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Let $E/\mathbb{Q}$ be an elliptic curve with potential good supersingular reduction at $p$. Thus, there is a finite extension $K/\mathbb{Q}_p$ such that $E/K$ has good supersingular reduction. Let us choose $K/\mathbb{Q}_p$ of minimal degree such that $E/K$ has good reduction, and let us assume $E/K$ is given by a minimal model. Let $A$ be the ring of integers of $K$, let $\pi$ be a uniformizer for $A$, let $\nu$ be a valuation on $K$ with $\nu(\pi)=1$ and $\nu(p)=e$, and let $\hat{E}/A$ be the formal group associated to $E/K$.

Let $[p](X) = pX+\cdots=\sum_{k=1}^\infty a_kX^k$ be the formal power series for the multiplication-by-$p$ map on $\hat{E}$. Let $e_0=\nu(a_1)=\nu(p)=e$, and $e_1=\nu(a_p)$. Since $E/K$ is supersingular, we know that $e_2=\nu(a_{p^2})=0$.

If we define points in the plane by $P_0=(1,e)$, $P_1=(p,e_1)$ and $P_2=(p^2,0)$, then $N$, the Newton polygon of $[p](X)$ (for those roots with valuation $>0$) is given either by one single segment $P_0P_2$, or two segments $P_0P_1$ and $P_1P_2$, according to whether $ep/(p+1) \leq e_1$ or $ep/(p+1) > e_1$, respectively.

The number $e$ is a divisor of $12$, and if $p\geq 5$, then $e\leq 6$. If $e>1$, and $N$ has two segments, then $1\leq e_1 < e$. Suppose that we are in this case, i.e., $e_1 < e$.

Question: Are there any further constraints on the values of $e_1$?

More specifically:

Question 1: Are there any divisibility conditions on $e_1$, or further relations between $e$ and $e_1$? Can $e_1$ take any of the values $1\leq e_1< e$ ?

Question 2: Is there a more conceptual way to think about $e_1$?

In all examples I have computed, I get that $e_1=1$, $2$ or $4$ (but I haven't done an exhaustive search either). For instance:

$E=27a4,\quad p=3, \quad [K:\mathbb{Q}]=12, \quad e=12, \quad e_1=2;$

$E=121a2,\quad p=11, \quad K=\mathbb{Q}(\sqrt[6]{11}), \quad e=6, \quad e_1 = 4;$

$E=14450p2,\quad p=17, \quad K=\mathbb{Q}(\sqrt[3]{17}), \quad e=3, \quad e_1=1.$

Thank you in advance for any answers!

Alvaro

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  • $\begingroup$ I am confused about this question. Firstly, you don't seem to make any assumptions on $K$ other than it's a field over which $E$ gets good reduction, so I don't see why $e$ has to divide 12. But more importantly, "the" formal group attached to $E$ depends on a choices of parameter $X$, and there is a huge amount of choice for $X$ as far as I can see, so it's not even clear to me that $e_1$ is well-defined. I think that if $e_1<e$ then it's well-defined, but if it isn't (e.g. if $E$ has good ss reduction over $\mathbf{Q}_p$ then $e_1$ can be pretty much anything. $\endgroup$ Aug 13, 2011 at 16:38
  • $\begingroup$ For more positive comments -- you could look in Katz' Antwerp paper, where he really uses the power series you talk about and explains some things that really are canonical about it (e.g. he shows that $v(a_p)$ is independent of the choice of $X$ if (1) $X$ satisfies some mild conditions ($[\zeta]X=\zeta X$ for all $\zeta\in\mu_{p-1}$) and (2) $v(a_p)<e$). $\endgroup$ Aug 13, 2011 at 16:40
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    $\begingroup$ But there is no canonical Weierstrass equation for $E/K$ right? $\endgroup$ Aug 13, 2011 at 20:03
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    $\begingroup$ In my mind, "the formal group associated to $E/K$ is just: take the Neron model, and then complete along the identity section, getting a ring isomorphic to $R[[X]]$ (because of smoothness -- but with no canonical isomorphism) equipped with a comultiplication. For me, there is no natural choice of $X$ -- what is canonical is only the ideal $(X)$, which has an awful lot of generators. $\endgroup$ Aug 13, 2011 at 20:09
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    $\begingroup$ Does the following paper of Coleman help? math.berkeley.edu/~coleman/Canonical/Canonical.pdf $\endgroup$ Aug 14, 2011 at 0:47

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