Modern proof of Serre's open image theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:49:14Z http://mathoverflow.net/feeds/question/108169 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108169/modern-proof-of-serres-open-image-theorem Modern proof of Serre's open image theorem? Adam Harris 2012-09-26T15:02:32Z 2013-03-12T23:43:00Z <p>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 $GL_2(\mathbb{Z}_l)$. </p> <p>Is there a modern proof of this written down somewhere using Faltings' Theorem (i.e. the Tate conjecture) or other methods?</p> <p>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.</p> http://mathoverflow.net/questions/108169/modern-proof-of-serres-open-image-theorem/108237#108237 Answer by Adam Harris for Modern proof of Serre's open image theorem? Adam Harris 2012-09-27T12:10:48Z 2012-09-29T14:33:39Z <p>Here is Ribet's proof (expanding on Ulrich's comment):</p> <p>Let $G_K:=Gal(\bar{K} / K)$ and <code>$V_l:=T_l(E)\otimes \mathbb{Q}_l$</code>.</p> <p>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 <code>$\mathfrak{g}_l$</code> 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. </p> <p>(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). </p> <p>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 $K^{cycl}$).</p> <p>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 $\mathbb{Q}_l$).</p> <p>Faltings proved two important facts about represenations $\rho_{l,E}$: <ul></p> <li><p>$\rho_{l,E}$ is a semisimple representation of $G_K$ over $\mathbb{Q}_l$</p></li> <li><p><code>$\text{End}(E)\otimes \mathbb{Q}_l \cong \text{End}_{\mathfrak{g}_l}(V_l)$</code>.</p></li> </ul> <p>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.</p>