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It follows from the modularity theorem for elliptic curves over $\mathbb{Q}$ that there are finitely many elliptic curves of a given conductor $N$. Moreover, one can algorithmically enumerate them. [Edit: As Emerton comments below, without further argument, this is only true for elliptic curves up to isogeny!]

Was the finiteness and/or algorithmic enumeration of elliptic curves of a given conductor known before the modularity theorem?

Every elliptic curve over $\mathbb{Q}$ can be written in the form $y^2 = x^3 + ax + b$ where $a, b \in \mathbb{Z}$ with discriminant $\Delta = -16(4a^3 +27b^2) \neq 0$. So the number of elliptic curves of discriminant $D$ is bounded above by number of nontrivial pairs $(a, b) \in \mathbb{Z}^2$ such that $D = -16(4a^3 +27b^2)$.

Let $D \in{\mathbb{Z}}, D \neq 0$ be given. Because $D \neq 0$, the cubic equation $b^2 = \frac{-D}{16\cdot27} - \frac{4a^3}{27}$ is nonsingular, so by Siegel's theorem there are finitely many solutions. It follows that there are finitely many elliptic curves of a given discriminant. Silverman's book says that Baker even gave an explicit upper bound in this case, which was refined by Stark.

However, a priori there is no bound on the size of the discriminant of elliptic curves of a given conductor, so it doesn't immediately follow that there are finitely many elliptic curves of a given conductor. Szpiro's conjecture implies that if one fixes the conductor there are only finitely many discriminants that give that conductor. However, this conjecture is open (or not, depending on the status of Mochizuki's work).

Is there a weaker form of Szpiro's conjecture that has been proved giving an upper bound on the discriminant of an elliptic curve of a given conductor? If so, what's the minimum amount of technology needed to get the results?

Of course there are also issues of effectivity as well, which I also welcome comments on.

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Jonah Sinick
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    Finiteness was known long before modularity.
    Given $N$, there are finitely many possible $D$ up to sixth powers, so finitely many equations $x^3-y^2 = D$ in $S$-units where $S$ is the finite set of prime factors of $2D$. The desired result now follows by applying Siegel's theorem for $S$-integral points on an elliptic curve.     – Noam D. Elkies Sep 15 '12 at 21:43
  • Perhaps I'm being dense, but why are there only finitely many discriminants for a given conductor (up to sixth powers)? – Jonah Sinick Sep 15 '12 at 22:06
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    I forgot to discuss effectivity in my answer. Baker's linear forms in logs allows one to give an effective bound for the largest solution in $S$-integers to an equation of the form $x^3-y^2=D$, so as sketched by Noam, one gets an effective version of Siegel's theorem, which in turn gives an effective (but quite large) bound for the heights of the coefficients of all elliptic curves of conductor $N$. – Joe Silverman Sep 15 '12 at 22:06
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    @Jonah: Look at the proof of Shafarevich's theorem, for example in my Arithmetic of Elliptic Curves book. It says something much stronger than the fact that there are only finitely many elliptic curves of a given conductor. – Joe Silverman Sep 15 '12 at 22:08
  • @ Joe - thanks. Why do you say that Shafarevich's theorem is much stronger than the finiteness of elliptic curves of a given conductor? Doesn't the latter imply the former up to powers of 2 & 3? Are the powers of 2 and 3 what you have in mind? – Jonah Sinick Sep 15 '12 at 22:24
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    @Jonah: All I meant was that the set of curve of conductor $N$ may be smaller than the set of curves with good reduction outside the set of primes dividing $N$. So for example, if you just use Wiles' theorem (and not the work done by Breuil et.al. later), then all you'll get is that there are finitely many elliptic curves of conductor $N$ for square-free conductors $N$, because Wiles' original proof only covered the semi-stable case. That would be weaker than Shafarevich's theorem. (Powers of 2 and 3 don't matter, since their power in $N$ is bounded respectively by $2^5$ and $3^3$.) – Joe Silverman Sep 16 '12 at 02:16
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    -- continued -- But you're right in saying that with a little work about 2 and 3, Shafarevich's theorem is equivalent to the statement that there are finitely many curves of conductor $N$ for all $N$. But for any particular $N$, it maybe be easier to prove that finiteness than it is to prove Shafarevich's theorem for the set of primes dividing $N$. – Joe Silverman Sep 16 '12 at 02:18
  • [I think you mean $2^8$ and $3^5$] – Noam D. Elkies Sep 16 '12 at 05:07
  • @Jonah Sinick: I guess it would have been clearer had I written "$S$ is the finite set of prime factors of $2N$" (instead of prime factors of $D$). It is then clear that there are at most $2 \cdot 6^{|S|}$ possibilities up to sixth powers (with a factor of $2$ for the choice of $\pm1$ sign), and the sixth powers can be absorbed into $x$ and $y$ while keeping them $S$-integral. – Noam D. Elkies Sep 16 '12 at 11:45
  • @Noam: Right. $2^8$ and $3^5$. Thanks for correcting my faulty memory. – Joe Silverman Sep 16 '12 at 12:12
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    @Joe Silverman: The way I remember it is that N_tame | N | 1728*N_tame, where N_tame is the tame conductor. – James Weigandt Sep 18 '12 at 15:51
  • Dear Jonah, Modularity and finite-dimensionality of spaces of cuspforms of a given level directly only imply that the number of isogeny classes of bounded conductor is finite. (Because a given cuspform parameterizes an isogeny class of elliptic curves, not an isomorphism class.) It takes an extra argument to show that a given isogeny class of elliptic curves over $\mathbb Q$ contains only finitely many isomorphism classes. Regards, Matthew – Emerton Oct 06 '12 at 21:22
  • @ Emerton - of course you're right, thanks for the catch. I edited the question accordingly. Note that I did get this right in my question about the modularity of abelian varieties ;). – Jonah Sinick Oct 06 '12 at 22:57

2 Answers2

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As Noam says, this was well-known long before Wiles. The proof that he sketches can be used to prove Shafarevich's theorem for elliptic curves, i.e., given a finite set of primes $S$, there are only finitely many elliptic curves over $\mathbb{Q}$ with good reduction outside of $S$. So if you have a particular $N$ in mind, you can take $S$ to be the set of primes dividing $N$.

And there were even some fairly explicit upper bounds for the number of elliptic curves of conductor $N$. See for example my paper with Brumer, "The number of elliptic curves over $\mathbb{Q}$ with conductor $N$," Manuscripta Math. 91 (1996), 95-102. Turning things around, we can then use Wiles' theorem and the previously proven upper bound for the number of elliptic curves of conductor $N$ to immediately deduce a non-trivial upper bound for the number of elliptic factors of $J_0(N)$.

Joe Silverman
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  • Re: The juxtaposition of knowledge of the conductor of elliptic curves together with pairing between elliptic curves and elliptic factors of the Jacobian - the two in conjunction imply that e.g. $J_0(p^3)$ has no elliptic factors for a prime $p > 3$. Can one see this directly on the automorphic side? – Jonah Sinick 0 secs ago – Jonah Sinick Sep 15 '12 at 23:17
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    No new factors of $J_0(p^3)$, you mean. [$J_0(p)$ is contained in $J_0(p^2)$ which is contained in $J_0(p^3)$.] – Noam D. Elkies Sep 16 '12 at 05:10
  • @Noam - Yes, that's what I meant. – Jonah Sinick Sep 16 '12 at 16:57
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I just want to mention a couple of things to complement what Joe Silverman and Noam Elkies have already said.

As Brian Conrad pointed out in response to this ill-posed question of mine, you need the Shafarevich conjecture (for abelian varieties of all dimensions as proved by Faltings) to prove modularity of elliptic curves.

I also want to mention this paper of Cremona and Lingham:

Finding all elliptic curves with good reduction outside a given set of primes

which deals with the algorithmic problems of finding all elliptic curves of a given conductor as discussed in Noam's first comment. They compute the $S$-integral points on $y^2 - x^3 = D$ assuming that a Mordell-Weil basis has been computed. The ideas are implemented in Sage and can be used to find the elliptic curves with everywhere good reduction outside a sufficiently simple finite set of primes $S$.

For example, if you wanted to find all elliptic curves with everywhere good reduction outside $37$ you would use the following code

sage: egros = EllipticCurves_with_good_reduction_outside_S
sage: time curves = egros([37])
Time: CPU 1.49 s, Wall: 1.50 s
sage: for E in curves:
....:     print E.a_invariants()
....:     
(0, 0, 1, -1, 0)
(0, 1, 1, -23, -50)
(0, 1, 1, -1873, -31833)
(0, 1, 1, -3, 1)
(0, 1, 1, -4563, 116200)
(0, 1, 1, -31943, -2138543)
(0, 1, 1, -2564593, -1581651042)
(1, -1, 0, 3166, -59359)
(1, -1, 0, -2276219, -1321241558)
(1, -1, 1, 2, -2)
(1, -1, 1, -1663, -25680)
(0, 0, 1, -1369, 12663)
(0, 1, 1, -12, -17)
(0, 1, 1, -2602, 50229)
(0, 1, 1, -16884, -647702)
(0, 1, 1, -3562594, 2587011456)
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James Weigandt
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