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There is a well-known theorem of Shafarevich that given a finite set $S$ of primes the number of isomorphism classes of elliptic curves over $\Bbb Q$ with everywhere good reduction outside $S$ is finite.

One way to prove this, which Cremona and Lingham use here to compute all such curves, is to use Siegel's theorem that an elliptic curve over $Q$ has only a finite number of $S$-integral points.

Here's a proof with overkill:

Given $S$ there are a finite number of possible conductors $N$ for elliptic curves with everywhere good reduction outside $S$. They must all be divisors of $2^8 3^5 d^2$ where $d$ is the product of those primes in $S$ different from 2 and 3.

The corresponding spaces $S_2(\Gamma_0(N))$ of cuspforms for each of our finite list of $N$ is finite dimensional.

By the modularity theorem, there is hence finite number isogeny classes of elliptic curves with everywhere good reduction outside $S$.

By Mazur's Modular Curves and the Eisenstein Ideal there are only a finite number of isomorphism classes of elliptic curves in a given isogeny class.

Question 1: Does any of this machinery rely on Siegel's theorem?

Question 2: If the answer to question 1 is no, can this proof of Shafarevich's theorem be "cheaply extended" to deduce Siegel's Theorem from these seemingly unrelated powerful results?

By "cheaply extended" I mean without the use of techniques with the diophantine flavor of Baker's theory of linear forms in logarithms.

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You mean to refer to "Rat'l isogenies of prime degree". The Eisenstein paper addresses isogenies whose kernel is constant (i.e., all pts are rat'l). Siegel's thm is irrelevant in Mazur's work. Anyway, your argument seems circular: the "modularity thm" relates (certain) eigenforms to $\ell$-adic Galois representations, and to know an isogeny class of elliptic curves is captured by Galois rep'n one needs Tate's isogeny conjecture for elliptic curves...first proved in general for elliptic curves over # fields by Faltings when he proved it in all dim's...via his proof of Shaf. conj. in all dim! –  BCnrd Aug 31 '10 at 4:30
    
Cool! Thanks Brian. I definitely learned something. I can see now that my question sort of springs from a great ignorance of the work of Faltings. –  Jamie Weigandt Aug 31 '10 at 5:06
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Faltings work doesn't use transcendence or dioph approximation and you can deduce Siegel's thm from the Mordell conjecture. –  Felipe Voloch Aug 31 '10 at 11:04
    
Ha! I feel silly. This is question is essentially Remark 6.5 in on page 295 of Silverman's AEC (2nd Edition). –  Jamie Weigandt Sep 23 '10 at 17:48

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