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Let $E$ be an elliptic curve, $\mathbb{Q}$ the field of rational numbers, $\mathbb{Z}[i]$ the ring of Gaussian integers and let $K$ be a number field that is some field extension of $\mathbb{Q}$.

  1. Is it possible for an elliptic curve $E/K$ to be modular when $K$ is some field $\not =$ $\mathbb{Q}$ but is instead some field extension of $\mathbb{Q}$? If yes, then what are the conditions for which this would hold?

  2. Is it possible for some elliptic curve $E/\mathbb{Q}[i]$ to be modular while the corresponding curve $E/\mathbb{Q}$ would not be modular? If yes, then what are the conditions for which this would hold?

  3. If the answer to the first question in (2) is yes, then I would like to know how does one define such a modular elliptic curve $E/\mathbb{Q}[i]$ as being reduced modulo a prime? Is it possible for $E/\mathbb{Q}[i]$ to have good reduction or multiplicative reduction modulo a prime? Or does one have to use p-adic integers in some way, when dealing with field extensions of $\mathbb{Z}[i]$?

Edit:

Thanks for the insights.

By "modular" I meant in the sense defined in descriptive articles on FLT and from expository articles by B. Mazur, K. Ribet, and Gouvea (AMM) on the work Wiles, Frey and Ribet did on FLT in 1994-1995. Loosely put, my "assumed curve" $E/\mathbb Q[i]$ (i.e. I never did think it could be modular; I was trying to find a way to show this lack of modularity by finding a contradiction) would be modular if somehow it was to have an $L$-function that equals the corresponding complex series expansion that is a cusp form and an eigenform on the upper complex plane, so that the coefficients in the complex series expansion are the $p$-defects $a_p$ in the $L$-function.

But I was wondering how it could be possible one could have an $L$-function for an elliptic curve over $\mathbb Q[i]$, by getting reductions modulo a prime, but by doing this with a curve that is over a complex rational field extension $\mathbb Q[i]$ of the rationals (i.e. with elements $a + bi$, $a, b \in\mathbb Q$, and above the real line on $\mathbb C$). I mean how could one define an element $$ a + bi \mod p? $$ I saw nothing about this in the related literature.

I was thinking there was a theorem by B. Mazur that showed if this assumed curve $E/\mathbb Q[i]$ was not modular over $\mathbb Q[i]$, where this is a complex field extension of $\mathbb Q$, then it certainly could not be modular as $E/\mathbb Q$.

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    $\begingroup$ What is your definition of modular? Do you mean "coming from modular/automorphic forms" or "quotient of a modular curve"? The two notions are not the same over an arbitrary number field -- the second one is more restrictive. $\endgroup$
    – François Brunault
    Commented Apr 21, 2013 at 21:31
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    $\begingroup$ Just to echo François Brunault's comment, the ``modularity theorem" over, e.g., a totally real field, doesn't refer to the existence of a covering by a modular curve, but by the equality of the Hasse-Weil $L$-function with the $L$-function of a Hilbert modular eigenform of parallel weight $2$. Under certain hypotheses on the elliptic curve, this implies the existence of a covering by a Shimura curve, analogous to a modular curve, but these hypotheses do not always hold (they fail for example when the curve has good reduction everywhere and the field has even degree). $\endgroup$
    – Keenan Kidwell
    Commented Apr 21, 2013 at 21:57
  • $\begingroup$ @Robert: It sounds like you are looking for, among other things, the definition of the Hasse-Weil Zeta Function of a variety defined over a number field. Knowing the name will allow you to see that there is quite a large literature here. Also Felipe is right: based on what you write, you will need to learn some basic algebraic number theory in order to make sense of all this. (If you do not have this background, you can feel free to ask very basic questions on the related site math.stackexchange.com.) $\endgroup$
    – Pete L. Clark
    Commented Apr 23, 2013 at 3:21
  • $\begingroup$ You should only post an answer if you have an answer; what you posted as an answer should have been an edit to the original question. I took the liberty of adding this text into the original question. Feel free to edit it. (It may help if you register your account.) $\endgroup$
    – Joonas Ilmavirta
    Commented Oct 22, 2015 at 20:37

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An elliptic curve is modular if and only if it is a quotient of the Jacobian of a modular curve and this is independent of the field of definition so the answer to 2. is no.

It is now known, I believe, that an elliptic curve over a number field is modular if and only if it is isogenous to all its conjugates (this is known as a $\mathbb{Q}$-curve). These things certainly exist over number fields other than the rationals.

For your last question, open a book on algebraic number theory and study completions of number fields.

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  • $\begingroup$ Concerning the first sentence: is it completely obvious that one cannot have an elliptic curve $E$ with rational $j$-invariant which splits off as an isogeny factor of some $J_0(N)$ only over some larger number field? I remember this can't happen when $N$ is squarefree, because by a 1975 theorem of Ribet the geometric endomorphism ring of $J_0(N)$ is equal to its $\mathbb{Q}$-rational endomorphism ring. What about the general case? (I suspect I once knew the answer to this but have forgotten...) $\endgroup$
    – Pete L. Clark
    Commented Apr 23, 2013 at 3:15
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    $\begingroup$ @Pete: It's not obvious. See the appendix to Mazur's paper "Number Theory as gadfly". $\endgroup$
    – Felipe Voloch
    Commented Apr 23, 2013 at 4:29

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