Elliptische Kurven mit Primzahlführer

Abstract

A besitze keinen Q-rationalen Punkt der (genauen) Ordnung 2 und habe den Fuhrer f (A) = p . ( V )

Elliptic curves of conductor 11

“…While the general question seems shrouded in mystery and quite inaccessible at present, one can at least try to verify the conjecture for small numerical values of N. A considerable amount of work has already been done in this direction (cf. [4], [5], [19] - [24], [29]). However, the first nontrivial case of the conjecture, namely TV = 11, has not previously been settled.…”

“…Setzer [29], and subsequently Boiling [5], consider the case where E has no rational point of order 2 but has prime conductor p. Then the 2-division field Q(E2) generated by the coordinates of the 2-division points of E over Q, is a Galois extension of Q with Galois group S3, and is unramified at all primes distinct from 2 and p. This yields only finitely many possibilities for Q(E2) and yields elliptic curves/? («, v), where («, v) is an integer solution of a diophantine equation /(«, v) = ± 2ep* for certain cubic forms / (depending, as does e, only on p).…”

“…Of course it is easier to show that a diophantine equation has no solutions (or, as in the present case, no solutions satisfying certain additional conditions) than it is to find all solutions when nontrivial solutions exist. Thus Boiling [5] has completed work of Setzer [29] and has found all primes p < 401 for which there are no elliptic curves defined over Q of conductor p. On the other hand, in those cases where there are curves without a rational 2-division point it has not proved possible to determine whether all solutions have been detected. However, we know (see Coates [7] or [30]) that a Thue-Mahler equation has at most finitely many solutions and that all solutions can in principle be effectively found by using Baker's method.…”

“…Let E be an elliptic curve over Q with conductor 11, and minimal model (1) with discriminant ± llr. Setzer [29] (or see [5]) shows that the 2-division field Q(E2) contains a sub field K cubic over Q, with discriminant Z) = ± 11 or ±44. A table of cubic fields (see for example [1]) shows that D = -44 is the only possibility, and that K is generated over Q by a zero of 3x3 -x2 +x + 1.…”