7
$\begingroup$

Suppose $E/\mathbb Q$ is a non CM elliptic curve and we look at the number field $K_d$ generated by the $d$-torsion of $E$. What is known about the (complete) splitting of small primes in $K_d$?

More precisely, since $|E(\mathbb F_p)| \sim p+1$, if $p$ were a completely split prime, we would need $d^2 < p+1$ (approximately). What can say about primes in, say, the range $d^2 < p < d^4$? What proportion of them split completely (with asymptotic dependence on $d$)?

(I would also be interested in the analogous question for $\mathbb G_m$.)

$\endgroup$
8
$\begingroup$

Let's discuss the $\mathbb G_m$ question first. For $p$ to split completely in the field generated by the $d$-torsion of $\mathbb G_m$, i.e. the field generated by the $d$th roots of unity, a necessary and sufficient condition is that $\mathbb F_p$ contains the $d$th roots of unity, i.e. $p \equiv 1\mod d$. This requires $p>d$ so I guess the analogous question would be to count primes congruent to $1$ mod $d$ between $d$ and $d^2$.

The usual heuristics suggest that this number should be roughly $d^2 / (2\phi(d) \log d)$. But a provable asymptotic is way too much to hope for. We don't even know a lower bound that this is nonzero - one would have to enlarge the range to $d < p < O(d^5)$ and apply Xylouris's strengthening of Linnik's theorem. Even under GRH the $d^2$ case is unknown.

Probably one can get upper bounds that are reasonably close to the truth using sieve methods.

For elliptic curves, the situation is similar, but more complicated. Some necessary conditions are that $p \equiv 1 \mod d$, since from two independent $d$-torsion points we can generate a $d$th root of unity by the Weil pairing, and $a_p \equiv p+1\mod d^2$, writing $E(\mathbb F_p) = p+1-a_p$.

Are the sufficient? The right perspective is to think of Frobenius as a $2 \times 2$ matrix acting on the torsion points, $p$ as the determinant, and $a_p$ as the trace. Knowing the determinant $p$ is congruent to $1$ mod $d$ and the trace $a_p$ is congruent to $1+p$ mod $d^2$ does not suffice to guarantee that the matrix is congruent to the identity mod $d$ as the counterexample $\begin{pmatrix} 1 & 1 \\ 0 & p \end{pmatrix} $ shows.

However, they do imply that the elliptic curve is congruent to an elliptic curve with full $d$-torsion, as any counterexample must be more-or-less congruent to that one modulo $d^2$.

Because these conditions aren't quite sufficient, I don't know a criterion simpler then the claim that the Frobenius conjugacy class in $GL_2(\mathbb Z/d)$ is equal to the identity matrix.

We expect this to hold for a proportion of primes equal to $1$ divided by the image of the Galois group in $GL_2(\mathbb Z/d)$. For a non-CM elliptic curve, this will typically be almost as large as $|GL_2(\mathbb Z/d)|\approx d^4$, and so we expect no or very few such primes $<d^4$. So lower bounds are hopeless, but perhaps some upper bounds exist, although they will surely be much harder than the case where all you have is a congruence condition.

For a CM elliptic curve, the situation is better, and you can express it using CM theory as a congruence condition on the primes lying over $p$ in the CM field, so the problem will only be as hard as an imaginary quadratic variant of the previous case (i.e., too hard to give an asymptotic.)

$\endgroup$
6
  • $\begingroup$ Let $K$ be the field generated by the $d$-torsion points of an elliptic curve $E/\mathbb{Q}$; assume for discussion that $E/\mathbb{Q}$ does not have complex multiplication. Chebotarev type arguments would optimistically suggest that the least unramified prime $p$ that splits completely in $K$ has size in the range $\log D_K$ up to $(\log D_K)^2$, where $D_K$ is the absolute discriminant of $K$. To go further than that, you would need to use other ideas. Note that each prime dividing $D_K$ also divides $dN$, and conversely. The power to which each of these primes is raised is another issue. $\endgroup$ – 2734364041 Mar 15 at 10:26
  • $\begingroup$ @2734364041 Shouldn't the lower bound of your range have more to do with the degree of the field than the size of the discriminant? It's not like quadratic fields of discriminant $D$ have no split primes smaller than $\log D$ for $D$ large, for example. $\endgroup$ – Will Sawin Mar 15 at 13:22
  • $\begingroup$ It depends on the Galois extension. See Fiori's "Lower bounds for the least prime in Chebotarev" for examples where least split prime is naturally bounded in terms of the discriminant and Sardari's "The least prime number represented by a binary quadratic form" for examples where the least split prime is expected to be bounded naturally in terms of the degree. $\endgroup$ – 2734364041 Mar 15 at 14:48
  • $\begingroup$ @2734364041 Fiori's example is for a very specific type of extension. Saying it "has size in the range" implies that the lower bound holds for all extensions. $\endgroup$ – Will Sawin Mar 15 at 15:51
  • $\begingroup$ When I say "Chebotarev type arguments", I mean "what do zeros of Hecke/Artin $L$-functions indicate?" In certain situations (like splitting conditions in ring class fields / representation of primes by BQFs), you can use geometry or other "low-tech" means to handle the problem, and the zeros are incapable of detecting these super-small split primes. I'm not trying to suggest that the zeros handle everything. I'm suggesting that in this problem, the conductor of the elliptic curve might come into play. Maybe it won't. But it's something to think about, that's all. $\endgroup$ – 2734364041 Mar 15 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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