22
$\begingroup$

As the title says. In particular, every elliptic curve over $\mathbb{Q}$ is modular; but what is the current state of the art for general totally real number fields? I assume the answer is extractable from some papers of, say, Kisin, but I am not an expert in this material and hesitate to try that myself.

$\endgroup$
4
  • 1
    $\begingroup$ All of them are potentially modular (this is because there exists a rational prime l such that E has good ordinary reduction at every prime in F above l). To prove that a specific example is actually modular might be hard, as all the works I am familiar with assume either residual modularity or large image (but I am definitely no expert either). $\endgroup$
    – Olivier
    Sep 16, 2010 at 5:24
  • 1
    $\begingroup$ But is one of those things enough? For example, if the curve has an irreducible mod-3 representation, then you can get started via Langlands-Tunnell - but are other things required? $\endgroup$ Sep 26, 2010 at 18:04
  • 5
    $\begingroup$ It is known (potential modularity + Solomon's induction theorem) that the $L$-function of any elliptic curve over any totally real field is meromorphic on ${\mathbb C}$ and satisfies the expected functional equation. For many applications this is as good as modularity, but I don't whether whether this is of help to you. $\endgroup$ Nov 18, 2010 at 23:11
  • $\begingroup$ I've recently asked a similar question in case you are still interested. mathoverflow.net/questions/96289/… $\endgroup$
    – Eugene
    May 12, 2012 at 1:10

1 Answer 1

9
$\begingroup$

Elliptic curves over real quadratic fields were proven to be modular very recently by Freitas, Le Hung and Siksek:

Some progress was alredy present in the thesis of Richard Taylor's student Le Hung.

Potentially modular is known unconditionally over arbitrary totally real fields, that is:

Theorem. Let $E/F$ be an elliptic curve over a totally real field. Then there is some totally real extension $F'/F$ such that $E/F'$ is modular.

As mentioned in the comments, for practical purposes this is almost as good as modularity.

The attribution of this theorem is complicated. There's some information about this in section 7 of our own Kevin Buzzard's very interesting survey on modularity:

You can also consult:

For arbitrary totally real fields, modularity is still not known.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.