橢圓曲線的二次扭變

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(1)Quadratic Twists of Elliptic Curves Liao, Hung-Min.

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(3) Contents Chapter I. Introduction. 5. Chapter II. Preliminary Results 1. The Cassels-Tate Pairing 2. The 2-division Points 3. The Local Pairing 4. Computing Sel2 (E/K) 5. The Global Duality. 11 11 13 14 18 20. Chapter III. The Independent Field Extensions 1. The Norm Map 2. The Map nL/Q 3. The Selmer Groups 4. The First Main Result. 23 23 24 25 26. Chapter IV. The Local Symbols 1. The Archimedean Place 2. The Split Multiplicative Places 3. The Non-Split Multiplicative Places 4. The Potential Good Additive Places 5. The map s 6. The d-Twist of E for d = ±p or d = ±pq 7. An Example. 29 29 29 30 31 32 33 34. Chapter V. The Proof of the Main Theorem 1. The Injectivity of s(q) 2. The Non-surjectivity of s(q) 3. The examples of the Congruent Curve. 39 39 39 43. Bibliography. 47. 3.

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(5) CHAPTER I. Introduction Let E be an elliptic curve defined over Q, and for each square-free rational integer d, let Ed denote the the quadratic twist of E by d (in brief, the d-twist). The question concerning the rank rk(Ed (Q)) of the Mordell-Weil group Ed (Q) (the rank of Ed over Q) can be traced back to the ancient Greek congruent number problem for which the involved elliptic curve is nowadays called the congruent curve defined by y 2 = x3 − x. A square free integer d is a congruent number if and only if the d-twist of the congruent curve has positive rank over Q (see [Kob84]). Based on various results in the past decades on quadratic twists of the congruent curve, it is known that modulo 8, prime numbers congruent to 5 or 7 are a congruent numbers, while those congruent to 3 are not ([Tun83], [Mon90], [Elk94], [ElkWeb]). In addition to the congruent curve, previous works also provide examples of elliptic curves for which the p-twists always have positive rank over Q for all prime number p contained in certain congruent classes modulo an integer (see [Mon90], [Bir69], [Bir70], and [Hee52]). As for the quadratic twists by composite numbers, [Ono97] gives a family of elliptic curves so that for each curve E in this family there Q is an associated infinite set S of prime numbers with density 13 for which if d = pj is a square free integer where pj ∈ S, then Ed has rank 0 over Q. In particular Ep has rank 0 for every p ∈ S. Also, it is proved in [M-M94 ] that for each E with rank 0 and conductor N , then there exists infinitely many square-free integers d congruent to 1 modulo 4N such that Ed has rank 0 over Q. Evidences show that for a given E usually the ranks of Ed over Q are rather small (see [CKRS], [ElkWeb]). However, it is also believed that the rank is not always small and sometimes it could become arbitrary large as d varies. Goldfeld’s conjecture [Gol79], saying that for a given E, the average rank over Q of all its quadratic twists is 12 , seems explains this philosophy. Recently there have been a number of investigations regarding the distribution of ranks of elliptic curves in various families (see [GouM91], [SteT95], [OS98], [RubS01], [RubS02] ). Conjecture. ([Ono97]) If E/Q is an elliptic curve, then there are infinitely many primes p for which Ep has rank 0 over Q, and there are infinitely many primes l for which El has positive rank over Q. The main aim of this thesis is to verify this conjecture for a large family of elliptic curves by giving an algorithm to find the suitable primes. For the technical reason, we need to assume that the subgroup of 2-torsion points, E[2], is contained 5.

(6) 6. I. INTRODUCTION. in E(Q), or equivalently, the defining equation of E can be written as y 2 = x(x − a)(x − b), a, b, ∈ Z, 0 < a < b, and (a, b) square free.. (1). From now on, let E be defined by (1). Then the integers a, b are unique, and we let m(E) denote the number of prime divisors of ab(b − a). For a square-free integer d, the defining equation of d-twist Ed of E can be written as y 2 = x(x − da)(x − db).. (2). Our main result is the following: Main Theorem I. Suppose E has semi-stable reduction at each odd prime and E[2] ⊂ E(Q). Then there exist two constants ε0 , ε1 of values 1 or −1 together with two sets S0 , S1 of primes, each of density at least ( 12 )m(E)+1 , defined by computable congruent relations, such that (i ) Eε0 p has rank 0 over Q for every p ∈ S0 and (ii ) if p is an element in S1 so that the Parity Conjecture holds for Eε1 p /Q, then Eε1 p has rank 1 over Q. The elliptic curve E has semi-stable reduction at every odd prime if and only if the greatest common divisor (a, b) = 1 or 2.. (3). In general, if (a, b) = 2l · N , with l = 0 or 1 and N odd, then the quadratic twist EN has semi-stable reduction at every odd prime, and we have m(EN ) = m(E) − m(N ), where m(N ) is the number of prime divisiors of N . Thus, the following is just another version of the main theorem: Main Theorem II. Suppose E/Q is an elliptic curve with E[2] ⊂ E(Q). Let N be the unique positive integer N so that the quadratic twist EN has semi-stable reduction at every odd prime. Then there exist two constants ε0 , ε1 of values 1 or −1 together with two sets S0 , S1 of primes, each of density at least ( 21 )m(E)−m(N )+1 , defined by computable congruent relations, such that (i ) Eε0 pN has rank 0 over Q for every p ∈ S0 and (ii ) if p is an element in S1 so that the Parity Conjecture holds for Eε1 pN /Q, then Eε1 pN has rank 1 over Q. Next, we shall explain our method. To do so, let us first recall the definition of the 2m -Selmer group. For a number field K and a positive integer m, the 2m -Selmer group Sel2m (E/K) is defined as the kernel of the composition M loc jK : H1 (K, E[2m ]) −→ H1 (K, E) −→ H1 (Kv , E), (4) v. ¯ and the where the first arrow is induced from the natural map E[2m ] ,→ E(K) second is the localization map sending the global cohomology group to the direct sum of local cohomology groups over all places of K. The Kummer’s exact sequence m. [2 ] ¯ −→ ¯ −→ 0, 0 −→ E[2m ] −→ E(K) E(K).

(7) I. INTRODUCTION. 7. where [2m ] denotes the multiplication by 2m on E, induces the exact sequence: 0 −→ E(K)/2m E(K) −→ Sel2m (E/K) −→ X(E/K)[2m ] −→ 0.. (5). Here, as usual, X(E/K) denotes the Tate-Shafarevich group of E over K. Most of the previous works compute the rank of Ed over Q by computing √ Sel2m (Ed /Q) for certain m. Our approach is slightly different, we consider L = Q( d) and determine the rank of Ed over Q by computing the rank of E over L and then using the equality: rk(E(L)) = rk(E(Q)) + rk(Ed (Q)). The advantage of doing so is that the Selmer group Sel2 (E/L) becomes controllable if the extension L/Q satisfies certain condition that can be easily formulated via local data. We shall describe this condition in the following paragraph. The local and global Kummer’s exact sequence induces the commutative diagram of exact sequences: i. E(Q)/2E(Q) −→ H1 (Q, E[2]) −→ H1 (Q, E)[2] ↓ Q 1 ↓ loc Q 1 ↓ Q H (Q , E[2]) −→ E(Q )/2E(Q ) −→ 0 −→ v v v v H (Qv , E)[2]. v v 0 −→. Since loc ◦ i = jQ , by the diagram chasing, we get a homomorphism R : Sel2 (E/Q) −→ E(Q)/2E(Q). Q Here, for a number field K, we let E(K) denote the group of adelic points w E(Kw ), by letting w run through all places of K. For the quadratic extension L/Q, consider the norm map NL/Q : E(L) −→ E(Q). Since 2E(Q) is contained in NL/Q (E(L)), we have a natural map from E(Q)/2E(Q) to E(Q)/ NL/Q (E(L)). Then we consider the composition R. R : Sel2 (E/Q) −→ E(Q)/2E(Q) −→ E(Q)/ NL/Q (E(L)).. (6). Let T := T0 ∪ T1 , where T0 is the set consisting of 2, ∞ and all odd prime divisors of ab(b − a), while T1 is consisting of prime divisors of d not contained in T0 . Note that T1 = ∅ means that d is a divisor of 2ab(b − a), and this holds only for a finite number, and hence a negligible amount, of d. Therefore, we assume that T1 6= ∅. Definition I.1. For a given elliptic curve E, we say that a quadratic extension L/Q is independent, if T1 = 6 ∅ and the homomorphism R is injective. The concept of independence turns out very useful. To illustrate its implications, we need to introduce some X(E/K)2,∞ be the 2-divisible S∞ notations. Let ∞ m subgroup of X(E/K)[2 T ] := m=1 X(E/K)[2 ] and let X2,0 (E/K) denote the intersection X(E/K)2,∞ X(E/K)[2]. Let Sel2,0 (E/K) denote the pre-image of X2,0 (E/K) under the map Sel2 (E/K) −→ X(E/K)[2]. Then (5) induces the exact sequence of vector spaces over F2 , the finite field of order 2, 0 −→ E(K)/2E(K) −→ Sel2,0 (E/K) −→ X2,0 (E/K) −→ 0.. (7). Note that we have sK,0 := dimF2 X2,0 (E/K) = rkZ2 (Hom(X(E/K)2,∞ , Q2 /Z2 )).. (8).

(8) 8. I. INTRODUCTION. Also, since dimF2 E(K)/2E(K) = rk(E(K)) + 2, we have dimF2 Sel2,0 (E/K) = rk(E(K)) + 2 + sK,0 .. (9). Consider the restriction map resL/Q : H1 (Q, E[2]) −→ H1 (L, E[2]) and write S2 (L/Q) for the pre-image of Sel2 (E/L) ⊂ H1 (L, E[2]). Suppose L/Q is independent. Then the homomorphism res. S2 (L/Q) −→ Sel2 (E/L). (10). induced from the restriction map is surjective (see Proposition III.1). Denote S2,0 (L/Q) = res−1 (Sel2,0 (E/L)). Since ker(res) = H1 (Gal(L/Q), E[2]) = Hom(Gal(L/Q), E[2]), we have dimF2 (S2,0 (L/Q)) = dimF2 (Sel2,0 (E/L)) + 2.. (11). On the other hand, we have (see Proposition III.2) Sel2 (E/Q) ∩ S2,0 (L/Q) = Sel2,0 (E/Q), and this implies the inclusion S2,0 (L/Q)/ Sel2,0 (E/Q) ,→ S2 (L/Q)/ Sel2 (E/Q). Denote tL,0 = dimF2 S2,0 (L/Q)/ Sel2,0 (E/Q) and tL = dimF2 S2 (L/Q)/ Sel2 (E/Q).. (12). Then by equations (9) and (11), we have tL,0 − 2 = (rk(E(L)) − rk(E(Q))) + (sL,0 − sQ,0 ) ≥ 0.. (13). In general, the dimension tL,0 might not be easy to compute. However, in the case where tL = 2 or 3, (14) we can conclude that (see Corollary II.1) tL,0 = tL .. (15). Thus, if tL = 2, then we know that Ed has rank 0 over Q; if tL = 3 and the Parity Conjecture holds for Ed , then we can deduce that Ed has rank 1 over Q (see Theorem III.1). To compute the dimension tL , consider the commutative diagram: L 1 jQ 0 −→ Sel2 (E/Q) −→ H1 (Q, E[2]) −→ v H (Qv , E) (16) k ↑ ↑ L 1 0 −→ Sel2 (E/Q) −→ S2 (L/Q) −→ v H (Gal(Lv /Qv ), E(Lv )), where both up-arrows are inclusions. According to it, we can interpret tL as the F2 -dimension of the intersection: M Im(jQ ) ∩ H1 (Gal(Lv /Qv ), E(Lv )). v.

(9) I. INTRODUCTION. 9. The direct sum is indeed a finite sum, since H1 (Gal(Lv /Qv ), E(Lv )) is trivial for v 6∈ T (see Lemma II.3). To determine the above intersection, we apply Tate’s local duality theorem (see [Tat57/58], or [Mil86], I.3.4) as well as the Cassels-Tate exact sequence (see [Tat62] or [Mil86],I.6.13). In our situation, the computation can be made very explicit. We shall describe the result in the next paragraph. For a finite set T 0 of place of Q such that T0 ⊂ T 0 , write T 0 (L) = {w | w is a place of L sitting over v, for some v ∈ T 0 }. (17). and put ET 0 (Q) =. Y. E(Qv ), ET 0 (L) =. v∈T 0. Y. E(Lw ).. (18). w∈T 0 (L). Let NL/Q : ET 0 (L) −→ ET 0 (Q) be the norm map, and consider the composition: R. RT 0 : Sel2 (E/Q) −→ E(Q)/2E(Q) −→ ET 0 (Q)/2ET 0 (Q), where E(Q)/2E(Q) −→ ET 0 (Q)2ET 0 (Q) is the natural projection. Here we should remind the reader that we are interested in the case where T1 consists of a single prime number. The following lemma will be proved at the end of Section II.5. Lemma I.1. Suppose |T1 | = 1. Then the extension L/Q is independent if and only if Im(RT0 ) ∩ (NL/Q (ET0 (L))/2ET0 (Q)) = {0}. (19) Furthermore, if (19) holds, then tL − 2 = dimF2 ET0 (Q)/2ET0 (Q) − dimF2 (Im(RT0 ) + NL/Q (ET0 (L))/2ET0 (Q)). Write NL = NL/Q (ET0 (L))/2ET0 (Q), for simplicity. For the given elliptic curve E/Q, the space ET0 (Q)/2ET0 and the subspace Im(RT0 ) are fixed. Our √ task is to investigate the possibility of constructing quadratic extensions L = Q( d) over Q, with exactly one prime divisor of d not contained in T0 , so that inside ET0 (Q)/2ET0 (Q) the subspace NL is in the compliment of Im(RT0 ) and the subspace Im(RT0 ) + NL formed by them is of co-dimension 0 or 1. Here we note that the space NL depends sorely on quadratic residue of d modulo certain multiple of the conductor of E (under the restrain that |T1 | = 1). Hence, if d satisfies our requirement, then so is any other with the same quadratic residue. This is why we can give a lower bound of the density of S0 and S1 in the main theorem. The rest of the work relies mostly on local computations. We first consider the case where d = p (T1 = {p}), a prime number. In this situation, we should be able to verify the main theorem for every explicitly given curve. However, our information on the local behavior of E at 2 is inadequate for us to treat a curve defined by a formally given equation. To overcome the difficulty, we replace E by Eq for a carefully chosen prime number q and apply the above method by taking d = pq. By doing so, we still get result on Ep . since (Eq )d = Ep . In this situation, q is contained in T0 . Roughly speaking, we use the local information at q to substitute that at 2. Then we are able to complete the proof of the main theorem..

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(11) CHAPTER II. Preliminary Results We keep the notation in the previous chapter so that E/Q is an elliptic curve with E[2] ⊂ E(Q). For each field F containing Q, let GF denote the Galois group Gal(F¯ /F ). Let K denote a number field. 1. The Cassels-Tate Pairing Recall that sK,0 = dimF2 (X2,0 (E/K)). Also, denote ∫K = dimF2 (Sel2 (E/K)), ∫K,0 = dimF2 (Sel2,0 (E/K)), sK = dimF2 (X(E/K)[2]). . The Cassels-Tate pairing ·, · : X(E/K)[2∞ ] × X(E/K)[2∞ ] −→ Q/Z is alternating in the sense that x, x = 0 for each x in X(E/K)[2∞ ], and both its right and left kernels equal X(E/K)2,∞ (see [Mil86], I.6.13). Lemma II.1. We have sK ≡ sK,0 (mod 2). The lemma is well known. But we shall give a detailed proof after the following discussion on its application, as we are not able to find the existing reference. First, if (14) holds, then tL − tL,0 = 0, or 1 (by (13)). But, (12) and Lemma II.1 imply that tL − tL,0 = ∫L − ∫L,0 − (∫Q − ∫Q,0 ) = sL − sL,0 − (sQ − sQ,0 ) ≡ 0. (mod 2).. Corollary II.1. If tL = 2, or 3, then tL = tL,0 . Another application is on the Parity Conjecture: The Parity Conjecture for E over K. We have sK ≡ 0 (mod 2). Corollary II.2. If the 2-primary Tate-Shafarevich group X(E/K)[2∞ ] is finite, then the Parity Conjecture holds for E/K. Proof. (of Lemma II.1) For simplicity, denote X[2m ] := X(E/K)[2m ], X2 := X(E/K)[2∞ ], and let Am = 2m−1 X[2m ]. Then we have X[2] = A0 ⊇ A1 ⊇ A2 ⊇ · · · ⊇ A∞ = X2,0 (E/K). 11.

(12) 12. II. PRELIMINARY RESULTS. It is enough to show that dimF2 (Am /Am+1 ) is even for every m. Consider the map: 2m−1. jm : X[2m ] Am Am /Am+1 . If 2m−1 x is in Am+1 for some x ∈ X[2m ], then 2m−1 x = 2m y , for some y ∈ X[2m+1 ]. Consequently, (x − 2y) ∈ X[2m−1 ]. From this we see that ker(jm ) = X[2m−1 ] + 2X[2m+1 ]. For a, b ∈ Am /Am+1 , let x, y be elements of X[2m ] with jm (x) = a and jm (y) = b, and define a, b m := x, 2m−1 y = 2m−1 x, y . This is well-defined, since X[2m ], 2m X2 = 0. Suppose a, b m = 0 for all b ∈ Am /Am+1 . Then 2m−1 x, y = 0, for all y ∈ X[2m ]. But, as the annihilator of X[2m ] is 2m X2 , we can conclude that x ∈ X[2m−1 ] + 2X[2m+1 ], and hence a = 0. Therefore, the pairing , m is perfect and alternating. Then we complete the proof by using the following lemma. Lemma II.2. Let A be a finite dimensional vector space over F2 , and suppose that there exists an alternating, non-degenerate pairing 1 Γ : A × A −→ Z/Z. 2 Then the dimension of A over F2 is even. Proof. Suppose that dimF2 (A) = 2n + 1 for some integer n. We want to find some non-zero element y in A such that Γ(y, x) = 0 , for all x ∈ A. If n = 0, then the non-zero element y in A satisfies this property by that Γ(x, x) = 0 for each x ∈ A. Suppose that n > 0. Let{xi ∈ A | i = 1, 2, · · · , (2n + 1)} be a basis for A over F2 and for each i define fi (x) := Γ(xi , x) , for all x ∈ A. Since Γ is non-degenerate and bilinear, fi is a non-trivial F2 -linear map, and hence ker fi has dimension (2n) over F2 . Note that ker(fi ) also determines fi . If ker fi = ker fj for some i 6= j, then fi = fj , and for y = xi + xj , we have Γ(y, x) = Γ(xi + xj , x) = Γ(xi , x) + Γ(xj , x) = 0 ∀x ∈ A. This contradicts to the non-degeneracy of Γ. Therefore, for i 6= j, ker fi 6= ker fj and dimF2 (ker fi ∩ ker fj ) = 2n − 1. We arrange the order x1 , ..., x2n+1 so that Γ(x1 , x2 ) = 12 . Then A is spanned by {x1 , x2 } ∪ (ker f1 ∩ ker f2 ). By induction, there must be a non-zero element y in (ker f1 ∩ ker f2 ) such that Γ(y, x) = 0 , for all x ∈ (ker f1 ∩ ker f2 )..

(13) 2. THE 2-DIVISION POINTS. 13. Since y is in (ker f1 ∩ ker f2 ), we have that Γ(y, x1 ) = Γ(y, x2 ) = 0, a contradiction.. 2. The 2-division Points. Suppose K is an extension field of Q and P = (ξ, η) is a point in E(K). Then the 2-division points of P √ can √ be computed√as follow (see [CLT05], Lemma 2.3). First choose (the signs of) ξ, ξ − a and ξ − b so that p p p η = ξ · ξ − a · ξ − b. Then the point Q = (x, y), with p p p p x = ( ξ + ξ − a)( ξ + ξ − b), p p p p p p y = ( ξ + ξ − a)( ξ + ξ − b)( ξ − a + ξ − b), is a 2-division point of P . Choices of the square roots corresponds to choice of the 2-division points. Let v be a place of K and denote the completion of K at v by Kv . We call a subset {P1 , ..., Pk } of E(Kv ) a starter, if there exist points Q1 , ..., Qn ∈ E(Kv ) and non-negative integers m1 , ..., mn so that each 2mi Qi is contained in the sub-group ¯ 1 , ..., Q ¯ n form a F2 -basis of E(Kv )/2E(Kv ). < P1 , ..., Pk > and the residue classes Q Beginning with a starter, we can find a F2 -basis of E(Kv )/2E(Kv ) by repeatedly applying the above 2-division formula. To find a starter, we consider all possible cases as follow. If v is complex, then, because E(C)/2E(C) = 0,. (20). we choose {0} as a starter. If v is real, then, E(R)/2E(R) = Z/2Z,. (21). generate by the point (0, 0) ∈ E[2]. If v is odd, then E(Kv )/2E(Kv ) = Z/2Z × Z/2Z,. (22). and the group E[2] is a starter. If v is even and Kv is a degree kv extension over Q2 b v ), we see that E(Kv ) with Ov its ring of integers, then via the formal group E(O contains a finite index open subgroup that is isomorphic to Ov (see [Sil86], VII.6.3), and hence E(Kv ) ' Zk2v × Z/2Z × Z/2Z. Therefore, E(Kv )/2E(Kv ) = (Z/2Z)kv +2 . (23) By a routine computation using the formal group law, we can find points P1 , ..., Pkv b v ) independent over Z2 . Then the set {P1 , ..., Pkv } ∪ E[2] is a starter. We in E(O omit the details here. Using this algorithm, for each v ∈ T0 (K), we compute a basis Bv of E(Kv )/2E(Kv ). Consider the exact sequence 0 −→ E(Kv )/2E(Kv ) −→ H1 (Kv , E[2]) −→ H1 (Kv , E)[2] −→ 0. (24).

(14) 14. II. PRELIMINARY RESULTS. induced from the Kummer exact sequence. For each point P in E(Kv ), let ρP denote its image in H1 (Kv , E[2]). Namely, ρP (σ) = σQP − QP , for each σ ∈ GKv , where QP is a 2-division point of P . By (24), we view E(Kv )/2E(Kv ) as a subspace of H1 (Kv , E[2]) and also let ρP denote the residue class of P in E(Kv )/2E(Kv ). Thus, the aforementioned basis Bv are considered as a subset of H1 (Kv , E[2]). We denote [ B= Bv . (25) v∈T0 (K) 1. For an element ϕ ∈ H (Kv , E[2]), let [ϕ] denote its image in H1 (Kv , E)[2]. If [ϕ] is trivial, then ϕ = ρP for some P in E(Kv ). In this case, we write Pϕ , for the residue class of P modulo 2E(Kv ). Since E[2] is contained in E(K), we have H1 (Kv , E[2]) = Hom(GKv , E[2]). (ϕ). For each ϕ ∈ Hom(GKv , E[2]), let Kv. be the fixed field of ker ϕ.. Definition II.1. We say that ϕ is unramified (resp. ramified), if the extension is unramified (resp. ramified).. (ϕ) Kv /Kv. Lemma II.3. Suppose that E has good reduction at the place v not dividing 2 and ϕ ∈ Hom(GKv , E[2]). Then [ϕ] is trivial if and only if ϕ is unramified. Proof. It it well known that if E has good reduction at v and Lv /Kv is unramified, then H1 (Gal(Lv /Kv ), E(Lv )) is trivial (see [Mil86], I.3.8). Thus, if ϕ is (ϕ) (ϕ) unramified, then H1 (Gal(Kv /Kv ), E(Kv )) = 0 and so is [ϕ], which is in the image (ϕ) (ϕ) of the inflation map H1 (Gal(Kv /Kv ), E(Kv )) −→ H1 (Kv , E). (ϕ) On the other hand, if [ϕ] is trivial and Iv is the inertia subgroup of Gal(Kv /Kv ), (ϕ) then there is a point Q ∈ E(Kv ) such that ϕ(σ) = σQ − Q for every σ ∈ Iv . This means the reduction of ϕ(σ) ∈ E[2] at v is trivial, as σ induces the trivial automor¯ v , the algebraic closure of the residue field Fv . This implies ϕ(σ) = 0, as phism of F v is odd and E has good reduction. Consequently, we should have σ ∈ ker(ϕ), and (ϕ) hence σ must be the identity of Gal(Kv /Kv ). 3. The Local Pairing Recall that the Weil pairing e2 : E[2] × E[2] −→ {±1} is bi-linear, alternating and non-degenerate. In our situation, it is in fact very easy to compute. For P ∈ E[2], we have e2 (P, P ) = 1, by the alternating property, and we have e2 (P, Q) = −1 for Q not in the subgroup generated by P , for otherwise we would have e2 (P, Q) = 1 for every Q ∈ E[2]. Thus, no matter what E is, we have the following lemma..

(15) 3. THE LOCAL PAIRING. 15. Lemma II.4. For any two non-trivial elements P1 , P2 of E[2], we have 1 , if P1 = P2 , e2 (P1 , P2 ) = −1 ,if P1 6= P2 . Since E[2] ⊂ E(Kv ), we have H1 (Kv , E[2]) = Hom(GKv , E[2]). Recall that under the cup product (see [Sht72]) ∪ : H1 (Kv , E[2]) × H1 (Kv , E[2]) −→ H2 (Kv , E[2] ⊗ E[2]) a pair (ϕ1 , ϕ2 ) is sent to the cohomology class represented by the 2 co-cycle ϕ1 ∪ ϕ2 so that for and δ and σ in GKv , (ϕ1 ∪ ϕ2 )(δ, σ) = (−1)2 ϕ1 (δ) ⊗ δϕ2 (σ) = ϕ1 (δ) ⊗ ϕ(σ). Consider the composition ¯ ∗ ), H2 (Kv , E[2] ⊗ E[2]) −→ H2 (Kv , {±1}) −→ H2 (Kv , K v where the first map is induced from the Weil pairing, while the second is from the ¯ v∗ , and composite it with the cup product to form inclusion {±1} ,→ K ∪ ¯ v∗ ). ωv : H1 (Kv , E[2]) × H1 (Kv , E[2]) −→ H2 (Kv , E[2] ⊗ E[2]) −→ H2 (Kv , K. Then ωv (ϕ1 , ϕ2 ) is represented by the 2 co-cycle ωv (ϕ1 , ϕ2 )(δ, σ) = e2 (ϕ1 (δ), ϕ2 (σ)), δ, σ ∈ GKv . ¯ v∗ ) = Br(Kv ), the Brauer group, and we let Now, H2 (Kv , K ¯ ∗ ) −→ Q/Z invv : H2 (Kv , K v denote the Hasse invariant. Then the assignment 1 (ϕ1 , ϕ2 ) 7→< ϕ1 , ϕ2 >Kv := invv (ωv (ϕ1 , ϕ2 )) ∈ Z/Z ⊂ Q/Z 2 gives the local pairing 1 < ·, · >Kv : H1 (Kv , E[2]) × H1 (Kv , E[2]) −→ Z/Z ⊂ Q/Z. 2 It is known (see [Mil86], I.3.5) that if P1 , P2 ∈ E(Kv ), then < ρP1 , ρP2 >Kv = 0, and since this implies that the pairing < ϕ, ρP >Kv depends only on the class [ϕ], we have the induced pairing (·, ·)Kv : H1 (Kv , E)[2] × E(Kv )/2E(Kv ) −→ 21 Z/Z ([ϕ], P ) 7→ < ϕ, ρP >Kv The paring can be extended to a pairing (·, ·)Kv : H1 (Kv , E) × E(Kv ) −→ Q/Z, which, by the local Tate’s duality theorem, is non-degenerated and induces the duality between H1 (Kv , E) and E(Kv ) (op. cit. I.3.4). Therefore, we have the following:.

(16) 16. II. PRELIMINARY RESULTS. Proposition II.1. The pairing 1 (·, ·)Kv : H1 (Kv , E)[2] × E(Kv )/2E(Kv ) −→ Z/Z 2 is non-degenerated and induces an isomorphism ∼. H1 (Kv , E)[2] −→ Hom(E(Kv )/2E(Kv ), Q/Z). Corollary II.3. The pairing < ·, · >Kv is non-degenerated. Via the pairing, the annihilator of E(Kv )/2E(Kv ) is itself. Proof. Suppose ϕ is a non-trivial element of H1 (Kv , E[2]). If [ϕ] 6= 0, then there is a point P ∈ E(Kv ) so that < ϕ, ρP >Kv 6= 0; if [ϕ] = 0 and ϕ = ρP for some P ∈ E(Kv ), then, since P 6∈ 2E(Kv ), there is some ϕ0 such that < ϕ0 , ϕ >Kv 6= 0. This proves the non-degeneracy. By (24) and Proposition II.1, we have dimF2 H1 (Kv , E[2]) = 2 dimF2 E(Kv )/2E(Kv ). Therefore, The F2 -spaces E(Kv )/2E(Kv ) and its annihilator have the same dimension. Since E(Kv )/2E(Kv ) is a subspace of its annihilator, they must be equal. The pairing < ·, · >Kv can be easily computed in the following way. First, for each α ∈ Kv∗ \(Kv∗ )2 , P ∈ E[2], we let √ ρ(√α,P ) : Gal(Kv ( α)/Kv ) −→ E[2] (26) √ denote the group homomorphism that sends the generator of Gal(Kv ( α)/Kv ) to the point P , and we view it as an element in H1 (Kv , E[2]) via the inflation map. Proposition II.2. For α, β ∈ Kv∗ and nontrivial A, B ∈ E[2], we have √ 0 if A = B, or β is in NKv (√α)/Kv (Kv ( α)∗ ) √ √ < ρ( α,A) , ρ( β,B) >Kv = 1 otherwise, 2 where NKv (√α)/Kv is the norm map. Proof. If A = B, then ωv (ρ(√α,A) ∪ ρ(√β,B) )(δ, σ) = 1, for all δ, σ ∈ GKv (as the Weil pairing is alternating), and hence represents √ class. √ the trivial Suppose A 6= B. We first assume that Kv ( α) 6= Kv ( β), and, for simplicity, write f = ωv (ρ(√α,A) , ρ(√β,B) ). Let σα (resp. σβ ) be the generator of the Galois √ √ √ √ √ √ group Gal(Kv ( α, β)/K ( α)) (resp. Gal(K ( α, β)/K ( β))). Then the v v v √ √ Galois group Gal(Kv ( α, β)/Kv ) is generated by σα and σβ . Again, because the Weil pairing is alternating, we have √ p √ f (δ, σ) = 1, for all δ, σ ∈ Gal(Kv ( α, β)/Kv ( α)), and hence the class [f ] of f is sent to 0 by the restriction map √ p √ p √ p √ √ p H2 (Gal(Kv ( α, β)/Kv ), Kv ( α, β)∗ ) −→ H2 (Gal(Kv ( α, β)/Kv ( α), Kv ( α, β)∗ ). Then the inflation-restriction exact sequence tells us that [f ] is contained in the image of the inflation map √ √ √ p √ p H2 (Gal(Kv ( α)/Kv ), Kv ( α)∗ ) −→ H2 (Gal(Kv ( α, β)/Kv ), Kv ( α, β)∗ )..

(17) 3. THE LOCAL PAIRING. 17. √ √ Let [¯ g ] ∈ H2 (Gal(Kv ( α)/Kv ), Kv ( α)∗ ) be represented by the unique 2 cocycle g¯ with √ g¯(¯ σβ , σ ¯β ) = β, where < σ ¯β >= Gal(Kv ( α)/Kv ), √ √ and let g√be the pull-back of g¯ through the projection Gal(Kv ( α, β)/Kv ) −→ Gal(Kv ( α)/Kv ). Then we have g(τ, δ) = τy(δ)y(τ )y(τ δ)−1 f (τ, δ), √ √ where y is the 1 co-chain on Gal(Kv ( α, β)/Kv ) with y(id) = y(σα ) = 1, and y(σβ ) = y(σα σβ ) =. p β.. Therefore, the class [f ] is just the inflation of [¯ g ]. In particular, the Hasse invariant of [f ] is the same as that of [¯ g ]. Now, the crossed product algebra given by g¯ is the cyclic algebra of the form √ (Kv ( α)/Kv , σβ , β) (see [Rei75], √ page 259). Therefore, it is in the trivial class if and only if β is in NKv (√α)/Kv (Kv ( α)∗ ) (op.cit., Theorem 30.4). √ √ To complete the proof, we need to consider the case where Kv ( α) = Kv ( β). We only need to note that f = ωv (ρ(√α,A) , ρ(√β,B) ) is the unique 2 co-cycle with √ f (σ, σ) = −1, where < σ >= Gal(Kv ( α)/Kv ), and the crossed product algebra given by f is the cyclic algebra of the form (op.cit., page 259) √ √ (Kv ( α)/Kv , σ, −1) = (Kv ( α)/Kv , σ, α). Corollary II.4. Let Fv /Kv be a unramified extension of degree 2. Then for any ψ, ϕ ∈ H1 (Gal(Fv /Kv ), E[2]), we have < ψ, ϕ >Kv = 0. Proof. Since √ Fv /Kv is unramified of∗ degree 2, there exists an unit α ∈ Kv such that Fv = Kv ( α) and α is in NFv /Kv (Fv ). Lemma II.5. Suppose that Fv /Kv is a finite Galois extension. Then the following diagram is commutative H1 (Fv , E[2]) × H1 (Fv , E[2]) −→ Q/Z ↓cor ↑res k < ·, · >Kv : H1 (Kv , E[2]) × H1 (Kv , E[2]) −→ Q/Z. < ·, · >Fv :. Proof. The lemma is proved by the composition of the diagrams ∪. H1 (Fv, E[2]) × H1 (Fvx, E[2]) −→ H2 (Fv , E[2]  ⊗ E[2]) ycor  ycor res ∪ H1 (Kv , E[2]) × H1 (Kv , E[2]) −→ H2 (Kv , E[2] ⊗ E[2]) (see [CaE56], XII.8 (12)) and inv H2 (Fv , E[2] ⊗ E[2]) −→ H2 (Fv , F¯v∗ ) −→v Q/Z ↓cor ↓cor k invv 2 2 ∗ ¯ H (Kv , E[2] ⊗ E[2]) −→ H (Kv , Fv ) −→ Q/Z,.

(18) 18. II. PRELIMINARY RESULTS. where the first square is induced from the Weil pairing and the second is a wellknown property of the Hasse invariants (see [Ser79], XI.2.1). Corollary II.5. Suppose Fv /Kv is a quadratic extension. Then, via the local pairing (·, ·)Kv : H1 (Kv , E) × E(Kv ), the subgroups H1 (Gal(Fv /Kv ), E(Fv )) ⊂ H1 (Kv , E) and NFv /Kv (E(Fv )) ⊂ E(Kv ) are the annihilators to each other. Proof. We have H1 (Gal(Fv /Kv ), E(Fv )) ⊂ H1 (Kv , E)[2]. Let [ψ] ∈ H1 (Kv , E)[2] be represented by some ψ ∈ H1 (Kv , E[2]). Then [ψ] is contained in H1 (Gal(Fv /Kv ), E(Fv )) if and only if res(ψ) is contained in E(Fv )/2E(Fv ), or equivalently (Corollary II.3), < res(ψ), ϕQ >Fv = 0, ∀Q ∈ E(Fv ). Lemma II.5 says this is equivalent to that ([ψ], P )Kv =< ψ, ϕP >Kv = 0, ∀P ∈ NFv /Kv (E(Fv )), since cor(ϕQ ) = ϕP , for P = NFv /Kv (Q).. . 4. Computing Sel2 (E/K) As before, let T 0 be a finite set of places of Q such that T0 ⊂ T 0 and let T 0 (K) (2) denote the set consisting of places of K sitting over T 0 . Let KT 0 /K be the maximal abelian extension unramified outside T 0 (K) such that the Galois group is isomorphic to (Z/2Z)r for some r, and denote (2). (2). GK,T 0 = Gal(KT 0 /K). Since E has good reduction outside T0 , by Lemma II.3, an element of the 2Selmer group Sel2 (E/K) ⊂ H1 (K, E[2]) = Hom(GK , E[2]) is unramified outside T0 (K). Therefore, it factors through the natural projection (2) GK −→ GK,T 0 , and hence we can say that (2). Sel2 (E/K) ⊂ Hom(GK,T 0 , E[2]).. (27). For the rest of this section, we assume that K is a number field with odd class number. Lemma II.6. Suppose K is a number field with odd class number. Then (2). dimF2 Hom(GK,T 0 , E[2]) = dimF2 ET 0 (K)/2ET 0 (K). Proof. The assumption implies that for each non-archimedean place v there √ is ∗ an element fv ∈ K , whose divisor equals an odd multiple of v. Suppose K( δ) (2) is a quadratic intermediate extension of KT 0 /K. Then the valuation of δ at every v 6∈ T 0 (K) is even, and by replacing δ with δ m f 2 , for some odd m and some f ∈ K ∗ , we can assume all such valuations are zero. This means δ is contains in OT∗ 0 (K) , the group of T 0 (K)-units. Thus, we have (2). dimF2 Hom(GK,T 0 , E[2]) = 2 dimF2 OT∗ 0 (K) /(OT∗ 0 (K) )2 = 2|T 0 (K)|..

(19) 4. COMPUTING Sel2 (E/K). 19. On the other hand, by (20), (21), (22) and (23), we know that 0 dimF2 ET 0 (K)/2ET 0 (K) = r1 + |K : Q| + 2|T0,0 (K)|, 0 where r1 is the number of real places of K and T0,0 = T 0 \ {∞}. Then we apply the Dirichlet theorem to complete the proof. √ As in (26), let ρ(√α,P ) : Gal(K( α)/K) −→ E[2] denote the group homomor√ phism that sends the generator of Gal(K( α)/K) to the point P . The above proof also show that if A, B is an F2 -basis of E[2] and δi , i = 1, ..., |T 0 (K)|, is a basis of OT∗ 0 (K) modulo (OT∗ 0 (K) )2 , then [ AT 0 := {ρ(√δi ,A) | i = 1, ..., |T 0 (K)|} {ρ(√δi ,B) | i = 1, ..., |T 0 (K)|} (2). is a basis of Hom(GK,T 0 , E[2]). For each ϕ ∈ H1 (K, E[2]), let ϕv denote its image under the localization map to H1 (Kv , E[2]). For simplicity, take T 0 = T0 . By (27), we know that Sel2 (E/K) (2) (2) is an F2 -subspace of Hom(GK,T0 , E[2]). Since every element ϕ ∈ Hom(GK,T0 , E[2]) is unramified outside T0 (K), by Lemma II.3, the class [ϕv ] is trivial for v 6∈ T0 (K). Thus, by Corollary II.3, the element ϕ is contained in Sel2 (E/K) if and only if for each v ∈ T0 (K), we have (see (25)) < ϕv , ψv >Kv = 0, ∀ψv ∈ Bv , v ∈ T0 (K). To compute Sel2 (E/K), we first apply Proposition II.2 to compute the values: < ϕv , ψv >Kv , ∀ ϕ ∈ AT0 , ψv ∈ Bv , v ∈ T0 (K). Then we determine the set {(xϕ )ϕ∈AT0 |. X. xϕ ϕ ∈ Sel2 (E/K)}. ϕ∈AT0. by solving the associated system of linear equations over F2 : X < ϕv , ψv >Kv ·xϕ = 0, ψv ∈ Bv , v ∈ T0 (K). ϕ∈AT0. If ϕ ∈ Sel2 (E/K), then for each v, we have ϕ = ρPv for some Pv ∈ E(Kv ) and hence ϕ determines a class Pϕv ∈ E(Kv )/2E(Kv ). The assignment ϕ 7→ {Pϕv }v∈T 0 (K) defines the following root map: RT 0 (K) : Sel2 (E/K) −→ ET 0 (K)/2ET 0 (K), which coincides the map RT 0 in the previous chapter, if K = Q. Lemma II.7. Suppose K is a number field with odd class number. Then RT 0 (K) is injective. Proof. If RT 0 (K) (ϕ) = 0, then ϕv is the trivial homomorphism, for v ∈ T 0 (K). (ϕ ) Therefore the fixed field Kv v of ker(ϕv ) is just the field Kv . The Lemma II.3 implies that ϕ is unramified everywhere. Since the fixed field K (ϕ) of ker(ϕ) over K has degree a power of 2 and K has odd class number, K (ϕ) is just the field K. Then ϕ must be trivial. .

(20) 20. II. PRELIMINARY RESULTS. 5. The Global Duality We have the commutative diagram: ω. ¯ ∗) H2 (K, K ↓ L L 1 ⊕ω v 2 1 ¯∗ v H (Kv , Kv ) v H (Kv , E[2]) × H (Kv , E[2]) −→ H1 (K, E[2]) × H1 (K, E[2]) ↓. −→. −→ P. 0 ↓. invv. v −→. Q/Z,. where the first row is also induced from the cup product, the Weil pairing and the ¯ ∗ , while both down-arrows are localization map to the direct inclusion {±1} ,→ K sum over all places. That the last square commutes is from the global class field theory (see [Tat67], Theorem B). Therefore, for ρ1 , ρ2 ∈ H1 (K, E[2]), we have X < ρ1v , ρ2v >Kv = 0. (28) υ. The assignment ϕ 7→ ([ϕv ])v∈T 0 (K) defines the map jT 0 (K) that fits into the following exact sequence extending (27) Y jT 0 (K) (2) 0 −→ Sel2 (E/K) −→ Hom(GK,T 0 , E[2]) −→ H1 (Kv , E)[2]. (29) v∈T 0 (K) (2). By Lemma II.3, if ϕ ∈ Hom(GK,T 0 , E[2]), then [ϕv ] = 0, for v 6∈ T 0 (K). We have global duality. P For simplicity write A = ET 0 (K)/2ET 0 (K), B = Q the following 1 v∈T 0 (K) H (Kv , E)[2] and (·, ·) = v∈T 0 (K) (·, ·)Kv . Proposition II.3. Q Suppose1 K is a number field with odd class number. An element (ξv )v∈T 0 (K) ∈ v∈T 0 (K) H (Kv , E)[2] is in the image of jT 0 (K) , if and only if the sum of local pairings X (ξv , Pϕv )Kv = 0, for all ϕ ∈ Sel2 (E/K). v∈T 0 (K). Proof. The groups A and B are dual to each other via (·, ·) (by Proposition II.1). In particular, we have (2). dimF2 B = dimF2 A = dimF2 Hom(GK,T 0 , E[2]) (Lemma II.6). Lemma II.7 says that, via the root map RT 0 (K) , we can view Sel2 (E/K) as a subspace of A. We need to show that Im(jT 0 (K) ) equals Sel2 (E/K)⊥ , the annihilator of Sel2 (E/K). By (28), we see that Im(jT 0 (K) ) ⊂ Sel2 (E/K)⊥ . But, (29) implies that dimF2 Im(jT 0 (K) ) is the same as that of Sel2 (E/K)⊥ . Now we prove Lemma I.1. Proof. (of Lemma I.1) If v 6∈ T , then L/Q is unramified at v. By Lemma II.3, we have H1 (Gal(Lv /Qv ), E(Lv )) = 0, for v 6∈ T. (30) Then Corollary II.5 says that E(Qv )/ NLv /Qv (E(Lv )) = 0, for v 6∈ T,. (31).

(21) 5. THE GLOBAL DUALITY. 21. as well. Consequently, using the projection E(Q) −→ ET (Q), we can identify E(Q)/ NL/Q (E(L)) with ET (Q)/ NL/Q (ET (L)). On the other hand, as T1 is consisting of some odd place v1 at which L/Q is ramified, we have NLv1 /Qv1 (E(Lv1 )) = 2E(Qv1 ), and hence ET (Q)/ NL/Q (ET (L)) = ET0 (Q)/ NL/Q (ET0 (L)) × E(Qv1 )/2E(Qv1 ). By Lemma II.7, we know that RT0 is injective and so is RT . Therefore, if (19) holds, then R is certainly injective. To show the converse holds, we first note that the pairing ([ψv1 ], Q)Qv1 , for ψ ∈ Hom(Gal(L/Q), E[2]) and Q ∈ E(Qv1 )/2E(Qv1 ), gives rise to a duality between Hom(Gal(L/Q), E[2]) and E(Qv1 )/2E(Qv1 ) (see Corollary II.5). If ϕ is an element in Sel2 (E/Q) such that RT0 (ϕ) ∈ NL/Q (ET0 (L))/2ET0 (Q), then, for ψ ∈ Hom(Gal(L/Q), E[2]), we have < ψv , ϕv >Qv = ([ψv ], Pϕv )Qv = 0 for v ∈ T0 (by Corollary II.5) and v 6∈ T (by Lemma II.3). This implies X X < ψv , ϕv >Qv = ([ψv ], Pϕv )Qv = ([ψv1 ], Pϕv1 )Qv1 . v. v. But, by (28), this sum is zero, and consequently, Pϕv1 ∈ NLv1 /Qv1 (E(Lv1 )). Thus, if R is injective, then we must have ϕ = 0 and hence (19) holds. To prove the last statement we take T 0 = T and recall the notation used in the previous paragraphs. Then we note that since L/Q is unramified outside T , (2) Lemma II.3 implies that S2 (L/Q) ⊂ Hom(GQ,T , E[2]) and it is indeed the pre-image Q of v∈T H1 (Gal(Lv /Qv ), E(Lv )) under the map jT in (29). By Proposition II.3, an element b = (ξv )v∈T ∈ B is in the image of jT if and only if it annihilates Sel2 (E/Q), via (·, ·). Also, by Corollary II.5, every class [ξv ] ∈ H1 (Qv , E)[2], v ∈ T , is contained in H1 (Gal(Lv /Qv ), E(Lv )) if and only if b annihilates NL/Q (ET (L))/2ET (Q) ⊂ A. This shows that S2 (L/Q)/ Sel2 (E/Q) = jT (S2 (L/Q)) ⊂ B is the annihilator of RT (Sel2 (E/Q)) + NL/Q ET (L)/2ET (Q). In particular, we have tL = dimF2 A − dimF2 Sel2 (E/Q) − dimF2 NL/Q ET (L)/2ET (Q). Then we complete the proof by noting that dimF2 A = dimF2 ET0 (Q)/2ET0 (Q) + 2 and NL/Q ET (L)/2ET (Q) = NL/Q ET0 (L)/2ET0 (Q). .

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(23) CHAPTER III. The Independent Field Extensions √ We keep the previous notation and assume that L = Q( d)/Q is an independent extension. In particular, T1 6= ∅. Let Γ denote the Galois group Gal(L/Q) and write Γ = {γ, id}. 1. The Norm Map Recall the norm map NL/Q : E(L) −→ E(Q). Since E has good reduction outside T , Lemma II.3 and Lang’s theorem together imply that Y E(Qv ) ⊂ Im(NL/Q ), v6∈T. and hence E(Q)/ NL/Q (E(L)) = ET (Q)/ NL/Q (ET (L)).. (32). Lemma III.1. Suppose that L/Q is independent. Then the co-restriction map Sel2 (E/L) −→ Sel2 (E/Q) is the trivial map. In particular, if ψ ∈ Sel2 (E/L), then γ ψ = ψ. Proof. The second statement follows from the first, since we have 0 = res(cor(ψ)) = γ ψ + ψ = γ ψ − ψ. The commutative diagram ET (L)/2E  T (L) −→ y NL/Q ET (Q)/2ET (Q) −→. L. 1 H  (Lw , E[2]) y cor H1 (Qv , E[2]). w∈T (L). L. v∈T. (33). implies RT (L) : Sel2 (E/L) −→ ET (L)/2ET (L) ↓ cor ↓ NL/Q RT : Sel2 (E/Q) −→ ET (Q)/2ET (Q) From (32), we see that Im(cor) is contained in the kernel of the injection R. Proposition III.1. Suppose L/Q is independent. Then the restriction map res : S2 (L/Q) −→ Sel2 (E/L) is surjective. 23.

(24) 24. III. THE INDEPENDENT FIELD EXTENSIONS. Proof. For an element ψ ∈ Sel2 (E/L) ⊆ Hom(GL , E[2]), let H = ker(ψ) and let K (ψ) denote the fixed field of H. Lemma III.1 says that if γ˜ is a lift of γ to GQ , then γ˜ H γ˜ −1 = H and γ˜ gH γ˜ −1 = gH for every g ∈ GL . In particular, H is a normal subgroup of GQ and we have the exact sequence {1} −→ GL /H −→ GQ /H −→ GQ /GL −→ {1} with both GL /H and GQ /GL abelian so that GQ /GL commutes with GL /H in the sense that the lifting of each element of GQ /GL to GQ /H commutes with GL /H. This together with the fact that GQ /GL = Γ is cyclic imply that GQ /H is abelian. Now, K (ψ) /L is an abelian extension with Gal(K (ψ) /L) = GL /H, and ψ can be view as an element in Hom(Gal(K (ψ) /L), E[2]). Since ψ ∈ Sel2 (E/L), Lemma II.3 implies that K (ψ) /L is unramified at every w ∈ T1 (L). Consequently, if we choose a place v ∈ T1 , then the inertia subgroup Iv of Gal(K (ψ) /Q) has trivial intersection with Gal(K (ψ) /L). On the other hand, the quadratic extension L/Q is totally ramified at v, and hence the composition Iv −→ Gal(K (ψ) /Q) −→ Gal(L/Q) is an isomorphism. Therefore, the group Gal(K (ψ) /Q) is indeed the direct product of Iv and Gal(K (ψ) /L). Let ψ˜ be the composition ∼. ψ. Gal(K (ψ) /Q) −→ Gal(K (ψ) /Q)/Iv −→ Gal(K (ψ) /L) −→ E[2]. ˜ = ψ. Then ψ˜ ∈ Hom(Gal(K (ψ) /Q), E[2]) and res(ψ). . 2. The Map nL/Q Suppose ψ ∈ Sel2 (E/L) ⊆ Hom(GL , E[2]) and let K (ψ) be the fixed field of ker(ψ). Then there is an η ∈ E(K (ψ) ) such that ψ(δ) = δη − η. (34). for each δ ∈ GL . Writing Pψ = (Pψv )v ∈ E(L), we have Pψ = 2η. Choose a lifting γ˜ of γ to GQ and put ξ = γ˜η + η. Then 2ξ = NL/Q (2η) and the commutative diagram (33) says if we define ϕ ∈ Hom(GQ , E[2]) so that ϕ(σ) := σξ − ξ, for each σ ∈ GQ , then ϕ = cor(ψ), the image of the co-restriction. If ψ = res(ψ0 ) for some ψ0 ∈ S2 (L/Q), then ϕ = cor(ψ) = 2ψ0 = 0, and consequently, ξ ∈ E(Q). The point ξ depends on the choice of η and γ˜ . A straightforward computation show that a different choice will only cause a difference by an element in E[2]+NL/Q (E(L)). This gives rise to a norm map (see (32)) nL/Q : S2 (L/Q) −→ E[2]\ET (Q)/ NL/Q (ET (L)). If in the above, ψ0 ∈ Sel2 (E/Q), then η can be chosen so that (34) holds for every δ ∈ GQ . From this, we deduce that the following lemma:.

(25) 3. THE SELMER GROUPS. 25. Lemma III.2. The diagram: R. Sel2 (E/Q) −→ ET (Q)/ NL/Q (ET (L)) ∩ ↓ nL/Q S2 (L/Q) −→ E[2]\ET (Q)/ NL/Q (ET (L)) is commutative. 3. The Selmer Groups Let K be a number field. The Kummer exact sequences induce the commutative diagram of exact sequences: in+m. n+m ] −→ 0 0 −→ E(K)/2n+m  E(K) −→ Sel2n+m  (E/K) −→ X(E/K)[2  y∗ y[2m ]∗ y2m. 0 −→. E(K)/2n E(K). −→. Sel2n (E/K). i. n −→. X(E/K)[2n ]. −→ 0,. where the first down-arrow, induced from the identity map E(K) → E(K), is surjective and the second is induced from [2m ]. Therefore, from the snake lemma, we have the exact sequence coker(∗) k 0. −→ Sel2n (E/K)/ Im([2m ]∗ ) X(E/K)[2n ]/2m X(E/K)[2n+m ].. This implies the following lemma. Lemma III.3. Every element ψ ∈ Sel2,0 (E/K) is 2-divisible in the sense that for each positive integer m, there is an element ψm ∈ Sel2m (E/K) so that [2m−1 ]∗ (ψm ) = ψ. Lemma III.4. Suppose ψ0 is an element in S2 (L/Q) such that res(ψ0 ) = [2]∗ (ψ2 ) for some ψ2 ∈ Sel4 (E/L) satisfying cor(ψ2 ) = 0. Then nL/Q (ψ0 ) = 0. Proof. Suppose ψ2 is obtained from H1 (Gal(L0 /L), E[4]) for some Galois extension L0 /L and let ρ be a 1 co-cycle on Gal(L0 /L) representing ψ2 . Since ψ2 ∈ Sel4 (E/L), there is an η2 ∈ E(L0 ) so that ρ(δ) = δη2 − η2 , for every δ ∈ GL . In particular, η := 2η2 satisfies (34). The commutative diagram (similar to (33)) L 1 ET (L)/4E (L) −→ T w∈T (L) H   (Lw , E[4]) y NL/Q y cor (35) L 1 ET (Q)/4ET (Q) −→ v∈T H (Qv , E[4]), implies that if we choose a lifting γ˜ of γ to GQ and put ξ2 = γ˜η2 + η2 , then the class ϕ2 := [ρ0 ] ∈ H1 (Q, E[4]) of the 1 co-cycle ρ0 (σ) := σξ2 − ξ2 ∈ E[4], for all σ ∈ GQ ,.

(26) 26. III. THE INDEPENDENT FIELD EXTENSIONS. is nothing but cor(ψ2 ). Because cor(ψ2 ) = 0, there is a point Q ∈ E[4] such that ξ2 ∈ Q + E(Q). This implies that ξ := 2ξ2 ∈ E[2] + 2E(Q) and hence, by the definition, nL/Q (ψ0 ) = 0. Recall that S2,0 (L/Q) is the pre-image of Sel2,0 (E/L) under the restriction map. Proposition III.2. Suppose L/Q is independent. Then we have Sel2 (E/Q) ∩ S2,0 (L/Q) = Sel2,0 (E/Q). Proof. Let ψ0 ∈ Sel2 (E/Q) ∩ S2,0 (L/Q) and let ψ = ψ1 = res(ψ0 ). By Lemma III.3, for each positive m, there is an ψm ∈ Sel2m (E/L) such that [2]∗ (ψm ) = ψm−1 . Write ϕm−1 = cor(ψm ) ∈ Sel2m (E/Q). Then ϕ−1 = cor(res(ψ0 )) = 0, and hence [2m−1 ]∗ (ϕm−1 ) = 0. By the exact sequence [2m−1 ]∗. H1 (Q, E[2m−1 ]) −→ H1 (Q, E[2m ]) −→ H1 (Q, E[2]) induced from [2m−1 ]. 0 −→ E[2m−1 ] −→ E[2m ] −→ E[2] −→ 0, we can assume that each ϕm−1 is contained inside Sel2m−1 (E/Q). In particular, we have ϕ0 ∈ Sel2,0 (E/Q). In view of this, we see that to prove the proposition, we are allowed to replace ψk with ψk − ϕk , for k = 0, ..., at the beginning, and by doing so, we can assume that ϕ1 = 0. Then Lemma III.2 and Lemma III.4 imply that R(ψ0 ) = (P ), for some P ∈ E[2]. Since L/Q is independent, we must have ψ0 ∈ E(Q)/2E(Q) ⊂ Sel2,0 (E/Q) ⊂ Sel2 (E/Q). 4. The First Main Result In this section, we put the previous results together to establish our first main result. Recall that tL = dimF2 S2 (L/Q)/ Sel2 (E/Q) (see (12)). √ Theorem III.1. Suppose L = Q( d) is an independent extension over Q. If tL = 2, then the quadratic twist Ed has trivial rank over Q. If tL = 3 and the Parity Conjecture holds for Ed over Q, then rk(Ed (Q)) = 1. Proof. By Corollary II.1, we have tL,0 = tL . Therefore, if tL = 2, then (13) implies that rk(E(L)) = rk(E(Q)), and hence rk(Ed (Q)) = 0. The 2-divisible group X(E/L)2,∞ equals the direct sum M X(E/L)+ X(E/L)− 2,∞ 2,∞ , − where X(E/L)+ 2,∞ (resp. X(E/L)2,∞ ) is the 1 (resp. −1) eigen-space of the action of Γ = Gal(L/Q) in the sense that it is the maximal subgroup such that the action of Γ is just the multiplication by 1 (resp. −1). Since Ed is the quadratic twist, we have ∓1 X(Ed /L)±1 2,∞ = X(E/L)2,∞ ..

(27) 4. THE FIRST MAIN RESULT. 27. Also, from the exact sequence (which is a part of the Hochschild-Serre spectral sequence, see [CaE56]) inf. res. H1 (Γ, E(L)) −→ H1 (Q, E) −→ H1 (L, E)Γ −→ H2 (Γ, E(L)) and the finiteness of the groups Hi (Γ, E(L)), i = 1, 2, we see that the 2-divisible group X(E/L)+ X(E/L)Γ2,∞ must be obtained from H1 (Q, E). Since the local 2,∞ = L 1 cohomology group v H (Γv , E(Lv )) is finite (see Lemma II.3) and X(E/L)+ 2,∞ is 2-divisible, we can deduce that the Z2 co-rank of X(E/L)+ (that is, the Z -rank of 2 2,∞ + Hom(X(E/L)2,∞ , Q/Z)) is the same as that of X(E/Q)2,∞ . By the same argument, we can also say that Z2 co-rank of X(E/L)− 2,∞ is the same as that of X(Ed /Q)2,∞ . If tL = 3, then from (13) again, we deduce that either rk(Ed (Q)) = 1 or the Z2 co-rank of X(Ed /Q)2,∞ equals 1 (see (8)). Thus, if the Parity Conjecture for Ed over Q holds, then we must have rk(Ed (Q)) = 1. .

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(29) CHAPTER IV. The Local Symbols In this chapter, we shall define, for each v ∈ T0 \ {2}, a group homomorphism Φv : E(Qv )/2E(Qv ) −→ fv , where fv ' E(Qv )/ NLv /Qv (E(Lv )) for a particular quadratic extension Lv /Qv . Then we define the local symbol {ϕ}υ := Φv (Pϕv ), for each ϕ ∈ Sel2 (E/Q). Also, we denote Nv = NLv /Qv (E(Lv )).. (36). 1. The Archimedean Place Suppose v = ∞. Then Qv = R and we take Lv = C. Also, since E[2] ⊂ E(Qv ), the topological group E(Qv ) = E(R) has two components and NC/R (E(C)) is just the identity component of E(R). Therefore, E(Qv )/ NLv /Qv (E(Lv )) ' Z/2Z. Let Φv be the composition ∼. E(Qv )/2E(Qv ) −→ E(Qv )/ NLv /Qv (E(Lv )) −→ F2 . Suppose an element [P ] ∈ E(Qv )/2E(Qv ) is obtained from a point P = (ξ, η) ∈ E(Qv ). Then we have ( 0 ∈ F2 , if ξ ≥ b, Φv ([P ]) = (37) 1 ∈ F2 , otherwise. From this, we see that Φv ([P ]) = 0 if and only if P ∈ 2E(Qv ). Therefore, we have the following: Lemma IV.1. An element ϕ ∈ Sel2 (E/Q) has local symbol {ϕ}∞ = 0 if and only if ϕ∞ is unramified. 2. The Split Multiplicative Places The main reference of this as well as next section is [Sil86], section C.14. Suppose v is an odd place at which E has split multiplicative reduction. Then over Qv , E is isomorphic to the Tate curve with some period Qv ∈ Q∗v so that ¯ v) ' Q ¯ ∗ /QZ E(Q (38) v. v. ¯ ∗ , let Pz ∈ E(Q ¯ v ) denote the point corresponding to as GQv -modules. For an z ∈ Q v Z z modulo Qv (the Tate’s uniformization). 29.

(30) 30. IV. THE LOCAL SYMBOLS. Let Lv /Qv be the unique unramified quadratic extension. As E[2] ⊂ E(Qv ), we have p Qv ∈ Q∗v . √ In particular, Qv = ( Qv )2 ∈ NLv /Qv (L∗v ), and hence NLv /Qv E(Lv ) ' NLv /Qv (L∗v )/QZv . This shows that E(Qv )/ NLv /Qv E(Lv ) ' Q∗v / NLv /Qv (L∗v ) ' Z/2Z. Also, since Lv /Qv is unramified, the local unit group Z∗v is contained in NLv /Qv (L∗v ). There(L∗v ) if and only if the valuation ordv (z) is fore, an z ∈ Q∗v is contained in NLv /Qv√ even, or equivalently, √ the extension Qv ( z)/Qv is unramified. Since, by the isomorphism (38), Qv ( z) is the same as the field generated by the coordinates of one (and hence all) of the 2-division points of Pz , we see that Pz ∈ E(Qv ) is in the norm NLv /Qv E(Lv ) if and only if some of its 2-division points is rational over an unramified extension of Qv For a point P = (ξ, η) ∈ E(Qv ), we use the formula in Section II.2 to compute a 2-division point Q = (x, y) of P . Then we define Φv so that for the class [P ] ∈ E(Qv )/2E(Qv ), ( 0 ∈ F2 , if Qv (x, y)/Qv is unramified, Φv ([P ]) = (39) 1 ∈ F2 , otherwise. In view of this, we have the following lemma. Lemma IV.2. Suppose v is an odd place at which E has split multiplicative reduction. An element ϕ ∈ Sel2 (E/Q) has local symbol {ϕ}v = 0 if and only if ϕv is unramified. In this case the reduction at v of ϕ([v]) ∈ E[2], where [v] is the ¯ the reduction curve of E. Frobenius element at v, is a smooth point on E, Proof. To prove the last statement, we note that ϕ([v]) = Pz for some z ∈ Z∗v . That the reduction of such point is smooth is well known for Tate’s uniformization (see [Sil86], theorem C.14.1(c)). 3. The Non-Split Multiplicative Places Suppose v is an odd place at which E has non-split multiplicative reduction √ and let Lv /Qv be the unique unramified quadratic extension. Then Lv = Qv ( µ) for some non-square local unit µ ∈ Z∗v \ (Z∗v )2 and the quadratic twist Eµ has split multiplicative reduction at v. Thus, as an abstract group, we have E(Lv ) = Eµ (Lv ). (40) √ Also, since Eµ is the µ-twist of E and Lv = Qv ( µ), for the natural action of the Galois group Gal(Lv /Qv ) =: {τ, id} on Eµ (Lv ), we have E(Qv ) = {P ∈ Eµ (Lv ) | P + τ P = 0},. (41). and NLv /Qv (E(Lv )) = {P ∈ Eµ (Lv ) | P = τ Q − Q, for some Q ∈ Eµ (Lv )}.. (42).

(31) 4. THE POTENTIAL GOOD ADDITIVE PLACES. 31. This shows that E(Qv )/ NLv /Qv (E(Lv )) = H1 (Gal(Lv /Qv ), Eµ (Lv )) which, by Corollary II.5, is the dual group of Eµ (Qv )/ NLv /Qv (Eµ (Lv )). Therefore, from the discussion in the previous section, we can deduce that E(Qv )/ NLv /Qv (E(Lv )) ' Z/2Z. We can explicitly describe the group NLv /Qv (E(Lv )) as follow. Let P = (ξ, η) ∈ E(Qv ). Then P 0 = (µξ, µ3/2 η) ∈ Eµ (Lv ) is the corresponding point under the identification (40). Here as usual, the defining equation for Eµ is y 2 = x(x − µa)(x − µb). Lemma IV.3. The following conditions are equivalent: (a) P ∈ NLv /Qv (E(Lv )). (b) P 0 = Pz with z = τ t/t for some t ∈ L∗v . (c) P 0 = Pz with z ∈ OL∗ v . (d) The reduction P¯ 0 is a smooth point on E¯µ ¯ (e) The reduction P¯ is a smooth point on E. Proof. That (a) ⇐⇒ (b) is from (42), and (b) ⇐⇒ (c) is from (41) as well as Hilbert’s Theorem 90. Also, (c) ⇐⇒ (d) is by [Sil86], theorem C.14.1(c)), and (d) ⇐⇒ (e) is obvious. Then we define Φv so that for the class [P ] ∈ E(Qv )/2E(Qv ), ( ¯ 0 ∈ F2 , if the reduction P¯ is a smooth point on E, Φv ([P ]) = 1 ∈ F2 , otherwise.. (43). Lemma IV.4. Suppose v is an odd place at which E has non-split multiplicative reduction. An element ϕ ∈ Sel2 (E/Q) has local symbol {ϕ}v = 0 if and only if ϕv is unramified and the point ϕ([v]) ∈ E[2], where [v] is the Frobenius element at v, has smooth reduction. Proof. Write Pϕ0 = Pz ∈ Eµ (Lv ) for the point corresponding to Pϕ . The local symbol {ϕ}v = 0 if and only if z can be chosen in OL∗ v , or equivalently, the field √ √ √ extension Qv ( z)/Qv is unramified and [v] z · z = u × Qm v where u is a local unit and m ∈ Z. 4. The Potential Good Additive Places Suppose v is an odd place at which E has additive reduction so that its πv twist Eπv , for some prime element πv , has good reduction. Since, for every u ∈ Z∗v , Euπv = (Eπv )u also has good reduction, we see that if Lv /Qv is a ramified √ quadratic extension with Lv = Qv ( µ), then Eµ has good reduction. In this case, the equalities (40), (41), (42) also hold, and hence the group E(Qv )/ NLv /Qv (E(Lv )) is dual to the cohomology group H1 (Gal(Lv /Qv ), Eµ (Lv )). We claim that in this case, E(Qv )/ NLv /Qv (E(Lv )) ' E[2]..

(32) 32. IV. THE LOCAL SYMBOLS. bµ (OLv ) is divisible by 2, and from the exact Since v is odd, the formal group E sequence bµ (OLv ) −→ Eµ (Lv ) −→ E¯µ (Fv ) −→ 0 0 −→ E induced from the reduction map, where Fv is the residue field, we deduce that H1 (Gal(Lv /Qv ), Eµ (Lv )) ' H1 (Gal(Lv /Qv ), E¯µ (Fv )). The Galois group Gal(Lv /Qv ) acts trivially on Fv , as Lv /Qv is totally ramified. Therefore, H1 (Gal(Lv /Qv ), E¯µ (Fv )) is isomorphic to E[2] and the claim is proved. Then we make another claim that E(Qv )/2E(Qv ) ' E[2], too. To see this, we apply the exact sequence ([Sil86], VII, §2) b v ) −→ E0 (Qv ) −→ E¯ns (Fv ) −→ 0, 0 −→ E(Z where E¯ns (Fv ) ' Fv is the subgroup consisting of non-singular points. It implies that E0 (Qv ) is 2-divisible. On the other hand, the Kodaira and N´eron’s theorem (see [Sil86],p. 183 Theorem 6.1) implies that E(Qv )/E0 (Qv ) is a finite group of order at most 4. Since E[2] is a subset of E(Qv ), we have E0 (Qv ) = 2E(Qv ) and the claim is proved. The above two claims together imply E(Qv )/2E(Qv ) ' E(Qv )/ NLv /Qv (E(Lv )), if Lv /Qv is ramified. Suppose Lv /Qv is the unramified quadratic extension. Then over Lv , our elliptic curve E also has additive reduction, and an argument similar to the above one shows that E0 (Lv ) is two divisible and E(Lv )/2E(Lv ) ' E[2]. Then we also have E(Qv )/ NLv /Qv (E(Lv )) = E(Qv )/2E(Qv ). We summarize as follow. Lemma IV.5. If Lv /Qv is a quadratic extension, then NLv /Qv (E(Lv )) = E0 (Qv ) = 2E(Qv ) and the composition E[2] −→ E(Qv ) −→ E(Qv )/ NLv /Qv (E(Lv )) is an isomorphism. We let fv = E[2] and define Φv so that for the class [P ] ∈ E(Qv )/2E(Qv ), ¯ Φv ([P ]) = A ∈ E[2], if the reduction P − A is a smooth point on E. Thus, for ϕ ∈ Sel2 (E/Q), the local symbol {ϕ}v = 0 if and only if Pϕv ∈ 2E(Qv ). Hence, we have the following: Lemma IV.6. Suppose v is an odd place at which E has potential good additive reduction. An element ϕ ∈ Sel2 (E/Q) has local symbol {ϕ}v = 0 if and only if ϕv = 0. 5. The map s For simplicity, denote T00 = T0 \ {2}..

(33) 6. THE d-TWIST OF E FOR d = ±p OR d = ±pq. 33. In the previous sections, we have defined, for each v ∈ T00 , the groups Nv ⊂ E(Qv ), fv ' E(Qv )/Nv and the local symbol {ϕ}υ . Now, we set M f= fv v∈T00. and define s : Sel2 (E/Q) −→ f, by putting s(ϕ) = ({ϕ}υ )v∈T00 ∈ f for each ϕ ∈ Sel2 (E/Q). Lemma IV.1, Lemma IV.2, Lemma IV.4, and Lemma IV.6 together imply the following: Lemma IV.7. If ϕ ∈ Sel2 (E/Q) with {ϕ}v0 = 0 for some v0 ∈ T00 , then ϕ is unramified at v0 . √ Corollary IV.1. The kernel of s is contained in Hom(Gal(Q( 2)/Q), E[2]). 6. The d-Twist of E for d = ±p or d = ±pq In this section, we assume that the integers a and b in the defining equation (1) have at most one odd common prime factor. Let q denote such prime, if it exists. We shall consider the d-twist of E with either d = ±p or d = ±pq for some p 6∈ T0 . Note that if q exists, then E has potential good additive reduction at q and we are in the situation of Section IV.4. In particular, the map that sends each P ∈ E[2] to the local symbol {ϕP }q is an isomorphism E[2] ' E(Qq )/2E(Qq ). Write Sel2 (E/Q)◦ = {ϕ ∈ Sel2 (E/Q) | {ϕ}q = 0} and ◦ T00 = T00 \ {q}. ◦ If q does not exist, we denote Sel2 (E/Q)◦ = Sel2 (E/Q) and T00 = T00 . In either ◦ case, let s denote the composition Y s Sel2 (E/Q)◦ −→ f −→ f◦ := fv ◦ v∈T00. where the last arrow is the projection. ◦ Assume that s is injective. Then so is s◦ . For each subset Θ ⊂ T00 , set Y fΘ := fv ⊂ f◦ . v∈Θ ◦ By the Gaussian elimination, we can find a subset Θ0 ⊂ T00 so that fΘ0 , considered ◦ as a subspace of f , forms a direct summand of the image Im(s◦ ). Namely, we have. Im(s◦ ) ∩ fΘ0 = {0}, and Im(s◦ ) + fΘ0 = f◦ .. (44) √ Let L = Q( d). In this case, we have T = T0 ∪ {p}. Suppose d is chosen so that Θ0 ∪ {2} = {v ∈ T | L/Q splits at v}.. (45).

(34) 34. IV. THE LOCAL SYMBOLS. For each v ∈ T0 \ (Θ0 ∪ {2}), we recall the notation in the previous sections and put Lv = Lv . Then NLv /Qv (E(Lv )) = Nv . For v ∈ Θ0 ∪ {2}, we have NLv /Qv (E(Lv )) = E(Qv ). Thus, the condition (44) actually implies that the equality (19) holds so that Im(RT0 ) + (NL/Q (ET0 (L))/2ET0 (Q)) = ET0 (Q)/2ET0 (Q). By Lemma I.1, the extension L/Q is independent and tL = 2, and Theorem III.1 says that Ed has rank zero over Q. For v ∈ T00 , x ∈ Z, let ( xv ) denote the Lengendre symbol in the sense that if v = ∞, then ( xv ) = 1 for x > 0 and ( xv ) = −1 for x < 0. Then the condition (45) is equivalent to d d ◦ \ Θ0 . (46) = 1, for v ∈ Θ0 , and = −1, for v ∈ T00 d ≡ 1 (mod 8), v v d ) determines the sign of d, In the above condition, the value of the symbol ( ∞ and others give congruent conditions for p. Obviously, the density of positive prime numbers p satisfying these congruent conditions is ( 12 )m(E)+1 . Assume that s is injective but not surjective. Then our Θ0 is non-empty, and we choose a subset Θ1 ⊂ Θ√ 0 so that |Θ1 | = |Θ0 | − 1. An argument similar to the above shows that if L = Q( d) with d satisfying the condition d d ◦ d ≡ 1 (mod 8), = 1, for v ∈ Θ1 , and = −1, for v ∈ T00 \ Θ1 , (47) v v. then L/Q is an independent extension and we have tL = 3. Hence, by Theorem III.1, if the Parity Conjecture holds for Ed over Q, then then rk(Ed (Q)) = 1. Again, the condition (47) not only determines the sign of d but also defines a set of congruent conditions for positive prime numbers of density ( 12 )m(E)+1 . We summarize the above discussion as follows. Proposition IV.1. Suppose s is injective and Θ0 is chosen so that (44) holds. Then for d = ±p or d = ±pq satisfying the condition (46), the d-twist Ed has zero rank over Q. Furthermore, if s is not surjective and Θ1 is a subset of Θ0 with cardinality less by one, then for d = ±p or d = ±pq satisfying the condition (47), the d-twist Ed should have rank one over Q unless the Parity Conjecture fails to hold for Ed over Q. In both cases, the set of prime number p satisfying the corresponding conditions has density ( 12 )m(E)+1 . 7. An Example Let the elliptic curve E/Q given by the Weierstrass equation E : y 2 = x(x − 15)(x − 77). Then T0 = {∞, 2, 3, 5, 7, 11, 31} E[2] = {O, P0 = (0, 0), P15 = (15, 0), P77 = (77, 0)}. For finite place v, let pv be the corresponding prime number. For infinite place √ (2) ∞, let p∞ = −1. The field QT0 defined in the Section II.4 is generated by pv.

(35) 7. AN EXAMPLE. 35 (2). (2). where v ∈ T0 . Let GQ,T0 be the Galois group of QT0 over Q. For v ∈ T0 we choose (2) σv ∈ GQ,T0 such that √ σv ( p) =. . √ − p √ p. , if p = pv , if p 6= pv .. (2). Then GQ,T0 is generated by {σv | v ∈ T0 }. As in (26), we choose the set AT0 := {ρ(√pv ,P0 ) , ρ(√pv ,P15 ) | v ∈ T0 }. Then ρ(√v,P0 ) (σp ). =. P0 O. , if p = pv , if p 6= pv .. In order to compute Sel2 (E/Q), we need to find B in (25). For each v ∈ T0 , let (2) be the maximal abelian extension of Qv with exponent 2 and Gv the Galois (2) (2) group of Qv /Qv . If v = ∞, then G∞ is generated by σ∞ . When v is odd, let τv √ (2) (2) be the non-trivial element in Gv which fixes Qv ( v). Then Gv is generated by σv (2) and τv . When v is even, let δ−1 (resp. δ3 ) be the non-trivila element in G2 fixed √ √ √ √ (2) 2 and 3 (resp. 2 and −1). Then G2 is generated by σ2 , δ−1 and δ3 . Since (2) Bv is contained in Hom(Gv , E[2]), we use the method described in the Section II.2 and list in the following tabel. (2) Qv. Table IV.1 Bv No.. σv. τv. 3.1 3.2 5.1 5.2 7.1 7.2 11.1 11.2 31.1 31.2. O P77 O P77 O P15 O P15 O P0. P77 O P77 P0 P15 P0 P15 P0 P0 P15. Table IV.2 B2 No.. σ2. δ−1. δ3. 2.1 2.2 2.3. P0 O P0. P15 P77 O P15 P77 O. Now, we use the Proposition II.2 to compute the values of the pairings and list it in the following table..

(36) 36. IV. THE LOCAL SYMBOLS. Table IV.3 2.1 2.2 2.3 3.1 3.2 5.1 5.2 7.1 7.2 11.1 11.2 31.1 31.2 ρ(√−1,P0 ) ρ(√−1,P15 ) ρ(√2,P0 ) ρ(√2,P15 ) ρ(√3,P0 ) ρ(√3,P15 ) ρ(√5,P0 ) ρ(√5,P15 ) ρ(√7,P0 ) ρ(√7,P15 ) ρ(√11,P0 ) ρ(√11,P15 ) ρ(√31,P0 ) ρ(√31,P15 ). 1 2. 0 1 2 1 2 1 2. 0 1 2. 0 0 0 1 2. 1 2 1 2. 0. 0 0. 1 2. 1 2. 0 0 0. 0 0. 0 0 0. 1 2. 1 2 1 2 1 2 1 2. 0. 0. 1 2. 1 2. 0. 1 2 1 2. 1 2. 0 1 2. 0. 0 0 0 0 1 2 1 2. 0 0 0 0 0 0 0 0. 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2. 0 0 0 0 0 0. 0 0. 1 2 1 2. 0. 0 0. 0 0 0 0 0 0. 1 2 1 2. 0 0. 1 2 1 2 1 2 1 2 1 2 1 2 1 2. 0 0 0 0. 1 2. 0 0 0 0 0 0 0 0. 0 0 0 1 2. 0 1 2. 0. 1 2. 1 2 1 2. 0 0 0 0 0. 1 2. 0 0 0 0 0 0 0 0 0 0. 1 2. 0 0 0 0 0 1 2. 0. 1 2. 0 0. 1 2 1 2. 0 0 0. 1 2. 0. 0 0 0 0 0 0 0 0 0 0 0 0 0. 0. 0 0. 1 2. 0 1 2. 0 0 0 1 2. 0 0 0 0 0 1 2 1 2 1 2. Let V = {(xϕ )ϕ∈AT0 |. X. xϕ ϕ ∈ Sel2 (E/Q)} ⊂ F14 2. ϕ∈AT0. be the solutions to the system of linear equations over F2 X < ϕv , ψv >Qv ·xϕ = 0, ψv ∈ Bv , v ∈ T0 . ϕ∈AT0. A basis for V and the corresponding element in Sel2 (E/Q) are listed in following. Table IV.4 No. ρ(√−1,P0 ) ρ(√−1,P15 ) ρ(√2,P0 ) ρ(√2,P15 ) ρ(√3,P0 ) ρ(√3,P15 ) ρ(√5,P0 ) ρ(√5,P15 ) ρ(√7,P0 ) ρ(√7,P15 ) ρ(√11,P0 ) ρ(√11,P15 ) ρ(√31,P0 ) ρ(√31,P15 ). 1 2 3 4 1 0 1 0 0 0 1 1 0 0 0 0 0 0. 1 0 1 0 0 0 0 0 0 1 0 0 0 0. 1 0 0 0 1 1 0 0 0 0 0 1 0 0. 0 0 0 0 1 1 0 0 0 0 0 0 1 0. Table IV.5 No.. 1. σ∞ σ2 σ3 σ5 σ7 σ11 σ31. P0 P0 O P77 O O O. 2. 3. 4. P0 P0 O P0 O O O P77 P77 O O O P15 O O O P15 O O O P0.

(37) 7. AN EXAMPLE. 37. Now, we need to determine the local symbols for the generators listed above. For an odd place v in T0 , the Frobenius element [v] at v is given by X [v] = σp , p where p runs through T0 with Legendre symbol vp = −1. Thus we may use the Lemma IV.1, IV.2 and IV.4 in the above sections to determine the local symbols. Table IV.6 No.. ∞. 3 5 7 11 31. 1 2 3 4. 1 1 1 0. 0 0 1 1. 1 0 0 0. 0 1 0 0. 0 0 1 0. 0 0 0 1. From the above table, we see that map s is injective but not surjective and Im(s◦ ) is a subspace of f with dimension 4. Thus the set Θ0 satisfying condition (44) contains two elements. The above table tell us that the choice of Θ0 is not unique, but an arbitrary choice of two elements of T00 might not work. If we choose Θ0 consisting of ∞ and 31, then the condition (46) tell us that S0 in Main Theorem I consists of the primes p satisfying p p = p5 = p7 = 11 3 = −1 p =1 31 p ≡ 1 mod 8, and ε0 = 1. When Θ0 consists of ∞ and 3, we may choose that S0 consists of the primes p satisfying p p p p = = = = −1 5 7 11 31 p =1 3 p ≡ 1 mod 8, and ε0 = 1. Thus the density of S0 is at least ( 12 )m(E)+1 . For the set S1 in Main Theorem I, we choose Θ1 so that Θ1 consists of ∞. Then the condition (47) tell us that S1 consists of the primes p satisfing p p p = p5 = p7 = 11 = 31 = −1 3 (48) p ≡ 1 mod 8, and ε1 = 1. The Main Theorem I implies that if p is an element in S1 so that the Parity Conjecture holds for Ep /Q, then Ep has rank 1 over Q. For example, we find three primes 17, 1097 and 4457 satisfying the condition (48) and list the curves and points with infinite order in the following. • p = 17 – E17 : y 2 = x(x − 255)(x − 1309) – (17493, 2209116) • p = 1097 – E1097 : y 2 = x(x − 16455)(x − 84469).

(38) 38. IV. THE LOCAL SYMBOLS. – ( 19874174293991588311797 , 2274823994119540883954733398081412 ) 70557598472000809 18741989309475259908083627 • p = 4457 – E4457 : y 2 = x(x − 66855)(x − 343189) – ( 50492119267754085453 , 27953176195024774075593726564 ) 146019164977921 1764472080507635918719.

(39) CHAPTER V. The Proof of the Main Theorem In this chapter, we complete the proof of our Main Theorem I. According to Proposition IV.1, if the local symbols map s is injective but not surjective, then the theorem is proved. But this condition does not always hold, as shown in the case of congruent curve. Our strategy is to find a prime number q 6∈ T0 , so that for the q-twist Eq , the associated local symbols map, s(q) : Sel2 (Eq /Q) −→ f(q) , is injective but not surjective. Then we can complete the proof of the Main Theorem I, by applying Proposition IV.1 again, taking the ±pq twist of Eq , which is the same as E±p . Thus, it is enough for us to prove the following: Proposition V.1. Let E/Q be an elliptic curve defined by the Weierstrass equation y 2 = x(x − a)(x − b), with a, b ∈ Z, 0 < a < b and (a, b) = 1 or 2. Then there exists an odd prime number q, relatively prime to ab(b − a), so that the local symbols map s(q) is injective but not surjective. For the rest of this Chapter, we assume that E is defined as in the proposition. 1. The Injectivity of s(q). √ By Corollary IV.1, the kernel of s(q) is contained in Hom(Gal(Q( 2)/Q), Eq [2]). On the other hand, since Eq has potential good additive reduction at q, an element ϕ ∈ Sel2 (Eq /Q) has local symbol {ϕ}q = 0 if and only if Pϕq ∈ 2Eq (Qq ) (see Section (ϕ) IV,4), √ or equivalently, the extension K /Q splits at q. However, the extension Q( 2)/Q only splits at odd primes that are congruent to ±1 modulo 8. Therefore, we have proved the following: Lemma V.1. If q is a prime number relatively prime to ab(b − a) with q ≡ ±3 (q). then the map s. (mod 8),. is injective. 2. The Non-surjectivity of s(q). Let q be congruent to ±3 modulo 8 so that by Lemma V.1, the map s(q) is Q (q) (q) injective. Write T00 = T00 ∪ {q} and f(q) = v∈T (q) fv . 00. Suppose that s(q) is surjective and let Λ ⊂ Sel2 (Eq /Q) denote the pre-image of (q) fq considered as a subspace of f(q) . Then by Lemma IV.5, we have Λ ' Eq [2], and 39.

(40) 40. V. THE PROOF OF THE MAIN THEOREM. hence |Λ| = 4. Lemma IV.7 implies that each element in Λ is unramified outside {2, q}, and hence √ √ Λ ⊆ Hom(Gal(Q( 2, q)/Q), Eq [2]). √ √ Suppose the extension Q( 2, q)/Q is inert at a finite place v ∈ T00 and let √ √ √ √ [v] ∈ Gal(Q( 2, q)/Q be the Frobenius element. Let K ⊂ Q( 2, q) be the fixed √ √ field of the decomposition subgroup < [v] > of Gal(Q( 2, q)/Q). Lemma V.2. Let the notation be as above. Then there exists a proper subgroup D ⊂ E[2] such that for each ϕ ∈ Λ, we have ϕ([v]) ∈ D. Proof. Since Eq has multiplicative reduction at v, by Lemma IV.2 and Lemma IV.4, for each ϕ ∈ Λ, the point ϕ([v]) has smooth reduction at v. Let D be the subgroup of Eq [2] consisting of points with smooth reduction. Then |D| = 2. Since |Λ| = 4 and |D| = 2, there exists a non-trivial element ϕ ∈ Λ such that ϕ([v]) =√O. Hence we have a non-trivial ϕ ∈ Λ ∩ Hom(Gal(K/Q), Eq [2]). Note that √ √ K = Q( 2), Q( q) or Q( 2q). ¯ we have Since E has good reduction at q, for the reduction curve E, ¯ q )| = 1 + q − α, |E(F √ for some integer α of absolute value not greater than 2 q. Let E¯ 0 /Fq denote the ¯ Then we have quadratic twist E. |E¯ 0 (Fq )| = 1 + q + α. Consider the composition jq : Hom(Gal(K/Q), Eq [2]) ,→ H1 (Q, Eq [2]) −→ H1 (Q, Eq ) −→ H1 (Qq , Eq ), where the first arrow is the natural embedding, the second is induced from the ¯ and the last is the localization map. inclusion Eq [2] ⊂ Eq (Q) Lemma V.3. Let the notation be as above. Then the following are true: √ (a) If K = Q( 2), then jq is injective. √ ¯ q )| ≡ 4 mod 8, then jq is injective. (b) If K = Q( q) and |E(F √ (c) If K = Q( 2q) and |E¯ 0 (Fq )| ≡ 4 mod 8, then jq is injective. Proof. For a local field K of residual characteristic q, the topological group Eq (K) = A × B where A is a compact q-group and B is a finite discrete group. We will simply call the Sylow √ 2-subgroup of B the 2-part of Eq (K). Suppose K = Q( 2). Then the extension K/Q is inert at q. Therefore, Eq /Kq has additive reduction and the the 2-part of Eq (Kq ) is Eq [2] (Lemma IV.5). In particular, the natural map Eq (Qq )/2Eq (Qq ) −→ Eq (Kq )/2Eq (Kq ) is an isomorphism. This implies the injectivity of jq . √ Suppose K = Q( q). Then the extension K/Q is totally ramified at q and Eq /Kq has good reduction and as groups Eq (Kq ) and E(Kq ) are isomorphic. Thus, ¯ q ) are isomorphic. The condition of the lemma the 2-parts of Eq (Kq ), E(Kq ) and E(F implies that the 2-part of Eq (Kq ) equals Eq [2]. Then we also deduce that the natural.