2
$\begingroup$

In page 109 of de Weger's paper, he says that for coprime $A, B, C$ the conductor $N$ of the Frey-Hellegouarch curve $ E: y^2 = x(x - A)(x + B) $ equals $N(A,B,C)$ (product of primes dividing $ABC$ without multiplicity, and where $A + B = C$) times an absolutely bounded power of $2$. Why is this the case?

Also, on page 114, he says that the conductor $N_q$ of the twisted curve

$ E_q : qy^2 = x(x - A)(x+B) $

where $q$ is a squarefree integer (ie the quadratic twist of $E$), is $lcm(N,q^2)$ and the difference in the power of $2$ is at most $2^8$. Why is this the case?

Thanks

1 Answers 1

4
  • The computation of the conductor of the Frey curve can be found in one of the early chapters (maybe the first) of Cornell--Silverman--Stevens.

  • The highest power of $2$ that can divide the conductor of an elliptic curve is $8$, if I remember correctly.

  • The formula for the conductor $N_q$ (which is valid provided that $q$ is coprime to $N$) can be checked directly from the definition of the conductor in terms of $\ell$-adic Tate modules. You can also think of it in terms of how the conductor of a newform changes when you make a twist. This is discussed in classical language in the article of Atkin and Lehner. It is also easily verified using representation-theoretic language.

  • 0
    @Eugene: Dear Eugene, Thank you for asking; I don't mind at all. Best wishes,2012-05-31