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Let $p$ be a prime number, $K/\mathbf{Q}_p$ a finite extension, with integers $O_K$, valuation ideal $\mathfrak{p}$, and residue field $k_\mathfrak{p}$. Let $E$ be an elliptic curve over $K$ with good reduction $E_\mathfrak{p}$ over $k_\mathfrak{p}$.

If $\ell$ is a prime $\neq p$, then $T_\ell(E)$ is identified with $T_\ell(E_\mathfrak{p})$ in a natural way, by the good reduction of $E$. As it turns out such a Galois representation is determined, up to isomorphism, by the characteristic polynomial $f_{E_\mathfrak{p}}(x)=x^2-a_{E_\mathfrak{p}}x+|k_\mathfrak{p}|$ associated to $E_\mathfrak{p}$ and by $j_E$ mod $\mathfrak{p}=j_{E_\mathfrak{p}}$ UNLESS we are in the following (very) ${\it special}$ case:

$p\equiv 3$ mod $4$; $|k_\mathfrak{p}|=p^{2m+1}$; $a_{E_\mathfrak{p}}=0$; $\ell=2$; and $j_E\equiv 1728$ mod $\mathfrak{p}$.

If the first three conditions hold, then $E_\mathfrak{p}$ is supersingular and its endomorphisms ring over $k_\mathfrak{p}$ is "only" isomorphic to an order in $\mathbf{Q}(\sqrt{-p})$ containing $\sqrt{-p}$, and thus isomorphic to either $\mathbf{Z}[\sqrt{-p}]$ or to $\mathbf{Z}[(1+\sqrt{-p})/2]$. The second case occurs precisely when all the two torsion is defined over $k_\mathfrak{p}$, the first case when $E_\mathfrak{p}[2]$ has only two $k_\mathfrak{p}$-points. Both cases do arise and give rise to non-isomorphic $T_2(E_\mathfrak{p})$.

Essentially by Deuring's Lifting Lemma one can decide which of the two possibilities occurs by looking at the $j$-invariant of $E_\mathfrak{p}$ UNLESS this is equal to 1728. The point is that if $j_{E_\mathfrak{p}}\neq 1728$ then the two $k_\mathfrak{p}$-forms of $E_\mathfrak{p}\otimes_{k_\mathfrak{p}}\bar k_\mathfrak{p}$ lying in the $k_\mathfrak{p}$-isogeny class $a_{E_\mathfrak{p}}=0$ have the same ring of endomorphisms over $k_\mathfrak{p}$, out of the two possibilities listed above. The opposite being true when $j_{E_\mathfrak{p}}=1728$ (this fact is very related to the analysis of the mod $p$ reduction of Hilbert Class Polynomials associated to discriminants $-p$ and $-4p$ done by Gross and Elkies (cf. $\S 2$, Proposition, in Elkies' "The existence of infinitely many supersingular primes for every elliptic curve over $\mathbf{Q}$", Inventiones 89 (1987))).

In other words, in the special case the pair $(f_{E_{\mathfrak{p}}}(x), j_E$ mod $\mathfrak{p})$ does ${\it not}$ determine $T_2(E_\mathfrak{p})$.

Here is the question then: in the special case can we determine what is the endomorphism ring of $E_\mathfrak{p}$ (and hence $T_2(E)$) from congruences of $j_E$ mod a higher power of $\mathfrak{p}$ (or of $p$)?

It is not even clear to me whether this should be possible, let alone what power of $p$ we would need to tell one case from the other. The hope behind this is that the $j$-invariant of $E$ be "close" to that of the CM lift of $E_\mathfrak{p}$ and of its endomorphisms ring over $k_\mathfrak{p}$. Thanks.

PS: I do not know if the question above has anything to do with Is there a "classical" proof of this $j$-value congruence?

[EDIT: I realize that for clarity of exposition I should have probably recalled that $T_\ell(E_\mathfrak{p})$ for $\ell\neq p$, in the above notation, is a free ${\rm End}(E_\mathfrak{p})\otimes\mathbf{Z}_\ell$-module of rank one. Therefore, roughly, the knowledge of either of the two is equivalent to that of the other]

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  • $\begingroup$ Maybe I am seriously misunderstanding, but isn't Theorem 3.1 in Silverman's "Arithmetic of elliptic curves" saying that $End(E_{\mathfrak{p}})$ is an order in a quarternion algebra in supersingular case? $\endgroup$ – i707107 Mar 10 '12 at 20:04
  • $\begingroup$ That is the case once you enlarge your finite field $k$ to, say, an algebraic closure of it $\bar k$, so as to allow the supersingular elliptic curve $E_\mathfrak{p}$ to have all of its endomorphisms defined. In the special case described above, $End_k(E_\mathfrak{p})$ really is just an order in $\mathbf{Q}(\sqrt{-p})$. Since this order has to contain $\sqrt{-p}$ (it has to be maximal at $p$), there are only the two possibilities mentioned above for it. We cannot tell one from the other just by looking at the $j$-invariant of $E_\mathfrak{p}$. That's the problem! $\endgroup$ – Tommaso Centeleghe Mar 10 '12 at 23:30
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For simplicity assume $p \neq 3$. Given an elliptic curve $E$ in Weierstrass form with one two-torsion point defined over its field of definition, the other 2-torsion points are defined over the field of definition if and only if the discriminant $g_2^3 - 27 g_3^2$ is a perfect square. (Clear from properties of the discriminant of a cubic).

So we need to determine whether the discriminant is a perfect square mod $\mathfrak p$.

We have $$j = 1728 \frac{g_2^3}{ g_2^3 - 27 g_3^2} = 1728 + \frac{ (216g_3)^2}{ g_2^3 - 27 g_3^2} $$

so $j-1728$ is a perfect square divided by the discriminant.

To determine whether the discriminant is a perfect square mod $\mathfrak p$, it is sufficient to evaluate the $\mathfrak p$-adic leading term of $j$, then divide by any perfect square with equal $\mathfrak p$-adic valuation, obtaining a residue mod $\mathfrak p$, which is either a quadrati residue or nonresidue. In the first case the order is maximal and the second case it is nonmaximal.

This works unless $j$ is exactly 1728, in which case it is clear that the $j$-invariant is not sufficient, as then there are multiple curves with distinct $2$-torsion but the same $j$-invariant.

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