Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear MO,

Let $p$ be a prime and let $E/\mathbb{Q}_p$ be an elliptic curve. Suppose that $E/\mathbb{Q}_p$ has good ordinary reduction at $p$. In his 1972 paper ``Propriétés galoisiennes des points d'ordre fini des courbes elliptiques'' (more specifically, see Corollaire, in p. 274), Serre shows along the way that the inertia subgroup of $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ is, with respect to a suitable basis of $E[p]$, isomorphic to either a matrix group of the form {$[\ast\ 0; 0\ 1]$} or {$[\ast\ \ast; 0\ 1]$} as a subgroup of $\operatorname{GL}(2,\mathbb{F}_p)$. After this result, Serre remarks that he doesn't know of any simple criterion that would determine whether one is in the first case or the second case.

Question: Nowadays, do we know of a criterion to tell whether one is in the first case or the second case?

A more concrete question: Here is the particular example that I am working with: Let $E/\mathbb{Q}$ be ``1225h1'' in Cremona's tables, given by $$E : y^2 + xy + y = x^3 + x^2 - 8x + 6. $$ This curve has a rational $37$-isogeny and therefore $\operatorname{Gal}(\mathbb{Q}(E[37])/\mathbb{Q})$ is a Borel subgroup of $\operatorname{GL}(2,\mathbb{F}_{37})$. The curve $E$ has good ordinary reduction at $p=37$ and I am trying to find out whether the ramification index of $37$ in the extension $\mathbb{Q}(E[37])/\mathbb{Q}$ is just $\varphi(37)$ or rather $\varphi(37)\cdot 37$, where $\varphi$ is the Euler phi function.

The $37$th division polynomial of $E/\mathbb{Q}$ has degree $684$ and it factors (over $\mathbb{Q}[x]$) as a product of $4$ polynomials of degrees $6$, $6$, $6$ and $666$, respectively. The extension of degree $666$ is, well, diabolically large and I can't find the ramification at $37$ computationally... or at least I don't know how to!

Thanks for your help!

share|improve this question
    
You won't need to factor your diabolic polynomial; the slopes of the Newton polygon in $\mathbb{Q}_p[x]$ are enough to determine the ramification. In your case there are 666 unit roots and 18 of valuation $−1/18$. –  Chris Wuthrich Jul 9 '11 at 8:05
    
Chris, if I am not mistaken, the slopes of the Newton polygon will determine the valuations of the x-coordinates of the 37-torsion points. But even if the valuations of the roots of the polynomial of degree $666$ are $0$, that doesn't mean that the extension generated by the roots is unramified, does it? For instance, the slopes of $1+x+x^2$ are $[0,0]$ for $p=3$ but, of course, the prime $3$ ramifies in $\mathbb{Q}(\zeta_3)/\mathbb{Q}$. –  Álvaro Lozano-Robledo Jul 9 '11 at 14:20
    
Absolutely correct. I am embarrassed about my mistake. The only thing it shows is that the reduction is ordinary. –  Chris Wuthrich Jul 10 '11 at 15:18
add comment

1 Answer

up vote 10 down vote accepted

Assume $p \ne 2$. The condition for the representation to be tamely ramified (i.e $* = 0$ in the upper right entry of the matrix) is that $j(E) \equiv j_0 \mod p^2$ where $j(E)$ is the $j$-invariant of $E$ and $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$. This is proved in Gross "A tameness criterion for galois representations..." Duke J. 61 (1990) on page 514. For $p=2$ you need the congruence modulo $8$. Serre gives an algorithm for computing $j_0$ in Lubin-Serre-Tate.

share|improve this answer
    
+1. Could you indicate which reference is meant by "Lubin-Serre-Tate"? –  user1594 Jul 8 '11 at 23:03
4  
@JT: It's a very old famous paper that remained unpublished for a long time. When the web was young, I took upon myself to make it available online. Now, you can probably just google for it. Anyway, here is the link: ma.utexas.edu/users/voloch/lst.html –  Felipe Voloch Jul 8 '11 at 23:41
    
Wonderful, thank you! –  user1594 Jul 9 '11 at 12:50
    
Thank you, Felipe! I will try to calculate $j_0$ in this case and report back. –  Álvaro Lozano-Robledo Jul 9 '11 at 14:22
1  
@Álvaro: Careful, $s(x)$ is not $x^2$, only congruent to it modulo $2$. It is the Frobenius on Witt vectors which squares the Witt coordinates. I doubt there is a typo, Serre would have picked it up. Most of the corrections I have listed on that page were his. –  Felipe Voloch Jul 12 '11 at 11:22
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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