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!

  • $\begingroup$ 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 $−1/18$. $\endgroup$ Jul 9, 2011 at 8:05
  • $\begingroup$ 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 $\mathbb{Q}(\zeta_3)/\mathbb{Q}$. $\endgroup$ Jul 9, 2011 at 14:20
  • $\begingroup$ Absolutely correct. I am embarrassed about my mistake. The only thing it shows is that the reduction is ordinary. $\endgroup$ Jul 10, 2011 at 15:18

1 Answer 1


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.

  • $\begingroup$ +1. Could you indicate which reference is meant by "Lubin-Serre-Tate"? $\endgroup$
    – user1594
    Jul 8, 2011 at 23:03
  • 5
    $\begingroup$ @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 $\endgroup$ Jul 8, 2011 at 23:41
  • $\begingroup$ Thank you, Felipe! I will try to calculate $j_0$ in this case and report back. $\endgroup$ Jul 9, 2011 at 14:22
  • $\begingroup$ Ok, if I am understanding everything correctly, there is a typo in the very last line of Lubin-Serre-Tate. When $p=2$ and $\lambda=1$, one should have $j_0 = 3375/31 = 3^3\cdot 5^3/31$. The polynomial $\phi_2(x,x^2)$ factors as $2^4(31x-3375)(3x^4 + 3x^3 + 82531x^2 + 26622000x + 2916000000)$, where $\phi_2(x,y)$ is the classical modular polynomial for $N=2$. The only root in $\mathbb{Q}_2$ congruent to $1$ mod $2$ is $3375/31$. $\endgroup$ Jul 12, 2011 at 4:09
  • 1
    $\begingroup$ @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. $\endgroup$ Jul 12, 2011 at 11:22

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