35
$\begingroup$

Let $V$ be a projective variety, defined over $\mathbb{Q}$. Suppose that for every number field $K \neq \mathbb{Q}$, there is a $K$-point of $V$. Does it follow that $V$ has a $\mathbb{Q}$-point?

More generally, what are the restrictions on the possible sets $S$ of number fields $K \supset k$ such that for some projective variety $V$ defined over $k$, we have that $V$ has a $K$-point if and only if $K \in S$? Note that there has to be some nontrivial restriction here, since there are only countably many projective varieties defined over number fields, but uncountably many sets of number fields closed under extensions.

$\endgroup$
25
$\begingroup$

I have often asked myself similar questions. As far as I recall, your question is wide open.

One can however say some things about such a variety:

1) $V$ must have a rational divisor of degree $1$. [Thus a curve with your property must have genus at least $2$.]

Indeed, it has points over both quadratic and cubic fields.

2) $V$ must have points everywhere locally.

Indeed, for all $p \leq \infty$, there exist number fields $K$ strictly larger than $\mathbb{Q}$ which embed in $\mathbb{Q}_p$.

Let me also note that given any finite field $\mathbb{F}_q$, it is not difficult to construct curves which have points over all proper algebraic extensions of $\mathbb{F}_q$ but not over $\mathbb{F}_q$. As a first exercise in this direction, one should determine the finite list of $q$ such that there exists such a genus $2$ curve over $\mathbb{F}_q$ (I am pretty sure that I once wrote down an example to show that this list is nonempty).

Here is a variant of your question that I have thought about enough to be quite sure that I don't know what to do with it: let $[C]: y^2 = P_4(x)$ be a hyperelliptic quartic curve over $\mathbb{Q}$ which has points everywhere locally but no $\mathbb{Q}$-rational points. Thus $C$ represents an order $2$ element of the Shafarevich-Tate group of its Jacobian elliptic curve. Clearly $C$ has many quadratic points: take any $x \in \mathbb{Q}$ and extract the square root of $P_4(x)$. I'm no analytic number theorist, but I believe it is pretty routine to see that $P_4(\mathbb{Q})$ hits infinitely many square classes in $\mathbb{Q}$, so there are infinitely many distinct quadratic splitting fields. If $C$ failed to have a point over $\mathbb{Q}_p$ for some $p$, we could also build infinitely many quadratic *non*splitting fields. But since $C$ has points everywhere locally, I don't know how to rule out the possibility that $C$ has points over every quadratic field $\mathbb{Q}(\sqrt{d})$. Does there actually exist such a curve? It seems reasonable to guess no, but who knows? Not me.

Finally, your question reminds me of this paper of Jordan Rizov, in which he proves a somewhat similar -- but weaker -- result. I think his paper is nice: he rather cleverly found something along these lines that one can actually prove.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ If the mod-2 representation of the Jacobian $E$ of $C$ is surjective, then there are no $2$-torsion points on $E(K)$ for quadratic $K$. So the rank of $E$ has to grow in every quadratic $K/\mathbb{Q}$ and so must the analytic rank. But that is not possible by non-vanishing results, I believe. Do I miss something ? $\endgroup$ – Chris Wuthrich Feb 4 '11 at 10:52
  • $\begingroup$ @Chris: well, it seems to me that you are imposing one (very mild) additional hypothesis about the $2$-torsion on $E(K)$ and then using some form of BSD to get from the rank to the analytic rank. But notwithstanding those cavils, you have a very nice idea here! I'll think about it and get back to you... $\endgroup$ – Pete L. Clark Feb 4 '11 at 14:16
  • $\begingroup$ You won't need BSD: If the analytic rank of the twist of $E$ corresponding to $K$ were zero, then, by Kolyvagin, the rank of this twist would be zero, too. $\endgroup$ – Chris Wuthrich Feb 8 '11 at 16:47
  • $\begingroup$ @Chris: very true. $\endgroup$ – Pete L. Clark Feb 18 '11 at 5:44
7
$\begingroup$

A simple case where the answer to your question is positive is provided by Heegner's Lemma, an observation made by Heegner in his solution of the class number 1 problem and extracted from his proof by Birch: if $y^2 = f_4(x)$, where $f_4$ is a quartic polynomial whose leading coefficient is not a square, has a solution in a number field of odd degree, then it has a rational point. This lemma has a cohomological interpretation, where the claim essentially boils down to the observation that if $a^n$ is a square for some odd $n$, then so is $a$.

I expect that there are other instances of Heegner's lemma for other types of curves, but I don't know anything specific.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ @Franz: since the (smooth projective model) of a hyperelliptic quartic is a genus one curve of index $1$ or $2$, this comes under the heading of my 1) above. $\endgroup$ – Pete L. Clark Feb 4 '11 at 9:23
  • 1
    $\begingroup$ D'accord - so I spelled it out for those of us who look at algebraic geometry as a mountain rather than a valley -) $\endgroup$ – Franz Lemmermeyer Feb 4 '11 at 18:32

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.