Infinite rank of elliptic curves over Q ab and quadratic twists with positive rank

Infinite rank of elliptic curves over Q ab and quadratic twists with positive rank Bo-Hae Im Chung-Ang University The 3rd East Asian Number Theory Conference National Taiwan University, Taipei January 16-19, 2012 1 / 34

Mordell-Weil Theorem Theorem (Mordell-Weil Theorem) If E/K is an elliptic curve over a number field K, then the set E(K) is a finitely generated abelian group, i.e. E(K) = E(K) tor Z r, where E(K) tor is finite and a nonnegative integer r is called the rank of E(K). 2 / 34

Birch and Swinnerton-Dyer Conjecture and Parity Conjecture (B-SD Conjecture) rank(e(q)) = ord s=1 L(E, s). Recall that the functional equation of L(E, s) determines the root number w E of E: Λ(s) = w E Λ(2 s), where Λ(s) = N s/2 (2π) s Γ(s)L(E, s). Then, w E = ( 1) ord s=1l(e,s). So the B-SD conjecture predicts the following Parity Conjecture: (Parity Conjecture) The rank(e(q)) is even, if w E = 1, and odd, if w E = 1. There are some examples of elliptic curves that give an affirmative answer to the Parity Conjecture. 3 / 34

Examples satisfying the Parity Conjecture Example (Heegner, Birch et al.) For an odd prime p, let E p /Q : y 2 = x 3 p 2 x. Then w Ep = +1 and rank(e p (Q)) = 0, if p 1 or 3 (mod 8). w Ep = 1 and rank(e p (Q)) = 1, if p 5 or 7 (mod 8). Note that E p /Q : y 2 = x 3 p 2 x is isomorphic (over Q)) to E. So it would be useful to study twists. 4 / 34

Rank over Q and infinite algebraic extension of Q It is conjectured that there is an elliptic curve E/Q with arbitrary rank over Q and it has been known so far that there is an elliptic curve E/Q with rank(e(q)) 32. Let K be a number field. What about for the rank of an abelian variety A over an algebraic closure K of K? Answer : Frey & Jarden, Rosen : The rank of A(K) is infinite. But the rank of E over an infinite algebraic extension of Q may not be infinite. Mazur : For an abelian variety A over K, if A(K) and III(A, K)[p ] are finite, then the rank of A(F ) is finite, where F is a Z p -extension over K. 5 / 34

Infinite rank results over other infinite algebraic extensions For a number field K, Mazur & Kurchanov: There are examples of non-cm elliptic curves over Q which have infinite rank over anticyclotomic extensions. Harris: If the p-division field of E/Q has Galois group GL 2 (Z p ), then E has infinite rank over this p-division field. Rosen & Wong: Any d-dimensional A/K with a degree n projective embedding over K has infinite rank over the compositum of all extensions of K of degree < n(4d + 2). Petersen: Any A/K with a degree n projective embedding over K for n 2, has infinite rank over the compositum of all extensions of K of degree n. 6 / 34

Infinite rank results over the fixed subfields under Gal ( K/K) I-. & Larsen: (1) Any A/K over an infinite cyclic field K with char(k) 2 has infinite rank over K. (2) Any A/K over a number field K has infinite rank over the fixed subfield K σ of K under each σ Gal ( K/K). The idea of proving this is to find a group G and an abelian variety V on which G acts and each element in G has a nontrivial fixed element in a G-representation and to find a rational curve in the quotient V /G. For example, V = E n and G = A n for even n. X V G P 1 K V /G 7 / 34

Conjecture over K (σ 1,...,σ n ) Conjecture (Larsen) Any A/K over a number field K has infinite rank over the fixed subfield K (σ 1,...,σ n) of K under each n-tuple (σ 1,..., σ n ) Gal ( K/K) n. Lozano-Robledo: Under the parity conjecture, for an infinite set S of rational primes, if Q (p) S denotes the compositum of all extensions unramified outside S of the form Q(µ p, d p) for p S and d Q, there is a family of elliptic curves E over Q such that the rank of E over the fixed subfield of Q (p) S under any n-tuple σ = (σ 1,..., σ n ) Gal (Q/Q) n is infinite. 8 / 34

Rank over the maximal abelian extension of a number field Is the rank of A over the maximal abelian extension K ab of K infinite? Conjecture (Frey & Jarden) For a number field K and A/K, the rank of A(K ab ) over K ab is infinite. Note that for a number field K and E/K, the rank of E(K ab ) is infinite. 9 / 34

Ample fields Definition An ample field (or a large field or a pseudo algebraically closed field) K is a field satisfying that every smooth curve over K has infinitely many K-rational points provided it has at least one K-rational point. Conjecture For a number field K, K ab is an ample field. Fehm & Petersen : Every nonzero abelian variety A over an ample field F which is not algebraic over a finite field has infinite rank. 10 / 34

Known results for the conjecture over the maximal abelian extension of a number field K For a number field K, Rosen, Wong: the Jacobian variety over K of a cyclic cover with positive genus of P 1 has infinite rank over K ab. Petersen: the Jacobian variety over K of a Galois cover with positive genus of P 1 with group Γ has infinite rank over some infinite Galois extension Ω/K with group i=1 Γ. If Γ is abelian, the rank of the Jacobian over K ab is infinite. E. Kobayashi: for an elliptic curve E/K which can be defined over an abelian extension K of Q of odd degree (so KQ ab = Q ab ), the rank of E(Q ab ) is infinite under the assumption of the BSD conjecture. 11 / 34

Question The Main Theorem over KQ ab For a general number field K (not necessarily contained in Q ab ), is every elliptic curve E over K of infinite rank over KQ ab? Theorem (I-. & Larsen) 1. Let E/K be an elliptic curve defined over a quadratic extension K of Q. If the j-invariant of E is not 0 or 1728, then E(KQ(2)), hence E(Q ab ) has infinite rank. 2. Let K be a cubic extension of Q and λ K. Then E : y 2 = x(x 1)(x λ) has infinite rank over KQ(2) KQ ab. 12 / 34

The strategy of the proof of the main Theorem Our strategy for proving this entails looking for a Q-rational curve on the Kummer variety Res K/Q E/(±1). When K is a quadratic field, Res K/Q E is an abelian surface isomorphic, over C, to a product of two elliptic curves. Our construction of a curve on the Kummer surface Res K/Q E/(±1) is modelled on the construction of a rational curve on (E 1 E 2 )/(±1) due to J-F. Mestre and to M. Kuwata and L. Wang. When K is a cubic field, we construct a curve of genus 0 on (E 1 E 2 E 3 )/(±1). 13 / 34

The strategy of the proof of the main Theorem For elliptic curves E 1, E 2,..., E n over K (non-isomorphic over K), let (±1) act on E 1 E 2 E n diagonally. Lemma For n 2, there exists a curve of genus g n in (E 1 E 2 E n )/(±1) defined over K, where g n = 2 n 3 (n 4) + 1. In particular, if n = 2 or n = 3, g n = 0. By Lemma, our task is reduced to finding a hyperelliptic curve over Q in the product of elliptic curves. 14 / 34

For i = 1, 2, 3, 4, E i : yi 2 = x i (x i 1)(x i λ i ) over K. z 12 := y 1 y 2, z 13 := y 1 y 3,..., z 1n := y 1 y n are fixed under (±1). A model in (E 1 E 2 E n )/(±1) of the inverse image of P 1 can be z12 2 = x 2 (x 1) 2 (x λ 1 )(x λ 2 ) C :=. z1n 2 = x 2 (x 1) 2 (x λ 1 )(x λ n ) u1 2 = (w λ 1 )(w λ 2 ). un 1 2 = (w λ 1 )(w λ n ). via u 1 z 12 x(x 1),..., u n 1 z 1n x(x 1), w x. By the Riemann-Hurtwitz formula, 2g n 2 = 2 n 1 ( 2) + n2 n 2. 15 / 34

The idea of the proof for the quadratic case Let K = Q( m) and E : y 2 = P(x) := x 3 + αx + β and assume that α = a + c m, β = b + d m, a, b, c, d Q and cd 0, since the j-invariant is not 0 or 1728. Then, for x 0 = d c Q, we have P(x 0) Q and (x 0, ( P(x 0 )) E Q( ) P(x 0 ) E ( Q ab) E ( KQ ab). For each γ = u + v m K such that γ 4 α, γ 6 β / Q, let E γ : y 2 = P γ (x) := x 3 + γ 4 αx + γ 6 β( = K E). Then, for x γ Q such that P γ (x γ ) Q, ) ( ( (γ 2 x γ, γ 3 P γ (x γ ) E K P γ (x γ )) ) ( E KQ ab). Now, we show that there are infinitely many such quadratic Pγ ) fields L = Q( (x γ ) for γ K. 16 / 34

The idea of the proof for the quadratic case - continued Expand P γ (x) = x 3 + γ 4 αx + γ 6 β as R + I m where γ = u + v m K and R, I Q[x]. Then, I = xt 1 (u, v) + S 1 (u, v) and R = x 3 + xt 2 (u, v) + S 2 (u, v), where T i and S i are homogeneous polynomials over Q of degree 4 and 6 respectively and they satisfy the relations T i (mu, v) = m 2 T i (v, u), S i (mu, v) = m 3 S i (v, u). Let x γ = S 1(u, v). (i.e. Solve I = 0 for x.) T 1 (u, v) Then, after replacing x by x γ, P γ (x γ ) = R is reduced to Q(u, v) := T 1 (S 3 1 + S 1 T 2 1 T 2 S 2 T 3 1 ). Q 0 is homogeneous of degree 22 over Q and satisfies Q(mu, v) = m 11 Q(v, u). 17 / 34

The idea of the proof for the quadratic case - continued Lemma Let k be a non-negative integer and Q(u, v) Q[u, v] a homogeneous polynomial of degree 2(2k + 1) satisfying the functional equation Q(mu, v) = m 2k+1 Q(v, u) for a fixed squarefree integer m 1. Then Q(u, v) cannot be a perfect square in C[u, v]. By Lemma, for the homogeneous polynomial Q of degree 22, y 2 Q(u, v) is irreducible over C. 18 / 34

The idea of the proof-continued For f (t) Q[t] which is a dehomogenization of Q and for a finite extension L of Q, H(f, L) := {t Q : y 2 f (t ) is irreducible over L} the intersection with the Hilbert set of f over L, which is infinite by Hilbert irreducibility theorem. Inductively, we get ( ) ( ) L k = Q P γk (x γk ) = Q Q(u γk, v γk ) which are all linearly disjoint. Let V be the set ) V := {(γ 2 x γk, γ 3 P γk (x γk ) ) E (KL } k. k=0 Then V contains all but finitely many non-torsion points over linearly disjoint fields KL k KQ ab over K. 19 / 34

The idea of the proof for the cubic case We may assume that Q(λ) = K, where E/K : y 2 = x(x 1)(x λ). Let L(t) = t 3 at 2 + bt c = min(λ, Q). ( ) b t 2 2 2 + (t a)λ + λ 2 = M(t) L(t)λ, where M(t) := t4 2bt 2 + 8ct + b 2 4ac. 4 ( ) (x t, y t ) = M(t) b t 2 L(t), +(t a)λ+λ 2 2 L(t) N(t) 2 E(K( N(t))), where N(t) = L(t)M(t)(M(t) L(t)) is of degree 11. By specializing t in Q and applying Hilbert irreducibility to w 2 N(t), there are infinitely many points (x t, y t ) E(KQ( N(t))) E(KQ(2)) E(KQ ab ). 20 / 34

An application to quadratic twists It would be useful to consider special families of elliptic curves to understand the growth of the rank of the Mordell-Weil groups. For a given E/K : y 2 = x 3 + ax + b over a number field K and d K x, the quadratic twist of E by d is defined by E d /K : dy 2 = x 3 + ax + b. Theorem rk ( E(K( ) d)) = rk(e(k)) + rk(e d (K)) 21 / 34

Goldfeld s conjecture over Q Let S(X ) = {square free d Z : d X }. (Goldfeld s conjecture) the average rank lim X d S(X ) rank(e d) = 1/2 #S(X ) Goldfeld s conjecture + Parity Conjecture imply : The set of elliptic curves with rank 0 has density 1/2. The set of elliptic curves with rank 1 has density 1/2. The set of elliptic curves with rank 2 has density 0. 22 / 34

Root number of products of Quadratic Twists Theorem (Rohrlich) For E/Q and for distinct square-free integers d, d relatively prime to the conductor of E, w Edd = w Ed w Ed w E. So by Goldfeld s conjecture over Q and the Parity Conjecture, if rank(e(q)) is even, then even if rank(e d (Q)) = 1 =rank(e d (Q)), we might expect that rank(e dd (Q)) = 0 for many square-free d, d Z. But, there is no 1/2-average-rank conjecture over arbitrary number fields, in general. T. Dokchitser & V. Dokchitser : There exists an elliptic curve over a number field K ( Q) all of whose quadratic twists have positive rank over K under the B-SD conjecture. 23 / 34

Products of two twists (Wong) If E/K has non-zero j-invariant, then for any δ K x, there exist infinitely many d K x with pairwise distinct modulo (K x ) 2 such that rke d (K) > 0 and rke δd (K) > 0. How many pairs (d, d ) satisfy that E d, E d, E dd are of positive rank? (I-. & Lozano-Robledo) (1) If the Parity Conjecture holds, then for any E/Q, there are infinitely many square-free (d, d, dd ) Q Q Q component-wise distinct modulo (Q x ) 2 such that rke d (Q) > 0, rke d (Q) > 0, and rke dd (Q) > 0. (2) For a number field K, without the Parity Conjecture, there exist wide various families of elliptic curves E/K such that there are infinitely many (d, d, dd ) K K K component-wise distinct modulo (K x ) 2 such that rke d (K) > 0, rke d (K) > 0, and rke dd (K) > 0. 24 / 34

Products of more than two twists A generalization from E d1 d 2 of positive rank to E d1 d 2 d n of positive rank: Theorem (I-. & Larsen) Let E be an elliptic curve over Q for which the set of quadratic twists with positive rank has positive density. Then for every n N there exists a d Q /Q 2 and an n-dimensional subspace V of Q /Q 2 such that for all d V, the quadratic twist E d d = (E d ) d has positive rank. (V. Vatsal) For E = X 0 (19), the set of positive integers d such that E d has rank 1 has positive density. 25 / 34

The Main ingredients of the proof the lower density d(s) = lim inf n the upper density D(S) = lim sup T S [1, n] n ( ) 1 1 e nt dt. T 1/T n S Lemma For a subset S N, D(S) d(s). Lemma If S is a subset of N with D(S) > 0 and N N, there exist primes p > q > N such that for S(p, q) = {n N : pn, qn S}, D(S(p, q)) > 0. 26 / 34

The Main ingredients of the proof - continued Theorem For each n N, any subset of N of positive (lower) density contains a subset consisting of c a i for every I {1, 2,..., n} i I for some c, a i Q such that a 1,..., a n are linearly independent in Q /(Q ) 2. 27 / 34

Related results on pairs of two elliptic curves For two non-isogenous elliptic curves E 1 /Q and E 2 /Q, (Kuwata, Wang) if j-invariants of E 1 and E 2 are not equal to 1728 or 0, then there exist infinitely many square-free integers d such that rk((e 1 ) d (Q)) > 0 and rk((e 2 ) d (Q)) > 0. (Coogan, Jiménez-Urroz) if 2-torsion points of E 1 and E 2 are not Q-rational, then there exist infinitely many square-free integers d such that rk((e 1 ) d (Q)) = 0 =rk((e 2 ) d (Q)). (Coogan, Jiménez-Urroz) if the conductors of E 1 and E 2 are coprime, then assuming the B-SD conjecture, there exist infinitely many square-free integers d such that rk((e 1 ) d (Q)) = 0 and rk((e 2 ) d (Q)) > 0. 28 / 34

An application of the main theorem to quadratic twists Simultaneous twists of more than two elliptic curves Theorem (1) For i = 1, 2, 3, 4 let E i be an elliptic curve defined over a number field K. Then there exists a number field L containing K such that there are infinitely many square-free d L x /(L x ) 2 such that all ((E i )( d )(L) for i = 1, 2, 3, 4 are of positive rank, equivalently, rank E i (L( ) d)) >rank(e i (L)) for all i = 1, 2, 3, 4. Theorem (2) For i = 1, 2, 3 let E i be an elliptic curve defined over a cubic extension K of Q. If every point of order 2 of E i ( K) lies in E i (K), then there are infinitely many square-free integers d such that all ((E i ) d )(K) for i = 1, 2, 3 are of positive rank. 29 / 34

Proof of Theorem 1 Let (±1) act on E 1 E 2 E 3 E 4 diagonally. Recall the following lemma. Lemma For n 2, there exists a curve of genus g n in (E 1 E 2 E n )/(±1) defined over K, where In particular, if n = 4, g n = 1. g n = 2 n 3 (n 4) + 1. 30 / 34

Proof of Theorem 1 By Lemma, there exists an elliptic curve C in (E 1 E 2 E 3 E 4 )/(±1) defined over a finite extension L of K, moreover C with a L-rational non-torsion point P. Since the quotient map π by (±1) is of degree 2 and the inverse image C of C under π has genus > 1, by Falting s theorem, for P k := kp for each k Z and R := {π 1 (P k ) : k Z}, there are infinitely many k Z such the fiber π 1 (P k ) R is not defined over L but defined over a quadratic extension L( d k ) for some d k L. Therefore, there exist infinitely many d k L such that ( rank E i (L( ) d k )) > rank(e i (L)), ( equivalently, rank E d k i ) (L) 1 for i = 1, 2, 3, 4. 31 / 34

Proof of Theorem 2 For E i : y 2 = x(x 1)(x λ i ) for i = 1, 2, 3, Case 1: Q(λ i ) = K for all i Case 2: Q(λ i ) = Q for all i. Case 3: Q(λ 1 ) = Q and Q(λ i ) = K for i = 2, 3, Case 4: Q(λ 1 ) = K and Q(λ i ) = Q for i = 2, 3. Let L(t) be the minimal polynomial of λ 1 over Q for Case 1, L(t) = (t λ 1 )(t λ 2 )(t λ 3 ) over Q for Case 2, L(t) = (t λ 1 ) 2 f (t) for Case 3, and L(t) := (t λ 2 )(t λ 3 )f (t) for Case 4, where f (t) is the minimal polynomial of λ 2 over Q. 32 / 34

Proof of Theorem 2 Then in all cases, there exists M(t) Q[t] and g i (t) K(t) such that d(t) = L(t)M(t)(L(t) M(t)) Q[t] and (x i (t), y i (t)) = ( M(t) L(t), g ) i(t) L(t) 2 E d(t) i (K(t)). By specializing at t, we obtain an infinite sequence of square-free integers d t such that the rank of E dt i (K) is positive. 33 / 34

Thank you very much! 34 / 34