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Title Selmer Parity of Quadratic Twists of Elliptic Curves

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Author Su, Heng

Publication Date 2015

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Selmer Parity of Quadratic Twists of Elliptic Curves

DISSERTATION submitted in partial satisfaction of the requirements for the degree of

DOCTOR OF PHILOSOPHY

in Mathematics

Heng Su

Dissertation Committee: Professor Karl Rubin, Chair Professor Alice Silverberg Professor Daqing Wan

2015 c 2015 Heng Su Dedication

To my parents, for their care in my pursuit of my dreams.

ii Table of Contents

Acknowledgments iv

Curriculum Vitae v

Abstract of the Dissertation vi

1 Introduction and Background 1 1.1 Introduction ...... 1 1.2 Background ...... 3

2 Selmer Parity Densities of Quadratic Twists 7 2.1 Counting by elements ...... 7 2.2 Counting by fair counting functions ...... 17

3 Cokernel of Local Norm Map 22 3.1 Introduction ...... 22 3.2 Calculating |E(L)/NE(F )| ...... 23 3.3 Proof of Theorem 3.1.1 ...... 30 3.4 Examples and remarks ...... 36

A Convex Bodies 39

iii Acknowledgments

First and foremost, I would like to thank Professor Karl Rubin for his advice and support over the years. However busy he was, he met with me every week and patiently explained mathematical concepts to me. He guided me to develop the theories in my thesis step by step. Every time I submitted a draft of paper, he read word by word. He can always give me valuable advice and detailed explanation, including grammar mistakes. Without his tremendous eﬀort and patience, the completion of my thesis is not possible. I would also thank Professor Daqing Wan and Professor Alice Silverberg for giving me advice on my mathematical career in UC Irvine. I am also grateful to Professors Vladimir Baranovsky, Christopher Davis, Dennis Eichhorn and Liang Xiao for introducing me to the various aspects of algebra and number theory and giving me comprehensive advice on every aspect of mathematics. I would not ﬁnd algebra and number theory so interesting without their guidance. I would like to express my special thanks to my academic brother Myungjun Yu. To me he is both brother and mentor. He is always trying to be helpful and we spent many memorable days and nights working together and encouraging each other. I would thank Rufei Ren, Hayan Nam other friends and fellows in the UCI math circle for their encouragement and support. I appreciate the accommodation and valuable lectures provided by the Arizona Winter School. My works in Section 2.2 were inspired by the talks given by Melanie Wood during the 2014 Arizona Winter School. Lastly, I would express my deepest appreciation to my family and friends for their spiritual support. Without their care, support and most importantly love, I would not be able to complete my Ph.D. study on the other side of the planet.

iv Curriculum Vitae

Heng Su

B.S. in Mathematics and Applied Mathematics, Beihang University, 2011

Ph.D. in Mathematics, University of California, Irvine, 2015

v Abstract of the Dissertation Selmer Parity of Quadratic Twists of Elliptic Curves

Heng Su

Doctor of Philosophy in Mathematics

University of California, Irvine, 2015

Professor Karl Rubin, Chair

Inspired by the paper of Klagsbrun, Mazur and Rubin [5], this thesis investigates the disparity of 2-Selmer ranks of quadratic twists of an arbitrary elliptic curve E over an arbitrary number field K. In the first part, we calculate the density of quadratic twists of E with even 2-Selmer ranks under two different counting methods. First we count twists by elements inside a large convex body of the Euclidean space that contains the integer lattice of K. The second counting method is counting quadratic twists EL by the norms of the finite part of conductors of quadratic extensions L/K. Under both counting methods we give an explicit formula for the densities, which are finite products of local factors. In the second part of the paper we give a method that uses Tate’s algorithm to calculate the size of the cokernel of the local norm maps of E at places over 2, assuming that E has good reduction. With this method we can extend Kramer’s early work on the cokernel of the local norm maps, and compute the local factors mentioned above in some additional cases.

vi Chapter 1

Introduction and Background

1.1 Introduction

We investigate the distribution of the parities of 2-Selmer ranks in the quadratic twist family of an arbitrary elliptic curve E over an arbitrary number field K. Let Eχ (resp. Ed) be the quadratic twist (Definition 1.2.3) of E by the quadratic character χ (resp. an element d ∈ OK , d 6= 0) of K, denote r(χ) (resp. r(d)) the 2-Selmer rank (Definition 1.2.4) of Eχ (resp. Ed). In Chapter 2, we will give an explict formula for the density of quadratic

d twists E with even 2-Selmer ranks, as d varies over a large subset of OK , and a formula for the density of quadratic twists Eχ with even 2-Selmer ranks counting by the ﬁnite part of conductor of χ.

To be precise, in Section 2.1, we identify the ring of integers OK of K with an N- dimensional lattice in RN under the map:

d 7→ (σ1d, ..., σr1 d, Re(τ1d), Im(τ1d), ..., Re(τr2 d), Im(τr2 d)) ,

where N is the extension degree [K : Q], σ1, ··· , σr1 are the real embeddings of K and N τ1, ··· , τr2 are the complex embeddings of K. Let C be a convex body in R . For real numbers λ > 0, let λC := {λv : v ∈ C} ⊂ RN . Deﬁne

|{d ∈ O ∩ λC : r(d) is even}| ρ(E/K, C) := lim K , λ→∞ |OK ∩ λC|

if the limit exists. We get the following theorem that evaluates ρ(E/K, C).

1 Theorem A. Suppose that E is an elliptic curve deﬁned over K. Then for any convex body C in RN 1 ξ(C) ρ(E/K, C) = + . 2 2

r(1K ) Q Here ξ(C) = (−1) ι(C) v∈Σ ξv is a finite product of local data (see Definition 2.1.5). In Section 2.2, we count quadratic twists Eχ of E by “fair” counting functions c of χ. In particular, c could be the norm of finite part of the conductor or norm of product of ramified primes of χ (see definition of fairness in Definition 2.2.2). Define

|{quadratic characters χ over K : r(χ) is even, c(χ) < X}| Pc(E/K) := lim , X→∞ |{quadratic characters χ over K : c(χ) < X}|

if the limit exists. We will show

Theorem B. Suppose that E is an elliptic curve deﬁned over a number ﬁeld K. Then

1 β P (E/K) = + c . c 2 2

r(1K ) Q Here βc = (−1) v∈Σ βc,v is a finite product of local probabilities (see Definition 2.2.7). Theorem A is based on the joint work of Klagsbrun, Mazur and Rubin [5], who proved an analogue of Theorem A with a different ordering of characters and different values of ξ (see Theorem 1.2.9 below). They showed that

Y (1.1.1) r(χ) ≡ r(1K ) (mod 2) ⇐⇒ ωv(χv) = 1, v∈Σ where Σ is a ﬁnite set of places depending only on E and K, χv’s are localizations of χ and

ωv’s are local factors (see Deﬁnition 1.2.6). Theorem B is a combination of [5] and work of Wood [13]. Chapter 3 deals with the problem of calculating the size of the cokernel of local norm map N : E(F ) → E(L) of an arbitrary elliptic curve E over an arbitrary ﬁnite extension

L of Q2 and F/L quadratic, inspired by the work of Kramer [6][Proposition 4]. If we add condition that the minimal model of E has no xy term, then we have the following theorem

Theorem C. Let L be a ﬁnite extension of Q2 with residue ﬁeld k and valuation v. Let √ F = L( d) be a quadratic extension of L. Suppose E/L has good reduction and has the

2 following minimal Weierstrass model

2 3 2 y + a3y = x + a2x + a4x + a6.

Then

dimF2 E(L)/NE(F ) ≡ [k : F2] · ordvd (mod 2)

The above theorem is Theorem 3.1.1. This will allow us to calculate ωv in (1.1.1) and deal with the case when E has good reduction and v|2, which is not given in the table of [5][Propostion 7.9]. More generally, Theorem 3.2.1 shows that using Tate’s algorithm we can eﬀectively calculate the size of the cokernel E(L)/NE(F ) assuming that E has good reduction over L and F/L ramiﬁed.

1.2 Background

The notations here are all from [5]. Let E be an elliptic curve defined over a number field K. Fix a finite set Σ of places of K, containing all places above 2, places of bad reduction and all archimedean places. Let Kv denote the completion of K at place v of K. Fix an algebraic closure K¯ of K.

Deﬁnition 1.2.1. For every place v of K, let

1 1 Hf (Kv,E[2]) = image(E(Kv)/2E(Kv) ,→ H (Kv,E[2]))

be the image of Kummer map obtained from GKv -cohomology of the exact sequence

2 0 / E(Kv)[2] / E(Kv) / E(Kv) / 0.

Deﬁne the 2-Selmer group of E/K as

1 1 Sel2(E/K) := {c ∈ H (K,E[2]) : resv(c) ∈ Hf (Kv,E[2]) for all v of K},

1 1 where resv : H (K,E[2]) → H (Kv,E[2]) is the restriction map at v.

3 Definition 1.2.2. If L is a field, let GL be the absolute Galois group of L, define

C(L) := Hom(GL, {±1}).

If L is a global ﬁeld and v is a place of L, then for any χ ∈ C(L) denote χv the localization of χ at v.

2 3 2 Definition 1.2.3. Let E be defined over a field L given by E : y = x +a2x +a4x+a6. Let χ ∈ C(L) be a quadratic character and let F/L be a quadratic extension corresponding to χ √ given by F = L( d) where d 6= 0. The quadratic twist of E by d (resp. by F or χ), denoted

d F χ 2 3 2 2 3 E (resp. E or E ), is the elliptic curve given by the model y = x + da2x + d a4x + d a6.

The 2-Selmer group of an elliptic curve is a ﬁnitely generated Z/2Z-module.

Deﬁnition 1.2.4. Deﬁne the 2-Selmer rank E as

χ r(E/K) := dimF2 Sel2(E ).

Let χ ∈ C(K), denoted r(χ) the 2-Selmer rank of Eχ.

∼ χv Note that for any χv ∈ C(Kv) there is a canonical GKv -isomorphism E [2] −→ E[2], given √ by (x, y) 7→ (x/d, y/d d), where d is a non-square element that induces χv. We then have an isomorphism

1 χv ∼ 1 φχv : H (Kv,E [2]) −→ H (Kv,E[2]),

1 χv 1 so that we can view H (Kv,E [2]) as a subgroup of H (Kv,E).

Deﬁnition 1.2.5. For any χv ∈ C(Kv). Let

φ χv χv 1 χv χv 1 αv(χv) = image(E (Kv)/2E (Kv) ,→ H (Kv,E [2]) −−→ H (Kv,E[2]))

be the image of the composition of Kummer map obtained from GKv -cohomology of the exact sequence

χv χv 2 χv 0 / E (Kv)[2] / E (Kv) / E (Kv) / 0,

and φχv . Then we deﬁne

hv(χv) := dimF2 αv(χv)/(αv(χv) ∩ αv(1v)).

4 Deﬁnition 1.2.6. Let χ ∈ C(K). For every v ∈ Σ, deﬁne a map ωv : C(Kv) → {±1} by

hv(χv) ωv(χv) := (−1) χv(∆), where ∆ is the discriminant of (any model of) E.

In this paper we will frequently use the following results from [5].

Theorem 1.2.7 (Proposition 7.2 [5]). Suppose χ ∈ C(K). Then

Y r(χ) ≡ r(1K ) (mod 2) ⇐⇒ ωv(χv) = 1. v∈Σ Theorem 1.2.8 (Proposition 7.9 [5]). Suppose E is an elliptic curve over K. For every

± × × 2 c ∈ Σ deﬁne mv := |{γ ∈ C(Kv): ωv(γ) = ±1}|, and let cv := |Kv /(Kv ) |, so cv = 4 if v - 2∞. Then we have the following table, where if v - ∞ then“type” denotes the Kodaira type of the Ne´ron Model.

+ − type of v mv mv

real 1 1

complex 1 0

split multiplicative 1 cv − 1

∗ type Iν or Iν, ν > 0, not split multiplicative cv − 1 1

∗ good reduction or type I0, v - 2 4 0

∗ ∗ × 2 type II, IV, II , IV , ∆ ∈ (Kv ) , v - 2 4 0

∗ ∗ × 2 type II, IV, II , IV , ∆ ∈/ (Kv ) , v - 2 2 2

∗ × 2 type III, III , −1 ∈ (Kv ) , v - 2 4 0

∗ × 2 type III, III , −1 ∈/ (Kv ) , v - 2 2 2 Using Theorem 1.2.7, Z. Klagsbrun, B. Mazur and K. Rubin studied the disparity of Selmer ranks of quadratic twists. More precisely, they proved

Theorem 1.2.9 (Theorem 7.6 [5]). If X > 0, denote

C(K,X) := {χ ∈ C(K): if χ is ramiﬁed at q, then N(q) < X}.

Then for all suﬃciently large X, we have

5 |{χ ∈ C(K,X): r(χ) is even}| 1 + δ = , |C(K,X)| 2

r(1K )Y where δ := (−1) δv, and v∈Σ

+ − 1 X mv − mv δv := ωv(χ) = , |C(Kv)| |C(Kv)| χ∈C(Kv)

± where mv are as in Theorem 1.2.8.

Remark 1.2.10. These δv’s are diﬀerent from the ξv’s and βc,v’s in Theorem A and B

(see Remark 2.1.14 and Example 2.2.11). However ξv = βc,v except for places over 2 (see Proposition 2.2.10).

6 Chapter 2

Selmer Parity Densities of Quadratic Twists

2.1 Counting by elements

Fix an arbitrary number ﬁeld K of degree N over Q. We identify its ring of integers OK as an N-dimensional lattice in RN via the map

(2.1.1) d 7→ (σ1d, ..., σr1 d, Re(τ1d), Im(τ1d), ..., Re(τr2 d), Im(τr2 d)) ,

for any d ∈ OK , where σ1, ··· , σr1 are the real embeddings of K and τ1, ··· , τr2 are the complex embeddings of K. This map gives a bijection from the family of integral bases of

K to the family of fundamental parallelograms of OK . To be precise, given an integral basis

{α1, ..., αN } of K, the corresponding fundamental parallelogram is n X o F = ciαi : 0 ≤ ci < 1 .

We define a convex body in RN to be a compact convex set with non-empty interior. In this chapter we fix E as an elliptic curve defined over K. Let r(d) be the 2-Selmer rank

d of the quadratic twist E of E by d ∈ OK , d 6= 0. As mentioned in Section 1.2, Σ is the ﬁnite set of places of K, containing all places above 2, places of bad reduction and all archimedean √ places. Denote χd the quadratic character of K( d)/K and χd,v the localization of χd at place v.

Deﬁnition 2.1.1. For any convex body C ⊂ RN . Deﬁne |{d ∈ O ∩ λC : r(d) is even}| ρ(E/K, C) := lim K . λ→∞ |OK ∩ λC|

7 Our goal is to evaluate ρ(E/K, C). We will show that

1 ξ(C) ρ(E/K, C) = + , 2 2 where ξ(C) is given by an explicit ﬁnite product of local factors (see Deﬁnition 2.1.5)

Theorem 1.2.7 shows that r(χ) ≡ r(1K ) (mod 2) if and only if the condition below holds

Y ωv(χv) = 1. v∈Σ Consider the composition of maps

Q ω Q v∈Σ v p : OK −→ C(K) −→ v∈Σ C(Kv) −−−−−→ {±1} Y d 7−→ χd 7−→ (ωv(χd,v))v∈Σ 7−→ ωv(χd,v). v∈Σ

We have r(d) is even if and only if p(d) = (−1)r(1). Complex places have no nontrivial characters, thus p(d) is determined by the ﬁnite places and real places. Denote by Σ0 the subset of ﬁnite places of Σ. Next we will study how √ 0 Kv( d)/Kv depends on the residue class of d modulo a high power of v for v ∈ Σ . To be precise, we have the following proposition.

0 Proposition 2.1.2. Let Ov be the ring of integers of Kv, d, d ∈ Ov, d 6= 0. If d ≡ √ √ 0 0 d (mod 4vd), then Kv( d) = Kv( d ).

0 0 × 2 Proof. By Hensel’s lemma, if d /d ≡ 1 (mod 4v), then d /d ∈ (Kv ) , and the proposition follows.

0 k Corollary 2.1.3. Under the same notation as in Proposition 2.1.2, suppose d, d ∈ Ov − v . √ √ 0 k+2ordv2 0 If d ≡ d (mod v ), then Kv( d) = Kv( d ). √ k ¯ Corollary 2.1.3 shows that if d ∈ Ov − v , then Kv( d)/Kv is determined by d the k congruence class d (mod 4v ). So we denote by ωv(χd,v¯ ) the common value of ωv(χd0,v) when d0 ∈ d¯.

Deﬁnition 2.1.4. Deﬁne hyperoctants of RN to be the collection of points in RN whose

r1 signs of the ﬁrst r1 coordinates follow the same pattern. So there are 2 hyperoctants of

8 N R . Any two hyperoctants only possibly intersect at the boundaries. Let ψ(Hi) = ±1 be the product of signs of the ﬁrst r1 coordinates of a hyperoctant Hi. For any convex body C, deﬁne r P2 1 ψ(H )Vol(C ∩ H ) ι(C) := i=1 i i . Vol(C)

Next we deﬁne ξ and local factors ξv that are needed in the theorems. We ﬁx the notation N(I) = [O : I].

+ ¯ 2 2 2 Deﬁnition 2.1.5. For each v - ∞, let λv be the number of cosets d in Ov/4v − v /4v such

that ωv(χd,v¯ ) = 1. Deﬁne 2λ+ ξ := v − 1. v N(4v2) − N(4) For any convex body C, deﬁne

r(1K ) Y ξ(C) := (−1) ι(C) ξv. v∈Σ0

1 ξv From the deﬁnition of ξv’s, one easily checks that when v is ﬁnite 2 + 2 is the proportion ¯ 2 2 2 of cosets d in Ov/4v − v /4v such that ωv(χd,v¯ ) = 1. Therefore it is the proportion of

2 2+2ordv2 residue classes in OK − v modulo v such that ωv(χd,v) equals 1 if d is contained in

these residue classes. Note that ξv = 1 if v 6∈ Σ by the chart in Theorem 1.2.8.

Theorem 2.1.6. Suppose C is an N-dimensional convex body in RN . Then

|{d ∈ O ∩ λC : r(χ ) is even}| 1 ξ(C) lim K d = + . λ→∞ |OK ∩ λC| 2 2 Before proving the Theorem, we ﬁx some notations and give some deﬁnitions and lemmas.

We ﬁx an integral basis α1, ..., αN of OK . Denote by F the fundamental parallelogram

induced by this basis. Let l be the diameter of F . For any nonzero ideal J of OK , we

identify J as an N-dimensional sublattice of OK .

0 kv For each v ∈ Σ we ﬁnd a positive integer kv such that v is a principal ideal and ﬁx a

generator γv ∈ OK . We can assume kv’s are all divisors of the class number of K.

Deﬁnition 2.1.7. For any tuple of positive integers n = (··· , nv, ··· )v∈Σ0 , deﬁne

2n Y 2nv γ := γv . v∈Σ0

9 Deﬁnition 2.1.8. Deﬁne

[ 2 Γ := OK − (γv ). v∈Σ0

0 From Corollary 2.1.3, each ωv of v ∈ Σ induces a map (we use the same symbol as ωv)

2kv+2ordv2 2kv 2kv+2ordv2 ωv : OK /v − v /v −→ {±1} ¯ d 7−→ ωv(χd,v¯ ).

Q 2kv Q 2kv+2ordv2 0 Let I be the ideal v∈Σ0 v , then v∈Σ0 v = 4I, since Σ contains all the places over 2. Under the restriction maps of the Chinese remainder theorem there is a bijection of sets

n o [ 2 Y 2kv+2ordv2 2kv 2kv+2ordv2 (φv)v∈Σ0 : OK /4I − (γv )/4I −→ OK /v − v /v . v∈Σ0 v∈Σ0

r(1K ) Q Composing (φv)v∈Σ0 and (−1) v∈Σ0 ωv we get a map

S 2 ω : OK /4I − v∈Σ0 (γv )/4I −→ {±1}

¯ r(1K ) Q ¯ b 7−→ (−1) v∈Σ0 ωv(φv(b)).

Proposition 2.1.9. For every ﬁnite place v,

−1 2|ωv (1)| ξv = − 1. N(v2kv+2ordv2) − N(v2ordv2)

Proof. Let v be a ﬁnite place. by Deﬁnition 2.1.5 of ξv, it is enough to show

−1 ¯ 2 2 2 |ωv (1)| |{d ∈ Ov/4v − v /4v : ωv(χd¯) = 1}| 2k +2ord 2 2ord 2 = 2 2 2 . N(v v v ) − N(v v ) |{Ov/4v − v /4v }|

2kv This is true because we can partition Ov − v into the disjoint union

kv−1 a 2j 2 (π )(Ov − v ), j=0

2 where πv is a uniformizer for Ov. The disjoint sets in the union all diﬀer from Ov − v by a square. Hence passing to quadratic local ﬁeld extensions, the proportion of classes in

2kv 2kv+2ordv2 Ov − v modulo v which induce local quadratic characters χ such that ωv(χ) = 1

2 2+2ordv2 is the same as the proportion of elements in Ov − v modulo v which induce local

quadratic characters such that ωv(χ) = 1.

10 Lemma 2.1.10. Under the above notations,

−1 r(1K ) Q |ω (±1)| 1 (−1) 0 ξ = ± v∈Σ v . S 2 |OK /4I − v∈Σ0 (γv )/4I| 2 2 Proof. We prove by induction on the cardinality s of Σ0. When s = 1 the statement is true by Proposition 2.1.9. Suppose the statement is true for s = n − 1, i.e. when Σ0 has n − 1 places, −1 r(1K ) Q |ω (±1)| 1 (−1) 0 ξ = ± v∈Σ v . S 2 |OK /4I − v∈Σ0 (γv )/4I| 2 2 Now we add one more place v0 to the set Σ0. Note that ω(¯b) = 1 if and only if either ¯ ¯ ¯ ωv0 (φv0 (b)) = 1 and the product of the other ωv(φv(d))’s is 1 or ωv0 (φv0 (b)) = −1 and the ¯ product of the other ωv(φv(d))’s is −1. By Proposition 2.1.9,

−1 1 ξ 0 |ω 0 (±1)| ± v = v . 2 2 N(v02kv0 +2ordv0 (2)) − N(v02ordv0 (2))

02kv0 +2ordv0 (2) Q 2kv+2ordv2 Since v is comaximal with v∈Σ0,v6=v0 v , by the Chinese remainder theorem,

−1 r(1K ) Q ! |ω (±1)| 1 (−1) 0 0 ξv 1 ξ 0 = + v∈Σ ,v6=v ± v S 2 |OK /4I − v∈Σ0 (γv )/4I| 2 2 2 2

r(1K ) Q ! 1 (−1) 0 0 ξv 1 ξ 0 + − v∈Σ ,v6=v ∓ v 2 2 2 2

r(1K ) Q 1 (−1) 0 ξ = ± v∈Σ v . 2 2 Thus we have proved the lemma.

Proof of Theorem 2.1.6. We will first prove Theorem 2.1.6 in the case when C is contained in a hyperoctant H of RN . By Theorem 1.2.8, the contribution of infinite places to p(d) is ψ(H), the product of signs of H over the real places. From Definition 2.1.8, Γ is the set of

2kv 2kv algebraic integers in K away from v ’s. Thus Γ ⊂ Ov − v canonically. By Corollary √ 2.1.3, if d is contained in Γ, then the local ﬁeld extension Kv( d)/Kv is determined by d (mod v2kv+2ordv2). By the Chinese remainder theorem, the product p(d) is determined by

Y d (mod v2kv+2ordv2 = 4I) v∈Σ0

Q 2kv (recall I = v∈Σ0 v ) and ψ(H).

11 However, the set Γ is exactly the union of the classes modulo I such that d∈ / v2kv for

Q 2kv+2ordv2 2ordv2 any ﬁnite v ∈ Σ. Passing to 4I, Γ is the union of v∈Σ0 (N(v ) − N(v )) classes

modulo 4I. We view the ideal 4I as an N-dimensional sublattice of OK . By Corollary A.0.13, when λ is suﬃciently large, each of these residue classes occurs

Vol(C) λN + O(λN−1) N(4I)Vol(F )

times in λC. The error term O(λN−1) is bounded by cλN−1, where

2N lA(∂C) c = Vol(F ) depends only on C and F . Because |ω−1(ψ(H))| is the number of residue classes modulo 4I such that p(d) = (−1)r(1K ) if d ∈ Γ is contained in these residue classes, the number of such d ∈ Γ is Vol(C) |ω−1(ψ(H))| λN + O(λN−1) . N(4I)Vol(F )

2n Q 2nv Let us observe the set Γ. Recall γ := v∈Σ0 γv for n = (··· , nv, ··· )v∈Σ0 of Deﬁnition

2.1.7. We have a partition of OK − {0} as follows:

a 2n OK − {0} = γ Γ. n √ 2n 2n 2kv+2ordv2 For any d ∈ γ Γ, the local ﬁeld extensions Kv( d)/Kv are determined by d/γ (mod v ), since d/γ2n diﬀers from d only by a square. Therefore the number of d ∈ γ2nΓ ∩ λC such that p(d) = (−1)r(1K ) is

1 Vol(C) |ω−1(ψ(H))| · λN + O(λN−1) . N(γ2n) N(4I)Vol(F )

Again the error term inside the expression is bounded by cλN−1. The total number of d ∈ λC

such that p(d) = (−1)r(1K ) has a main term

X |ω−1(ψ(H))| Vol(C) · λN . N(γ2n) N(4I)Vol(F ) n,γ2nΓ∩λC6=∅

The error term is at most the number of summands times c|ω−1(ψ(H))|λN−1. Let

q q n n 2 2 2 2 oo M = sup sup |x1|, ··· , |xr1 |, xr1+1 + xr1+2, ··· , xN−1 + xN , (x1,··· ,xN )∈C

12 N then for any d ∈ OK such that |N(d)| > (λM) , d is not contained in λC, for at least one N of the coordinates of d in R has absolute value greater than M. Noticing that |N(γv)| ≥ 2 N 2n for all v, if nv > log2(λM) for some v, then γ ∩ λC = ∅. Hence the number of summands is an O(logsλ) where s is the number of places in Σ0. The error term of the summation is an O(λN−1logsλ).

Vol(C) N N−1 Note that |OK ∩ λC| = Vol(F ) λ + O(λ ). We have

P |ω−1(ψ(H))| Vol(C) N n,γ2nΓ∩λC6=∅ 2n · λ ρ(E/K, C) = lim N(γ ) N(4I)Vol(F ) λ→∞ Vol(C) N Vol(F ) λ X 1 1 |ω−1(ψ(H))| Y 1 = |ω−1(ψ(H))| · · = (1 − )−1 N(γ2n) N(4I) N(4I) N(γ2) n v∈Σ0 v |ω−1(ψ(H))| |ω−1(ψ(H))| = = . Q 2 S 2 N(4) v∈Σ0 (N(γv ) − 1) |OK /4I − v∈Σ0 (γv )/4I|

By Lemma 2.1.10 and Deﬁnition 2.1.5,

r(1K ) Q 1 (−1) ψ(H) 0 ξ 1 ξ(C) ρ(E/K, C) = + v∈Σ v = + . 2 2 2 2

Now we prove the general case in which C is an arbitrary convex body in RN . The

N hyperoctants of R are H1, ··· ,H2N , so C is the union of convex bodies H1 ∩C, ··· ,H2N ∩C, where any pair of these sets intersect only at the boundary. Each Hi ∩ C is contained in the hyperoctant Hi. So for suﬃciently large λ, the density of elements d such that

r(1K ) p(d) = (−1) in Hi ∩ λC is

r(1K ) Q 1 (−1) ψ(H ) 0 ξ + i v∈Σ v . 2 2

N−1 s The error term of counting in Hi ∩ λC is O(λ log λ) and it comes from the boundary. On the other hand, the number of intersecting boundaries is at most N2N−1 which is a ﬁxed number, hence the error term of counting in λC is still O(λN−1logsλ).

13 Hence the overall density ρ(E/K, C) is

N 2 r(1K ) Q ! 1 X 1 (−1) ψ(Hi) 0 ξv + v∈Σ Vol(H ∩ C) Vol(C) 2 2 i i=1 N r(1K ) Q P2 1 (−1) 0 ξ ψ(H )Vol(H ∩ C) = + v∈Σ v i=1 i i 2 2 Vol(C)

r(1K ) Q 1 (−1) ι(C) 0 ξ = + v∈Σ v 2 2 1 ξ(C) = + . 2 2

Recall that |{d ∈ O ∩ λC : r(d) is even}| ρ(E/K, C) = lim K . λ→∞ |OK ∩ λC| we now have the following corollary.

Corollary 2.1.11. Suppose K has a real embedding and the convex body C is symmetric

in the sense that if (x1, ··· , xN ) ∈ C, then (±x1, ··· , ±xN ) are also contained in C. Then 1 ρ(E/K) = 2 .

Proof. In this case ξ(C) = ι(C) = 0.

Remark 2.1.12. Suppose the convex body C is contained in the union of hyperoctants of RN with all positive signs at the real places. Then ρ(E/K, C) is independent of C, and we write ρ(E/K)+ in this case to denote the density of even quadratic twists for all such C. The following example shows the value of ρ(E/K)+ is dense in [0, 1] as K varies.

Example 2.1.13. Let E be the elliptic curve labelled 50B1 in [2]:

y2 + xy + y = x3 + x2 − 3x − 1

√ and let K be a ﬁnite extension of Q( −2), unramiﬁed at 5. Let N be the degree of K/Q.

Then √ 1 (−1)[K:Q( −2)] Y 1 N(v) ρ(E/K)+ = + (1 − ). 2 2 2[Kv:Q2] N(v) + 1 v|2 As K varies, these values are dense in the interval [0, 1].

14 Proof. There is no confusion if we write ρ(E/K)+ since there are no real embeddings of K. √ According to [5, Example 7.11], r(1K ) ≡ n∞ + n2 (mod 2), where n∞ = [K : Q( −2)] and n2 is the number of places in K above 2. The ﬁeld K has no real embeddings, and E has good reduction at all primes of K not dividing 10. Over Q2, E has multiplicative reduction,

so E has split multiplicative reduction at every prime of K over 2. Over Q5, E has Kodaira type II, and since K/Q is unramiﬁed at 5, E has Kodaira type II at all primes of K above

n∞+n2 Q 1 ξv 5. By Proposition 1.2.8, ξ = (−1) v|2 ξv, where 2 + 2 is the fraction of classes in

2+2ordv2 2 2+2ordv2 Ov/v − v /v that induce trivial quadratic characters over Kv. √ 2+2ordv2 2 2+2ordv2 2 For d ∈ Ov/v − v /v , Kv( d)/Kv is trivial if and only if x − d has

a root in Ov. It follows that ordvd is even, and equals 0, so ordvx = 0. By Hensel’s

lemma (see for example [7, (II, §2) Proposition 2]), if there exists an x ∈ Ov such that 2 2 ordv(x − d) > 2ordv(2x), then x − d is solvable in Ov. Thus the existence of such x is 2 2e+1 equivalent to the existence of solution of x −d ≡ 0 (mod v ) , where e = ordv2. According 2e+1 × 2e to [4, VI. Section 6, Chapter 15.], (Ov/v ) has order N(v )(N(v) − 1) isomorphic to a

direct sum Z/(N(v) − 1)Z ⊕ G, the group G is a direct product of [Kv : Q2] + 1 cyclic groups of 2nd power order. The number of squares in this group is N(v2e)(N(v) − 1)/2[Kv:Q2]+1.

2+2ordv2 2 2+2ordv2 In Ov/v − v /v the number of elements that induce trivial character is then N(v2e+1)(N(v) − 1)/2[Kv:Q2]+1. Thus

1 ξ 1 N(v2e+1)(N(v) − 1) 1 N(v) + v = = . 2 2 2[Kv:Q2]+1 N(v2e+2) − N(v2e) 2[Kv:Q2]+1 N(v) + 1

Hence 1 N(v) ξv = − 1, 2[Kv:Q2] N(v) + 1 and ρ(E/K)+ is in the desired form. To prove the final assertion, suppose L is a finite extension of Q, unramified at 5, in √ which 2 splits completely, and let t := [L : Q]. Let K := L( 2m −2) with m ≥ 1. Then K is √ unramified at 5, [K : Q( −2)] = tm, n2 = [L : Q] = t, and if v | 2 then [Kv : Q2] = 2m and N(v) = 2. In this case the quantity in the formula of the theorem is

1 (−1)tm 1 + (1 − )t. 2 2 3 · 22m−1

15 As m and t vary, the sets 1 1 {log((1 − )t): tm even}, {log((1 − )t): tm odd} 3 · 22m−1 3 · 22m−1

are both dense in R≤0. It follows from the continuity of the exponential function that the tm 1 t set {(−1) (1 − 3·22m−1 ) } is dense in [−1, 1]. This completes the proof.

Remark 2.1.14. The ξv’s are diﬀerent from the δv’s in Theorem 1.2.9 ([5, Theorem 7.6]). In √ √ Theorem 1.2.9, the local factors δv ∈ [−1, 1]∩Z[1/2]. However if we take K = Q( 13, −2) in Example 2.1.13, then K is unramiﬁed at 5 and there is a unique prime v|2. In this case

√ 4 [K : Q( −2)] = 2, [Kv : Q2] = 2 and N(v) = 4. Hence ξ = ξv = − 5 6∈ Z[1/2].

Example 2.1.15. This example shows that even if the ﬁeld K has a real embedding but we count the twist by elements in a given quadrant, the distribution of 2-Selmer rank parity may be skew. Let E/Q be the elliptic curve 722C1 in [2]. In fact, E is the quadratic twist √ of the elliptic curve 38B1 in [2] by the quadratic character of Q( −19)/Q. We are going to calculate the following limit: |{d ∈ Z : d ≤ X and r(χ ) is even}| lim ≥0 d . X→∞ |{d ∈ Z≥0 : d ≤ X}| Over Z, E has 2-Selmer rank 2. The places where E has bad reduction are 2 and 19. At 2, E has non-split multiplicative reduction. By the table in Theorem 1.2.8, there is only one local quadratic character of Q2 that will change the parity of 2-Selmer rank, namely the nontrivial unramiﬁed character. The residue classes in Z/16Z − 4Z/16Z that induce the

+ 2(2) 2 nontrivial unramiﬁed character are 5 and 13. Thus λ2 = 2 and ξ2 = 12 − 1 = − 3 . At ∗ 19, E has additive reduction with Kodaira type I1. Since E is the twist of elliptic curve √ 38B1 by −19, and 38B1 has non-split multiplicative reduction at 19 over Z, the only local

quadratic character of Q19 that change the parity of 2-Selmer rank of E is the product of √ the nontrivial unramified character and the character of Q19( −19)/Q19, namely the other ramified character. The number of residue classes in Z/192Z − 0 that induce this ramified

+ 1 2(9) 19 character is λ19 = 2 (19 − 1) = 9. Hence ξ19 = 192−1 − 1 = − 20 . Now the convex set in R is the interval [0,X], the numbers all have the same sign, so

2 19 ξ = (−1) ξ2ξ19 = 30 . Hence |{d ∈ Z : d ≤ X and r(χ ) is even}| 1 19 lim ≥0 d = ρ(E/Q) = + . X→∞ |{d ∈ Z≥0 : d ≤ X}| 2 60

16 2.2 Counting by fair counting functions

Recall C(K) := Hom(GK , {±1}), is the group of quadratic characters of the absolute Galois

group of K. In this section will ﬁrst deﬁne fairness of a counting function c : C(K) → Z≥0, then we evaluate the following density:

|{χ ∈ C(K): r(χ) is even, c(χ) < X}| Pc(E/K) := lim , X→∞ |{χ ∈ C(K): c(χ) < X}|

when the limit exists, for a fair counting function c. We want the norm of finite part of conductor of a quadratic character to be a fair counting function. One can also find the definition of fairness and the computation of density of quadratic characters with local specifications in Section 2 of Wood’s paper [13].

We deﬁne a quadratic Kv-algebra to be either a quadratic extension of Kv or the direct

product of two copies of Kv.

Remark 2.2.1. Every quadratic Kv-algebra comes from quadratic extensions of the global

ﬁeld K. If L/K is a quadratic extension, then for each place v of K, Lv = L ⊗K Kv is a

quadratic Kv-algebra. If L/K induces nontrivial extension Lw/Kv, then Lv is Lw. Otherwise

Lv is Kv ⊕ Kv.

Let S be a finite set of places of K, a local specification Θ is a collection {Θv}v∈S of quadratic Kv-algebra for each v ∈ S. If there exist a quadratic extension L/K such that ∼ Lv = Θv for all v ∈ S, then we call Θ viable. Since we are working with quadratic Kv- algebras, every local specification is viable (see [1, Chapter X]). If we regard Kv as a trivial quadratic extension of itself, then every quadratic Kv-algebra Θv corresponds to a local quadratic extension Lw/Kv thus a local quadratic character χLw . Therefore if we define a ∼ counting function c : {quadratic extensions L/K : Lv = Θv for all v ∈ S} → Z≥0, we can ask if the following limit exists, ∼ |{L/K :[L : K] = 2,Lv = Θv for all v ∈ S, c(L) < X}| Pc(K, Θ) := lim . X→∞ |{L/K :[L : K] = 2, c(L) < X}|

Given any global character χ, denote by c(χ) := c(K¯ ker χ). Recall that Σ is the ﬁnite set of places of K, containing all places above 2, places of bad reduction and all archimedean

17 places. If S = Σ, we can ask if the limit deﬁning Pc(E/K) converges, since the parity of

r(χ) is determined by the local factors ωv(χv), v ∈ Σ. We deﬁne fairness as the following.

Deﬁnition 2.2.2. A fair counting function c : {quadratic extensions L/K} → Z≥0 is a function that sends Y L 7→ Nvev(Lv), v where N(v) is the absolute norm of the place v over Q and by convention 1 at v|∞

ev : {isomorphism classes of quadratic Kv-algebras} → Z≥0

is defined by   0 if v - 2∞, Θv/Kv is unramified;  ev(Θv) = 1 if v - 2∞, Θv/Kv is ramified;   0  ev(Θv) if v|2∞, 0 where ev : {isomorphism classes of quadratic Kv-algebras} → Z≥0, for v|2∞ is an arbitrary function. (For a more general definition of fair counting functions see [13, Section 2.1].)

Example 2.2.3. The norm of ﬁnite part of the conductor and the norm of the product of

0 ramified primes are all fair counting functions. In the first case, for all v|2 we can let ev n × be the smallest number n such that the higher unit group v is contained in NL/K (L ). Let 0 0 ev = 0 for places v|∞. In the second case, for v|2 we let ev be equal to 1 if L/K is ramified 0 at v and 0 otherwise. For v|∞, let ev = 0. Note that in general the discriminant is not a fair

counting function if we replace quadratic Kv-algebra by a so called G-structured Kv-algebra as in [13].

Proposition 2.2.4 (Corollary 2.2 [13]). Let c be a fair counting function, Θi, i = 1, 2 are

local speciﬁcations on a ﬁnite set of places S, then Pc(K, Θi) exist and

1 1 −ev(Θ ) Pc(K, Θ ) Y Nv v = . 2 −e (Θ2 ) Pc(K, Θ ) Nv v v v∈S We then have the following corollary.

Corollary 2.2.5. Let Θ be a local speciﬁcation on a ﬁnite set S. For v ∈ S, let

Nv−ev(Θv) Pc(K, Θv) = P −ev(Lv) Lv: quad. Nv Kv-algebra

18 then Y Pc(K, Θ) = Pc(K, Θv). v∈S

Lemma 2.2.6. Suppose 1, ··· , n are independent events where i = ±1. Denote by βi = Qn 1 i=1 βi 2P (i = 1) − 1. Let = 1 ··· n. Then P ( = ±1) = 2 ± 2 .

The proof of Lemma 2.2.6 is similar to Lemma 2.1.10.

1 i Deﬁnition 2.2.7. Let c be a fair counting function. For each place v ∈ Σ, let Θv, ··· , Θv be

the local speciﬁcations at v such that their corresponding local quadratic character χj,v satisfy + Pi j + ωv(χj,v) = 1. Denote by Pc,v(E/K) = j=1 Pc(K, Θv). We deﬁne βc,v := 2Pc,v(E/K) − 1,

r(1K ) Q and βc := (−1) v∈Σ βc,v.

Theorem 2.2.8. Let c be a fair counting function, then

1 β P (E/K) = + c . c 2 2

Proof. By Theorem 1.2.7, Pc(E/K) is the sum of the probabilities Pc(K, Θ) where Θ runs

over the speciﬁcations on Σ such that if χv corresponds to Θv, then

Y r(1K ) (2.2.1) ωv(χv) = (−1) . v∈Σ

By Corollary 2.2.5, the local speciﬁcations at diﬀerent places are independent, thus if v is

the value of ωv(χv) for the localization of global character χ, then the probability of v = 1 is + Pi j Pc,v(E/K) = j=1 Pc(K, Θv). By lemma 2.2.6, the probability that (2.2.1) holds is βc.

1 Corollary 2.2.9. If K has a real place v, then Pc(E/K) = 2 .

1 Proof. By Corollary 2.2.5, Pc(L/K is ramiﬁed at v) = Pc(L/K is unramiﬁed at v) = 2 .

Thus by Deﬁnition 2.2.7, βc,v = 0.

Proposition 2.2.10. Let c be any fair counting function, we have βc,v = ξv for all places v - 2∞, v ∈ Σ.

Proof. By Deﬁnition 2.1.5, we also have ξv = 0. If v is complex, then βc,v = 1 and by

Deﬁnition 2.1.5, ξv = 1. If v is odd, |C(Kv)| = 4, the local characters are the trivial

19 character χ1, the nontrivial unramiﬁed character χ2 and the ramiﬁed characters χ3 and χ4. ¯ 2 2 2 The fraction of elements d ∈ Ov/4v − v /4v that induces each of these characters are

1 N(v) 1 N(v) 1 1 1 1 , , and , 2 N(v) + 1 2 N(v) + 1 2 N(v) + 1 2 N(v) + 1 respectively. Let Θi be the corresponding local condition of χi. Note that Pc(K, Θi) is in proportion to the inverse of norm of conductor of local characters at v, and the local conductors of the four characters are 1, 1, v and v respectively. Thus the probabilities

Pc(K, Θi) are identical to the above fractions of elements. Consequently βc,v = ξv at the places away from 2.

Example 2.2.11. This example shows that the fraction of even ranks given by Theorem 2.1.6 does depend on the way we have chosen to order the twists. Let E be the elliptic curve 38B1 in [2] y2 + xy + y = x3 + x2 + 1 and K = Q(i). Then r(1K ) = 0, and E has split multiplicative reduction at the primes (1 + i) and (19), and good reduction everywhere else. Let C(K,X) be as in Deﬁnition 1.2.9, [5, Example 7.13] shows that for all suﬃciently large X,

|{χ ∈ C(K,X): r(χ) is even}| 1 7 = + . |C(K,X)| 2 32

Now consider also another natural ordering

0 C (K,X) := {χd : d ∈ OK : |N(d)| < X}.

Again [5, Example 7.13] shows that

|{χ ∈ C0(K,X): r(χ) is even}| 361 1331 2177 1 5 lim = + = = + . X→∞ |C0(K,X)| 8688 2896 4344 2 4344

2 p 2 2 Noticing that {d ∈ Z[i]: |N(d)| < X} = {(x1, x2) ∈ R : x1 + x2 < X} ∩ Z[i] and N p 2 2 2 {(x1, x2) ∈ R : x1 + x2 < X} is a convex body in R , the above right hand side should be equal to ρ(E/K). Let us verify this. The ﬁnite set Σ consists of complex places and (1+i) and (19). Since E has split multiplicative reduction at both ﬁnite places, by Theorem

1.2.8, ωv(χd,v) = 1 if and only if χd,v is trivial character.

20 √ 2 For (1 + i), let d ∈ Z[i] − (1 + i) , by Hensel’s lemma K1+i( d) = K1+i if and only if d 5 is a unit in the ring of integers of K1+i and is a square mod (1 + i) . By [4, VI. Section 6, Chapter 15.], (Z[i]/(1 + i)5)× ∼= Z/2Z ⊕ Z/2Z ⊕ Z/4Z and the number of squares in it is 2.

6 2 6 + 2·4 5 Passing to Z[i]/(1 + i) − (1 + i) /(1 + i) , λ1+i = 2N(1 + i) = 4 and ξ1+i = 26−24 − 1 = − 6 . √ 2 For (19), let d ∈ Z[i] − (19) , then K19( d) = K19 if and only if d is a unit in the ring of

integers of K19 and is a square mod 19. The number of squares in Z[i]/(19) is 180. Passing 2 2 2 + 2 2·180·192 1 to Z[i]/(19 ) − (19 )/(19 ), λ19 = 180N(19) = 180 · 19 and ξ19 = 194−1 − 1 = − 362 . Hence 0 5 1 5 ξ = (−1) · (− 6 ) · (− 362 ) = 2172 . Hence 1 5 ρ(E/K) = + . 2 4344

Let us evaluate Pcond(E/K) when c is the conductor. Note that K1+i has two quadratic characters of conductor 1, two of conductor (1 + i)2, four of conductor (1 + i)4 and eight of conductor (1 + i)5, thus by Corollary 2.2.5,

1 1 Pcond(K,K1+i ⊕ K1+i) = 2 4 8 = 2 + 22 + 24 + 25 3

1 1 and β1+i = − 3 6= ξ1+i. For v = (19), by Proposition 2.2.10, β19 = ξ19 = − 362 . Therefore, by Theorem 2.2.8,

1 1 1 1 1 1 P (E/K) = + (−1)0(− )(− ) = + . cond 2 2 3 362 2 2172

Note that this result is diﬀerent from counting by elements and largest norm of ramiﬁed primes.

21 Chapter 3

Cokernel of Local Norm Map

3.1 Introduction

In this chapter, L is a finite extension of Q2 with valuation v. Let mL be the maximal ideal of the integers OL in L. Denote k the residue field and denote q the cardinality of k. Let √ E be an elliptic curve defined over L. For each d ∈ L×/(L×)2, d 6= 1,¯ F = L( d) is a quadratic extension of L. Let σ be the nontrivial element in Gal(F/L). Define the local norm map N : E(F ) → E(L) be such that for each P ∈ E(F ), N(P ) = P + P σ. Because 2E(L) is contained in the image of the local norm map, the cokernel of the local norm map

E(L)/NE(F ) is a ﬁnite F2-vector space. Denote the dimension of the cokernel of the local norm map by i(F/L).

Recall that in Chapter 1, K is a number ﬁeld and v is any place of K, Kv is the completion of K with respect to v. Let χ be the quadratic character corresponding to the extension √ Kv( d)/Kv. The local factors ωv(χ) determine the parity of 2-Selmer rank of the twist. A description of the disparity of ωv when E has diﬀerent reduction types over v is given in [5, Proposition 7.9] (or Theorem 1.2.8). But for v|2, only the multiplicative reduction case is

hv(χ) treated. Recall that ωv(χ) = (−1) χ(∆), where hv(χ) is defined in Definition 1.2.5 and √ ∆ is the discriminant of E. By [5, Lemma 5.5], hv = i(Kv( d)/Kv). Motivated by this, we will study the size of E(L)/NE(F ) and parity of i(F/L) when E has good reduction over L. Concretely, we will provide an effective algorithm based on Tate’s algorithm to calculate the dimension i(F/L). We will also prove the following theorem when the elliptic curve has

22 a Weierstrass integral model without the xy term. √ Theorem 3.1.1. Let L be a ﬁnite extension of Q2 with residue ﬁeld k. Let F = L( d) be a quadratic extension of L. Suppose E/L has good reduction and has the following minimal Weierstrass model

2 3 2 y + a3y = x + a2x + a4x + a6.

Then

i(F/L) ≡ [k : F2] · ordvd (mod 2)

3.2 Calculating |E(L)/NE(F )|

Let E be deﬁned over L. Suppose E has a minimal model

2 3 2 (3.2.1) y + a1xy + a3y = x + a2x + a4x + a6,

with discriminant ∆. Let E0(L) be the points of E over L with nonsingular reduction. So if

E has good reduction, E0(L) = E(L). For any n ≥ 1, let En(L) be the collection of points

(x, y) in E(L) such that ordvx ≤ −2n, together with the point at inﬁnity. The En(L)’s are additive subgroups of E(L) under elliptic curve addition, with a ﬁltration

E1(L) ⊃ E2(L) ⊃ E3(L) ⊃ · · · .

ˆ ˆ Let E be the formal group associated with E. Denote E(mL) the set mL with the formal ˆ group addition from E. Let w(z) ∈ Z[a1, ··· , z6][[z]] be as in [12][IV. 1.1 (a)]. We have ˆ ∼ E(mL) −→ E1(L) given by the map

z 1 z 7→ , − . w(z) w(z)

ˆ n ˆ n ∼ Deﬁne E(mL) to be the subgroup of E(mL) with underlying set mL. It follows that En(L) = ˆ n ˆ n ˆ n+1 ∼ n n+1 E(mL) for n ≥ 1. Moreover E(mL)/E(mL ) = mL/mL under the identity map of sets. So

En(L)/En+1(L) is a ﬁnite set of size q = |OL/mL| if n ≥ 1. ˆ n n Although in general each E(mL) is not necessarily isomorphic to the additive group mL, ˆ n ∼ n if n is large enough, the formal logarithm logE gives an isomorphism E(mL) −→ mL.

23 From now on we assume E has good reduction over L. Let us calculate i(F/L). When F/L is unramified, i(F/L) = 0 by [8, Corollary 4.4]. Next we treat the case when F/L is ramified. In this case, the residue field of F is isomorphic to k. We will need the following assumptions for the calculation of i(F/L).

∼ n • n is an integer large enough so that En(L) = mL;

∼ m m n • m is an integer large enough so that Em(F ) = mF and Tr(mF ) ⊂ mL.

This makes the following commutative diagram well-deﬁned,

∼ m (3.2.2) Em(F ) −→ mF

N Tr

∼ n En(L) −→ mL.

This means the image of Em(F ) under the norm map is contained in En(L). Now consider the following commutative diagram where the rows are exact and the vertical maps are norm maps.

(3.2.3) 0 / Em(F ) / E(F ) / E(F )/Em(F ) / 0

α β γ 0 / En(L) / E(L) / E(L)/En(L) / 0. The snake lemma gives rise to the following exact sequence,

(3.2.4) 0 / ker(β)/ ker(α) / ker(γ)

. coker(α) / coker(β) / coker(γ) / 0. Here coker(β) is the target set, namely the cokernel of the local norm map. The equation of the twist of E can be derived in the following way. Note that E is isomorphic to the following curve over L a2 a a a2 E0 : y2 = x3 + (a + 1 )x2 + (a + 1 3 )x + (a + 3 ), 2 4 4 2 6 4 under the change of variables   x = x0, (3.2.5) a x0 + a  y = y0 + 1 3 , 2 24 and replacing x0, y0 by x, y again. The quadratic twist of E0 by F has the following equation

a2 a a a2 (3.2.6) E0F : y2 = x3 + d(a + 1 )x2 + d2(a + 1 3 )x + d3(a + 3 ). 2 4 4 2 6 4

Note that E0F is not an integral model and is singular over L. The discriminant ∆(E0F ) of

0F E has order 6ordvd.

Theorem 3.2.1. Let L be a finite extension of Q2 with ramification degree e. Suppose elliptic curve E has good reduction over L with minimal model given by equation (3.2.1) and √ E0F is the curve in (3.2.6). Let F = L( d) be a ramified quadratic extension. Let

ord ∆ (E0F ) − 6ord d t = v min v 12

Remark 3.2.2. In Theorem 3.2.1, if ordvd is even, without loss of generality, we assume ∗ 2k+1 ∗ d ∈ OL of the form d = 1 + uπ for some u ∈ OL and 0 ≤ k < ordv2.

We will ﬁrst prove two lemmas that give us components that we need to prove Theorem 3.2.1.

Proof. The following two elliptic curves are isomorphic over L,

2 3 2 E : y1 + a1x1y1 + a3y1 = x1 + a2x1 + a4x1 + a6 a2 a a a2 E0 : y2 = x3 + (a + 1 )x2 + (a + 1 3 )x + (a + 3 ), 2 2 2 4 2 4 2 2 6 4 under the change of variables   x1 = x2, (3.2.7) a1x2 + a3  y1 = y2 + . 2

25 The kernel of norm map ker(N : E(F ) → E(L)) is isomorphic to ker(N : E0(F ) → E0(L))

0 under this translation. Moreover, the corresponding subgoup of Em(F ) in E (F ) is the point 0 at inﬁnity and those points (x2, y2) ∈ E (L) such that ordvF x2 ≤ −2m. Hence

∼ 0 0 0 0 ker(β)/ ker(α) = ker(N : E (F ) → E (L))/ ker(N : Em(F ) → E (L)).

Let E0F be the quadratic twist of E0 by F given by the following equation, a2 a a a2 E0F : y2 = x3 + d(a + 1 )x2 + d2(a + 1 3 )x + d3(a + 3 ), 3 3 2 4 3 4 2 3 6 4 The kernel of N : E0(F ) → E0(L) is isomorphic to E0F (L) under the map

x3 y3 (3.2.8) (x2, y2) 7→ ( , √ ). d d d Next we will show that once we transform E0F into its minimal integral model EF over L,

0 0 we will find that ker(N : Em(F ) → E (L)) is transformed into a layer of the elliptic curve F filtration of E (L). Because minimal integral models of a curve are all isomorphic over OL, we can assume the translation is done by twisting this curve by 2 to get an integral model first, then go through Tate’s algorithm to arrive at a minimal model. Suppose the following translation is used to derive a minimal integral model of E0,  2  x3 = c x4, (3.2.9) 3 2  y3 = c y4 + c sx4 + t.

2 0 0 Then ordvF x4 = ordvF x2 + ordvF d − ordvF c , and the image of ker(N : Em(F ) → E (L)) in EF (L) is

F (3.2.10) {(x4, y4) ∈ E (L) : ordvF x4 ≤ −2m + ordvF d − 2ordvF c} ∪ ∞.

If ordvd is even, then we can assume d is a unit in OL. Note that if (x2, y2) is contained in √ 0 0 the kernel of N : E (F ) → E (L) and ordvF x2 = −2a for some a, then x2 ∈ L, y2 ∈ dL √ and ordvy2 = −3a. Since d is a unit in OL, d is a unit in OF and ordvF y2 is even. Thus a must be even. So the set (3.2.10) is the same as

F {(x4, y4) ∈ E (L) : ordvx4 ≤ −2dm/2e − 2ordvc} ∪ ∞, namely EF (L). dm/2e+ordvc

26 If ordvd is odd, then we can assume d is a uniformizer π of OL. Again if (x2, y2) is 0 0 contained in the kernel of N : E (F ) → E (L) and ordvF x2 = −2a for some a, then x2 ∈ L, √ √ y2 ∈ dL and ordvy2 = −3a. Since d is a uniformizer in OL, d is a uniformizer in OF and

ordvF y2 is odd. Thus a must be odd. So the set (3.2.10) is

F {(x4, y4) ∈ E (L) : ordvx4 ≤ −2bm/2c − 2ordvc} ∪ ∞, namely EF (L). bm/2c+ordvc Under the composition of the change of variables (3.2.7), (3.2.8) and (3.2.9), ker(β) is isomorphic to EF (L). At the same time  F  E (L) if ordvd is even, ker(α) ∼= dm/2e+ordvc EF (L) if ord d is odd.  bm/2c+ordvc v

0F 0F Since the discriminant of E is ∆(E ) = 6ordvd, and the translation (3.2.9) gives rise to the minimal model, we have ord ∆ (E0F ) − 6ord d ord c = v min v = t. v 12 Hence  F F  E (L)/E (L) if ordvd is even, ker(β)/ ker(α) ∼= dm/2e+t F F  E (L)/Ebm/2c+t(L) if ordvd is odd. Note that for any l ≥ 1,

F F F F l−1 = |E (L)/E0 (L)| · |E0 (L)/E1 (L)| · q .

Since F/L is ramiﬁed, EF has bad reduction over L. Moreover since EF has good reduction F F ˜F + over F , it has additive reduction over L. Thus |E0 (L)/E1 (L)| = |E (k)| = |k | = q. So

F F F F l |E (L)/El (L)| = |E (L)/E0 (L)| · q .

27 n m The following lemma allows us to calculate the size of mL/Tr(mF ). √ Lemma 3.2.4. Let F/L be a ramiﬁed quadratic extension given by F = L( d). We have  √ ordv2−k+bm/2c−n 2k+1 n m  q if ordvd is even and F = L( 1 + uπ ), |mL/Tr(mF )| = ordv2+dm/2e−n  q if ordvd is odd.

Proof. For i ≥ −1, deﬁne the i-th ramiﬁcation group of F/L by

Gi = {σ ∈ Gal(F/L)|ordvF (σa − a) ≥ i + 1 for all a ∈ OF }

Let s be the integer such that Gs = Gal(F/L) and Gs+1 = 1. By [3, Proposition. p. 71], let j(m) = s + 1 + b(m − 1 − s)/2c, then

m j(m) Tr(mF ) = mL . √ If ord d is even and F = L( 1 + uπ2k+1) for some u ∈ O∗ . We can choose the uniformizer v √ L 1+ 1+uπ2k+1 of F to be πF = πk . We have √ 2 1 + uπ2k+1 ord (σπ − π ) = ord ( ) = 2ord 2 − 2k. vF F F vF πk v

So s = 2ordv2 − (2k + 1). Hence j(m) = ordv2 − k + bm/2c. If ordvd is odd, we can assume √ d is the uniformizer π. Thus π is a uniformizer of F . We have

√ ordvF (σπF − πF ) = ordvF (2 π) = 2ordv2 + 1.

Thus s = 2ordv2 and j(m) = ordv2 + 1 + b(m − 1)/2c = ordv2 + dm/2e. Hence we have the desired result.

n m Proof of Theorem 3.2.1. By diagram (3.2.2), coker(α) is isomorphic to mL/Tr(mF ), which is a ﬁnite set. The last vertical map γ of diagram (3.2.3) induces the following exact sequence,

γ 0 / ker(γ) / E(F )/Em(F ) / E(L)/En(L) / coker(γ) / 0.

So ker(γ) and coker(γ) are both ﬁnite and

28 Hence

|coker(γ)| (3.2.11) |E(L)/NE(F )| = | ker(β)/ ker(α)| · |mn /Tr(mm)| · L F | ker(γ)|

n m Next we calculate |mL/Tr(mF )|. This is done by Lemma 3.2.4,  √ ordv2−k+bm/2c−n 2k+1 n m  q if ordvd is even and F = L( 1 + uπ ), |mL/Tr(mF )| = ordv2+dm/2e−n  q if ordvd is odd.

Finally we need to calculate |E(L)/En(L)|/|E(F )/Em(F )|. Since E is in its minimal integral model and it has good reduction over L, E(L) = E0(L). So E(L)/En(L) = E0(L)/En(L) and by the elliptic curve ﬁltration,

|E0(L)/En(L)| = |E0(L)/E1(L)| · |E1(L)/E2(L)| · · · · · |En−1(L)/En(L)|

2 n−1 n n−1 = |E0(L)/E1(L)| · |mL/mL| · · · · · |mL /mL| = |E0(L)/E1(L)| · q .

For elliptic curves of good reduction over local fields with residue field k, we have the exact sequence ˜ 0 / E1(L) / E0(L) / E(k) / 0, ˜ ˜ where E(k) is the reduction of E over the residue field. Hence |E0(L)/E1(L)| = |E(k)|. So

˜ n−1 (3.2.12) |E0(L)/En(L)| = |E(k)| · q .

Similarly, since E has good reduction, the minimal integral model of E/L is also the minimal integral model over F . Consequently,

m−1 ˜ m−1 (3.2.13) |E0(F )/Em(F )| = |E0(F )/E1(F )| · |N(mF )| = |E(k)| · q .

Combining (3.2.12) and (3.2.13) the quotient

|E(L)/E (L)| |E˜(k)| · qn−1 n = = qn−m. |E(F )/Em(F )| |E˜(k)| · qm−1

29 In conclusion,

We completed the proof of the theorem.

F F Remark 3.2.5. The term |E (L)/E0 (L)| in Theorem 3.2.1 can be eﬀectively calculated using Tate’s algorithm. Because we have the equation of the twisted curve (3.2.6), we can

0F F F go through Tate’s algorithm to obtain ordv∆min(E ) and |E (L)/E0 (L)| simultaneously. So we can subsequently obtain |E(L)/NE(F )| using Theorem 3.2.1.

3.3 Proof of Theorem 3.1.1

Suppose L has ramification degree e. If F/L is unramified, by [8, Corollary 4.4], i(F/L) = 0. So we are concerned with the situation when F/L is ramified. Suppose F/L is ramified. It is

enough to consider the case when ordvd is 0 and the case when ordvd is 1. When ordvd = 0, √ d is a unit, F can be written as L( 1 + uπ2k+1) for some k < e, where π is the uniformizer

2k+1 and u is a suitable unit in OL. We assume d = 1 + uπ . When ordvd = 1, we assume d is the uniformizer π. Case (1). d = 1 + uπ2k+1. To apply Theorem 3.2.1, we need to ﬁnd the isomorphism that transforms the following curve

a2 E0F : y2 = x3 + da x2 + d2a x + d3(a + 3 ) 2 4 6 4

into its minimal integral model EF . Using the change of variables   x = x0, a3  y = y0 + d, 2 we get the model E00F isomorphic to E0F given below

π2k+1 E00F : y2 + da y = x3 + da x2 + d2a x + d3a + d2a2u . 3 2 4 6 3 4

30 Note that we still use variables x, y for convenience. We will follow this convention for the rest of this section. The Weierstrass coeﬃcients of this new model are   a˜ = 0,  1   a˜ = da ,  2 2  a˜3 = da3,  2  a˜4 = d a4,   π2k+1  a˜ = d3a + d2a2u ,  6 6 3 4 and that  ord a˜ = ord da = 0,  v 3 v 3  ordva˜2 = ordva2, ordva˜4 = ordva4,  π2k+1  ord a˜ = ord = 2k + 1 − 2e < 0. v 6 v 4

The reason a3 is a unit is that otherwise E will have bad reduction over L. Note thata ˜6 is 00F 2e−2k−1 the only non-integral coeﬃcient of E . Let l = d 6 e. If we apply the following change of variables  1  x = x0,  π2l (3.3.1) 1  y = y0,  π3l

and go through Tate’s algorithm, it is easy to see that ordva˜6 will determine the reduction type of EF . It is also easy to check that the term t in Theorem 3.2.1 is, in this case, −l. Concretely, we have the following proposition that explicitly evaluates |E(L)/NE(F )|.

Proposition 3.3.1. Suppose E has good reduction over L with the following Weierstrass integral model

2 3 2 y + a3y = x + a2x + a4x + a6. √ Let F = L( 1 + uπ2k+1) be ramiﬁed. Then the reduction type of EF is   II, if e − k ≡ 0 (mod 3),  II∗, if e − k ≡ 1 (mod 3),   ∗  I0, if e − k ≡ 2 (mod 3).

31 3 −3 Let n(˜a6) be the number of roots of T ≡ π a˜6 (mod π), where a˜6 is the constant term of EF . We have  2 (e−k)  q 3 , if e − k ≡ 0 (mod 3),  2 (e−k−1) |E(L)/NE(F )| = q 3 , if e − k ≡ 1 (mod 3),   2 (e−k−2)+1  (1 + n(˜a6)) · q 3 , if e − k ≡ 2 (mod 3).

Proof. If e − k ≡ 0 (mod 3), then 2k + 1 − 2e ≡ 1 (mod 6). After the change of variables (3.3.1), the valuation of the constant term is 1 and other coeﬃcients are suﬃciently divisible

F F by π. By Tate’s algorithm, the reduction type is II. Thus |E (L)/E0 (L)| = 1. Note that in the change of variables (3.3.1),

2e − 2k − 1 e − k l = d e = . 6 3

By Theorem 3.2.1,

−l+e−k 2 (e−k) |E(L)/NE(F )| = q = q 3 .

If e − k ≡ 1 (mod 3), then 2k + 1 − 2e ≡ 5 (mod 6). So the constant term of EF has valuation 5 and every other coeﬃcients are suﬃciently divisible by π. By Tate’s algorithm

∗ F F the reduction type is II , thus |E (L)/E0 (L)| = 1. In this case 2e − 2k − 1 e − k − 1 l = d e = + 1. 6 3

By Theorem 3.2.1,

2 (e−k−1) |E(L)/NE(F )| = q 3 .

Finally, if e − k ≡ 2 (mod 3), then 2k + 1 − 2e ≡ 3 (mod 6). So the constant term of EF has valuation 3 and every other coeﬃcients are suﬃciently divisible by π. By Tate’s

∗ F F algorithm, the reduction type is I0. The group structure of E (L)/E0 (L) is determined by the factorization pattern of the polynomial

3 −1 2 −2 −3 P (T ) = T + π a˜2T + π a˜4T + π a˜6 (mod π),

F wherea ˜2, a˜4, a˜6 are coeﬃcients from E . Since

3 −3 P (T ) ≡ T + π a˜6 (mod π),

32 F F By Tate’s algorithm, |E (L)/E0 (L)| = 1 + n(˜a6) where n(˜a6) is the number of roots of 3 −3 T ≡ π a˜6 (mod π). Note that 2e − 2k − 1 e − k − 2 l = d e = + 1. 6 3 By Theorem 3.2.1,

2 (e−k−2)+1 |E(L)/NE(F )| = (1 + n(˜a6)) · q 3

This completes the proof.

Corollary 3.3.2. Under the assumption of Proposition 3.3.1, i(F/L) is even.

Proof. In the cases e − k ≡ 0, 1 (mod 3) the exponent of q is even. So the F2-dimension of 2 E(L)/NE(F ) is even. Suppose e − k ≡ 2 (mod 3). Then 3 (e − k − 2) + 1 is always odd.

Note that n(ã6) = 1 if and only if 2 - [k : F2], and n(ã6) = 0 or 3 if 2|[k : F2]. In either case 2 (e−k−2)+1 (1 + n(ã6)) · q 3 is an even power of 2.

Case (2). d = π. In this case we will follow the same step as we did in the previous case, using Theorem 3.2.1. We start with the curve a2 E0F : y2 = x3 + πa x2 + π2a x + π3(a + 3 ). 2 4 6 4 The Weierstrass coeﬃcients of this new model are   a˜1 =a ˜3 = 0,    a˜2 = πa2, 2  a˜4 = π a4,   a2  a˜ = π3(a + 3 ), 6 6 4 and that  ord a˜ = ord a + 1,  v 2 v 2  ordva˜4 = ordva4 + 2,  a2  ord a˜ = ord 3 + 3 = 3 − 2e. v 6 v 4 To get an integral model, it is enough to apply the following change of variables  1  x = x0,  π2l (3.3.2) 1  y = y0,  π3l

33 2e−3 where l = d 6 e. The valuation of the new constant terma ˜6 of the new integral model is either 1, 3 or 5. Other coeﬃcients are suﬃciently divisible by π. If one goes through Tate’s algorithm, this is the minimal model EF . The reduction type of EF over L is determined

by the valuation of the constant terma ˜6. Moreover the term t in Theorem 3.2.1 in this case is −l. To be precise, we have the following proposition.

Proposition 3.3.3. Suppose E has good reduction over L with the following Weierstrass integral model

2 3 2 y + a3y = x + a2x + a4x + a6. √ Let F = L( π). Then the reduction type of EF is  ∗  I0, if e ≡ 0 (mod 3),  II, if e ≡ 1 (mod 3),   II∗, if e ≡ 2 (mod 3).

3 −3 Let n(˜a6) be the number of roots of T ≡ π a˜6 (mod π), where a˜6 is the constant term of EF . We have

 2e  (1 + n(˜a )) · q 3 , if e ≡ 0 (mod 3),  6  2e+1 |E(L)/NE(F )| = q 3 , if e ≡ 1 (mod 3),  2e−1  q 3 , if e ≡ 2 (mod 3).

Proof. If e ≡ 0 (mod 3), then 3 − 2e ≡ 3 (mod 6). In particular, 3 − 2e < 0 since e ≥ 3. So

after the change of variables (3.3.2), the valuation of the constant terma ˜6 is 3, while other F ∗ coeﬃcients are suﬃciently divisible by π. The reduction type of E is I0. The structure of F F E (L)/E0 (L) is determined by the number of roots of

3 −1 2 −2 −3 P (T ) = T + π a˜2T + π a˜4T + π a˜6 (mod π).

3 −3 A closer look at the coefficients yields that P (T ) ≡ T +π a˜6 (mod π). By Tate’s algorithm, F F 3 −3 |E (L)/E0 (L)| = 1 + n(ã6) where n(ã6) is the number of roots of T ≡ π a˜6 (mod π). On the other hand we have 2e − 3 e l = d e = . 6 3 By Theorem 3.2.1,

2e |E(L)/NE(F )| = (1 + n(˜a6)) · q 3 .

34 Next we consider the case when e ≡ 1 (mod 3). We have 3 − 2e ≡ 1 (mod 6). So the

F F reduction type is II, |E (L)/E0 (L)| = 1. In this case 2e − 3 e − 1 l = d e = . 6 3

Consequently, by Theorem 3.2.1,

2e+1 |E(L)/NE(F )| = q 3 .

Finally, when e ≡ 2 (mod 3). We have 3 − 2e ≡ 5 (mod 6). The reduction type is II∗,

F F |E (L)/E0 (L)| = 1. A quick calculation shows 2e − 3 e + 1 l = d e = . 6 3

By Theorem 3.2.1,

2e−1 |E(L)/NE(F )| = q 3 .

Proposition 3.3.3 implies that, in some cases, i(F/L) can be odd if we twist the curve √ with no xy term by π. For example if e = 3 and 2 - [k : F2], then n(˜a6) = 1, i(F/L) is

odd. Another situation is, for example, when e ≡ 1 (mod 3) and 2 - [k : F2], then q is an 2e+1 odd power of 2 and the exponent 6 of q is also odd, thus i(F/L) is odd. The following corollary tells us the equivalent condition of i(F/L) being odd or even.

Corollary 3.3.4. Under the assumption of Proposition 3.3.3, i(F/L) ≡ [k : F2] (mod 2).

Proof. If e ≡ 1 or 2 (mod 3), the exponent of q in Proposition 3.3.3 is odd, so i(F/L) ≡

[k : F2] (mod 2). Suppose e ≡ 0 (mod 3). By Proposition 3.3.3, the exponent of q is even,

so the parity of i(F/L) depends on the exponent of 1 + n(˜a6) as a power of 2. Note that

n(˜a6) = 1 if 2 - [k : F2] and n(˜a6) = 0 or 4 otherwise. We then have i(F/L) ≡ [k : F2] (mod 2).

Proof of Theorem 3.1.1. If F/L is unramiﬁed, this is [8, Corollary 4.4]. If F/L is ramiﬁed

and ordvd is even, this is Corollary 3.3.2. If F/L is ramiﬁed and ordvd is odd, this is Corollary 3.3.4.

35 3.4 Examples and remarks

Proposition 3.3.1 and Proposition 3.3.3 together is a generalization of [6, Proposition 4], which can be stated as the following.

Proposition 3.4.1 (Kramer, 1981). Assume that L is an unramiﬁed extension of Q2 and that F over L is ramiﬁed. If E has supersingular reduction modulo 2 then   0 if ordvd is even, i(F/L) =  [L : Q2] if ordvd is odd.

Proof. For supersingular reduction curves, π|a1. Since L/Q2 is unramiﬁed, we have 2|a1.

Thus we can make a1 = 0 by a suitable translation of y to meet the requirement on minimal Weierstrass model in Proposition 3.3.1 and Proposition 3.3.3.

Since L/Q2 is unramiﬁed, e = 1. If d is a unit, we can assume d = 1 + 2u for some unit u that makes non-square. This mean k = 0. So by Proposition 3.3.1, i(F/L) = 0. So when

ordvd is even i(F/L) = 0. Suppose ordvd = 1. We have i(F/L) = [k : F2] by Proposition

3.3.3. So when ordvd is odd, i(F/L) = [k : F2] = [L : Q2].

Let K be a number ﬁeld of degree N. Suppose E is an elliptic curve over K and E† is a twist of E. Recall the counting method given in Section 2.1, i.e. counting by elements with respect to any ﬁxed convex body C ⊂ RN , and the counting method in Section 2.2 with respect to fair counting functions. The following example shows that ρ(E/K, C) and

† † ρ(E /K, C) can be different, and Pcond(E/K) and Pcond(E /K) can be different, even if E and E† have the same set places of bad reduction. √ Example 3.4.2. Let K be the number field Q( −3). Let E and E† be the curves 121D1 and 11A3 in [2], E : y2 + y = x3 − x2 − 40x − 221,

E† : y2 + y = x3 − x2. √ Then E† is the twist of E by −11. Both E and E† have bad reduction only at the primes above 11. Then for a ﬁxed convex body C ⊂ R2, we have 143 ρ(E/K, C) = P (E/K) = , cond 288 36 while 145 ρ(E†/K, C) = P (E†/K) = . cond 288

7 Proof. The 2-Selmer rank of E/K is 1. The discriminant of E is ∆E = −11 . Note that 11 ∼ splits in K. Hence if v1 and v2 are the places above 11, we have Kvi = Q11. Since (2) is √ ∼ unramified, we have K2 = Q2( −3), the unique unramified quadratic extension of Q2. So the finite set of places Σ = {∞, (2), v1, v2}, where ∞ is the complex infinite place. Note that there is no real embedding. The density ρ(E/K, C) is independent of C, so can be written as ρ(E/K). The complex place has no effect on ρ(E/K). So we only study the local factors

at (2), v1 and v2. At (2), since E has good reduction and its Weierstrass model has no xy √ term, by Theorem 3.1.1, i(F/K2) is always even for all quadratic extension F = K2( d) of

K2. Thus

ω2(χd,2) = χd,2(∆E),

for all quadratic character χd ∈ C(K). Note that ∆E ≡ −3 (mod 8) and −3 is a square in

K, we have χd,2(∆E) = 1. Thus ω2(χd,2) ≡ 1 and ξ2 = β2 = 1. Next we consider v1 and v2. † Since E has split multiplicative reduction over Q11 and Kv1 and Kv2 are unramiﬁed over † Q11, E has split multiplicative reduction over Kv1 and Kv2 . Thus, by Theorem 1.2.8, only † trivial character will make ωvi equals 1. Since E is the quadratic twist of E by a ramiﬁed

character, ωvi equals −1 only if χ is the same ramiﬁed character. Thus

+ 1 qv1 1 11 13 λv1 = 1 − · = 1 − · = , 2 qv1 + 1 2 12 24

2·13 1 1 and ξv1 = 24 − 1 = 12 . Since v1 is over an odd prime, by Proposition 2.2.10, βv1 = ξv1 = 12 . 1 Similarly, ξv2 = βv2 = 12 . Thus by Theorem 2.1.6 and Theorem 2.2.8, 1 (−1)1ξ ξ ξ 143 ρ(E/K) = P (E/K) = + 2 v1 v2 = . cond 2 2 288

For E†, we will add an index † for the local factors. The 2-Selmer rank is 0 and the

† † † discriminant of E is −11. So the ﬁnite set Σ is also {∞, (2), v1, v2}. Since E has no xy term in its Weierstrass model and has good reduction at 2, we have

† ω2(χd,2) = χd,2(∆E† ) = χd,2(−11) = 1.

37 † † for all quadratic character χd ∈ C(K). Hence ξ2 = β2 = 1. For places v1 and v2, since only trivial character will make ω† equals 1, vi

†+ 1 qv1 11 λv1 = · = , 2 qv1 + 1 24 and ξ† = 2·11 −1 = − 1 . By Proposition 2.2.10, β† = ξ† = − 1 . Similarly, ξ† = β† = − 1 . v1 24 12 v1 v1 12 v2 v2 12 Thus by Theorem 2.1.6 and Theorem 2.2.8,

1 (−1)0ξ†ξ† ξ† 145 ρ(E†/K) = P (E†/K) = + 2 v1 v2 = . cond 2 2 288

38 Appendix A

Convex Bodies

A convex body in N-dimensional Euclidean space RN is a compact convex set with non-empty interior. Let C be a convex body in RN , then it has positive N-dimensional volume Vol(C). The boundary ∂C of C has ﬁnite (N − 1)-dimensional area A(∂C)(for example see page 7 [11]). For any real number λ > 0, deﬁne

λC := {λv : v ∈ C} ⊂ RN to be the cone generated by C of multiple λ. Therefore A(∂(λC)) = λN−1A(∂C) and Vol(λC) = λN Vol(C).

Deﬁnition A.0.1. Let C be a compact set in RN , let P be a point on the boundary of C, and H is a hyperplane that passes through P . We call H a supporting hyperplane for C at P when all the points of C are on one side of H.

Theorem A.0.2 (Supporting hyperplane theorem). Let C be a closed set in RN with nonempty interior, then C is convex if and only if every point on the boundary has a supporting hyperplane.

Proof. See for example [9, Page 129].

Lemma A.0.3. Let C ⊆ D be two compact subsets of RN and C is a convex body, then there is a surjective map π : ∂D → ∂C such that for any point x ∈ ∂D, π(x) is the unique point that is closest to x.

39 Proof. By convexity of C, given a point x outside C, there is a unique point y ∈ ∂C that is closest to x. On the other hand, given a point y ∈ ∂C, by Theorem A.0.2, there exists a hyperplane H passing through y, and on one side of H there is no point of C. Let x ∈ ∂D on a diﬀerent side of H from C, which is on the line perpendicular to H and passing through y. Then π(x) = y. The map π is surjective.

Lemma A.0.4. If C ⊆ D are two convex bodies in RN , then A(∂C) ≤ A(∂D).

Proof. Let π : ∂D → ∂C be the same as in the Lemma A.0.3. Given x ∈ ∂D, let l be the line connecting x and π(x), then the hyperplane H passing through π(x) and perpendicular to l is a supporting hyperplane for C at π(x). For any y ∈ ∂D, since x and C are on diﬀerent sides of H, we have (x − π(x)) · (π(x) − π(y)) ≥ 0 and (y − π(y)) · (π(y) − π(x)) ≥ 0. Adding the two inequalities and applying Cauchy-Schwarz inequality yields

|π(x) − π(y)|2 ≤ (x − y) · (π(x) − π(y)) ≤ |x − y||π(x) − π(y)|.

Thus |π(x) − π(y)| ≤ |x − y|.

Combining with the subjectivity of π, the Hausdorﬀ measure of D is greater than C for any dimension.

Definition A.0.5. Let S be a subset of RN , for any r ≥ 0, define the r-neighborhood of S to be [ Out(S, r) := B(x, r) x∈S and define the inner parallel body of S at distance r to be

[ Inn(S, r) := S\ B(x, r) x∈∂S

where B(x, r) is the N-dimensional open ball centered at x with radius r.

Lemma A.0.6. If C is a convex body in RN , then

(i) set Out(C, r) is a convex body for any r ≥ 0;

40 (ii) set Inn(C, r) is a convex body for any r ≥ 0 such that Inn(C, r) has non-empty interior.

Proof. We denote by [x1, x2] the line segment with ends x1 and x2 for any two points x1 and N x2 in R . Suppose C is a convex body. For (i), it is obvious that Out(C, r) has non-empty interior. We ﬁrst prove that Out(C, r) is convex. By Theorem A.0.2, it is suﬃcient to show every x ∈ ∂Out(C, r) has a supporting hyperplane.

0 By convexity of C, there is a unique point x ∈ C closest to x with distance r. Let Hx 0 and Hx0 be the hyperplanes passing through x and x respectively that is perpendicular to 0 0 [x, x ], then dist(Hx,Hx0 ) = r and Hx0 is a supporting hyperplane for C at x . Thus C is on one side of Hx and dist(C,Hx) = r. Noticing that Out(C, r) is exactly the collection of points with distance to C equal to r, it is completely on the same side of Hx as C. Therefore

Hx is a supporting hyperplane for Out(C, r) at x. For Inn(C, r), assume r is small enough such that Inn(C, r) has non-empty interior. We will prove convexity of Inn(C, r) by showing every point y ∈ ∂Inn(C, r) has a supporting hyperplane. Let y ∈ ∂Inn(C, r), by deﬁnition of Inn(C, r), the closed ball B(y, r) is properly contained

0 in C and the surface of B(y, r) touches ∂C at some point y . Let Hy and Hy0 be the hyperplanes perpendicular to the line segment [y, y0] that passes through y and y0 respectively.

0 Hence dist(Hy,Hy0 ) = |[y, y ]| = r. If we can show C is on one side of Hy0 , then since

dist(∂Inn(C, r),C) = r, we have the distance from any point of Inn(C, r) to Hy0 is at least

r. Thus Inn(C, r) is on one side of Hy. Therefore Hy is a supporting hyperplane of Inn(C, r) at y. We will prove this by contradiction.

Now suppose there is a point z ∈ C that is on the diﬀerent side of Hy0 from y. By connecting z with points on B(y, r) we obtain a convex set that contains y0 in its interior. Noticing that B(y, r) and z are all contained in C, y0 is in the interior of C. This contradicts the choice of y0.

Lemma A.0.7. Let C be a convex body in RN , then Inn(Out(C, r), r) = C for any r ≥ 0.

Proof. We ﬁrst prove C ⊆ Inn(Out(C, r), r). This is true because any point in C is not contained in the open ball B(x, r) for any x ∈ ∂Out(C, r).

41 For the other direction, noticing that Out(C, r) is convex by Lemma A.0.6, let π : ∂Out(C, r) −→ ∂C be the as map in Lemma A.0.3. Then π is onto. For any y ∈ ∂C, let y0 be in the pre-image of y, then dist(y, y0) is the minimal distance from y0 to C, namely r. But the distance from Inn(Out(C, r), r) to ∂Out(C, r) is at least r, thus Inn(Out(C, r), r) ⊆ C.

Lemma A.0.8. Let C be a convex body in RN , then for any r ≥ 0

Vol(Out(C, r)) − Vol(Inn(C, r)) ≤ 2r · A(∂Out(C, r)).

Proof. This is an application of [10, Theorem 23], which implies that Vol(C)−Vol(Inn(C, r)) ≤ rA(∂C). First we have

Vol(Out(C, r)) − Vol(Inn(C, r))

= (Vol(Out(C, r)) − Vol(C)) + (Vol(C) − Vol(Inn(C, r))).

By [10, Theorem 23] and Lemma A.0.7,

Vol(Out(C, r)) − Vol(C) ≤ rA(∂Out(C, r)).

By [10, Theorem 23] and Lemma A.0.4,

Vol(C) − Vol(Inn(C, r)) ≤ rA(∂C) ≤ rA(∂Out(C, r)).

Combining the equality and inequalities above we complete the proof.

For the rest of this section we ﬁx an N-dimensional lattice L in RN , ﬁx a fundamental parallelogram F of L. Suppose the diameter of the circumscribed hypersphere of F , or for simplicity the diameter of F is l.

Deﬁnition A.0.9. For any convex body C ∈ RN , deﬁne n(C,F ) to be the number of translations of F by lattice points of L, which are contained in the interior of C, and m(C,F ) be the number of translations of F by lattice points of L that intersect C.

Lemma A.0.10. Under the notation of Deﬁnition A.0.9 we have

(i) n(C,F )Vol(F ) ≤ Vol(C) ≤ m(C,F )Vol(F );

42 (ii) the number f of lattice points in C satisﬁes n(C,F ) ≤ f ≤ m(C,F ).

Proof. The inequality of (i) comes directly from Deﬁnition A.0.9. The number of lattice points in C is at most n(C,F ) plus the number of translations of F by elements of L that intersect the boundary ∂C, which is m(C,F ) − n(C,F ) by Deﬁnition A.0.9. Hence we get (ii).

N Proposition A.0.11. For any convex body C ∈ R , there exists a constant eC such that if

λ > eC then 2N lA(∂C) m(λC, F ) − n(λC, F ) ≤ λN−1. Vol(F )

Proof. Let v0 be a point in the interior of C. For any λ > 0, let 2λC−λv0 be the translation of 1 2λC by −λv0. By convexity of λC, for any λv ∈ λC, 2 (λv+λv0) ∈ λC. Thus λv ∈ 2λC−λv0.

So λC ⊂ 2λC − λv0 for all λ > 0. All the translations that intersect ∂(λC) are contained in Out(λC, l)\Inn(λC, l), thus by Lemma A.0.8

(A.0.1) m(λC, F ) − n(λC, F ) Vol(Out(λC, l)) − Vol(Inn(λC, l)) A(∂Out(λC, l)) · 2l ≤ ≤ Vol(F ) Vol(F )

Note that C ∩ ∂(2C − v0) = ∅. Let δ be the distance between ∂C and ∂(2C − v0), let

eC = l/δ. When λ > eC , the distance between λC and 2λC − λv0 is greater than l, hence

Out(λC, l) ⊂ 2λC − λv0. Noticing that both sets are convex bodies, by Lemma A.0.4,

A(∂Out(λC, l)) ≤ A(∂(2λC − λv0)) = A(∂(2λC)). Combining this with (A.0.1) we obtain:

(A.0.2) m(λC, F ) − n(λC, F ) A(∂Out(λC, l)) · 2l A(∂(2λC)) · 2l 2N lA(∂C) ≤ ≤ = λN−1. Vol(F ) Vol(F ) Vol(F )

Remark A.0.12. The value of eC depends on C and the choice of fundamental parallelogram of the lattice L, but it is invariant under translation of C.

43 Corollary A.0.13. All the notations are the same as in Proposition A.0.11. If M is a

sublattice of L, then when λ > eC , the number of elements of each residue class of L/M occurs Vol(λC) + O(λN−1) (L : M)Vol(F ) times in λC where (L : M) is the index of M as a subgroup of L. The absolute value of the error term O(λN−1) is at most 2N lA(∂C) λN−1. Vol(F )

Proof. We ﬁx a fundamental parallelogram FM of M which is a union of translations of F .

In each translation of FM there are (L : M) lattice points of L. By Lemma A.0.10 (ii), the number of lattice points in λC that belong to a given coset of L/M is n(λC, FM ) plus a non-negative error term at the boundary ∂(λC). The non-negative error term is at most

m(λC, FM ) − n(λC, FM ). Note that FM is a union of translations of F . If λ > eC , by Proposition A.0.11,

2N lA(∂C) (A.0.3) m(λC, F ) − n(λC, F ) ≤ m(λC, F ) − n(λC, F ) ≤ λN−1. M M Vol(F )

Combining (A.0.3) and the fact that Vol(FM ) = (L : M)Vol(F ) with the following inequality

Vol(λC) 0 ≤ − n(λC, FM ) ≤ m(λC, FM ) − n(λC, FM ), Vol(FM )

the corollary follows.

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