10 Minute Postdoc Introductions ICERM Computational Aspects of the Langlands Program September 11, 2015
The Mordell-Weil Problem and its Inverse Let E be an abelian variety defined over a number field K. Theorem There exists a finite abelian group E(K ) tors and a non-negative integer r = r(e, K ) such that E(K ) E(K ) tors Z r.
The Mordell-Weil Problem and its Inverse Let E be an abelian variety defined over a number field K. Theorem There exists a finite abelian group E(K ) tors and a non-negative integer r = r(e, K ) such that E(K ) E(K ) tors Z r. Problem Given E/K, compute E(K ) tors and r = r(e, K ).
The Mordell-Weil Problem and its Inverse Let E be an abelian variety defined over a number field K. Theorem There exists a finite abelian group E(K ) tors and a non-negative integer r = r(e, K ) such that E(K ) E(K ) tors Z r. Problem Given E/K, compute E(K ) tors and r = r(e, K ). Given dim E and [K : Q], which pairs (E(K ) tors, r) occur?
The Mordell-Weil Problem and its Inverse Let E be an abelian variety defined over a number field K. Theorem There exists a finite abelian group E(K ) tors and a non-negative integer r = r(e, K ) such that E(K ) E(K ) tors Z r. Problem Given E/K, compute E(K ) tors and r = r(e, K ). Given dim E and [K : Q], which pairs (E(K ) tors, r) occur? Examples Do there exist infinitely many elliptic curves over Q with r(e, Q) at least 20? At least 22?
The Mordell-Weil Problem and its Inverse Let E be an abelian variety defined over a number field K. Theorem There exists a finite abelian group E(K ) tors and a non-negative integer r = r(e, K ) such that E(K ) E(K ) tors Z r. Problem Given E/K, compute E(K ) tors and r = r(e, K ). Given dim E and [K : Q], which pairs (E(K ) tors, r) occur? Examples Do there exist infinitely many elliptic curves over Q with r(e, Q) at least 20? At least 22? Does there exist a single elliptic curve over Q with E(Q) Z/2Z Z/8Z Z 4?
Bounding r(e, K ) via Selmer Groups For each integer m 2, we have r(e, K ) = log m [E(K ) : me(k )] [E(K ) tors : me(k ) tors ].
Bounding r(e, K ) via Selmer Groups For each integer m 2, we have r(e, K ) = log m [E(K ) : me(k )] [E(K ) tors : me(k ) tors ]. Galois cohomology gives an exact sequence 0 E(K )/me(k ) δ Sel (m) (E/K ) III(E/K )[m] 0.
Bounding r(e, K ) via Selmer Groups For each integer m 2, we have r(e, K ) = log m [E(K ) : me(k )] [E(K ) tors : me(k ) tors ]. Galois cohomology gives an exact sequence 0 E(K )/me(k ) δ Sel (m) (E/K ) III(E/K )[m] 0. There is a finite set Σ = Σ(E, m, K ) of places of K so that Sel (m) (E/K ) = ker [ H 1 (K, E[m]; Σ) ν Σ ] H 1 (K ν, E[m]) [ δ ν E(Kν ) ].
Number Field Method of Descent Find a ring R and a group S (m) (E/K ) such that H 1 (K, E[m]; Σ) S (m) (E/K ) S (m) (R), in terms of the group [ Quot(R) S (m) (R) = ker I(R) (Quot(R) ) m I(R) m ].
Number Field Method of Descent Find a ring R and a group S (m) (E/K ) such that H 1 (K, E[m]; Σ) S (m) (E/K ) S (m) (R), in terms of the group [ Quot(R) S (m) (R) = ker I(R) (Quot(R) ) m I(R) m The group S (m) (R) fits in the exact sequence ]. 1 R /(R ) m S (m) (R) Cl(R)[m] 1.
Number Field Method of Descent Find a ring R and a group S (m) (E/K ) such that H 1 (K, E[m]; Σ) S (m) (E/K ) S (m) (R), in terms of the group [ Quot(R) S (m) (R) = ker I(R) (Quot(R) ) m I(R) m The group S (m) (R) fits in the exact sequence ]. 1 R /(R ) m S (m) (R) Cl(R)[m] 1.
Number Field Method of Descent Find a ring R and a group S (m) (E/K ) such that H 1 (K, E[m]; Σ) S (m) (E/K ) S (m) (R), in terms of the group [ Quot(R) S (m) (R) = ker I(R) (Quot(R) ) m I(R) m The group S (m) (R) fits in the exact sequence ]. 1 R /(R ) m S (m) (R) Cl(R)[m] 1. Example Suppose m = 2 and E : Y 2 = X 3 + AX + B. Denote the K -algebra L = K [t]/(t 3 + At + B). If L is a field, then take R = O L,Σ and S (2) (E/K ) = {d S (2) (R) : N L/K (d) (K ) 2 }.
Sel (m) (E/K ) H 1 (G K, E[m]; Σ) S (m) (E/K ) δ E(K ) me(k ) δ ν Res Gν ν Σ E(K ν ) me(k ν ) ν δν ν Σ H 1 (G ν, E[m]) [ δ ν E(Kν ) ] α (Z/mZ) N ν Σ Sel (m) (E/K ) ker(β) [ im(α) im(β) ] hc β
Other directions Computation of Sel (2) (E/Q) for all E/Q with E[2] E(Q) and good reduction outside {2, l} for any odd prime l.
Other directions Computation of Sel (2) (E/Q) for all E/Q with E[2] E(Q) and good reduction outside {2, l} for any odd prime l. Heegner Point Challenge: The elliptic curve E : y 2 = x(x 2)(x 2 32 ) has rank 1. Find a generator to (probably) break a record.
Other directions Computation of Sel (2) (E/Q) for all E/Q with E[2] E(Q) and good reduction outside {2, l} for any odd prime l. Heegner Point Challenge: The elliptic curve E : y 2 = x(x 2)(x 2 32 ) has rank 1. Find a generator to (probably) break a record. (with Davis and Goins) Explicitly visualizing elements of H 1 (K, E[2]).
Other directions Computation of Sel (2) (E/Q) for all E/Q with E[2] E(Q) and good reduction outside {2, l} for any odd prime l. Heegner Point Challenge: The elliptic curve E : y 2 = x(x 2)(x 2 32 ) has rank 1. Find a generator to (probably) break a record. (with Davis and Goins) Explicitly visualizing elements of H 1 (K, E[2]). (with Balakrishnan, Ho, Kaplan, Spicer, and Stein) A database of elliptic curves by naive height.
Other directions Computation of Sel (2) (E/Q) for all E/Q with E[2] E(Q) and good reduction outside {2, l} for any odd prime l. Heegner Point Challenge: The elliptic curve E : y 2 = x(x 2)(x 2 32 ) has rank 1. Find a generator to (probably) break a record. (with Davis and Goins) Explicitly visualizing elements of H 1 (K, E[2]). (with Balakrishnan, Ho, Kaplan, Spicer, and Stein) A database of elliptic curves by naive height. (with Bober and Goins) Search for E(Q) Z/2 Z/8 Z 4.
Other directions Computation of Sel (2) (E/Q) for all E/Q with E[2] E(Q) and good reduction outside {2, l} for any odd prime l. Heegner Point Challenge: The elliptic curve E : y 2 = x(x 2)(x 2 32 ) has rank 1. Find a generator to (probably) break a record. (with Davis and Goins) Explicitly visualizing elements of H 1 (K, E[2]). (with Balakrishnan, Ho, Kaplan, Spicer, and Stein) A database of elliptic curves by naive height. (with Bober and Goins) Search for E(Q) Z/2 Z/8 Z 4. Let m, n Z satisfy 0 < m < n 5000. Assume BSD + GRH. The rank of E m,n : Y 2 = X(X + (2mn) 4 )(X + (n 2 m 2 ) 4 ) is 3.
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