This is a sequel to the following previous MathOverflow posts, but the question itself appears to be of a different flavor, being limited to rational curves:

Are most curves over Q pointless?

How many curves in a family possess a rational point?

Are most cubic plane curves over the rationals elliptic?

Let $B$ be a geometrically irreducible quasi-projective variety over $\mathbb{Q}$ such that $B(\mathbb{Q})$ is infinite, and consider a flat family $X \to B$ of geometrically irreducible (projective) rational curves. Let $h : B(\bar{\mathbb{Q}}) \to \mathbb{R}$ be a Weil height associated to a projective embedding of $B$, and consider $$ c := \limsup_{T \to +\infty} \frac{\big\{ b \in B(\mathbb{Q}) \mid h(b) \leq T, \, X_b(\mathbb{Q}) = \emptyset \big\}}{\big\{ b \in B(\mathbb{Q}) \mid h(b) \leq T \big\}} \in [0,1]. $$ [Most likely, the limit actually exists. ]

The values of $c$ which arise in this way (with some $X/B$ and $h$) form a certain countable subset of $[0,1]$ which includes the two endpoints.

Question. What is this set $\{c\} \subset [0,1]$?

[Remark. Obviously this question has many variants, although it is conceivable that the answer could be the same in all of them. For example, we may restrict to families over $B = \mathbb{A}_{\mathbb{Q}}^1$. ]

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    $\begingroup$ One observation: you specify that the curves should be affine. I don't think that makes any difference. As soon as a (smooth, projective) rational curve has a single rational point, then it has a Zariski dense set of rational points. Thus, excluding finitely many points in the "boundary" will not change the problem. $\endgroup$ – Jason Starr Sep 28 '13 at 17:16
  • $\begingroup$ Thanks for spotting this, I will edit. I had considered the curves to be affine in relation to the other problem (in genus $> 1$, where the density should be $0$ unless there is a section); its purpose there was to remove a finite number of sections from a family of projective curves. But as you write, this has no significance in the genus zero case considered here. $\endgroup$ – Vesselin Dimitrov Sep 28 '13 at 17:25
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    $\begingroup$ Do you have an example where $c>0$ but the family doesn't have a section? $\endgroup$ – Felipe Voloch Sep 28 '13 at 20:22
  • $\begingroup$ Actually no, I asked this question as a follow-up to #138581, which I only saw today. I have to admit that I do not even know the density of $q \in \mathbb{Q}$ for which $qy^2 = x^2 +1$ has a solution... (although this should be easy). I will think about this and other examples. But I was motivated by this basic observation: if we let $q$ run through the primes alone, then the density is $1/2$. $\endgroup$ – Vesselin Dimitrov Sep 28 '13 at 20:48
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    $\begingroup$ Daniel has given a complete answer below. But for $x^2+y^2=n$ we can be more precise. This has a solution for $O(N/(\log N)^{1/2})$ of integers $n\le N$. I think this is due to Landau. $\endgroup$ – Felipe Voloch Sep 29 '13 at 0:40

Edit: I have clarified a bit more the relationship between conic bundles and quaternion algebras and the relationship to weak approximation, and tidied up some typos.

I have myself recently started studying problems of this type, and it turns out that there is much to be said and such questions lead to very rich research problems. As far as I am aware, the first person to study such problems in any kind of generality was Serre [1].

I work in the following setting: Let $B$ be a projective variety over $\mathbb{Q}$ with a Zariski dense set of rational points and let $X$ be a non-singular variety together with a flat morphism $\pi:X \to B$ such that every fibre is a non-singular projective curve of genus zero. Then it is well-known that such a curve is isomorphic to a plane conic, and flatness implies that all the fibres are plane conics. Hence we obtain a conic bundle. To such a conic bundle one may asssociate an element $Q_\pi$ of $Br_2(\mathbb{Q}(B))$. If for example the generic fibre of $\pi$ is given by $$ax^2 + by^2 = z^2, (*)$$ where $a,b \in \mathbb{Q}(B)$, then $Q_\pi$ is the class given by the quaternion algebra $(a,b)$. This construction is essentially a bijection, namely given a quaternion algebra $(a,b)$ over $\mathbb{Q}(B)$ one may associate to it in an obvious way the conic bundle $(*)$. This element controls most of the arithemtic of the family. One can show that $Q_\pi$ is trivial if and only if $\pi$ admits a rational section (we will use this fact later to address Felipe's question). Following Serre, I instead consider the limit $$C(B,\pi,H) = \limsup_{T \to +\infty} \frac{\big\{ b \in B(\mathbb{Q}) \mid H(b) \leq T, \, X_b(\mathbb{Q}) \neq \emptyset \big\}}{\big\{ b \in B(\mathbb{Q}) \mid H(b) \leq T \big\}},$$ where $H$ is a choice of height function on $B$.

Serre [1] basically only considers the case where $B = \mathbb{P}^n$. Let $H$ be the usual naive height on $\mathbb{P}^n$. Then Serre shows that $$C(\mathbb{P}^n, \pi, H)=0, $$ as soon as there exists a divisor $D$ in $\mathbb{P}^n$ such that the residue of $Q_\pi$ at $D$ is non-trivial (he in fact shows something more precise, see [1]). In particular in such cases, "most" conics in the family do not contain a rational point. Moreover, one knows that an element $A \in Br(\mathbb{Q}(T_1,..,T_n))$ admits a non-zero residue if and only if $A \not \in Br(\mathbb{P}^n_{\mathbb{Q}}) = Br(\mathbb{Q})$. In particular, we see that $C(\mathbb{P}^n,\pi,H) >0$ if and only if $Q_\pi=0$, which by the above remark happens if and only if $\pi$ obtains a rational section (which in particular implies that we must have $C(\mathbb{P}^n,\pi,H)=1$). I believe this answers your question in this case. Your specific conic bundle $$qx^2 -y^2 = z^2, (**)$$ corresponds to the class of the quaternion algebra $(t,-1)$ over $Br(\mathbb{Q}(t))$. In this case, Serre's more general result in fact shows that $$\#\{q \in \mathbb{Q}: H(q) \leq T, (**) \text{ has a solution}\} \ll \frac{T^2}{\log T}.$$

Here is how one may obtain conic bundles which do not admit rational sections, but for which $C(B,\pi,H) =1$. Suppose that $B$ admits a Brauer group element $A \in Br(B)$ which is the class of a non-trivial quaternion algebra, for which $A \otimes_{\mathbb{Q}_v} \mathbb{Q}_v = 0$ for every place $v$ of $\mathbb{Q}$ (such elements may e.g. be given by elements of the Tate-Shaferavich groups of algebraic tori, see Sansuc's classical paper [2]). As by assumption $A \otimes_{\mathbb{Q}_v} \mathbb{Q}_v = 0$ for all $v$, we see that every conic in the family is everywhere locally soluble. The Hasse-Minkowski theorem therefore implies that every conic in the family contains a rational point, hence $C(B,\pi,H) =1$. However, since $A$ was non-trivial we see that $\pi$ does not admit a rational section.

Your question in the general case is highly non-trivial. Assuming that the results proved by Serre hold in any kind of generality, one would expect that $C(B,\pi,H)=0$ if $Q_\pi \not \in Br(B)$. To obtain positive values of $C(B,\pi,H)$ which are not $1$, one would need a quaternion algebra $A$ over $\mathbb{Q}(B)$ for which there exists some place $v$ of $\mathbb{Q}$ and some $b_1,b_2 \in B(\mathbb{Q})$ such that $A_v(b_1) \neq 0 \in Br(\mathbb{Q}_v)$ but $A_v(b_2) = 0 \in Br(\mathbb{Q}_v)$. This would force some conics in the family to fail to have $\mathbb{Q}_v$-points (hence not have $\mathbb{Q}$-points), but at least make sure that there are also some conics in the family which still have $\mathbb{Q}_v$-points (hence could still have $\mathbb{Q}$-points). Such elements are exactly the ones which give rise to a failure of weak approximation for $B$ via the Brauer-Manin obstruction. So the values of $C(B,\pi,H)$ which may be obtained should intimately depend on the extent to which $B$ fails weak approximation, at least if the Brauer-Manin obstruction is the only one to weak approximation (for more details on the Brauer-Manin obstruction, see [2]).

Problems of this type naturally admit numerous generalistions e.g. to Brauer-Severi varieties and to number fields, where many of the above remarks still hold true.

[1] Serre - Spécialisation des éléments de $Br_2(\mathbb{Q}(T_1,..,T_n))$.

[2] Sansuc - Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.

  • $\begingroup$ Thank you very much! So it appears, after all, that the set of possible $C(B,\pi,H) \in [0,1]$ is (countably) infinite. I was wondering what sort of set it might be (topologically), but as you write, this appears to be a difficult question. $\endgroup$ – Vesselin Dimitrov Sep 29 '13 at 14:46

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