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Does there exist a variety $X$ over $\mathbb{Q}$ (or a number field) such that it has no rational points over $\mathbb{Q}$ but acquires points over any quadratic extension $\mathbb{Q}(\sqrt{d})$?

If there is an example, how often can this happen? Can we generalize to an extension of fixed degree?

(This might be a stupid question and I'm guessing answer is no, but I couldn't show it)

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    $\begingroup$ Very possibly there is no such $X$, but the question is apparently a difficult open problem. A variant of it has been discussed here: mathoverflow.net/questions/54263/… . $\endgroup$ – Vesselin Dimitrov Mar 6 '15 at 2:04
  • $\begingroup$ I see. It would be nice to know some cases where this has a negative answer. For example equations of the form $y^2=f(x)$? $\endgroup$ – Gazerun Mar 6 '15 at 2:55
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    $\begingroup$ For some classes of varieties one can prove that this cannot happen. For example, if a smooth cubic hypersurface over $\mathbb{Q}$ has a point over a single quadratic extension of $\mathbb{Q}$, then it already has a point over $\mathbb{Q}$. $\endgroup$ – Daniel Loughran Mar 6 '15 at 9:22
  • $\begingroup$ I think if one replaces Q with Z the answer is positive. $\endgroup$ – joro Mar 6 '15 at 9:42
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    $\begingroup$ Another comment is that there will be no such variety in any situation where the Hasse principal holds. $\endgroup$ – John Binder Mar 6 '15 at 15:50
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Per Gazerun's comment answer to related question which leads to relaxation of the OP.

This is possible over the integers for all $d$ and the question can be relaxed by allowing finitely many points over the rationals.

First we define the set $\{1,-1\}$ with the equation $(m-1)(m+1)=0$.

The only integers solutions to $x^2-m n^2y^2=1$ are $x=\pm 1, ny=0$ and $y=\pm 1,x=0$ and $n=\pm 1,x=0$.

Over $\mathbb{Z}[\sqrt{d}]$ for $n=\sqrt{d}$ this is Pell equation $x^2 \pm ny^2=1$ with infinitely many solutions and if $d$ is square we are in the above case.

So we must get rid of the bad points $x=\pm 1,0$.

Use the following cheap trick: $ (x^2-1)x z = 1$. This is linear in $z$ unless $(x^2-1)x=0$ which leads to $0=1$.

So our final system of equations is $(m-1)(m+1)=0,x^2-m n^2y^2=1, (x^2-1)x z = 1$ which doesn't have integer solutions but have infinitely many over $\mathbb{Z}[\sqrt{d}]$ for all $d$. If $d$ is negative chose $m=-1$ otherwise $m=1$


Partial result, some one could try to extend it.

It is possible to define variety the complement of $x=y^2$, that is $x \ne y^2$:

$$ zt=1 \qquad (1)$$ $$ (x-y^2) z=1 \qquad (2)$$.

In (1) and (2) $z$ can be any nonzero rational. $x\ne y^2$ for obvious reasons and $x-y^2=1/z$.

Appears to me this allows to describe the complement of variety given by single equation.

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According to this answer by Laurent Moret-Bailly, the set of rational non-squares is diophantine over the rationals (a result of Bjorn Poonen): there is a polynomial $P(a,x_1,\dots,x_n)$ which for $a\in\mathbb{Q}$ has a rational solution iff $a$ is a rational non-square. For the variety, take the hypersurface with equation $P^2+(x_{n+1}^2-a)^2=0$ in $n+2$ coordinates $(a,x_1,\dots,x_{n+1})$.

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  • $\begingroup$ I guess you mean the variety "$P^2+(x_{n+1}^2-a)^2=0$", viewed as hypersurface of the $(n+2)$-dimensional affine space with coordinates $(a,x_1,\dots,x_{n+1})$. Also it seems you want, for $a\in\mathbf{Q}$, that $P(a,x_1,\dots,x_n)=0$ have a rational solution iff $a$ is not a rational square (which is covered by the linked post, actually easier than the half-integral version). $\endgroup$ – YCor Oct 16 '19 at 10:14
  • $\begingroup$ @Ycor Thanks. I edited adding equality. $\endgroup$ – joro Oct 16 '19 at 10:54
  • $\begingroup$ The linked answer does not give a polynomial that has a rational solution iff $a$ is integer non-square. It gives a polynomial that has a rational solution iff $a$ is not an integer square. That is, the polynomial always has a rational solution when $a$ is a non-integer rational. Whether the set of integer non-squares is diophantine over the rationals is a hard open problem, as it easily implies that the set of integers itself is diophantine over the rationals. $\endgroup$ – Emil Jeřábek Oct 16 '19 at 15:02
  • $\begingroup$ @EmilJeřábek: Integers are irrelevant here. Just replace "integer non-squares" by "rational non-squares" in the answer and everything is OK. $\endgroup$ – Laurent Moret-Bailly Oct 16 '19 at 16:46
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    $\begingroup$ By the way, the result of Dittmann mentioned in the linked answer suggests that the following generalization is also true: for any $n$, there exists a variety over $\mathbb Q$ that has no rational points over $\mathbb Q$, but it has rational points over every proper extension of $\mathbb Q$ of degree at most $n$. $\endgroup$ – Emil Jeřábek Oct 16 '19 at 19:54
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Consider a hyperelliptic curve $y^2=f(x)$. I think by Falting's theorem on subvarieties of abelian varieties, we know that all but finitely many points of this over quadratic subfields have $x \in \mathbb Q$.

So this curve is such a variety if and only if the equation $y^2 D = f(x)$ has solutions for all but finitely many squarefree $D$.

Heuristically this shouldn't happen as the squarefree part of a random number $n$ is rarely much smaller than $n$, so the squarefree parts of $f(x)$ should grow quickly with $x$, because $f(x)$ has degree at least $3$, and so most numbers should not appear. In other words, because most hyperelliptic curves have no rational points, we expect the family $y^2 D = f(x)$ to mostly have no rational points, unless there is something strange going on with $f$.

But can this property actually be established for hyperelliptic curves?

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    $\begingroup$ Your first claim is false in general for curves of genus 2: if the Mordell-Weil group is infinite, then its elements will give you infinitely many quadratic points with irrational $x$-coordinate. (Let $P$ be a rational point on the Jacobian, then $P = [P_1 + P_2] - W$ where $P_1$ and $P_2$ are either two rational points or two conjugate quadratic points and $W$ is the canonical class. There are only finitely many rational points, so the second possibility occurs infinitely often, and for $P \neq 0$, $x(P_1) \notin \mathbb Q$.) $\endgroup$ – Michael Stoll Mar 6 '15 at 20:42
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    $\begingroup$ For curves of genus 3 or larger, you can consider the image of the symmetric square of the curve inside the Jacobian and apply Faltings' result to it. If the curve is not a doule cover of an elliptic curve of positive rank (this would give you a copy of the elliptic curve inside the symmetric square), then this tells you that there are only finitely many points on the symmetric square outside the "anti-diagonal", which means only finitely many quadratic points with irrational $x$-coordinate. $\endgroup$ – Michael Stoll Mar 6 '15 at 20:46
  • $\begingroup$ @MichaelStoll Yes. The proof (which isn't completely straightforward, but not hard) is in "Bielliptic curves and symmetric products." Proc. Amer. Math. Soc. 112 (1991). It was greatly generalized by Abramovich and Harris in "Abelian varieties and curves in $W_d(C)$." Compositio Math. 78 (1991). $\endgroup$ – Joe Silverman Mar 6 '15 at 21:24
  • $\begingroup$ @JoeSilverman OK, I see. Thanks for the references! $\endgroup$ – Michael Stoll Mar 6 '15 at 22:18
  • $\begingroup$ @MichaelStoll - Good point! So let me take $g \geq 3$. $\endgroup$ – Will Sawin Mar 7 '15 at 4:16
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Will Sawin and Michael Stoll have noted that, as a consequence of Faltings's "Big Theorem," a hyperelliptic equation $y^2 = f(x)$ with $\deg{f} > 6$ (genus $> 2$) and not admitting a degree $2$ non-constant map to an elliptic curve, has all but finitely many of its quadratic solutions $x, y \in \mathbb{Q}(\sqrt{d}), \, d \in \mathbb{Z}$, satisfy $x \in \mathbb{Q}$. We may add to this an argument due to Granville in Rational and integral points of quadratic twists of a given hyperelliptic curve to show:

Claim. The ABC conjecture implies, for a fixed $f$ having $\deg{f} > 6$ and no repeated roots, that the number of squarefree $d$ in $|d| \leq D$ for which the equation $dy^2 = f(x)$ has a rational solution with $y \neq 0$, is $O(D^{2/3+o(1)})$.

This answers Will's question under ABC. Hence ABC takes care of the problem for most $y^2 = f(x)$ - save for the ones of genus one or two or those doubly covering an elliptic curve. (For rational curves $C/\mathbb{Q}$ the problem is easy: Hasse's theorem shows that $C(\mathbb{Q}) = \emptyset$ is only possible when $C(\mathbb{Q}_p) = \emptyset$ for some prime $p$, but then $C$ will not have points in any quadratic field $\mathbb{Q}(\sqrt{d})$ split by $p$.) It seems to me that the case of hyperelliptic curves of genus $> 2$ doubly covering an elliptic curve can be settled under ABC by similar methods (Faltings's theorem and a modification of Granville's argument), whereas the genus one case should be solved unconditionally by Kolyvagin's theorem and non-vanishing results, cf. Chris Wurthrich's comment in the linked question. I am not sure about the genus two case though - it is barely missed by the argument below.

Proof of the claim. (Granville). M. Langevin has noted (cf. Thm. 12.2.12 in Heights in Diophantine Geometry by Bombieri and Gubler) that Elkies's construction for "ABC $\Rightarrow$ Roth" via Belyi maps yields in fact much more than Roth's theorem:

Lemma. Let $\varepsilon > 0$ and let $F \in \mathbb{Z}[X,Y]$ be a homogeneous polynomial with distinct linear factors over $\mathbb{C}$. Then for all co-prime $m,n$ with $F(m,n) \neq 0$, the (strong) ABC conjecture implies $\mathrm{rad}(F(m,n)) \gg_{\varepsilon,F} \max(|m|,|n|)^{\deg{F}-2-\varepsilon}$.

The ABC conjecture is recovered as the special case $F(X,Y) = XY(X+Y)$, whereas Roth's theorem is just the weakening of this statement dropping the radical.

We apply this is follows. Consider the $\asymp T^2$ rational values $x = m/n \in \mathbb{Q}$ with $T \leq \max(|m|, |n|) < 2T$, $(m,n)=1$, $n > 0$, and $f(x) \neq 0$. Writing $du^2 = F(m,n)$ with $F$ the homogenization of $f$, the above quoted form of ABC yields $$ (DT^{\deg{f}})^{1/2} \gg_f |dF(m,n)|^{1/2} = |du| \geq \mathrm{rad}(F(m,n)) \gg_{\varepsilon,F} T^{\deg{f} - 2 - \varepsilon}, $$ or $D > T^{\deg{f} - 4 - o(1)}$, in $|d| \leq D$. As the irreducible fraction $x$ uniquely determines the squarefree part $d$, we get what we want by splitting the range $T < D^{1/(\deg{f}-4-o(1))}$ into dyadic intervals.

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    $\begingroup$ Did you see my answer which appears to give positive answer? Some of the comments suggest results about algebraic geometry. $\endgroup$ – joro Oct 17 '19 at 11:01
  • $\begingroup$ @joro: Great, thanks! This solves the question for varieties of a sufficiently high dimension. For curves, it should still be negative. $\endgroup$ – Vesselin Dimitrov Oct 17 '19 at 11:52
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    $\begingroup$ Честита нова година! Здраве и щастие! $\endgroup$ – joro Jan 2 '20 at 7:32
  • $\begingroup$ Честит Нова година, Жоро! $\endgroup$ – Vesselin Dimitrov Jan 2 '20 at 16:19

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