5
$\begingroup$

Let $K$ be a number field, $E$ be an elliptic curve over $K$, $l \neq p$ be two different prime numbers and $v$ be a place of $K$ above $l$.

I am trying to understand the proof of proposition I.6.7 in the the book Euler Systems by Rubin (which you can find here : http://swc.math.arizona.edu/aws/1999/99RubinES.pdf)

I think that at some point he uses the fact $T_p(E)^{G_{K_v}}= 0$. Is that true and if yes why ? (we write $T_p(E) = \varprojlim E(\overline{K})[p^n]$, the Tate module of $E$ at $p$).

$\endgroup$
  • $\begingroup$ Good reduction at $\ell$? $\endgroup$ – Lubin Aug 28 '15 at 16:54
  • $\begingroup$ Well it doesn't seem to be an hypothesis in the book but if you know an answer in that case i'd be very interested. But (maybe i'm being silly), isn't it "harder" to prove in the good reduction case since it that case we know that the inertia acts trivially and we are left with only the action of Frobenius ? An other thought I had was that maybe we need that $G_K \to End(T_p(E))$ is surjective but I haven't been able to prove the result under this assumption either (an it's not an hypothesis in the book). Anyway any kind of help would be very much appreciated. $\endgroup$ – anton Aug 28 '15 at 17:19
  • $\begingroup$ Let me give an unsatisfactory answer below, which would easily be annihilated by a short answer from an expert, I’m sure. $\endgroup$ – Lubin Aug 28 '15 at 18:05
3
$\begingroup$

We’re looking in the local, complete situation above $\ell$ at the $p^m$-torsion points of $E$ for all $m$. What does it mean to say that $T_p(E)^{G_v}\ne0$, where $G_v=G_{K_v}$, the Galois group of an algebraic closure of $K_v$ over $K_v$? It would mean that there was a consistent sequence of $p^m$-torsion points of $E$, in particularly infinitely many of them, that are rational over a finite extension of $K_v$. But since we have the exact sequence $$ 0\>\rightarrow\>\widehat E(\mathfrak m)\>\rightarrow E(\mathfrak o)\>\rightarrow\>\tilde E(\kappa)\rightarrow\>0\,, $$ this can’t happen. Here, $\widehat E(\mathfrak m)$ is the points of the formal group of $E$ with values in the maximal ideal $\mathfrak m$ of the ring $\mathfrak o$ of $v$-integers of $K_v$ (or some finite extension if necessary); $E(\mathfrak o)$ is the $\mathfrak o$-points of $E$ (same as the $K_v$-points), and $\tilde E(\kappa)$ is the group of points of the reduced curve $\tilde E$ rational over the finite field $\kappa$ of characteristic $\ell$. But the points of the formal group are uniquely divisible by any prime different from $\ell$, so there’s no $p$-torsion there; and there are only finitely many points over the finite field. So no good.

$\endgroup$
  • $\begingroup$ Thank you very much, that's great :) $\endgroup$ – anton Aug 28 '15 at 19:58
  • 1
    $\begingroup$ I think I have found another way of answering the question. Indeed saying that the invariants are $0$ is equivalent to saying that the Frobenius has no fixed point on the invariants of the Inertia which in turn is equivalent to $det(1-Frob)_{|V_p(E)^{I}} \neq 0$. But we have very precise formulae for this (see arxiv.org/pdf/1009.5389v2.pdf page 18 and 19) which shows that this happens only if $E$ has split multiplicative reduction. $\endgroup$ – anton Aug 30 '15 at 21:56
  • $\begingroup$ So the result is not true in the generality claimed by Rubin (unless i'm mistaken) but it's true for good reduction (which we already knew by your proof). Since I only need Rubin's result in the good reduction case after all, everything is ok :) Thanks again for you help. $\endgroup$ – anton Aug 30 '15 at 21:56
  • $\begingroup$ Indeed, bad reduction is one of my many blind spots and areas of ignorance. I was definitely thinking of good reduction. $\endgroup$ – Lubin Aug 31 '15 at 1:16
  • $\begingroup$ I think that my answer, and yours, also works if you assume additive reduction as well since you can pass to a finite extension where it obtains good reduction. $\endgroup$ – Alex Youcis May 8 at 21:44
1
$\begingroup$

$\newcommand{\E}{\overline{E}}$Here's a pretty simple explanation in the good reduction case. Let $\mathcal{E}$ denote the unique elliptic curve model of $E$ over $\mathcal{O}$(:=$\mathcal{O}_{K_v}$) and let $\E$ denote the reduction of $\mathcal{E}$ over the residue field $\mathbb{F}_q$ of $K_v$ (where we denote by $q=\ell^r$ for some $r$). We know then that there is a $G_{K_v}$-equivariant isomorphism $T_p E\cong T_p \E$ where $G_{K_v}$ acts on the latter via its surjection $G_{K_v}\to G_{\mathbb{F}_q}$. A similar statement holds with $T_p$ replaced by $V_p$.

Now, if $T_p \E$ has a Galois fixed point then the matrix representation of of any $\rho(g)$, with $g\in G_{\mathbb{F}_q}$, acting on $V_p \E$ has the form $\begin{pmatrix}1 & \ast\\ 0 & \chi(g)\end{pmatrix}$ for some character $\chi:G_{\mathbb{F}_q}\to \mathbb{Z}_p^\times$. Moreover, since we know that $\det V_p \E$ is isomorphic to the $p$-adic cyclotomic character of $G_{\mathbb{F}_q}$ we deduce that $\chi=\chi_p$. Note then that for any $g\in G_{\mathbb{F}_q}$ we'd have that $\mathrm{tr}\rho(g)=1+\chi_p(g)$. In particular, we get that $\mathrm{tr}\rho(\mathrm{Frob}_{q})=1+q$ where $q$ is the size of the residue field of $\mathcal{O}$. Note though that $\mathrm{tr}\rho(\mathrm{From}_{q})=q+1-\# E(\mathbb{F}_{q})$. So, we deduce that $\# E(\mathbb{F}_q)=0$ which is preposterous.

It's also clear that this works for additive reduction since this is the same thing as potentially good reduction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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