Let $\mathcal{E}/\mathbb{Q}(t)$ be given by $$y^2=x^3+A(T)x+B(T)$$ for some $A(T),B(T)\in\mathbb{Q}[T]$ and assume $\mathcal{E}$ is non-isotrivial (the $j$-invariant $\frac{6912 A(T)^3}{4A(T)^3 + 27B(T)^2}$ is not constant). Does there necessarily exist $t\in\mathbb{Q}$ such that the Mordell-Weil group of $\mathcal{E}_t/\mathbb{Q}:y^2=x^3+A(t)x+B(t)$ has positive rank?

An unconditional proof or explicit counterexample would be wonderful, but if that's not possible, I would be okay with a conditional proof assuming standard conjectures (e.g. BSD), or a proof for as wide a class of curves as possible, or a discussion of some properties a hypothetical counterexample would have.

Here are my thoughts so far:

  • We expect that for "most" families, at least $50\%$ of all specializations (ordering $t$ by height) are positive rank. One result in this direction is by Helfgott, who shows (assuming some standard conjectures) that if $\mathcal{E}$ has a finite place of multiplicative reduction, then half of the specializations $\mathcal{E}_t$ ordered by height have root number $-1$, and therefore have positive rank assuming the parity conjecture.
  • In contrast to Helfgott's work, there exist non-isotrivial families of curves with constant root number: for example, $W(\mathcal{E}_t)\equiv -1$ due to Rizzo and $W(\mathcal{E}_t)\equiv 1$ due to Bettin, David, and Delaunay (EDIT: These results only hold for all $t\in\mathbb{Z}$, not $t\in\mathbb{Q}$, so are not directly relevant; see comments for discussion). It turns out that in both of these families, $100\%$ of the specializations have positive rank (again assuming the parity conjecture), but it certainly may be possible to have a family with generic rank $0$ and constant root number $1$. Even in this case, though, I would still expect some rank $\geq 2$ specializations.
  • Joe Silverman conjectures that in fact any non-isotrivial family should have infinitely many positive rank specializations, but notes that it's not clear how one would prove this. My question is weaker (I'm only asking for a single specialization), and perhaps naively I would hope this makes it more tractable.
  • For any particular family, it is often possible to explicitly construct a rank $1$ subfamily (as Joe mentions in the answer I cited above, and Siksek demonstrates). There likely isn't any way to turn this into a general construction guaranteed to work for all families (if there were, it would prove Joe's conjecture).
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    $\begingroup$ I had long wondered this myself! Note that if true, it would realize the Suzuki group of order 29120 as a Galois group over $\mathbb{Q}$ (in the sense of the inverse Galois problem). This would probably also work for a number of other groups. See for example $\S4.3$ in arxiv.org/pdf/1510.05687.pdf $\endgroup$
    – Will Chen
    Oct 8 at 23:12
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    $\begingroup$ What you call “constant root number” examples is not really so: those only occur for parameter values $t$ in $\mathbf Z$, but that is not geometrically meaningful: let $t$ run over $\mathbf Q$ and you will find root numbers of both signs. Better examples are due to Cassels and Rohrlich. See the introduction of arxiv.org/abs/math/0408153 for those examples and what is known or expected. $\endgroup$
    – KConrad
    Oct 17 at 18:49

I suspect that this is unknown in general. I would guess that any method which produces at least one elliptic curve of positive rank should also produce infinitely many of positive rank, which as you already mention is an open problem.

Anyway, here is some discussion around this problem and a particular open case which would need to be resolved.

I first recall some facts that you already know to fix notation.

A generalisation of the Mordell-Weil theorem says that $\mathcal{E}(\mathbb{Q}(t))$ is a finitely generated abelian group. We denote by $r$ its rank, which is called the generic rank of the family. A theorem of Silverman says that for all but finitely many $t \in \mathbb{Q}$, the rank of $\mathcal{E}_t$ is at least equal to the generic rank.

Thus if $r \geq 1$ then the answer to your question yes. So the crucial case to study is when $r= 0$, and you are looking for infinitely many specialisations for which the rank is larger.

There is a very nice way to interpret this problem in terms of the associated surface. Namely, associated to $\mathcal{E}$ there is a unique smooth projective relatively minimal elliptic surface $\pi: X \to \mathbb{P}^1$. There key result is now the following:

Theorem There are infinitely many fibres of $\pi$ with a rational point if and only if $X(\mathbb{Q})$ is Zariski dense.

The proof is not too difficult. It relies in a crucial way on Mazur's torsion theorem. You can find a more general statement in [1, Lemma 3.2].

So you just have to show that the rational points on your surface are Zariski dense! There are various conjectures about Zariski density in the literature due to e.g. Lang, Vojta, Campana, and others. These say in the first instance for $X(\mathbb{Q})$ to be Zariski dense we must have that $X$ is not of general type (which does indeed hold in our case).

In any case, it suffices to find some elliptic surface for which it's unknown whether $X(\mathbb{Q})$ is Zariski dense. These already exist amongst geometrically rational surfaces (these satisfy $\deg A(T) \leq 4$ and $\deg B(T) \leq 6$). Such a class is provided by Del Pezzo surfaces of degree $1$. These have a special rational point given the base-locus of the anticanonical linear system, and the blow-up of this point is an elliptic surface. It is a big open problem to prove that these always have a Zariski dense set of rational points. I suspect even in this case it isn't known whether there is a fibre of positive rank.

[1] Julian Lawrence Demeio - Elliptic Fibrations and Hilbert Property

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    $\begingroup$ In the spirit of "as hard to show one as to show infinitely many", see mathoverflow.net/questions/383580/… and mathoverflow.net/questions/226794/…. The key point I am making is that while it is often easy in practice to find one example in individual cases, trying to prove there is one example in all cases is where proofs appear to be or are really known to be basically just as hard as proving there are infinitely many examples in all cases. $\endgroup$
    – KConrad
    Oct 17 at 20:07
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    $\begingroup$ Sometimes finding infinitely many examples in an individual case is subtle. For example, it's easy to show there are infinitely many primes $p \equiv 1 \bmod 5$ using the 5th cyclotomic polynomial and a modification of Euclid's proof of infinitude of the primes, but I've never seen a proof that there are infinitely many primes $p \equiv 2 \bmod 5$ that is not essentially like the general proof of Dirichlet's theorem on primes in arithmetic progression. Of course it's pretty easy to see there is at least one prime $p \equiv 2 \bmod 5$ (or at least two primes, or at least three primes, or ...). $\endgroup$
    – KConrad
    Oct 17 at 20:10
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    $\begingroup$ @KConrad In fact, shouldn't we be able to make a rigorous such implication in this case by taking a family and base changing it along a map $\mathbb P^1 \to \mathbb P^1$ whose induced map $\mathbb P^1(\mathbb Q) \to \mathbb P^1(\mathbb Q)$ misses any given finite set of points? $\endgroup$
    – Will Sawin
    Oct 19 at 15:00
  • $\begingroup$ @WillSawin Ha this is a very funny trick. Why not add it as an answer? $\endgroup$ Oct 19 at 16:45

This question is actually equivalent to Silverman's conjecture, not just morally.

The implication in one direction, that if every non-isotrivial family has infinitely many positive rank specializations then it has at least one, is obvious.

For the other implication, assume that $\mathcal E / \mathbb Q(T)$ is a family with only finitely many positive rank specializations, say at points $s_1,\dots, s_n \in \mathbb Q$. We will show there is another family that has no positive rank specializations at all.

Let $u \in \mathbb Q$ be such that $s_1- u, \dots, s_n-u$ are not perfect squares in $\mathbb Q$ (e.g. take $u$ sufficiently large) and then define a family $\mathcal E'/ \mathbb Q(S)$ by base-changing $\mathcal E$ along the map $T = S^2+u$.

For any positive rank specialization $S^2=s$ of $\mathcal E'$, then $T = s^2+u$ is a positive rank specialization of $\mathcal E$, so $s^2+u$ is one of $s_1,\dots, s_n$, contradicting our assumption on $u$.

So there is no positive rank specialization of $\mathcal E$.

This idea was inspired by a comment of KConrad.


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