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]

  • $\begingroup$ 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? $\endgroup$ Mar 10, 2012 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! $\endgroup$ Mar 10, 2012 at 23:30

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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