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).
Best Answer
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