# Formal groups in the supersingular reduction case

Dear MO,

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:

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

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

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

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. –  Kevin Buzzard Aug 13 '11 at 16:38
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) $v(a_p)<e$). –  Kevin Buzzard Aug 13 '11 at 16:40
Thank Kevin. I've modified the question so $K$ is chosen to be of minimal degree such that $E/K$ has good reduction. –  Álvaro Lozano-Robledo Aug 13 '11 at 17:17
But there is still this more important issue on how one chooses $X$. The formal group is something isomorphic to $R[[X]]$ with $R$ the integers of $K$. But there's no canonical $X$, so there's no canonical $a_p$ etc. For example one can change $X$ to $X'=X+\pi X^2$ and this will completely change $a_p$. –  Kevin Buzzard Aug 13 '11 at 18:10