# Serre's Open Image Theorem Without Shafarevich's Theorem

In Abelian l-adic Representations and Elliptic Curves (1968), J. P. Serre showed that the adelic representation $$\rho_{E}\colon G_K \to \mathrm{GL}(\hat{\mathbb{Z}}^2)$$ associated to an elliptic curve $E/K$ over a number field $K$ has open image. To do it, he uses Shafarevich's Theorem on the finiteness of isomorphism classes of elliptic curves in a given isogeny class to show that the $\ell$-adic representation $$\rho_{E,\ell}\colon G_K \to \mathrm{GL}(T_\ell(E))$$ is irreducible for all $\ell$ and that the mod $\ell$ representation $$\bar{\rho}_{E,\ell}\colon G_K \to \mathrm{GL}(E[\ell])$$ is irreducible for almost all $\ell$.

My question is, do we now have a method of proving this theorem without using Shafarevich's Theorem? The latter depends on Siegel's Theorem, which depends on Roth's Theorem in Diophantine Geometry.

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This is an interesting question and I suspect there is a way using p-adic Hodge theory. In the meanwhile, I thought I'd point out that Shafarevich's theorem here requires only Siegels theorem for the `discriminant elliptic curves,' something like 4a3−27b2=c. These have CM, and a proof of finiteness that doesn't use Diophantine approximations at all can be found in annals.math.princeton.edu/wp-content/.../annals-v172-n1-p16-p.pdf (By the way, I had mistakenly posted this as an answer earlier.) – Minhyong Kim Jul 18 '12 at 22:28
Oh sorry, I should also say that the remark above applies only to elliptic curves over $\mathbb{Q}$. – Minhyong Kim Jul 18 '12 at 22:29
@Davidac897: I asked a somewhat similar question a while back out my own ignorance of Falting's work: mathoverflow.net/questions/37212 . I wanted to deduce Shafarevich's theorem over $\Bbb Q$ from modularity without using Siegel's Theorem, but my argument was cyclic because I unknowingly assumed Tate's Isogeny Conjecture, which was proved by Faltings by proving Shafarevich in all dimensions. It was mentioned in the comments there that you can deduce Siegel's Theorem from Faltings Theorem (Mordell's Conjecture) which doesn't use Diophantine Approximation. Not sure if that will help you. – Jamie Weigandt Jul 19 '12 at 0:09

First, you forgot to assume that $E$ does not have CM. However, this actually suggests a difficulty in a Shafarevich-free proof.

Let $K = \mathbf{Q}(\sqrt{-1})$, and let $C/K$ be an elliptic curve with CM by $\mathbf{Z}[\sqrt{-1}]$. Now consider the following thought experiment. Can you rule out the existence of an elliptic curve $E/K$ without CM such that $\rho_{E,\ell} = \rho_{C,\ell}$ for all primes $\ell$?

This is certainly implied by the Tate conjecture (in a case proved by Faltings), but not only is this harder than the original proof, it also really uses/implies a (generalization of) Shafarevich's conjecture. Certainly $E$ admits isogenies $E \rightarrow E'$ of degree $p$ for any prime $p$ which splits in $K$, but ruling this out is exactly Shafarevich again. I'm not sure you can overcome this obstacle.

On the other hand, it is elementary to (essentially) reduce to this case, basically using Serre's original argument. Namely, one reduces to the case that the $\rho_{E,\ell}$ are abelian, and a classification of crystalline characters of the right weight (plus purity) essentially reduces to this CM-like case.

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Masser and Wüstholz have given an effective proof that the representation $\bar{\rho}_{E,\ell}\colon G_K \to \mathrm{GL}(E[\ell])$ is irreducible for all $\ell$ greater than some constant $c_E$, see their paper Some effective estimates for elliptic curves. They use isogeny bounds coming from transcendence theory to prove Shafarevich's Theorem without Siegel's theorem. They show that $c_E$ can be chosen to be less than $C h^4$ where $h$ is some naive height attached to $E/K$ and $C$ is a constant that can in principle be computed.

(The isogeny bounds have since been repeated improved. The state of the art might be the paper Théorème des périodes et degrés minimaux d'isogénies of Gaudron and Rémond.)

Added afterwards: The surjectivity of $\bar{\rho}_{E,\ell}$ for $\ell$ sufficiently large is also discussed by Masser and Wüstholz in Galois properties of division fields of elliptic curves. It is effective and again does not require Siegel's theorem.

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I thought that Davidac897 wanted a proof of Serre's theorem that does use Shafarevich. If you allow Shafarevich, then Faltings original proof of the Shafarevich conjecture (as a consequence of the Tate conjecture) does not rely on Siegel's theorem either. – Damian Rössler Jul 19 '12 at 21:49
I originally said without Shafarevich, though what I implicitly hoped for was something without using Sigel's Theorem either. It sounds a bit overkill to go to Faltings's Theorem, but I wonder whether Faltings's proof is simpler in dimension $1$. – David Corwin Jul 19 '12 at 22:14
That there are only fin. many AV in an isog class is actually a step in the proof of the Tate conjecture (which is then used to show that there are only fin many isog classes of AV of a given dimension and with good reduction outside a finit set of primes) Falting's proof that there are only fin many AV in an isog class involves introducing the Faltings height of an AV and 1. showing there are only finitely many AV with bounded Faltings height, and 2. understanding how height changes under isogeny (using Tate's results on p-divisible groups and Raynaud's results on finite flat group schemes.) – user18237 Jul 19 '12 at 23:16
Step 1 probably simplifies a lot for elliptic curves as moduli of EC is much easier than moduli of AV. I'm not sure how much step 2 simplifies. – user18237 Jul 19 '12 at 23:18
@gb: I think this description is not quite acurate. The proof of the Tate conjecture proceeds by studying the variation of the Faltings height in a p-divisible group (and there using Raynaud's results etc.). After that you deduce Tate's conjecture and semi-simplicity, which is necessary to prove Shafarevich. I don't think any of this simplifies in dim. 1 (see also Silverman's remarks in his book on the arithmetic of elliptic curves about this). I agree that step 1 can probably be simplified, though. – Damian Rössler Jul 20 '12 at 7:06