On the Solutions

of Quadratic Diophantine Equations

Takashi Yoshinaga

Received: December 13, 2009 Revised: January 9, 2010

Communicated by Don Blasius

Abstract. We determine a finite set of representatives of the set of local solutions in a maximal lattice modulo the stabilizer of the lattice in question for a quadratic Diophantine equation. Our study is based on the works of Shimura on quadratic forms, especially [AQC] and [IQD]. Indeed, as an application of the result, we present a criterion (in both global and local cases) of the maximality of the lattice of (11.6a) in [AQC]. This gives an answer to the question (11.6a). As one more global application, we investigate primitive solutions contained in a maximal lattice for the sums of squares on each vector space of dimension 4, 6, 8, or 10 over the field of rational numbers.

2010 Mathematics Subject Classification: 11D09, 11E08, 11E12 Keywords and Phrases: Maximal lattices, Quadratic Diophantine equations

1 Introduction

In this paper we study quadratic forms over global and local fields of characteristic zero, i.e. over number fields and their p-adic completions. Let F be a field of one of these two types. We let 𝔤 denote the ring of all integers in F (in both global and local cases). We denote by 𝔭 the maximal ideal of 𝔤 in the local case. Throughout the paper we mainly follow the notion and the notation in Shimura’s book [AQC] and the paper [IQD]. We denote by V an n-dimensional vector space over F. Let φ : V × V → F be a nondegenerate symmetric F -bilinear form. We denote by φ[x] the quadratic form φ(x, x) on V. By a maximal lattice L in V with respect to φ, we understand a 𝔤-lattice L in V, which is maximal among 𝔤-lattices on which the values φ[x] are contained in 𝔤. For simplicity, when φ is fixed on V, we will often refer to a maximal lattice in V, omitting reference to the φ needed to define it. All results in the paper concern only maximal lattices in V. Let SO^{φ} be the special orthogonal group of φ. In this paper we consider the set of the solutions of the quadratic Diophantine equation φ[x] = q in L, that is

L[q] = {x ∈ L | φ[x] = q},

and

L[q, 𝔟] = {x ∈ V | φ[x] = q, φ(x, L) = 𝔟},

where q ∈ 𝔤 ∩ F^{×} and a fractional ideal 𝔟 of F. Assume now that F is local, put C(L) = {γ ∈ SO^{φ} | Lγ = L}, and take h ∈ L such that φ[h] ≠ 0. It was shown by Shimura that there exists a finite subset A of SO^{φ} such that

L[φ[h]] = ⨆_{α∈A} hαC(L)

([AQC, Theorem 10.3]) and

#{L[q, 𝔟]/C(L)} ≤ 1 if n > 2

([IQD, Theorem 1.3]). Note that [AQC, Theorem 10.3] is true even when L is not maximal. In Theorem 3.5 we shall obtain, using the proof of [AQC, Theorem 10.3], an explicit complete set {hα}_{α∈A} of representatives for L[φ[h]]/C(L). Also, we show that

L[φ[h]] = \begin{cases}L[φ[h], 2^{−1}𝔭^{τ(φ[h])}] & if φ is anisotropic, \\ ⨆_{i=0}^{τ(φ[h])} L[φ[h], 2^{−1}𝔭^{i}] & if φ is isotropic, \end{cases}

with the value τ(φ[h]); see Theorem 3.5. As a result of this theorem we prove Theorem 5.3: Suppose F is local and n ≥ 2. Then

L ∩ (F h)^{⊥} is maximal in (F h)^{⊥} if and only if h ∈ L[φ[h], 2^{−1}𝔭^{τ(φ[h])}]

for h ∈ L such that φ[h] ≠ 0. Here (F h)^{⊥} = {x ∈ V | φ(x, h) = 0}. We also obtain the global version of the maximality of the lattice L ∩ (F h)^{⊥} in (F h)^{⊥} in Theorem 6.3. This theorem answers the question raised in [AQC, (11.6a)]. As a global application of Theorem 3.5, in Theorem 7.5 we give the criterion of the existence of solutions contained in L[q, Z] and L[q, 2^{−1}Z] in both cases when q is a squarefree positive integer, by taking V = Q_{n}^{1} (4 ≤ n ≤ 10, n even), the sums of squares as φ, and a maximal lattice L in V. It is known that L[q] = L[q, 2^{−1}Z] ⊔ L[q, Z]; see [AQC, (12.17)]. For example, when n = 6, the set L[q, Z] = ∅ if and only if q − 1 ∈ 4Z. When n = 10, the genus of L consists of two SO^{φ}-classes L_{10}SO^{φ} and ΛSO^{φ} (cf. [CGQ, §3.2]). In this case,

L[q, Z] = ∅ if and only if \begin{cases}L ∈ L_{10}SO^{φ}, q = 1 or q − 3 ∈ 4Z; or \\ L ∈ ΛSO^{φ}, q − 3 ∈ 4Z. \end{cases}

On the Solutions of Quadratic Diophantine Equations 349

We summarize the contents of the paper. In Section 2 we recall the notion of Shimura [AQC] and [IQD] and introduce the basic facts of a local Witt decomposition with respect to φ. In Sections 3 through 5 we treat local cases. In Section 3 we introduce the result obtained from the proof of [AQC, Theorem 10.3] and state the first result. In Section 4 we prove Theorem 3.5. In Section 5 we shall give a criterion of the maximality of the lattice L ∩ (F h)^{⊥} in (F h)^{⊥} in the local case. In Section 6 we prove the global version of Theorem 5.3. In Section 7 we prove Theorem 7.5.

I would like to express my sincere thanks to Professor Koji Doi, who read the manuscript and made several corrections and comments. I wish to thank Dr. Manabu Murata for useful suggestions during the preparation of the paper.

Notations and Conventions As usual, Z (resp. Z_{p}) is the ring of rational (resp. p-adic) integers, Q (resp. Q_{p}) the field of rational (resp. p-adic) numbers. In this paper we consider the base field F in two cases. One is a global field and the other is a local field. When we do not need to specify the case of F, we call it only “a field”.

If R is an associative ring with identity element, then R^{×} is the group of units of R. If K is a finite algebraic extension of a field F, then D_{K/F} denotes the relative discriminant of K over F. Let 𝔡_{K/F} be the different of K relative to F.

If F is a local field, then for x ∈ F^{×}, put

ξ(x) = \begin{cases}1 & if \sqrt{x} ∈ F, \\ −1 & if F (\sqrt{x}) is an unramified quadratic extension of F, \\ 0 & if F (\sqrt{x}) is a ramified quadratic extension of F \end{cases}

as in [NRQ, (3.3.1)].

If F is the field of quotients of a Dedekind domain 𝔤 and V an n-dimensional vector space over F, then by a 𝔤-lattice in V, we understand a finitely generated 𝔤-module in V that spans V over F. In particular, if 𝔞 is a 𝔤-lattice in F, we call 𝔞 a 𝔤-ideal of F. We write dim_{F} (V) for the dimension of V over F. We let GL(V, F) denote the group of all F -linear automorphisms of V. If R = F or 𝔤, then we write R_{n}^{m} for the ring of all m × n-matrices with entries in R and let GL_{n}(R) = (R_{n}^{n})^{×}.

If X is a set, then #X denotes the cardinality of X. If X is a disjoint union of its subsets Y_{1}, ⋯, Y_{m}, we write X = ⨆_{i=1}^{m} Y_{i} or X = Y_{1} ⊔ ⋯ ⊔ Y_{m}. For a subgroup H of a group G, we let [G : H] = #(H \ G).

We denote by δ_{ij} Kronecker’s delta. For a real number a, we let [a] denote the greatest integer not greater than a.

2 Preliminaries

2.1. Let F be a field and we consider the pair (V, φ) as in the introduction. Define

SO^{φ}(V) = {α ∈ GL(V, F) | det(α) = 1, φ[xα] = φ[x] for all x ∈ V }.

We understand that GL(V, F) acts on V on the right. Let φ_{0} = [φ(x_{i}, x_{j})]_{i,j=1}^{n} for an F -basis {x_{i}}_{i=1}^{n} of V, then φ_{0} ∈ GL_{n}(F) such that φ_{0} = ^{t}φ_{0}. Define the discriminant of (V, φ) by

(2.1)  δ(φ) = δ(V, φ) = (−1)^{n(n−1)/2} det(φ_{0})F^{×2}.

Let A(V) = A(V, φ) be the Clifford algebra of φ (cf. [AQC, Chap. I Section 2]). We say that (V_{1}, φ_{1}) is isomorphic to (V_{2}, φ_{2}) if there is an F -linear isomorphism f of V_{1} onto V_{2} such that φ_{1}[x] = φ_{2}[xf] for any x ∈ V_{1}. If W is a subspace of V, then we always consider (W, ψ), where ψ is the restriction of φ to W (ψ[x] = φ[x] for x ∈ W). For a 𝔤-lattice Λ in V, put

Λ̃ = Λ = {x ∈ V | φ(x, Λ) ⊂ 2^{−1}𝔤}, (2.2) C(Λ) = {γ ∈ SO^{φ}(V) | Λγ = Λ}. (2.3)

By an integral lattice L in V (with respect to φ), we understand a 𝔤-lattice L in V such that φ[x] ∈ 𝔤 for every x ∈ L. We call L maximal (with respect to φ) if it is maximal among integral lattices in V. We note that L ⊂ L̃ when L is an integral lattice in V.

2.2. Here we assume that F is a local field and L is a maximal lattice in V with respect to φ. Considering the maximality of L, we have a Witt decomposition by [AQC, Lemma 6.5];

(2.4)  V = Z + ∑_{i=1}^{r}(F f_{i} + F e_{i}), L = N + ∑_{i=1}^{r}(𝔤f_{i} + 𝔤e_{i}),

where

(2.5)  φ(e_{i}, e_{j}) = φ(f_{i}, f_{j}) = 0, φ(e_{i}, f_{j}) = 2^{−1}δ_{ij},
(2.6)  Z = {z ∈ V | φ(e_{i}, z) = φ(f_{i}, z) = 0 for all i},
(2.7)  N = {z ∈ Z | φ[z] ∈ 𝔤}.

Here the restriction of φ to Z is anisotropic and N is a unique maximal lattice in Z by [AQC, Lemma 6.4]. We say that Z is a core subspace of V with respect to φ. Until the end of Section 5, we fix these decompositions. Put t = dim_{F} (Z) then n = 2r + t. We have t ≤ 4 by [AQC, Theorem 7.6(ii)]. We call t the core dimension of (V, φ).

2.3. We introduce here the basic notions of (Z, φ) and of N, which play an important role in this paper. Note that we use the same letters c and δ, for simplification, in the following different cases (I) (2.10), (II) (2.13), and (III) (2.15). (I) Assume t = 1 (cf. [AQC, §7.1 and §7.7(I)] and [IQD, §1.5(A)]). Take g ∈ Z such that

N = 𝔤g (2.8)

and put

c = φ[g]. (2.9)

Then Z = F g and φ[xg] = cx^{2} for x ∈ F. Furthermore we obtain c ∈ 𝔤^{×} (resp. c𝔤 = 𝔭) if δ(φ) ∩ 𝔤 ≠ ∅ (resp. δ(φ) ∩ 𝔤 = ∅) by (2.7). Put

(2.10)  c𝔤 = 𝔭^{δ} with δ ∈ Z.

By (2.2) and (2.8), we easily see that

(2.11)  Ñ = 2^{−1}𝔭^{−δ}g.

(II) Next suppose t = 2 (cf. [AQC, §7.2 and §7.7(II)]). We can take g_{1}, g_{2} ∈ Z such that Z = F g_{1} + F g_{2} and φ(g_{1}, g_{2}) = 0 by [EPE, Lemma 1.8]. Put

(2.12)  b = φ[g_{1}] and c = φ[g_{2}].

Put K = F + F g_{1}g_{2} in A(Z). Then K is a quadratic extension of F, which is isomorphic to F (\sqrt{−bc}), Z = Kg_{2}, and φ[xg_{2}] = cN_{K/F} (x) for x ∈ K. We may assume c ∈ 𝔤^{×} or c𝔤 = 𝔭. Moreover when K is a ramified extension of F, we can take c ∈ 𝔤^{×}. Then by (2.7) we have N = 𝔯g_{2} if K is either unramified or ramified, where 𝔯 is the valuation ring of K. We put

(2.13)  c𝔤 = 𝔭^{δ} with δ ∈ Z.

(III) Suppose t = 3 (cf. [AQC, §7.3 and §7.7(III)] and [IQD, §1.5(B)]). There exist g_{i} ∈ Z such that Z = F g_{1} +F g_{2} +F g_{3} and φ(g_{i}, g_{j}) = 0 if i ≠ j by [EPE, Lemma 1.8]. Put

(2.14)  c = φ[g_{1}]φ[g_{2}]φ[g_{3}].

Then we can take c ∈ 𝔤^{×} (resp. c𝔤 = 𝔭) if δ(φ) ∩ 𝔤 ≠ ∅ (resp. δ(φ) ∩ 𝔤 = ∅). We put

(2.15)  c𝔤 = 𝔭^{δ} with δ ∈ Z.

Put ζ = g_{1}g_{2}g_{3}, T = F g_{1}g_{2} + F g_{2}g_{3} + F g_{3}g_{1}, and B = F + T in A(Z). Then B is a division quaternion algebra over F and T = {x ∈ B | x + x^{ι} = 0}, where ι is the main involution of B. Moreover we have Z = T ζ and φ[xζ] = cxx^{ι} for x ∈ T. Then by (2.7),

(2.16)  N = (T ∩ 𝔓^{−δ})ζ,

where 𝔓 = {x ∈ B | xx^{ι} ∈ 𝔭}. By [AQC, Theorem 5.13], there exist an unramified quadratic extension K over F and an element ω ∈ B such that B = K + Kω, aω = ωa^{ι} for each a ∈ K, and ω^{2} ∈ π𝔤^{×}. Here π is a prime element of F. Let 𝔯 be the valuation ring of K. There exists u ∈ 𝔯 such that 𝔯 = 𝔤[u] and u − u^{ι} ∈ 𝔯^{×} by [AQC, Lemma 5.7]. Put v = u − u^{ι}. Then T = F v + Kω. For a, α ∈ 𝔤 and b, β ∈ 𝔯,

(2.17)  φ[(av + bω^{1−2δ})ζ] = −c(a^{2}v^{2} + ω^{2(1−2δ)}N_{K/F} (b)),
φ((av + bω^{1−2δ})ζ, (αv + βω^{1−2δ})ζ) = −2^{−1}c(2aαv^{2} + ω^{2(1−2δ)}T r_{K/F} (bβ^{ι})). (2.18)

From (2.16) and (2.17),

(2.19)  N = (𝔤v + 𝔯ω^{1−2δ})ζ = (𝔤v + 𝔤ω^{1−2δ} + 𝔤uω^{1−2δ})ζ.

From (2.2) and (2.19),

(2.20)  Ñ = (2^{−1}𝔭^{−δ}v + 𝔯ω^{−1})ζ.

Put T r_{B/F} (x) = x + x^{ι} and N_{B/F} (x) = xx^{ι} for x ∈ B. (IV) Finally assume t = 4 (cf. [AQC, Theorem 7.5 and §7.7(IV)]). There exist a division quaternion algebra B over F and an F -linear isomorphism γ : B → Z such that φ[xγ] = xx^{ι} for x ∈ B, where ι is the main involution of B. Then N = 𝔒γ. Here 𝔒 is the unique maximal order of B.

3 A complete set of representatives for L[q]/C(L)

Until the end of Section 5, we assume that F is a local field and L is a maximal lattice in V with respect to φ. In this section, we first introduce the facts obtained from the proof of [AQC, Theorem 10.3]. After that, we state our first main theorem.

3.1. We suppose that V and L are represented as in (2.4). If r ≥ 1, put M = N + ∑_{i=2}^{r}(𝔤f_{i} + 𝔤e_{i}). We consider M = N if r = 1. Then

(3.1)  L = 𝔤f_{1} + M + 𝔤e_{1}

for every r ≥ 1. For 0 ≤ i ∈ Z and q ∈ 𝔤 ∩ F^{×}, put

(3.2)  X_{i}(q) = {x ∈ M | φ[x] − q ∈ 𝔭^{i}}.

Note that X_{i}(q) ⊃ X_{i+1}(q). Hereafter we take a prime element π of F and fix it. We obtain the following theorem from the proof of [AQC, Theorem 10.3].

3.2 Theorem. (Shimura) Let the notation be as above. Let h ∈ L such that φ[h] ≠ 0 and ν ∈ Z such that φ[h]𝔤 = 𝔭^{ν}. Put C = C(L) in the notation of (2.3). Let t, e_{1}, and f_{1} be as in §2.2.

(1) Suppose r = 0. Then

L[φ[h]] = \begin{cases}hC ⊔ (−h)C, C = \{1\} & if t = 1, \\ hC & if t > 1. \end{cases}

(2) Suppose n = 2r = 2. Then L[φ[h]] = ⨆_{i=0}^{ν}(π^{i}f_{1} + φ[h]π^{−i}e_{1})C.

(3) Suppose n > 2, r > 0, and M[φ[h]] = ∅. Put

(3.3)  κ_{0} = min({k ∈ Z | X_{k}(φ[h]) = ∅}).

Then

L[φ[h]] = ⋃_{i=0}^{κ_{0}−1} ⋃_{b∈X_{i}(φ[h])/𝔭^{i}M}[π^{i}f_{1} + b + π^{−i}(φ[h] − φ[b])e_{1}]C.

Here b runs over all elements of X_{i}(φ[h])/𝔭^{i}M.

(4) Suppose that n > 2, r > 0, M[φ[h]] ≠ ∅, and that there exists a finite subset B of M[φ[h]] such that M[φ[h]] = ⊔_{b∈B}bC(M). Then

L[φ[h]] = ⋃_{b∈B} ⋃_{y∈𝔤/2φ(b,M)}(b + ye_{1})C.

3.3 Lemma. Let the notation be the same as in Theorem 3.2. We let q ∈ 𝔤 ∩ F^{×}. Assume r ≥ 2. If there are a finite number of elements x_{0}, ⋯, x_{τ} of M such that

(3.4)  M[q] = ⊔_{i=0}^{τ}x_{i}C(M) and φ(x_{i}, M) = 2^{−1}𝔭^{i},

then we have L[q] = ⊔_{i=0}^{τ}x_{i}C and φ(x_{i}, L) = 2^{−1}𝔭^{i}.

Proof. From (3.4) and Theorem 3.2(4),

(3.5)  L[q] = ⋃_{i=0}^{τ} ⋃_{y∈𝔤/𝔭^{i}}(x_{i} + ye_{1})C.

We fix 0 ≤ i ≤ τ. By (2.5), (2.6), (3.1), and (3.4),

φ(x_{i} + ye_{1}, L) = φ(x_{i}, M) + 2^{−1}y𝔤 = \begin{cases}2^{−1}𝔭^{i} & if y ∈ 𝔭^{i}, \\ 2^{−1}y𝔤 & if y ∉ 𝔭^{i}. \end{cases}

From this and [IQD, Theorem 1.3],

(x_{i} + ye_{1})C = L[q, φ(x_{i} + ye_{1}, L)] = \begin{cases}L[q, 2^{−1}𝔭^{i}] & if y ∈ 𝔭^{i}, \\ L[q, 2^{−1}y𝔤] & if y ∉ 𝔭^{i}. \end{cases} (3.6)

For y ∈ 𝔤 such that y ∉ 𝔭^{i}, if y𝔤 = 𝔭^{j} then 0 ≤ j ≤ i − 1. Thus we see that ∪_{y∈𝔤/𝔭^{i}}(x_{i} + ye_{1})C = ⊔_{j=0}^{i}L[q, 2^{−1}𝔭^{j}] and L[q, 2^{−1}𝔭^{i}] = x_{i}C by (3.6). From this and (3.5) we obtain

L[q] = ⋃_{i=0}^{τ}[ ⨆_{j=0}^{i} L[q, 2^{−1}𝔭^{j}]] = ⨆_{i=0}^{τ} L[q, 2^{−1}𝔭^{i}] = ⨆_{i=0}^{τ} x_{i}C.

Clearly φ(x_{i}, L) = 2^{−1}𝔭^{i} by (2.5), (2.6), (3.1), and (3.4). This completes the proof.

3.4 Lemma. In the Witt decomposition of V of (2.4), let N be as in (2.7). Let q be an element of 𝔤 ∩ F^{×} and ξ as in Notation. Let t and c be as in §2.2 and §2.3, respectively. Then we obtain the following assertions:

(1) If t = 1, then N[q] ≠ ∅ if and only if ξ(cq) = 1.

(2) Assume t = 2. Let K, 𝔯, and δ be as in §2.3(II). Let ν ∈ Z such that q𝔤 = 𝔭^{ν}. Then N[q] ≠ ∅ if and only if c^{−1}q ∈ N_{K/F} (𝔯). Moreover if K is unramified over F, then this is the case if and only if ν ≡ δ (mod 2).

(3) If t = 3, then N[q] ≠ ∅ if and only if ξ(−cq) ≠ 1.

(4) If t = 4, then we have N[q] ≠ ∅ for all q ∈ 𝔤 ∩ F^{×}.

(5) Let L be a maximal lattice in V and r as in (2.4). If r > 0, then we have L[q] ≠ ∅ for all q ∈ 𝔤 ∩ F^{×}.

Proof. We may assume that:

if t = 1, then Z = F, N = 𝔤, and φ[x] = cx^{2} for x ∈ F; if t = 2, then Z = K, N = 𝔯, and φ[x] = cN_{K/F} (x) for x ∈ K; if t = 3, then Z = T, N = T ∩ 𝔓^{−δ}, and φ[x] = cN_{B/F} (x) = −cx^{2} for x ∈ T; if t = 4, then Z = B, N = 𝔒, and φ[x] = N_{B/F} (x) for x ∈ B

in (2.4); see §2.3. Then (1) and the first statement of (2) are trivial. We prove the second assertion of (2). Assume that t = 2 and K is unramified over F, then π𝔯 = 𝔮 and N_{K/F} (𝔯^{×}) = 𝔤^{×} by [BNT, Chapter VIII, Proposition 3]. Here 𝔮 is the maximal ideal of 𝔯. From these, we obtain the second assertion of (2). Assume t = 3. Noticing that B is division, the “only if”-part of (3) is immediate. If ξ(−cq) ≠ 1, then F (\sqrt{−c^{−1}q}) is a quadratic extension of F, and hence there exists z ∈ B such that z ∉ F and z^{2} = −c^{−1}q by [AQC, Proposition 5.15(ii)]. We easily see that z ∈ T ∩ 𝔓^{−δ}, and hence N[q] ≠ ∅. Assume t = 4. Then we see that N_{B/F} (𝔒) = 𝔤 from [AQC, Theorem 5.13] and [AQC, Proposition 5.15(i)]. This implies (4). Finally we prove (5). Assume r > 0. Since L can be represented as in (2.4), we have f_{1} + qe_{1} ∈ L[q] for every q ∈ 𝔤 ∩ F^{×} with e_{1} and f_{1} in (2.4). This completes the proof.

Now, our first main result in this paper can be stated as follows:

3.5 Theorem. Let L be a maximal lattice in V and put C = C(L) as in Theorem 3.2. Let ξ be as in Notation. Let q ∈ 𝔤 ∩ F^{×} and ν ∈ Z such that q𝔤 = 𝔭^{ν}. Let κ ∈ Z such that 2𝔤 = 𝔭^{κ}. Let r, t, e_{r}, f_{r}, and N be as in §2.2. For 0 ≤ i ∈ Z and x ∈ N, put

(3.7) h_{i,x} = x + π^{i}e_{r}, k_{i,x} = π^{i}f_{r} + x + \frac{q − φ[x]}{π^{i}}e_{r}, ℓ_{i} = π^{i}f_{r} + qπ^{−i}e_{r}.

Then we have

L[q] = ⨆_{u∈R} uC = \begin{cases}L[q, 2^{−1}𝔭^{τ(q)}] & if r = 0, \\ ⨆_{i=0}^{τ(q)} L[q, 2^{−1}𝔭^{i}] & if r ≥ 1. \end{cases} (3.8)

Here the set R and the index τ(q) are defined as follows:

(i) Suppose t = 0 and r ≥ 1. Then

R = \begin{cases}\{ℓ_{i}\}_{i=0}^{ν} & if r = 1, \\ \{ℓ_{i}\}_{i=0}^{τ(q)} & if r ≥ 2, \end{cases}
(3.9)  τ(q) = [ν/2].

Moreover

(3.10)  L[q,2^{−1}𝔭^{i}] = \begin{cases}ℓ_{i}C ⊔ ℓ_{ν−i}C & if r = 1 and 0 ≤ i < ν/2, \\ ℓ_{ν/2}C & if r = 1, ν ∈ 2Z, and i = ν/2, \\ ℓ_{i}C & if r ≥ 2. \end{cases}

(ii) Suppose t = 1. Let c be as in (2.9) and δ as in (2.10). Let us define an integer d ∈ Z as follows: D_{F (\sqrt{cq})/F} = 𝔭^{d} when ξ(cq) = 0 (in the ordinary sense) and d = 1 when ξ(cq) = −1 (This is only for a simplification of the following statements (3.11) and (3.12)). When ξ(cq) = 1, we take any element y of N[q] and fix it (By Lemma 3.4(1), N[q] ≠ ∅). When 2 ∈ 𝔭, ξ(cq) ≠ 1, and ν ≡ δ (mod 2), take any element z of N[sq] and fix it, with

(3.11)  s ∈ 1 + π^{2κ+1−d}𝔤^{×} such that c^{−1}qπ^{δ−ν} ∈ s^{−1}𝔤^{×2}.

(As for the existence of s and z, see (4.30) and (4.31), respectively.) Then R and τ(q) are given as follows:

R = \begin{cases}\{±y\} & if r = 0 and ξ(cq) = 1, \\ \{y\} ⊔ \{h_{i,y}\}_{i=0}^{τ(q)−1} & if r ≥ 1 and ξ(cq) = 1, \\ \{k_{i,z}\}_{i=0}^{τ(q)} & if r ≥ 1, ξ(cq) ≠ 1, \\ & ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ \{ℓ_{i}\}_{i=0}^{τ(q)} & otherwise, \end{cases}

(3.12) τ(q) = \begin{cases}κ + \frac{ν+δ}{2} & if ξ(cq) = 1, \\ κ + [\frac{ν+1−d}{2}] & if ξ(cq) ≠ 1, ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ [\frac{ν}{2} ] & otherwise. \end{cases}

Moreover

(3.13) L[q,2^{−1}𝔭^{i}] = \begin{cases}yC ⊔ (−y)C & if r = 0, ξ(cq) = 1, and i = τ(q), \\ yC & if r ≥ 1, ξ(cq) = 1, and i = τ(q), \\ h_{i,y}C & if r ≥ 1, ξ(cq) = 1, and i < τ(q), \\ k_{i,z}C & if r ≥ 1, ξ(cq) ≠ 1, \\ & ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ ℓ_{i}C & otherwise. \end{cases}

(iii) Suppose t = 2. Let c and δ be as in (2.12) and (2.13), respectively. Let K and 𝔯 be as in §2.3(II). Let 𝔡 be the different of K relative to F. Let d ∈ Z such that D_{K/F} = 𝔭^{d} when 𝔡 ≠ 𝔯. Put d = 1 when 𝔡 = 𝔯 (This is the same simplification as in (ii)). When c^{−1}q ∈ N_{K/F} (𝔯), we take any element y of N[q] and fix it (By Lemma 3.4(2), N[q] ≠ ∅). When c^{−1}q ∉ N_{K/F} (𝔯) and d > 1, we take any element z of N[sq] and fix it, with

(3.14)  s ∈ 1 + π^{d−1}𝔤^{×} such that c^{−1}qπ^{−ν} ∈ s^{−1}N_{K/F} (𝔯^{×}).

(As for the existence of s and z, see (4.32) and (4.33), respectively.) Then R and τ(q) are given as follows:

R = \begin{cases}\{y\} & if r = 0 and c^{−1}q ∈ N_{K/F} (𝔯), \\ \{y\} ⊔ \{h_{i,y}\}_{i=0}^{τ(q)−1} & if r ≥ 1 and c^{−1}q ∈ N_{K/F} (𝔯), \\ \{k_{i,z}\}_{i=0}^{τ(q)} & if r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and d > 1, \\ \{ℓ_{i}\}_{i=0}^{τ(q)} & otherwise, \end{cases}

(3.15)  τ(q) = \begin{cases}\frac{ν+δ}{2} & if c^{−1}q ∈ N_{K/F} (𝔯) and 𝔡 = 𝔯, \\ [\frac{ν+d}{2}] & if c^{−1}q ∈ N_{K/F} (𝔯) and 𝔡 ≠ 𝔯, \\ [\frac{ν+d−1}{2}] & otherwise. \end{cases}

Moreover

(3.16) L[q,2^{−1}𝔭^{i}] = \begin{cases}yC & if c^{−1}q ∈ N_{K/F} (𝔯) and i = τ(q), \\ h_{i,y}C & if c^{−1}q ∈ N_{K/F} (𝔯) and i < τ(q), \\ k_{i,z}C & if c^{−1}q ∉ N_{K/F} (𝔯) and d > 1, \\ ℓ_{i}C & otherwise. \end{cases}

(iv) Suppose t = 3. Let c and δ be as in (2.14) and (2.15), respectively. When ξ(−cq) ≠ 1, we take any element y of N[q] and fix it (By Lemma 3.4(3), N[q] ≠ ∅). When ξ(−cq) = 1 and 2 ∈ 𝔭, we take any element z of N[sq] and fix it, with

(3.17)  s ∈ 1 + 4𝔤 such that s ∉ 𝔤^{×2}.

(As for the existence of s and z, see (4.34) and (4.35), respectively.) Then R and τ(q) are given as follows:

R = \begin{cases}\{y\} & if r = 0 and ξ(−cq) ≠ 1, \\ \{y\} ⊔ \{h_{i,y}\}_{i=0}^{τ(q)−1} & if r ≥ 1 and ξ(−cq) ≠ 1, \\ \{k_{i,z}\}_{i=0}^{τ(q)} & if r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭, \\ \{ℓ_{i}\}_{i=0}^{τ(q)} & if r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔤^{×}, \end{cases}

(3.18) τ(q) = \begin{cases}κ + [\frac{ν}{2} ] & if ξ(−cq) = 1, \\ \frac{ν−δ+1}{2} & if ξ(−cq) ≠ 1 and ν ≢ δ (mod 2), \\ κ + 1 + \frac{ν−δ−d}{2} & if ξ(−cq) = 0, ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ κ + \frac{ν+δ}{2} & otherwise, \end{cases}

where d ∈ Z such that D_{F (\sqrt{−cq})/F} = 𝔭^{d}. Moreover

(3.19) L[q, 2^{−1}𝔭^{i}] = \begin{cases}yC & if ξ(−cq) ≠ 1 and i = τ(q), \\ h_{i,y}C & if ξ(−cq) ≠ 1 and i < τ(q), \\ k_{i,z}C & if ξ(−cq) = 1 and 2 ∈ 𝔭, \\ ℓ_{i}C & if ξ(−cq) = 1 and 2 ∈ 𝔤^{×}. \end{cases}

(v) Suppose t = 4. Take any element y of N[q] and fix it. Then R and τ(q) are given as follows:

R = \begin{cases}\{y\} & if r = 0, \\ \{y\} ⊔ \{h_{i,y}\}_{i=0}^{τ(q)−1} & if r ≥ 1, \end{cases}
(3.20)  τ(q) = [(ν + 1)/2].

Moreover

(3.21)  L[q,2^{−1}𝔭^{i}] = \begin{cases}yC & if i = τ(q), \\ h_{i,y}C & if i < τ(q). \end{cases}

The proof of this theorem will be given in the following Section 4. Here we insert one elementary lemma:

3.6 Lemma. Let F be a local field and L a maximal lattice in V. Let κ ∈ Z such that 2𝔤 = 𝔭^{κ}. Then for q ∈ 𝔤 and i ∈ Z, we have L[q, 𝔭^{i}] ⊂ L[q] if and only if i ≥ −κ.

Proof. From (2.2), clearly i ≥ −κ if and only if L[q, 𝔭^{i}] ⊂ L̃[q]. Here [AQC, Lemma 6.2(3)] implies L̃[q] = L[q]. This proves the lemma.

3.7 Corollary. Let the notation be the same as in Theorem 3.5. Assume L[q] ≠ ∅ for q ∈ 𝔤 ∩ F^{×}. Then for every i ∈ Z,

L[q, 2^{−1}𝔭^{i}] ≠ ∅ ⟺ \begin{cases}i = τ(q) & if n = t, \\ i ≤ τ(q) & otherwise. \end{cases}

Proof. For 0 ≤ i ∈ Z, the result follows from Theorem 3.5 and Lemma 3.6. Assume i < 0. Clearly L[q, 2^{−1}𝔭^{i}] = π^{i} · L[π^{−2i}q, 2^{−1}𝔤]. Here Lemma 3.6 implies L[π^{−2i}q, 2^{−1}𝔤] ⊂ L[π^{−2i}q]. Since π^{i} · L[π^{−2i}q] = (π^{i}L)[q] ⊃ L[q], we obtain L[π^{−2i}q] ≠ ∅. Applying Theorem 3.5 to (V, φ), L, and π^{−2i}q, we find that if n = t then L[π^{−2i}q] = L[π^{−2i}q, 2^{−1}𝔭^{τ(π^{−2i}q)}] and τ(π^{−2i}q) = τ(q) − i > 0. Therefore L[π^{−2i}q, 2^{−1}𝔤] = ∅, and hence L[q, 2^{−1}𝔭^{i}] = ∅. If n ≠ t, then L[π^{−2i}q, 2^{−1}𝔤] ≠ ∅ by Theorem 3.5, and hence L[q, 2^{−1}𝔭^{i}] ≠ ∅. This completes the proof.

4 Proof of Theorem 3.5

4.1. We first prove Theorem 3.5(i). Assume t = 0. Then L[q] ≠ ∅ for all q ∈ 𝔤 ∩ F^{×} by Lemma 3.4(5). First suppose r = 1. Then L = 𝔤f_{1} + 𝔤e_{1} by (2.4). We obtain

(4.1)  L[q] = ⨆_{i=0}^{ν} ℓ_{i}C

by Theorem 3.2(2) with ℓ_{i} in (3.7). We have clearly

φ(ℓ_{i}, L) = 2^{−1}π^{i}𝔤 + 2^{−1}qπ^{−i}𝔤 = \begin{cases}2^{−1}𝔭^{i} & if 0 ≤ i ≤ [ν/2], \\ 2^{−1}𝔭^{ν−i} & if i > [ν/2] \end{cases} (4.2)

from (2.5). Assume ν ∈ 2Z. Then

(4.3)  φ(ℓ_{i}, L) = φ(ℓ_{ν−i}, L) = 2^{−1}𝔭^{i}

for 0 ≤ i ≤ (ν − 2)/2 and φ(ℓ_{ν/2}, L) = 2^{−1}𝔭^{ν/2} by (4.2). Thus ℓ_{i}C ⊔ ℓ_{ν−i}C ⊂ L[q, 2^{−1}𝔭^{i}] for 0 ≤ i ≤ (ν − 2)/2 and ℓ_{ν/2}C ⊂ L[q, 2^{−1}𝔭^{ν/2}]. On the other hand, we have L[q, 2^{−1}𝔭^{i}] ⊂ L[q] for 0 ≤ i ≤ ν/2 by Lemma 3.6. Hence ℓ_{i}C ⊔ ℓ_{ν−i}C = L[q, 2^{−1}𝔭^{i}] for 0 ≤ i ≤ (ν − 2)/2 and ℓ_{ν/2}C = L[q, 2^{−1}𝔭^{ν/2}]. From this and (4.1) we obtain the assertion in the case r = 1, t = 0, and ν ∈ 2Z. Next assume ν ∉ 2Z. Then

(4.4)  φ(ℓ_{i}, L) = φ(ℓ_{ν−i}, L) = 2^{−1}𝔭^{i}

for 0 ≤ i ≤ (ν − 1)/2. Thus ℓ_{i}C ⊔ ℓ_{ν−i}C = L[q, 2^{−1}𝔭^{i}] for 0 ≤ i ≤ (ν − 1)/2, in the same manner as in the case ν ∈ 2Z. This proves the assertion when r = 1 and t = 0. Next suppose r = 2. Then L = 𝔤f_{1} + M + 𝔤e_{1} by (3.1). We obtain

(4.5)  ∅ ≠ M[q] = ⊔_{i=0}^{ν}ℓ_{i}C(M)

by Theorem 3.2(2) and Lemma 3.4(5). In this case we can not apply Lemma 3.3 since we have (4.3) and (4.4) in the notation of (4.1). By (4.5) and Theorem 3.2(4),

L[q] = ⋃_{i=0}^{ν} ⋃_{a∈𝔤/2φ(ℓ_{i},M)}(ℓ_{i} + ae_{1})C.

Since

2φ(ℓ_{i}, M) = \begin{cases}𝔭^{i} & if 0 ≤ i ≤ [ν/2], \\ 𝔭^{ν−i} & if i > [ν/2] \end{cases}

by (4.2), we have

(4.6)  L[q] = [ ⋃_{i=0}^{[ν/2]} ⋃_{a∈𝔤/𝔭^{i}}(ℓ_{i} + ae_{1})C] ⋃[ ⋃_{j>[ν/2]}^{ν} ⋃_{b∈𝔤/𝔭^{ν−j}}(ℓ_{j} + be_{1})C].

For 0 ≤ i ≤ [ν/2] and a ∈ 𝔤/𝔭^{i},

φ(ℓ_{i} + ae_{1}, L) = φ(ℓ_{i}, M) + 2^{−1}a𝔤 = \begin{cases}2^{−1}𝔭^{i} & if a ∈ 𝔭^{i}, \\ 2^{−1}a𝔤 & if a ∉ 𝔭^{i} \end{cases}

by (2.5) and (3.1). Therefore by [IQD, Theorem 1.3],

(4.7)  (ℓ_{i} + ae_{1})C = L[q, φ(ℓ_{i} + ae_{1}, L)] = \begin{cases}L[q, 2^{−1}𝔭^{i}] & if a ∈ 𝔭^{i}, \\ L[q, 2^{−1}a𝔤] & if a ∉ 𝔭^{i}. \end{cases}

Similarly we have

(4.8)  (ℓ_{j} + be_{1})C = \begin{cases}L[q, 2^{−1}𝔭^{ν−j}] & if b ∈ 𝔭^{ν−j}, \\ L[q, 2^{−1}b𝔤] & if b ∉ 𝔭^{ν−j} \end{cases}

for j > [ν/2] and b ∈ 𝔤/𝔭^{ν−j}. From (4.7) and (4.8), the argument in the proof of Lemma 3.3 shows that

(4.9)  ⋃_{i=0}^{[ν/2]} ⋃_{a∈𝔤/𝔭^{i}}(ℓ_{i} + ae_{1})C = ⨆_{i=0}^{[ν/2]} ℓ_{i}C, ℓ_{i}C = L[q, 2^{−1}𝔭^{i}],

(4.10)  ⋃_{j>[ν/2]}^{ν} ⋃_{b∈𝔤/𝔭^{ν−j}}(ℓ_{j} + be_{1})C = ⨆_{j>[ν/2]}^{ν} ℓ_{j}C, ℓ_{j}C = L[q, 2^{−1}𝔭^{ν−j}].

Combining (4.6), (4.9), and (4.10), we have

L[q] = ( ⨆_{i=0}^{[ν/2]} ℓ_{i}C) ∪ ( ⨆_{j>[ν/2]}^{ν} ℓ_{j}C) = ⨆_{i=0}^{[ν/2]} L[q, 2^{−1}𝔭^{i}] = ⨆_{i=0}^{[ν/2]} ℓ_{i}C

and L[q, 2^{−1}𝔭^{i}] = ℓ_{i}C. This proves our theorem in the case r = 2 and t = 0. As for the case r ≥ 3, we apply (repeatedly, if necessary) Lemma 3.3 and [IQD, Theorem 1.3] to this case, we can reduce the proof to the case r = 2. This completes the proof of (i).

4.2 Lemma. Let F be a local field. Assume 2 ∈ 𝔭 and let κ ∈ Z such that 2𝔤 = 𝔭^{κ}. Let ξ be as in Notation. Put

(4.11)  ε(a) = max({e ∈ Z | e ≤ 2κ + 1 and a ∈ (1 + 𝔭^{e})𝔤^{×2}})

for a ∈ 𝔤^{×}. Then we obtain the following assertions:

(1) For a ∈ 𝔤^{×},

ε(a) = \begin{cases}2κ + 1 − d & if ξ(a) = 0, \\ 2κ & if ξ(a) = −1, \\ 2κ + 1 & if ξ(a) = 1, \end{cases} (4.12)

where d ∈ Z such that D_{F (\sqrt{a})/F} = 𝔭^{d}.

(2) If ξ(a) = 0, then we have 2κ > ε(a) ∉ 2Z.

(3) If 0 < ℓ < 2κ and ℓ ∈ Z, ∉ 2Z, then 𝔤^{×2} ∩ (1 + π^{ℓ}𝔤^{×}) = ∅.

(4) If a ∈ (1 + π^{ℓ}𝔤^{×})𝔤^{×2} with 0 < ℓ ∈ Z, ∉ 2Z and ε(a) < 2κ, then ε(a) = ℓ.

(5) If a ∈ (1 + π^{ℓ}𝔤^{×})𝔤^{×2} with 0 < ℓ ∈ Z, ∉ 2Z and ε(a) = 2κ + 1, then ε(a) ≤ ℓ.

Proof. Assertions (1) and (2) are in [NRQ, Lemma 3.5]. (3): If there exists an element b ∈ 𝔤^{×} such that b^{2} ∈ 1 + π^{ℓ}𝔤^{×}, then b ∈ Z_{ℓ}. Here Z_{ℓ} = {x ∈ 𝔤^{×} | x^{2} − 1 ∈ 𝔭^{ℓ}} as in [NRQ, §3.4]. By [NRQ, (3.5.1)], Z_{ℓ} = 1 + 𝔭^{(ℓ+1)/2}, and hence we can take y ∈ 𝔤 such that b = 1 + π^{(ℓ+1)/2}y. Then b^{2} = 1 + π^{ℓ+1}y(2π^{−(ℓ+1)/2}+y) ∈ 1+𝔭^{ℓ+1} since 2^{−1}(ℓ+1) ≤ κ. This gives a contradiction. Thus we obtain (3). (4): We find ℓ ≤ ε(a) from (4.11), (4.12), and [NRQ, Lemma 3.2(1)]. Clearly

(4.13)  (1 + π^{ℓ}𝔤^{×})𝔤^{×2} ∩ (1 + 𝔭^{ε(a)})𝔤^{×2} ≠ ∅.

If ℓ < ε(a), then (4.13) contradicts (3) settled above, and hence ℓ = ε(a). (5): By (4.12), we have ξ(a) = 1. Thus

(4.14)  𝔤^{×2} ∩ (1 + π^{ℓ}𝔤^{×}) ≠ ∅.

If ℓ < ε(a) = 2κ + 1, then (4.14) contradicts (3). This completes the proof.

4.3. Now we prove (ii), (iii), (iv), and (v) of Theorem 3.5. We may assume that:

if t = 1, then Z = F, N = 𝔤, and φ[x] = cx^{2} for x ∈ F; if t = 2, then Z = K, N = 𝔯, and φ[x] = cN_{K/F} (x) for x ∈ K; if t = 3, then Z = T, N = 𝔤v + 𝔯ω^{1−2δ}, and φ[x] = cN_{B/F} (x) for x ∈ T; if t = 4, then Z = B, N = 𝔒, and φ[x] = N_{B/F} (x) for x ∈ B.

Then for x, w ∈ Z,

(4.15) φ(x, w) = \begin{cases}cxw & if t = 1, \\ 2^{−1}cT r_{K/F} (xw^{ρ}) & if t = 2, \\ 2^{−1}cT r_{B/F} (xw^{ι}) & if t = 3, \\ 2^{−1}T r_{B/F} (xw^{ι}) & if t = 4. \end{cases}

Here ρ ∈ Gal(K/F) such that ρ ≠ 1. In this §4.3 we prove the theorem in the case r = 0 and t > 0. Note that L = N in this case. If L[q] ≠ ∅, then

(4.16) L[q] = \begin{cases}yC ⊔ (−y)C & if t = 1, \\ yC & otherwise \end{cases}

by Theorem 3.2(1). Here, y is any element of L[q] and fix it until the end of §4.3. This proves the first equality of (3.8) in this case. From Lemma 3.6 and (4.16),

(4.17)  L[q] = L[q, φ(y, L)]

for 1 ≤ t ≤ 4. Note that φ(y, L) ⊂ 2^{−1}𝔤 since L is an integral lattice in V. To prove the second equality of (3.8) we determine the ideal φ(y, L) as the next step.

(4.18)  We let μ denote the normalized order function of F.

First suppose t = 1, then C = {1}. Here Lemma 3.4(1) implies that L[q] ≠ ∅ if and only if ξ(cq) = 1. Since y^{2} = c^{−1}q, we have φ(y, L) = cy𝔤 = 𝔭^{(ν+δ)/2} by (4.15). Next suppose t = 2. We have L[q] ≠ ∅ if and only if c^{−1}q ∈ N_{K/F} (𝔯) by Lemma 3.4(2). From [BNT, Chapter VIII, Propositon 4] and (4.15), we see that

φ(y, L) = 2^{−1}cT r_{K/F} (y𝔯) = \begin{cases}2^{−1}𝔭^{(ν+δ)/2} & if 𝔡 = 𝔯, \\ 2^{−1}𝔭^{[(ν+d)/2]} & if 𝔡 ≠ 𝔯. \end{cases}

Note that we take c ∈ 𝔤^{×} if K is ramified over F; see §2.3. Suppose t = 3. By Lemma 3.4(3),

(4.19)  L[q] ≠ ∅ if and only if ξ(−cq) ≠ 1.

Take m ∈ Z such that φ(y, L) = 2^{−1}𝔭^{m}. Let us determine m. Let μ_{K} be the normalized order function of K. Since L = 𝔤v + 𝔯ω^{1−2δ} = v(𝔤 + 𝔯ω^{1−2δ}), we can put y = v(a + bω^{1−2δ}) with a ∈ 𝔤 and b ∈ 𝔯. Then by (2.18), φ(y, L) = 2^{−1}c(2a𝔤 + ω^{2(1−2δ)}T r_{K/F} (b𝔯)), and hence

(4.20)  m = min(κ + δ + μ(a), 1 + μ_{K}(b) − δ).

We have also

(4.21)  q = φ[y] = −cv^{2}(a^{2} − ω^{2(1−2δ)}N_{K/F} (b))

by (2.17), and hence

(4.22)  ν − δ = min(2μ(a), 1 + 2(μ_{K}(b) − δ)).

Assume ν ≢ δ (mod 2). Then ν − δ = 1 + 2(μ_{K}(b) − δ) < 2μ(a) by (4.22), and hence 2^{−1}(ν−δ+1) = 1+μ_{K}(b)−δ ≤ κ+δ+μ(a). Thus we find m = 2^{−1}(ν−δ+ 1) by (4.20). Next assume ν ≡ δ (mod 2). Then ν−δ = 2μ(a) < 1+2(μ_{K}(b)−δ) by (4.22). If 2 ∈ 𝔤^{×}, then κ + 2^{−1}(ν + δ) = κ + δ + μ(a) ≤ 1 + μ_{K}(b) − δ, and hence m = κ + 2^{−1}(ν + δ) from (4.20). If 2 ∈ 𝔭, then Lemma 4.2 implies ε(−c^{−1}qπ^{δ−ν}) ≤ 2κ since ξ(−c^{−1}qπ^{δ−ν}) ≠ 1 by (4.19). Put β = −c^{−1}qπ^{δ−ν}, then

(4.23)  ε(v^{−2}β) = \begin{cases}ε(β) & if ε(β) < 2κ, \\ 2κ + 1 & if ε(β) = 2κ; \end{cases}

see the proof of [NRQ, Lemma 4.5]. By (4.21), (4.24) v^{−2}β = (π^{(δ−ν)/2}a)^{2}(1−ω^{2(1−2δ)}a^{−2}N_{K/F} (b)) ∈ (1−ω^{2(1−2δ)}a^{−2}N_{K/F} (b))𝔤^{×2}.

Hence if ε(β) < 2κ, that is ξ(−cq) = 0 from Lemma 4.2, then Lemma 4.2(4) and (4.23) imply ε(β) = 1 − 2δ − 2μ(a) + 2μ_{K}(b), and hence 1 + μ_{K}(b) − δ = 2^{−1}(ε(β)+1)+2^{−1}(ν −δ) ≤ κ+μ(a)+δ. Thus m = 2^{−1}(ε(β)+1)+2^{−1}(ν −δ) from (4.20). Here Lemma 4.2(1) implies ε(β) = 2κ + 1 − d, where d ∈ Z such that D_{F (\sqrt{−cq})/F} = 𝔭^{d}. Therefore m = κ + 1 + (ν − δ − d)/2. If ε(β) = 2κ, then Lemma 4.2(5), (4.23), and (4.24) imply 2κ + 1 ≤ 1 − 2δ − 2μ(a) + 2μ_{K}(b), and hence 1 + μ_{K}(b) − δ ≥ κ + μ(a) + δ. Thus m = κ + μ(a) + δ = κ + (ν + δ)/2 from (4.20). Consequently we obtain

φ(y, L) = \begin{cases}2^{−1}𝔭^{(ν−δ+1)/2} & if ν ≢ δ (mod 2), \\ 𝔭^{1+(ν−δ−d)/2} & if ν ≡ δ (mod 2), 2 ∈ 𝔭, ξ(−cq) = 0, \\ 𝔭^{(ν+δ)/2} & otherwise \end{cases} (4.25)

under the assumption t = 3, r = 0, and ξ(−cq) ≠ 1. Here d ∈ Z such that D_{F (\sqrt{−cq})/F} = 𝔭^{d}. We see the second equality of (3.8) by (4.17) and (4.25). Moreover combining this with (4.16), we obtain the theorem in the case r = 0 and t = 3. Finally suppose t = 4. Then Lemma 3.4(4) implies L[q] ≠ ∅ for every q ∈ 𝔤 ∩ F^{×}. From [AQC, Theorem 5.9(2), (6), (7)] and (4.15), we see that

(4.26)  φ(y, L) = 2^{−1}T r_{B/F} (𝔓^{μ(N_{B/F} (y))}) = 2^{−1}𝔭^{[(ν+1)/2]},

where 𝔓 = {x ∈ 𝔒 | N_{B/F} (x) ∈ 𝔭}, with μ of (4.18). This completes the proof of our theorem in the case r = 0 and t > 0.

4.4. Here we prove the theorem in the case r ≥ 1 and t > 0. First we note that when r ≥ 2 we apply (repeatedly, if necessary) Lemma 3.3 and [IQD, Theorem 1.3] to this case, we can reduce the proof to the case r = 1. Thus hereafter until the end of §4.7, we assume r = 1. Then L = 𝔤f_{1} +N +𝔤e_{1} by (2.4). Assume 1 ≤ t ≤ 4. We recall here Lemma 3.4. We know that L[q] ≠ ∅ for all q ∈ 𝔤 ∩ F^{×} and

N[q] ≠ ∅ ⟺ \begin{cases}ξ(cq) = 1 & if t = 1, \\ c^{−1}q ∈ N_{K/F} (𝔯) & if t = 2, \\ ξ(−cq) ≠ 1 & if t = 3. \end{cases} (4.27)

When t = 4, we have N[q] ≠ ∅ for every q ∈ 𝔤 ∩ F^{×}.

4.5. In this § we prove the theorem in the case r = 1, 1 ≤ t ≤ 4, and N[q] ≠ ∅. Applying Theorem 3.2(1) and (4), we find

L[q] = \begin{cases}[⋃_{a∈𝔤/2φ(y,N)}(y + ae_{1})C] ⋃[⋃_{a∈𝔤/2φ(−y,N)}(−y + ae_{1})C] & if t = 1, \\ ⋃_{a∈𝔤/2φ(y,N)}(y + ae_{1})C & otherwise, \end{cases}

where y is any element of N[q] and fix it. Since φ(y +ae_{1}, L) = φ(−y +ae_{1}, L), we obtain (y + ae_{1})C = (−y + ae_{1})C by [IQD, Theorem 1.3]. Thus L[q] = ⋃_{a∈𝔤/2φ(y,N)}(y + ae_{1})C for 1 ≤ t ≤ 4. We have already obtained our theorem in the case r = 0 and t > 0. Thus φ(y, N) = 2^{−1}𝔭^{τ(q)}. Put simply τ = τ(q). The same argument as in the proof of Lemma 3.3 shows that

(y + ae_{1})C = \begin{cases}L[q, 2^{−1}𝔭^{τ}] & if a ∈ 𝔭^{τ} , \\ L[q, 2^{−1}𝔭^{μ(a)}] & if a ∉ 𝔭^{τ} \end{cases}

with μ of (4.18). If τ ≥ 1, then 0 ≤ μ(a) ≤ τ − 1 for a ∈ 𝔤 such that a ∉ 𝔭^{τ}. Thus

L[q] = ⋃_{a∈𝔤/𝔭^{τ}}(y + ae_{1})C = ⨆_{i=0}^{τ} L[q, 2^{−1}𝔭^{i}],

L[q, 2^{−1}𝔭^{τ}] = yC, and L[q, 2^{−1}𝔭^{i}] = (y + π^{i}e_{1})C for 0 ≤ i ≤ τ − 1. If τ = 0, then it is clear that L[q] = yC = L[q, 2^{−1}𝔤]. This proves the theorem in the case r = 1, t > 0, and N[q] ≠ ∅.

4.6. In §§4.6 and 4.7 we assume r = 1, t > 0, and N[q] = ∅. Then 1 ≤ t ≤ 3 since N[q] ≠ ∅ for all q ∈ 𝔤 ∩ F^{×} if t = 4. By Theorem 3.2(3),

(4.28)  L[q] = ⋃_{i=0}^{κ_{0}−1} ⋃_{b∈X_{i}(q)/𝔭^{i}N} k_{i,b}C

with k_{i,b} in (3.7). Let us determine κ_{0} in this §4.6. With the notation of (3.2)

(4.29)  0 ∈ X_{ν}(q)

since q ∈ 𝔭^{ν}, and hence κ_{0} > ν with the notation of (3.3). Suppose t = 1, then ξ(cq) ≠ 1 by (4.27) and the assumption N[q] = ∅. First assume 2 ∉ 𝔭 or ν ≢ δ (mod 2). If there exists x ∈ X_{ν+1}(q), then x^{2} ∈ c^{−1}q(1 + 𝔭) by (3.2). This implies ν ≡ δ (mod 2), and hence 2 ∉ 𝔭. Then by [NRQ, Lemma 3.2(1)], we have 1+𝔭 ⊂ 𝔤^{×2}, which contradicts ξ(cq) ≠ 1. Thus κ_{0} = ν +1 by (4.29). Next assume 2 ∈ 𝔭 and ν ≡ δ (mod 2). Then ε(c^{−1}qπ^{δ−ν}) ≤ 2κ from ξ(cq) ≠ 1 and Lemma 4.2. Hereafter we put ε = ε(c^{−1}qπ^{δ−ν}). There exist

(4.30)  s ∈ 1 + π^{ε}𝔤^{×} and α ∈ 𝔤^{×} such that c^{−1}qπ^{δ−ν} = s^{−1}α^{2}

by (4.11). From this we have c(π^{(ν−δ)/2}α)^{2} = sq, and hence

(4.31)  N[sq] ≠ ∅.

Since N[sq] ⊂ X_{ν+ε}(q), we obtain X_{ν+ε}(q) ≠ ∅, and hence κ_{0} > ν + ε in the notation of (3.3). If there exists x ∈ X_{ν+ε+1}(q), then we can take a ∈ 𝔭^{ν+ε+1} such that cx^{2} + a = q. Thus c^{−1}qπ^{δ−ν} = (1 + c^{−1}x^{−2}a)(π^{(δ−ν)/2}x)^{2} ∈ (1 + 𝔭^{ε+1})𝔤^{×2}, which contradicts (4.11). Hence κ_{0} = ν + ε + 1. Moreover Lemma 4.2(1) implies that: if ξ(cq) = −1, then ε = 2κ, and hence κ_{0} = 2κ + ν + 1; if ξ(cq) = 0 and D_{F (\sqrt{cq})/F} = 𝔭^{d}, then ε = 2κ+1−d, and hence κ_{0} = 2κ+ν+2−d. Next suppose t = 2, then c^{−1}q ∉ N_{K/F} (𝔯) by (4.27). Let d be as in Theorem 3.5(iii). If X_{ν+d}(q) ≠ ∅, then c^{−1}q ∈ N_{K/F} (𝔯)(1 + 𝔭^{d}) ⊂ N_{K/F} (𝔯) by [BNT, Chapter VIII, Proposition 3] or the conductor-discriminant theorem according as 𝔡 = 𝔯 or 𝔡 ≠ 𝔯. This contradicts c^{−1}q ∉ N_{K/F} (𝔯). Thus κ_{0} ≤ ν + d. In particular if 𝔡 = 𝔯 or 𝔮, that is d = 1, then κ_{0} = ν + 1 from (4.29). Here 𝔮 is the maximal ideal of 𝔯. Assume 𝔡 = 𝔮^{d} and d > 1. Take a prime element π_{K} of K such that N_{K/F} (π_{K}) = π. We see that 2 ∈ 𝔭 by [BNT, Chapter VIII, Corollary 3 of Proposition 7]. By local class field theory, we have (1 + 𝔭^{d−1})N_{K/F} (𝔯^{×}) = 𝔤^{×}. Thus there exist

(4.32)  s ∈ 1 + π^{d−1}𝔤^{×} and α ∈ 𝔯^{×} such that c^{−1}qπ^{−ν} = s^{−1}N_{K/F} (α).

Note that c ∈ 𝔤^{×} since K is ramified over F; see §2.3. Then cN_{K/F} (π_{K}^{ν}α) = sq, and hence

(4.33)  N[sq] ≠ ∅.

We obtain N[sq] ⊂ X_{ν+d−1}(q) by the definition of s. Thus κ_{0} = ν + d. Finally suppose t = 3. Then ξ(−cq) = 1 by (4.27), and hence ν ≡ δ (mod 2). For b ∈ 𝔤 and 0 ≤ m ∈ Z, put

Y_{m}(b) = {y ∈ N | y^{2} − b ∈ 𝔭^{m}}

as in [NRQ, (4.1.3)]. Then X_{i}(q) = π^{(ν−δ)/2}Y_{i−ν}(−c^{−1}qπ^{δ−ν}) if i > ν. The proof of [NRQ, Lemma 4.2(2)] shows that Y_{m}(−c^{−1}qπ^{δ−ν}) = ∅ for m ≥ 2κ + 1 even for c𝔤 = 𝔭. Thus κ_{0} ≤ ν + 2κ + 1. If 2 ∈ 𝔤^{×}, then κ_{0} = ν + 1 by (4.29). Assume 2 ∈ 𝔭. There exists

(4.34)  s ∈ 1 + 4𝔤 such that s ∉ 𝔤^{×2}

by [NRQ, Lemma 3.2(1)], then ξ(s) = −1 from Lemma 4.2(1). Thus ξ(−csq) = −1, and hence, by Lemma 3.4(3),

(4.35)  N[sq] ≠ ∅.

We find N[sq] ⊂ X_{ν+2κ}(q), and hence κ_{0} = ν + 2κ + 1. Consequently we have

κ_{0} = \begin{cases}ν + 2κ + 2 − d & if t = 1, ξ(cq) ≠ 1, ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ ν + d & if t = 2, c^{−1}q ∉ N_{K/F} (𝔯), and d > 1, \\ ν + 2κ + 1 & if t = 3, ξ(−cq) = 1, and 2 ∈ 𝔭, \\ ν + 1 & otherwise. \end{cases}

This completes the determination of the number κ_{0}.

4.7. Now, in (4.28) we have φ(k_{i,b}, L) = 2^{−1}𝔭^{i} + φ(b, N) + 2^{−1}(q − φ[b])𝔭^{−i} by (2.5), (2.6), and (2.7) for 0 ≤ i ≤ κ_{0} − 1 and b ∈ X_{i}(q). Let m(i, b) ∈ Z such that φ(k_{i,b}, L) = 2^{−1}𝔭^{m(i,b)}. Here [IQD, Theorem 1.3] implies k_{i,b}C = L[q, 2^{−1}𝔭^{m(i,b)}]. We have 0 ≤ m(i, b) ≤ μ(q − φ[b]) − i ≤ (κ_{0} − 1) − i in the notation of (3.3), with μ of (4.18). From this and m(i, b) ≤ i, we see that

(4.36)  0 ≤ m(i, b) ≤ [(κ_{0} − 1)/2]

for 0 ≤ i ≤ κ_{0} − 1 and b ∈ X_{i}(q). On the other hand, when κ_{0} = ν + 1, put z = 0; when κ_{0} > ν + 1, take any element z ∈ N[sq] and fix it. Here s is of (4.30), (4.32), or (4.34) according as t = 1, 2, or 3. Then z ∈ X_{κ_{0}−1}(q). We assert that

(4.37)  m(i, z) = i

for 0 ≤ i ≤ [(κ_{0} −1)/2]. Indeed, if z = 0, then it is obvious. Suppose z ∈ N[sq]. Then μ(q − φ[z]) = κ_{0} − 1. From the theorem in the case r = 0 and t > 0, we find that

φ(z, N) = \begin{cases}𝔭^{(ν+δ)/2} & if t = 1, 3, \\ 2^{−1}𝔭^{[(ν+d)/2]} & if t = 2. \end{cases}

Therefore

m(i, z) = \begin{cases}min(i, κ + (ν + δ)/2, (κ_{0} − 1) − i) & if t = 1, 3, \\ min(i, [(ν + d)/2], (κ_{0} − 1) − i) & if t = 2. \end{cases}

From this, we obtain (4.37). As a consequence, from (4.36) and (4.37),

L[q] = ⋃_{i=0}^{κ_{0}−1} ⋃_{b∈X_{i}(q)/𝔭^{i}N} k_{i,b}C = ⨆_{i=0}^{[(κ_{0}−1)/2]} L[q, 2^{−1}𝔭^{i}]

and L[q, 2^{−1}𝔭^{i}] = k_{i,z}C. This completes the proof.

5 The maximality of L ∩ (F h)^{⊥}

In this section still the field F is local and we prove the second main Theorem 5.3 as an application of Theorem 3.5. We first prepare two lemmas.

5.1 Lemma. Let ξ be as in Notation. Let α ∈ F^{×} such that ξ(α) ≠ 1. Let 𝔬 be the valuation ring of F (\sqrt{α}), 𝔮_{𝔬} the maximal ideal of 𝔬, 𝔡_{F (\sqrt{α})/F} the different of F (\sqrt{α}) relative to F, and μ the normalized order function of F. Then we obtain the following assertions:

(1) If 2 ∈ 𝔤^{×} and μ(α) ∈ 2Z, then 𝔡_{F (\sqrt{α})/F} = 𝔬.

(2) If μ(α) ∉ 2Z, then 𝔡_{F (\sqrt{α})/F} = 2𝔮_{𝔬}.

(3) If 2 ∈ 𝔭, μ(α) ∈ 2Z, ξ(α) = 0, and 𝔡_{F (\sqrt{α})/F} = 𝔮_{𝔬}^{d}, then d ∈ 2Z.

Proof. The first two assertions are well known. Assertion (3) follows from Lemma 4.2(1), (2).

5.2 Lemma. Let H be an integral lattice in V. Let t be the core dimension of V. Let δ(φ) be defined as in (2.1). Assume n ∉ 2Z and δ(φ) ∩ 𝔤^{×} = ∅. Then we have H is maximal in V if and only if [H̃ : H] = [𝔤 : 2𝔭]. Here H̃ is defined as in (2.2).

Proof. Assume that H is maximal in V. Then [AQC, Lemma 6.9] implies [H̃ : H] = [L̃ : L] with L of (2.4). Since L̃ = Ñ + ∑_{i=1}^{r}(𝔤f_{i} + 𝔤e_{i}), we have [L̃ : L] = [Ñ : N]. By (2.8), (2.11), (2.19), and (2.20), we obtain [Ñ : N] = [𝔤 : 2𝔭] since δ = 1. Thus we obtain the “only if”-part of the assertion. Conversely, we assume that H is an integral lattice in V such that

(5.1)  [H̃ : H] = [𝔤 : 2𝔭].

By [AQC, Lemma 6.2(1)], there exists a maximal lattice H_{0} in V such that H ⊂ H_{0}. Then

(5.2)  H ⊂ H_{0} ⊂ H̃_{0} ⊂ H̃.

From the “only if”-part of the lemma, which is settled above, we obtain [H̃_{0} : H_{0}] = [𝔤 : 2𝔭]. Combining this with (5.1) and (5.2), we obtain H = H_{0}, and hence H is maximal in V.

We remark that the index [H̃ : H] of a maximal lattice H in V is given in [AQC, Lemma 8.4(iv)] when n ∈ 2Z or δ(φ) ∩ 𝔤^{×} ≠ ∅. Now, for h ∈ L such that φ[h] ≠ 0, put

(5.3)  (F h)^{⊥} = {x ∈ V | φ(x, h) = 0}.

5.3 Theorem. Let L be a maximal lattice in V and τ(q) as in (3.8) for a given q ∈ 𝔤 ∩ F^{×}. Assume n ≥ 2. Then for h ∈ L such that φ[h] ≠ 0, we have

L ∩ (F h)^{⊥} is maximal in (F h)^{⊥} ⟺ h ∈ L[φ[h], 2^{−1}𝔭^{τ(φ[h])}].

Hereafter we prove this theorem until the end of §5.14.

5.4. Before stating the proof let us recall the basic notion and terminology in the previous subsections, which will be needed in the next arguments. Put q = φ[h], then h ∈ L[q]. Put simply τ = τ(q). We have

L[q] = \begin{cases}L[q, 2^{−1}𝔭^{τ}] & if r = 0, \\ ⨆_{i=0}^{τ} L[q, 2^{−1}𝔭^{i}] & if r ≥ 1 \end{cases} (5.4)

by (3.8). Hence h ∈ L[q, 2^{−1}𝔭^{i}] for some 0 ≤ i ≤ τ. Put W = (F h)^{⊥}. Our aim is to show that L ∩ W is maximal in W if and only if i = τ. We have a Witt decomposition of V with respect to φ

(5.5)  V = Z + ∑_{j=1}^{r}(F f_{j} + F e_{j}), L = N + ∑_{j=1}^{r}(𝔤f_{j} + 𝔤e_{j})

as in (2.4). Let t be the core dimension of (V, φ). Let ξ be as in Notation. Assume t = 1. Let c and δ be as in (2.9) and (2.10), respectively. For q, let d and s be as in Theorem 3.5(ii). Assume t = 2. Let b and c be as in (2.12) and δ as in (2.13). Let K and 𝔯 be as in §2.3(II). Let 𝔮 be the maximal ideal of 𝔯 and ρ ∈ Gal(K/F) such that ρ ≠ 1. Let 𝔡 and d be as in Theorem 3.5(iii). Then

(5.6) K is isomorphic to F (\sqrt{−bc}),

𝔡 = 𝔡_{K/F} = 𝔮^{d} when 𝔡 ≠ 𝔯. (5.7)

For q, let s be as in Theorem 3.5(iii). Assume t = 3. Let c and δ be as in (2.14) and (2.15), respectively. For q, let s be as in Theorem 3.5(iv). Let δ(φ) be as in (2.1). We may assume that: if t = 1 or 3 and δ(φ) ∩ 𝔤^{×} ≠ ∅, then c ∈ 𝔤^{×}; if t = 1 or 3 and δ(φ) ∩ 𝔤^{×} = ∅, then c𝔤 = 𝔭; if t = 2, then b, c ∈ 𝔤^{×} ⊔ π𝔤^{×}; if t = 2 and K is ramified over F, then c ∈ 𝔤^{×}; see §2.3.

5.5. First suppose that:

t = 1, r ≥ 1, and ξ(cq) = 1; or t = 2 and c^{−1}q ∈ N_{K/F} (𝔯); or t = 3 and ξ(−cq) ≠ 1; or t = 4.

This assumption is equivalent to N[q] ≠ ∅ by Lemma 3.4. Here 0 ≠ q = φ[h], h ∈ L. Then h ∈ L[q, 2^{−1}𝔭^{i}] for some 0 ≤ i ≤ τ by (5.4). We obtain

L[q, 2^{−1}𝔭^{i}] = \begin{cases}yC & if i = τ, \\ h_{i,y}C & if i < τ \end{cases} (5.8)

by (3.13), (3.16), (3.19), and (3.21). Here, y is any element of N[q] and fix it.

We first prove that L ∩ W is maximal in W when i = τ. In this case, h ∈ L[q, 2^{−1}𝔭^{τ}] = yC by (5.8). Therefore there exists γ ∈ C such that y = hγ. We have L ∩ (F y)^{⊥} = (L ∩ W)γ since Lγ = L and (F y)^{⊥} = W γ. Therefore L ∩ (F y)^{⊥} is maximal in (F y)^{⊥} if and only if L ∩ W is maximal in W. Hence we assume h = y and W = (F y)^{⊥}. Now, we have

W = (Z ∩ W) + ∑_{j=1}^{r}(F f_{j} + F e_{j})

since h ∈ N, (5.3), and (5.5). This is a Witt decomposition of W with respect to the restriction of φ to W. Moreover we obtain

(5.9) L ∩ W = (N ∩ W) + ∑_{j=1}^{r}(𝔤f_{j} + 𝔤e_{j}), N ∩ W = {x ∈ Z ∩ W | φ[x] ∈ 𝔤}

from (2.7) and (5.5). Thus by [AQC, Lemma 6.5], L ∩ W is maximal in W when i = τ. Next suppose i < τ. We shall show that L ∩ W is not a maximal lattice in W in this case. We obtain h ∈ h_{i,y}C by (5.8). Thus we may assume h = h_{i,y}. By (3.8), N[q] = N[q, 2^{−1}𝔭^{τ}]. From this,

(5.10)  φ(y, N) = 2^{−1}𝔭^{τ} .

We see that

W = X + ∑_{j=1}^{r−1}(F f_{j} + F e_{j}), X = {ae_{r} + x − 2π^{−i}φ(y, x)f_{r} | a ∈ F, x ∈ Z}, (5.11)

L ∩ W = H + ∑_{j=1}^{r−1}(𝔤f_{j} + 𝔤e_{j}), H = {ae_{r} + x − 2π^{−i}φ(y, x)f_{r} | a ∈ 𝔤, x ∈ N}

by the definition of h_{i,y} (in (3.7)), (5.3), (5.5), and (5.10). Take

(5.12)  w ∈ N such that φ(y, w) = −2^{−1}π^{τ}

and fix it. Put u = π^{τ−i}f_{r} + w − π^{i−τ} φ[w]e_{r}, v = π^{i−τ} e_{r}, and

Y = {x − 2π^{i−τ} φ(x, w)e_{r} | x ∈ Z such that φ(x, y) = 0}.

Then we find that X = Y + F u + F v is a Witt decomposition of X by a straightforward calculation. Here X is defined as in (5.11). Put Λ = {k ∈ Y | φ[k] ∈ 𝔤}, then Λ + 𝔤u + 𝔤v is maximal in X by [AQC, Lemma 6.5]. We assert that H ⊊ Λ + 𝔤u + 𝔤v. Indeed, it is clear that v ∉ H since i − τ < 0. For any ℓ = ae_{r} + x − 2π^{−i}φ(y, x)f_{r} ∈ H, put ξ = −2π^{−τ} φ(y, x) and η = aπ^{τ−i} + 2π^{−τ} φ(y, x)φ[w] + 2φ(x, w) with w in (5.12). Then a straightforward computation shows that ξ, η ∈ 𝔤 and ℓ − ξu − ηv ∈ Λ by (5.10). Therefore H ⊊ Λ + 𝔤u + 𝔤v. Thus H is not maximal in X, and hence L ∩ W is not maximal in W by [AQC, Lemma 6.3]. This completes the proof in the case N[q] ≠ ∅.

5.6. Let us now suppose that:

t = 0 and r ≥ 1; or t = 1, r ≥ 1, ξ(cq) ≠ 1, and 2 ∈ 𝔤^{×}; or t = 1, r ≥ 1, ξ(cq) ≠ 1, and ν ≢ δ (mod 2); or t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and 𝔡_{K/F} = 𝔯; or t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and 𝔡_{K/F} = 𝔮; or t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔤^{×}.

Here ν ∈ Z such that q𝔤 = 𝔭^{ν} with 0 ≠ q = φ[h], h ∈ L. In this case we obtain

L[q] = \begin{cases}⊔_{j=0}^{τ}ℓ_{j}C & if t ≥ 1 or r ≥ 2, \\ ⊔_{j=0}^{ν}ℓ_{j}C & if t = 0 and r = 1 \end{cases}

with ℓ_{j} in (3.7) and τ = [ν/2] of Theorem 3.5. Moreover

(5.13)  N[q] = ∅

by Lemma 3.4. We have h ∈ L[q, 2^{−1}𝔭^{i}] for some 0 ≤ i ≤ τ by (5.4). Hereafter until the end of §5.7 we prove the theorem in the case t ≥ 1 or r ≥ 2. Then

L[q, 2^{−1}𝔭^{i}] = ℓ_{i}C for 0 ≤ i ≤ τ

by (3.10), (3.13), (3.16), and (3.19). Thus we may assume h = ℓ_{i} and W = (F ℓ_{i})^{⊥} since h ∈ ℓ_{i}C. In this §5.6 we determine [(L ∩ W) : L ∩ W]. Put

(5.14)  w = f_{r} − qπ^{−2i}e_{r}

with e_{r} and f_{r} in (5.5). Then

(5.15)  W = (F w + Z) + ∑_{j=1}^{r−1}(F f_{j} + F e_{j})

from the definition of ℓ_{i} (in (3.7)), (5.3), and (5.5). We understand that: ∑_{j=1}^{r−1}(F f_{j} + F e_{j}) = {0} when t > 0 and r = 1; Z = {0} when t = 0 and r ≥ 2. We assert that (5.15) is a Witt decomposition. Indeed, it is clear when t = 0 and r ≥ 2. Assume t ≥ 1. If φ[aw + x] = 0 for a ∈ F and x ∈ Z, then φ[x] = qπ^{−2i}a^{2}. If a ≠ 0, then this is the case if and only if φ[a^{−1}π^{i}x] = q, and hence N[q] ≠ ∅. This contradicts (5.13). Thus a = 0, and hence x = 0. Therefore the restriction of φ to F w + Z is anisotropic. Combining this with (2.5), (2.6), and (5.14), we see that (5.15) is a Witt decomposition. Now, we have

(5.16)  L ∩ W = (𝔤w + N) + ∑_{j=1}^{r−1}(𝔤f_{j} + 𝔤e_{j})

by (5.5) and (5.15). Therefore a straightforward computation shows that

(5.17)  (L ∩ W) = 2^{−1}𝔭^{2i−ν}w + Ñ + ∑_{j=1}^{r−1}(𝔤f_{j} + 𝔤e_{j})

from (2.2) with (5.16). Combining (5.16) with (5.17), we have

(5.18)  [(L ∩ W) : L ∩ W] = [𝔤 : 2𝔭^{ν−2i}] · [Ñ : N].

Here we obtain the index [Ñ : N] by [AQC, Lemma 8.4(iv)] and Lemma 5.2. Combining this with (5.18), we have

[(L ∩ W) : L ∩ W] (5.19)

= \begin{cases}[𝔤 : 2𝔭^{ν−2i}] & if t = 0 and r ≥ 2, \\ [𝔤 : 4𝔭^{ν−2i+δ}] & if t = 1, r ≥ 1, ξ(cq) ≠ 1, and 2 ∈ 𝔤^{×}, \\ [𝔤 : 4𝔭^{ν−2i+δ}] & if t = 1, r ≥ 1, ξ(cq) ≠ 1, and ν ≢ δ (mod 2), \\ [𝔤 : 2𝔭^{ν−2i+2δ}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and 𝔡 = 𝔯, \\ [𝔤 : 2𝔭^{ν−2i+1}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and 𝔡 = 𝔮, \\ [𝔤 : 𝔭^{ν−2i+2−δ}] & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔤^{×}. \end{cases}

Note that: if t = 2, r ≥ 1, and 𝔡 = 𝔯, then c^{−1}q ∉ N_{K/F} (𝔯) if and only if ν ≢ δ (mod 2) by Lemma 3.4(2); if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔤^{×}, then ν ≡ δ (mod 2).

5.7. In this § we still assume t ≥ 1 or r ≥ 2. For an integral lattice R in W, the following assertion holds:

R is maximal in W (5.20)

⟺ [R̃ : R] = \begin{cases}[𝔤 : 2𝔭^{ν−2τ}] & if t = 0 and r ≥ 2, \\ [𝔤 : 𝔭^{2δ}] & if t = 1, r ≥ 1, ξ(cq) ≠ 1, \\ & and ν ≡ δ (mod 2), \\ [𝔤 : 4𝔭] & if t = 1, r ≥ 1, ξ(cq) ≠ 1, \\ & and ν ≢ δ (mod 2), \\ [𝔤 : 2𝔭^{1+δ}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and 𝔡 = 𝔯, \\ [𝔤 : 2𝔭^{ν−2τ+1}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and 𝔡 = 𝔮, \\ [𝔤 : 𝔭^{2}] & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔤^{×}. \end{cases}

Note that if t = 1, r ≥ 1, ξ(cq) ≠ 1, and ν ≡ δ (mod 2), then 2 ∈ 𝔤^{×}. If (5.20) holds, then combining (5.19) with (5.20), we obtain our theorem in the case t ≥ 1 or r ≥ 2 since L ∩ W is an integral lattice in W. Let us prove (5.20). We observe a core subspace F w + Z of W with w of (5.14). If t = 0 and r ≥ 2, then dim_{F} (F w + Z) = 1 and δ(F w + Z, φ) = φ[w]F^{×2} = −qF^{×2}. Here δ(F w + Z, φ) is defined as in (2.1). This implies δ(F w + Z, φ) ∩ 𝔤^{×} ≠ ∅ if ν ∈ 2Z and δ(F w + Z, φ) ∩ 𝔤^{×} = ∅ if ν ∉ 2Z. Therefore we have (5.20) by [AQC, Lemma 8.4(iv)] and Lemma 5.2 when t = 0 and r ≥ 2. Assume t = 1, r ≥ 1, and ξ(cq) ≠ 1. Since φ[w] = −qπ^{−2i} and Z = F g with g of (2.8), (F w + Z, φ) is isomorphic to (F (\sqrt{cq}), ψ), where ψ[x] = cN_{F (\sqrt{cq})/F} (x) for x ∈ F (\sqrt{cq}), as explained in §2.3(II). Suppose ν ≡ δ (mod 2), then 2 ∈ 𝔤^{×}. Thus Lemma 5.1(1) implies that F (\sqrt{cq}) is unramified. Therefore we have (5.20) in this case by [AQC, Lemma 8.4(iv)]. Next suppose ν ≢ δ (mod 2), then 𝔡_{F (\sqrt{cq})/F} = 2𝔮_{F (\sqrt{cq})} by Lemma 5.1(2), where 𝔮_{F (\sqrt{cq})} is the maximal ideal of the valuation ring of F (\sqrt{cq}). Therefore [AQC, Lemma 8.4(iv)] implies (5.20) when t = 1, r ≥ 1, ξ(cq) ≠ 1, and 2 ∈ 𝔤^{×} or ν ≢ δ (mod 2). Assume t = 2, r ≥ 1, and c^{−1}q ∉ N_{K/F} (𝔯), then dim_{F} (F w + Z) = 3. Since Z = F g_{1} + F g_{2} with g_{1} and g_{2} in (2.12) and φ[w] = −qπ^{−2i}, we have

(5.21)  δ(F w + Z, φ) = bcqF^{×2}.

Suppose 𝔡 = 𝔯. Then ν ≢ δ (mod 2) by Lemma 3.4(2). Since K is unramified over F, we have b𝔤 = c𝔤(= 𝔭^{δ}) by Lemma 5.1(2). Thus (5.21) implies δ(F w + Z, φ)∩𝔤^{×} = ∅ if δ = 0 and δ(F w +Z, φ)∩𝔤^{×} ≠ ∅ if δ = 1. From this we obtain (5.20) by [AQC, Lemma 8.4(iv)] and Lemma 5.2 in this case. Next suppose 𝔡 = 𝔮. Here Lemma 5.1 implies b𝔤 = 𝔭 since c ∈ 𝔤^{×}. Therefore (5.21) implies δ(F w + Z, φ) ∩ 𝔤^{×} = ∅ if ν ∈ 2Z and δ(F w + Z, φ) ∩ 𝔤^{×} ≠ ∅ if ν ∉ 2Z. Hence we have (5.20) by [AQC, Lemma 8.4(iv)] and Lemma 5.2 when t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and 𝔡_{K/F} = 𝔯 or 𝔮. Finally if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔤^{×}, then dim_{F} (F w + Z) = 4, and hence [AQC, Lemma 8.4(iv)] implies (5.20). This proves the theorem in the case when N[q] = ∅, L[q] = ⊔_{j}ℓ_{j}C, and t ≥ 1 or r ≥ 2.

5.8. In this § we prove the theorem in the case t = 0 and r = 1. Then

L[q, 2^{−1}𝔭^{i}] = \begin{cases}ℓ_{i}C ⊔ ℓ_{ν−i}C & if t = 0, r = 1, and i ≠ ν/2, \\ ℓ_{i}C & if t = 0, r = 1, and i = ν/2 \end{cases}

for 0 ≤ i ≤ τ, by (3.10). Since h ∈ L[q, 2^{−1}𝔭^{i}] for some 0 ≤ i ≤ τ by (5.4), we may assume that h = ℓ_{i} or h = ℓ_{ν−i} when i ≠ ν/2 and h = ℓ_{i} when i = ν/2. If h = ℓ_{i} for 0 ≤ i ≤ τ (including the case h = ℓ_{ν/2}), we can obtain the assertion in the same way as §§5.6 and 5.7. Assume h = ℓ_{ν−i} for 0 ≤ i < ν/2. Put here w = f_{1} − qπ^{−2(ν−i)}e_{1}. Then we see that

W = (F ℓ_{ν−i})^{⊥} = F w, L ∩ W = 𝔭^{ν−2i}w, and (L ∩ W) = 2^{−1}𝔤w

in a similar way as §5.6. Thus [(L ∩ W) : L ∩ W] = [𝔤 : 2𝔭^{ν−2i}]. Therefore we obtain the theorem in the same way as in the case when t = 0 and r ≥ 2 since φ[w]F^{×2} = −qF^{×2}. This completes the proof in the case when N[q] = ∅ and L[q] = ⊔_{j}ℓ_{j}C.

5.9. Finally we suppose that:

t = 1, r ≥ 1, ν ≡ δ (mod 2), ξ(cq) ≠ 1, and 2 ∈ 𝔭; or t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and d > 1; or t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭.

Note that c ∈ 𝔤^{×} when t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and d > 1. Then we have

(5.22)  L[q] = ⨆_{j=0}^{τ} k_{j,z}C and k_{j,z}C = L[q, 2^{−1}𝔭^{j}]

with k_{j,z} in (3.7), from (3.8), (3.13), (3.16), and (3.19). Here z is any element of N[sq] with s of (3.11), (3.14), or (3.17) of Theorem 3.5 according as t = 1, 2, or 3. We fix z. We obtain (5.23)

τ = \begin{cases}κ + [\frac{ν+1−d}{2}] & if t = 1, r ≥ 1, ξ(cq) ≠ 1, ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ [\frac{ν+d−1}{2}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and d > 1, \\ κ + [\frac{ν}{2} ] & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭 \end{cases}

from (3.12), (3.15), and (3.18), where κ ∈ Z such that 2𝔤 = 𝔭^{κ}. Moreover, by Lemma 3.4,

(5.24)  N[q] = ∅.

Applying Theorem 3.5 to (Z, φ), N, and sq, we have N[sq] = N[sq, 2^{−1}𝔭^{τ(sq)}] by (3.8). Here

τ(sq) = \begin{cases}κ + \frac{ν+δ}{2} & if t = 1, r ≥ 1, ξ(cq) ≠ 1, \\ & ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ [\frac{ν+d}{2}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and d > 1, \\ κ + \frac{ν+δ}{2} & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭 \end{cases} (5.25)

from (3.12), (3.15), and (3.18). Thus

(5.26)  φ(z, N) = 2^{−1}𝔭^{τ(sq)}.

Assume h ∈ L[q, 2^{−1}𝔭^{i}] for some 0 ≤ i ≤ τ. Then by (5.22), h ∈ k_{i,z}C, and hence we may assume h = k_{i,z}. Hereafter we show that L ∩ W is maximal in W if and only if i = τ. Put

(5.27)  x_{1} = π^{i−τ} [f_{r} − π^{−2i}(q − φ[z])e_{r}]

with e_{r} and f_{r} in (5.5). Then

(5.28)  φ[x_{1}] = π^{−2τ} (s − 1)q and π^{τ−i}x_{1} ∈ L.

By a straightforward computation, we obtain (5.29)

W = X + ∑_{j=1}^{r−1}(F f_{j} + F e_{j}), X = {ax_{1} + x − 2π^{−i}φ(x, z)e_{r} | a ∈ F, x ∈ Z}

with x_{1} of (5.27). Then (5.29) is a Witt decomposition. Indeed, if φ[ax_{1} + x − 2π^{−i}φ(x, z)e_{r}] = 0 for a ∈ F and x ∈ Z, then φ[x − π^{−τ} az] = (π^{−τ} a)^{2}q. Assuming a ≠ 0, we have φ[π^{τ} a^{−1}x − z] = q, and hence N[q] ≠ ∅. This contradicts (5.24). Thus a = 0, and hence x = 0. Therefore the restriction of φ to X is anisotropic. Combining this with (2.5), (2.6), and (5.27), we see that (5.29) is a Witt decomposition. Now, we easily see that 2π^{−i}φ(x, z) ∈ 𝔤 for x ∈ N from (5.23), (5.25), (5.26), and 0 ≤ i ≤ τ. From this and (5.28), (5.30)

L∩W = H+∑_{j=1}^{r−1}(𝔤f_{j} +𝔤e_{j}), H = {ax_{1}+x−2π^{−i}φ(x, z)e_{r} | a ∈ 𝔭^{τ−i}, x ∈ N}.

By [AQC, Lemma 6.3(1)], L ∩ W is maximal in W if and only if H is maximal in X. Thus we consider the lattice H in X instead of L ∩ W in W.

5.10. In this § we first determine the structure of N under the assumption of §5.9. Now we put

(5.31)  Y = {k ∈ Z | φ(k, z) = 0} and z_{1} = π^{−[ν/2]}z.

Then Z = F z_{1} + Y. First assume t = 1, r ≥ 1, ν ≡ δ (mod 2), ξ(cq) ≠ 1, and 2 ∈ 𝔭. Since φ[z_{1}]𝔤 = 𝔭^{δ},

(5.32)  N = 𝔤z_{1}.

Next assume t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), and d > 1. Then Lemma 5.1(1) and (2) imply 2 ∈ 𝔭. Take y ∈ Y such that Y = F y, then

(5.33)  Z = F z_{1} + F y.

Thus from (2.1) and (2.12), we have φ[z_{1}]φ[y]F^{×2} = −δ(Z, φ) = bcF^{×2}, and hence we may assume

φ[z_{1}]φ[y](bc)^{−1} = \begin{cases}1 & if b𝔤 = 𝔭, \\ π^{2λ} & if b ∈ 𝔤^{×}, \end{cases} (5.34)

where

(5.35)  λ = ν − 2[ν/2].

From (5.6) and (5.34), we see that F + F z_{1}y ⊂ A(Z, φ) is isomorphic to K. Here we identify F + F z_{1}y with K. Suppose b𝔤 = 𝔭. Then

(5.36)  d = 2κ + 1

by c ∈ 𝔤^{×}, Lemma 5.1(2), and (5.7). Put

z_{ν} = \begin{cases}z_{1} & if ν ∈ 2Z, \\ y & if ν ∉ 2Z. \end{cases}

Then φ[z_{ν}] ∈ 𝔤^{×} from (5.31) and (5.34), and hence Z = Kz_{ν} and N = 𝔯z_{ν} by (5.33). We find z_{1}y ∈ 𝔯 and (z_{1}y − (z_{1}y)^{ρ})𝔯 = 2𝔮 = 𝔡, and hence 𝔯 = 𝔤[z_{1}y] by [AQC, Lemma 5.6(ii)]. Thus

(5.37)  N = 𝔤z_{1} + 𝔤y.

Next suppose b ∈ 𝔤^{×}. Then bc ∈ 𝔤^{×} and 2 ∈ 𝔭 from c ∈ 𝔤^{×}, d > 1, and Lemma 5.1(1). Thus there exist

(5.38)  α, β ∈ 𝔤^{×} such that − bc = α^{−2}(1 + π^{2κ+1−d}β)

by d > 1 and Lemma 4.2(1). Put η = π^{(d−2κ)/2}(1 + απ^{−λ}z_{1}y) in K with λ of (5.35). Then η is a root of an Eisenstein equation x^{2} − 2π^{(d−2κ)/2}x − πβ = 0, and hence η is a prime element of K and (η − η^{ρ})𝔯 = (2η − 2π^{(d−2κ)/2})𝔯 = 𝔡. Here x is an indeterminate. Thus 𝔯 = 𝔤[η] by [AQC, Lemma 5.6(ii)]. By (5.31) and (5.34), we have φ[y]𝔤 = 𝔭^{λ}. From this and η^{−1} = −π^{−1+(d−2κ)/2}β^{−1}(1 − απ^{−λ}z_{1}y), we obtain

(5.39)  N = 𝔯η^{−λ}y = 𝔤y + 𝔤π^{−λ+(d−2κ)/2}[y + (−1)^{ν}απ^{−λ}φ[y]z_{1}].

Finally assume t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭. Then ν ≡ δ (mod 2), and hence φ[z_{1}]𝔤 = 𝔭^{δ}. We can take

(5.40)  y_{1}, y_{2} ∈ Y so that Y = F y_{1} + F y_{2} and φ(y_{1}, y_{2}) = 0

by [EPE, Lemma 1.8]. Then we may assume

(5.41)  φ[y_{1}]φ[y_{2}] ∈ 𝔤^{×}

since

(5.42)  −φ[z_{1}]φ[y_{1}]φ[y_{2}]F^{×2} = δ(Z, φ) = −cF^{×2}

by (2.14), (5.31), and (5.40). Put

(5.43) T = F y_{1}y_{2} +F z_{1}y_{1} +F z_{1}y_{2}, K_{Y} = F +F y_{1}y_{2}, B = F +T, ζ = z_{1}y_{1}y_{2}

in A(Z). Moreover put c_{1} = φ[z_{1}]φ[y_{1}]φ[y_{2}]. Then Z = T ζ, Y = K_{Y} y_{2}, B is a division quaternion algebra over F, c_{1}𝔤 = c𝔤, and φ[xζ] = c_{1}N_{B/F} (x) for x ∈ T. From ξ(−cq) = 1 and (5.42), we have (y_{1}y_{2})^{2}F^{×2} = sF^{×2}. Thus K_{Y} is an unramified quadratic extension of F by (3.17). We may assume

(5.44)  φ[y_{1}]𝔤 = φ[y_{2}]𝔤 = 𝔤 or φ[y_{1}]^{−1}𝔤 = φ[y_{2}]𝔤 = 𝔭

by (5.41). Then we see that

(5.45) φ[y_{2}]𝔤 = 𝔭^{1−δ} = \begin{cases}𝔭 & if ν ≡ 0 (mod 2), \\ 𝔤 & if ν ≡ 1 (mod 2) \end{cases}

as shown below. Put

ω = z_{1}y_{1} and v = y_{1}y_{2}.

Then B = K_{Y} + K_{Y} ω, ωv = −vω, ω^{2}𝔤 = 𝔭^{2δ−1}, v ∈ K_{Y} ∩ T, and v^{2} ∈ 𝔤^{×} from (5.31), (5.40), (5.41), (5.42), (5.43), and (5.45). Therefore

(5.46)  N = (𝔤v + 𝔯_{Y} ω^{−1})ζ = 𝔤z_{1} + 𝔯_{Y} y_{2};

see §2.3. Here 𝔯_{Y} is the valuation ring of K_{Y}. Now we assert (5.45). Indeed, if this is not the case, then

φ[y_{2}]𝔤 = 𝔭^{δ} = \begin{cases}𝔤 & if ν ≡ 0 (mod 2), \\ 𝔭 & if ν ≡ 1 (mod 2) \end{cases}

by (5.44). Since K_{Y} is unramified over F, there exists θ ∈ B such that B = K_{Y} + K_{Y} θ, θ^{2}𝔤 = 𝔭, and θy_{1}y_{2} = −y_{1}y_{2}θ by the proof of [AQC, Theorem 5.13]. Then we easily see that θ ∈ F z_{1}y_{1} + F z_{1}y_{2}. Put J = F z_{1}y_{1} + F z_{1}y_{2} and ψ[x] = N_{B/F} (x) for x ∈ J. We consider the Clifford algebra A(J, ψ) of ψ. Put K_{J} = F + F z_{1}y_{1} · z_{1}y_{2}, then J = K_{J}z_{1}y_{2}, ψ[z_{1}y_{2}]𝔤 = 𝔭^{2δ}, and K_{J} is isomorphic to F (\sqrt{s}) which is an unramified quadratic extension of F. Thus Lemma 3.4(2) implies Λ_{J}[ψ[θ]] = ∅, where Λ_{J} is a maximal lattice in J. Since θ^{2}𝔤 = 𝔭, this gives a contradiction, and hence φ[y_{2}]𝔤 = 𝔭^{1−δ}.

5.11. Put

(5.47)  x_{2} = π^{−[ν/2]}(z − 2π^{−i}φ[z]e_{r}).

Then φ[x_{2}] = φ[z_{1}] with z_{1} in (5.31). From (5.30), (5.32), (5.37), (5.39), and (5.46),

(5.48) H =
𝔭^{τ−i}x_{1} + 𝔤x_{2}  if t = 1, r ≥ 1, ξ(cq) ≠ 1,
ν ≡ δ (mod 2), and 2 ∈ 𝔭,
𝔭^{τ−i}x_{1} + 𝔤x_{2} + 𝔤y  if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1,
and b𝔤 = 𝔭,
𝔭^{τ−i}x_{1} + 𝔤x_{3} + 𝔤y  if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1,
and b ∈ 𝔤^{×},
𝔭^{τ−i}x_{1} + 𝔤x_{2} + 𝔯_{Y} y_{2} if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭

with x_{1} of (5.27), y ∈ Y satisfying (5.34), y_{2} of (5.40), and 𝔯_{Y} in (5.46). Moreover x_{3} is given as

(5.49)  x_{3} = π^{−λ+(d−2κ)/2}[y + (−1)^{ν}απ^{−λ}φ[y]x_{2}],

with λ of (5.35).

5.12. On the other hand for the space X in (5.29) we put

(5.50)  Λ = {x ∈ X | φ[x] ∈ 𝔤}.

Then [AQC, Lemma 6.4] implies that Λ is a unique maximal lattice in X. Here we put

(5.51)  w = −π^{i}(q − φ[z])^{−1}φ[z]f_{r} + z − π^{−i}φ[z]e_{r}

with e_{r} and f_{r} in (5.5). Then we find that

x_{2} = π^{−[ν/2]}[w + π^{τ} (q − φ[z])^{−1}φ[z]x_{1}], (5.52)
φ[w] = (1 − s)^{−1}sq, (5.53)
X = F x_{1} + F w + Y, φ(x_{1}, w) = 0, Y = {k ∈ X | φ(k, x_{1}) = φ(k, w) = 0}. (5.54)

Here Y is given in (5.31) and x_{2} is of (5.47).

5.13. In §§5.13 and 5.14 we determine the structure of Λ in the above (5.50). In §5.13 we suppose that: t = 1, r ≥ 1, ν ≡ δ (mod 2), 2 ∈ 𝔭, and ξ(cq) ≠ 1; or t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭; or t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭. (The case when t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b ∈ 𝔤^{×} will be treated in §5.14.) To prove the theorem in this case, it suffices to show that

Λ = (5.55)
𝔤x_{1} + 𝔤x_{2}  if t = 1, r ≥ 1, ξ(cq) ≠ 1, ν ≡ δ (mod 2), and 2 ∈ 𝔭,
𝔤x_{1} + 𝔤x_{2} + 𝔤y  if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭,
𝔤x_{1} + 𝔤x_{2} + 𝔯_{Y} y_{2} if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭

by (5.48). First we shall show that

(5.56)  𝔤x_{1} + 𝔤x_{2} is maximal in F x_{1} + F w.

For the purpose, we consider the Clifford algebra A(F x_{1}+F w) of the restriction of φ to F x_{1} + F w. Put E = F + F wx_{1} in A(F x_{1} + F w). Then we obtain that

(5.57)  F x_{1} + F w = Ex_{1}, φ[xx_{1}] = φ[x_{1}]N_{E/F} (x) for x ∈ E,

and E is isomorphic to F (\sqrt{s}) since (wx_{1})^{2}F^{×2} = sF^{×2} by (5.28) and (5.53). First we suppose that:

t = 1, r ≥ 1, ξ(cq) = 0, ν ∉ 2Z, 2 ∈ 𝔭, and δ = 1; or t = 1, r ≥ 1, ξ(cq) = −1, ν ≡ δ (mod 2), and 2 ∈ 𝔭; or t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭; or t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭.

Then

D_{E/F} = \begin{cases}𝔭^{d} & if t = 1, r ≥ 1, ξ(cq) = 0, ν ∉ 2Z, 2 ∈ 𝔭, and δ = 1, \\ 𝔤 & otherwise. \end{cases} (5.58)

Indeed, if t = 1, r ≥ 1, ξ(cq) = 0, ν ∉ 2Z, 2 ∈ 𝔭, and δ = 1, then cqF^{×2} = sF^{×2}. Thus we have (5.58) by the definition of d; see Theorem 3.5(ii). If t = 1, r ≥ 1, ξ(cq) = −1, ν ≡ δ (mod 2), and 2 ∈ 𝔭, then from (3.11), Lemma 4.2(1), and cq ∉ F^{×2}, we have ε(s) = 2κ with ε of (4.11). Thus we obtain (5.58). If t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭, then ε(s) = 2κ from (3.14), (5.36), and c^{−1}q ∉ N_{K/F} (𝔯). If t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭, then clearly we have (5.58) by (4.34) and Lemma 4.2(1). Therefore we have (5.58) for all cases. Now, by (5.23), (5.36), and Lemma 5.1(3),

τ = \begin{cases}κ + (ν + 1 − d)/2 & if t = 1, r ≥ 1, ξ(cq) = 0, ν ∉ 2Z, 2 ∈ 𝔭, and δ = 1, \\ κ + [ν/2] & otherwise. \end{cases}

Thus by (3.11), (3.14), (3.17), and (5.28), we have (5.59)

φ[x_{1}]𝔤 = \begin{cases}𝔤 & if t = 1, r ≥ 1, ξ(cq) = 0, ν ∉ 2Z, 2 ∈ 𝔭, and δ = 1, \\ 𝔭^{δ} & if t = 1, r ≥ 1, ξ(cq) = −1, ν ≡ δ (mod 2), and 2 ∈ 𝔭, \\ 𝔭^{λ} & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭, \\ 𝔭^{δ} & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭. \end{cases}

Therefore 𝔯_{E}x_{1} is a unique maximal lattice in Ex_{1}, where 𝔯_{E} is the valuation ring of E. Since N_{E/F} (x_{2}x_{1}^{−1}) = φ[x_{2}]φ[x_{1}]^{−1} ∈ 𝔤, we have x_{2}x_{1}^{−1} ∈ 𝔯_{E}. By (5.52) and (5.58), N_{E/F} (x_{2}x_{1}^{−1} − (x_{2}x_{1}^{−1})^{ρ_{E}})𝔤 = D_{E/F}, where 1 ≠ ρ_{E} ∈ Gal(E/F). Thus [AQC, Lemma 5.6(ii)] implies 𝔯_{E} = 𝔤[x_{2}x_{1}^{−1}], and hence 𝔤x_{1} + 𝔤x_{2} is a maximal lattice in Ex_{1} in this case. Next suppose that t = 1, r ≥ 1, ξ(cq) = 0, ν ∈ 2Z, 2 ∈ 𝔭, and δ = 0. Then d ∈ 2Z by Lemma 5.1(3). From this and (5.23), τ = κ + 2^{−1}(ν − d). Thus φ[x_{1}]𝔤 = 𝔭 from (3.11) and (5.28). Put η = π^{2^{−1}d−κ}(1 + π^{τ} φ[z]^{−1}wx_{1}) ∈ E, then η is a root of an Eisenstein equation x^{2} − 2π^{2^{−1}d−κ}x − π^{d−2κ}φ[z]^{−1}(q − φ[z]) = 0. Here x is an indeterminate. Therefore η is a prime element of E and (η − η^{ρ_{E}})𝔯_{E} = (2η − 2π^{2^{−1}d−κ})𝔯_{E} = 𝔡_{E/F}. Here 𝔯_{E} and ρ_{E} are the same symbols as above. Thus 𝔯_{E} = 𝔤[η] by [AQC, Lemma 5.6(ii)]. From φ[x_{1}]𝔤 = 𝔭, we have Λ = 𝔯_{E}η^{−1}x_{1} = 𝔤x_{1} + 𝔤η^{−1}x_{1} with Λ of (5.50). Since η^{−1} = −π^{κ−2^{−1}d}φ[z](q − φ[z])^{−1}(1 − π^{τ} φ[z]^{−1}wx_{1}), we obtain Λ = 𝔤x_{1} + 𝔤x_{2}. This proves (5.56). Now, we obtain (5.55) when t = 1, r ≥ 1, ν ≡ δ (mod 2), ξ(cq) ≠ 1, and 2 ∈ 𝔭 by (5.56). When t > 1, we have

[(𝔤x_{1} + 𝔤x_{2}) : 𝔤x_{1} + 𝔤x_{2}] = (5.60)
\begin{cases}[𝔤 : 𝔭^{2λ}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭, \\ [𝔤 : 𝔭^{2δ}] & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭. \end{cases}

by (5.56), (5.57), (5.58), (5.59), and [AQC, Lemma 8.4(iv)]. Put

Λ_{Y} = \begin{cases}𝔤y & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭, \\ 𝔯_{Y} y_{2} & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭. \end{cases}

Then Λ_{Y} is maximal in Y. Indeed, if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭, then dim_{F} (F y) = 1. We have φ[y]𝔤 = 𝔭^{1−λ} from (5.31) and (5.34), and hence 𝔤y is maximal in Y by [AQC, Lemma 6.4]. If t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭, then K_{Y} in (5.43) is unramified over F and Y = K_{Y} y_{2}. Also we have φ[ky_{2}] = φ[y_{2}]N_{K_{Y} /F} (k) for k ∈ K_{Y}. Thus 𝔯_{Y} y_{2} is maximal in Y by (5.45) and [AQC, Lemma 6.4]. Since Λ_{Y} is maximal in Y, [AQC, Lemma 8.4(iv)] and Lemma 5.2 imply

[ΛỸ : Λ_{Y}] = \begin{cases}[𝔤 : 2𝔭^{1−λ}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭, \\ [𝔤 : 𝔭^{2(1−δ)}] & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭. \end{cases}

Combining this with (5.60), we obtain

[(𝔤x_{1} + 𝔤x_{2} + Λ_{Y}) : 𝔤x_{1} + 𝔤x_{2} + Λ_{Y}] = [(𝔤x_{1} + 𝔤x_{2}) : 𝔤x_{1} + 𝔤x_{2}] · [ΛỸ : Λ_{Y}]

= \begin{cases}[𝔤 : 2𝔭^{1+λ}] & if t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, \\ & and b𝔤 = 𝔭, \\ [𝔤 : 𝔭^{2}] & if t = 3, r ≥ 1, ξ(−cq) = 1, and 2 ∈ 𝔭. \end{cases}

Therefore 𝔤x_{1} + 𝔤x_{2} + Λ_{Y} is maximal in X by [AQC, Lemma 8.4(iv)] and Lemma 5.2. Note that φ[x_{1}]φ[w]φ[y]𝔤 = 𝔭^{ν−2κ+1} by (5.34), (5.53), and (5.59), when t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b𝔤 = 𝔭. Therefore we obtain (5.55).

5.14. Finally, we suppose that t = 2, r ≥ 1, c^{−1}q ∉ N_{K/F} (𝔯), d > 1, and b ∈ 𝔤^{×}. We have already obtained

(5.61)  H = 𝔤π^{τ−i}x_{1} + 𝔤x_{3} + 𝔤y

by (5.48), with x_{1} of (5.27), y satisfying (5.34), and x_{3} of (5.49). Our aim is to show that H coincides with the unique maximal lattice Λ in X if and only if i = τ, with Λ of (5.50). For the purpose, we shall find a 𝔤-basis of Λ in a similar way as §2.3(III) (cf. (2.19)). We consider the Clifford algebra A(X) of the restriction of φ to X. Put

x_{0} = π^{λ−1}x_{1} and w_{0} = π^{−(ν−d+λ)/2}w

with λ of (5.35) and w of (5.51). Then {x_{0}, w_{0}, y} is an F -basis of X satisfying φ(x_{0}, w_{0}) = φ(w_{0}, y) = φ(y, x_{0}) = 0 by (5.54) and Y = F y with Y in (5.31). Thus we obtain

X = T_{X}ζ_{X} and φ[xζ_{X}] = c_{X}N_{B/F} (x) for x ∈ T_{X}.

Here

T_{X} = F x_{0}w_{0} + F x_{0}y + F w_{0}y, B = F + T_{X}, c_{X} = φ[x_{0}]φ[w_{0}]φ[y], ζ_{X} = x_{0}w_{0}y.

Note that B is a division quaternion algebra over F. Moreover we have c_{X}𝔤 = 𝔭^{λ} by (5.28), (5.34), and (5.53). Therefore

(5.62)  Λ = (T_{X} ∩ 𝔓^{−λ})ζ_{X},

where 𝔓 = {x ∈ B | N_{B/F} (x) ∈ 𝔭}. Put

(5.63)  ω_{X} = −x_{1}ζ_{X}^{−1}, u_{X} = x_{3}ζ_{X}^{−1}ω_{X}^{−1}, and v_{X} = u_{X} − u_{X}^{ι}

in B, where ι is the main involution of B. Then we assert that

(5.64)  Λ = [𝔤v_{X} + (𝔤 + 𝔤u_{X})ω_{X}]ζ_{X} = 𝔤x_{1} + 𝔤x_{3} + 𝔤v_{X}ζ_{X}.

Indeed, we have ω_{X}^{2}𝔤 = 𝔭^{1−2λ} by (5.34) and (5.53). Thus we obtain

N_{B/F} (u_{X}) = c_{X}^{−1}N_{B/F} (ω_{X})^{−1}π^{d−2κ−2λ}φ[y](−π^{2κ+1−d}β) ∈ 𝔤^{×},
N_{B/F} (v_{X}) = N_{B/F} (u_{X} − u_{X}^{ι})
= 4π^{−2λ+d−2κ}φ[x_{1}]^{−1}φ[y](s + π^{2κ+1−d}β)(s − 1)^{−1} ∈ 𝔤^{×}

by (5.28), (5.34), (5.38), (5.49), and (5.53). Note that 2κ + 1 > d by d > 1 and Lemma 4.2. Thus [AQC, Lemma 5.6(ii)] implies that F +F u_{X} is an unramified quadratic extension of F and 𝔤 + 𝔤u_{X} is the valuation ring of F + F u_{X}. Also we have v_{X} ∈ 𝔤[u_{X}]^{×}. We see that B = (F + F u_{X}) + (F + F u_{X})ω_{X} and v_{X}ω_{X} = −ω_{X}v_{X} by a straightforward caluculation. Combining these with (5.62), we obtain the first equality of (5.64) in the same way as §2.3(III). The second equality of (5.64) is trivial from (5.63). This proves (5.64). Now, we consider the 𝔤-base of both H and Λ, that is, {π^{τ−i}x_{1}, x_{3}, y} and {x_{1}, x_{3}, v_{X}ζ_{X}}; see (5.61) and (5.64) above. We see that

v_{X}ζ_{X} = 2π^{d−κ−1−(ν+λ)/2}φ[y][w + (−1)^{ν+1}π^{−(ν+λ)/2}αφ[w]y]
= A(π^{τ−i}x_{1}) + Bx_{3} + Cy

with

A = −2φ[y]φ[z](q − φ[z])^{−1}π^{i+d−κ−1−(ν+λ)/2} ∈ 𝔭^{i−τ} ,
B = (−1)^{ν}2α^{−1}π^{λ−1+2^{−1}d} ∈ 𝔤,
C = (−1)^{ν+1}2π^{d−κ−1}(α^{−1} + αφ[y]φ[w]π^{−ν−λ}) ∈ 𝔤^{×}

by (5.23), (5.34), (5.38), (5.53), and d > 1. Thus

(π^{τ−i}x_{1}, x_{3}, y) = (x_{1}, x_{3}, v_{X}ζ_{X})γ, γ = [ \begin{matrix}π^{τ−i} & 0 & −π^{τ−i}AC^{−1} \\ 0 & 1 & −BC^{−1} \\ 0 & 0 & C^{−1} \end{matrix} ] .

Since γ ∈ 𝔤_{3}^{3} and det(γ)𝔤 = 𝔭^{τ−i}, we obtain H = Λ if and only if i = τ, also in this case. This completes the proof.

6 Global results

In this section we assume that F is a global field and L is a maximal lattice in V with respect to φ. We state two global results which are derived from the local cases.

6.1. Let h be the set of nonarchmedean primes of F and fix v ∈ h. We let F_{v} denote the completion of F at v. Then F_{v} is a local field. Let 𝔤_{v} be the valuation ring of F_{v} and 𝔭_{v} the maximal ideal of 𝔤_{v}. We also write 𝔭_{v} for the prime ideal of 𝔤 corresponding to v. Put X_{v} = X ⊗_{F} F_{v} for a subspace X of V and Λ_{v} = Λ ⊗_{𝔤} 𝔤_{v} for a 𝔤-lattice Λ in V. Let φ_{v} be the F_{v}-bilinear extension of φ to V_{v} ×V_{v}. We consider (V_{v}, φ_{v}). By [AQC, Lemma 9.4(iii)], L_{v} is a maximal lattice in V_{v}. For q ∈ 𝔤 ∩ F^{×} such that L_{v}[q] ≠ ∅, put

(6.1)  τ_{v}(q) = max({i ∈ Z | L_{v}[q] ⊃ L_{v}[q, 2^{−1}𝔭_{v}^{i}] ≠ ∅}).

This is given by (3.9), (3.12), (3.15), (3.18), and (3.20) for every v ∈ h.

6.2 Proposition. Let the notation be as above. Let L be a maximal lattice in V and q ∈ 𝔤 ∩ F^{×}. Let t_{v} be the core dimension of (V_{v}, φ_{v}) for v ∈ h. Put n = dim_{F} (V). Then for a 𝔤-ideal 𝔞 = ∏_{v∈h} 𝔭_{v}^{i_{v}} such that 𝔞 ⊂ 𝔤, we have

L[q, 2^{−1}𝔞] ≠ ∅ ⟹ \begin{cases}i_{v} = τ_{v}(q) & if n = t_{v}, \\ i_{v} ≤ τ_{v}(q) & otherwise \end{cases}

for all v ∈ h.

Proof. Assume L[q, 2^{−1}𝔞] ≠ ∅. For every v ∈ h, we have L[q, 2^{−1}𝔞] ⊂ L_{v}[q, 2^{−1}𝔭_{v}^{i_{v}}] since φ(x, L)_{v} = φ_{v}(x, L_{v}) for any x ∈ V. Thus we obtain ∅ ≠ L_{v}[q, 2^{−1}𝔭_{v}^{i_{v}}] ⊂ L_{v}[q] by Lemma 3.6, and hence Corollary 3.7 implies the assertion.

6.3 Theorem. Let the notation be the same as in Proposition 6.2. Let L be a maximal lattice in V. Assume n ≥ 2. Then for h ∈ L such that φ[h] ≠ 0, we have

L ∩ (F h)^{⊥} is maximal in (F h)^{⊥} ⟺ h ∈ L[φ[h], 2^{−1} ∏_{v∈h} 𝔭_{v}^{τ_{v}(φ[h])}].

Here (F h)^{⊥} = {x ∈ V | φ(x, h) = 0}.

Proof. Put W = (F h)^{⊥}. Then we see that W_{v} = (F_{v}h)^{⊥} in V_{v} for all v ∈ h. By [AQC, Lemma 9.4(iii)], L ∩ W is maximal in W if and only if L_{v} ∩ W_{v} = (L ∩ W)_{v} is maximal in W_{v} for every v ∈ h; Moreover Theorem 5.3 shows that this is the case if and only if h ∈ L_{v}[φ[h], 2^{−1}𝔭_{v}^{τ_{v}(φ[h])}] for all v ∈ h. Since φ(h, L)_{v} = φ_{v}(h, L_{v}), the assertion holds.

This theorem answers the question raised in [AQC, (11.6a)].

7 Sums of squares

7.1. Put V = Q_{n}^{1} and φ(x, y) = x·^{t}y for x, y ∈ V. Let L be a maximal lattice in V and {e_{i}}_{i=1}^{n} the standard Q-basis of V in this section. Then φ[x] = ∑_{i=1}^{n} x_{i}^{2} for x = ∑_{i=1}^{n} x_{i}e_{i} ∈ V. Hereafter we assume that q is a squarefree positive integer. By [AQC, (12.17)],

L[q] = L[q, 2^{−1}Z] ⊔ L[q, Z].

Here we apply our results on L[q] in this case and investigate the sets L[q, 2^{−1}Z] and L[q, Z] when 4 ≤ n ≤ 10 and n ∈ 2Z. As for the case n ∉ 2Z, we can refer to [AQC, Section 12].

7.2 Lemma. Assume n ≥ 4. Let L be a maximal lattice in V and q a squarefree positive integer. Then

L[q, Z] = ∅ if \begin{cases}n ≡ 0 (mod 8); or \\ n ≡ ±1 (mod 8) and (−1)^{(n−1)/2}q ≢ 1 (mod 4); or \\ n ≡ ±2 (mod 8) and (−1)^{(n−2)/4}q ≡ 3 (mod 4); or \\ n ≡ 4 (mod 8) and q ≡ 1 (mod 2) \end{cases}

and L[q, 2^{−1}Z] = ∅ if n = 4 and q ≡ 0 (mod 2).

Proof. Let p be a rational prime number. The core dimension t_{p} of (V_{p}, φ_{p}) is given by [AQC, (7.12a) and (7.12b)]. Let c_{p} be as in §2.3 when 1 ≤ t_{p} ≤ 3. By a Witt decomposition of V_{p} as in (2.4), we have (−1)^{(n−t_{p})/2}c_{p}Q_{p}^{×2} = δ(V_{p}, φ_{p}) = Q_{p}^{×2} for t_{p} = 1, 3. From this and [AQC, §7.15], we can take c_{p} so that c_{p} ∈ Z_{p}^{×} when p ≠ 2 and

c_{2} = \begin{cases}(−1)^{(n−t_{2})/2} & if t_{2} = 1, 3, \\ (−1)^{(n−2)/4} & if t_{2} = 2 \end{cases}

when p = 2. Let τ_{p}(q) be as in (6.1). Then we see that

τ_{p}(q) = \begin{cases}1 & if n ≡ ±1 (mod 8), p = 2, and (−1)^{(n−1)/2}q ≡ 1 (mod 4); \\ & or n ≡ ±2 (mod 8), p = 2, and (−1)^{(n−2)/4}q ≢ 3 (mod 4); \\ & or n ≡ ±3 (mod 8) and p = 2; \\ & or n ≡ 4 (mod 8), p = 2, and q ≡ 0 (mod 2), \\ 0 & otherwise \end{cases} (7.1)

by Theorem 3.5 and Lemma 4.2(1). Note that N_{Q(\sqrt{−1})_{2}/Q_{2}}(Z[\sqrt{−1}]_{2}^{×}) = (1 + 4Z_{2})Z_{2}^{×2}. Combining (7.1) with Proposition 6.2, the assertion holds.

If n ≡ ±1 (mod 8), then this lemma is a restatement of [AQC, Lemma 12.13(ii)].

7.3. Put

(7.2)  L_{n} = Ze_{1} + Ze_{2} + ∑_{i=2}^{n/2}(Ze_{2i−1} + Zg_{2i})

for 4 ≤ n ≡ 2, 4, 6 (mod 8), where g_{2i} = 2^{−1}(e_{2i−3} + e_{2i−2} + e_{2i−1} + e_{2i}). Then L_{n} is maximal in V by [CGQ, Lemma 3.1]. When n = 10, we put

(7.3)  Λ = H + M

with a maximal lattice H (resp. M) in ∑_{i=1}^{8} Qe_{i} (resp. Qe_{9} + Qe_{10}). Then Λ is a maximal lattice in V by [NRQ, §6.8]. Hereafter we suppose 4 ≤ n ≤ 10 and n ∈ 2Z. By [AQC, §12.12], if n < 10, then the genus LSO_{A}^{φ} of L (cf. [AQC, §§9.3 and 9.7]) equals to the SO^{φ}-class LSO^{φ}. Here SO_{A}^{φ} is the adelization of SO^{φ}. If n = 10, then [CGQ, §3.2] and [AQC, Lemma 9.23(i)] imply LSO_{A}^{φ} = L_{10}SO^{φ} ⊔ ΛSO^{φ}.

7.4 Lemma. Let L_{n} be as in (7.2) and q a squarefree positive integer. Assume n = 4, 6, or 10. Then we obtain the following assertions:

(1)Assume n = 4. Then we have L_{4}[q, 2^{−1}Z] = ∅ if and only if q ≡ 0 (mod 2) and L_{4}[q, Z] = ∅ if and only if q ≡ 1 (mod 2).

(2) If n > 4, then L_{n}[q, 2^{−1}Z] ≠ ∅.

(3) If n > 4 and q ≡ 0 (mod 2), then L_{n}[q, Z] ≠ ∅.

(4) If n = 6 and q ≡ 3 (mod 4), then L_{6}[q, Z] ≠ ∅.

(5) Assume n = 10 and q ≡ 1 (mod 4). Then we have L_{10}[q, Z] = ∅ if and only if q = 1.

Proof. (1): Assume n = 4. Since L_{4} is maximal in V,

L_{4}[q] = \begin{cases}L_{4}[q, 2^{−1}Z] & if q ≡ 1 (mod 2), \\ L_{4}[q, Z] & if q ≡ 0 (mod 2) \end{cases}

by Lemma 7.2. We have ∑_{i=1}^{4} Ze_{i} ⊂ L_{4}, and hence

(7.4)  L_{4}[q] ≠ ∅ for any squarefree positive integer q.

This proves (1). (2): Assume n > 4. We can take x ∈ L_{4} so that φ[x] = q or q − 1 according as q ≡ 1 (mod 2) or q ≡ 0 (mod 2) by (7.4). If q ≡ 1 (mod 2), then put h = x; if q ≡ 0 (mod 2), then put h = x + e_{5}. By (1) settled above, h ∈ L_{n}[q, 2^{−1}Z] in both cases. This proves (2). In the proof of (3) and (4) we take

(7.5)  y = ∑_{i=1}^{4} y_{i}e_{i} ∈ ∑_{i=1}^{4} Ze_{i} such that φ[y] = q

for a given q. (3): Suppose n > 4 and q ≡ 0 (mod 2). Since q is even and squarefree, at least two of y_{1}, y_{2}, y_{3}, and y_{4} are even. We may assume y_{3}, y_{4} ∈ 2Z. Then y ∈ L_{n}[q, Z] from (1). This proves (3). Now, for h = ∑_{i=1}^{n} h_{i}e_{i} ∈ V such that φ[h] = q, we have

(7.6) h ∈ L_{n}[q, Z] ⟺ h ∈ ∑_{i=1}^{n} Ze_{i} and ∑_{k=0}^{3} h_{2j−k} ∈ 2Z for 2 ≤ j ≤ n/2.

(4): Suppose n = 6 and q ≡ 3 (mod 4). Then one and only one of y_{1}, y_{2}, y_{3}, and y_{4} in (7.5) is even. We may assume y_{1} ∈ 2Z. Put h = ∑_{i=1}^{3} y_{i}e_{i} + y_{4}e_{5}, then h ∈ L_{6}[q, Z] by (7.6). This proves (4). (5): Assume n = 10 and q ≡ 1 (mod 4). Then L_{10}[1, Z] = ∅ by (7.6). If q > 1, then there exists z = ∑_{i=1}^{4} z_{i}e_{i} ∈ ∑_{i=1}^{4} Ze_{i} such that ∑_{i=1}^{4} z_{i}^{2} = 4^{−1}(q − 5). Put h = ∑_{i=1}^{4} 2z_{i}e_{2i} + ∑_{j=1}^{5} e_{2j−1}. Then h ∈ L_{10}[q, Z] by (7.6). This completes the proof.

7.5 Theorem. Let L be a maximal lattice in V and q a squarefree positive integer. We assume 4 ≤ n ≤ 10 and n ∈ 2Z. Then

L[q, Z] = ∅ if and only if \begin{cases}n = 4 and q ≡ 1 (mod 2); or \\ n = 6 and q ≡ 1 (mod 4); or \\ n = 8; or \\ n = 10, L ∈ L_{10}SO^{φ}, q = 1 or q ≡ 3 (mod 4); or \\ n = 10, L ∈ ΛSO^{φ}, q ≡ 3 (mod 4) \end{cases}

and L[q, 2^{−1}Z] = ∅ if and only if n = 4 and q ≡ 0 (mod 2). Here L_{10} (resp. Λ) is of (7.2) (resp. (7.3)).

Proof. If n = 4, 6, or, n = 10 and LSO^{φ} = L_{10}SO^{φ}, then we have LSO^{φ} = L_{n}SO^{φ}. Therefore we obtain the assertion by Lemma 7.2 and Lemma 7.4. Assume n = 8. Then Lemma 7.2 implies L[q] = L[q, 2^{−1}Z]. By [AQC, Lemma 6.2(1)], we may assume ∑_{i=1}^{4} Ze_{i} ⊂ L, and hence L[q] ≠ ∅. This proves our theorem in the case n = 8. Next assume n = 10 and LSO^{φ} = ΛSO^{φ}. Then we may put L = Λ. For x = h + m ∈ H + M = Λ, we have φ[x] = φ[h] + φ[m] and φ(x, Λ) = φ(h, H) + φ(m, M). Thus we obtain H[q, 2^{−1}Z] ⊂ Λ[q, 2^{−1}Z]. From this and the result of the case n = 8, we have Λ[q, 2^{−1}Z] ≠ ∅. Next we consider Λ[q, Z]. We see that L_{6} + Zf_{7} + Zg_{8} (resp. Ze_{9} + Ze_{10}) is maximal in ∑_{i=1}^{8} Qe_{i} (resp. Qe_{9} + Qe_{10}) by [CGQ, Lemma 3.1]. Here L_{6} and g_{8} are given in (7.2) and f_{7} = 2^{−1}(e_{1} + e_{3} + e_{5} + e_{7}). Thus we can put

Λ = H + M = L_{6} + Zf_{7} + Zg_{8} + Ze_{9} + Ze_{10}.

Then, for x = ∑_{i=1}^{10} x_{i}e_{i} ∈ V such that φ[h] = q, we have (7.7)

x ∈ Λ[q, Z] ⟺ x ∈ ∑_{i=1}^{10} Ze_{i}, ∑_{k=0}^{3} x_{2k+1} ∈ 2Z, ∑_{k=0}^{3} x_{2j−k} ∈ 2Z for 2 ≤ j ≤ 4.

Assuming q ≢ 3 (mod 4), we take y so that (7.5). Then at least two of y_{1}, y_{2}, y_{3}, and y_{4} are even. We may assume y_{1}, y_{2} ∈ 2Z. Put h = y_{1}e_{1}+y_{2}e_{2}+y_{3}e_{9}+y_{4}e_{10}, then h ∈ Λ[q, Z] by (7.7). Therefore if Λ[q, Z] = ∅, then q ≡ 3 (mod 4). Combining this with Lemma 7.2, we obtain our theorem.

References

[CGQ] T. Hiraoka, On the class number of the genus of Z-maximal lattices with respect to quadratic form of the sum of squares, J. Math. Kyoto Univ. 46-2 (2006), 291-302.

[EPE] G. Shimura, Euler Products and Eisenstein Series, CBMS Regional Conference Series in Mathematics, No.93, Amer. Math. Soc. , 1997.

[NRQ] G. Shimura, The number of representations of an integer by a quadratic form, Duke Math. J. 100 (1999), 59-92.

[AQC] G. Shimura, Arithmetic and Analytic Theories of Quadratic Forms and Clifford Groups, Mathematical Surveys and Monographs, vol 109 ,Amer. Math. Soc. (2004).

[IQD] G. Shimura, Integer-Valued Quadratic Forms and Quadratic Diophantine Equations, Documenta Math. 11 (2006), 333-367

[BNT] A. Weil, Basic Number Theory, Springer-Verlag, Berlin, Heidelberg, and New York, 1967.

Takashi Yoshinaga

Department of Mathematics Ritsumeikan University Kusatsu Shiga 525-8577 Japan tyt24349@pl.ritsumei.ac.jp 386

Documenta Mathematica 15 (2010)