NNON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTICCURVES
ZHANGJIE WANG
Abstract.
We characterize quadratic twists of y = x ( x − a )( x + b ) with Mordell-Weil groups and 2-primary part of Shafarevich-Tate groups being isomorphic to (cid:0) Z / Z (cid:1) under certain conditions. We also obtain the distribution result of these elliptic curves. Keywords
Shafarevich-Tate groups, Full 2-torsion, Cassels pairing, Gauss genus the-ory, independence property, residue symbol
MSC (2010) 11G05, 11R11, 11R29, 11N99 Introduction
In our previous paper [18], we use Cassels pairing to characterize congruent ellipticcurves y = x − n x with Mordell-Weil ranks zero and 2-primary parts of Shafarevich-Tategroups being isomorphic to (cid:0) Z / Z (cid:1) provided that all prime divisors of n are congruent to1 modulo 4. On the other hand side, we use the independence property of residue symbolsto obtain corresponding distribution results in [19]. These tools play an important role inthe proof of the breakthrough of Smith [2] on the distribution of 2-Selmer groups. Thegoal of this paper is to generalize these methods to quadratic twist family of elliptic curveswith full rational 2-torsion points.Let a and b be coprime integers. Denote by E = E a,b the elliptic curve E : y = x ( x − a )( x + b ) . Then the quadratic twist family of E consists of the elliptic curves(1.1) E ( n ) : y = x ( x − a n )( x + b n ) , where n runs over all non-zero square-free integers. Note that if a = b = 1, these are thecongruent elliptic curves. To state our main theorem, we introduce some notation. Fora positive square-free integer m and a positive integer k , the 2 k -rank h k ( m ) of the idealclass group A = A m of Q ( √− m ) is defined to bedim F k − A / k A . Here the group operation of A is written additively. Theorem 1.
Let ( a, b, c ) be any positive primitive integer solution to a + b = 2 c suchthat the dimension of the -Selmer group of E is two. Denote by n a positive square-free integer such that all prime factors of n are congruent to ± modulo and quadraticresidues modulo any prime divisor p of abc . If n ≡ , then the following areequivalent: (1) E ( n ) ( Q ) and X ( E ( n ) / Q )[2 ∞ ] are isomorphic to ( Z / Z ) , (2) h ( n ) = 1 and h ( n ) = 0 . From elementary number theory, if ( a, b, c ) is any positive primitive integer solution to a + b = 2 c , then a, b and c are the absolute values of 4 k − k − , k + 4 k − k + 1 respectively, where k is an integer. For example, if k = 2 , , , , , , , · · · , a r X i v : . [ m a t h . N T ] M a r ZHANGJIE WANG then the dimensions of the 2-Selmer groups of the corresponding E a,b are two. Among allpositive integers no larger than 50, there are 19 such E a,b .To state another theorem, we have to use Gauss genus theory (refer to § § m is odd and h ( m ) = 1, then there are exactly twodivisors d and d of 2 m which correspond to the non-trivial element of 2 A m ∩ A m [2],where A m [2] denotes the ideal classes with trivial squares. Furthermore, the product ofthe odd parts of d and d is m . Theorem 2.
Let ( a, b, c ) be any positive primitive integer solution to a + b = 2 c suchthat the dimension of the -Selmer group of E is two. Assume that n is a positive square-free integer such that all prime factors of n are quadratic residues modulo p with p anyprime divisor of abc . If n is congruent to modulo , then the following are equivalent: (1) E ( n ) ( Q ) and X ( E ( n ) / Q )[2 ∞ ] are isomorphic to ( Z / Z ) , (2) h ( n ) = 1 and h ( n ) ≡ d − (mod 2) . Here d denotes the odd part of d which corresponds to the non-trivial element of A∩A [2] . Now we explain the idea of the proof of Theorem 1 and 2. Via the exact sequence(1.2) 0 → E ( n ) ( Q ) / E ( n ) ( Q ) → Sel ( E ( n ) ) → X ( E ( n ) / Q )[2] → , we derive that (1) implies s ( n ) = 2. Here s ( n ) is the pure 2-Selmer rankdim F Sel ( E ( n ) ) /E ( n ) ( Q )[2] . By Gauss genus theory, h ( n ) is closely related to the R´edei matrix R n ; we have parallelresults between s ( n ) and the generalized Monsky matrix M n by Proposition 4. Then weget that s ( n ) = 2 if and only if h ( n ) = 1. Cassels [11] introduced a skew-symmetricpairing on the pure 2-Selmer group Sel ( E ( n ) ) /E ( n ) ( Q )[2]. We can show that (1) is equiv-alent to the non-degeneracy of the Cassels pairing provided that h ( n ) = 1 . According toCassels pairing and Gauss genus theory, the non-degeneracy of the Cassels pairing underthis condition is equivalent to (2).To give the distribution result on the elliptic curves in Theorem 2, we first introducesome notation. Let a, b and c be coprime positive integers such that a + b = 2 c and thedimension of the 2-Selmer group of E is two. Let k be a fixed positive integer. We denoteby Q k ( x ) the set of positive square-free integers n = p · · · p k ≤ x satisfying • n is congruent to 1 modulo 8, and • all p l are quadratic residues modulo 4 p with p any prime divisor of abc .We define P k ( x ) to be all n ∈ Q k ( x ) such that(1.3) rank Z E ( n ) ( Q ) = 0 and X ( E ( n ) / Q )[2 ∞ ] (cid:39) (cid:0) Z / Z (cid:1) . Denote by C k ( x ) the set of positive square-free integers n ≤ x with exactly k prime factors.Then the independence property of Legendre symbols of Rhoades [14] implies C k ( x ) ∼ k − x (log log x ) k − log x . Here the symbol ” ∼ ” and many other symbols ” (cid:28) , O (cid:0) · (cid:1) , o (cid:0) · (cid:1) , Li( x )” are standard nota-tion in analytic number theory, it can be found in many references such as Iwaniec-Kowalski[13]. Let (cid:98) k (cid:99) be the maximal integer no larger than k/
2. We define (cid:8) u k : k ∈ N (cid:9) to bethe decreasing sequence (cid:110) (cid:98) k (cid:99) (cid:89) i =1 (1 − − i ) : k ∈ N (cid:111) with limit u ≈ . Theorem 3.
Let a, b and c be coprime positive integers such that a + b = 2 c and thedimension of the -Selmer group of E is two. Then for any positive integer k , P k ( x ) ∼ − kk (cid:48) − k − (cid:16) u k + (2 − − − k ) u k − (cid:17) · C k ( x ) . ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 3
Here k (cid:48) is the number of different prime factors of abc . The key ingredient of the proof of Theorem 3 is the independence property of residuesymbols (Theorem 4), which reduces counting P k ( x ) to counting certain symmetric k × k matrices over F .In the end of this introduction, we introduce the arrangement of this paper. In Section2, we introduce some preliminary results and several residue symbols. Section 3 is focusedon the matrix representation of 2-Selmer group. We prove that E ( n )tor ( Q ) is isomorphic to (cid:0) Z / Z (cid:1) is Section 4. We devote Section 5 to prove Theorem 1 and 2. We use the methodof Cremona-Odoni [10] to prove the independence property of residue symbols (Theorem4) in Section 6. In the last section, we prove the distribution result (Theorem 3).2. Preliminary section
Identification of -Selmer Group. Let E/ Q be an elliptic curve with full rational 2-torsion points defined by(2.1) E : y = ( x − a )( x − a )( x − a ) . Then we can identify (see Cassels [11]) the 2-Selmer group Sel ( E ) of E with (cid:110) Λ = ( d , d , d ) ∈ ( Q × / Q × ) (cid:12)(cid:12)(cid:12) d d d ∈ Q × , D Λ ( A ) (cid:54) = ∅ (cid:111) . Here A is the adele ring of Q and D Λ is a genus one curve defined by(2.2) H :( a − a ) t + d u − d u = 0 ,H :( a − a ) t + d u − d u = 0 ,H :( a − a ) t + d u − d u = 0 . Moreover, E ( Q ) / E ( Q ) can be embedded into Sel ( E ). If ( x, y ) (cid:54)∈ E ( Q )[2] is a rationalpoint on E , the embedding is given by ( x, y ) (cid:55)→ ( x − a , x − a , x − a ). The 2-torsionpoint ( a ,
0) corresponds to (cid:0) ( a − a )( a − a ) , a − a , a − a (cid:1) . Similar correspondences are defined for the 2-torsion points ( a ,
0) and ( a , Cassels Pairing.
For a general elliptic curve E/ Q , Cassels [11] defined a skew-symmetric bilinear pair-ing (cid:104)· , ·(cid:105) on the pure 2-Selmer group Sel (cid:48) ( E ), which is an F -vector space defined bySel ( E ) /E ( Q )[2]. We assume that E is defined by the equation (2.1), and we use the iden-tification of Sel ( E ) in § d , d , d ) be any element of Sel ( E ) and D Λ thecorresponding genus one curve associated to Λ. Since H i is locally solvable everywhere,there is a Q i ∈ H i ( Q ) by Hasse-Minkowski principle. We define L i to be a linear form inthree variables (all of t, u , u , u except u i ) such that L i = 0 defines the tangent plane of H i at Q i . Then we call L i the tangent linear form of H i at Q i . Moreover, we consider it as alinear form in u , u , u and t with the coefficient of u i being zero. As D Λ is locally solvableeverywhere, there are enough points on D Λ ( A ) such that we may choose P = ( P p ) ∈ D Λ ( A )with all (cid:81) i =1 L i ( P p ) non-vanishing. Given any Λ (cid:48) = ( d (cid:48) , d (cid:48) , d (cid:48) ) ∈ Sel ( E ), the local Cas-sels pairing (cid:104) Λ , Λ (cid:48) (cid:105) p is defined to be (cid:89) i =1 (cid:16) L i ( P p ) , d (cid:48) i (cid:17) p . Here p is any rational prime or infinity, and (cid:0) · , · (cid:1) p denotes the Hilbert symbol at Q p ( Q ∞ = R if p = ∞ ). Then the Cassels pairing (cid:104) Λ , Λ (cid:48) (cid:105) is given by (cid:89) p (cid:104) Λ , Λ (cid:48) (cid:105) p . ZHANGJIE WANG
Here p runs over all places of Q .Cassels [11] proved that this pairing is well-defined, namely it is independent of thechoices of P, Q i and the representatives of the cosets of Λ and Λ (cid:48) . Since skew-symmetryover F is also symmetry, the left kernel and the right kernel of the Cassels pairing are thesame. To show its kernel, we first introduce some notation. From the short exact sequence0 → E [2] → E [4] × → E [2] → , we can derive the long exact sequence0 → E ( Q )[2] / E ( Q )[4] → Sel ( E ) → Sel ( E ) → ImSel ( E ) → . Cassels showed that the kernel of this pairing is ImSel ( E ) /E ( Q )[2]. The following lemmashows that almost all the local Cassels pairings are trivial. Lemma 1 (Cassels [11] Lemma 7.2) . The local Cassels pairing (cid:104) Λ , Λ (cid:48) (cid:105) p = +1 if p satisfies (1) p (cid:54) = 2 , ∞ ; (2) The coefficients of H i and L i are all integral at p for ≤ i ≤ ; (3) Modulo D Λ and L i by p , they define a curve of genus over F p together withtangents to it. Residue Symbols.
In this subsection, we will introduce several residue symbols. The first residue symbolis the additive Jacobi symbol. Let d be a positive odd integer and m an integer coprimeto d . Then we define the additive Jacobi symbol (cid:2) md (cid:3) = 1 if the Jacobi symbol (cid:0) md (cid:1) = − Z [ i ], we first recall someconceptions related to Z [ i ]. A prime element λ of Z [ i ] is called Gaussian if it is not arational prime. An integer θ ∈ Z [ i ] is called primary if θ ≡ i ). In particular,any primary integer can be written uniquely as the product of primary primes. We use N to denote the norm of an element or an ideal of Z [ i ].The second residue symbol is the general Legendre symbol over Z [ i ]. Let p be a primeideal of Z [ i ] coprime to (1 + i ). The general Legendre symbol (cid:16) α p (cid:17) is the unique element of (cid:8) , ± (cid:9) such that α N p − ≡ (cid:16) α p (cid:17) (mod p ) . We refer to Page 196 of Hecke [3]. In particular,if λ is the unique primary prime in p , we put (cid:0) αλ (cid:1) = (cid:16) α p (cid:17) . If λ has a factorization (cid:81) kl =1 λ l of primary primes, then we define (cid:0) αλ (cid:1) to be (cid:81) kl =1 (cid:16) αλ l (cid:17) . The third residue symbol is the quartic residue symbol. We refer to Ireland-Rosen [13].Assume that λ is a prime element coprime to (1 + i ). For a Gaussian integer α , thequartic residue symbol (cid:0) αλ (cid:1) is defined to be the unique element of (cid:8) , ± , ± i (cid:9) such that α Nλ − ≡ (cid:0) αλ (cid:1) (mod λ ). Let λ and λ be two coprime Gaussian primes. We have thequartic reciprocity law (cid:18) λ λ (cid:19) = (cid:18) λ λ (cid:19) ( − Nλ − Nλ − . Assume that λ has a factorization (cid:81) kl =1 λ l of primary primes. We define (cid:0) αλ (cid:1) to be (cid:81) kl =1 (cid:16) αλ l (cid:17) . The last residue symbol is the rational quartic residue symbol. Let p be a rational primecongruent to 1 modulo 4. So there are exactly two primitive primes λ and ¯ λ lying above p . Here λ is the complex conjugate of λ and p = λλ . If q a rational integer such that (cid:16) qp (cid:17) = 1, then the two quartic residue symbols (cid:0) qλ (cid:1) and (cid:16) q ¯ λ (cid:17) take the same value, andwe use the symbol (cid:16) qp (cid:17) to denote any of them. Moreover, if d is a positive integer such ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 5 that all prime factors of d are congruent to 1 modulo 4, then (cid:0) qd (cid:1) is defined to be (cid:89) p | d (cid:18) qp (cid:19) v p ( d )4 provided that q is a rational integer satisfying (cid:16) qp (cid:17) = 1 for any p | d . Here v p ( d ) is the p -adic valuation of d .2.4. Gauss Genus Theory.
In this subsection, we briefly summarize Gauss genus theory. One can refer to § K be an imaginary quadratic number field with ideal class group A . We write themultiplication of ideal classes additively. Then the 2 i -rank h i ( A ) of A is defined to bedim F i − A / i A with i any positive integer. By classical Gauss genus theory, the 2-rank h ( A ) = t − t the number of different prime factors of the fundamental discriminant D of K . In fact, h ( A ) equals to the dimension of A [2], which is the set consisting of idealclasses killed by 2. In addition, A [2] is an elementary abelian 2-group generated by ( p, α )with p any prime factor of D and 2 α = D + √ D .As to the 4-rank, we can easily deduce that h ( A ) = dim F A ∩ A [2]. Therefore, thestudy of the 4-rank is reduced to that of 2
A ∩ A [2]. The key tool to study 2
A ∩ A [2] isthe R´edei matrix. They are closely tied via D ( K ) ∩ N K/ Q ( K × ). Here D ( K ) is the set ofpositive square-free divisors of D and N K/ Q is the norm map from K to Q . In fact, wehave a two to one epimorpism θ : D ( K ) ∩ N K/ Q ( K × ) −→ A ∩ A [2]with θ ( d ) = [( d, α )]. Note that D ( K ) is a group under the group operation d (cid:12) d = d d ( d ,d ) . Moreover, the kernel of θ is { , D (cid:48) } with D (cid:48) the square-free part of D .To connect D ( K ) ∩ N K/ Q ( K × ) with the R´edei matrix, we first introduce some notation.Let p , · · · , p t be the different prime divisors of D . We assume that p t = 2 if 2 | D . TheR´edei matrix R = ( r ij ) ( t − × t of K is an F matrix defined by r ii = (cid:104) D/p ∗ i p i (cid:105) and r ij = (cid:104) p j p i (cid:105) if i (cid:54) = j . Here p ∗ i = ( − pi − p i . Lemma 2.
Assume that w is a positive odd integer satisfying ( w, D ) = 1 and (cid:16) Dp (cid:17) = 1 forany prime divisor p of w . Let W = (cid:16)(cid:104) wp (cid:105) , · · · , (cid:104) wp t − (cid:105)(cid:17) T . Then we have an isomorphism (cid:8) d ∈ D ( K ) (cid:12)(cid:12) dw ∈ N K/ Q ( K × ) (cid:9) −→ (cid:8) Y ∈ F t (cid:12)(cid:12) RY = W (cid:9) , where the map is given by d (cid:55)→ Y d := (cid:0) v p ( d ) , · · · , v p t ( d ) (cid:1) T and its inverse is ( y , · · · , y t ) T (cid:55)→ (cid:81) t p y i i . Choose w = 1 in this lemma, we obtain the following isomorphism D ( K ) ∩ N K/ Q ( K × ) −→ (cid:8) Y ∈ F t (cid:12)(cid:12) RY = 0 (cid:9) . Consequently, h ( A ) = t − − rank F R .Like the 4-rank, the study of the 8-rank h ( A ) is equivalent to that of 4 A ∩ A [2].This is equivalent to determine which [ a ] ∈ A ∩ A [2] still lies in 4 A . We assume that K = Q ( √− n ), where n = p · · · p k is a positive square-free odd integer. Assume that 2 r d lies in D ( K ) ∩ N K/ Q ( K × ) such that d is a non-trivial divisor of n and r equals to 0 , dx + nd y = 2 r z has a non-trivial integer solution. ZHANGJIE WANG
Lemma 3.
Assume that n, d, r, R are as above and ( u, v, w ) is a positive primitive integersolution to (2.3). Let W = (cid:16)(cid:104) wp (cid:105) , · · · , (cid:104) wp k (cid:105)(cid:17) T . Then [ a ] ∈ A if and only if there is a Y ∈ F k +12 such that RY = W , where a = (2 r d, α ) . Analytic Results.
Given a number field F , let n and O be its degree and ring of algebraic integers re-spectively. We define ∆ and N F to be the discriminant of F and the norm from F to Q respectively. We call a non-zero element γ ∈ F totally positive if it is positive under allreal embedding provided that F has a real embedding. If F has no real embedding, allnon-zero elements of F are totally positive.Let † be an integral ideal of O . We denote by I ( † ) the group of all the fractional idealsthat are coprime to † . We use P † to denote the group consisting of the principal fractionalideals ( γ ) such that γ is totally positive and γ ≡ † ). Here the notation γ ≡ † ) denotes γ ∈ O p and γ ≡ p v p ( † ) ) for every prime ideal p | † , where O p isthe integer ring of F p .If χ is a character of I ( † ) /P † with † an integral ideal, then we view it as a character on I ( † ) call χ a character modulo † . In addition, if a fractional ideal a is not coprime to † ,we define χ ( a ) = 0. Let Λ( a ) be the Mangoldt function defined by (cid:40) log N F p if a = p m with m ≥ , . Then ψ ( x, χ ) is defined to be ψ ( x, χ ) = (cid:88) N F a ≤ x χ ( a )Λ( a ) . The following explicit formula (Proposition 1) of ψ ( x, χ ) is proved in P114 of Iwaniec-Kowalski [7]. Proposition 1. If χ is a non-principal character modulo an integral ideal † and ≤ T ≤ x , then (2.4) ψ ( x, χ ) = − (cid:88) | Im ρ |≤ T x ρ − ρ + O F (cid:16) xT − · log x · log( x n · N F † ) (cid:17) . Here ρ runs over all the zeros of L ( s, χ ) with ≤ Re ρ ≤ and O F means the impliedconstant only depends on F . Note that the first term of the formula (2.4) is not estimated. It can be estimated bythe same way as the classical case, and we omit its proof. We derive the explicit formula(2.5) ψ ( x, χ ) = − x β (cid:48) β (cid:48) + R ( x, T )with R ( x, T ) (cid:28) x · log ( xN F † ) · exp (cid:16) − c log x log | T N F †| (cid:17) + xT − log x · log (cid:12)(cid:12) x n N F † (cid:12)(cid:12) + x log x. Here c is a positive constant and the term − x β (cid:48) β (cid:48) occurs only if χ is a real character suchthat L ( s, χ ) has a zero β (cid:48) satisfying β (cid:48) > − c log N F † with c a positive constant.For further application, we introduce Siegel Theorem and Page Theorem over F . Thefollowing Proposition 2 is Siegel Theorem over F , and the references are Fogels [4, 5, 6]. Proposition 2.
Let χ be a character modulo an integral † and D = | ∆ | N F † > D > . ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 7 (i)
There is a positive constant c = c ( n ) such that in the region Re( s ) > − c log D (1 + | Im( s ) | ) > there is no zero of L ( s, χ ) if χ is complex, and for at most one real character χ (cid:48) there maybe a simple zero β (cid:48) of L ( s, χ (cid:48) ) . (ii) Let β (cid:48) be the exceptional zero of the exceptional character χ (cid:48) modulo † . Then forany (cid:15) > there exists a positive constant c = c ( n, (cid:15) ) such that − β (cid:48) > c ( n, (cid:15) ) D − (cid:15) . Proposition 3 is Page Theorem over F . One can refer to Hoffstein-Ramakrishnan [9]. Proposition 3.
For any Z ≥ and c a suitable constant, there is at most a realprimitive character χ to a modulus † with N F † ≤ Z such that L ( s, χ ) has a real zero β satisfying β > − c log Z . Representation of -Selmer Group Now we apply the result of § E a,b : y = x ( x − a )( x + b )with a and b positive odd integers. For positive square-free integer n , we consider theelliptic curve E ( n ) a,b = E ( n ) E ( n ) : y = x ( x − an )( x + bn ) . Choosing a = an, a = − bn and a = 0 in (2.2), we get the following identificationSel ( E ( n ) ) = (cid:110) Λ = ( d , d , d ) ∈ ( Q × / Q × ) (cid:12)(cid:12)(cid:12) d d d ∈ Q × , D Λ ( A ) (cid:54) = ∅ (cid:111) with D Λ the genus one curve defined by H : − bnt + d u − d u = 0 ,H : − ant + d u − d u = 0 ,H : 2 cnt + d u − d u = 0 . Here a + b = 2 c . If ( x, y ) ∈ E ( n ) ( Q ) is not a 2-torsion point, it corresponds to ( x − an, x + bn, x ). Moreover, the four elements ( an, , ( − bn, , (0 ,
0) and O of E ( n ) ( Q )[2] correspondto (cid:0) ac, cn, an (cid:1) , (cid:0) − cn, bc, − bn (cid:1) , ( − an, bn, − ab ) and (1 , ,
1) respectively.In this section, we always assume that a, b and c are coprime positive odd integers suchthat(3.1) a + b = 2 c. Local Solvability Conditions on D Λ . We denote by a = a a , b = b b and c = c c with a , b and c square-free integers. Lemma 4.
Assume that n is a positive square-free integer coprime to abc . Let Λ =( d , d , d ) with d i non-zero square-free integers such that d d d is a square. (1) If p (cid:45) abcn , then D Λ ( Q p ) (cid:54) = ∅ if and only if p (cid:45) d d d . (2) D Λ ( R ) is non-empty if and only if d > . (3) If D Λ ( Q ) is non-empty, then d and d have the same parity. (4) If d and d are odd, then D Λ ( Q ) is non-empty if and only if either | d − d , | d − d or | d + an, | d − d + 2 cn . ZHANGJIE WANG
Proof. (1) Let p be a prime such that p (cid:45) abcn . If p | d d d , then p divides exactly twoof d , d and d . We assume that p | d and p | d . Since we are dealing with homogeneousequations, we may assume that u , u , u and t are p -adic integers and at least one of themis a p -adic unit. By comparing p -adic valuations of both sides of H , we get that p | t .Similarly, from H we derive that p | u . Then via H and H we infer that p | u and p | u . So p | ( t, u , u , u ), which is impossible. Thus D Λ ( Q p ) is empty provided that p | d d d .Now we assume that p (cid:45) d d d . We will use Weil conjecture for curve (see Silverman[12] P134) to show that D Λ ( Q p ) is non-empty. Note that D Λ modulo p gives rise to asmooth projective curve over F p by p (cid:45) nabcd d d . Weil conjecture implies that Z ( D Λ , T ) = P ( T )(1 − T )(1 − pT )with P ( T ) ∈ Z [ T ] factoring as P ( T ) = (1 − αT )(1 − αT )over C . Here α has norm p and Z ( D Λ , T ) is the zeta function of D Λ over F p given by Z ( D Λ , T ) = exp (cid:16) ∞ (cid:88) m =1 D Λ ( F p m ) T m m (cid:17) , where D Λ ( F p m ) denotes the number of points of D Λ over F p m . Therefore, ∞ (cid:88) m =1 D Λ ( F p m ) T m m = log(1 − αT ) + log(1 − αT ) − log(1 − T ) − log(1 − pT ) . Comparing the coefficients of T in above equation, we have D Λ ( F p ) = 1 + p − ( α + α ) ≥ p − √ p > . Therefore, D Λ ( Q p ) is non-empty by Hensel’s Lemma.(2) If d <
0, we see that H ( R ) is not solvable for the coefficients on its left hand sideare all negative. Conversely, if d >
0, then D Λ ( R ) is obviously solvable.(3) Now we assume that D Λ ( Q ) is non-empty. Assume that d and d have differentparity. We first treat the case that d is even and d is odd. Then d is even by d d d ∈ Q × . Considering the 2-adic valuations of both sides of H and H , we get that t and u are even. From similar considerations on H and H , we derive that u and u are alsoeven. So u , u , u and t are even, which is impossible. Similar arguments shows that thecase that d is odd and d is even is also impossible.(4) Let d and d be odd integers. First, we assume that D Λ ( Q ) is non-empty. Since weare dealing with homogeneous equations, we may assume that u , u , u and t are 2-adicintegers and at least one of them is odd. Viewing H as a congruence modulo 4, we seethat u and u are odd. From H we infer that exactly one of t and u is even. Now wedivide this into two subcases according to the parity of t .Case (i): t is odd. Then u is even. Considering H and H as congruences modulo 4 and8 respectively, we get that 4 | d + an and 8 | d − d + 2 cn .Case (ii): t is even. Then u is odd. Viewing H and H as congruences modulo 4 and 8respectively, we obtain that 4 | d − d and 8 | d − d .Finally, we assume that either 4 | d − d , | d − d or 4 | d + an, | d − d + 2 cn .According to these conditions, we divide it into two subcases.Case (iii): 4 | d − d and 8 | d − d . We have d ≡ d d ≡ | d − d ,we get d ≡ d ≡ d i has square root in Z and we choose t = 0 and u i = d − i for 1 ≤ i ≤
3. If d ≡ d + 4 an ≡ ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 9 d + 8 cn ≡ d (mod 8); so ( d + 8 cn ) d − and ( d + 4 an ) d − have square roots in Z ; inaddition, we choose t = 2 , u = 1 ,u = (cid:113) ( d + 8 cn ) d − and u = (cid:113) ( d + 4 an ) d − . Case (iv): 4 | d + an and 8 | d − d + 2 cn . If 8 | d + an , then we choose t = 1 and u = 0;so − and − and ( − an + 2 cn ) d − are congruent to 1 modulo 8, and they have square rootsin Z ; let u and u be any square roots of − and − and ( − an + 2 cn ) d − respectively. If4 (cid:107) d + an , we choose t = 1 and u = 2; so (4 − an ) d − and (2 cn + 4 − an ) d − arecongruent to 1 modulo 8, and they have square roots in Z ; we define u and u to be anysquare roots of (4 − an ) d − and (2 cn + 4 − an ) d − respectively. Therefore, in any case wehave D Λ ( Q ) (cid:54) = ∅ .This completes the proof of the lemma. (cid:3) Assume that n is a positive square-free integer coprime to 2 abc . By Lemma 4, any ele-ment of Sel ( E ( n ) ) / E ( n ) ( Q )[2] has a unique representative Λ = ( d , d , d ) with d i positivesquare-free integers satisfying(3.2) d d d ∈ Q × , d > , d | nabc and d | nabc. Lemma 5.
Assume that n is a positive square-free integer coprime to abc and Λ =( d , d , d ) with d i positive square-free integers such that (3.2) holds. Let p be a primedivisor of n . • If p (cid:45) d and p (cid:45) d , then D Λ ( Q p ) (cid:54) = ∅ if and only if (cid:16) d p (cid:17) = (cid:16) d p (cid:17) = 1 . • If p (cid:45) d and p | d , then D Λ ( Q p ) is non-empty if and only if (cid:16) d p (cid:17) = (cid:16) acp (cid:17) and (cid:16) N/d p (cid:17) = (cid:16) abp (cid:17) , where N = abcn . • If p | d and p (cid:45) d , then D Λ ( Q p ) is non-empty if and only if (cid:16) N/d p (cid:17) = (cid:16) − abp (cid:17) and (cid:16) d p (cid:17) = (cid:16) bcp (cid:17) . • If p | d and p | d , then D Λ ( Q p ) is non-empty if and only if (cid:16) N/d p (cid:17) = (cid:16) − bcp (cid:17) and (cid:16) N/d p (cid:17) = (cid:16) acp (cid:17) .Proof. Assume that p (cid:45) d d . So p (cid:45) d . If D Λ ( Q p ) is non-empty, then by H and H weget (cid:16) d d p (cid:17) = (cid:16) d d p (cid:17) = 1, namely (cid:16) d p (cid:17) = (cid:16) d p (cid:17) = 1. Conversely, if (cid:16) d p (cid:17) = (cid:16) d p (cid:17) = 1,then we set t = 0 and u i = d − i for 1 ≤ i ≤
3. So ( t, u , u , u ) lies in D Λ ( Q p ). Theremained cases can be proved similarly. This completes the proof of the lemma. (cid:3) Lemma 6.
Assume that n is a positive square-free integer coprime to abc and Λ =( d , d , d ) with d i positive square-free integers such that (3.2) holds. • Let p be a prime divisor of a . Then D Λ ( Q p ) (cid:54) = ∅ implies that p (cid:45) d . Under thecondition p (cid:45) d , we have – if p (cid:45) d , then D Λ ( Q p ) is non-empty if and only if (cid:16) d p (cid:17) = 1 ; – if p | d , then D Λ ( Q p ) is non-empty if and only if (cid:16) bnd p (cid:17) = 1 provided that p | a with p (cid:45) a and (cid:16) d p (cid:17) = (cid:16) bnp (cid:17) = 1 provided that p | a . • Let p be a prime divisor of b . Then D Λ ( Q p ) (cid:54) = ∅ implies that p (cid:45) d . Under thecondition p (cid:45) d , we have – if p (cid:45) d , then D Λ ( Q p ) (cid:54) = ∅ if and only if (cid:16) d p (cid:17) = 1 ; – if p | d , then D Λ ( Q p ) is non-empty if and only if (cid:16) − and p (cid:17) = 1 provided that p | b with p (cid:45) b and (cid:16) − anp (cid:17) = (cid:16) d p (cid:17) = 1 provided that p | b . • Let p be a prime divisor of c . Then D Λ ( Q p ) (cid:54) = ∅ implies that p (cid:45) d . Under thecondition p (cid:45) d , we have – if p (cid:45) d d , then D Λ ( Q p ) is non-empty if and only if (cid:16) d p (cid:17) = 1 ; – if p | d and p | d , then D Λ ( Q p ) is non-empty if and only if (cid:16) and p (cid:17) = 1 provided that p | c with p (cid:45) c and (cid:16) anp (cid:17) = (cid:16) d p (cid:17) = 1 provided that p | c .Proof. Let p be a prime divisor of a . Assume that D Λ ( Q p ) is non-empty. If p | d , then p divides exactly one of d and d . We may assume that p | d . So p (cid:45) d . Via H and H , we obtain that p | u and p | t . Then H and H imply that p | u and p | u . So p | ( t, u , u , u ), which is impossible. Therefore, D Λ ( Q p ) (cid:54) = ∅ implies that p (cid:45) d .Now we assume that p (cid:45) d d d . If D Λ ( Q p ) (cid:54) = ∅ , then H implies that (cid:16) d d p (cid:17) = 1,namely (cid:16) d p (cid:17) = 1. Conversely, if (cid:16) d p (cid:17) = 1, then we choose u = d ( d ,d ) . We have u = d + and − t = d + O (cid:0) p (cid:1) ,u = d + 2 cnd − t , where O (cid:0) p (cid:1) is wp with w some p -adic integer. The first equation is solvable for anygiven t by (cid:16) d p (cid:17) = 1. Denote by R the quadratic residue classes modulo p . Then R and d + 2 cnd − R have non-empty intersections for their cardinalities are p +12 . Thus wecan choose t such that d + 2 cnd − t is a quadratic residue. So D Λ ( Q p ) is non-empty byHensel’s Lemma.Now we consider the case p | d and p | d . Then p (cid:45) d . First, we treat the subcase p | a . If D Λ ( Q p ) is non-empty, then (cid:16) d bnp (cid:17) = 1 by H . In addition, dividing H by p we derive that (cid:16) d /d p (cid:17) = 1, namely (cid:16) d p (cid:17) = 1. Conversely, if (cid:16) d p (cid:17) = (cid:16) bnp (cid:17) = 1, then wechoose u = d ( d ,d ) . We get u = d + a d − · a nt = d + O (cid:0) p (cid:1) ,u = d + 2 cnd − t = bnd − t + O (cid:0) p (cid:1) . These equations are solvable for (cid:16) d p (cid:17) = (cid:16) bnp (cid:17) = 1. So D Λ ( Q p ) is non-empty. Finally, wetreat the subcase p | a with p (cid:45) a . If D Λ ( Q p ) is non-empty, we get (cid:16) d bnp (cid:17) = 1 by H .Conversely, if (cid:16) d bnp (cid:17) = 1, then we choose u = d ( d ,d ) . We obtain u = d + a d − · a nt ,u = d + 2 cnd − t = bnd − t + O (cid:0) p (cid:1) . The second equation is solvable for any t by (cid:16) d bnp (cid:17) = 1. Like the case p (cid:45) d d d , we knowthat there is a t such that d + a d − a nt is a quadratic residue modulo p . So D Λ ( Q p ) isnon-empty. Hence, we finish the proof of the case p | a .For the case p | bc , we can prove similarly. This completes the proof of the lemma. (cid:3) Lemma 7.
Assume that n is a positive square-free integer coprime to abc and Λ =( d , d , d ) with d i positive square-free odd integers such that (3.2) holds. If D Λ ( Q p ) isnon-empty for all odd primes p and p = ∞ , then D Λ ( Q ) is also non-empty. ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 11
Proof.
We only prove the case that a, b and c are squares, the general case can be provedsimilarly but the process is much more complicate. Now we assume that a, b and c aresquares. Define A = ( a, d ). Then Lemma 6 implies that (cid:0) bnA (cid:1) = (cid:16) d A (cid:17) = 1. In addition, (cid:0) cnA (cid:1) = 1 for a + b = 2 c . Since b and c are squares, we have(3.3) (cid:16) nA (cid:17) = 1 , (cid:18) d A (cid:19) = 1 , (cid:18) A (cid:19) = 1 . Denote by B = ( b, d ) and C = ( c, d ) = ( c, d ). Similarly we have(3.4) (cid:18) − nB (cid:19) = 1 , (cid:18) d B (cid:19) = 1 , (cid:18) B (cid:19) = 1and(3.5) (cid:16) nC (cid:17) = 1 , (cid:18) d C (cid:19) = 1 , (cid:18) − C (cid:19) = 1 . Put n = n n n n with ( d , n ) = n n , ( d , n ) = n n and n = ( d , d , n ). Then we getthat d = ACn n , d = BCn n and d = ABn n . By Lemma 5, we have (cid:16) ACn (cid:17) = (cid:16) n n n (cid:17) , (cid:16) BCn (cid:17) = (cid:16) n n n (cid:17) , (cid:16) ACn (cid:17) = (cid:16) n n n (cid:17) , (cid:16) BCn (cid:17) = (cid:16) n n n (cid:17) , ( ∗ ) (cid:16) ACn (cid:17) = (cid:16) − n n n (cid:17) , (cid:16) BCn (cid:17) = (cid:16) n n n (cid:17) , (cid:16) ACn (cid:17) = (cid:16) − n n n (cid:17) , (cid:16) BCn (cid:17) = (cid:16) n n n (cid:17) . From the second identity of equation (3.3) we have (cid:104) n n A (cid:105) = (cid:20) BCA (cid:21) = (cid:20) AB (cid:21) + (cid:20) AC (cid:21) + (cid:20) − A (cid:21)(cid:20) − B (cid:21) by the quadratic reciprocity law and (cid:0) − C (cid:1) = 1 of (3.5). Via the second identities of (3.4)and (3.5), we get (cid:104) n n A (cid:105) = (cid:104) n n B (cid:105) + (cid:104) n n C (cid:105) + (cid:20) − A (cid:21)(cid:20) − B (cid:21) , = (cid:20) Bn n (cid:21) + (cid:20) Cn n (cid:21) + (cid:20) − A (cid:21)(cid:20) − B (cid:21) + (cid:20) − B (cid:21)(cid:20) − n n (cid:21) . Here we have used (cid:0) − C (cid:1) = 1 and the quadratic reciprocity law. The third, sixth, seventhand the last identities of (*) imply that (cid:104) n n A (cid:105) = (cid:20) − n (cid:21) + (cid:20) n n (cid:21) + (cid:20) An n (cid:21) + (cid:20) n n n (cid:21) + (cid:20) n n n (cid:21) + (cid:20) n n n (cid:21) + (cid:20) − A (cid:21)(cid:20) − B (cid:21) + (cid:20) − B (cid:21)(cid:20) − n n (cid:21) . By the quadratic reciprocity law we get (cid:20) n n (cid:21) + (cid:20) − n (cid:21) = (cid:20) − A (cid:21)(cid:20) − B (cid:21) + (cid:20) − A (cid:21)(cid:20) − n n (cid:21) + (cid:20) − B (cid:21)(cid:20) − n n (cid:21) + (cid:20) − n (cid:21)(cid:20) − n (cid:21) + (cid:20) − n (cid:21)(cid:20) − n (cid:21) + (cid:20) − n (cid:21)(cid:20) − n (cid:21) . Expanding the residue symbols (cid:104) − n i n (cid:105) = (cid:104) − n i (cid:105) + (cid:104) − n (cid:105) for i = 2 and 3, we have(3.6) (cid:20) n n (cid:21) + (cid:20) − n (cid:21) = (cid:20) − An (cid:21)(cid:20) − Bn (cid:21) + (cid:20) − n (cid:21)(cid:20) − ABn n (cid:21) . Similarly, starting from the first identity of (*) we derive(3.7) (cid:20) − n (cid:21) = (cid:20) − An (cid:21)(cid:20) − Bn (cid:21) + (cid:20) − n (cid:21)(cid:20) − ABn n (cid:21) , and starting from the second identity of (*) we obtain(3.8) (cid:20) − Bn (cid:21) = (cid:20) − An (cid:21)(cid:20) − Bn (cid:21) + (cid:20) − n (cid:21)(cid:20) − ABn n (cid:21) . Now we use the identities (3.6), (3.7) and (3.8) to prove the lemma. By Lemma 4, itsuffices to show that either 8 | d − d , | d − d or 8 | d − d + 2 cn, | d + an . Notethat d − d = Cn ( An − Bn ) and d − d + 2 cn ≡ Cn ( An − Bn + 2 n n n ) (mod 8).Moreover, d − d ≡ An ( n − Bn ) (mod 4) and d + an ≡ n n ( A + n n ) (mod 4). Sowe reduce to showing that either 8 | An − Bn , | n − Bn or 8 | An − Bn +2 n n n , | A + n n .Since A, B, n and n are odd, we know that either 4 | An − Bn or 4 | An − Bn +2 n n n . First, we consider the case 4 | An − Bn . Then(3.9) (cid:20) − An (cid:21) = (cid:20) − Bn (cid:21) . Substituting this into (3.6) we get (cid:20) n n (cid:21) = (cid:20) − An n (cid:21) = (cid:20) − Bn n (cid:21) . Moreover, substituting (3.9) into (3.7) and (3.8) we have0 = (cid:20) − An n (cid:21)(cid:20) − Bn n (cid:21) = (cid:20) − Bn n (cid:21)(cid:20) − Bn n (cid:21) , (cid:20) − ABn n (cid:21)(cid:20) − Bn n (cid:21) = (cid:20) − n (cid:21)(cid:20) − Bn n (cid:21) . Adding these we deduce that (cid:104) n n (cid:105) = (cid:104) − Bn n (cid:105) = 0. From the third identities of (3.3)and (3.4) we see that (cid:16) An (cid:17) = (cid:16) Bn (cid:17) . Thus 8 | An − Bn . Since (cid:104) − Bn n (cid:105) = 0, we have (cid:16) − n (cid:17) = (cid:16) − Bn (cid:17) . So 4 | n − Bn .Finally, we consider the case 4 | An − Bn + 2 n n n . Then we have(3.10) (cid:20) − ABn n (cid:21) = 1 . Substituting this into (3.6) we obtain (cid:104) n n (cid:105) = 0 . From (3.7) we derive that (cid:20) − ABn n (cid:21)(cid:20) − Bn n (cid:21) = 0 . Via (3.8) we obtain (cid:20) − Bn n (cid:21) = (cid:20) − ABn n (cid:21)(cid:20) − Bn n (cid:21) . Adding these two equations we get (cid:104) − Bn n (cid:105) = 0 by noting that (cid:104) − ABn n (cid:105) = 1 + (cid:2) − n (cid:3) . Toprove 8 | An − Bn + 2 n n n , it suffices to show that (cid:104) An +2 n n n (cid:105) = (cid:104) Bn (cid:105) . By thesupplementary law of the quadratic reciprocity law, we have (cid:20) An + 2 n n n (cid:21) = (cid:20) An (cid:21) + (cid:20) − An n (cid:21) + 1 = (cid:20) An (cid:21) + (cid:20) − Bn n (cid:21) = (cid:20) An (cid:21) . ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 13
Noting that (cid:104) n n (cid:105) = 0 and (cid:2) A (cid:3) = (cid:2) B (cid:3) = 0, we have 8 | An − Bn + 2 n n n . Via (cid:104) − Bn n (cid:105) = 0 and (3.10), we have (cid:104) − An n (cid:105) = 1, namely 4 | A + n n .This completes the proof of the lemma. (cid:3) Matrix Representation of -Selmer Group. From Lemma 4, 5, 6 and 7, there is a matrix representation of the pure 2-Selmer groupSel (cid:48) ( E ( n ) ) = Sel ( E ( n ) ) / E ( n ) ( Q )[2]. For our purpose, we only give this matrix representa-tion under the condition that a, b, c are squares and n is a positive square-free odd integersuch that all prime factors p of n satisfy(3.11) (cid:18) pq (cid:19) = 1with q any prime divisor of abc .To give the matrix representation of Sel (cid:48) ( E ( n ) ), we first introduce some notation. Denoteby n = p · · · p k . Assume that a , b and c have the prime decompositions(3.12) q t · · · q t k k , q t k k +1 · · · q t k k and q t k k +1 · · · q t k k respectively. Here k ≤ k ≤ k are non-negative integers, and all the t i are positive evenintegers. Note that every element of Sel (cid:48) ( E ( n ) ) has a unique representative ( d , d , d )with d i positive integers satisfying (3.2). Since d | nac and d | nbc , we put d = p x · · · p x k k q z · · · q z k k q z k k +1 · · · q z k k and d = p y · · · p y k k q w k k +1 · · · q w k k . Here x, y, z and w are row vectors over F given by ( x , · · · , x k ), ( y , · · · , y k ), ( z , · · · , z k , z k +1 , · · · , z k )and ( w k +1 , · · · , w k ) respectively.First, we define the matrix M which gives rise to the matrix representation of Sel (cid:48) ( E ).To this purpose, we first introduce some F matrices. Let F = ( f ij ) k × k be the matrixdefined by f ii = 0 and f ij = (cid:104) q j q i (cid:105) if i (cid:54) = j . We write F in the following block matrix form F = F F F F F F F F F . Here F and F have sizes k × k and ( k − k ) × ( k − k ) respectively. Let ∆ and ∆ (cid:48) bethe diagonal matrices given by∆ = diag (cid:0) , · · · , (cid:1) , ∆ (cid:48) = diag (cid:18)(cid:20) − q k +1 (cid:21) , · · · , (cid:20) − q k (cid:21)(cid:19) , where the size of ∆ is ( k − k ) × ( k − k ).Now we define M to be the (2 k − k ) × (2 k − k ) matrix F F F F F F ∆ (cid:48) ∆ ∆ . By Lemma 4, 6 and 7, we know that the map ( d , d , d ) (cid:55)→ ( z, w ) induces an isomorphismSel (cid:48) ( E ) −→ (cid:26) ( z, w ) : M (cid:18) z T w T (cid:19) = 0 , z ∈ F k + k − k , w ∈ F k − k (cid:27) . In fact, this can be verified by block matrices. Taking the first block row of M (cid:18) z T w T (cid:19) =0, we get (cid:0) F F (cid:1) w T = 0. This is (cid:80) k k +1 w j (cid:104) q j q i (cid:105) = 1 for all i ≤ k . From this we obtain that (cid:16) d q i (cid:17) = 1 for all i ≤ k , which is compatible with the case of q | a in Lemma 6. Theremaining block rows can be checked similarly. We have to remak on the last block row of M (cid:18) z T w T (cid:19) = 0. From this we obtain that ∆( z (cid:48) T + w (cid:48) T ) = 0 with z (cid:48) = ( z k +1 , · · · , z k )and w (cid:48) = ( w k +1 , · · · , w k ). Then z i = w i for i > k , which is equivalent to ( c, d ) = 1.Now we use M to give the matrix representation of Sel (cid:48) ( E ( n ) ). We first introduce somenotation. Let A = A n = ( a ij ) k × k be the F matrix given by a ii = (cid:80) l (cid:54) = i a il and a ij = (cid:104) p j p i (cid:105) if i (cid:54) = j . Denote by G = ( g ij ) k × k the F matrix defined by g ij = (cid:104) q j p i (cid:105) . We write G in theblock matrix form G = (cid:0) G G G (cid:1) with the sizes of G and G being k × k and k × ( k − k ) respectively. Let M n be theMomsky matrix (see Appendix of Heath-Brown [15]) (cid:18) A + D − D D A + D (cid:19) , where D u = diag (cid:16)(cid:104) up (cid:105) , · · · , (cid:104) up k (cid:105)(cid:17) . We define M n to be the F matrix (cid:18) M n G M (cid:19) with G = (cid:18) G G G G (cid:19) . Proposition 4.
Let a, b, c be odd squares with factorization (3.12) and n a positive square-free integer satisfying (3.11). Then the following is an isomorphism Sel (cid:48) ( E ( n ) ) −→ ker M n , ( d , d , d ) (cid:55)→ ( x, y, z, w ) T , where ( d , d , d ) is the representative of an element of Sel (cid:48) ( E ( n ) ) such that (3.2) holds.Here x = (cid:0) v p ( d ) , · · · , v p k ( d ) (cid:1) , y = (cid:0) v p ( d ) , · · · , v p k ( d ) (cid:1) , w = (cid:0) v q k ( d ) , · · · , v q k ( d ) (cid:1) and z = (cid:0) v q ( d ) , · · · , v q k ( d ) , v q k ( d ) , · · · , v q k ( d ) (cid:1) . For any ( d , d , d ) above, we put d = d (cid:48) d (cid:48)(cid:48) with d (cid:48) = ( d , n ) and d (cid:48)(cid:48) = ( d , abc ).Similarly, we set d = d (cid:48) d (cid:48)(cid:48) and d = d (cid:48) d (cid:48)(cid:48) . Since n satisfies (3.11), those d (cid:48) , d (cid:48) , d (cid:48) and n occurred in the local solvability conditions for p | abc (Lemma 6) vanish. Note that d (cid:48)(cid:48) and d (cid:48)(cid:48) correspond to z and w respectively. Combing these, we have M (cid:18) z T w T (cid:19) = 0.From a, b and c being squares, those a, b and c occurred in the local solvability conditionsfor p | n (Lemma 5) also vanish. Note that d (cid:48) and d (cid:48) correspond to x and y respectively.Observe the identity x i (cid:88) l (cid:54) = i (cid:20) p l p i (cid:21) + (cid:88) j (cid:54) = i x j (cid:20) p j p i (cid:21) = x i (cid:20) n/d (cid:48) p i (cid:21) + (1 − x i ) (cid:20) d (cid:48) p i (cid:21) . From Lemma 5, we get x i (cid:20) n/d (cid:48) p i (cid:21) + (1 − x i ) (cid:20) d (cid:48) p i (cid:21) = (cid:20) d (cid:48)(cid:48) p i (cid:21) + x i (cid:20) − p i (cid:21) + y i (cid:20) p i (cid:21) . Similar result also holds for y . From these we can derive that M n (cid:18) x T y T (cid:19) + G (cid:18) z T w T (cid:19) = 0 . So the map is well-defined. Its injectivity is obvious; by Lemma 4, 5, 6 and 7, it issurjective.
ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 15 Torsion Subgroup
In this section, we will prove that E ( n )tor ( Q ) (cid:39) (cid:0) Z / Z (cid:1) , where E ( n ) is defined by (1.1)with ( a, b, c ) any positive primitive integer solution to a + b = 2 c . Let E be an ellipticcurve with full 2-torsion points. According to Mazur’s classification theorem on torsionsubgroup of elliptic curves over Q (see [12]), E tor ( Q ) is isomorphic to Z / Z × Z / m Z forsome m = 1 , , ,
4. Ono [17] has the following characterization of E tor ( Q ) . Lemma 8 (Ono) . Let E : y = x ( x − a )( x + b ) be an elliptic curve over Q with a, b integers.Then E [2]( Q ) (cid:39) (cid:0) Z / Z (cid:1) . (1) E tor ( Q ) contains a point of order if and only if one of the three pairs [ − a, b ] , [ a, a + b ] and [ − b, − a − b ] consists of squares of integers. (2) E tor ( Q ) has a point of order if and only if there exist a positive integer d andpairwise coprime integers u, v and w such that u + v = w and [ d u , d v ] isone of the three pairs in (1). (3) E tor ( Q ) has a point of order if and only if there exist a positive integer d andpairwise coprime integers u and v such that a = − ( u + 2 u v ) d , b = ( v +2 v u ) d and uv (cid:54)∈ (cid:110) − , − , − , , (cid:111) . Proposition 5.
For any square-free integer n , E ( n ) ( Q ) (cid:39) (cid:0) Z / Z (cid:1) .Proof. From the definition of E ( n ) , we see that E ( n )tor ( Q ) contains a subgroup isomorphicto (cid:0) Z / Z (cid:1) . The proposition will be proved by showing that E ( n )tor ( Q ) contains no point oforder 4 and 3. Here we have used Mazur’s classification theorem on torsion subgroup ofelliptic curves over Q .Now we show that E ( n )tor ( Q ) has no point of order 4. Note that none of the three pairs[ − a n, b n ] , [ a n, a n + b n = 2 c n ] and [ − b n, − a n − b n ] consists of squares of integers.So (1) of Lemma 8 implies that E ( n )tor ( Q ) contains no point of order 4.So we remain to prove that E ( n )tor ( Q ) contains no point of order 3. By (3) of Lemma 8,it suffices to show that the equation (cid:40) d u ( u + 2 v ) = − a n,d v ( v + 2 u ) = b n ( ∗ )with d a positive integer and u, v pairwise coprime integers is not solvable. We will dividethe proof of this into several steps.First, as a and b are coprime integers, we get that the greatest common divisor ( − a n, b n )of − a n and b n is | n | , which is square-free. From equation (*), we get that d | ( − a n, b n ) = | n | . So d = 1 by noting that d is a positive integer.Second, we claim that | n | = ( u + 2 u v, v + 2 v u ) = (3 , u − v ). Since ( u, v ) = 1, wehave ( u, v + 2 u ) = 1. Thus ( u , v ( v + 2 u )) = 1. Similarly, ( u + 2 v, v ) = 1. Therefore,( u + 2 u v, v + 2 v u ) equals to( u + 2 v, v ( v + 2 u )) = ( u + 2 v, v + 2 u ) = ( u + 2 v, u − v ) = ( u − v, u ) = (3 , u − v ) . So n divides u + 2 v and v + 2 u . Therefore, the equation (*) is reduced to (cid:40) v · v +2 un = b , ( ∗ − u · u +2 vn = a . ( ∗ v > v is coprime to v + 2 u , from (*1) we derive that there are integers b and b which divide b and satisfy v = b , v + 2 u = nb . The first equation implies that b = b for some b | b . Since b is odd, there are oddintegers b and b such that(4.1) v = b , v + 2 u = nb . Now we claim that u <
0. Otherwise u >
0, from v + 2 u = nb we get n > na = − u · ( u + 2 v ) < a <
0, which is impossible. Hence, we get u <
0. Like (*1), from(*2) we know that there are odd integers a and a such that u = − a , u + 2 v = na . Combing this with equation (4.1), we derive that there are odd integers a , a , b and b such that b − a = nb , b − a = na . Viewing these equations as congruences modulo 8, we see that n ≡ n ≡ − v > v <
0. This can be provedsimilarly as the case v > E ( n )tor ( Q ) contains no point of order 3. This finishes the proof of the proposi-tion. (cid:3) Non-trivial Shafarevich-Tate Group
In this section, we always assume that ( a, b, c ) is a positive primitive integer solutionto a + b = 2 c and n = p · · · p k ≡ (cid:16) p i q (cid:17) = 1 for any 1 ≤ i ≤ k and prime divisor q of abc . Note that E = E a,b = E a ,b and E ( n ) ( Q )[2] consists of(2 , n, n ) , ( − n, , − n ) , ( − n, n, −
1) and (1 , , . We assume that Sel ( E ) has dimension two.5.1. Proof of Theorem 1.
In this subsection, we always assume that all prime divisors p i of n are congruent to ± M n is of the formdiag (cid:0) A + D − , A (cid:1) . Lemma 9.
Assume that all prime divisors p i of n are congruent to ± modulo . Let x =( x , · · · , x k ) T and x = (1 , · · · , T be two column vectors in F k . Denote by d = (cid:81) ki =1 p x i i . (1) Ax = 0 if and only if x T ( A + D − ) = 0 . (2) If ( A + D − ) x = 0 , then x T A = 0 if d ≡ and ( x − x ) T A = 0 if d ≡ − . (3) The dimension of the pure -Selmer group Sel (cid:48) ( E ( n ) ) is two if and only if h ( n ) =1 . If this is satisfied, then Sel (cid:48) ( E ( n ) ) is generated by Λ = (2 , , , Λ (cid:48) = ( d, , d ) . Proof.
For notational simplicity, we assume that p ≡ · · · ≡ p l ≡ p l +1 ≡· · · ≡ p k ≡ − n ≡ k − l is even. In addition,we can divide the matrices A, D − , x and x into block matrices A = (cid:18) A A A A (cid:19) , D − = (cid:18) I (cid:19) , x = (cid:18) zw (cid:19) , x = (cid:18) z w (cid:19) . ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 17
Here the sizes of A and I = diag(1 , · · · ,
1) are ( k − l ) × ( k − l ), and the sizes of z and z are l ×
1. We hope that the matrices A , A , A and A are not confused with A n when n = 1 , , A and D − , we obtain that(5.1) A T1 = A , A T2 = A and A T4 + A + I = E. Here E is a ( k − l ) × ( k − l ) matrix with all components being 1, and we have used thequadratic reciprocity law. Since k − l is even, we get Ew = 0.(1) If Ax = 0, then by the block forms of A and x we get that A z + A w = 0 and A z + A w = 0. While x T ( A + D − ) = (cid:0) z T w T (cid:1) (cid:18) A A A A + I (cid:19) = (cid:0) z T A + w T A z T A + w T ( A + I ) (cid:1) . Note that ( z T A + w T A ) T = A z + A w = 0 and (cid:2) z T A + w T ( A + I ) (cid:3) T = A z + ( A T4 + I ) w . We claim that w has even many non-zero components. From this claim we deducethat Ew = 0, namely ( A T4 + I ) w = A w by (5.1). Therefore, (cid:2) z T A + w T ( A + I ) (cid:3) T = A z + A w = 0 . Thus x T ( A + D − ) = 0. Moreover, the above process is invertible.So the proof of (1) is reduced to proving the claim. From the definition of A , we knowthat Ax = 0, namely(5.2) A z + A w = 0 and A z + A w = 0 . Like the expansion of x T ( A + D − ), we get x T0 A = (cid:0) w T0 (cid:1) . Therefore, 0 = x T0 Ax = (cid:0) w T0 (cid:1) x = w T0 w , which is equivalent to the claim.(2) If d ≡ w also has even many non-zero components and the proofis the same as that of (1). We assume that d ≡ − w has odd manynon-zero components. Then(5.3) Ew = w = Iw . Denote by ¯ z = z − z and ¯ w = w − w . So ( x − x ) T = (cid:0) ¯ z T ¯ w T (cid:1) . From ( A + D − ) x = 0we obtain that A z + A w = 0 and A z + ( A + I ) w = 0. Observe that( x − x ) T A = (cid:0) ¯ z T A + ¯ w T A ¯ z T A + ¯ w T A (cid:1) . But we have (cid:0) ¯ z T A + ¯ w T A (cid:1) T = A ¯ z + A ¯ w = A z + A w − ( A z + A w ) = 0 by (5.2).Moreover, (cid:0) ¯ z T A + ¯ w T A (cid:1) T = A ¯ z + A T4 ¯ w = A z + A T4 w + A z + A T4 w = A z + ( A + I ) w + A z + ( A + I ) w + Ew = Iw + Ew = 0 . Here we have used (5.2) and (5.3). Consequently, ( x − x ) T A = 0.(3) By Proposition 4, to find all the elements of Sel (cid:48) ( E ( n ) ) is equivalent to compute thekernel of M n . This is equivalent to find all ( X, Y, Z, W ) such that M n (cid:18) XY (cid:19) + G (cid:18) ZW (cid:19) = 0 and M (cid:18) ZW (cid:19) = 0 . Here
X, Y ∈ F k , Z ∈ F k + k − k and W ∈ F k − k . Since Sel ( E ) has dimension two,we know that ker M = { } by Proposition 4. This implies that Z and W are zerovectors. Thus we reduce to finding X and Y in F k such that M n (cid:18) XY (cid:19) = 0. Thendim F Sel (cid:48) ( E ( n ) ) = 2 k − rank M n . From (1) we derive that rank A = rank( A + D − ). Hence, rank M n = 2rank A . Butthe R´edei matrix R of Q ( √− n ) takes the form (cid:0) A (cid:1) . Therefore, rank R = rank A = k − h ( n ). Thus h ( n ) = 1 if and only if rank M n = 2 k −
2. Hence, the dimension ofSel (cid:48) ( E ( n ) ) is two if and only if h ( n ) = 1.Now we assume that h ( n ) = 1 to find representatives of Sel (cid:48) ( E ( n ) ). From aboveargument, it suffices to find X, Y ∈ F k such that M n (cid:18) XY (cid:19) = 0, namely ( A + D − ) X = 0and AY = 0. As the rank of A is k − Ax = 0, we get Y = 0 or x . Similarly, X = 0or x . Hence, Sel (cid:48) ( E ( n ) ) is generated by (1 , n, n ) and ( d, , d ). Noting that Λ = (2 , ,
1) =(1 , n, n )(2 , n, n ), the proof of (3) is complete.This completes the proof of the lemma. (cid:3) Now we can prove Theorem 1.
Proof of Theorem 1.
From the exact sequence (1.2), we see that the a necessary conditionfor (1) is dim F Sel (cid:48) ( E ( n ) ) = 2. This is equivalent to h ( n ) = 1 by Lemma 9. Now weassume this and use the notation of Lemma 9. Since h ( n ) = 1, from (3) of Lemma 9 weknow that Sel (cid:48) ( E ( n ) ) is generated by Λ = (2 , ,
1) and Λ (cid:48) = ( d, , d ).First, we compute the Cassels pairing (cid:104) Λ , Λ (cid:48) (cid:105) . The genus one curve D Λ is H : − b nt + 2 u − u = 0 ,H : − a nt + u − u = 0 ,H : c nt + u − u = 0 . From the definition of the Cassels pairing, we have to choose points on H i ( Q ). For H wechoose Q = (0 , , L of H at Q is L : u − u . For H we will use Gauss genus theory to select a point. As R (cid:18) (cid:19) = (cid:0) A (cid:1) (cid:18) (cid:19) =0, by Gauss genus theory (Lemma 2) 2 is a norm element, namely there is a positiveprimitive integer solution ( α, β, γ ) to(5.4) α + nβ = 2 γ . Let Q = ( β, γb, αb ). Then Q lies in H ( Q ) and the corresponding tangent linear form is L : βbnt − γu + αu . So by Lemma 1, (cid:104) Λ , Λ (cid:48) (cid:105) equals to (cid:89) p | nabc ∞ (cid:16) L L ( P p ) , d (cid:17) p . Here P p is any point on D Λ ( Q p ) such that L i ( P p ) is non-vanishing. For any prime divisor p of abc , we have (cid:16) p i p (cid:17) = 1; so we have (cid:16) dp (cid:17) = 1. The local Cassels pairing is trivial at all p | abc . In addition, it is also trivial at p = ∞ by d >
0. So we only need to compute itat p = 2 and p | n .For p | n , we choose the local point ( t, u , u , u ) such that t = 0 , u = 1 , u = − , u = 2 . As (2 γ + αu )(2 γ − αu ) = 4 γ − α = 2 nβ , we choose u such that p | γ − αu . Then p (cid:45) γ + αu by the primitivity of ( α, β, γ ). So (cid:16) L L ( P p ) , d (cid:17) p = (cid:16) γ + αu ) , d (cid:17) p = (cid:18) γp (cid:19) δ . Here δ is 1 if p | d and 0 otherwise. ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 19
For p = 2, we note that the local Cassels pairing is trivial if d ≡ d ≡ − t, u , u , u ) be the local point satisfying t = 1 , u = 0 , u = c n, u = a n. We observe that( βbn + αu )( βbn − αu ) = β b n − α a n = b n (2 γ − α ) − α a n = 2 γ b n − α c n = 2 b n (cid:16) γ − α c b (cid:17) ≡ . So we may choose u such that 8 | βbn + αu . Then (cid:16) L L ( P p ) , d (cid:17) = (cid:16) − u ( βbn − γu + αu ) , − (cid:17) = (cid:16) γu , − (cid:17) = (cid:18) − γ (cid:19) . So we get (cid:104) Λ , Λ (cid:48) (cid:105) = (cid:16) γd (cid:17) or (cid:16) γd (cid:17)(cid:18) − γ (cid:19) according to d ≡ h ( n ) = 1 and the non-degeneracy of the Casselspairing on Sel (cid:48) ( E ( n ) ). From the following short exact sequence0 −→ E ( n ) [2] −→ E ( n ) [4] × −→ E ( n ) [2] −→ , we obtain the derived long exact sequence0 −→ E ( n ) ( Q )[2] / E ( n ) ( Q )[4] −→ Sel ( E ( n ) ) −→ Sel ( E ( n ) ) −→ ImSel ( E ( n ) ) −→ . By Proposition 5, we see that (1) is equivalent to dim F Sel (cid:48) ( E ( n ) ) = 2 and ( E ( n ) ) = ( E ( n ) ), which are equivalent to dim F Sel (cid:48) ( E ( n ) ) = 2 and ( E ( n ) ) = 4 bythe long exact sequence. Noting that the kernel of the Cassels pairing on Sel (cid:48) ( E ( n ) )is ImSel ( E ( n ) ) /E ( n ) ( Q )[2], we infer that (1) is equivalent to h ( n ) = 1 and the non-degeneracy of the Cassels pairing on Sel (cid:48) ( E ( n ) ) by Lemma 9.Finally, we connect the Cassels pairing with h ( n ). We first assume that d ≡ | n is a norm element satisfying (5.4), h ( n ) = 1 if and only if W ∈ Im R with W = (cid:16)(cid:104) γp (cid:105) , · · · , (cid:104) γp k (cid:105)(cid:17) T by Lemma 3. Via (2) of Lemma 9, we see that x T A = 0. As A has rank k − R = (cid:0) A (cid:1) , we obtain that W ∈ Im R if and only if x T W = 0,namely (cid:0) γd (cid:1) = 1. Hence, the Cassels pairing is non-degenerate if and only if h ( n ) = 0provided that d ≡ d ≡ − x − x ) T A = 0 by (2) of Lemma 9. Likethe above case, h ( n ) = 1 if and only if ( x − x ) T W = 0, namely (cid:16) γn/d (cid:17) = 1. Viewing (5.4)as a congruence modulo γ , we have (cid:16) − nγ (cid:17) = 1. Since n ≡ (cid:16) − γ (cid:17) = (cid:0) γn (cid:1) from the quadratic reciprocity law. So the Cassels pairing (cid:104) Λ , Λ (cid:48) (cid:105) is (cid:16) γn/d (cid:17) in this case.Hence, the Cassels pairing is non-degenerate if and only if h ( n ) = 0.In summary, we have proved that (1) is equivalent to (2). This completes the proof ofthe theorem. (cid:3) Proof of Theorem 2.Lemma 10.
Assume that all prime divisors of n are congruent to modulo . Then thedimension of the pure -Selmer group Sel (cid:48) ( E ( n ) ) is two if and only if h ( n ) = 1 .Proof. Like (3) of Lemma 9, it suffices to show that rank M n = 2 k − h ( n ) = 1. Note that the Monsky matrix takes the form M n = (cid:18) A + D D D A + D (cid:19) . Here D = diag (cid:16)(cid:104) p (cid:105) , · · · , (cid:104) p k (cid:105)(cid:17) . To relate M n to the R´edei matrix R of n , we performsome elementary linear transforms on the block matrix M n . Adding the first block rowto the second, we have (cid:18) A + D D A A (cid:19) . Then adding the second block column to the first, we get (cid:18)
A D A (cid:19) . Summating all the last ( k −
1) columns to the ( k + 1)-th column, we derive (cid:18) R D (cid:48) A (cid:48) (cid:19) . Here D (cid:48) and A (cid:48) denote the matrices obtained from D and A respectively by deleting theirfirst columns. Adding all the first ( k −
1) rows to the k -th row and then moving the k -throw as the last row, we yield(5.5) (cid:18) R k ∗ R T1 (cid:19) . Here R i is the matrix obtained from the R´edei matrix R by deleting its i -th row. Sinceevery p i is congruent to 1 modulo 4 and n is congruent to 1 modulo 8, we see that thecolumn sum of R is zero, namely the sum of any of R ’s given column is zero. Thusrank R i = rank R = k − h ( n ) . From this and (5.5) we get2rank R ≤ rank M n ≤ k − R. Therefore, rank R = k − M n = 2 k −
2. Then the lemma follows fromrank R = k − h ( n ). (cid:3) Lemma 11.
Assume that all prime divisors of n are congruent to modulo and h ( n ) =1 . (1) If the rank of A is k − , then Sel (cid:48) ( E ( n ) ) is generated by ( n, n, and ( d, d, ,where d = (cid:81) k p x i i with x = ( x , · · · , x k ) T (cid:54) = 0 , x = (1 , · · · , T ∈ F k satisfying Ax = 0 . (2) If the rank of A is k − , then Sel (cid:48) ( E ( n ) ) is generated by ( n, n, and ( d, n/d, n ) ,where d = (cid:81) k p x i i with x = ( x , · · · , x k ) T and b = (cid:16)(cid:104) p (cid:105) , · · · , (cid:104) p k (cid:105)(cid:17) T satisfying Ax = b .Proof. Like (3) of Lemma 9, we reduce to finding X and Y in F k such that M n (cid:18) XY (cid:19) = 0,namely(5.6) (cid:40) AX + D ( X + Y ) = 0 ,AY + D ( X + Y ) = 0 . Adding these two equations, we get A ( X + Y ) = 0. So X + Y lies in ker A . According torank A n , we can divide this into two cases.First, we deal with the case rank A = k −
1. Then ker A = (cid:8) , x (cid:9) . If X + Y = 0, then X = Y ∈ ker A by (5.6). These give rise to two elements X = Y = 0 or X = Y = x . So(1 , ,
1) and ( n, n,
1) lie in Sel (cid:48) ( E ( n ) ). If X + Y = x , then (5.6) implies that AX = D x = b . In fact, b is indeed in the image of A . From Gauss genus theory, we know thatrank (cid:0) A b (cid:1) = k − h ( n ) = k − A. ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 21
Let x be such that Ax = b . Then X = x or x − x . So these give rise to the remainingtwo elements ( d, n/d, n ) and ( n/d, d, n ) of Sel (cid:48) ( E ( n ) ).Finally, we consider the case rank A = k −
2. If X + Y = 0, then X = Y ∈ ker A by(5.6). Thus there are four elements in Sel (cid:48) ( E ( n ) ) given by( n, n, , ( d, d, , ( n/d, n/d,
1) and (1 , , d defined in the lemma. This completes the proof of the lemma. (cid:3) Now we can prove Theorem 2.
Proof of Theorem 2.
Like the proof of Theorem 1, a necessary condition for (1) is h ( n ) = 1.So we may assume that h ( n ) = 1. Then there are two cases according to the rank of A .First, we consider the case rank A = k −
2. We use the notation defined in (1) of Lemma11. Then we have R (cid:18) x (cid:19) = (cid:0) A b (cid:1) (cid:18) x (cid:19) = Ax = 0. Here b = (cid:16)(cid:104) p (cid:105) , · · · , (cid:104) p k (cid:105)(cid:17) T .So by Gauss genus theory (Lemma 2 and the epimorphism θ in § d = (cid:81) k p x i i corre-sponds to the non-trivial element of 2 A ∩ A [2]. So d = d = (cid:81) k p x i i . We claim that d ≡ h ( n ) = 1, we know that rank (cid:0) A b (cid:1) = k − A = k −
2, we see that b is not in the image of A . As A is symmetric andker A is generated by x and x , a column vector y ∈ F k is in the image of A if and onlyif x T0 y = x T y = 0. For the vector b , we have x T0 b = (cid:2) n (cid:3) = 0 for n ≡ x T b (cid:54) = 0. Note that x T b = (cid:2) d (cid:3) . Therefore, d is congruent to 5 modulo 8.By Lemma 10, the pure 2-Selmer group Sel (cid:48) ( E ( n ) ) has dimension two. Moreover, viaLemma 11 it is generated by Λ = ( d, d, , Λ (cid:48) = ( − , , − . Note that Λ (cid:48) = ( n, n, − n, n, − (cid:104) Λ , Λ (cid:48) (cid:105) . ForΛ = ( d, d, D Λ associated to Λ is given by H : − b nt + du − u = 0 ,H : − a nt + u − du = 0 ,H : 2 c nt + du − du = 0 . By Cassels pairing, we have to choose global points on H i . For H , we choose the globalpoint Q = (0 , , ∈ H ( Q ). Then the corresponding tangent linear form L of H at Q is L : u − u . Now we are ready to choose a point on H ( Q ). Since d corresponds to the non-trivialelement of 2 A ∩ A [2], Gauss genus theory implies that d is a norm element, namely thereis a positive primitive integer solution ( α, β, γ ) to(5.7) dα + d (cid:48) β = γ with dd (cid:48) = n . We may assume that α is even. If not, α is odd and β is even. We can geta new solution ( ¯ α, ¯ β, ¯ γ ) = (cid:16) d (cid:48) α − d (cid:48) β − dα, dβ − dα − d (cid:48) β, ( d + d (cid:48) ) γ (cid:17) to (5.7). Thus 4 | ¯ α and 2 (cid:107) ¯ β . Dividing the solution ( | ¯ α | , | ¯ β | , | ¯ γ | ) by gcd( ¯ α, ¯ β, ¯ γ ), we derivea positive primitive integer solution to (5.7) with corresponding α being even. From (5.7),we know that ( β, γb, dαb ) lies in H ( Q ) with the corresponding tangent linear form L : d (cid:48) βbt − γu + αu . By Lemma 1, the Cassels pairing (cid:104) Λ , Λ (cid:48) (cid:105) equals to (cid:89) p | nabc ∞ (cid:16) L L ( P p ) , − (cid:17) p . Here P p is any point of D Λ ( Q p ) such that L i ( P p ) is non-vanishing. Since any prime divisor p of n is congruent to 1 modulo 4, we know that the local Cassels pairing is trivial at p | n .This pairing is also trivial at p | c for − a = b − c ≡ b (mod p ). Thus we reduce tocomputing the local Cassels pairing at p | ab ∞ .For p = ∞ , we choose the local point P ∞ = ( t, u , u , u ) = (0 , , − , √ d ). Then (cid:16) L L ( P ∞ ) , − (cid:17) ∞ = (cid:16) γ + α √ d ) , − (cid:17) ∞ = 1 . For p = 2, we choose the local point P = ( t, u , u , u ) such that t = 2 , u = 1 , u = 1 + 8 c d (cid:48) , u = d − b n. Here we used the fact that d ≡ u = 1 + 8 c d (cid:48) ≡ u ≡ α is even, we have 2 (cid:107) α by (5.7). Let u be anysquare root of d − b n . Then 2 | d (cid:48) βb + α u . The local Cassels pairing at p = 2 is (cid:16) L L ( P ) , − (cid:17) = (cid:16) ( u − d (cid:48) βb + αu − γ ) , − (cid:17) = (2 , − (cid:16) (cid:0) d (cid:48) βb + α u (cid:1) − γ, − (cid:17) = ( − γ, − = − (cid:18) − γ (cid:19) . For p | ab , we choose the local solution t = 0 , u = 1 , u = − , u = d. By (5.7), ( γ + αu )( γ − αu ) = γ − α u = d (cid:48) β . If p | β , then we choose u such that p | γ − αu ; so p (cid:45) γ + αu . If p (cid:45) β , we choose u to be any square root of d and have p (cid:45) γ − αu . Thus in any case γ + αu is a p -adic unit. So the local pairing is (cid:16) L L ( P p ) , − (cid:17) p = (cid:0) γ + αu ) , − (cid:1) p = 1 . Therefore, the Cassels pairing (cid:104) Λ , Λ (cid:48) (cid:105) is (cid:104) Λ , Λ (cid:48) (cid:105) = − (cid:18) − γ (cid:19) . Like Theorem 1, (1) is equivalent to h ( n ) = 1 and the non-degeneracy of the Casselspairing.Now we claim that (cid:16) − γ (cid:17) = 1 if and only if h ( n ) = 1 provided that h ( n ) = 1. As d = d | n corresponds to the non-trivial element of 2 A ∩ A [2] and ( dα ) + nβ = dγ by (5.7),we see that h ( n ) = 1 if and only if W ∈ Im R by Lemma 3. Here W = (cid:16)(cid:104) γp (cid:105) , · · · , (cid:104) γp k (cid:105)(cid:17) T and R is the R´edei matrix (cid:0) A b (cid:1) . Since the rank of R is k − x T0 R = 0, a vector z is in the image of R if and only if x T0 z = 0. We have x T0 W = (cid:2) γn (cid:3) . Consequently, h ( n ) = 1if and only if (cid:0) γn (cid:1) = 1. Viewing (5.7) as a congruence modulo γ , we see that (cid:16) − nγ (cid:17) = 1.By n ≡ (cid:16) − γ (cid:17) = (cid:0) γn (cid:1) . Therefore, h ( n ) = 1 if and only if (cid:16) − γ (cid:17) = 1.So in the case rank A = k −
2, (1) is equivalent to (2) by noting that d ≡ A = k −
1. We use the notation of (2) of Lemma11. Note that R (cid:18) x (cid:19) = (cid:0) A b (cid:1) (cid:18) x (cid:19) = Ax + b = 0. By Gauss genus theory, ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 23 d = 2 (cid:81) k p x i i corresponds to the non-trivial element of 2 A ∩ A [2]. Then d = (cid:81) k p x i i . FromLemma 11 we derive that Sel (cid:48) ( E ( n ) ) is generated byΛ = (2 d, d, , Λ (cid:48) = ( − , , − . Here we used the fact that Λ = ( d, n/d, n )(2 , n, n ) and Λ (cid:48) = ( − n, n, − n, n, (cid:104) Λ , Λ (cid:48) (cid:105) . The genus one curve D Λ is definedby H : − b nt + 2 du − u = 0 ,H : − a nt + u − du = 0 ,H : c d (cid:48) t + u − u = 0 . Here d (cid:48) = n/d . According to the definition of the Cassels pairing, we first choose globalpoints Q i on H i ( Q ). Let Q = (0 , , Q lies in H ( Q ) and the correspondingtangent linear form L is L : u − u . Since 2 d corresponds to the non-trivial element of 2 A ∩ A [2], by Gauss genus theory thereis a positive primitive integer solution ( α, β, γ ) to(5.8) dα + d (cid:48) β = 2 γ . Then Q = ( β, γb, αbd ) lies in H ( Q ). In addition, the tangent linear form L of H at Q is L : d (cid:48) βbt − γu + αu . Like the case rank A = k −
2, the Cassels pairing (cid:104) Λ , Λ (cid:48) (cid:105) equals to (cid:89) p | ab ∞ (cid:16) L L ( P p ) , − (cid:17) p with P p any point on D Λ ( Q p ) such that L i ( P p ) is non-vanishing. For p = ∞ , we choosethe local point ( t, u , u , u ) = (0 , , − , √ d ). Then (cid:16) L L ( P ∞ ) , − (cid:17) ∞ = (cid:16) α √ d + 2 γ ) , − (cid:17) ∞ = 1 . For p = 2, we choose a point P = ( t, u , u , u ) such that t = 1 , u = 2 (cid:20) d (cid:21) , u = c d (cid:48) + u , u = a n + 2 du with γu ≡ d (cid:48) βb + αu )( d (cid:48) βb − αu ) = d (cid:48) β b − α ( a n + 2 du )= d (cid:48) b (2 γ − dα ) − α ( a n + 2 du ) = 2 d (cid:48) b (cid:104) γ − α d (cid:48) b (cid:0) c n + 2 du (cid:1)(cid:105) = 2 d (cid:48) b (cid:104) γ − α (cid:16) c b d + db d (cid:48) u (cid:17)(cid:105) ≡ . So we may choose u such that 8 | d (cid:48) βb + αu . We have (cid:16) L L ( P ) , − (cid:17) = (cid:16) ( u − u )( d (cid:48) βb + αu − γu ) , − (cid:17) = (cid:16) − γu ( u − u ) , − (cid:17) = (2 , − ( − γ, − ( u γ − , − = (cid:18) − γ (cid:19)(cid:18) d (cid:19) . For p | a , we put P p = ( t, u , u , u ) with t = 1 , u = 0 , u = c d (cid:48) , u = a √ n. Observe that ( d (cid:48) βb − γu )( d (cid:48) βb + 2 γu ) = d (cid:48) β b − γ c d (cid:48) ≡ d (cid:48) c ( d (cid:48) β − γ ) ≡ − c nα (mod p ) . If p | α , we choose u such that p | d (cid:48) βb + 2 γu ; in addition, p (cid:45) d (cid:48) βb − γu , otherwise p | β which contradicts that ( α, β, γ ) is a positive primitive integer solution to (5.8). If p (cid:45) α , we have p (cid:45) d (cid:48) βb ± γu . So we can always choose u such that p (cid:45) d (cid:48) βb − γu . Since p | a , we get (cid:16) L L ( P p ) , − (cid:17) p = (cid:16) − u ( d (cid:48) βb − γu + αu ) , − (cid:17) p = (cid:0) d (cid:48) βb − γu , − (cid:1) p = 1 . Similarly, for p | b , we have (cid:16) L L ( P p ) , − (cid:17) p = 1 . In summary, we have (cid:104) Λ , Λ (cid:48) (cid:105) = (cid:18) − γ (cid:19)(cid:18) d (cid:19) . Like the case rank A = k −
2, (1) is equivalent to h ( n ) = 1 and (cid:104) Λ , Λ (cid:48) (cid:105) = −
1, and Gaussgenus theory implies that (cid:16) − γ (cid:17) = 1 if and only if h ( n ) = 1 provided that h ( n ) = 1.Therefore, (1) is equivalent to h ( n ) = 1 and h ( n ) ≡ (cid:2) d (cid:3) ≡ d − (mod 2) . This completesthe proof of the theorem. (cid:3) Independence Property of Residue Symbols
In this section, we assume that a, b and c are coprime positive integers satisfying a + b =2 c . Let q , · · · , q k (cid:48) be all the prime divisors of abc .We first introduce some notation. Given k ≥
2, let α = ( α , · · · , α k ) with all α j ∈ (cid:8) , , , (cid:9) and (cid:81) kj =1 α j ≡ B = ( B lj ) k × k is a symmetric F matrix with rank k − Bz = 0. Here z = (1 , · · · , T ∈ F k . So there is a unique z = ( z , · · · , z k ) T (cid:54) = 0 , z such that Bz = 0 and z = 1. Our goal in this section is toestimate the number of C k ( x, α, B ). Here C k ( x, α, B ) consists of all n = p · · · p k ≤ x satisfying • p < · · · < p k and p l ≡ α l (mod 16) for all 1 ≤ l ≤ k , • (cid:104) p j p l (cid:105) = B lj for all 1 ≤ l < j ≤ k , • (cid:16) p l q j (cid:17) = 1 for all 1 ≤ l ≤ k and 1 ≤ j ≤ k (cid:48) , and • (cid:16) d (cid:48) d (cid:17) (cid:0) dd (cid:48) (cid:1) = − d = (cid:81) kl =1 p z l l and d (cid:48) = (cid:81) kl =1 p − z l l .Due to the existence of the quartic residue symbols, we can’t estimate C k ( x, α, B )directly. This problem can be solved by identifying C k ( x, α, B ) with a set counting corre-sponding integers over Z [ i ]. To this purpose, we first introduce some notation. Denote by P the set of primary primes of Z [ i ] with positive imaginary part. We define N to be thenorm map from Z [ i ] to Z . Let C (cid:48) k ( x, α, B ) be all η = λ · · · λ k satisfying • N η ≤ x and N λ < · · · < N λ k , • λ l ∈ P and N λ l ≡ α l (mod 16) for all 1 ≤ l ≤ k , • (cid:104) Nλ j Nλ l (cid:105) = B lj for all 1 ≤ l < j ≤ k , • (cid:16) Nλ l q j (cid:17) = 1 for all 1 ≤ l ≤ k and 1 ≤ j ≤ k (cid:48) , and • (cid:16) λ (cid:48) λ (cid:17) = − λ = (cid:81) kl =1 λ z l l and λ (cid:48) = (cid:81) kl =1 λ − z l l . ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 25
Lemma 12.
The following map is a bijection C (cid:48) k ( x, α, B ) −→ C k ( x, α, B ) , η (cid:55)→ N η.
Proof.
Let η = λ · · · λ k be an element of C (cid:48) k ( x, α, B ) satisfying N λ < · · · < N λ k and λ l ∈ P for all 1 ≤ l ≤ k . Denote by p l = N λ l for all 1 ≤ l ≤ k . To show that N η ∈ C k ( x, α, B ), we only need to verify (cid:16) d (cid:48) d (cid:17) (cid:0) dd (cid:48) (cid:1) = − d = N λ and d (cid:48) = N λ (cid:48) .Note that (cid:18) p l p j (cid:19) (cid:18) p j p l (cid:19) = (cid:18) λ l λ l λ j (cid:19) (cid:18) λ j λ j λ l (cid:19) = (cid:18) λ l λ j (cid:19) (cid:18) λ l λ j (cid:19) (cid:18) λ j λ l (cid:19) (cid:18) λ j λ l (cid:19) = (cid:18) λ j λ l (cid:19) (cid:18) λ j λ l (cid:19) (cid:18) λ j λ l (cid:19) (cid:18) λ j λ l (cid:19) . Here we have used the quadratic reciprocity law for (cid:16) λ l λ j (cid:17) and (cid:16) λ l λ j (cid:17) . From the definitionof the quartic residue symbol, we have (cid:16) λ j λ l (cid:17) (cid:16) λ j λ l (cid:17) = 1. Consequently, (cid:18) p l p j (cid:19) (cid:18) p j p l (cid:19) = (cid:18) λ j λ l (cid:19) . Therefore, (cid:18) d (cid:48) d (cid:19) (cid:18) dd (cid:48) (cid:19) = (cid:18) λ (cid:48) λ (cid:19) = − . So N η lies in C k ( x, α, B ). The map is obviously injective. The map is surjective byobserving that for every rational prime p ≡ P with norm p . This completes the proof of the lemma. (cid:3) To use the idea of Cremona-Odoni [10], we introduce another set T ( x ). Here T ( x ) isthe set of positive integers n = p · · · p k − ≤ x satisfying • p < · · · < p k − , • p l ≡ α l (mod 16) for all 1 ≤ l ≤ k − • (cid:104) p j p l (cid:105) = B lj for all 1 ≤ l < j ≤ k −
1, and • (cid:16) p l q j (cid:17) = 1 for all 1 ≤ l ≤ k − ≤ j ≤ k (cid:48) .The independence property of Legendre symbols of Rhoades [14] implies(6.1) T ( x ) ∼ − ( k (cid:48) +3)( k − − ( k − ) · C k − ( x ) , where (cid:0) k (cid:1) is the binomial coefficient and C k ( x ) is the set of all positive square-free integers n ≤ x with exactly k prime factors. Like C k ( x, α, B ), we have to identify T ( x ) with anotherset T (cid:48) ( x ). Here T (cid:48) ( x ) is the set of η = λ · · · λ k − satisfying • N η ≤ x and N λ < · · · < N λ k − , • λ l ∈ P and N λ l ≡ α l (mod 16) for all 1 ≤ l ≤ k − • (cid:104) Nλ j Nλ l (cid:105) = B lj for all 1 ≤ l < j < k , and • (cid:16) Nλ l q j (cid:17) = 1 for all 1 ≤ l < k and 1 ≤ j ≤ k (cid:48) .Similarly, we have the following lemma. Lemma 13.
The following map is a bijection T (cid:48) ( x ) −→ T ( x ) , η (cid:55)→ N η.
Now we can use the idea of Cremona-Odoni [10] to prove the independence property ofresidue symbols.
Theorem 4.
For k an integer greater than , let α = ( α , · · · , α k ) with (cid:81) kj =1 α j ≡ and all α j ∈ (cid:8) , , , (cid:9) . Assume that B = ( B lj ) k × k is a symmetric F matrixwith rank k − and the sum of its any given row being zero. Then C k ( x, α, B ) = 1 + o (1)2 k + k (cid:48) k +1+ ( k ) · C k ( x ) . Proof.
Like Cremona-Odoni [10], we consider the comparison map f : C (cid:48) k ( x, α, B ) −→ T (cid:48) ( x ) , η (cid:55)→ η/ ˜ η, where ˜ η lies in P such that its norm is the maximal prime divisor of N η . According to z k = 0 or not, the proof can be divided into two cases.We first consider the case z k = 0. The next step is to consider the fiber of the comparisonmap f . Let (cid:15) = (cid:81) k − j =1 λ j ∈ T (cid:48) ( x ) with N λ < · · · < N λ k − and all λ j ∈ P . Then (cid:15) lies inIm f if and only if there exists a λ ∈ P with N λ lying in ( N λ k − , x/N (cid:15) ] such that(1) N λ ≡ α k (mod 16) and (cid:104) Nλ Nλ l (cid:105) = B lk for all 1 ≤ l ≤ k − (cid:16) Nλ q j (cid:17) = 1 for all 1 ≤ j ≤ k (cid:48) and(3) (cid:16) λ λ (cid:17) = − (cid:16) (cid:15)/λλ (cid:17) with λ = (cid:81) k − l =1 λ z l l .Then there is a unique subset A = A (cid:15) of (cid:16) Z [ i ] / (cid:15)(cid:15) (cid:15) Z [ i ] (cid:17) × such that for any prime θ theinteger θ(cid:15) lies in C (cid:48) k ( x, α, B ) if and only if θ ∈ P and θ ∈ A satisfying N θ ∈ ( N λ k − , N (cid:15) ].Here (cid:15) is the product of all primary primes lying in P and lying above p with p | abc and p ≡ (cid:15) is the product of all prime factors p of abc with p ≡ A is evaluated in the following lemma. Lemma 14.
Let ϕ ( (cid:15) ) be the cardinality of G = (cid:16) Z [ i ] / c (cid:17) × with c = c (cid:15) the ideal (cid:15)(cid:15) (cid:15) Z [ i ] .Then A = 2 − k − k (cid:48) − ϕ ( (cid:15) ) . Proof of Lemma 14.
From the definition of A , we see that A represents those primaryclasses β of G such that(a) N β ≡ α k (mod 16) and (cid:104) NβNλ l (cid:105) = B lk for all 1 ≤ l ≤ k − (cid:16) Nβq j (cid:17) = 1 for all 1 ≤ j ≤ k (cid:48) , and(c) (cid:16) βλ (cid:17) = (cid:16) (cid:15)/λλ (cid:17) with λ = (cid:81) k − l =1 λ z l l .Via Chinese Remainder Theorem we obtain the following isomorphism G (cid:39) (cid:16) Z [ i ] / Z [ i ] (cid:17) × × (cid:18) k − (cid:89) l =1 (cid:16) Z [ i ] /λ l Z [ i ] (cid:17) × (cid:19) × (cid:16) Z [ i ] /(cid:15) Z [ i ] (cid:17) × × (cid:16) Z [ i ] /(cid:15) Z [ i ] (cid:17) × , where the map is given by β (cid:55)→ ( β , · · · , β k − , β (cid:48) , β (cid:48) ). Here β l denotes the class β (mod λ l Z [ i ]) if 1 ≤ l ≤ k − β (mod 16 Z [ i ]) if l = 0, and β (cid:48) j denotes the class β (mod (cid:15) j Z [ i ]) for j = 1 ,
2. Note that the residue symbol (cid:0) · λ (cid:1) is only non-trivial on those (cid:16) Z [ i ] /λ l Z [ i ] (cid:17) × -component with z l = 1. For any 1 ≤ l ≤ k −
1, the condition (cid:104) Nβ l Nλ l (cid:105) = B lk takes up a half of the (cid:16) Z [ i ] /λ l Z [ i ] (cid:17) × -component. Here we have used the isomorphism (cid:16) Z [ i ] /λ l Z [ i ] (cid:17) × (cid:39) (cid:16) Z /N λ l Z (cid:17) × by λ l ∈ P . The condition (cid:16) βλ (cid:17) = − (cid:16) (cid:15)/λλ (cid:17) selects another ON-TRIVIAL SHAFAREVICH-TATE GROUPS OF ELLIPTIC CURVES 27 half of the (cid:81) k − l =1 (cid:16) Z [ i ] /λ l Z [ i ] (cid:17) × -component. Similarly, we have the following isomorphisms (cid:16) Z [ i ] /(cid:15) Z [ i ] (cid:17) × (cid:39) (cid:89) λ (cid:48) j | (cid:15) (cid:16) Z [ i ] /λ (cid:48) j Z [ i ] (cid:17) × and (cid:16) Z [ i ] /(cid:15) Z [ i ] (cid:17) × (cid:39) (cid:89) q j | (cid:15) (cid:16) Z [ i ] /q j Z [ i ] (cid:17) × . Like above, (cid:16) Nβ (cid:48) q j (cid:17) = 1 selects a half of the (cid:16) Z [ i ] /λ (cid:48) j Z [ i ] (cid:17) × -component provided that λ (cid:48) j | (cid:15) . For q j | (cid:15) , the condition (cid:16) Nβ (cid:48) q j (cid:17) = 1 also selects a half of the (cid:16) Z [ i ] /q j Z [ i ] (cid:17) × -component by the composition of the homomorphisms (cid:16) Z [ i ] /q j Z [ i ] (cid:17) × (cid:16) (cid:16) Z /q j Z (cid:17) × → (cid:8) ± (cid:9) . Here the last homomorphism is given by the Legendre symbol.Now we consider the G (cid:48) -component, where G (cid:48) = (cid:16) Z [ i ] / Z [ i ] (cid:17) × . The primary conditionselects G = 1 + (2 + 2 i ) Z [ i ] of G (cid:48) and G (cid:48) = 4 G . Then N β ≡ α k (mod 16) selects afourth of G . So A = 2 − k − k (cid:48) − ϕ ( (cid:15) ) . This completes the proof of the lemma. (cid:3)
For any (cid:15) ∈ T (cid:48) ( x ), let h ( (cid:15) ) be the number of primes θ ∈ P such that θ + 16 (cid:15)(cid:15) (cid:15) Z [ i ] liesin A and N θ lies in (
N λ k − , N (cid:15) ]. Then we have(6.2) C (cid:48) k ( x, α, B ) = (cid:88) (cid:15) ∈ T (cid:48) ( x ) h ( (cid:15) ) . We will divide the sum in (6.2) into several parts according to the norm of η . To this pur-pose, we introduce some notation. We define µ and ν to be (log x ) and exp (cid:16) log x (log log x ) (cid:17) respectively. For a set M consisting of positive integers and a function g on Z [ i ], we define ∗ (cid:88) Nδ ∈ M g ( δ ) = (cid:88) δ ∈ T (cid:48) ( x ) Nδ ∈ M g ( δ ) . Like Lemma 3.1 of Cremona-Odoni, we have the following lemma (the proof is the sameas that of Cremona-Odoni).
Lemma 15. If m = 20 and n = µ , then ∗ (cid:88) m 20, we get h ( (cid:15) ) ≤ π ( x/N (cid:15) ) by noting that every primeideal corresponds to at most one primary prime element. Here π ( y ) denotes the numberof those prime ideals with norm no larger than y , and the prime ideal theorem says that π ( y ) ∼ Li( y ) . So we have ∗ (cid:88) N(cid:15) ≤ h ( (cid:15) ) = O (cid:0) Li( x ) (cid:1) . Next, if (cid:15) ∈ T (cid:48) ( x ) and 20 < N (cid:15) ≤ µ , then we also have h ( (cid:15) ) = O (cid:0) Li( x/N (cid:15) ) (cid:1) . Thus Lemma15 implies that ∗ (cid:88) N (cid:15) > x k − k , then N λ k − > x k . So x/N (cid:15) < x k < N λ k − < N θ . Thus from thedefinition of h ( (cid:15) ) we get h ( (cid:15) ) = 0 in this case. Therefore, ∗ (cid:88) x k − k Here ψ ( y, T , c ) is given by (cid:88) N a ≤ y a P c ∈ T Λ( a ) . Via the orthogonality of characters and the exact sequence (6.4), we get ψ ( y, T (cid:15) , c (cid:15) ) = 4 ϕ ( (cid:15) ) (cid:88) χ mod c (cid:15) ψ ( y, χ ) (cid:88) [ a ] ∈ T (cid:15) χ ( a ) . Here χ runs over all characters of I ( c (cid:15) ) /P c (cid:15) and ψ ( y, χ ) = (cid:88) N a ≤ y Λ( a ) χ ( a ) . Applying this formula to (6.7), we derive ∗ (cid:88) µ III ) (cid:28) ∗ (cid:88) µ Similarly, we get ( V I ) (cid:28) x (log x ) µ − (cid:28) x (log x ) − , ( V II ) (cid:28) x log x · ν (cid:28) x . Consequently, the sum ( IV ) is an error term. This completes the proof of the theorem. (cid:3) Distribution Result In this section, we assume that a, b and c are coprime positive integers such that a + b =2 c and the dimension of the 2-Selmer group of E is two. Let q , · · · , q k (cid:48) be all the primedivisors of abc . Denote by k a fixed positive integer.Let n = p · · · p k ∈ Q k ( x ) with p < · · · < p k . By Theorem 2, n lies in P k ( x ) if andonly if h ( n ) = 1 and h ( n ) ≡ d − (mod 2). Here d is a certain divisor of n . From theproof of Theorem 2, the characterization of n ∈ P k ( x ) is divided into two cases (accordingto the rank of A = A n being k − k − A = k − 2. Then h ( n ) = 1 is equivalent to b (cid:54)∈ Im A , where b = (cid:16)(cid:104) p (cid:105) , · · · , (cid:104) p k (cid:105)(cid:17) T . Since rank A = k − Az = 0 with z = (1 , · · · , T ∈ F k ,there is a unique column vector z = ( z , · · · , z k ) T (cid:54) = 0 , z with z = 1 such that Az = 0.Then d = (cid:81) kl =1 p z l l is congruent to 5 modulo 8. Jung-Yue [8] (Theorem 3.3 (ii)) showedthat in this case h ( n ) = 1 is equivalent to(7.1) (cid:18) dd (cid:48) (cid:19) (cid:18) d (cid:48) d (cid:19) = − , where d (cid:48) = (cid:81) kl =1 p − z l l .Now we assume that rank A = k − 1. Then h ( n ) = 1 and b ∈ Im A with b = (cid:16)(cid:104) p (cid:105) , · · · , (cid:104) p k (cid:105)(cid:17) T . Let z = ( z , · · · , z k ) T be a column vector with Az = b . Then d = (cid:81) kl =1 p z l l . Jung-Yue [8] (Theorem 3.3 (iii) and (iv)) proved that h ( n ) = 1 if andonly if (cid:0) dd (cid:48) (cid:1) (cid:16) d (cid:48) d (cid:17) = ( − n − . Here d (cid:48) = (cid:81) kl =1 p − z l l .Now we begin to prove Theorem 3. Proof of Theorem 3. Like Theorem 2, we also divide the proof into two cases.First, we count the number N ( x ) of those n ∈ P k ( x ) with rank A n = k − 2. Obviously,we have k ≥ B be the set of k × k symmetric matrices B over F with rank B = k − Bz = 0, where z =(1 , · · · , T ∈ F k . We define I to be all the α = ( α , · · · , α k ) with (cid:81) kl =1 α l ≡ α l ∈ { , , , } for 1 ≤ l ≤ k . Given B ∈ B , we denote by I B the set of α ∈ I suchthat b α does not lie in the image of B . Here b α = (cid:16)(cid:104) α (cid:105) , · · · , (cid:104) α k (cid:105)(cid:17) T . Note that b T α z = 0by (cid:81) kl =1 α l ≡ B ∈ B and α ∈ I B , by (7.1) those n = p · · · p k ∈ P k ( x )satisfying • p < · · · < p k and A n = B , • (cid:16) p l q j (cid:17) = 1 for all 1 ≤ l ≤ k and 1 ≤ j ≤ k (cid:48) , and • p l ≡ α l (mod 16) for all 1 ≤ l ≤ k consist the set C k ( x, α, B ). Moreover, given B ∈ B and α ∈ I − I B , the intersection of C k ( x, α, B ) and P k ( x ) is empty. Therefore, the number N ( x ) of those n ∈ P k ( x ) withrank A n = k − N ( x ) = (cid:88) B ∈ B (cid:88) α ∈ I B C k ( x, α, B ) ∼ − k − kk (cid:48) − − ( k ) · C k ( x ) · (cid:88) B ∈ B I B . Here we have used Theorem 4. Now we count the number of I B with B given. First, given b = ( b , · · · , b k ) T with b (cid:54)∈ Im B and b T z = 0, we count the number of α such that b = b α . As b = b α , we get (cid:104) α l (cid:105) = b l for all 1 ≤ l ≤ k . So any α l has exactly two choices. Thus the number of α suchthat b = b α is 2 k . Next, we count the number of column vectors b such that b T z = 0 andrank (cid:0) B b (cid:1) = k − 1. Since (cid:0) B b (cid:1) z = 0 and B is symmetric, we get rank B (cid:48) = k − (cid:0) B (cid:48) b (cid:48) (cid:1) = k − . Here B (cid:48) is the matrix obtained from B by deleting its lastrow and column, and b (cid:48) is the vector obtained from b by deleting its last component. Thus b (cid:48) does not lie in the image of B (cid:48) . So there are 2 k − many such b (cid:48) and b . Consequently, I B = 2 k − . Then (7.2) implies that N ( x ) ∼ − k − k (cid:48) k − − ( k ) · C k ( x ) · B . The number of B can be obtained from the following result of Brown et al [1]. Proposition 6. For positive integers r ≤ k , we denote by B k,r the set of k × k symmetricmatrices over F with rank r . Then B k,r = u r +1 r +12 ) · k − r − (cid:89) l =0 k − l k − r − l with u l defined above Theorem 3. Note that the map sending B to B (cid:48) induces a bijection between B and B k − ,k − . 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