4
$\begingroup$

Given a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ with integers and random coefficients, what is more probable? To find a rational point on the conic or not?

$\endgroup$
  • $\begingroup$ It depends on the distribution on integers. $\endgroup$ – Fedor Petrov Dec 23 '15 at 18:24
  • 2
    $\begingroup$ @fedor The standard way to order is as a six tuple of of rational numbers treated as a point in projective space ordered by height. $\endgroup$ – Joe Silverman Dec 23 '15 at 19:49
11
$\begingroup$

For any given prime number $p > 2$ the probability that there is no $p$-adic point is $\ge c/p$ for some constant $c > 0$ indpendent of $p$. Since the sum over $1/p$ diverges, this implies that the `probability' (or rather, density) of conics with a rational point is zero.

EDIT: I assume that the `probability' is meant in the sense of density, i.e. $$ \lim_{N \to \infty} \frac{\#\{(A,B,C,D,E,F) \in ({\mathbb Z} \cap [-N,N])^6 : \exists \text{ rational point}\}}{(2N+1)^6}. $$ The probability of $p$-adic solubility (with respect to the standard measure on ${\mathbb Z}_p^6$) comes down to the $p$-adic density of pairs of $p$-adic integers $a, b$ such that the Hilbert symbol $(a,b)_p = -1$. This is the case (e.g.) when $v_p(a) = 1$, $v_p(b) = 0$ and $b$ is a nonsquare mod $p$; the probability for this is $\frac{p-1}{p^2} \cdot \frac{p-1}{2p} \ge \frac{2}{9p}$.

EDIT 2: I realized that the map that produces $a,b$ from $A,B,C,D,E,F$ very likely does not preserve $p$-adic measure. So here is an alternative and more direct argument. We work with projective coordinates, as suggested by Joe Silverman. By Hensel's Lemma, the non-existence of a smooth ${\mathbb F}_p$-point on the reduction mod $p$ of the conic is a necessary condition for the non-existence of $p$-adic points, and this is also sufficient when the conic is regular over ${\mathbb Z}_p$. So whenever the reduction mod $p$ is the product of two conjugate and distinct linear forms with coefficients in ${\mathbb F}_{p^2}$ and the intersection point of the two corresponding lines is regular, then there is no $p$-adic point on the conic. The $p$-adic measure (as a subset of ${\mathbb P}^5({\mathbb Q}_p)$) of the corresponding set is $$\frac{1}{2} \frac{\#{\mathbb P}^2({\mathbb F}_{p^2}) - \#{\mathbb P}^2({\mathbb F}_p)}{\#{\mathbb P}^5({\mathbb F}_p)} \frac{p-1}{p} = \frac{(p-1)^2}{2(p^3+1)} \ge \frac{3}{14p} .$$ (The factor $(p-1)/p$ is the probability that the intersection point is regular.) The paper by Bhargava, Cremona, Fisher, Jones and Keating linked in post.as.a.guest's answer gives the precise value $p/(2(p+1)^2)$.

$\endgroup$
  • $\begingroup$ Thanks Michael Stoll. Why the constant $c$ is independent of $p$? $\endgroup$ – user84475 Dec 23 '15 at 18:20
  • $\begingroup$ It is more than $c/p$ for the natural measure on $p$-adics. But there is no probabilistic measure on integers which is uniform modulo any prime. $\endgroup$ – Fedor Petrov Dec 23 '15 at 18:25
  • 1
    $\begingroup$ @MichaelJoyce : thanks, I've fixed the omission. $\endgroup$ – Michael Stoll Dec 23 '15 at 20:05
  • $\begingroup$ It would seem to be more natural to restrict to 6-tuples with $\gcd(A,B,\ldots,F)=1$, since the quadratic form is determined by the homogeneous coordinates $[A,B,\ldots,F]$. Certainly this is the right thing to do if one wishes to generalize from $\mathbb Q$ to a number field. Of course, this won't affect your conclusion that the density is 0. $\endgroup$ – Joe Silverman Dec 23 '15 at 22:43
13
$\begingroup$

Problems of this type are considered by Serre in the paper:

Serre - Spécialisation des éléments de $\mathrm{Br}_2(\mathbb{Q}(T_1,\ldots, T_n))$

The case relevant to you is Exemple 4. Here Serre shows that

\begin{align*} &\#\{|A|,|B|,|C|,|D|,|E|,|F| \leq N : \\ & Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \text{ has a rational point} \} \ll \frac{N^6}{(\log N)^{1/2}}. \end{align*}

This gives a more precise quantitative version of "probability $0$" mentioned by Michael Stoll.

Hooley has in fact shown that Serre's bound is sharp. This result is the object of the paper:

Hooley - On ternary quadratic forms that represent zero II.

$\endgroup$
3
$\begingroup$

There is a paper, that answers this in more generality.

https://www.dpmms.cam.ac.uk/~taf1000/papers/isotropic.html

In your case, the probability is 0 (asymptotically) under conditions that the distribution is piecewise smooth and rapidly decaying, though likely can be generalized, in your case.

$\endgroup$
  • $\begingroup$ The linked paper is definitely helpful. $\endgroup$ – Sebastian Goette Dec 24 '15 at 10:07

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.