Let $E$ be an elliptic curve defined over a number field $K$ without complex multiplication. Serre's open image theorem (which appears in his book 'Abelian $l$-Adic Representations and Elliptic Curves') says that the image of the representation of $Gal(\bar{K} / K)$ on the $l$-adic Tate module $T_l(E)$ is open in $GL_2(\mathbb{Z}_l)$.

Is there a modern proof of this written down somewhere using Faltings' Theorem (i.e. the Tate conjecture) or other methods?

Edit: I've just found Ribet's review of Serre's book, which contains fairly detailed sketch of the kind of proof I was after, so I included it below.

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    $\begingroup$ You must add that $E$ does not have complex multiplication (else the representation is abelian, and the conclusion clearly fails). This is rather more elementary than Faltings' theorem (which I don't think would actually help here). See also Serre's Inventiones paper, "Proprietes galoisiennes des points d'ordre fini des courbes elliptiques," for a complement showing that the image equals $\mathrm{GL}_2(\mathbb{Z}_l)$ for all but finitely many $l$. $\endgroup$ Sep 26, 2012 at 15:21
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    $\begingroup$ When E has CM, the Galois representation is not generally abelian. It is only abelian when restricted to the field by which it has CM. $\endgroup$
    – Rob Harron
    Sep 27, 2012 at 2:08
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    $\begingroup$ Faltings's theorem does help. When combined with semisimplicity it implies that the image of Galois contains an open subgroup of $SL_2(\mathbb{Z}_l)$ (when $E$ does not have CM). Using the fact that the action on the determinant of the Tate module is given by the cyclotomic character, it follows that the image of Galois contains an open subgroup of $GL_2(\mathbb{Z}_l)$. $\endgroup$
    – naf
    Sep 27, 2012 at 4:14
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    $\begingroup$ Faltings' theorem is also useful in proving generalizations of Serre's open image theorem for elliptic curves to abelian varieties of higher dimension. You may take at look at Serre's letters to Ribet and Vigneras (if I remember correctly) in his collected works vol. 4. $\endgroup$
    – Nicolas B.
    Sep 27, 2012 at 8:45
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    $\begingroup$ @RH, Ulrich, NB: Thanks for the corrections! $\endgroup$ Sep 27, 2012 at 12:01

1 Answer 1


Here is Ribet's proof (expanding on Ulrich's comment):

Let $G_K:=Gal(\bar{K} / K)$ and $V_l:=T_l(E)\otimes \mathbb{Q}_l$.

The image $\rho_{l,E}(G)$ is a closed subgroup of the $l$-adic Lie group $\text{Aut}(V_l(E)) \cong \text{GL}_{2}(\mathbb{Q}_l)$ and is therefore a Lie subgroup of $\text{Aut}(V_l(E))$. Its Lie algebra $\mathfrak{g}_l$ is a subalgebra of $\mathfrak{gl}_{2}(\mathbb{Q}_l)$. We want to show that $\mathfrak{g}_l=\mathfrak{gl}(V_l)\cong \mathfrak{gl}_2(\mathbb{Q}_l)$ and the result follows.

(Note that the Lie algebra of the image $\rho_{l,E}(G_K)$ is the tangent space of the identity component of the Zariski closure of $\rho_{l,E}(G_K)$ in $\text{GL}_{2}(\mathbb{Q}_l)$. So $\mathfrak{g}_l$ `measures the representation up to finite extensions of the base field $K$', since a finite index subgroup of an algebraic group has the same identity component).

Now $V_l$ is irreducible as a $\mathfrak{g}_l$-module (this is a theorem of Shafarevich, and depends on Siegel's theorem on the finiteness of integral points on curves). Secondly, $\mathfrak{g}_l$ can't be contained in the subalgebra $\mathfrak{sl}(V_l)$ of $\mathfrak{gl}(V_l)$ since $\det(\rho_{l,E})=\chi_l$ (where $\chi_l$ is the cyclotomic character giving the action of Galois on $K^{cycl}$).

This leaves two possibilities for $\mathfrak{g}_l$: either $\mathfrak{g}_l$ is $\mathfrak{gl}_2(\mathbb{Q}_l)$ and we're done, or $\mathfrak{g}_l$ is a non-split Cartan subalgebra of $\mathfrak{gl}_2( \mathbb{Q}_l)$ (an abelian semisimple algebra coming from a quadratic field extension of $\mathbb{Q}_l$).

Faltings proved two important facts about represenations $\rho_{l,E}$:

  • $\rho_{l,E}$ is a semisimple representation of $G_K$ over $\mathbb{Q}_l$

  • $\text{End}(E)\otimes \mathbb{Q}_l \cong \text{End}_{\mathfrak{g}_l}(V_l)$.

Faltings results then rule out the possibility that $\mathfrak{g}_l$ is a non-split Cartan subalgebra of $\mathfrak{gl}_2( \mathbb{Q}_l)$ and we're done.

  • $\begingroup$ Can one show in a similar way that the image is not only open, but even the whole group for almost all $\ell$? $\endgroup$
    – user19475
    Sep 29, 2012 at 14:44
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    $\begingroup$ @Timo:I think there must be more modern proofs than the one appearing in Serre's inventiones paper. For example have a look at `Galois properties of division fields of elliptic curves' by Masser and Wustholz $\endgroup$ Sep 29, 2012 at 15:11
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    $\begingroup$ @Timo:There is also `An open image theorem for a general class of abelian varieties' by Chris Hall (which may be the state of the art in the adelic case, although I'm not sure): here's a link if you're interested arxiv.org/pdf/0803.1682.pdf $\endgroup$ Oct 6, 2012 at 11:28
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    $\begingroup$ @Adam Harris: The point with Masser and Wüstholz's series of papers is to provide an alternate, effective, proof of Faltings's theorem and of its consequences, such as Serre's open image theorem. $\endgroup$
    – ACL
    Mar 13, 2013 at 9:03
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    $\begingroup$ @Adam Harris: The image of the $2g$-dimensional Galois representation is contained in the $\ell$-adic Mumford-Tate group which is a subgroup of the general symplectic group $GSp_{2g}(\mathbf Z_\ell)$. The Mumford-Tate conjecture asserts that the image is open in the M-T group. Hall's paper provides a criterion which asserts that the M-T group is the full symplectic group and the Galois image is open. $\endgroup$
    – ACL
    Mar 13, 2013 at 9:09

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