5
$\begingroup$

Dear MO,

Let $K/\mathbb{Q}_2$ be a finite extension, and let $E/K$ be an elliptic curve with good ordinary reduction, and such that $\mathbb{Q}_2(j(E))=K$. Let $\rho:\operatorname{Gal}(\overline{K}/K)\to \operatorname{GL}(2,\mathbb{Z}_2)$ be the $2$-adic representation associated to $T_2(E)$, the $2$-adic Tate module of $E$, and let $I_K$ be the image of the inertia subgroup via $\rho$. The theory tells us that there is a $\mathbb{Z}_2$-basis $\{P,Q\}$ of $T_2(E)$, such that $I_K$ is a subgroup of $$ \left\{\left( \begin{array}{cc} \ast & \ast \\\ 0 & 1\\\ \end{array}\right) \right\} \subset \operatorname{GL}(2,\mathbb{Z}_2),$$ and, moreover, if $\sigma\in I_K$, then the upper-left corner is $\chi_K(\sigma)$, where $\chi_K$ is the $2$-adic cyclotomic character $\chi_K:\operatorname{Gal}(\overline{K}/K)\to (\mathbb{Z}_2)^\times$. Let $\Phi$ be the image of $\chi_K$.

My question is the following: is $I_K$ necessarily a conjugate in $\operatorname{GL}(2,\mathbb{Z}_2)$ of a subgroup of the form $$ B(\Phi,\Psi)=\left\{\left( \begin{array}{cc} a & b \\\ 0 & 1\\\ \end{array}\right) : a\in \Phi, b \in \Psi \right\} \subset \operatorname{GL}(2,\mathbb{Z}_2),$$ for some fixed additive subgroup $\Psi$ of $\mathbb{Z}_2$? If so, what is the reason?

This would rule out some weird subgroups of $\operatorname{GL}(2,\mathbb{Z}_2)$ to appear as inertia subgroups. For example, can the following group $I$ appear as an inertia subgroup $I_K$ in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$? $$I=\left\langle \left( \begin{array}{cc} 5 & 2 \\\ 0 & 1\\\ \end{array}\right) \right\rangle =\left\{ \left( \begin{array}{cc} 1 & 0 \\\ 0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 5 & 2 \\\ 0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 1 & 4 \\\ 0 & 1\\\ \end{array}\right),\ \left( \begin{array}{cc} 5 & 6 \\\ 0 & 1\\\ \end{array}\right)\right\}\subset \operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z}).$$

Edit: The example group $I$ may appear as inertia in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$ if we do not assume that $K$ is the minimal field of definition of $E/K$, i.e., $K=\mathbb{Q}_2(j(E))$, for trivial reasons. For instance, suppose $E / \Bbb Q_2$ has inertia subgroup $B((\mathbb{Z}/8\mathbb{Z})^\times,\mathbb{Z}/8\mathbb{Z})$ in $\mathbb{Q}_2(E[8])/\mathbb{Q}_2$, and let $K$ be the fixed field $\Bbb{Q}_2(E[8])^I$, where $I$ is as in the example above modulo $8$. Then the inertia subgroup for $E/K$ in $\operatorname{GL}(2,\mathbb{Z}/8\mathbb{Z})$ is as indicated in the example.

Thank you!

Improve this question
$\endgroup$
3
  • $\begingroup$ Please tell me where you are confused, as my response completely answers your problem. $\endgroup$
    – Socky
    Apr 17, 2013 at 14:59
  • $\begingroup$ @Rabelais, if I understand your comment correctly, you say that any finite subgroup of $(\ast \ast; 0 1) \bmod 8$ may appear as inertia for some $E/K$ in $\text{Gal}(K(E[8])/K)$. But what about the $2$-adic $I_K$? Is it necessarily a subgroup of finite index of a $B(\Phi,\Psi)$ as I defined above? $\endgroup$
    – Álvaro Lozano-Robledo
    Apr 18, 2013 at 1:30
  • $\begingroup$ Why are you changing the original question? You asked whether $I_K$ was equal to a conjugate of $B(\Phi,\Psi)$, now you are asking whether it is a finite index subgroup. The answer to the new question is obviously yes. $\endgroup$
    – Socky
    Apr 19, 2013 at 16:02

1 Answer 1

Reset to default
4
$\begingroup$

The assumption that $K$ is not the minimal field of definition is no assumption at all. By Krasner's Lemma, the $E[8]$-torsion is locally constant in any family of elliptic curves over $K$, so simply deform your example slightly (e.g. by $2$-adically peturbing the coefficients of the Weierstrass equation) so that the $j$-invariant of the resulting curve cuts out $K$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ One needs to do more than Krasner's Lemma in order to control the structure of the Galois representation: Krasner's Lemma only controls very coarse information, the splitting field. One has to use the Henselian property of analytic local rings or other input from non-archimedean analytic geometry to rigorously justify the local constancy assertion being made here. $\endgroup$
    – user30180
    Apr 19, 2013 at 12:58
  • $\begingroup$ It is clear that a basis $P$, $Q$ will vary continuously in the coefficients of the Weierstrass equation (considered as elements of the field $K$) because $[8][x,y] = 0$ is reduced - this is the inverse function theorem. But perhaps we are not disagreeing on mathematics, rather only on what is obvious or not. $\endgroup$
    – Socky
    Apr 19, 2013 at 16:10
  • $\begingroup$ I agree that an essential input is the analytic inverse function theorem (applied to the finite etale cover of trivializations of the $n$-torsion). I agree that Krasner's Lemma provides initial intuition for believing the result, and one could certainly use the analytic inverse function theorem to give a proof of Krasner's Lemma that is a bit more advanced than the usual proof, but I don't think the proof of the actual result in question here ("local constancy" for the Galois representation in the rational-point fibers of a finite etale group over an analytic space) uses Krasner's Lemma. :) $\endgroup$
    – user30180
    Apr 19, 2013 at 16:58
  • $\begingroup$ If you take it as a straightforward analytic fact that $P = [x,y]$ deforms to $P' = [x',y']$ and $Q = \sigma P = [X,Y]$ deforms to $Q' = [X',Y']$, then, by Krasner's lemma, $P'$ and $P$ are defined over the same field, and $\sigma P'$ is sufficiently close (by continuity of Galois actions and etaleness) to $Q'$ to deduce that $\sigma P' = Q'$. This pretty much is the same argument as the usual first application of Krasner's Lemma, namely that the Galois groups of separable polynomials are locally constant. I guess one may prove this simultaneously with Krasner's lemma itself... $\endgroup$
    – Socky
    Apr 19, 2013 at 17:18
  • $\begingroup$ ...but I will still maintain that this local constancy "is" Krasner's lemma. $\endgroup$
    – Socky
    Apr 19, 2013 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.