Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

An explicit variant (although not quite a special case as it stands). Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero. Alternatively, drop the properness and generic smoothness assumptions, including instead the assumption that the smooth projective model of the generic fibre has genus $> 1$.

  • $\begingroup$ I guess the answer is trivially yes, because the hypotheses are never fulfilled. If $B(\mathbb{Z})$ contains an element $b$, then $X_b$ must be a proper smooth curve over $\operatorname{Spec}(\mathbb{Z})$ of genus $>1$, and such curves don't exist. $\endgroup$ – RP_ Feb 7 '13 at 21:56
  • $\begingroup$ Edited: what I really wanted to say is that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ was a smooth proper family of curves. So I have replaced "smooth" by "generically smooth." $\endgroup$ – Vesselin Dimitrov Feb 7 '13 at 22:01
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    $\begingroup$ That's expected but has not been proved. There has been some recent progress for the universal family of hyperelliptic curves by Bharghava, Gross, Poonen and Stoll. $\endgroup$ – Felipe Voloch Feb 7 '13 at 22:54
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    $\begingroup$ Although the meaning of this question has been clear from the start (and I haven't got a clue what the answer is), there are still issues with the precise formulation. If $B$ is a variety over $\mathbf{Q}$ then $B(\mathbf{Z})$ is empty. $\endgroup$ – user30035 Feb 8 '13 at 7:15
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    $\begingroup$ For "explicit variant" of your question you may consult the paper by Andrew Granville "Rational and integral points on quadratic twists of a given hyperelliptic curve" IMRN (2007), in particular Corollary 1 (ii) and Conjecture 1 (ii). I am wondering what is known (or conjectured) concering the same question for degree 4 (the mentioned paper deals with integer points for degree 4, but not with rational points). $\endgroup$ – duje Feb 10 '13 at 16:52

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