) Any BONG of L may be obtained in this way from a certain Jordan decomposition if L has property A.

Proof. (i) We proceedby induction over t: If rank L1 ¼ 1 then the BONG for L1 has only one element x1 for which L1 ¼ Ox1: But nL1 ¼ nL so x1 is a norm generator for L: It follows that it is the first element of a BONG for L: Now pr > L ¼ L2>y>Lt x1 andby the induction step the sequence x2; y; xn obtainedby putting together the BONGs of L2; y; Lt is a BONG for L2>y>Lt: It follows that the sequence x1; y; xn obtainedby putting together the BONGs of L1; y; Lt is a BONG for L: If rank L1 ¼ 2 then let x1; x2 be a BONG for L1: We have that Qðx1ÞO ¼ nL1 ¼ nL so x1 is a norm generator for L so it is the first element in a BONG of L: We have pr > L ¼ðpr > L1Þ>L2>y>Lt ¼ Ox2>L2>y>Lt: Since L1 ¼ !x1; x2g is x1 x1 modular we have ord Qðx2Þpord Qðx1Þ¼ord nL: It follows that x2 is a norm generator for Ox2>L2>y>Lt andit is a secondelement in a BONG of L: We have pr > ðOx2>L2>y>LtÞ¼L2>y>Lt: Now the sequence x3; y; xn obtainedby x2 putting together the BONGs of L2; y; Lt is a BONG for L2>y>Lt by the induction step. It follows that the sequence x1; y; xn obtainedby putting together the BONGs of L1; y; Lt is a BONG for L: For (ii) we use again induction. Let us suppose first that nL ¼ sL: If x1; y; xn is a BONG for L we have that Qðx1ÞO ¼ nL ¼ sL so we can split L ¼ Ox1>L : We have that x2; y; xn is a BONG for L : Now if sL CsL we have that Ox1 is the first Jordan component of L for a certain Jordan composition. It follows that L also has ARTICLE IN PRESS

C.N. Beli / Journal of Number Theory 102 (2003) 125–182 149 property A. (If L ¼ L1>y>Lt has property A then obviously L2>y>Lt also has property A.) By the induction step we have a Jordan decomposition L ¼ L2>y>Lt s.t. the BONG x2; y; xn of L is obtainedby putting together the BONGs of L2; y; Lt: Hence x1; y; xn is obtainedby putting together the BONGs of L1; y; Lt where L1 ¼ Ox1: If sL ¼ sL then the sL-modular component of L must be unary since otherwise L ¼ Ox1>L wouldbe split by an sL-modular lattice of rank X3: (Ox1 is also sL- modular.) It follows that nL ¼ sL ¼ sL: Since x2 is a norm generator of L we have Qðx2ÞO ¼ sL andthere is a splitting L ¼ Ox2>L with sL CsL: We also have that x3; y; xn is BONG for L : Now L1 ¼ Ox1>Ox2 is sL-modular so it is the first Jordan component of L and L ¼ L1>L : Hence L has also the property A and by induction step there is a Jordan splitting L ¼ L2>y>Lt s.t. the BONG x3; y; xn is obtainedby putting together the BONGs of L2; y; Lt: Since also x1; x2 is a BONG for L1 we have that x1; y; xn is made by putting together the BONGs of L1; y; Lt: If nLCsL then the first Jordan component of L is binary. Let L1 be the first Jordan component of L (in a certain decomposition). We write L ¼ L1>L : Since L 0 has property A we have that nL1 ¼ nL and nL CnL: We write x1 ¼ x þ x with 0 0 xAL1 and x AL : We have Qðx1Þ¼QðxÞþQðx Þ: But ord Qðx1Þ¼ord nL and 0 ord Qðx ÞXord nL 4ord nL so ord QðxÞ¼ord nL which implies that xAL1 is primitive. Hence L1 ¼ Ox þ Oy for some yAL1: Since nL1CsL1 we must have 0 ord Bðx; yÞ¼ord sL1 ¼ ord sL: But Bðx1; yÞ¼Bðx þ x ; yÞ¼Bðx; yÞ: Hence ord Bðx1; yÞ¼ord sL andwe also have Qðx1Þ; QðyÞAnLCsL so Ox1 þ Oy is sL- modular. Hence by replacing L1 with Ox1 þ Oy we reduce to the case when x1AL1: Let pr > L1 ¼ Ox: We have pr > L ¼ Ox>L : Since L1 ¼ !x1; xg is modular we x1 x1 have ord QðxÞpord Qðx1Þoord nL so x is a norm generator for Ox>L : Now x2 is 0 also a norm generator for Ox>L so ord Qðx2Þ¼ord QðxÞ: We have x2 ¼ ax þ x 0 2 0 0 with aAO and x AL : Since Qðx2Þ¼a QðxÞþQðx Þ andord Qðx ÞX Â ord nL 4ord QðxÞ¼ord Qðx2Þ we must have aAO : Let L1 ¼ Ox1 þ Oy: Since Ox ¼ pr > ðOx1 þ OyÞ¼pr > Oy we may suppose after multiplying y by a unit that x1 x1 pr > y ¼ x: Since L1 ¼ Ox1 þ Oy and nL1CsL1 we must have ord Bðx1; yÞ¼ x1 0 Â ord sL1 ¼ ord sL: Let y1 ¼ ay þ x : We have Bðx1; y1Þ¼aBðx1; yÞ: But aAO so ord Bðx1; y1Þ¼ord Bðx1; yÞ¼ord sL andwe also have Qðx1Þ; Qðy1ÞAnLCsL which implies that Ox1 þ Oy1 is sL-modular. Hence we can replace L1 by Ox1 þ Oy1: We 0 0 have pr > y1 ¼ pr > ðay þ x Þ¼ax þ x ¼ x2: Hence pr > L1 ¼ pr > ðOx1 þ Oy1Þ¼ x1 x1 x1 x1 pr > Oy1 ¼ Ox2: It follows that x1; x2 is a BONG for L1: We write L ¼ L1>L x1 andwe have that x3; y; xn is a BONG for L : Since L has property A andhas t À 1 Jordan components by the induction step there is a decomposition L ¼ L2>y>Lt s.t. x3; y; xn is obtainedby putting together the BONGs of L2; y; Lt: Hence x1; y; xn is obtainedby putting together the BONGs of L1; y; Lt: &

Corollary 4.2. Let L ¼ !x1; y; xng with Ri ¼ ord QðxiÞ: We have:

(i) If L has property A then RioRiþ2 for any 1pipn À 2: ARTICLE IN PRESS

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(ii) Suppose that L ¼ L1>y>Lt is a maximal norm splitting and suppose that the BONG x1; y; xn is obtained from the BONGs of L1; y; Lt: Then RipRiþ2 for any 1pipn À 2:

Proof. (i) From 4.1 we know that the succession x1; y; xn is obtainedby putting together BONGs of L1; y; Lt for a certain Jordan decomposition L ¼ L1>y>Lt: r u x ux k k k *?* Let sLk ¼ p ; nLk ¼ p and nLk ¼ p : The inclusions nL1 nLt and xC?C x ? x ? x nL1 nLt can be written as u1o out and u14 4ut : Let us note that in x the case when Lk ¼ !xig ¼ Oxi we have rk ¼ uk ¼ Ri and uk ¼ÀRi so Ri ¼ uk ¼ x Àuk: In the case when Lk ¼ !xi; xiþ1g we have from 3.4(iii) that RiXRiþ1 , rk ¼ x x ðRi þ Riþ1Þ=2; uk ¼ Ri and uk ¼ÀRiþ1 so uk ¼ RiXRiþ1 ¼Àuk: In conclusion in both cases when Lk is unary or binary we have that the ﬁrst element of the BONG of x Lk has length of order uk andthe last element of the BONG has length of order Àuk x andwe have ukX À uk: Let xiþ1 be in the BONG of Lk: If Lk ¼ !xiþ1g then xi is the last element of the BONG of LkÀ1 and xiþ2 is the ﬁrst element in the BONG of Lkþ1: Hence we have x ! g Ri ¼ÀukÀ1pukÀ1oukþ1 ¼ Riþ2 so we are done. If Lk ¼ xi; xiþ1 then uk ¼ Ri and ukþ1 ¼ Riþ2: Since ukoukþ1 we are done. Similarly if Lk ¼ !xiþ1; xiþ2g then x x x x Riþ2 ¼Àuk and Ri ¼ÀukÀ1: But ukÀ14uk which implies ÀRi4 À Riþ2; i.e. RioRiþ2: (ii) Same proof as for (iii) with the difference that this time the inequalities x x xX x ukoukþ1 and uk4ukþ1 are replacedby ukpukþ1 and uk ukþ1 so all the inequalities are non-strict. &

Deﬁnition 9. A BONG x1; y; xn with ord QðxiÞ¼Ri is called good if RipRiþ2 for any 1pipn À 2:

From Corollary 4.2 we see that a BONG is goodif the lattice has property A or if this BONG is obtainedfrom a maximal norm splitting.

Lemma 4.3. If x1; y; xn is an orthogonal basis of a quadratic space V with ord QðxiÞ¼Ri then:

(i) x1; y; xn is a BONG for a lattice having property A iff !xi; xiþ1g exists for any 1pipn À 1 and Riþ24Ri for any 1pipn À 2: (ii) x1; y; xn is a good BONG iff !xi; xiþ1g exists for any 1pipn À 1 and Riþ2XRi for any 1pipn À 2: (iii) Any good BONG can be obtained from a maximal norm splitting. Moreover the maximal norm splitting can be chosen s.t. all the binary components are improper.

Proof. If !x1; y; xng is a BONG then from 2.8 we have that !xi; xiþ1g exists for any 1pipn À 1: The conditions Riþ24Ri from (i) and Riþ2XRi from (ii) follow ARTICLE IN PRESS

C.N. Beli / Journal of Number Theory 102 (2003) 125–182 151 from 4.2(i), resp. from the definition of a good BONG. Hence we have the ‘‘only if’’ part of (i) and(ii). ‘‘If’’ for (i): We prove by induction over n that for any x1; y; xn that satisfies conditions of (i) we have some unary and modular binary lattices L1; y; Lt s.t. L1>y>Lt is the Jordan splitting of a lattice L with property A andthe sequence x1; y; xn is obtainedby putting together some BONGs of L1; y; Lt: This will imply that x1; y; xn is a BONG for L andwe will be done. For n ¼ 1 this statement is trivial. For the general case we have two situations: If R1oR2 we define L1 ¼ !x1g andif R1XR2 we define L1 ¼ !x1; x2g: (!x1; x2g exists by hypothesis.) In the secondcase R1XR2 implies that L1 is modular. Since the sequence x2; y; xn; resp. x3; y; xn; satisfies the conditions of (i) we have by the induction hypothesis that there are some unary andmodularbinary lattices L2; y; Lt s.t. L2>y>Lt is the 0 Jordan splitting of a lattice L with property A andthe sequence x2; y; xn; resp. x3; y; xn; is obtainedby putting together some BONGs of L2; y; Lt: We define L ¼ L1>y>Lt: Obviously, the sequence x1; y; xn is obtainedby putting together the BONGs of L1; y; Lt: We have to prove that L has property A and L ¼ L1>y>Lt is its Jordan decomposition, i.e. that sL1*?*sLt; nL1*?*nLt and xC?C x *?* *?* xC?C x nL1 nLt : We already have sL2 sLt; nL2 nLt and nL2 nLt * * xC x so we only have to prove that sL1 sL2; nL1 nL2 and nL1 nL2: We keep the rk uk x x notations sLk ¼ p ; nLk ¼ p and nLk ¼ uk from the proof of 4.2. We want to x x prove that r1or2; u1ou2 and u14u2: There are four cases: rank L1 ¼ rank L2 ¼ 1: We have R1oR2 and L1 ¼ !x1g; L2 ¼ !x2g: We x x x x have R1 ¼ u1 ¼Àu1 and R2 ¼ u2 ¼Àu2 so R1oR2 implies u1ou2 and u14u2: rank L1 ¼ 1; rank L2 ¼ 2: We have R1oR2 and R1oR3 (from hypothesis) and x L1 ¼ !x1g; L2 ¼ !x2; x3g: We have u1 ¼ R1oR2 ¼ u2 and Àu1 ¼ R1oR3 ¼ x x x Àu2 so u14u2: rank L1 ¼ 2; rank L2 ¼ 1: We have R1XR2 (L1 is modular) and R1oR3 (from x hypothesis) and L1 ¼ !x1; x2g; L2 ¼ !x3g: Hence u1 ¼ R1oR3 ¼ u2 and Àu1 ¼ x x x R2pR1oR3 ¼Àu2 which implies u14u2: rank L1 ¼ rank L2 ¼ 2: We have from hypothesis R1oR3 and R2oR4 and L1 ¼ x x !x1; x2g; L2 ¼ !x3; x4g: Hence u1 ¼ R1oR3 ¼ u2 and Àu1 ¼ R2oR4 ¼Àu2 x x which implies u14u2: x x x x x Hence we have proved u1ou2 and u14u2 which imply u1 À u1ou2 À u2: But uk ¼ x x x uk À 2rk so uk À uk ¼ 2rk: Hence u1 À u1ou2 À u2 implies 2r1o2r2; i.e. r1or2 so we are done. For (iii) andthe ‘‘if’’ part of (ii) we will show that for any sequence x1; y; xn satisfying the hypothesis of (ii) there are some unary andbinary improper modular lattices L1; y; Lt s.t. L1>y>Lt is a maximal norm splitting of a lattice L and x1; y; xn is obtainedby putting together some BONGs of L1; y; Lt: This will imply that x1; y; xn is a goodBONG for L so we have (ii). For (iii) if x1; y; xn is a good BONG for L it will satisfy the conditions of (ii) (because of the ‘‘only if’’ part). As seen above this will imply that x1; y; xn is a goodBONG obtainedfrom a maximal % % norm splitting L1>y>Lt of a lattice L: But x1; y; xn is a BONG for both L and L ARTICLE IN PRESS

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% so we must have L ¼ L ¼ L1>y>Lt: Moreover by the choice of Lk’s all the binary components of the maximal norm splitting L ¼ L1>y>Lt are improper so we have (iii). The construction of L1; y; Lt is similar as for (i). The difference is that this time we only have RipRiþ2 insteadof RioRiþ2: However this time we only have to prove that L1>y>Lt is a maximal norm splitting which means D?D D?D x+?+ x sL1 sLt; nL1 nLt and nL1 nLt so all we have to do is replace strict inclusions andinequalities by non-strict ones. Also in order to make sure that all binary Lk’s are improper we will choose this time L1 ¼ !x1g if R1pR2 and L1 ¼ !x1; x2g if R14R2 (insteadof R1oR2; resp. R1XR2). In the secondcase R14R2 will imply that L1 is improper modular. &

Corollary 4.4. Let x1; y; xn be a good BONG for a lattice L with ord QðxiÞ¼Ri:

(i) If RipRiþ1 then !x1; y; xng ¼ !x1; y; xig>!xiþ1; y; xng: (ii) If RiXRiþ1 then !x1; y; xng¼!x1; y; xiÀ1g>!xi; xiþ1g>!xiþ2; y; xng: (iii) If Ri4Riþ1 and the BONG x1; y; xn is obtained from a maximal norm splitting L ¼ L1>y>Lt then !xi; xiþ1g ¼ Lk for some k: R ðR þR Þ=2 (iv) If R1pR2 then sL ¼ p 1 and if R1XR2 then sL ¼ p 1 2 : For short ord sL ¼ minfR1; ðR1 þ R2Þ=2g: If x1; y; xi and xiþ1; y; xn are good BONGs then: (v) x1; y; xn is a good BONG for !x1; y; xig>!xiþ1; y; xng iff RiÀ1pRiþ1; RipRiþ2 and RipRiþ1:

Proof. (iii) We assume that xi; xiþ1 is not a BONG for any of the Lk’s. It follows that xi and xiþ1 are the last, resp. the ﬁrst, element in the BONGs of two consecutive x components Lk; Lkþ1: Hence Ri ¼Àukpukpukþ1 ¼ Riþ1: Contradiction. Hence !xi; xiþ1g ¼ Lk for some k: For (i) and(iv) we take a maximal norm splitting L ¼ L1>y>Lt with all the binary components improper s.t. x1; y; xn is obtainedby putting together some BONGs of L1; y; Lt: (i) If RipRiþ1 then !xi; xiþ1g cannot be one of Lk’s since otherwise Ri4Riþ1: (Lk wouldbe improper modular.) Hence xi and xiþ1 are the last, resp. the ﬁrst, element in the BONGs of two consecutive components Lk; Lkþ1: This implies that !x1; y; xig ¼ L1>y>Lk and !xiþ1; y; xng ¼ Lkþ1>y>Lt: Hence L ¼ !x1; y; xig>!xiþ1; y; xng: (iv) If R1pR2 then !x1; x2g cannot be improper modular so we cannot have R1 L1 ¼ !x1; x2g: It follows that L1 ¼ !x1g and sL ¼ sL1 ¼ p : If R14R2 we ðR1þR2Þ=2 have from (i) that L1 ¼ !x1; x2g so sL ¼ sL1 ¼ p : (If R1 ¼ R2 then R1 ¼ðR1 þ R2Þ=2:) Condition (ii) follows from (i). We have RiÀ1pRiþ1pRipRiþ2: Since Riþ1pRiþ2 we have !x1; y; xng ¼ !x1; y; xiþ1g>!xiþ2; y; xng: But ARTICLE IN PRESS

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RiÀ1pRi so !x1; y; xiþ1g ¼ !x1; y; xiÀ1g>!xi; xiþ1g: Hence !x1; y; xng ¼ !x1; y; xiÀ1g>!xi; xiþ1g>!xiþ2; y; xng: For (v) ‘‘only if’’ the conditions RiÀ1pRiþ1; RipRiþ2 follow from the deﬁnition of a goodBONG. If Ri4Riþ1 then we have Riþ2XRi4Riþ1XRiÀ1: Since Riþ24Riþ1 and Ri4RiÀ1 we have by (i) that !x1; y; xig ¼ !x1; y; xiÀ1g>Oxi and !xiþ1; y; xng ¼ Oxiþ1>!xiþ2; y; xng: On the other side Ri4Riþ1 so by (ii) we have !x1; y; xng ¼ !x1; y; xiÀ1g>!xi; xiþ1g>!xiþ2; y; xng: Since !x1; y; xng ¼ !x1; y; xig>!xiþ1; y; xng we get !x1; y; xiÀ1g> !xi; xiþ1g > !xiþ2; y; xng ¼ !x1; y; xiÀ1g> Oxi>Oxiþ1>!xiþ2; y; xng which implies that !xi; xiþ1g ¼ Oxi>Oxiþ1: But this is impossible since Ri4Riþ1 so !xi; xiþ1g is improper modular. Hence RipRiþ1 as claimed. For the ‘‘if’’ part we use 4.3(ii) in order to prove that x1; y; xn is a goodBONG. The condition RjpRjþ2 for 1pjpi À 2 andfor ipjpn À 2 andthe conditionthat !xj; xjþ1g exists for 1pjpi À 1 andfor ipjpn À 1 follow from the fact that x1; y; xi and xiþ1; y; xn are goodBONGs. By hypothesis RjpRjþ2 also holds for j ¼ i; i þ 1 so we only have to show that !xi; xiþ1g exists. But ord Qðxiþ1Þ=QðxiÞ¼ Riþ1 À RiX0 which implies that Qðxiþ1Þ=QðxiÞAA so !xi; xiþ1g exists. Hence x1; y; xn is a goodBONG andsince RipRiþ1 we have from (i) !x1; y; xng ¼ !x1; y; xig>!xiþ1; y; xng: &

Lemma 4.5. Let L ¼ L1>L2 where L1; L2 are modular and L1 is binary.

0 0 (i) If sL1+sL2 and nL1CnL2 there is another decomposition L ¼ L1 >L2 where 0 0 0 0 L1 ; L2 are modular of the same scales and ranks as L1; L2 and nL1 ¼ nL2 ¼ nL2: D xC x 0> 0 (ii) If sL1 sL2 and nL1 nL2 there is another decomposition L ¼ L1 L2 where 0 0 0x 0x L1 ; L2 are modular of the same scales and ranks as L1; L2 and nL 1 ¼ nL 2 ¼ x nL2:

r u Proof. (i) We denote sLi ¼ p i and nLi ¼ p i : We have r1pr2pu2ou1: We have r u sL ¼ sL1 ¼ p 1 and nL ¼ nL2 ¼ p 2 : Suppose ﬁrst that u2 ¼ r1: This implies r r that r1 ¼ r2 ¼ u2 so L is p 1 -modular with nL ¼ sL ¼ p 1 so L ¼ Ox1>y>Oxn 0 0 with ord QðxiÞ¼r1: We take L1 ¼ Ox1>Ox2 and L2 ¼ Ox3>y>Oxn andwe are done. If u24r1 let L1 ¼ Ox þ Oy andlet zAL2 be a norm generator. Since L1 ¼ Ox þ Oy r1 0 is p -modular and nL1CsL1 (r1ou1) we must have ord Bðx; yÞ¼r1: We take L1 ¼ 0 Oðx þ zÞþOy: We have L1 DL andord Bðx þ z; yÞ¼ord Bðx; yÞ¼r1 ¼ ord sL so 0 r 0 u r 0 0 r sL1 ¼ p 1 : Since also nL1 DnL ¼ p 2 Cp 1 ¼ sL1 we must have that L1 is p 1 - 0 0 0 r modular. This implies that we have a splitting L ¼ L1 >L2 where L2 is p 2 -modular 0 0 of the same rank as L2: We must now prove that nL1 ¼ nL2 ¼ nL2: Since nL ¼ u2 0 0 nL2 ¼ p it is enough to prove that both L1 and L2 represent numbers of order u2: 0 For L1 we have ord QðxÞXu14u2 andord QðzÞ¼u2 andsince Qðx þ zÞ¼QðxÞþ 0 0 0 QðzÞ we have ord Qðx þ zÞ¼u2: For L2 let us note that L1 CL1>Oz and sL1 ¼ r1 0 0 0 sðL1>OzÞ¼p so we have another splitting L1>Oz ¼ L1 >Oz where ord Qðz Þ¼ ARTICLE IN PRESS

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0 0 0 0 ord QðzÞ¼u2: But we have L1 >Oz ¼ L1>OzDL ¼ L1 >L2 which implies 0 0 0 0 z AL2 : Hence both L1 and L2 represent numbers of order u2 andwe are done. (ii) Follows trivially from (i) by dualization. &

Lemma 4.6. Any lattice admits a maximal norm splitting and hence a good BONG.

Proof. Let L1>y>Lt be an arbitrary decomposition into unary and modular binary lattices with sL1+?+sLt: Let us suppose that this is not a maximal norm sL sL splitting. Then there is i s.t. nLiCnL i : In this case Li must be binary. Since L i ¼ > sLi sLi sLi a jLj we must have nLiCnL ¼ nLj for certain j i: + sLi sLi > If j4i then sLi sLj so nLiCnL ¼ nLj ¼ nLj: By 4.5(i) we have Li Lj ¼ 0 0 0 0 0 Li >Lj where Li ; Lj are modular of the same scales and ranks as Li; Lj and nLi ¼ 0 sL nLj ¼ nLj ¼ nL i :

sLi sLi x x If joi then sLiDsLj so nLiCnL ¼ nLj ¼ nððsLiÞLj Þ: It follows that nLi ¼ À1 x 0> 0 0 0 nððsLiÞ LiÞCnLj : By 4.5(ii) we have L ¼ Li Lj where Li ; Lj are modular of the 0x 0x x 0 same scales andranks as Li; Lj and nL i ¼ nL j ¼ nLj : We have sLj ¼ sLj and 0x x 0 0 0x x sLi nL j ¼ nLj so nLj ¼ nLj: We also have nLi ¼ nððsLiÞL i Þ¼nððsLiÞLj Þ¼nL : In both cases if we replace the components Li; Lj in the splitting L ¼ L1>y>Lt 0 0 by Li ; Lj then we get a new splitting of L where the components have the same scales andranks andthe only component that changedits norm is Li: We have sL 0 nLiCnL i ¼ nLi so by this procedure we decreased the number of indices k s.t. sL sL nLkCnL k : By repeating the procedure we get a splitting s.t. nLi ¼ nL i for any 1pipt; i.e. a maximal norm splitting. &

Lemma 4.7. Let x1; y; xn be a good BONG of a lattice L with ord QðxiÞ¼Ri and let L ¼ L1>y>Lt be a Jordan splitting. Then the numbers R1; y; Rn are in 1–1 sLk correspondence with the invariants t; rank Lk; sLk and nLk (i.e. they uniquely determine each other). In particular Ri’s do not depend on the choice of the good BONG x1; y; xn: P Proof. Let nk ¼ rank Lk; rk ¼ ord sLk and uk ¼ ord nLk: We put la ¼ koa nk: Let L ¼ L1>y>Ls be a maximal norm splitting where all the binary components are improper such that x1; y; xn is obtainedby putting together the BONGsP of 1 y s k k k k k k a k L ; ; L : We put n ¼ rank L ; r ¼ ord sL ; u ¼ ord nL and l ¼ koa n : 1 aÀ1 We have !x1; y; xla g ¼ L >y>L : By 4.4 we have that xi; xiþ1 belong to the 1 2 s BONGs of different Lk’s iff RipRiþ1: Hence the sequence 0 ¼ l ol o?ol is k made of all i’s s.t. RipRiþ1: It follows that s and l ’s are uniquely determined by Ri’s. kþ1 k k k k kþ1 If l ¼ l þ 1 then L ¼ !xlkþ1g andwe have sL ¼ nL ¼ Rlkþ1: If l ¼ k k k k l þ 2 then L ¼ !xlkþ1; xlkþ2g andwe have sL ¼ðRlkþ1 þ Rlkþ2Þ=2 and nL ¼ 1 s Rlkþ1: It follows that the ranks, scales andnorms of the modular lattices L ; y; L ARTICLE IN PRESS

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1 s are uniquely determined by Ri’s. Since L ¼ L >y>L this implies that the sLk fundamental invariants t; rank Lk; sLk and nLk are uniquely determined by Ri’s. ? Conversely, we will show that for any 1pkpt we have Rlkþ1 ¼ ¼ Rlkþ1 ¼ rk if ? ? rk ¼ uk and Rlkþ1 ¼ Rlkþ3 ¼ ¼ Rlkþ1À1 ¼ uk; Rlkþ2 ¼ Rlkþ4 ¼ ¼ Rlkþ1 ¼ 2rk À uk if rkouk: This will imply that Ri’s are uniquely determined by nk’s, rk’s and uk’s as claimed. By putting together the lattices with same scales from the splitting L ¼ 1 s L >y>L we obtain a Jordan splitting L ¼ L1>y>Lt: Let 1pkpt: We put g g 1 a a ¼ maxfgjr orkg and b ¼ maxfgjr prkg: We have L >y>L ¼ L1>y>LkÀ1 aþ1 b 1 a and L >y>L ¼ Lk: Since rankðL >y>L Þ¼rankðL1>y>LkÀ1Þ¼lk and 1 b rankðL >y>L Þ¼rankðL1>y>LkÞ¼lkþ1 we have that the sequence y aþ1 y b xlkþ1; ; xlkþ1 is obtainedby putting together the BONGs of L ; ; L : For any g r g sL u aogpb we have that sL ¼ sLk ¼ p k so nL ¼ nL k ¼ p k for aogpb: aþ1 b If rk ¼ uk then L ; y; L must be unary since all the binary components are improper. It follows that for any aogpb we have Lg ¼ !xig for some i andwe y have Ri ¼ rk: Since the sequence xlkþ1; ; xlkþ1 is obtainedfrom BONGs of Laþ1; y; Lb we must have Ri ¼ uk for lk þ 1piplkþ1 as claimed. g g If rkouk then L must be binary for aogpb so we have L ¼ !xi; xiþ1g for g rk g uk some i: But sL ¼ p and nL ¼ p so Ri ¼ uk and Riþ1 ¼ 2rk À uk: Since the y y sequence xlkþ1; ; xlkþ1 is obtainedfrom BONGs of Laþ1; ; Lb we must have that y & the sequence Rlkþ1; ; Rlkþ1 is made of copies of the pair uk; 2rk À uk as claimed.

Next we will show that goodBONGs have properties similar to orthogonal bases. One can check that Lemmas 4.8 and4.9 are trivial if we suppose that L ¼ Ox1>y>Oxn ¼ !x1; y; xng where ord Qðx1Þp?pord QðxnÞ:

Notation. If x is an anisotropic element of a quadratic space V we denote xx ¼ x QðxÞÀ1x: We have sðOxÞ¼QðxÞO so ðOxÞ ¼ QðxÞÀ1Ox ¼ Oxx which justiﬁes the notation. Obviously, QðxxÞ¼QðxÞÀ1 and ðxxÞx ¼ x:

x x Lemma 4.8. If x1; y; xn is a good BONG for a lattice L then xn; y; x1 is a good x BONG for L : As a consequence if LD!a1; y; ang relative to a good BONG then xD À1 À1 L !an ; y; a1 g relative to a good BONG.

Proof. If L ¼ Ox then Lx ¼ Oxx so if L is unary we are done. Let now L be modular binary with sL ¼ pr and nL ¼ pu: We have uXr and RðLÞ¼2r À 2u: If x; x0 is a BONG for L we have from 3.3(iii) that ord QðxÞ¼u and ord Qðx0Þ¼2r À u: We have L ¼ Ox þ Oy with ord Bðx; yÞ¼r: We have Ox0 ¼ 0 prx> ðOx þ OyÞ¼prx> Oy so by multiplying y with a unit we may assume that x ¼ À1 uÀr 0 uÀr uÀr À1 prx> y ¼ y À Bðx; yÞQðxÞ x: Now p x ¼ p y À p Bðx; yÞQðxÞ xAL because ord puÀr ¼ u À rX0 andord puÀrBðx; yÞQðxÞÀ1 ¼ u À r þ r À u ¼ 0: We also have that ord QðpuÀrx0Þ¼2ðu À rÞþ2r À u ¼ u ¼ ord nL so puÀrx0 is also a norm ARTICLE IN PRESS

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k uÀr 0 k generator for L: Now prx0> LCFx so there is k s.t. prx0> L ¼ p x: Hence p x ; p x is a BONG for L: Hence ord QðpkxÞ¼2r À u (3.3(iii)). Since ord QðxÞ¼u we get 2k þ u ¼ 2r À u so k ¼ r À u: So L ¼ !puÀrx0; prÀuxg: It follows that Lx ¼ pÀrL ¼ pÀr!puÀrx0; prÀuxg ¼ !puÀ2rx0; pÀuxg: But ord Qðx0ÞÀ1 ¼Àð2r À uÞ¼ord puÀ2r andord QðxÞÀ1 ¼Àu ¼ ord pÀu so Qðx0ÞÀ1x0; QðxÞÀ1x is also a BONG for L: (Because if we multiply the elements of a BONG by units we obtain also a BONG.) Hence L ¼ !Qðx0ÞÀ1x0; QðxÞÀ1xg ¼ !x0x; xxg: For the general case we have by 4.3(iii) that there is a maximal norm splitting L ¼ L1>y>Lt s.t. the BONG x1; y; xn of L is obtainedby putting together the x x> > x BONGs of L1; y; Lt: Now L ¼ Lt y L1 is also a maximal norm splitting so by x x x 4.1(i) if we put together any BONGs of Lt ; y; L1 we get a goodBONG for L : For x x any 1pkpt if the BONG of Lk was xi then we choose xi to be the BONG for Lk x x x while if the BONG of Lk was xi; xiþ1 we choose xiþ1; xi to be the BONG of Lk: By x x x putting together these BONGs of Lk we obviously get the sequence xn; y; x1 as desired. The secondclaim of the lemma is obvious since if QðxiÞ¼ai then x À1 Qðxi Þ¼ai : &

Lemma 4.9. (i) If x1; y; xn is a good BONG, resp. the BONG of a lattice with property A; then so is xi; y; xj for any 1pipjpn: Moreover Oð!xi; y; xjgÞDOð!x1; y; xngÞ: (ii) If x1; y; xn and yi; y; yj are good BONGs s.t. !xi; y; xjg ¼ !yi; y; yjg then !x1; y; xng ¼ !x1; y; xiÀ1; yi; y; yj; xjþ1; y; xng: As a consequence if LD!a1; y; ang and !ai; y; ajgD!bi; y; bjg relative to good BONGs then LD!a1; y; aiÀ1; bi; y; bj; ajþ1; y; ang:

Proof. (i) The fact that xi; y; xj is a goodBONG, resp. a BONG of a lattice with property A, follows from 4.3(i) and(ii). Hence we have the ﬁrst claim of (i). In the case j ¼ n the secondclaim of (i) follows from 2.7(i) and(ii) follows from 2.7(iv). For (i) in the general case we have Oð!xi; y; xngÞDOð!x1; y; xngÞ andby x x D x x x dualization we get Oð!xn; y; xi gÞ Oð!xn; y; x1gÞ: (In general OðM Þ¼ OðMÞ so if OðNÞDOðMÞ then OðNxÞDOðMxÞ:) Now if we apply the case j ¼ n of (i) x x x x x x to !xn; y; xi g we get Oð!xj ; y; xi gÞDOð!xn; y; xi gÞ: We dualize again andwe get Oð!xi; ?xjgÞDOð!xi; y; xngÞDOð!x1; y; xngÞ as claimed. For (ii) in general case we apply the case j ¼ n to !x1; y; xjg andwe get !x1; y; xiÀ1; yi; y; yjg ¼ !x1; y; xjg: Now xi; y; xj and yi; y; yj are both goodBONGs of the same lattice so by 4.7 we have that ord QðykÞ¼ord QðxkÞ¼ Rk: Hence the fact that x1; y; xj is a goodBONG implies that x1; y; xiÀ1; yi; y; yj is also a goodBONG. (The inequalities RkpRkþ2 are preserved.) Therefore we can x x apply 4.8 to both BONGs andwe get by dualization !xj ; y; x1g ¼ ! x y x x y xg ! x y xg yj ; ; yi ; xiÀ1; ; x1 : Now we apply the case j ¼ n to xn; ; x1 andwe ARTICLE IN PRESS

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! x y xg ! x y x x y x x y xg x y x have that xn; ; x1 ¼ xn; ; xjþ1; yj ; ; yi ; xiÀ1; ; x1 : Since xn; ; x1 x x x is a goodBONG andord QðykÞ¼ord QðxkÞ¼ÀRk we have that xn; y; x x y x x y x xjþ1; yj ; ; yi ; xiÀ1; ; x1 is also a goodBONG. Hence we can apply 4.7 to both BONGs andwe get by dualization !x1; y; xng ¼ !x1; y; xiÀ1; yi; y; yj; xjþ1; y; xng as claimed. &

Corollary 4.10. If LD!a1; y; ang relative to a good BONG x1; y; xn then: À (i) aiAyðO ðLÞÞ for any 1pipn: (ii) Gðaiþ1=aiÞDyðLÞ for 1pipn À 1: (iii) If ZAgðai=aiÀ1Þ or ZAgðaiþ2=aiþ1Þ then we have GðZaiþ1=aiÞDyðLÞ: ! g D A À A À Proof. (i) We have Oð xi Þ OðLÞ: Hence txi O ðLÞ so ai ¼ QðxiÞ yðO ðLÞÞ: (ii) We have Oð!xi; xiþ1gÞDOðLÞ so Gðaiþ1=aiÞ¼yð!ai; aiþ1gÞDyðLÞ: (iii) We have yð!aiÀ1; ai; aiþ1gÞDyðLÞ: But for any ZAgðai=aiÀ1Þ we have !aiÀ1; aigD!ZaiÀ1; Zaig: This implies !aiÀ1; ai; aiþ1gD!ZaiÀ1; Zai; aiþ1g: Hence GðZaiþ1=aiÞ¼yð!Zai; aiþ1gÞDyð!aiÀ1; ai; aiþ1gÞDyðLÞ: When ZAgðaiþ2=aiþ1Þ the proof is similar but this time we use the fact that !aiþ1; aiþ2gD!Zaiþ2; Zaiþ1g so !ai; aiþ1; aiþ2gD!ai; Zaiþ1; Zaiþ2g which implies that GðZaiþ1=aiÞ¼yð!ai; Zaiþ1gÞDyð!ai; aiþ1; aiþ2gÞDyðLÞ: &

R Deﬁnition 10. Let LD!a1; y; ang have property A with ai ¼ p i ei: We say that L has property B if whenever Riþ1 À Rip2e þ 1 is odd or Riþ1 À Ri is even and dðÀeieiþ1Þpe ÀðRiþ1 À RiÞ=2 we have both Ri À RiÀ1X2e þ 1 and Riþ2 À Riþ1X2e þ 1: (If i ¼ 1 we just ignore the condition Ri À RiÀ1X2e þ 1 andif i ¼ n À 1 we ignore Riþ2 À Riþ1X2e þ 1:)

Note: The deﬁnition of property B depends on how we write LD!a1; y; ang Hence whenever we refer to a lattice having property B we will always mean with respect to a speciﬁedBONG.

Property B is invariant under scaling. It is also invariant under dualization in the x sense that if L ¼ !x1; y; xngD!a1; y; ang has property B then L ¼ x x D À1 À1 !xn; y; x1g !an ; y; a1 g has property B. Also if L has property B with respect to a BONG x1; y; xn then the lattice !xi; y; xjg also has property B for any 1pipjpn: One can actually prove that property B does not depend on the choice of the BONG. Also if L ¼ L0>L00 has property B then both L0; L00 have property B. However this is beyondthe purpose of this paper.

Lemma 4.11. If a lattice L has property A but not property B then yðLÞ¼F’:

Ri ’ Proof. Let LD!a1; y; ang with ai ¼ p ei s.t. yðLÞaF we have to prove that L has property B. ARTICLE IN PRESS

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Riþ1ÀRi (i) If Riþ1 À Rip2e þ 1 is odd then yðLÞ+Gðaiþ1=aiÞ¼Gðp eieiþ1Þ¼ NðÀpeiþ1eiÞ (see Definition 4 andthe remark following it). If Ri À RiÀ1p2e then 2 2 RiÀRiÀ1 Â Â gðai=aiÀ1Þ¼gðp eieiÀ1ÞaO (see Definition 4). Let ZAgðai=aiÀ1ÞÀO : We Riþ1ÀRi have yðLÞ+GðZai=aiÀ1Þ¼Gðp eiþ1eiZÞ¼NðÀpeiþ1eiZÞ: Now NðÀpeiþ1eiÞ and ’ ’2 ’2 NðÀpeiþ1eiZÞ are subgroups of F=F of index 2 and they are different (ZeF ). Since they are both containedin yðLÞ we must have yðLÞ¼F’: Contradiction. In the case Â2 when Riþ2 À Riþ1p2e we have similarly gðaiþ2=aiþ1ÞaO so in this case we take Â2 ZAgðaiþ2=aiþ1ÞÀO : Hence we must have Ri À RiÀ1X2e þ 1 and Riþ2 À Riþ1X2e þ 1 as claimed. Let us suppose now that Riþ1 À Ri is even and dðÀeiþ1eiÞpe ÀðRiþ1 À RiÞ=2: This Riþ1ÀRi implies that yðLÞ+Gðp eieiþ1Þ¼NðÀeiþ1eiÞ: If dðÀeiþ1eiÞ¼2e then we have 2epe ÀðRiþ1 À RiÞ=2soRiþ1 À Rip À 2e: Since Riþ14RiÀ1 and Riþ24Ri this implies that RiÀ1 À Rip À 2e À 1 and Ri À Riþ2p À 2e À 1 so the conditions Ri À RiÀ1X2e þ 1 and Riþ2 À Riþ1X2e þ 1 are satisfiedanyways. Hence we may suppose that dðÀeiþ1eiÞo2e and Ri À RiÀ1p2e or Riþ2 À Riþ1p2e: If Ri À RiÀ1o2e þ 1 is odd or Riþ2 À Riþ1o2e þ 1 is odd then we are in the previous ’ situation. Since Riþ1 À Rio2e þ 1 we have yðLÞ¼F so we are done. If Ri À RiÀRiÀ1 RiÀ1p2e is even then DAgðp eieiÀ1Þ¼gðai=aiÀ1Þ so yðLÞ+GðDaiþ1=aiÞ¼ Riþ1ÀRi Gðp eiþ1eiDÞ: But dðDÞ¼2e4dðÀeiþ1eiÞ so dðÀeiþ1eiDÞ¼dðÀeiþ1eiÞpe À Riþ1ÀRi ðRiþ1 À RiÞ=2: This implies that Gðp eiþ1eiDÞ¼NðÀeiþ1eiDÞ: Now ’ ’2 NðÀeiþ1eiÞ; NðÀeiþ1eiDÞ are two different subgroups of F=F of index 2 that are ’ containedin yðLÞ: It follows that yðLÞ¼F: The case when Riþ2 À Riþ1p2e is even is similar but this time we use the fact that DAgðaiþ2=aiþ1Þ: Hence in order that ’ yðLÞaF we needagain Ri À RiÀ1X2e þ 1 and Riþ2 À Riþ1X2e þ 1: &

R1 Rn Remark 4.12. If LD!p e1; y; p eng has property B then Ri þ 2pRiþ2 for any 1pipn À 2:

Indeed from property A we have RioRiþ2: But if Ri þ 1 ¼ Riþ2 then ðRiþ2 À Riþ1ÞþðRiþ1 À RiÞ¼1 so one of Riþ2 À Riþ1; Riþ1 À Ri is even and the other is odd. If e.g. Riþ2 À Riþ1 is even and Riþ1 À Ri odd then we have Riþ2 À Riþ1X À 2e and Riþ1 À RiX1: This implies Riþ1 À Rip1 ÀðÀ2eÞ¼2e þ 1 and Riþ2 À Riþ1p1 À 1 ¼ 0: Since Riþ1 À Rip2e þ 1 is odd and Riþ2 À Riþ1o2e þ 1 L cannot have property B. Similarly, we get a contradiction if Riþ2 À Riþ1 is odd and Riþ1 À Ri even. Hence Ri þ 1aRiþ2 which implies Ri þ 2pRiþ2: &

5. Statement of the main result

The main result of this paper is Theorem 1. The proof is rather complicatedandit is divided into two parts. This section is devoted to the proof of the easier containment +: Section 6 will handle the reverse inclusion. ARTICLE IN PRESS

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Theorem 1. Let LD!a1; y; ang be a lattice that has the property A. Then:

a ’2 yðLÞ¼Gða2=a1ÞGða3=a2Þ?Gðan=anÀ1Þð1 þ p ÞF where a ¼ minf½ðord aiþ2 À ord aiÞ=2j1pipn À 2g:

Proof of the + inclusion. By 4.10(ii) we have Gðaiþ1=aiÞDyðLÞ for any 1pipn À 1: Thus it sufﬁces to prove that ð1 þ paÞF’2DyðLÞ which is equivalent to proving that ½ðord aiþ2Àord aiÞ=2 ’2 ð1 þ p ÞF DyðLÞ for any 1pipn À 2: Since yð!ai; aiþ1; aiþ2gÞD ½ðord aiþ2Àord aiÞ=2 ’2 yðLÞ it is enough to prove that ð1 þ p ÞF Dyð!ai; aiþ1; aiþ2gÞ: This reduces to the case when L is ternary. By scaling we may assume that r s ½s=2 ’2 LD!1; p e2; p e3g andwe want to prove that ð1 þ p ÞF DyðLÞ: From Corollary sÀr r r 4.10(iii) we have Gðp e2e3ZÞDyðLÞ for any ZAgðp e2Þ and Gðp e2ZÞDyðLÞ for any sÀr x Às Àr s ZAgðp e2e3Þ: Note that if we scale L D!p e3; p e2; 1g by p e3 we get the lattice sÀr s x sÀr s !1; p e2e3; p e3g: Hence yðLÞ¼yðL Þ¼yð!1; p e2e3; p e3gÞ: If L does not have property B then yðLÞ¼F’ which makes our statement trivial. Hence we may suppose that L has property B. We consider cases (i)–(iv): r (i) dðÀe2Þpe À r=2 and r is even. Then we have yðLÞ+Gðp e2Þ¼NðÀe2Þ: Since L has property B we must have s À rX2e þ 1: Now if dðÀe2Þþ½s=2X2e þ ½s=2 ’2 1; i.e. if dðÀe2Þþs=2X2e þ 1 then ð1 þ p ÞF DNðÀe2ÞDyðLÞ (1.1(iii)) so we are done. Hence we may suppose that dðÀe2Þþs=2o2e þ 1; i.e. so4e À r 2dðÀe2Þþ2orsp4e À 2dðÀe2Þþ1: In particular since rX À dðÀe2Þðp e1AAÞ we have s À rp4e À 2dðÀe2Þþ1 ÀðÀdðÀe2ÞÞ ¼ 4e À dðÀe2Þþ1p4e: We have two cases: r rþdðÀe2Þ ’2 (a) s À r42e is even. Then for any ZAgðp e2Þ¼ð1 þ p ÞF -NðÀe2Þ we have sÀr sÀr sÀrþdðZÞÀ2e ’2 yðLÞ+Gðp e2e3ÞGðp e2e3ZÞ+ð1 þ p ÞF (3.15(i)). Now if dðÀe2Þ¼e À r=2; i.e. if r ¼ 2e À 2dðÀe2Þ then s ¼ r þðs À rÞX2e À 2dðÀe2Þþ2e þ 2 ¼ 4e À 2dðÀe2Þþ2 contrary to our assumption that sp4e À 2dðÀe2Þþ1: Hence we may suppose that dðÀe2Þoe À r=2; i.e. ro2e À 2dðÀe2Þ: In particular this rules out the case dðÀe2Þ¼2e since we cannot have ro À 2e: We claim that there is ZANðÀe2Þ Â s.t. dðZÞ¼r þ dðÀe2Þ: Indeed if we take an ZAO arbitrary with dðZÞ¼r þ dðÀe2Þ if Â ZANðÀe2Þ we are done. If ZeNðÀe2Þ we take aAO with dðaÞ¼2e À dðÀe2Þ s.t. 2eÀdðÀe Þ ’2 aeNðÀe2Þ: (It is possible since ð1 þ p 2 ÞF D/NðÀe2Þ:) Hence aZANðÀe2Þ: But dðZÞ¼r þ dðÀe2Þo2e À 2dðÀe2ÞþdðÀe2Þ¼2e À dðÀe2Þ¼dðaÞ: Hence dðaZÞ¼ r þ dðÀe2Þ so aZ works. Now if dðZÞ¼r þ dðÀe2Þ and ZANðÀe2Þ we have r sÀrþdðZÞÀ2e ’2 ZAgðp e2Þ so ð1 þ p ÞF DyðLÞ: But s À r þ dðZÞÀ2e ¼ s À r þ r þ sþdðÀe ÞÀ2e ’2 dðÀe2ÞÀ2e ¼ s þ dðÀe2ÞÀ2e so ð1 þ p 2 ÞF DyðLÞ: Now dðÀe2Þþðs þ dðÀe2ÞÀ2eÞ¼s þ 2dðÀe2ÞÀ2ep4e À 2dðÀe2Þþ2dðÀe2ÞÀ2e ¼ 2eo2e þ 1: (We have sp4e À 2dðÀe2Þþ1 and s is even so sp4e À 2dðÀe2Þ:) Hence ð1 þ sþdðÀe ÞÀ2e ’2 sþdðÀe ÞÀ2e ’2 p 2 ÞF D/NðÀe2Þ (1.1(iii)). Since both ð1 þ p 2 ÞF andN ðÀe2Þ are DyðLÞ we must have yðLÞ¼F’*ð1 þ ps=2ÞF’2 so we are done. sÀr (b) s À r42e is odd. We have yðLÞ+Gðp e2e3Þ¼/pe2e3Sð1 þ sÀrÀ2e ’2 sÀrÀ2e ’2 p ÞF -NðÀpe2e3Þ: We also have DANðÀe2ÞDyðLÞ and DAð1 þ p ÞF but ARTICLE IN PRESS

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sÀrÀ2e ’2 sÀrÀ2e ’2 DeNðÀpe2e3Þ: Hence yðLÞ+/DSðð1 þ p ÞF -NðÀpe2e3ÞÞ ¼ ð1 þ p ÞF : sÀrÀ2e ’2 ’ We also have yðLÞ+NðÀe2Þ so if ð1 þ p ÞF D/NðÀe2Þ then yðLÞ¼F andwe are sÀrÀ2e ’2 done. Hence we must have ð1 þ p ÞF DNðÀe2Þ which implies by 1.2(iii) that 2e þ 1pdðÀe2Þþðs À r À 2eÞ p dðÀe2Þþ4e À 2dðÀe2Þþ1 À r À 2e ¼ 2e þ 1 À rÀ dðÀe2Þ: (We have s À rp4e À 2dðÀe2Þþ1:) This implies rp À dðÀe2Þ: But we have X r A D Â2 r A D Â2 r À dðÀe2Þ with equality only when p e2 À 4 O : It follows that p e2 À 4 O r Â ’2 sÀr which implies yðLÞ+Gðp e2Þ¼O F : But we also have pe2e3AGðp e2e3ÞDyðLÞ so yðLÞ¼F’ andwe are done. r (ii) ro2e is odd. In this case we have yðLÞ+Gðp e2Þ¼NðÀpe2Þ: If sX4e þ 2 then ½s=2X2e þ 1soð1 þ p½s=2ÞF’2 ¼ F’2DyðLÞ andwe are done. Hence we may assume that sp4e þ 1: sÀr If s À rp4e is even then DAGðp e2e3ÞDyðLÞ: Since DeNðÀpe2ÞDyðLÞ we have yðLÞ¼F’ so we are done. If s À rX4e þ 2 is even then s ¼ r þ s À rX1 þ 4e þ 2 ¼ 4e þ 3 which contradicts our assumption that sp4e þ 1: If s À r is odd then we have s À r42e since otherwise L does not have property B. r r ’2 We also have s À rp4e þ 1 À rp4e so for any ZAgðp e2Þ¼ð1 þ p ÞF -NðÀpe2Þ we sÀr sÀr sÀrþdðZÞÀ2e ’2 have yðLÞ+Gðp e2e3ÞGðp e2e3ZÞ+ð1 þ p ÞF (3.15(i)). We claim that r Â gðp e2Þ contains a unit Z with dðZÞ¼r: Indeed we can take ZAO arbitrary with dðZÞ¼r andif ZANðÀpe2Þ then we are done. If ZeNðÀpe2Þ then we also have DeNðÀpe2Þ so ZDANðÀpe2Þ: But dðDÞ¼2e4r ¼ dðZÞ so dðZDÞ¼r: Hence ZD r sÀrþdðZÞÀ2e ’2 works. If ZAgðp e2Þ with dðZÞ¼r we have yðLÞ+ð1 þ p ÞF ¼ð1 þ s 2e ’2 p À ÞF : But s À 2ep4e À 2e ¼ 2e (because s is even and p4e þ 1) so DAð1 þ sÀ2e ’2 p ÞF DyðLÞ: But we also have yðLÞ+NðÀpe2Þ and DeNðÀpe2Þ: Hence yðLÞ¼ F’*ð1 þ ps=2ÞF’2 andwe are done. (iii) rp2e is even and dðÀe2Þ4e À r=2: We have several cases: (a) If s À r42e then sXr þ 2e þ 1so½s=2X½r=2 þ e þ 1=2¼r=2 þ e which ½s=2 ’2 r=2þe ’2 r r implies ð1 þ p ÞF Dð1 þ p ÞF : But we have yðLÞ+Gðp e2Þ+gðp e2Þ¼ð1 þ 2 pr=2þeÞOÂ which implies ð1 þ pr=2þeÞF’2DyðLÞ: Hence ð1 þ p½s=2ÞF’2DyðLÞ: (b) If s À ro2e is odd or s À r is even and dðÀe2e3Þpe Àðs À rÞ=2 then L does not satisfy condition B ðro2e þ 1Þ so this case can be ruledout. (c) If s À rp2e is even and dðÀe2e3Þ4e Àðs À rÞ=2 then WLOG we may assume that either r=2 þ e is odd or ðs À rÞ=2 þ e is even since otherwise we can replace r s sÀr s r r=2þe Â2 LD!1; p e2; p e3g by !1; p e2e3; p e3g: Now gðp e2Þ¼ð1 þ p ÞO so r ZAgðp e2Þ iff dðZÞXr=2 þ e: But for such Z we have dðZÞXr=2 þ e4r=2 þ e À s=2 ¼ e Àðs À rÞ=2: Since also dðÀe2e3Þ4e Àðs À rÞ=2 we have dðÀe2e3ZÞ4e Àðs À rÞ=2: ðsÀrÞ=2þdðZÞÀe ’2 sÀr This implies by Lemma 3.15(ii) that ð1 þ p ÞF DGðp e2e3Þ sÀr Â Â Gðp e2e3ZÞDyðLÞ: If r=2 þ eAdðO Þ then we take ZAO with dðZÞ¼r=2 þ e and we get ðs À rÞ=2 þ dðZÞÀe ¼ s=2soð1 þ ps=2ÞF’2DyðLÞ andwe are done.Otherwise, i.e. if r=2 þ ep2e À 2 is even then r=2 þ e þ 1AdðOÂÞ: We choose in this case ZAOÂ with dðZÞ¼r=2 þ e þ 1 andwe have ðs À rÞ=2 þ dðZÞÀe ¼ s=2 þ 1soð1 þ ps=2þ1ÞF’2DyðLÞ: But r=2 þ e is even so ðs À rÞ=2 þ e must also be even. Hence s=2 ¼ ARTICLE IN PRESS

C.N. Beli / Journal of Number Theory 102 (2003) 125–182 161 r=2 þ e þðs À rÞ=2 þ e À 2e is also even. We also have r=2 þ ep2e À 2sor=2pe À 2 and ðs À rÞ=2p2e=2 ¼ e which implies that s=2pe þ e À 2 ¼ 2e À 2: It follows that ð1 þ ps=2ÞF’2 ¼ð1 þ ps=2þ1ÞF’2DyðLÞ andwe are done. (iv) rX2e þ 1: We have two cases: (a) If s À rX2e þ 1 then s ¼ s À r þ rX2e þ 1 þ 2e þ 1 ¼ 4e þ 2so½s=2X2e þ 1 which implies that ð1 þ p½s=2ÞF’2 ¼ F’2DyðLÞ andwe are done. sÀr s (b) If s À rp2e then the lattice !1; p e2e3; p e3g is in one of cases (i)–(iii) already studied. Again we are done. &

Remark. If we look at the proof of ð1 þ p½s=2ÞF’2DyðLÞ in the case when s is odd (i.e. cases (i)(b), (ii) with s À r even, (iii)(b) and, in part (a)) we notice that we actually ½s=2 ’2 r sÀr provedthat ð1 þ p ÞF DGðp e2ÞGðp e2e3Þ: This happens also in the case when L does not have property B. Indeed in this case we have that either rp2e þ 1is odd and s À rp2e is even or rp2e is even and s À rp2e þ 1 is odd. In the ﬁrst r sÀr case we have Gðp e2Þ¼NðÀpe2Þ and Gðp e2e3Þ{D which implies that r sÀr ’ ½s=2 ’2 Gðp e2ÞGðp e2e3Þ¼F+ð1 þ p ÞF : The secondcase is similar. ½ðord aiþ2Àord aiÞ=2 ’2 In general we have ð1 þ p ÞF DGðaiþ1=aiÞGðaiþ2=aiþ1Þ when- a ’2 ever ord aiþ2 À ord ai is odd. Now ð1 þ p ÞF is the product of the groups ½ðord ai 2Àord aiÞ=2 ’2 ð1 þ p þ ÞF andthese groups are alreadycontainedin Gða2=a1Þy Gðan=anÀ1Þ whenever ord aiþ2 À ord ai is odd. So we actually have

a ’2 a1 ’2 Gða2=a1ÞyGðan=anÀ1Þð1 þ p ÞF ¼ Gða2=a1ÞyGðan=anÀ1Þð1 þ p ÞF ; where a1 ¼ minfðord aiþ2 À ord aiÞ=2j1pipn À 2; ord aiþ2 À ord ai is eveng: a1 ’2 (If ord aiþ2 À ord ai is odd for all 1pipn À 2 we just ignore the factor ð1 þ p ÞF or, equivalently, we put a1 ¼ N:)

6. The reverse inclusion

The proof of the inclusion D is more complicatedandconsumes all of this section.

R1 Rn Lemma 6.1. Let LD!p e1; y; p eng relative to a good BONG x1; y; xn:

(i) If !px1; x2g exists and R1 þ 2pR3 then fvALjv not a norm generatorg is a 0 0 lattice L with L ¼ !px1; x2; y; xng: Moreover px1; x2; y; xn is a good BONG. k (ii) If kX0 is an integer s.t. !p x1; x2g exists and R1 þ 2kpR3 then k k fvALjord QðvÞXR1 þ 2k À 1g¼!p x1; x2; y; xng and p x1; x2; y; xn is a good BONG. (If no3 we just ignore the conditions R1 þ 2pR3; resp. R1 þ 2kpR3; from (i) and (ii).) (iii) If either R2 À R1X2e þ 1 or R2 À R14 À 2e is even and dðÀe1e2Þ4e ÀðR2 À R1Þ=2 then !px1; x2g exists. If moreover L has property B then L satisﬁes the conditions of (i). ARTICLE IN PRESS

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k Proof. We note that the condition that !p x1; x2g exists is equivalent to k R ÀR À2k Qðx2Þ=Qðp x1ÞAA; i.e. to p 2 1 e1e2AA: Also note that (i) is the case k ¼ 1of (ii). (i) Note first that if L ¼ J>K with nL ¼ nJ*nK then L0 ¼ J0>K where J0 ¼ fvAJjv not a norm generatorg: There are two cases: R1oR2: We have L ¼ J>K with J ¼ Ox1 and K ¼ !x2; y; xng: Since nJ ¼ R1 R2 0 0 p *p ¼ nK we have L ¼ J >K ¼ px1>K: Now !px1; x2g exists so R2ÀR1À2 À1 p e1e2AA: Since p e1e2eA we cannot have R2 ¼ R1 þ 1: Hence R2XR1 þ R1þ2 R2 0 2: It follows that npx1 ¼ p +p ¼ nK so px1 is a norm generator for L ¼ 0 0 px>K: Since pr > L ¼ K ¼ !x2; y; xng we have L ¼ !px1; x2; y; xng: px1 R1XR2: We consider first case n ¼ 2: By scaling we may assume that Qðx1Þ¼1 R R RÀ2 and Qðx2Þ¼p e: We have p eAA and Rp0soR is even. The condition p eAA implies that R À 2X À 2e and R À 2X À d whered ¼ dðÀeÞ: In particular R=2 þ D 1 pR=2 e40andR þ d40: Since Qðx1Þ¼1 we have L pR=2 ÀdpR with the usual notations relative to a basis x1; y: Now if v ¼ ax1 þ byAL the condition ord QðvÞ4ord nL ¼ 0 is equivalent to ord ða2 þ 2pR=2ab À dpRb2Þ40: But ord2 pR=2abXR=2 þ e40 and ord dpRb2XR þ d40 so ordða2 þ 2pR=2ab À dpRb2Þ40 is equivalent to ord a240; i.e. aAp: Hence fvALjord QðvÞ4ord nLg¼px þ Oy: Now the lattice L0 ¼ px þ 1 1 D p2 pR=2þ1 R X Oy pR=2þ1 ÀdpR relative to the basis px1; y: But since ord dp ¼ R þ d 2 ðR À 2X À dÞ andord2 pR=2þ1 ¼ R=2 þ e þ 1X2 ðR À 2X À 2eÞ: we have that ord nL0 ¼ 0 0 2sopx1 is a norm generator of L : Now since pr > L ¼ pr > ðpx1 þ OyÞ¼ px1 x1 0 pr > ðOx1 þ OyÞ¼pr > L ¼ Ox2 we have that px1; x2 is a BONG for L as claimed. x1 x1 For the general case we write L ¼ J>K with J ¼ !x1; x2g and K ¼ R1 R3 0 0 !x3; y; xng: Since nJ ¼ p *p ¼ nK we have L ¼ J >K: But 0 R R þ2 0 J ¼ !px1; x2g by the case n ¼ 2: Now R1 þ 2pR3 so nK ¼ p 3 Dp 1 ¼ nJ ¼ 0 0> 0 > OQðpx1Þ so px1 is a norm generator for L ¼ J K: We have prpx1 L ¼ Ox2 K: But 0 R2pR1oR3 so Ox2>K ¼ Ox2>!x3; y; xng ¼ !x2; y; xng so L ¼ !px1; x2; y; xng: (ii) The case k ¼ 1 is just (i). For k41 we use induction over k: Since R2ÀR1À2k R2ÀR1À2kþ2 kÀ1 p eAA we also have p AA so the lattice !p x1; x2g also exists. Also the inequality R1 þ 2kpR3 implies that R1 þ 2ðk À 1ÞpR3 so L satisfies the conditions of (ii) for k À 1: Let vAL s.t. ord QðvÞXR1 þ 2k À 1: Then also 0 ord QðvÞXR1 þ 2ðk À 1ÞÀ1 so the induction step k À 1 implies vAL ¼ kÀ1 kÀ1 k !p x1; x2; y; xngDL: But !pp x1; x2g ¼ !p x1; x2g exists and kÀ1 0 ord Qðp x1Þþ2 ¼ R1 þ 2kpR3 so L satisfies the conditions of (i). Now kÀ1 ord n!p x1; x2; y; xng ¼ R1 þ 2k À 2 so the condition ord QðvÞXR1 þ 2k À 1 means that vAL0 is not a norm generator which by (i) is equivalent to kÀ1 k vA!pp x1; x2; y; xng ¼ !p x1; x2; y; xng as claimed.

R2ÀR1À2 (iii) If R2 À R1X2e þ 1 then R2 À R1 À 2X2e À 140sop e1e2AA: If R2 À R14 À 2e is even and dðÀe1e2Þ4e ÀðR2 À R1Þ=2 then R2 À R142e À 2dðÀe1e2Þ: If ARTICLE IN PRESS

C.N. Beli / Journal of Number Theory 102 (2003) 125–182 163

R2ÀR2À2 dðÀe1e2Þp2e we get R2 À R1 À 2X2e À 2dðÀe1e2ÞX À dðÀe1e2Þ so p e1e2AA: If dðÀe1e2Þ¼N we have R2 À R14 À 2e so R2 À R1 À 2X À 2e so again R2ÀR1À2 p e1e2AA: But this implies that !px1; x2g exists. If L has property B the condition R1 þ 2pR3 follows from 4.12. &

Lemma 6.2.

À1 (i) If !x1; y; xng exists then !p x1; x2; y; xng also exists. 0 À1 (ii) If L ¼ !x1; y; xng is a lattice s.t. either L or L ¼ !p x1; x2; y; xng has property B then: R1 2 R2 (a) If R2XR1 then QðLÞDp e1O þ p : If R2 À R1p2e is even then: R1 2 R2þdðÀe1e2Þ (b) If dðÀe1e2Þpe ÀðR2 À R1Þ=2 then QðLÞDp e1O þ p : R1 2 ðR1þR2Þ=2þe (c) If dðÀe1e2ÞXe ÀðR2 À R1Þ=2 then QðLÞDp e1O þ p :

0 À1 0 À1 Proof. (i) Let L ¼ !x1; y; xng and L ¼ Op x1 þ L: We have L Dp L so 0 À2 À1 À1 0 nL Dp nL ¼ Qðp x1ÞO: Hence p x1AL must be a norm generator. But 0 À1 0 À1 pr À1 > L ¼pr > ðp x1 þ LÞ¼ pr > L ¼ !x2; y; xng so L ¼ !p x1; x2; y; xng: p x1 x1 x1 0 À1 (ii) Note that if L ¼ !p x1; x2; y; xng has property B then by 4.12 we have À1 R1 ¼ ord Qðp x1Þþ2pord Qðx3Þ¼R3: Since also RioRiþ2 for 2pipn À 2 the BONG x1; y; xn is good. In case (a) we have R1pR2 so L ¼ !x1g>!x2; y; xng ¼ Ox1>L with L ¼ R2 R1 2 !x2; y; xng: We have QðL ÞDnL ¼ p so QðLÞ¼QðOx1ÞþQðL ÞDp e1O þ pR2 so we are done. For (b) and(c) we use inductionover n: For the binary case we may assume by scaling that LD!1; pReg andour (b) and(c) become the (b) and(c) of 3.10. For the induction step note that in the case when dðÀe1e2Þ¼e ÀðR2 À R1Þ=2we have R2 À R1 þ d ¼ðR2 À R1Þ=2 þ e so the statements of (b) and(c) are equivalent. For this reason we will treat the case dðÀe1e2Þ¼e ÀðR2 À R1Þ=2 only for the statement (c). For the induction step for (b) let us note that the inequality dðÀe1e2Þoe ÀðR2 À R1Þ=2 implies dðÀe1e2Þpe ÀðR2 À R1Þ=2 À 1 ¼ e ÀðR2 ÀðR1 À 2ÞÞ=2: In both 0 cases when L or L has property B this implies that R3 À R2X2e þ 1: Since R2oR3 we have L ¼ !x1; x2g>!x3; y; xng so QðLÞ¼Qð!x1; x2gÞþ R1 2 R2þdðÀe1e2Þ Qð!x3; y; xngÞ: By the binary case Qð!x1; x2gÞDp e1O þ p and R3 R1 2 we also have Qð!x3; y; xngÞDn!x3; y; xng ¼ p : Hence QðLÞDp e1O þ R2þdðÀe1e2Þ R3 R1 2 p þ p : Since R3XR2 þ 2e þ 14R2 þ dðÀe1e2Þ we get QðLÞDp e1O þ pR2þdðÀe1e2Þ: For the induction step for (c) note that if R2 À R1 ¼ 2e then R2 ¼ðR1 þ R2Þ=2 þ R1 2 R2 R1 2 ðR1þR2Þ=2þe e: From (a) we get QðLÞDp e1O þ p ¼ p e1O þ p : Hence we may suppose that R2 À R1p2e À 2o2e þ 1soR2 ÀðR1 À 2Þp2eo2e þ 1: In both cases 0 when L or L has property B this implies that either R3 À R242e þ 1orR3 À R2p2e R2 is even and dðÀe2e3Þ4e ÀðR3 À R2Þ=2: We claim that Qð!x2; y; xngÞDp e2 þ ARTICLE IN PRESS

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ðR þR Þ=2þe p 1 2 : Indeed if R3 À R242e þ 1 we use (a) andwe have R R Qðx1; y; xngÞDp 2 e2 þ p 3 : But R3XR1 and R34R2 þ 2e so R34ðR1 þ R2 þ R ðR þR Þ=2þe 2eÞ=2 ¼ðR1 þ R2Þ=2 þ e so p 3 Cp 1 2 : If R3 À R2p2e is even with R 2 ðR þR Þ=2þe dðÀe2e3Þ4e ÀðR3 À R2Þ=2 then QðLÞDp 1 e1O þ p 2 3 by the induction step. ðR2þR3Þ=2þe ðR1þR2Þ=2þe But R1pR3 so p Dp : R1 2 R1 2 ðR1þR2Þ=2þe Let now vAL: If vAOx1 then QðvÞAp e1O Cp e1O þ p so we are done. Otherwise v belongs to J :¼ Ox þ Ov so it is enough to prove that R1 2 ðR1þR2Þ=2þe QðJÞDp e1O þ p : Now x1AJDL so x1AJ is a norm generator which implies J ¼ !x1; ug with uApr > JDpr > L ¼ !x2; y; xng: Since x1 x1 R2 ðR1þR2Þ=2þe R2þ2k 2 Qð!x2; y; xngÞDp e2 þ p we have QðuÞ¼p e2x þ a with kX0; 0 Â ðR1þR2Þ=2þe R2 0 xAO and aAp : Let QðuÞ¼p e2 : There are two cases: ðR1þR2Þ=2þe 0 If R1 þ 2kXðR1 þ R2Þ=2 þ e then QðuÞAp so R2 XðR1 þ R2Þ=2 þ 0 R1 2 R2 ðR1þR2Þ=2þe eXR1: We use (a) andwe get QðJÞDp e1O þ p Dp : 0 0 2 ÀðR þ2kÞ If R1 þ 2koðR1 þ R2Þ=2 þ e then R2 ¼ R2 þ 2k and e2 ¼ e2x þ p 2 a ¼ e2Z 2 ÀðR2þ2kÞ À1 ÀðR2þ2kÞ À1 where Z ¼ x þ p e2 a: But ord p e2 aX ÀðR2 þ 2kÞþðR1 þ R2Þ=2 þ e ¼ e ÀðR2 À R1Þ=2 À 2k so dðZÞXðR2 À R1Þ=2 þ e À 2k: Since also dðÀe1e2ÞXe À 0 ðR2 À R1Þ=2Xe ÀðR2 À R1Þ=2 À 2k we have dðÀe1e2 Þ¼dðÀe1e2ZÞXe ÀðR2 À 0 0 R1Þ=2 À 2k: If dðÀe1e2 ÞXe ÀðR2 À R1Þ=2 then we use the binary case of (c) and R 2 ðR þR 0Þ=2þe R 2 ðR þR Þ=2þe 0 we get QðJÞDp 1 e1O þ p 1 2 Dp 1 e1O þ p 1 2 : (We have R2 XR2:)If 0 0 R R 0þdðÀe e 0Þ dðÀe1e2 Þoe ÀðR2 À R1Þ=2 we use (c) andwe have QðJÞDp 1 e1 þ p 2 1 2 : But 0 0 R 0þdðÀe e 0Þ R2 þ dðÀe1e2 ÞXR2 þ 2k þ e ÀðR2 À R1Þ=2 À 2k ¼ðR1 þ R2Þ=2 þ e so p 2 1 2 D pðR1þR2Þ=2þe: &

2 Definition 11. We define the function g0 : A-SgpðOÂ=OÂ Þ as follows: For a ¼ pReAA with dðÀaÞ¼d define:

2 (I) g0ðaÞ¼OÂ when R42e (i) (II) When Rp2e: ( 2 1 Rþd Â if d e R=2 ii 0 ð þ p ÞO p À ð Þ g ðaÞ¼ 2 ð1 þ pR=2þeÞOÂ if R is even and dXe À R=2 ðiiiÞ

2 Note that for Ro2e odd we have d ¼ 0 so (ii) gives g0ðpReÞ¼ð1 þ pRÞOÂ :

Ri Lemma 6.3. (i) If L ¼ !x1; y; xng with QðxiÞ¼p ei is a lattice s.t. either L or 0 À1 R2ÀR1 L ¼ !p x1; x2; y; xng has property B, then gðp e1e2ÞDfQðvÞ=Qðx1ÞjvALisa 0 R ÀR norm generatorgDg ðp 2 1 e1e2Þ: (ii) Moreover if L has property B then fQðvÞ=Qðx1ÞjvAL is a norm gener- R2ÀR1 atorg¼gðp e1e2Þ:

Â Proof. We have fQðvÞ=Qðx1ÞjvAL is a norm generatorg¼QðLÞ=Qðx1Þ-O : We Â R2ÀR1 have Qð!x1; x2gÞ=Qðx1Þ-O ¼gðað!x1; x2gÞÞ¼gðp e1e2Þ: But !x1; x2gDL ARTICLE IN PRESS

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Â Â so Qð!x1; x2gÞ=Qðx1Þ-O DQðLÞ=Qðx1Þ-O which gives the ﬁrst inclusion of R 2 a (i). For the other inclusion we note that QðLÞDp 1 e1O þ p with a given by 6.2. It

Â R1 2 a R1 Â 2 aÀR1 Â follows that QðLÞ=Qðx1Þ-O Dðp e1O þ p Þ=ðp e1Þ-O ¼ðO þ p Þ-O ¼ aÀR Â2 aÀR Â2 0 R ÀR ð1 þ p 1 ÞO : Hence we have to prove that ð1 þ p 1 ÞO ¼ g ðp 2 1 e1e2Þ: If R2 À R1X2e þ 1orR2 À R1o2e is odd we have R24R1 so we can take a ¼ R2: 2 2 aÀR1 Â R2ÀR1 Â It follows that ð1 þ p ÞO ¼ð1 þ p ÞO : If R2 À R1o2e is odd then 2 0 R2ÀR1 R2ÀR1 Â g ðp e1e2Þ¼ð1 þ p ÞO so we are done. If R2 À R1X2e þ 1 then ð1 þ 2 2 R2ÀR1 Â Â 0 R2ÀR1 p ÞO ¼ O ¼ g ðp e1e2Þ so again we are done. If R2 À R1p2e is even we use 6.2(b) or (c). If dðÀe1e2Þpe ÀðR2 À R1Þ=2 we have 2 2 aÀR1 Â R2ÀR1þdðÀe1e2Þ Â 0 R2ÀR1 a ¼ R2 þ dðÀe1e2Þ so ð1 þ p ÞO ¼ð1 þ p ÞO ¼ g ðp e1e2Þ: If 2 aÀR1 Â dðÀe1e2ÞXe ÀðR2 À R1Þ=2 then a ¼ðR1 þ R2Þ=2 þ e so ð1 þ p ÞO ¼ 2 ðR2ÀR1Þ=2þe Â 0 R2ÀR1 ð1 þ p ÞO ¼ g ðp e1e2Þ so we are done. R2ÀR1 Â 0 R2ÀR1 (ii) We have from (i) that gðp e1e2ÞDQðLÞ=Qðx1Þ-O Dg ðp e1e2Þ: If either R2 À R1X2e þ 1orR2 À R1p2e is even and dðÀe1e2Þ4e ÀðR2 À R1Þ=2 then R ÀR 0 R ÀR gðp 2 1 e1e2Þ¼g ðp 2 1 e1e2Þ so (i) implies (ii). If R2 À R1o2e is odd or R2 À R1p2e is even and dðÀe1e2Þpe ÀðR2 À R1Þ=2 R2ÀR1 0 R2ÀR1 R2ÀR1 we only have gðp e1e2Þ¼g ðp e1e2Þ-NðÀp e1e2Þ (see Deﬁnitions 6 and Â R2ÀR1 11) so we needto prove that QðLÞ=Qðx1Þ-O DNðÀp e1e2Þ: In both cases property B implies R3 À R2X2e þ 1: In particular L ¼ !x1; x2g>!x3; y; xng R3 so QðLÞ¼Qð!x1; x2gÞþQð!x3; y; xngÞDQð!x1; x2gÞþp : Hence QðLÞ= R R ÀR Qðx1Þ D Qð!x1; x2gÞ=Qðx1Þþp 3 =Qðx1Þ¼Qð!x1; x2gÞ=Qðx1Þþp 3 1 so Â Â R ÀR R ÀR R ÀR QðLÞ=Qðx1Þ-O D Qð!x1; x2gÞ=Qðx1Þ-O þ p 3 1 ¼ gðp 2 1 e1e2Þþp 3 1 : R2ÀR1 R3ÀR1 R2ÀR1 Hence we needto prove that gðp e1e2Þþp DNðÀp e1e2Þ: For this let R2ÀR1 R3ÀR1 R2ÀR1 ZAgðp e1e2Þ and aAp : We needto prove that Z þ aANðÀp e1e1Þ: Since R2ÀR1 R2ÀR1 ZAgðp e1e2ÞCNðÀp e1e2Þ this is equivalent to x :¼ 1 þ a=Z ¼ðZ þ R2ÀR1 aÞ=ZANðÀp e1e2Þ: Now ordða=ZÞ¼ord aXR3 À R1 so dðxÞXR3 À R1XR2 À R1 þ 2e þ 1 (because R3XR2 þ 2e þ 1). If R2 À R1 is odd then it is positive so 2 Â R2ÀR1 dðxÞXR2 À R1 þ 2e þ 142e þ 1: It follows that xAO CNðÀp e1e2Þ andwe are done. If R2 À R1o2e even and dðÀe1e2Þpe ÀðR2 À R1Þ=2 then dðxÞþ dðÀe1e2ÞXR2 À R1 þ 2e þ 1 þ dðÀe1e2ÞX2e þ 1 (because R2 À R1X À dðÀe1e2Þ). R2ÀR1 By 1.2(iii) we have xANðÀe1e2Þ¼NðÀp e1e2Þ so we are done. &

Lemma 6.4.

r r (i) If L ¼ Ox1>K and sKCp then p Að0; 0Þ L: (ii) If L ¼ J>K where J is binary pr-modular, sKCpr and nKCnJ then prAð0; 0Þ L iff JDprAð0; 0Þ: R1 Rn (iii) Let LD!p e1; y; p eng relative to a good BONG x1; y; xn and let r ¼ r r ðR1 þ R2Þ=2: If R1oR3 then p Að0; 0Þ L iff !x1; x2gDp Að0; 0Þ:

Proof. If prAð0; 0Þ L then let HDL where HDprAð0; 0Þ relative to the basis x; y: ARTICLE IN PRESS

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(i) Let x ¼ ax1 þ z with aAO and zAK: We have 0 ¼ QðxÞ¼Qðax1ÞþQðzÞ a so Qðax1Þ¼ÀQðzÞAnK: Hence ord aXa where a is minimal s.t. Qðp xÞAnK: 0 0 a 0 0 It follows that xAL where L ¼ p x1>K: Similarly yAL so HDL : But a a 0 r 0 sðp x1Þ¼Qðp x1ÞODnKDsK so sL ¼ sKCp which makes the inclusion HDL impossible. (ii) We have that sL ¼ pr and J is a ﬁrst Jordan component of L: Since H is also binary and pr-modular we have that L ¼ H>K0 where sK0 ¼ sKCpr: We have r 1 K0CLp þ ¼ pJ>K: But nðpJÞCnJ and nKCnJ by hypothesis so we have nK0DnðpJ>KÞCnJ: On the other hand, nH þ nK0 ¼ nL ¼ nJ þ nK ¼ nJ so we must have nH ¼ nJ: Since HDprAð0; 0Þ and J have the same norm andscale we must have JDprAð0; 0Þ or prAð2; 2rÞ: Suppose that JDprAð2; 2rÞ: We have x ¼ z þ t with zAJ and tAK: But Bðx; LÞ¼pr ¼ sL and J is the ﬁrst Jordan component of L so zAJ must be primitive. Since JDprAð2; 2rÞ this implies that QðzÞO ¼ nJ: But QðzÞþQðtÞ¼QðxÞ¼0soQðtÞO ¼ QðzÞO ¼ nJ which is impossible since tAK and nKCnJ: Hence JDprAð0; 0Þ: (iii) ‘‘Only if’’ is trivial since !x1; x2gDL: For ‘‘if’’ we consider two cases: If R1oR2 then L ¼ Ox1>K where K ¼ !x2; y; xng: Now R1oR2 implies R24ðR1 þ R2Þ=2 ¼ r and R1oR3 implies ðR2 þ R3Þ=24ðR1 þ R2Þ=2 ¼ r hence r r ord sK ¼ minfR2; ðR2 þ R3Þ=2g4r so sKCp : By (i) p Að0; 0Þ L: r If R1XR2 then L ¼ J>K where J ¼ !x1; x2g is p -modular (r ¼ðR1 þ R2Þ=2) and K ¼ !x3; y; xng: We also have r ¼ðR1 þ R2Þ=2pR1oR3 and r ¼ðR1 þ r R2Þ=2oðR3 þ R4Þ=2(R1oR3 and R2pR4)soroord sK: Hence sKCp andwe also R3 R1 r have nK ¼ p Cp ¼ nJ: By (ii) we have that p Að0; 0Þ L iff !x1; x2g ¼ JDprAð0; 0Þ as claimed. &

Ri Lemma 6.5. Let L ¼ !x1; y; xng with QðxiÞ¼p ei be a lattice having property B such that !x1; x2g is not a hyperbolic plane. Let k be the smallest non-negative k integer with the property that !x1; p x2g can be spanned by two norm generators of the same length. Let xALwithQðxÞ¼Qðx1Þ: We have:

k (i) pr > xA!p x2; x3; y; xng with the possible exception of the case when x1 R2 À R1 ¼ 2e À 2; dðÀe1e2Þ41 and O=p has only two elements. In this case we have that k ¼ 2 and pr > xA!px2; x3; y; xng is not a norm generator. x1 k If R2 þ 2k À R1p2e and pr > x is a norm generator for !p x2; x3; y; xng x1 then: A (ii) x1 À x is not isotropic and tx1Àx OðLÞ: 0 0 0 k (iii) If x AL s.t. Qðx1Þ¼Qðx Þ and pr > x is not a norm generator for !p x2; x3; x1 0 k y; xng then pr > tx Àxx is a norm generator for !p x2; x3; y; xng: x1 1 If R2 þ 2k À R1X2e þ 1 then: A (iv) If ord Qðx1 À xÞ¼R1 þ 2e then tx1Àx OðLÞ:

Proof. By scaling we may assume that Qðx1Þ¼1: We denote d ¼ dðÀe2Þ: ARTICLE IN PRESS

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(i) If !x1; x2g is spannedby two norm generators then k ¼ 0 so (i) is trivial k since pr > xApr > L ¼ !x2; x3; y; xng: Let us note that að!x1; p x2gÞ¼ x1 x1 R2þ2k R2þ2k R2þ2k að!1; p e2gÞ¼p e2 so kX0 is minimal with the property that p e2 is in one of cases (i)–(iv) of 3.17(3). Let y ¼ pr > x: We have prx ðOx1 þ OxÞ¼ x1 1 ! g ! g prx1 Ox ¼ Oy so Ox1 þ Ox ¼ x1; y : Now x1; y is spannedby two norm generators of the same length (x1 and x) and að!x1; ygÞ¼að!1; QðyÞgÞ¼QðyÞ so QðyÞ must in one of cases (i)–(iv) of 3.17(3). We consider several cases: R2 (1) R2X2e þ 1: Then p e2 satisfies conditions of 3.17 so k ¼ 0; a case already discussed. (2) R2o2e is odd. Then the minimal k is the one for which R2 þ 2k ¼ 2e þ 1: Property B implies R3 À R2X2e þ 1: Since R3XR2 þ 2e þ 142e þ 1 ¼ R2 þ 2k we k k have ord Qðp x2Þ¼R2 þ 2koR3 ¼ ord n!x3; y; xng so p x2>!x3; y; xng ¼ k !p x2; x3; y; xng: We have y ¼ bx2 þ z with bAO and zA!x3; y; xng: If ord b ¼ lok then ord Qðbx2Þ¼R2 þ 2loR2 þ 2k ¼ 2e þ 1: We also have ord QðzÞXord n!x3; y; xng ¼ R342e þ 1: Since QðyÞ¼Qðbx2ÞþQðzÞ this implies that ord QðyÞ¼R2 þ 2lo2e þ 1: Since R2 þ 2l is also odd QðyÞ does not satisfy the conditions of 3.17. Contradiction. Hence ord bXk which implies y ¼ bx2 þ k k zAp x2>!x3; y; xng ¼ !p x2; x3; y; xng: (3) R2 is even and dpe À R2=2: We have that R2p2e À 2d is even so our k is given by R2 þ 2k ¼ 2e À 2d: Property B implies R3 À R2X2e þ 1 andwe also have R2X À R d ðp 2 e2AAÞ: Hence R3XR2 þ 2e þ 1X2e þ 1 À d42e À 2d ¼ R2 þ 2k: It follows k k that ord Qðp x1Þ¼R2 þ 2koR3 ¼ ord n!x3; y; xng so p x2>!x3; y; xng ¼ k !p x2; x3; y; xng: We have y ¼ bx2 þ z with bAO and zA!x3; y; xng: Suppose l Â that ord b ¼ lok: We have R2 þ 2loR2 þ 2k ¼ 2e À 2d: Let b ¼ p Z with ZAO : l R2þ2l 2 R2þ2l 2 We have QðyÞ¼Qðp Zx2ÞþQðzÞ¼p Z e2 þ QðzÞ¼p x where x ¼ e2ðZ þ R2þ2l 2 R2þ2l QðzÞ=ðp e2ÞÞ: But dðÀe2Þ¼d and dðZ þ QðzÞ=ðp e2ÞÞXord QðzÞ= R2þ2l ðp e2Þ¼ord QðzÞÀðR2 þ 2lÞXR3 ÀðR2 þ 2lÞ42e þ 1 À d Àð2e À 2dÞ¼d þ 1 ðR3X2e þ 1 À d and R2 þ 2lo2e À 2dÞ so dðÀxÞ¼d: Since R2 þ 2lo2e À 2d ¼ 2e À 2dðÀxÞ QðyÞ¼pR2þ2lx does not satisfy the conditions of 3.17. Contradiction. k k Hence ord bXk so y ¼ bx2 þ zAp x2>!x3; y; xng ¼ !p x2; x3; y; xng: (4) R2 even and2 e À 2doR2p2e if O=p has more than two elements, resp. 2e À R2 2doR2o2e À 2; if O=p has two elements. If R2=2 þ e is even then p e2 satisfies the conditions of Lemma 3.17 so k ¼ 0: If R2=2 þ e is odd then ðR2 þ 2Þ=2 þ e is even so R2 R2þ2 p e2 does not satisfy the conditions of Lemma 3.17 but p e2 does so k ¼ 1: If k ¼ 0 then we are done. If k ¼ 1 let us suppose that y is a norm generator in R2 Â !x2; y; xng: Then QðyÞ¼p x for some xAO : From 6.3(ii) we have that x=e2 ¼ R3ÀR2 QðyÞ=Qðx2ÞAgðp e2Þ: Since R2 À R1 ¼ R2p2e we have that either R3 À R2X2e þ 1orR3 À R2p2e is even and dðÀe2e3Þ4e ÀðR3 À R2Þ=2: In the first case 2 2 R3ÀR2 Â R3ÀR2 ðR3ÀR2Þ=2þe Â we have gðp e2Þ¼O andin the second gðp e2Þ¼ð1 þ p ÞO : In both cases we have dðx=e2ÞXðR3 À R2Þ=2 þ e ¼ R3=2 þ e À R2=24e À R2=2 (because R34R1 ¼ 0). We also have dðÀe2Þ¼d4e À R2=2 (because R242e À 2d). Hence dðÀxÞ4e À R2=2: It follows that R242e À 2dðÀxÞ: Since also R2p2e (resp. ARTICLE IN PRESS

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R R2o2e if O=p has only two elements) and R2=2 þ e is odd we have that QðyÞ¼p2 x does not satisfy the conditions of Lemma 3.17 which gives a contradiction. Hence ord QðyÞ4ord n!x2; y; xng: We have that either R3 À R242e þ 1orR3 À R2p2e is even and dðÀe2e3Þ4e ÀðR3 À R2Þ=2: Since R2p2e and R34R1 ¼ 0we also have R3 À R24 À 2e so by 6.1(iii) !x2; y; xng satisfies the conditions of 6.1(i). Since yA!x2; y; xng is not a norm generator this implies that yA!px2; y; xng as claimed. (5) O=p has two elements and R2 ¼ 2e or 2e À 2doR2 ¼ 2e À 2: 2e 2eþ2 If R2 ¼ 2e then p e2 does not satisfy the conditions of 3.17 but p e2 doessowe have k ¼ 1: Now y cannot be a norm generator of !x2; y; xng since otherwise ord QðyÞ¼2e andthis implies that QðyÞ does not satisfy the conditions of 3.17. By the same reasoning from case (4). !x2; y; xng satisfies the conditions of 6.1(i) and yA!x2; y; xng is not a norm generator so we have yA!px2; y; xng as claimed. 2eÀ2 If 2e À 2doR2 ¼ 2e À 2; i.e. if R2 ¼ 2e À 2 and d41 then p e2 does not satisfy 2e the conditions of 3.17 because ð2e À 2Þ=2 þ e ¼ 2e À 1 is odd and neither does p e2 2eþ2 but p e2 does so. Hence k ¼ 2: By the reasoning from case (4) we have that yA!x2; y; xng is not a norm generator and !x2; y; xng satisfies the conditions of 6.1(i) so yA!px2; x3; y; xng: Now yA!px2; x3; y; xng cannot be a norm generator since otherwise ord QðyÞ¼ord Qðpx2Þ¼2e so QðyÞ does not satisfy the conditions of 3.17.

We write x ¼ ax1 þ y with aAF: Note that Bðx1; xÞ¼Bðx1; ax1 þ yÞ¼a so Qðx17xÞ¼Qðx1ÞþQðxÞ72Bðx1; xÞ¼1 þ 172a ¼ 2ð17aÞ: 2 Let us suppose now that R2 þ 2kp2e: We have 1 ¼ QðxÞ¼Qðax1 þ yÞ¼a þ 2 k k QðyÞ so 1 À a ¼ QðyÞ: Now ord n!p x2; y; xng ¼ ord Qðp x2Þ¼R2 þ 2k so k ord QðyÞXR2 þ 2k with equality when yA!p x2; x3; y; xng is a norm generator. 2 We have 1 À a ¼ QðyÞ so ordð1 À aÞþordð1 þ aÞ¼ord QðyÞXR2 þ 2k: We claim that ordð1 À aÞ; ordð1 þ aÞXR2=2 þ k andord ð1 À aÞ¼ordð1 þ aÞ¼R2=2 þ k k iff yA!p x2; x3; y; xng is a norm generator. Indeed if we assume that e.g. ordð1 þ 1 aÞoR2=2 þ k then ordð1 þ aÞo2ðR2 þ 2kÞpð2eÞ=2 ¼ e: Since ordð1 þ aÞoord2 we 1 have ordð1 À aÞ¼ordð2 Àð1 þ aÞÞ ¼ ordð1 þ aÞo2ðR2 þ 2kÞ which contradicts ordð1 À aÞþordð1 þ aÞXR2 þ 2k: Similarly, we get a contradiction if we assume 1 X that ordð1 À aÞo2ðR2 þ 2kÞ: Now ordð1 À aÞ; ordð1 þ aÞ R2=2 þ k so ordð1 À aÞ¼ ordð1 þ aÞ¼R2=2 þ k is equivalent to ord QðyÞ¼ordð1 À aÞþordð1 þ aÞ¼ k 2ðR2=2 þ kÞ¼R2 þ 2k; i.e. to yA!p x2; x3; y; xng is a norm generator. Since ord Qðx17xÞ¼ord2 ð17aÞ¼ordð17aÞþe this implies that ord Qðx17xÞXR2=2 þ k k þ e and yA!p x2; x3; y; xng is a norm generator iff ord Qðx1 À xÞ¼ord Qðx1 þ xÞ¼R2=2 þ k þ e: We will use this in the proof of (ii), (iii). k 0 Now the fact that pr > xA!p x2; x3; y; xng implies by 2.7(ii) that xAL where x1 0 k k L ¼ !x1; p x2; x3; y; xng: Let us note that !x1; p x2g cannot be a hyperbolic plane. Indeed if k ¼ 0 this wouldcontradict the hypothesis andfor k40 we would k have R2 þ 2k ¼ Rð!x1; p x2gÞ¼À2e so Rð!x1; x2gÞ¼R2o À 2e which is impossible. ARTICLE IN PRESS

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Claim. (ii), (iii) are true if we prove that:

0 pR2=2þk (a) fvAL jord QðvÞ¼R2=2 þ k þ egDL and (b) pR2=2þkAð0; 0Þ L0:

0 0 Proof. For (ii) we have x1; xAL so x1 À xAL : We also have ord Qðx1 À xÞ¼ k R2=2 þ k þ e: (pr > x is a norm generator in !p x2; x3; y; xng:) By (a) this implies x1 R =2þk p 2 R2=2þk that x1 À xAL so Bðx1 À x; LÞDp which implies that ord2 Bðx1 À X A x; LÞ R2=2 þ k þ e ¼ ord Qðx1 À xÞ so tx1Àx OðLÞ (see [OM, Section 91B]). A 00 0A 00 0 For (iii) we have tx1Àx OðLÞ so x ¼ tx1Àxx L and Qðx Þ¼Qðx Þ¼1: Since 0 0 0 pr > x is not a norm generator we have that either ord Qðx1 À x Þ or ord Qðx1 þ x Þ is x1 0 4R2=2 þ k þ e: WLOG we may assume that ord Qðx1 À x Þ4R2=2 þ k þ e: (Otherwise we replace x0 by Àx0:) 00 k We assume that pr > x is not a norm generator of !p x2; x3; y; xng: This x1 00 implies that for a choice of the 7 sign we have ord Qðx17x Þ4R2=2 þ k þ e: But 7 00 7 0 7 00 7 0 07 tx1Àxðx1 x Þ¼x x so Qðx1 x Þ¼Qðx x Þ: Hence ord Qðx xÞ4R2=2 þ k þ e for a choice of 7: We also have ord Qðx17xÞ¼R2=2 þ k þ e since pr > x is a x1 k norm generator of !p x2; x3; y; xng: 0 0 Now ord Qðx 7xÞ; ord Qðx1 À x Þ4R2=2 þ k þ e andord Qðx17xÞ¼R2=2þ 0 0 0 k þ e and ðx 7xÞþðx1 À x Þ¼x17x: By 3.19 this implies Oðx 7xÞþ 0 R2=2þk R2=2þk 0 0 0 Oðx1 À x ÞDp Að0; 0Þ so p Að0; 0Þ L (x1; x; x AL ). But this contradicts (b). Let us prove now the hypothesis of our claim. Since R2 þ 2kp2e we are in one of cases (3) or (4) of (i). k In case (4) we have that k ¼ 0or1: It implies that the BONG x1; p x2; x3; y; xn is good. (For k ¼ 0 this follows from the fact that L has property A andfor k ¼ 1we have ord Qðpx2Þ¼R2 þ 2pR4 ¼ ord Qðx4Þ). We prove ﬁrst (a). There are two cases: If 0 ¼ R1XR2 then !x1; x2g is the ﬁrst Jordan component of L andwe have R =2 R =2þ1 pR2=2 sL ¼ s!x1; x2g ¼ p 2 and s!x3; y; xngDp 2 : If k ¼ 0 then L ¼ L so pR2=2þ1 statement (a) is trivial. If k ¼ 1 then L ¼ p!x1; x2g>!x3; y; xng: Now let 0 vAL with ord QðvÞ¼R2=2 þ 1 þ e: We have R2X À 2e so ord QðvÞ¼R2=2 þ 1 þ 0 0 eX140 ¼ ord nL : It follows that vAL ¼ !x1; px2; x3; y; xng is not a norm generator. But since !px1; px2g ¼ p!x1; x2g exists and R1 þ 2pR3 we can apply 6.1(i) andwe get vA!px1; px2; x3; y; xng: But ord Qðpx2Þ¼ R2 þ 2pR1 þ 2 ¼ ord Qðx1ÞpR3 ¼ ord Qðx3Þ so p!x1; x2g>!x3; y; xng ¼ !px1; px2g>!x3; y; xng ¼ !px1; px2; x3; y; xng so we are done. If 0 ¼ R1oR2 then R2X2 andwe have L ¼ Ox1>!x2; y; xng: We have ord sðOx1Þ¼0oR2=2 þ k andord s!x2; y; xng¼minfR2; ðR2 þ R3Þ=2gXR2=2 þ 1XR2=2 þ k: (We have R2X2soR2XR2=2 þ 1 and R3XR1 þ 2 ¼ 2soðR2 þ R3Þ= R =2þk p 2 R2=2þk 2XðR2 þ 2Þ=2 ¼ R2=2 þ 1:) It follows that L ¼ p x1>!x2; y; xng: Let 0 vAL with ord QðvÞ¼R2=2 þ 1 þ e: We have v ¼ ax1 þ y with aAO and ARTICLE IN PRESS

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k R =2þk yA!p x2; x3; y; xngD!x2; y; xng: We have to prove that aAp 2 : Now 1 X ord QðvÞ¼R2=2 þ k þ e ¼ 2ðR2 þ 2kÞþe R2 þ 2k (because R2 þ 2kp2e). We k also have ord QðyÞXord n!p x2; x3; y; xng ¼ R2 þ 2k: Since QðvÞ¼Qðax1Þþ 2 2 QðyÞ¼a þ QðyÞ this implies ord a XR2 þ 2k so ord aXR2=2 þ k as claimed. k 0 For (b) we just note that x1; p x2; x3; y; xn is a goodBONG for L with R1oR3 so k we can apply 6.4(iii). Since !x1; p x2g is not a hyperbolic plane we have r 0 p Að0; 0Þ L where r ¼ð0 þðR2 þ 2kÞÞ=2 ¼ R2=2 þ k so we are done. In case (3) ðdpe À R=2Þ we have R3 À R2X2e þ 1 and R2 þ 2k ¼ 2e À 2d so R2=2 þ k ¼ e À d and R2=2 þ k þ e ¼ 2e À d: Now R2X À d so R3XR2 þ 2e þ 1X2e þ 1 À d42e À d ¼ R2=2 þ k þ e: This implies that ord s!x3; y; xngXord n!x3; y; xng À e4R2=2 þ k: Also R3X2e þ 1 À d42e À 2d ¼ k k R2 þ 2k: Hence ord n!x3; y; xng4ord Qðp x2Þ so !p x2; x3; y; xng ¼ k p x2>!x3; y; xng: We have R2oR3 so L ¼ !x1; x2g>!x3; y; xng: Since s!x3; y; xngC R =2þk pR2=2þk pR2=2þk p 2 we have L ¼ !x1; x2g >!x3; y; xng: k k k Since !p x2; x3; y; xng ¼ p x2>!x3; y; xng we have !x1; p x2; x3; k y; xng ¼ !x1; p x2g>!x3; y; xng: (x1 is a norm generator for k k !x1; p x2g>!x3; y; xng andwe have pr > ð!x1; p x2g>!x3; y; xngÞ¼ x1 k k ðpr > !x1; p x2gÞ>!x3; y; xng ¼ p x2>!x3; y; xng: Note that the BONG x1 k x1; p x2; x3; y; xn is not necessarily good.) 0 Let now vAL with ord QðvÞ¼R2=2 þ k þ e: We write v ¼ y þ z with k pR2=2þk yA!x1; p x2g and zA!x3; y; xng: We want to prove that vAL which is pR2=2þk equivalent to yAð!x1; x2gÞ : We have ord QðzÞXord n!x3; y; xng ¼ R34R2=2 þ k þ e ¼ ord QðvÞ: Since QðvÞ¼QðyÞþQðzÞ we get that ord QðyÞ¼ k R2=2 þ k þ e: Hence we reduced (a) to proving that for any yA!x1; p x2g with pR2=2þk ord QðyÞ¼R2=2 þ k þ e we have yA!x1; x2g : There are two cases: R =2 pR2=2þk If 0 ¼ R1XR2 then !x1; x2g is p 2 -modular so !x1; x2g ¼ k p !x1; x2g: Let us note that since R2X À d; R2=2 þ k þ e ¼ 2e À d and R2 þ 2k ¼ 2e À 2d we have 2k ¼ 2e À 2d À R2p2e À 2d ÀðÀdÞ¼2e À d ¼ R2=2 þ k þ e: k Hence ord QðyÞ¼R2=2 þ k þ eX2k ¼ R1 þ 2k: Since yA!x1; p x2g and k k k k k !p x1; p x2g ¼ p !x1; x2g exists we have by 6.1(ii) that yA!p x1; p x2g ¼ k p !x1; x2g so we are done. k If 0 ¼ R1oR2 then !x1; x2g ¼ Ox1>Ox2: We have yA!x1; p x2g ¼ k k Ox1>p x2 so y ¼ ax1 þ bx2 with aAO and bAp : But ord Qðbx2Þ¼2 ord b þ X 1 X R2 R2 þ 2k andord QðyÞ¼R2=2 þ k þ e ¼ 2ðR2 þ 2kÞþe R2 þ 2k (because 2 2 R2 þ 2kp2e). Since QðyÞ¼Qðax1ÞþQðbx2Þ¼a þ Qðbx2Þ we get ord a XR2 þ R =2þk k pR2=2þk 2k so ord aXR2=2 þ k: Hence yAp 2 x1>p x2DðOx1>Ox2Þ andwe are R =2þk k R =2þk k done. (We have Bðp 2 x1>p x2; Ox1>Ox2Þ¼p 2 Qðx1Þþp Qðx2Þ¼ pR2=2þk þ pR2þk ¼ pR2=2þk). ARTICLE IN PRESS

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For (b) we have again two cases: k 0 If 0 ¼ ord Qðx1ÞXord Qðp x2Þ¼R2 þ 2k we have L ¼ J>K where J ¼ k !x1; p x2g is modular and K ¼ !x3; y; xng: Since ord sJ ¼ðR2 þ 2kÞ=2 ¼ R =2þk R R R2=2 þ k; sKCp 2 and nK ¼ p 3 Cp 1 ¼ nJ and J is not hyperbolic we have pR2=2þkAð0; 0Þ L0 (6.4(i)). k k 0 If 0oR2 þ 2k then !x1; p x2g ¼ Ox1>p x2 so L ¼ Ox1>K where K ¼ k k R2þ2k R2=2þk p x2>!x3; y; xng: But sðp x2Þ¼p Cp (because R2 þ 2k40so 1 ! y gC R2=2þk C R2=2þk R2 þ 2k42ðR2 þ 2kÞ¼R2=2 þ k) and s x3; ; xn p so sK p which implies pR2=2þkAð0; 0Þ L0 (6.4(ii)). A (iv) If ord Qðx1 À xÞ¼ord2 ð1 À aÞ¼2e; i.e. if ordð1 À aÞ¼e then tx1Àx OðLÞ iff ord2 Bðx1 À x; LÞXord Qðx1 À xÞ¼2e; i.e. iff ord Bðx1 À x; LÞXe: But x1 À x ¼ð1 À aÞx1 À y so Bðx1 À x; LÞ¼Bðð1 À aÞx1 À y; Ox1>!x2; y; xngÞ¼ ð1 À aÞsðOx1ÞþBðy; !x2; y; xngÞ: We have ordð1 À aÞsðOx1Þ¼ordð1 À aÞ¼e so we only needto prove that ord Bðy; !x2; y; xngÞXe: Now the only cases of (i) when R2 þ 2kX2e þ 1 are (1) (when R2X2e þ 1), (2) (when R2o2e is odd) and (5) (when R2 ¼ 2e or 2e À 2andO=p has two elements). With the exception of the cases when R2o2e is odd and when R2 ¼ 2e À 2 we have R2X2e so ord Bðy; !x2; y; xngÞXord s!x2; y; xngXord n!x2; y; xng À e ¼ R2 À eXe andwe are done. k k If R2o2e is odd we have !p x2; x3; y; xng ¼ p x2>!x3; y; xng: Hence k y ¼ bx2 þ z with bAp and zA!x3; y; xng: It follows that Bðy; !x2; y; xngÞ¼ Bðbx2 þ z; Ox2>!x3; y; xngÞDbsðOx2Þþs!x3; y; xng: We have ord bsOx2 ¼ ord b þ R2Xk þ R2 andord s!x3; y; xngXord n!x3; x4; y; xng À e ¼ R3 À e: But k þ R24R2=2 þ k ¼ðR2 þ 2kÞ=2 ¼ð2e þ 1Þ=24e and R3 À R2X2e þ 1(L has property B) so R3 À eXR2 þ 2e þ 1 À e4e þ 1(R240). Hence ord Bðy; !x2; y; xngÞ4e andwe are done. If R2 ¼ 2e À 2 then we may suppose that ord s!x2; y; xngoe since otherwise ord Bðy; !x2; y; xngÞXord s!x2; y; xngXe andwe are done. If R3pR2 then ord s!x2; y; x3g ¼ðR2 þ R3Þ=24R2=2 ¼ e À 1 so ord s!x2; y; x3gXe: If R2oR3 then we have ord s!x2; y; xng ¼ R2 ¼ 2e À 2Xe provided that eX2: If e ¼ 1 (which implies F ¼ Q2 since O=p has two elements) then we no longer have ord s!x2; y; xngXe: In this case we use the fact that yA!x2; y; xng is not a norm generator. Since R2 ¼ 2e À 2 ¼ 0 we have that !x1; x2g is unimodular so ord s!x3; y; xng4ord s!x1; x2g ¼ 0: Now property B implies that R4XR2 þ 2 ¼ ord Qðpx2Þ and R3XR1 þ 2 ¼ R2 þ 2 ¼ ord Qðpx2Þ so the BONG px2; y; xn is goodand !px2; y; xng ¼ !px2g>!x3; y; xng: Hence we can write y ¼ bx2 þ z with zA!x3; y; xng and bAp: We get Bðy; !x2; y; xngÞ¼Bðbx2 þ z; Ox2>!x3; y; xngÞ¼bsðOx2Þþs!x3; y; xng: But ord bsðOx2Þ¼ord bX1 andord s!x3; y; xngX1 so ord Bðy; !x2; y; xngÞX1 ¼ e andwe are done. &

Lemma 6.6. Let L ¼ !x1; y; xng; nX3 be a lattice satisfying property B with R þ QðxiÞ¼ai ¼ p i ei and let xAL s.t. QðxÞ¼Qðx1Þ: Then there is sAO ðLÞ s.t. sx1 ¼ x ðord a Àord a Þ=2 ’2 ðord a Àord a Þ=2 ’2 ’ and yðsÞAGða2=a1Þð1 þ p 3 1 ÞF if Gða3=a2Þð1 þ p 3 1 ÞF aF: ARTICLE IN PRESS

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ðR ÀR Þ=2 ’2 0 ðR ÀR Þ=2 ’2 Proof. Let H ¼ Gða2=a1Þð1 þ p 3 1 ÞF and H ¼ Gða3=a2Þð1 þ p 3 1 ÞF : þ 0 ’ We want to prove that there is sAO ðLÞ s.t. sx1 ¼ x andeither yðsÞAH or H ¼ F: We consider first the case when !x1; x2g is a hyperbolic plane. We have A 1 Â2 + Â ’2 ! gD r a2=a1 À 4 O so H Gða2=a1Þ¼O F : We have that H ¼ x1; x2 p Að0; 0Þ with r ¼ðR1 þ R2Þ=2 is the first Jordan component of L andwe write L ¼ H > K where K ¼ !x3; y; xng: Now xAL is also a norm generator so we can get a new 0 0 0 0 0 BONG x ¼ x1 ; x2 ; y; xn for L: Now both x1; y; xn and x1 ; y; xn are good 0 BONGs (L has property A) so by 4.7 we must have ord Qðxi Þ¼ord QðxiÞ¼Ri: r 0 0 r Since R1oR3 and p Að0; 0Þ L we must have !x1 ; x2 gDp Að0; 0Þ (6.4(iii)). 0 0 0 0 0 0 0 0 We write similarly L ¼ H >K where H ¼ !x1 ; x2 g and K ¼ !x3 ; y; xn g: Since HDH0DprAð0; 0Þ there is a product of Eichler transformations tAOþðLÞ s.t. 0 0 0 tH ¼ H : Since tx1 and x ¼ x1 belong to H and Qðtx1Þ¼Qðx1Þ¼QðxÞ þ 0 there is fAO ðFH Þ s.t. ftx1 ¼ x: But x ¼ fðtx1Þ is a common norm generator for H0 and fH0 andwe also have RðH0Þ¼RðfH0Þ: By 3.2(ii) we get 0 0 þ 0 þ H ¼ fH so fAO ðH ÞDO ðLÞ: Hence if we take s ¼ ft we have sx1 ¼ x and sAOþðLÞ: We also have yðtÞ¼1 and yðfÞAyðH0Þ¼OÂF’2 so yðsÞAOÂF’2DH and we are done. From now on we will assume that !x1; x2g is not hyperbolic so we can apply 6.5. By scaling we may assume that Qðx1Þ¼QðxÞ¼1soR1 ¼ 0: Let k be defined k like in 6.5, i.e. kX0 is minimal s.t. !x1; p x2g can be spannedby two norm generators of the same length. In order to make the notation similar to the one in the R proof of + part we put r ¼ R2 þ 2k and s ¼ R3: We have H+Gðp 2 e2Þ which by 3.8 R2þ2k r ðR3ÀR1Þ=2 ’2 implies that H+Gðp e2Þ¼Gðp e2Þ: We also have ð1 þ p ÞF ¼ð1 þ s=2 ’2 R2 s=2 ’2 0 s=2 ’2 p ÞF so H ¼ Gðp e2Þð1 þ p ÞF and H ¼ Gða3=a2Þð1 þ p ÞF : Also note that k R2þ2k r Qðx2Þ¼a2 ¼ a2=a1AGða2=a1ÞDH and Qðp x2Þ¼p e2 ¼ p e2: We consider first the case when r ¼ R2 þ 2kp2e: We prove that our lemma is true if we assume that Qðx1 À xÞAH whenever xAL with QðxÞ¼1andpr > x is a norm x1 k generator of !p x2; x3; y; xng: k Indeed if pr > x is a norm generator of !p x2; x3; y; xng we take s ¼ tx Àxtx : x1 1 2 A ! g D We have sx1 ¼ tx1Àxtx2 x1 ¼ tx1Àxx1 ¼ x: We also have tx2 Oð x2 Þ OðLÞ and A A þ A from 6.5(ii) we have tx1Àx OðLÞ so s O ðLÞ: Finally, Qðx1 À xÞ H by our assumption and Qðx2ÞAH so yðsÞ¼Qðx1 À xÞQðx2ÞAH: k 0 If pr > x is not a norm generator of !p x2; x3; y; xng we take x AL with x1 0 k 0 k Qðx Þ¼Qðx1Þ¼1 s.t. !x1; p x2g ¼ Ox1 þ Ox : It follows that Op x2 ¼ k 0 0 k Â 0 pr > !x1; p x2g ¼ pr > Ox so pr > x ¼ ap x2 for some aAO : Hence pr > x is a x1 x1 x1 x1 k norm generator for !p x2; x3; y; xng: From 6.5(ii) and(iii) we have that k 00 tx Àx0 AOðLÞ and pr > tx Àx0 x is a norm generator for !p x2; x3; y; xng: Let x ¼ 1 x1 1 00 k tx Àx0 x: Since pr > x is a norm generator of !p x2; x3; y; xng we have 1 x1 þ 00 A 0 00 A 0 00 tx1Àx OðLÞ by 6.5(ii). Hence s :¼ tx1Àx tx1Àx O ðLÞ: But sx1 ¼ tx1Àx tx1Àx x1 ¼ 00 2 0 0 0 tx1Àx x ¼ðtx1Àx Þ x ¼ x andfrom our assumption we have Qðx1 À x Þ; Qðx1 À 00 0 00 x ÞAH so yðsÞ¼Qðx1 À x ÞQðx1 À x ÞAH: ARTICLE IN PRESS

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Let now xAL with QðxÞ¼1 s.t. y ¼ pr > x is a norm generator of x1 k !p x2; x3; y; xng: We have that x1 is a norm generator for Ox1 þ OxDL and pr > ðOx1 þ OxÞ¼Oy so Ox1 þ Ox ¼ !x1; yg: Now xA!x1; yg with Qðx1Þ¼ x1 A ! g QðxÞ and x1 À x is not isotropic so by 3.18(i) we have tx1Àx Oð x1; y Þ and Qðx1 À xÞ¼Qðx1ÞQðx1 À xÞAGðQðyÞÞ: (We have að!x1; ygÞ¼QðyÞ=Qðx1Þ¼ QðyÞ:) Hence it is enough to prove that GðQðyÞÞDH: Since r ¼ R2 þ 2kp2e we are in one of case (3) or (4) of the proof of 6.5(i). We consider the two cases separately: If we are in case (3) then r ¼ R2 þ 2k ¼ 2e À 2dðÀe2Þ: As seen in the proof of 6.5(i) k k R2 we have !p x2; x3; y; xng ¼ p x2>!x3; y; xng: We also have H+Gðp e2Þ¼ NðÀe2Þ: There are two cases: If sp4e À 2dðÀe2Þ then s=2 þ dðÀe2Þpð4e À 2dðÀe2ÞÞ=2 þ dðÀe2Þ¼2e so ð1 þ s=2 ’2 s=2 ’2 ’ p ÞF D/NðÀe2Þ: Since NðÀe2Þ; ð1 þ p ÞF DH we have H ¼ F so the relation GðQðyÞÞDH is trivial. If sX4e À 2dðÀe2Þþ1 then sX2e À 2dðÀe2Þþ2e þ 1 ¼ r þ 2e þ 1: Now k k yA!p x2; x3; y; xng ¼ Op x2>!x3; y; xng is a norm generator so ord QðyÞ¼ k k ord Qðp x2Þ¼r andwe have y ¼ ap x2 þ z with aAO and zA!x3; y; xng: We 2 k have QðyÞ¼a Qðp x2ÞþQðzÞ: Now ord QðzÞXord n!x3; y; xng ¼ sXr þ 2e þ 2 k Â2 1 ¼ ord QðyÞþ2e þ 1soa Qðp x2Þ=QðyÞ¼1 À QðzÞ=QðyÞAO : But ord QðyÞ¼ k Â k Â2 ord Qðp x2Þ so aAO which implies QðyÞ=Qðp x2ÞAO so GðQðyÞÞ ¼ k r GðQðp x2ÞÞ ¼ Gðp e2ÞDH andwe are done. In case (4), i.e. when R2p2e (resp. R2o2e À 2 when O=p has only two elements) k and dðÀe2Þ4e À R2=2 we have k ¼ 0ork ¼ 1: In both cases !p x2; x3; y; xng k 0 sÀr satisﬁes the conditions of 6.3(i) so Z :¼ QðyÞ=Qðp x2ÞAg ðp e2e3Þ: We have QðyÞ¼ k r ZQðp x2Þ¼p e2Z: If we prove that dðZÞ4e À r=2 then we also have dðÀe2Þ4e À r r R2=2Xe À r=2sodðÀe2ZÞ4e À r=2: Hence p e2; p e2Z satisfy the conditions of r r=2þdðZÞÀe ’2 r 3.15(ii) andwe have GðQðyÞÞ ¼ Gðp e2ZÞDð1 þ p ÞF Gðp e2Þ: Since we r r=2þdðZÞÀe ’2 already have Gðp e2ÞDH it is enough to prove that ð1 þ p ÞF DH: Hence we have reduced to proving that dðZÞ4e À r=2 and ð1 þ pr=2þdðZÞÀeÞF’2DH for any 0 sÀr ZAg ðp e2e3Þ: We have several cases: 0 sÀr Â2 (a) s À rX2e þ 1: In this case ZAg ðp e2e3Þ¼O so dðZÞ¼N which makes both dðZÞ4e À r=2 and ð1 þ pr=2þdðZÞÀeÞF’2DH trivial. (b) s À ro2e is odd. We have 2e À 1Xs À r ¼ R3 ÀðR2 þ 2kÞXR3 À R2 À 2so R3 À R2p2e þ 1: Since R3 À R2p2e þ 1 is odd and R2 À R1 ¼ R2prp2e the lattice L does not have property B. Hence this case cannot happen. (c) s À rp2e is even. We have dðÀe2e3Þ4e ÀðR3 À R2Þ=2 since otherwise L does not have property B. (We have R2 À R1 ¼ R2p2e:) Since R3 ¼ s and R2 ¼ r À 2k we have dðÀe2e3Þ4e ÀðR3 À R2Þ=2 ¼ e Àðs À r þ 2kÞ=2 ¼ e Àðs À rÞ=2 À kXe Àðs À rÞ=2 À 1: This implies that dðÀe2e3ÞXe Àðs À rÞ=2so 0 sÀr ðsÀrÞ=2þe ’2 g ðp e2e3Þ¼ð1 þ p ÞF : Hence dðZÞXðs À rÞ=2 þ e4e À r=2 (we have s40) andwe also have r=2 þ dðZÞÀeXr=2 þðs À rÞ=2 þ e À e ¼ s=2 which implies ð1 þ pr=2þdðZÞÀeÞF’2Dð1 þ ps=2ÞF’2DH: ARTICLE IN PRESS

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Let us consider now the case when r ¼ R2 þ 2k42e: First let us note that if y ¼ p > x is isotropic andwe write x ¼ ax1 þ y we have x1 2 1 ¼ Qðax1 þ yÞ¼a so a ¼ 71: Hence x ¼ 7x1 þ y: If y is not isotropic then ord QðyÞ42e: Indeed with the exception of the case k R2 ¼ 2e À 2 we have ord QðyÞXord n!p x2; x3; y; xng ¼ R2 þ 2k ¼ r42e while in the case R2 ¼ 2e À 2 we have that yA!px2; x3; y; xng is not a norm generator so ord QðyÞ4ord n!px2; x3; y; xng ¼ 2e: Like in the case rp2e we have that Ox1 þ Ox ¼ !x1; yg and x1; xA!x1; yg are norm generators with Qðx1Þ¼QðxÞ: Since að!x1; ygÞ¼QðyÞ=Qðx1Þ¼QðyÞ andord QðyÞ42e we have from 3.18(ii) that ord Qðx17xÞ¼2e for one choice of the 7 sign. Moreover if ord Qðx1 À xÞ¼ 0 2e then 3.18(iii) gives Qðx1 À xÞ¼Qðx1ÞQðx1 À xÞAG ðQðyÞÞ: We claim that our lemma is true if we assume that for any xAL with QðxÞ¼1 and ord Qðx1 À xÞ¼2e we have that Qðx1 À xÞAH: Indeed if xAL with QðxÞ¼1 then ord Qðx17xÞ¼2e for one choice of the sign 7 7 A andfor that choice of we have from 6.5(iv) that tx17x OðLÞ: If ord Qðx1 À xÞ¼ A A þ 2e then we take s ¼ tx1Àxtx2 : We have tx1Àx; tx2 OðLÞ so s O ðLÞ: We also have A sx1 ¼ tx1Àxtx2 x1 ¼ tx1Àxx1 ¼ x and yðsÞ¼Qðx1 À xÞQðx2Þ: But Qðx2Þ H andby the hypothesis of our claim we have that Qðx1 À xÞAH so yðsÞAH as desired. If A A þ ord Qðx1 þ xÞ¼2e we take s ¼ tx1þxtx1 : We have tx1þx; tx1 OðLÞ so s O ðLÞ: We also have sx1 ¼ tx1þxtx1 x1 ¼ tx1þxðÀx1Þ¼x and yðsÞ¼Qðx1 þ xÞQðx1Þ¼Qðx1 þ xÞ: But by the hypothesis of our claim we have that Qðx1 þ xÞAH so yðsÞAH andwe are done. Let us suppose now that xAL with QðxÞ¼1 andord Qðx1 À xÞ¼2e: If y is isotropic we cannot have x ¼ x1 þ y since otherwise Qðx1 À xÞ¼QðÀyÞ¼0: Hence ’2 x ¼Àx1 þ y so Qðx1 À xÞ¼Qð2x1 À yÞ¼4AF DH andwe are done.If y is not 0 isotropic then Qðx1 À xÞAG ðQðyÞÞ so we reduce the lemma to proving that 0 G ðQðyÞÞAH whenever y ¼ pr > x is not isotropic. x1 0 ’2 In the case R244e we have ord QðyÞ44e so G ðQðyÞÞ ¼ F DH so we are done. Hence we may assume that R2p4e: The strategy of proof is the following: We will ﬁnda number t42e independent of r 2 t 0 x s.t. QðyÞAp e2O þ p andwe reducethe inclusion G ðQðyÞÞDH to proving that ð1 þ ptÀ2eÞF’2DH: r 2l 2 Â We have QðyÞ¼p e2p e þ b for some integer lX0; eAO and bAF with 0 r0 2 ord bXt: If we denote r ¼ r þ 2l then QðyÞ¼p e2e þ b: We have two cases: If r0Xt then ord QðyÞXt so G0ðQðyÞÞDð1 þ ptÀ2eÞF’2: (In general we have G0ðpReÞDð1 þ pRÀ2eÞF’2 (3.12(i)).) Hence if ð1 þ ptÀ2eÞF’2DH then G0ðQðyÞÞDH: 0 r0 2 r0 2 Àr0 À1 If r ot we write QðyÞ¼p e2e þ b ¼ p e2Z where Z ¼ e þ p e2 b: But Àr0 À1 0 0 Â 0 0 ord p e2 b ¼ ord b À r Xt À r 40soZAO with dðZÞXt À r : If r 44e then 0 r0 ’2 0 0 r0 G ðp e2ZÞ¼F DH: If r p4e then 3.15(i) implies that G ðp e2ZÞDð1 þ r0þdðZÞÀ2e ’2 0 r0 0 0 0 r0 r0 p ÞF G ðp e2Þ: But r Xr and r r ðmod2 Þ so G ðp e2ÞDGðp e2ÞD r r0þdðZÞÀ2e ’2 0 Gðp e2ÞDH andwe only have to prove that ð1 þ p ÞF DH: But r þ dðZÞÀ 2eXr0 þðt À r0ÞÀ2e ¼ t À 2e so ð1 þ pr0þdðZÞÀ2eÞF’2Dð1 þ ptÀ2eÞF’2: Hence ð1 þ ptÀ2eÞF’2DH implies ð1 þ pr0þdðZÞÀ2eÞF’2DH andwe are done. ARTICLE IN PRESS

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Let us consider ﬁrst the case when R2 is odd so r ¼ R2 þ 2k is also odd. First we rÀ2e ’2 note that we may assume that DeH: Otherwise ð1 þ p ÞF -NðÀpe2Þ¼ 0 r rÀ2e ’2 rÀ2e ’2 G ðp e2ÞDH and DAH so ð1 þ p ÞF DH: (We have DAð1 þ p ÞF but k k DeNðÀpe2Þ:) But yA!p x2; x3; y; xng so ord QðyÞXord n!p x2; x3; y; xng ¼ k 0 rÀ2e ’2 ord Qðp x2Þ¼r which implies that G ðQðyÞÞDð1 þ p ÞF DH andwe are done. 2e ’2 s=2 ’2 Now if sp4e then s=2p2e so DAð1 þ p ÞF Dð1 þ p ÞF DH: Hence we must have sX4e þ 1: k k If R2o2e then r ¼ R2 þ 2k ¼ 2e þ 1 and !p x2; x3; y; xng ¼ Op x2> !x3; y; xng (see the proof of 6.5(i) case 2). But this holds also if R242e since we have k ¼ 0 and R2 ¼ ro4eos ¼ R3 so !x2; y; xng ¼ Ox2>!x3; y; xng: It k k follows that QðyÞAQðOp x2>!x3; y; xngÞ¼QðOp x2ÞþQð!x3; y; xngÞ: But k k 2 r 2 R QðOp x2Þ¼Qðp x2ÞO ¼ p e2O and Qð!x3; y; xngÞDn!x3; y; xng ¼ p 3 ¼ s r 2 s p : Hence QðyÞAp e2O þ p andwe can take t ¼ s: Since t ¼ sX4e þ 1 we have t À 2eX2e þ 1soð1 þ ptÀ2eÞF’2 ¼ F’2DH: If R2 is even so r is even, with the exception of the case when R2 ¼ 2e À 2; k dðÀe2Þ41andO=p has two elements, then yA!p x2; x3; y; xng so QðyÞA k Qð!p x2; x3; y; xngÞ: The exceptional case ðR2 ¼ 2e À 2Þ will be treated separately. Now k ¼ 0ifR242e and k ¼ 1ifR2 ¼ 2e and O=p has two elements. k Now !x2; y; xng has property B so in both cases !p x2; x3; y; xng satisﬁes the k r s conditions of 6.3(ii). Since Qðp x2Þ¼p e2 and Qðx3Þ¼p e3 we have from 6.3(ii) that k r 2 t QðyÞAQð!p x2; x3; y; xngÞDp e2O þ p where t can be chosen:

(i) t ¼ s if s4r; (ii) t ¼ðr þ sÞ=2 þ e if s À rp2e is even and dðÀe2e3ÞXe Àðs À rÞ=2; (iii) t ¼ s þ dðÀe2e3Þ if s À rp2e is even and dðÀe2e3Þpe Àðs À rÞ=2:

If s À rX2e þ 1 we use formula (i) andtake t ¼ s: Since rX2e þ 2 and s À rX2e þ 1 we have t ¼ sX4e þ 3sot À 2eX2e þ 3: Hence ð1 þ ptÀ2eÞF’2 ¼ F’2DH andwe are done. Now if s À ro2e is odd then R3 À R2 ¼ s Àðr À 2kÞp2e À 1 þ 2kp2e À 1 þ 2 ¼ 0 R ÀR 0 ’ 2e þ 1 which implies that H +Gðp 3 2 e2e3Þ¼NðÀpe2e3Þ: In order to have H aF we needto have DeH0 so Deð1 þ ps=2ÞF’2 so s=242e which implies that sX4e þ 14r: We use formula (i) andwe have t À 2e ¼ s À 2eX2e þ 1 which implies that ð1 þ ptÀ2eÞF’2 ¼ F’2DH: If s À rp2e is even and dðÀe2e3ÞXe Àðs À rÞ=2 we take t ¼ðr þ sÞ=2 þ e: We have t À 2e ¼ðr þ sÞ=2 À e4ð2e þ sÞ=2 À e ¼ s=2soð1 þ ptÀ2eÞF’2Dð1 þ ps=2ÞF’2DH: Finally, if dðÀe2e3Þoe Àðs À rÞ=2 we use formula (ii) andwe take t ¼ s þ dðÀe2e3Þ: Note that this cannot happen when R2 ¼ 2e: Indeed if R2 ¼ 2e then dðÀe2e3Þ4e ÀðR3 À R2Þ=2 since otherwise L does not have property B. (We have R2 À R1 ¼ 2eo2e þ 1:) Since R3 ¼ s and R2 ¼ 2e ¼ r À 2 we have e ÀðR3 À R2Þ=2 ¼ e Àðs À rÞ=2 À 1 which implies that dðÀe2e3Þ4e Àðs À rÞ=2 À 1so dðÀe2e3ÞXe Àðs À rÞ=2: ARTICLE IN PRESS

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Hence R2X2e þ 2sor ¼ R2 and s ¼ R3 so we have dðÀe2e3Þoe ÀðR3 À R2Þ=2: It 0 R ÀR s=2 ’2 0 follows that H +Gðp 3 2 e2e3Þ¼NðÀe2e3Þ: Now we also have ð1 þ p ÞF DH so 0 ’ s=2 ’2 in order to have H aF we needto have ð1 þ p ÞF DNðÀe2e3Þ which implies by 1.2(iii) that dðÀe2e3Þþs=242e so s=2 þ dðÀe2e3ÞÀ2e40: It follows that t À 2e ¼ tÀ2e ’2 s=2 ’2 s þ dðÀe2e3ÞÀ2e4s=2soð1 þ p ÞF Dð1 þ p ÞF DH andwe are done. In the exceptional case (R2 ¼ 2e À 2) we only have yA!px2; x3; y; xng: Now R2 2eÀ2 !px2; x3; y; xng satisﬁes the conditions of 6.2(ii). Since Qðx2Þ¼p e2 ¼ p e2 2e s so Qðpx2Þ¼p e2 and Qðx3Þ¼p e3 we have from 6.2(ii) that QðyÞA 2e 2 t Qð!px2; x3; y; xngÞ¼p e2O þ p where t42e can be chosen:

(i) t ¼ s if s42e; (ii) t ¼ð2e þ sÞ=2 þ e ¼ s=2 þ 2e if s À 2ep2e is even and dðÀe2e3ÞXe Àðs À 2eÞ=2 ¼ 2e À s=2; (iii) t ¼ s þ dðÀe2e3Þ if s À 2ep2e is even and dðÀe2e3Þpe Àðs À 2eÞ=2 ¼ 2e À s=2:

2e 2 t Since ord QðyÞ42e and t42e we have that QðyÞAp e2O þ p implies 2eþ2 2 t r 2 t QðyÞAp e2O þ p ¼ p e2O þ p as desired. If s ¼ R3 is odd we must have R3 À R2X2e þ 3 since otherwise L does not have property B. (We have R2 À R1 ¼ 2e À 2o2e þ 1:) In this case as well as in the case when R3 À R2 is even and X2e þ 2 we use (i) andwe have t ¼ s ¼ R3XR2 þ 2e þ 2 ¼ 4e: If t ¼ s ¼ 4e then t À 2e ¼ s=2 ¼ 2e so ð1 þ ptÀ2eÞF’2 ¼ð1 þ ps=2ÞF’2DH and we are done. If tX4e þ 1 then t À 2eX2e þ 1soð1 þ ptÀ2eÞF’2 ¼ F’2DH: If s À 2ep2e is even and dðÀe2e3ÞX2e À s=2 we use formula (ii) andwe have t ¼ s=2 þ 2e so ð1 þ ptÀ2eÞF’2 ¼ð1 þ ps=2ÞF’2DH andwe are done. Finally, if s is even and dðÀe2e3Þo2e À s=2 then we wouldhave to use (iii). However this case cannot happen. Indeed if s ¼ R3 is even then dðÀe2e3Þ4e À ðR3 À R2Þ=2 ¼ e Àðs Àð2e À 2ÞÞ=2 ¼ 2e À s=2 À 1 since otherwise L does not have property B. (We have R2 À R1 ¼ 2e À 2o2e þ 1:) But this implies dðÀe2e3ÞX 2e À s=2: &

Remark. Since ðord a3 À ord a1Þ=2X½ðord a3 À ord a1Þ=2 we have ð1 þ ord a ord a =2 ’2 ord a ord a =2 ’2 pð 3À 1Þ ÞF Dð1 þ p½ð 3À 1Þ ÞF so Lemma 6.6 remains true if we replace ord a ord a =2 ’2 ord a ord a =2 ’2 ð1 þ pð 3À 1Þ ÞF with ð1 þ p½ð 3À 1Þ ÞF which is enough for the purpose of this paper. However the slightly stronger statement in Lemma 6.6 might be more useful in future applications.

a ’2 ’ Lemma 6.7. Using the notations of Theorem 1 we have Gðaiþ1=aiÞð1 þ p ÞF ¼ F for some i if L does not have property B.

Proof. If L does not have property B we have two cases: (1) There is i s.t. Riþ1 À Rip2e þ 1 is odd but either Riþ2 À Riþ1 or Ri À RiÀ1 is p2e: In the ﬁrst case we have Riþ2 À Ri ¼ðRiþ2 À Riþ1ÞþðRiþ1 À RiÞp2e þ 2e þ 1 ¼ 4e þ 1: In the secondcase Riþ1 À RiÀ1 ¼ðRiþ1 À RiÞþðRi À RiÀ1Þp4e þ 1: In ARTICLE IN PRESS

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a ’2 both cases ap½ð4e þ 1Þ=2¼2e so DAð1 þ p ÞF : But we also have Gðaiþ1=aiÞ¼ Riþ1ÀRi a ’2 ’ Gðp eieiþ1Þ¼NðÀpeieiþ1Þ which implies that Gðaiþ1=aiÞð1 þ p ÞF ¼ F: (2) There is i s.t. Riþ1 À Ri is even and dðÀeieiþ1Þpe ÀðRiþ1 À RiÞ=2 but either Riþ2 À Riþ1p2e or Ri À RiÀ1p2e: The inequality dðÀeieiþ1Þpe ÀðRiþ1 À RiÞ=2 implies that Riþ1 À Rip2e À 2dðÀeieiþ1Þ: If Riþ2 À Riþ1p2e then Riþ2 À Ri ¼ðRiþ2 À Riþ1ÞþðRiþ1 À RiÞp2e þ 2e À 2dðÀeieiþ1Þ¼4e À 2dðÀeieiþ1Þ: If Ri À RiÀ1p2e then Riþ1 À RiÀ1 ¼ðRiþ1 À RiÞþðRi À RiÀ1Þp2e À 2dðÀeieiþ1Þþ 2e ¼ 4e À 2dðÀeieiþ1Þ: In both cases we have apð4e À 2dðÀeieiþ1ÞÞ=2 ¼ 2e À a ’2 dðÀeieiþ1Þ so a þ dðÀe2e3Þp2e which implies that ð1 þ p ÞF D/NðÀeieiþ1Þ: Riþ1ÀRi a ’2 But Gðaiþ1=aiÞ¼Gðp eieiþ1Þ¼NðÀeieiþ1Þ so Gðaiþ1=aiÞð1 þ p ÞF ¼ a ’2 ’ ð1 þ p ÞF NðÀeieiþ1Þ¼F: &

a ’2 Proof of the reverse inclusion. Let G :¼ Gða2=a1ÞGða3=a2ÞyGðan=anÀ1Þð1 þ p ÞF where a ¼ minf½ðRiþ2 À RiÞ=2j1pipn À 2g: We want to prove that yðLÞDG: If L does not have property B then Lemmas 6.7 and 4.11 give G ¼ F’ ¼ yðLÞ: If L has property B we induct over n: The binary case ðn ¼ 2Þ this is handled by Lemma 3.7. Write G ¼ Gða1; y; anÞ: By the induction hypothesis, yð!x2; y; xngÞ¼Gða2; y; anÞ: But Gða1; y; anÞ is the product of the groups ½ðRiþ2ÀRiÞ=2 ’2 Gðaiþ1=aiÞ with 1pipn À 1 and ð1 þ p ÞF for 1pipn À 2: Hence ½ðR3ÀR1Þ=2 ’2 Gða1; y; anÞ¼HGða2; y; anÞ where H :¼ Gða2=a1Þð1 þ p ÞF : 0 ½ðR ÀR Þ=2 ’2 ’ 0 ’ Note that if H :¼ Gða3=a2Þð1 þ p 3 1 ÞF ¼ F then G+H ¼ F: Therefore we 0 ’ þ may assume that H aF: For any sAO ðLÞ we have Qðsx1Þ¼Qðx1Þ: By Lemma 6.6 þ andthe remark following it there is fAO ðLÞ with fx1 ¼ sx1 andalso yðfÞAH 0a ’ À1 À1 A þ À1 A þ since H F: As f sx1 ¼ x1 and f s O ðLÞ we see that f s O ðprx1 LÞ¼ þ À1 O ð!x2; y; xngÞ by Lemma 2.3. Induction hypothesis gives yðf sÞA À1 Gða2; y; anÞ; whence yðsÞ¼yðfÞyðf sÞAHGða2; y; anÞ¼G: &

7. Lattices not having property A

If a lattice L does not have property A then yðLÞ¼F’ or OÂF’2: We want to distinguish between these two cases.

Lemma 7.1.

Â ’2 (i) If L ¼ L1>y>Lt is any orthogonal splitting and yðLÞDO F then all ord nLi’s have the same parity. (ii) If L ¼ H > L0 where H is a hyperbolic plane and yðL0ÞDOÂF’2 then yðLÞ¼OÂF’2 if ord nH ord nL0 ðmod2 Þ and yðLÞ¼F’ otherwise.

Proof. (i) Let 1pi; jpt be arbitrary andlet xALi and yALj be norm generators. We þ have that txAOðLiÞ and tyAOðLjÞ so s ¼ txtyAO ðLÞ: It follows that QðxÞQðyÞ¼ ARTICLE IN PRESS

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Â ’2 yðsÞAyðLÞDO F : This implies that ord QðxÞord QðyÞðmod2 Þ; i.e. ord nLi ord nLj ðmod2 Þ: (ii) Since yðHÞ¼OÂF’2 we have that yðLÞ+OÂF’2 so yðLÞ can only be OÂF’2 or F’: If ord nHcord nL0 ðmod2 Þ then (i) implies yðLÞD/ OÂF’2 so we must have yðLÞ¼F’: If ord nH ord nL0 ðmod2 Þ let sAOþðLÞ: There is tAOþðLÞ a product of Eicher transformations s.t. tH ¼ sH: Since the Eichler transformations are rotations with spinor norm 1 we t must also be a rotation with yðtÞ¼1: Now since tÀ1sH ¼ Hwe À1 0 À1 must have t s ¼ s1s2 where and s1AOðHÞ and s2AOðL Þ: Since t s is a rotation we have that s1; s2 are either both rotations or both reﬂexions. If they Â ’2 0 Â ’2 are both rotations then yðs1ÞAyðHÞ¼O F and yðs2ÞAyðL ÞDO F : Hence Â ’2 yðs1Þyðs2ÞAO F : If both of them are reﬂexions let x; y be norm generators of H, 0 À þ À resp. L : Now we have s1AO ðHÞ¼txO ðHÞ so yðs1ÞAyðO ðHÞÞ ¼ QðxÞyðHÞ¼ Â ’2 0 Â ’2 QðxÞO F : Similarly yðs2ÞAQðyÞyðL ÞDQðyÞO F : Since x; y are norm generators of H and L0 andord nH ord nL0 ðmod2 Þ we have that ord QðxÞQðyÞ is even. It Â ’2 Â ’2 follows that yðs1Þyðs2ÞAQðxÞQðyÞO F ¼ O F : Hence in any case we have À1 Â ’2 Â ’2 yðt sÞ¼yðs1s2ÞAO F andsince yðtÞ¼1 we get that yðsÞAO F andwe are done. If ord nHcord nL0 ðmod2 Þ we have from (i) that yðLÞD/ OÂF’2 so we must have yðLÞ¼F’: &

0 ri Theorem 2. If L ¼ H > L where H ¼ H1>y>Ht with HiDp Að0; 0Þ and 0 Ri L D!a1; y; ang with ai ¼ p ei is a lattice not splitting any hyperbolic plane then we have that yðLÞDOÂF’2 iff the following two conditions are satisﬁed:

(i) L0 has property A and yðL0ÞDOÂF’2; 0 (ii) The orders of nHi for 1pipt and of nL have the same parity.

Proof. Let us suppose that yðLÞDOÂF’2: This implies yðL0ÞDOÂF’2: Since L0 does not split any hyperbolic planes and yðLÞaF’; L0 must have property A. This prove the necessity of (i). Condition (ii) follows from 7.1(i). For sufﬁciency we will show by induction on 0pipt that 0 Â ’2 yðH1>y>Hi>L ÞDO F : For i ¼ 0 we get condition (i). For the induction step 0 Â ’2 i À 1-i we note that yðH1>y>HiÀ1>L ÞDO F andsince the orderof nHi 0 andthe ordersof the norms of H 1; y; HiÀ1; L have the same parity we have 0 0 Â ’2 ord nHi ord nðH1>y>HiÀ1>L Þ: By 7.1(ii) we get yðH1>y>Hi>L Þ¼O F : (Obviously the inclusion yðLÞDOÂF’2 becomes an equality if tX1). &

We want now to ﬁndthe spinor norm group for a lattice not satisfying conditionA in terms of goodBONGs.

Lemma 7.2.

R A R D Â ’2 R A D Â2 (i) If p e A then Gðp eÞ O F iff R is even and either p e À 4 O or dðÀeÞ4e À R=2: In both cases we have dðÀeÞXe À R=2: ARTICLE IN PRESS

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R R Â ’2 (ii) If p i eiAA and Gðp i eiÞDO F for 1pipk then for any R even with RXRi we kÀ1 R kÀ1 R Â ’2 have ðÀ1Þ p e1yekAA and GððÀ1Þ p e1yekÞDO F :

Proof. Condition (i) follows from the deﬁnition of G andthe fact that N ðÀeÞDOÂF’2 2 iff eA À DOÂ : (ii) We have RiX À 2e so RX À 2e: If R ¼À2e then we must have Ri ¼À2e for R Â2 Â2 1pipk andsince p i eiAA we must have that Àei belongs to either O or DO : It k Â2 Â2 follows that ðÀ1Þ e1yek ¼ðÀe1ÞyðÀekÞ belongs to either O or DO : Hence kÀ1 R y kÀ1 À2e y 1 Â2 D Â2 ðÀ1Þ p e1 ek ¼ðÀ1Þ p e1 ek belongs to either À4 O or to À4 O : In kÀ1 R Â ’2 both cases GððÀ1Þ p e1yekÞ¼O F so we are done. k k If R4 À 2e we have to prove that dððÀ1Þ e1yekÞ4e À R=2: Since ðÀ1Þ e1yek ¼ ðÀe1ÞyðÀekÞ it is enough by the domination principle to show that dðÀeiÞ4e À R=2 2 Ri D Â ’2 Ri A D Â for 1pipk: But we have Gðp Þei O F so by (i) we have that either p ei À 4 O or dðÀeiÞ4e À Ri=2: But in the ﬁrst case we have dðÀeiÞ¼2e4e À R=2 (because R4 À 2e) while in the secondcase we have dðÀeiÞ4e À Ri=2Xe À R=2 so we are done. &

R1 Rn Lemma 7.3. Let LD!p e1; y; p eng relative to a good BONG x1; y; xn: If for Rjþ1ÀRj Â ’2 some index i we have Ri ¼ Riþ2 and Gðp ejejþ1ÞDO F for j ¼ i; i þ 1 then there ðRiþRiþ1Þ=2 is a splitting !xi; xiþ1; xiþ2g ¼ H > Ov where HDp Að0; 0Þ and QðvÞ¼ Ri Àeieiþ1eiþ2p : Moreover L ¼ H >!x1; y; xiÀ1; v; xiþ3; y; xng: In particular the BONG x1; y; xiÀ1; v; xiþ3; y; xn is also good.

Proof. We consider ﬁrst the case i ¼ 1; n ¼ 3 andwe suppose ﬁrst that R1XR2: R1 Since our statement is invariant to scaling we may suppose WLOG that p e1 ¼ 1 R (i.e. R1 ¼ 0; e1 ¼ 1). We denote R2 ¼ R so Rp0: Hence LD!1; p e2; e3g: Since R R=2 R1XR2 we have L ¼ !x1; x2g>Ox3 and !x1; x2gD!1; p eg is p -modular. 0 We may assume that e2 ¼À1 À d whereDðÀe2Þ¼dO and e3 ¼À1 À d where 0 ! gD 1 pR=2 DðÀe3Þ¼d O: We have x1; x2 pR=2 ÀdpR relative to a basis x; y: We R=2 denote z ¼ x3 andwe have L ¼ðOx þ OyÞ>Oz where QðxÞ¼1; Bðx; yÞ¼p ; R R=2 QðyÞ¼Àdp andQðzÞ¼e3: We have sðOx þ OyÞ¼p ¼ sL and sðOzÞ¼O: Now D Àd0 pR=2 Oðx þ zÞþOy pR=2 ÀdpR relative to the basis x þ z; y andwe want to prove that H :¼ Oðx þ zÞ>OyDpR=2Að0; 0Þ: We have to prove that nH ¼ 2sH ¼ 2pR=2 and 2 det HA À pROÂ : The relation nH ¼ 2pR=2 is equivalent to d0; dpRA2pR=2; i.e. to ord d0Xe þ R=2 and ord dXe þ R=2 À R ¼ e À R=2: Now det H ¼Àð1 þ dd0ÞpR so if at least one of the inequalities ord d0Xe þ R=2 andord dXe À R=2 is strict we have ord dd04e À R=2 þ 2 2 e þ R=2 ¼ 2e which implies 1 þ dd0AOÂ so det HA À pROÂ : So we have to prove that ord d0Xe þ R=2 andord dXe À R=2 andat least one inequality is strict. ARTICLE IN PRESS

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R ÀR Â ’2 We have by hypothesis that Gðp e2Þ; Gðp e2e3ÞDO F : This implies that ord d ¼ X R A D Â2 0 dðÀe2Þ e À R=2 with equality iff p e2 À 4 O (when R ¼À2e) andord d ¼ dðÀe2e3Þ4e ÀðÀRÞ=2 ¼ e þ R=2: (The secondinequality is strict since Rp0sowe cannot have ÀR ¼À2e:) Now dðÀe2e3Þ4e þ R=2; dðÀe2ÞXe À R=2Xe þ R=2 with equality only if R ¼ 0 and dðÀ1ÞXeXe þ R=2 with equality only if R ¼ 0: Since Â ðÀe2e3ÞðÀe2ÞðÀ1ÞA À e3O we have by the domination principle dðÀe3ÞXe þ R=2 andthe inequality is strict unless R ¼ 0: Since also dðÀe2ÞXe À R=2 we are done in the case Ro0: In the case R ¼ 0 the inequality dðÀe2ÞXe À R=2 is strict (Ra À 2e) so again we are done. Since HDL is sL-modular we have a splitting L ¼ H > Ov for some vAL: Since R R det L ¼ 1 Á p e2 Á e3 anddet H ¼Àp we can choose v s.t. QðvÞ¼Àe2e3 as claimed. x x x x ÀRj À1 If R14R2 we use dualization. We have L ¼ !x3; x2; x1g with QðxjÞ¼p ej : x x D Â ’2 We have GðQðxj Þ=Qðxjþ1ÞÞ ¼ GðQðxjþ1Þ=QðxjÞÞ O F for j ¼ 1; 2 and x ÀR3 ¼ÀR1o À R2: Hence we are in the previous case andwe have L ¼ H > Ov D ðÀR3ÀR2Þ=2 ÀðR1þR2Þ ÀR3 À1 À1 À1 with H p Að0; 0Þ¼p Að0; 0Þ and QðvÞ¼Àp e3 e2 e1 ¼ R1 À1 x ðR1þR2Þ=2 x R1 ðÀp e1e2e3Þ : It follows that H Dp Að0; 0Þ and Qðv Þ¼Àp e1e2e3: But L ¼ Lxx ¼ Hx>Ovx so we are done. For the general case the existence of the splitting !xi; xiþ1; xiþ2g ¼ H > Ov follows from case n ¼ 3 so we only have to prove that for any such splitting we have L ¼ H >!x1; y; xiÀ1; v; xiþ3; y; xng andthe BONG x1; y; xiÀ1; v; xiþ3; y; xn is good. Next we do the case when i ¼ 1; R1XR2 and n is arbitrary. The inequality R1XR2 ðR þR Þ=2 ðR þR Þ=2 implies sL ¼ p 1 2 : We have !x1; x2; x3g ¼ H > Ov with HDp 1 2 Að0; 0Þ R and QðvÞ¼Àp 1 e1e2e3: We want to prove that L ¼ H >!v; x4; y; xng: Since sH ¼ sL andH C!x1; x2; x3gDL we have a splitting L ¼ H > K: We have to show that K ¼ !v; x4; y; xng: Now vA!x1; x2; x3gDL ¼ H > K and v>Hso vAK: Since ord QðvÞ¼R1 we have that v is a norm generator in L andhence also in K: We have pr > K pr > H > K pr > L !x ; y; x g: Hence v ¼ ðH > OvÞ ¼ !x1;x2;x3g ¼ 4 n K ¼ !v; x4; y; xng: The BONG v; x4; y; xn is goodsince ord QðvÞ¼R1 ¼ R3: Now we do the case RiXRiþ1 for arbitrary i and n: We have RiÀ1pRiþ1pRi so L ¼ !x1; y; xiÀ1g>!xi; y; xng: Form the previous case appliedto !xi; y; xng we get !xi; y; xng ¼ H >!v; xiþ3; y; xng with D ðRiþRiþ1Þ=2 R > H p Að0; 0Þ and QðvÞ¼Àpi eieiþ1eiþ2: It follows that L ¼ H K where K ¼ !x1; y; xiÀ1g>!v; xiþ3; y; xng: Hence we only have to prove that x1; y; xiÀ1; v; xiþ3; y; xn is a goodBONG for K: But this follows from 4.4(v) since RiÀ2pRi ¼ ord QðvÞ; RiÀ1pRiþ3 (we have RiÀ1pRiþ1pRiþ3) and RiÀ1pRi ¼ ord QðvÞ: x For the case when RioRiþ1 we use again dualization. We have L ¼ ! x y xg x ÀRj À1 x x xn; ; x1 where Qðxj Þ¼p ej : We have GðQðxjþ1Þ=Qðxj ÞÞ ¼ Â ’2 x GðQðxjþ1Þ=QðxjÞÞDO F for j ¼ i; i þ 1 and ÀRiþ2 ¼ÀRiX À Riþ1: Hence L is ðRiþRiþ1Þ=2 in the previous case. If !xi; xiþ1; xiþ2g ¼ H > Ov with HDp Að0; 0Þ and Ri QðvÞ¼Àp eieiþ1eiþ2: then by dualization we get the corresponding splitting ARTICLE IN PRESS

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! x x xg x> x x xiþ2; xiþ1; xi ¼ H Ov : By the previous case we have L ¼ x>! x y x x x y xg >! y H xn; ; xiþ3; v ; xiÀ1; ; x1 which by dualization gives L ¼ H x1; ; xiÀ1; v; xiþ3; y; xng: &

Theorem 3. If L ¼ !a1; y; ang relative to a good BONG x1; y; xn then yðLÞDOÂF’2 iff:

Â ’2 (i) Gðaiþ1=aiÞDO F for any 1pipn À 1: (ii) For any 1pipn À 2 s.t. ord ai ¼ ord aiþ2 we have ðord aiþ1 À ord aiÞ=2 e ðmod2 Þ:

Ri Proof. Let ai ¼ p ei: Note that (i) implies that ord aiþ1=ai is even so Riþ1 Ri ðmod2 Þ so all Ri’s have the same parity. For the ‘‘only if’’ part we have yðLÞ+yð!xi; xiþ1gÞ¼Gðaiþ1=aiÞ which implies Â ’2 (i). If Ri ¼ Riþ2 then the inclusions Gðaiþ1=aiÞ; Gðaiþ2=aiþ1ÞDO F imply by 7.3 that there is a splitting L ¼ H > L0 where H pðRiþRiþ2Þ=2Að0; 0Þ and L0 ¼ Ri 0 !x1; y; xiÀ1; v; xiþ3yxng with QðvÞ¼Àp eieiþ1eiþ2: We have ord nL ¼ R1 and ord nH ¼ðRi þ Riþ1Þ=2 þ e so ðRi þ Riþ1Þ=2 þ e R1 Ri ðmod2 Þ by 7.1(i) which implies ðRiþ1 À RiÞ=2 e ðmod2 Þ: For the ‘‘if’’ part we use induction on n: If L has property A then from Theorem 1 a ’2 we have yðLÞ¼Gða2=a1ÞyGðan=anÀ1Þð1 þ p ÞF with a ¼ minf½ðRiþ2 À RiÞ=2j1pipn À 2g: But RioRiþ2 and Ri Riþ2 ðmod2 Þ so Riþ2 À RiX2soaX1: a ’2 Â ’2 Â ’2 This implies that ð1 þ p ÞF DO F andsince also Gðaiþ1=aiÞ’s are DO F we get yðLÞCOÂF’2: If L does not have property A then Ri ¼ Riþ2 for some 1pipn À 2: By 7.3 we 0 0 have that L ¼ H > L where L ¼ !x1; y; xiÀ1; v; xiþ3yxng with H ðRiþRiþ1Þ=2 Ri p Að0; 0Þ and QðvÞ¼Àp eieiþ1eiþ2: Since ðRiþ1 À RiÞ=2 e ðmod2 Þ we 0 have ðRi þ Riþ1Þ=2 þ e Ri R1 ðmod2 Þ; i.e. ord nH ord nL ðmod2 Þ: By 7.1(ii) it is enough to prove that yðL0ÞDOÂF’2: By the induction hypothesis we only have to prove that L0 satisﬁes the conditions of our lemma. We note that x1; y; xiÀ1; v; xiþ3; y; xn have lengths a1; y; aiÀ1; QðvÞ; aiþ3; y; an with orders R1; y; RiÀ1; Ri ¼ Riþ2; Riþ3; y; Rn so most part of conditions (i) and (ii) for L0 follows from the corresponding conditions for L: For (i) we only have to prove Â ’2 that GðQðvÞ=aiÀ1Þ; Gðaiþ3=QðvÞÞDO F andfor (ii) we have to prove that if RiÀ1 ¼ Riþ3 then ðRi À RiÀ1Þ=2 e ðmod2 Þ: RiÀRiÀ1 Â ’2 Riþ2ÀRiþ1 Â ’2 Now we have Gðp eiÀ1eiÞDO F and Gðp eiþ1eiþ2ÞDO F : But Ri À RiÀ1XRiþ2 À Riþ1 (because Ri ¼ Riþ2 and RiÀ1pRiþ1). By 7.2(ii) this implies that RiÀRiÀ1 Â ’2 GðQðvÞ=aiÀ1Þ¼GðÀp eiÀ1eieiþ1eiþ2ÞDO F : Riþ1ÀRi Â ’2 Riþ3ÀRiþ2 Â ’2 We also have Gðp eieiþ1ÞDO F and Gðp eiþ2eiþ3ÞDO F : But Riþ3 À RiXRiþ1 À Ri and Riþ3 À Ri ¼ Riþ3 À Riþ2 (because Riþ1pRiþ3 and Ri ¼ Riþ2). By Riþ3ÀRi Â ’2 7.2(ii) we get Gðaiþ3=QðvÞÞ ¼ GðÀp eieiþ1eiþ2eiþ3ÞDO F : For the last condition we just note that RiÀ1pRiþ1pRiþ3 so RiÀ1 ¼ Riþ3 implies RiÀ1 ¼ Riþ1 which by hypothesis implies ðRi À RiÀ1Þ=2 e ðmod2 Þ: & ARTICLE IN PRESS

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Acknowledgments

I thank professor J.S. Hsia who carefully readthe previous draftsof this paper. His many suggestions andhints considerablyimprovedthe quality of this paper.

References

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