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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!

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  • $\begingroup$ 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

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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 $K$.

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  • $\begingroup$ 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

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