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Dear MO Community,

Let $N$ be a prime, and let $X_0(N)$ be the classical modular curve over $\mathbb{Q}$. We know ([1]) that, if there are noncuspidal points in $X_0(N)(\mathbb{Q})$, then $N \in$ {${ \mbox{primes } \leq 19} $} $ \cup $ {37,43,67,163}.

The basic question of this post is:

Are there similar lists of primes when $\mathbb{Q}$ is replaced by a number field $K$? That is, if we fix a general number field $K$, can we determine the primes $N$ for which $X_0(N)(K)$ has noncuspidal points?

Perhaps in this generality the question is hard, so suppose we restrict from now on to imaginary quadratic $K$. Then [1] gives an approach to the question, but with the following snags:

  1. We need to construct an "optimal" quotient of $J_0(N)$, call it $A$, such that $A(K)$ has Mordell-Weil rank 0;

  2. We must restrict ourselves to primes $N$ which are inert in $K$.

Actually, I don't think (1) is a big problem; provided $N \> 48h(K)^3 + 1$, we can take $A = \widetilde{J}$, the Eisenstein quotient. [Speculation : if we took the "winding quotient" instead, maybe we can lower that bound...]

When $N$ splits, then we can construct "CM points" on $X_0(N)(\mathbb{C})$, but usually they will not be defined over $K$, and even if they are, there will only be a handful of them.

Question: For which $N$ that splits in $K$ do we have points on $X_0(N)(K)$ that are neither cuspidal nor CM? Is there a way to systematically find these points?

By "systematically", I guess I mean something like the 'isogeny character' approach of [1], where the hunt for the $N$s comes down to when certain congruences are satisfied mod $N$.

Many thanks.

[1]: Mazur, B. "Rational isogenies of prime degree", Inventiones Mathematicae, 1978

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up vote 10 down vote accepted

I'm glad to have stumbled on your question -- I'm actually working on something along the lines of what you're asking about now with Eric Larson. If what we think we've proven is true (and don't quote this yet, since we're not yet even done with the write-up,) then in fact you can say something stronger: you cannot have isogenies of order p for any prime p sufficiently large (where "sufficiently large" depends on K), as long as K does not contain the class field of an imaginary quadratic extension in which p splits (this exactly gets rid of primes which are isogenies of a CM curve defined and with CM over K). So this would mean that if K is quadratic imaginary, there can be no p-isogenies for any p sufficiently large unless K is one of the finitely many quadratic imaginary fields of class number one (in which case, if our arguments work, this would mean that all but finitely many possible prime degrees of isogeny p split over K.)

If you're interested in what happens to primes that split, you can ask whether, more generally, for any number field K, there are only finitely many primes that can be isogenies for a curve over K without CM. This seems harder to prove, and as far as we know, is an open problem (to prove it, you'd need to have a good way of distinguishing CM from non-CM curves, or carefully counting points on $X_0(N)$). But hypothetically this is also true: it fits into the general framework of Serre's hypothesis that "exceptional primes" of curves without CM are bounded by a constant depending only on $K$.

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Thanks for your post. I'm very interested in your result. Regarding your first paragraph, Momose's paper (see the answer by stankewicz) proves the quadratic case of your result. Have you used different ideas to the isogeny character and its 'rigidity' approach (Theorem A of loc. cit.)? Are your methods effective, and if so, are the bounds sharp or reasonable? In the quadratic case, how do your bounds compare with Momose's bounds? Regarding the imaginary quadratic case, about all but finitely many isogeny primes must split, this was known to Mazur (Proposition 8.1 in [1]) – Barinder Banwait Oct 19 '10 at 13:53
Sorry not to have got back to you sooner. Hopefully our paper will be up on arXiv soon, but so far I can give you the bounds. Conditionally (on the GRH) they're as follows: Suppose K is a field not containing the Hilbert class field of an imaginary quadratic field. Set S_K the set of primes $\ell$ such that there exists an $\ell$-isogeny. Then the product of all primes in S_K is bounded by something on the order of $\exp((12n)^n(R+h^2\log^2 \Delta)+n^2 h^2 \log^2\Delta)$ where $n$, $\Delta$, $h$, $R$ are the degree, discriminant, class number and regulator of $K$. – Dmitry Vaintrob Jan 8 '11 at 2:50

You can see also the recent paper of A. David on ArXiv: Critères d'irréductibilité pour une courbe elliptique semi stable sur un corps de nombres. A. David gives a uniform bound for irreducibility of mod p Galois representations of semi stable elliptic curves over number field K provided some condition (namely that there is not any CM elliptic curve with everywhere good reduction on K).

But the general question of isogenies over a number field is hard. The "Mazur-Kamienny" style method to study this should be to look at the morphism from the d-th symmetric power of X_0(p) to an optimal quotient of J_0(p). (See also Kamienny works for torsion points on number fields).

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Dear Barinder,

Are you familiar with Fumiyuki Momose's "Isogenies of prime degrees over number fields?" If not, you may find it here on NUMDAM In it he performs an analysis of the isogeny character and finds that if $k$ is a quadratic field which is not a class number one imaginary quadratic field there are only finitely many $p$ for which $X_0(p)$ has noncuspidal rational points.

Furthermore if $k$ is any number field, a noncuspidal point of $X_0(p)(k)$ must be one of 3 types($\theta_p$ is the $p$-th cyclotomic character and $\lambda$ is the isogeny character of the point so $\lambda^{12}$ is independent of the representative elliptic curve or isogeny defining the point):

Type 1: $\lambda^{12}$ or $(\lambda\theta_p^{-1})^{12}$ is unramified

Type 2: $\lambda^{12} = \theta_p^6$ and $p\equiv 3 \bmod 4$

Type 3: $k\supset H_L$, the hilbert class field of an imaginary quadratic field $L$ such that $p$ splits in $L$ and there are some further congruence conditions.

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I wasn't aware of that paper, but by the title, it seems like the kind of thing I'm looking for. It's late here now, so I'll read it in the morning. Thanks a lot! – Barinder Banwait Oct 16 '10 at 23:59

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