Let $K$ be a quadratic field, and $E/K$ a non-CM elliptic curve with a $K$-rational $p$-isogeny, for $p$ a prime. I would like to say the following:

For large enough $p$, the $j$-invariant $j(E)$ must be in $O_K$.

I would also like to know exactly how large $p$ must be. (This lower bound may very well not depend on $K$.)

The above is indeed true when $K$ is replaced with $\mathbb{Q}$; 37 is definitely large enough; (though the true bound may be as small as 17. EDIT: Actually, it's 19. See the comments.)

Corollary 4.3 in Mazur's article [1] gives me hope that the above may be true. Assuming condition A (which I'll describe presently), it says (for my set-up) that the only primes which can divide the denominator of the $j$-invariant are the primes above 2 and 3; (moreover, if 3 doesn't ramify in $K$, then it's only the primes above 2 that need concern us).

Condition A is that $J_0(p)$, the jacobian of the modular curve $X_0(p)$, possesses an "optimal quotient" whose Mordell-Weil rank over $K$ is zero.

So I guess I'm hoping for two things:

  1. That these small primes can be dealt with (i.e. for $p$ large enough, they don't arise in the denominator of $j$); and

  2. Condition A can be removed after all of these years (at least for $p$ large enough).

EDIT (after comments from Felipe Voloch and Noam Elkies): Merel's "winding quotient" has rank 0 over $\mathbb{Q}$; but I don't think it will have rank zero over every quadratic field. I also don't know over which quadratic fields the winding quotient has rank zero.

Noam Elkies is quite right that the desired result is "vacuously true", since it is known (by work of Momose, Theorem B in [2]) that there are only finitely many $p$ for which there is a $K$-rational $p$-isogeny. However, I'd still like a more direct approach to the lower bound question, not using Momose's much stronger result, if there is one…

[1]: Mazur, B. "Rational Isogenies of Prime Degree". Inventiones, 1978.

[2]: Momose, F. "Isogenies of Prime Degree over Number Fields". Compositio, 1995

  • $\begingroup$ Doesn't the work of Merel answer your question? At least it provides an optimal quotient. See mathoverflow.net/questions/62950/… $\endgroup$ Aug 30, 2012 at 19:41
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    $\begingroup$ If $E$ has an isogeny, which is not multiplication by an integer, then $E$ is CM (over $\bar K$). See for instance Silverman, III, Cor. 9.4 p. 102 of "Arithmetic of elliptic curves". But this seems to contradict your hypothesis. $\endgroup$ Aug 30, 2012 at 19:44
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    $\begingroup$ Well for large enough $p$ there shouldn't be such an isogeny at all, so the desired result should be vacuously true... Anyway over ${\bf Q}$ it is not quite enough to take $p=17$ (the 17-isogeny mentioned in modular.math.washington.edu/Tables/antwerp/table1/small_80.jpg involves curves with multiplicative reduction at $2$), but $p=19$ is sufficiently large. $\endgroup$ Aug 30, 2012 at 19:49
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    $\begingroup$ @ Damian Rössler the isogeny need not be between $E$ and $E$ itself. $\endgroup$ Aug 30, 2012 at 19:50
  • $\begingroup$ Sorry, I understand the meaning of $p$-isogeny now. Forget my comment. $\endgroup$ Aug 30, 2012 at 19:55

1 Answer 1


If you are given K, and you want to find a p such that J_0(p) has a rank-0 quotient over K, you might try proceeding as follows. Let $\chi$ be the quadratic twist whose kernel is the Galois group of K. Then what you really want is to show that there exists a newform f in S_2(p) with the property that both the corresponding quotient A_f and its twist by $\chi$ have rank 0. This follows if you know that both $L(f,1)$ and $L(f \times \chi, 1)$ are nonzero. Why should there be any such f? One way to get at this is by averaging the quantity

$L(f,1) L(f \times \chi, 1)$

over the an orthogonal basis of cuspforms f for S_2(p). Often there's a nice analytic way to estimate this average as

something that's not zero + something that decays with p

which means that, for sufficiently large p (usually explicit) the average is not 0, which means that one of the guys being averaged is not zero, which means that you have a quotient which does what you want. See the recent paper of Michel and Ramakrishnan, and the papers it cites, to see how the method works:



Momose's argument actually doesn't go via this route; he uses the same quotient A of J_0(p) that Mazur does. As you note, this quotient has rank 0 over Q but not over K. But that's OK; Momose's argument (which goes back to Kamienny) involves mapping the symmetric square of X_0(p), not X_0(p) itself, into A. And a point on X_0(p) over ANY quadratic field gives a Q-rational point on the symmetric square.

  • $\begingroup$ Thank you for your answer JSE, this approach is indeed very interesting. I've meanwhile realised that Mazur himself proved the existence of a rank zero quotient over imaginary quadratic fields at primes inert in the field. The approach you outline however would work for real quadratic fields also. Thanks again! $\endgroup$ Sep 7, 2012 at 8:50
  • $\begingroup$ The L-function part of this answer will not work. As is for example mentioned on page 1 of ams.org/journals/proc/2013-141-08/S0002-9939-2013-11532-7/…. If you twist a newform $f \in S_2(p)$ by a quadratic character $\chi$ of conductor then the sign changes by a factor of $\chi(p)$. In particular if $\chi(p)=-1$ then the functional equations of $f$ and $f \times \chi$ have opposing signs, and hence $L(f,1)L(f \times \chi,1)$ will be zero for all newfors in $S_2(p)$. Maybe that it will produce something if $\chi(p)=1$. The "But note" part however is great! $\endgroup$ Nov 3, 2021 at 19:08

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