Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X .
Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X . Idea: If E : y 2 = x 3 + Ax + B ,
Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X . Idea: If E : y 2 = x 3 + Ax + B , choose any x ∈ Q .
Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X . Idea: If E : y 2 = x 3 + Ax + B , choose any x ∈ Q . If x = u / v for coprime u , v ,
Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X . Idea: If E : y 2 = x 3 + Ax + B , choose any x ∈ Q . If x = u / v for coprime u , v , then v 4 y 2 = v ( u 3 + Auv 2 + Bv 3 ) .
Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X . Idea: If E : y 2 = x 3 + Ax + B , choose any x ∈ Q . If x = u / v for coprime u , v , then v 4 y 2 = v ( u 3 + Auv 2 + Bv 3 ) . Choosing | u | , | v | ≤ X 1 / 4
Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X . Idea: If E : y 2 = x 3 + Ax + B , choose any x ∈ Q . If x = u / v for coprime u , v , then v 4 y 2 = v ( u 3 + Auv 2 + Bv 3 ) . Choosing | u | , | v | ≤ X 1 / 4 = ⇒ | RHS | ≤ X .
Rank two quadratic twists Theorem (Gouvêa–Mazur) √ rk an ( E ) grows by ≥ 2 in ≫ X 1 / 2 − ǫ fields Q ( d ) with | d | ≤ X . Idea: If E : y 2 = x 3 + Ax + B , choose any x ∈ Q . If x = u / v for coprime u , v , then v 4 y 2 = v ( u 3 + Auv 2 + Bv 3 ) . Choosing | u | , | v | ≤ X 1 / 4 = ⇒ | RHS | ≤ X . Problem � v ( u 3 + Auv 2 + Bv 3 )) ? How do we distinguish the fields Q (
Distinguishing quadratic fields Problem � v ( u 3 + Auv 2 + Bv 3 )) ? How do we distinguish the fields Q (
Distinguishing quadratic fields Problem � v ( u 3 + Auv 2 + Bv 3 )) ? How do we distinguish the fields Q ( Gouvêa and Mazur show that: 1 v ( u 3 + Auv 2 + Bv 3 ) assumes ≫ X 1 / 2 squarefree values ≤ X ,
Distinguishing quadratic fields Problem � v ( u 3 + Auv 2 + Bv 3 )) ? How do we distinguish the fields Q ( Gouvêa and Mazur show that: 1 v ( u 3 + Auv 2 + Bv 3 ) assumes ≫ X 1 / 2 squarefree values ≤ X , 2 any particular value arises ≪ X ǫ times.
Distinguishing quadratic fields Problem � v ( u 3 + Auv 2 + Bv 3 )) ? How do we distinguish the fields Q ( Gouvêa and Mazur show that: 1 v ( u 3 + Auv 2 + Bv 3 ) assumes ≫ X 1 / 2 squarefree values ≤ X , 2 any particular value arises ≪ X ǫ times. ⇒ rk ( E ) grows in at least X 1 / 2 − ǫ fields K ∈ F 2 ( X ) . =
Distinguishing quadratic fields Problem � v ( u 3 + Auv 2 + Bv 3 )) ? How do we distinguish the fields Q ( Gouvêa and Mazur show that: 1 v ( u 3 + Auv 2 + Bv 3 ) assumes ≫ X 1 / 2 squarefree values ≤ X , 2 any particular value arises ≪ X ǫ times. ⇒ rk ( E ) grows in at least X 1 / 2 − ǫ fields K ∈ F 2 ( X ) . = Get growth ≥ 2 of rk an ( E ) by controlling the root number.
First parametrization in higher degree fields
First parametrization in higher degree fields If E : y 2 = f ( x ) and n is even,
First parametrization in higher degree fields If E : y 2 = f ( x ) and n is even, then ( x , tx n / 2 ) is a point on E ( K t ) ,
First parametrization in higher degree fields If E : y 2 = f ( x ) and n is even, then ( x , tx n / 2 ) is a point on E ( K t ) , where P f ( x , t ) := t 2 x n − f ( x ) . K t := Q ( t )[ x ] / P f ( x , t ) ,
First parametrization in higher degree fields If E : y 2 = f ( x ) and n is even, then ( x , tx n / 2 ) is a point on E ( K t ) , where P f ( x , t ) := t 2 x n − f ( x ) . K t := Q ( t )[ x ] / P f ( x , t ) , Proposition There is a model E : y 2 = f ( x ) s.t. Gal ( � K t / Q ( t )) ≃ S n .
First parametrization in higher degree fields If E : y 2 = f ( x ) and n is even, then ( x , tx n / 2 ) is a point on E ( K t ) , where P f ( x , t ) := t 2 x n − f ( x ) . K t := Q ( t )[ x ] / P f ( x , t ) , Proposition There is a model E : y 2 = f ( x ) s.t. Gal ( � K t / Q ( t )) ≃ S n . Get many K ∈ F n ( X ; S n ) in which rk ( E ) grows by specializing t
First parametrization in higher degree fields If E : y 2 = f ( x ) and n is even, then ( x , tx n / 2 ) is a point on E ( K t ) , where P f ( x , t ) := t 2 x n − f ( x ) . K t := Q ( t )[ x ] / P f ( x , t ) , Proposition There is a model E : y 2 = f ( x ) s.t. Gal ( � K t / Q ( t )) ≃ S n . Get many K ∈ F n ( X ; S n ) in which rk ( E ) grows by specializing t , provided we can control multiplicities!
First parametrization: Controlling multiplicities ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .)
First parametrization: Controlling multiplicities ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma If t = u / v , then Disc x ( P f ( x , u / v )) = u 2 n − 4 v 4 − 2 n H ( u , v ) for a not-squarefull sextic form H ( u , v ) .
First parametrization: Controlling multiplicities ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma If t = u / v , then Disc x ( P f ( x , u / v )) = u 2 n − 4 v 4 − 2 n H ( u , v ) for a not-squarefull sextic form H ( u , v ) . Theorem (Greaves) Any “not obstructed” form H ( u , v ) of degree ≤ 6 assumes ≫ T 2 squarefree values with | u | , | v | ≤ T .
First parametrization: Controlling multiplicities ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma If t = u / v , then Disc x ( P f ( x , u / v )) = u 2 n − 4 v 4 − 2 n H ( u , v ) for a not-squarefull sextic form H ( u , v ) . Theorem (Greaves) Any “not obstructed” form H ( u , v ) of degree ≤ 6 assumes ≫ T 2 squarefree values with | u | , | v | ≤ T . Each value occurs ≪ X ǫ times
First parametrization: Controlling multiplicities ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma If t = u / v , then Disc x ( P f ( x , u / v )) = u 2 n − 4 v 4 − 2 n H ( u , v ) for a not-squarefull sextic form H ( u , v ) . Theorem (Greaves) Any “not obstructed” form H ( u , v ) of degree ≤ 6 assumes ≫ T 2 squarefree values with | u | , | v | ≤ T . Each value occurs ≪ X ǫ times = ⇒ there are ≫ X 2 / ( n + 4 ) − ǫ fields K t with | Disc ( K t ) | ≤ X .
How do we control root numbers? ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .)
How do we control root numbers? ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma (V. Dokchitser) If K and K ′ ∈ F n ( X ; S n ) are such that
How do we control root numbers? ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma (V. Dokchitser) If K and K ′ ∈ F n ( X ; S n ) are such that • K ⊗ Q p ≃ K ′ ⊗ Q p for each p | N E ,
How do we control root numbers? ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma (V. Dokchitser) If K and K ′ ∈ F n ( X ; S n ) are such that • K ⊗ Q p ≃ K ′ ⊗ Q p for each p | N E , and • sgn ( Disc ( K )) = − sgn ( Disc ( K ′ )) ,
How do we control root numbers? ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma (V. Dokchitser) If K and K ′ ∈ F n ( X ; S n ) are such that • K ⊗ Q p ≃ K ′ ⊗ Q p for each p | N E , and • sgn ( Disc ( K )) = − sgn ( Disc ( K ′ )) , then w ( E , ρ K ) = − w ( E , ρ K ′ ) .
How do we control root numbers? ( Recall: P f ( x , t ) = x n t 2 − f ( x ) and K t = Q ( t )[ x ] / P f ( x , t ) .) Lemma (V. Dokchitser) If K and K ′ ∈ F n ( X ; S n ) are such that • K ⊗ Q p ≃ K ′ ⊗ Q p for each p | N E , and • sgn ( Disc ( K )) = − sgn ( Disc ( K ′ )) , then w ( E , ρ K ) = − w ( E , ρ K ′ ) . Theorem The number of K ∈ F n ( X ; S n ) s.t. rk ( E ( K )) > rk ( E ( Q )) and w ( E , ρ K ) = + 1 is ≫ X 1 / ( ⌈ n 2 ⌉ + 2 ) − ǫ .
Second Parametrization If n is even,
Second Parametrization If n is even, let F , G ∈ Z [ x ] have degree n / 2 and n / 2 − 2, resp.
Second Parametrization If n is even, let F , G ∈ Z [ x ] have degree n / 2 and n / 2 − 2, resp. Then ( x , F ( x ) G ( x ) ) is on E ( K F , G ) ,
Second Parametrization If n is even, let F , G ∈ Z [ x ] have degree n / 2 and n / 2 − 2, resp. Then ( x , F ( x ) G ( x ) ) is on E ( K F , G ) , where K F , G = Q [ x ] / ( F 2 − fG 2 ) .
Second Parametrization If n is even, let F , G ∈ Z [ x ] have degree n / 2 and n / 2 − 2, resp. Then ( x , F ( x ) G ( x ) ) is on E ( K F , G ) , where K F , G = Q [ x ] / ( F 2 − fG 2 ) . Lemma Gal ( � K F , G / Q ) ≃ S n for almost all F , G .
Second Parametrization If n is even, let F , G ∈ Z [ x ] have degree n / 2 and n / 2 − 2, resp. Then ( x , F ( x ) G ( x ) ) is on E ( K F , G ) , where K F , G = Q [ x ] / ( F 2 − fG 2 ) . Lemma Gal ( � K F , G / Q ) ≃ S n for almost all F , G . Proof. If F ( x ) = tx n / 2 and G ( x ) = 1,
Second Parametrization If n is even, let F , G ∈ Z [ x ] have degree n / 2 and n / 2 − 2, resp. Then ( x , F ( x ) G ( x ) ) is on E ( K F , G ) , where K F , G = Q [ x ] / ( F 2 − fG 2 ) . Lemma Gal ( � K F , G / Q ) ≃ S n for almost all F , G . Proof. If F ( x ) = tx n / 2 and G ( x ) = 1, then K F , G = K t .
Second Parametrization If n is even, let F , G ∈ Z [ x ] have degree n / 2 and n / 2 − 2, resp. Then ( x , F ( x ) G ( x ) ) is on E ( K F , G ) , where K F , G = Q [ x ] / ( F 2 − fG 2 ) . Lemma Gal ( � K F , G / Q ) ≃ S n for almost all F , G . Proof. If F ( x ) = tx n / 2 and G ( x ) = 1, then K F , G = K t . Now use Hilbert Irreducibility.
Second Parametrization: Controlling Multiplicities Question How do we control the multiplicity of K F , G = Q [ x ] / ( F 2 − fG 2 ) ?
Second Parametrization: Controlling Multiplicities Question How do we control the multiplicity of K F , G = Q [ x ] / ( F 2 − fG 2 ) ? Lemma Given f , H ∈ Z [ x ] , there are O n ( 1 ) solutions F , G to H = F 2 − fG 2 .
Second Parametrization: Controlling Multiplicities Question How do we control the multiplicity of K F , G = Q [ x ] / ( F 2 − fG 2 ) ? Lemma Given f , H ∈ Z [ x ] , there are O n ( 1 ) solutions F , G to H = F 2 − fG 2 . Question How do we make sure the same field isn’t cut out by lots of polynomials?
Multiplicities of fields Let S n ( Y ) := { g ( x ) = x n + a 1 x n − 1 + · · · + a n ∈ Z [ x ] : | a i | ≤ Y i } .
Multiplicities of fields Let S n ( Y ) := { g ( x ) = x n + a 1 x n − 1 + · · · + a n ∈ Z [ x ] : | a i | ≤ Y i } . ( Note: If g ∈ S n ( Y ) , then | Disc ( g ) | ≪ Y n ( n − 1 ) .)
Multiplicities of fields Let S n ( Y ) := { g ( x ) = x n + a 1 x n − 1 + · · · + a n ∈ Z [ x ] : | a i | ≤ Y i } . ( Note: If g ∈ S n ( Y ) , then | Disc ( g ) | ≪ Y n ( n − 1 ) .) Lemma (Ellenberg–Venkatesh + ǫ · (LO–Thorne)) If K ∈ F n ( X ) , then # { g ∈ S n ( Y ) : Q [ x ] / g ≃ K }
Multiplicities of fields Let S n ( Y ) := { g ( x ) = x n + a 1 x n − 1 + · · · + a n ∈ Z [ x ] : | a i | ≤ Y i } . ( Note: If g ∈ S n ( Y ) , then | Disc ( g ) | ≪ Y n ( n − 1 ) .) Lemma (Ellenberg–Venkatesh + ǫ · (LO–Thorne)) If K ∈ F n ( X ) , then � Y n Disc ( K ) − 1 / 2 , Y n / 2 � # { g ∈ S n ( Y ) : Q [ x ] / g ≃ K } ≪ max .
Multiplicities of fields Let S n ( Y ) := { g ( x ) = x n + a 1 x n − 1 + · · · + a n ∈ Z [ x ] : | a i | ≤ Y i } . ( Note: If g ∈ S n ( Y ) , then | Disc ( g ) | ≪ Y n ( n − 1 ) .) Lemma (Ellenberg–Venkatesh + ǫ · (LO–Thorne)) If K ∈ F n ( X ) , then � Y n Disc ( K ) − 1 / 2 , Y n / 2 � # { g ∈ S n ( Y ) : Q [ x ] / g ≃ K } ≪ max . = ⇒ # {| Disc ( K F , G ) | ≤ X } / iso .
Multiplicities of fields Let S n ( Y ) := { g ( x ) = x n + a 1 x n − 1 + · · · + a n ∈ Z [ x ] : | a i | ≤ Y i } . ( Note: If g ∈ S n ( Y ) , then | Disc ( g ) | ≪ Y n ( n − 1 ) .) Lemma (Ellenberg–Venkatesh + ǫ · (LO–Thorne)) If K ∈ F n ( X ) , then � Y n Disc ( K ) − 1 / 2 , Y n / 2 � # { g ∈ S n ( Y ) : Q [ x ] / g ≃ K } ≪ max . 4 − n 2 + 4 n − 2 1 2 n 2 ( n − 1 ) . = ⇒ # {| Disc ( K F , G ) | ≤ X } / iso . ≫ X
The limit of the method Theorem Let E / Q be an elliptic curve.
The limit of the method Theorem Let E / Q be an elliptic curve. If for each K ∈ F n ( X ; S n ) , • L ( s , E K ) is automorphic,
The limit of the method Theorem Let E / Q be an elliptic curve. If for each K ∈ F n ( X ; S n ) , • L ( s , E K ) is automorphic, • L ( s , E K ) satisfies GRH,
The limit of the method Theorem Let E / Q be an elliptic curve. If for each K ∈ F n ( X ; S n ) , • L ( s , E K ) is automorphic, • L ( s , E K ) satisfies GRH, and • L ( s , E K ) satisfies BSD,
The limit of the method Theorem Let E / Q be an elliptic curve. If for each K ∈ F n ( X ; S n ) , • L ( s , E K ) is automorphic, • L ( s , E K ) satisfies GRH, and • L ( s , E K ) satisfies BSD, then # { K ∈ F n ( X ; S n ) : rk ( E ( K )) ≥ 2 + rk ( E ( Q )) }
The limit of the method Theorem Let E / Q be an elliptic curve. If for each K ∈ F n ( X ; S n ) , • L ( s , E K ) is automorphic, • L ( s , E K ) satisfies GRH, and • L ( s , E K ) satisfies BSD, then 1 1 4 + 2 ( n 2 − n ) . # { K ∈ F n ( X ; S n ) : rk ( E ( K )) ≥ 2 + rk ( E ( Q )) } ≫ X
But what’s the truth?
But what’s the truth? Conjecture (Birch–Swinnerton-Dyer) If r = rk ( E ) , then r = ord s = 1 L ( s , E ) and L ( r ) ( 1 , E ) = | X ( E ) | Reg ( E ) Tam ( E )Ω R ( E ) . | E ( Q ) tors | 2 r !
But what’s the truth? Conjecture (Birch–Swinnerton-Dyer) If r = rk ( E ) , then r = ord s = 1 L ( s , E ) and L ( r ) ( 1 , E ) = | X ( E ) | Reg ( E ) Tam ( E )Ω R ( E ) . | E ( Q ) tors | 2 r ! Conjecture (Tate’s Séminaire Bourbaki) If K / Q has sig. ( r 1 , r 2 ) and r = rk ( E K ) ,
But what’s the truth? Conjecture (Birch–Swinnerton-Dyer) If r = rk ( E ) , then r = ord s = 1 L ( s , E ) and L ( r ) ( 1 , E ) = | X ( E ) | Reg ( E ) Tam ( E )Ω R ( E ) . | E ( Q ) tors | 2 r ! Conjecture (Tate’s Séminaire Bourbaki) If K / Q has sig. ( r 1 , r 2 ) and r = rk ( E K ) , then r = ord s = 1 L ( s , E K ) and L ( r ) ( 1 , E K ) = | X ( E K ) | Reg ( E K ) Tam ( E K )Ω R ( E ) r 1 Ω C ( E ) r 2 . r ! | Disc ( K ) | 1 / 2 | E ( K ) tors | 2
But what’s the truth? Conjecture (Birch–Swinnerton-Dyer) If r = rk ( E ) , then r = ord s = 1 L ( s , E ) and L ( r ) ( 1 , E ) = | X ( E ) | Reg ( E ) Tam ( E )Ω R ( E ) . | E ( Q ) tors | 2 r ! Conjecture (Tate’s Séminaire Bourbaki) If K / Q has sig. ( r 1 , r 2 ) and r = rk ( E K ) , then r = ord s = 1 L ( s , E K ) and L ( r ) ( 1 , E K ) = | X ( E K ) | Reg ( E K ) Tam ( E K )Ω R ( E ) r 1 Ω C ( E ) r 2 . r ! | Disc ( K ) | 1 / 2 | E ( K ) tors | 2 Idea: Pay attention to the case when rk ( E Q ) = rk ( E K ) .
Relative rank zero BSD Conjecture Let L ( s , E , ρ K ) = L ( s , E K ) / L ( s , E ) .
Relative rank zero BSD Conjecture Let L ( s , E , ρ K ) = L ( s , E K ) / L ( s , E ) . If E ( K ) = E ( Q ) , then Ω R ( E ) r 1 − 1 Ω C ( E ) r 2 L ( 1 , E , ρ K ) = | X ( E K ) | Tam ( E K ) . | Disc ( K ) | 1 / 2 | X ( E ) | Tam ( E )
Relative rank zero BSD Conjecture Let L ( s , E , ρ K ) = L ( s , E K ) / L ( s , E ) . If E ( K ) = E ( Q ) , then Ω R ( E ) r 1 − 1 Ω C ( E ) r 2 L ( 1 , E , ρ K ) = | X ( E K ) | Tam ( E K ) . | Disc ( K ) | 1 / 2 | X ( E ) | Tam ( E ) Expect: L ( 1 , E , ρ K ) , Tam ( E K ) ≪ ( ht ( E ) | Disc ( K ) | ) ǫ =: Q ǫ ,
Relative rank zero BSD Conjecture Let L ( s , E , ρ K ) = L ( s , E K ) / L ( s , E ) . If E ( K ) = E ( Q ) , then Ω R ( E ) r 1 − 1 Ω C ( E ) r 2 L ( 1 , E , ρ K ) = | X ( E K ) | Tam ( E K ) . | Disc ( K ) | 1 / 2 | X ( E ) | Tam ( E ) Expect: L ( 1 , E , ρ K ) , Tam ( E K ) ≪ ( ht ( E ) | Disc ( K ) | ) ǫ =: Q ǫ , so | Disc ( K ) | 1 / 2 | X ( E K ) | Ω R ( E ) r 1 − 1 Ω C ( E ) r 2 Q ǫ . | X ( E ) | ≪
Relative rank zero BSD Conjecture Let L ( s , E , ρ K ) = L ( s , E K ) / L ( s , E ) . If E ( K ) = E ( Q ) , then Ω R ( E ) r 1 − 1 Ω C ( E ) r 2 L ( 1 , E , ρ K ) = | X ( E K ) | Tam ( E K ) . | Disc ( K ) | 1 / 2 | X ( E ) | Tam ( E ) Expect: L ( 1 , E , ρ K ) , Tam ( E K ) ≪ ( ht ( E ) | Disc ( K ) | ) ǫ =: Q ǫ , so | Disc ( K ) | 1 / 2 | X ( E K ) | Ω R ( E ) r 1 − 1 Ω C ( E ) r 2 Q ǫ . | X ( E ) | ≪ Crude model: | X ( E K ) / X ( E ) | = m 2 uniformly with | Disc ( K ) | 1 / 4 Q ǫ m ≪ . r 1 − 1 r 2 2 Ω C ( E ) Ω R ( E ) 2
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