1
$\begingroup$

Let $C$ be an algebraic curve of genus $g \geq 2$, defined over $\overline{\mathbb{Q}} \subset \mathbb{C}$. It is then defined over a finite extension $K$ of $\mathbb{Q}$. We assume that $C(K) \ne \emptyset$ and thus we may embed $C$ into its Jacobian variety $J$, also definable over $K$. We abuse notation and continue to call this embedded image $C$. Then Mordell's conjecture, now a theorem of Faltings, asserts that $C(K) \cap J(K)$ is finite.

One point of view, presented in J.S. Milne's notes on abelian varieties (Heuristic arguments, page 131), is that "since there is no reason to expect any relation between $C(\mathbb{C})$ and $J(K)$ as subsets of $J(\mathbb{C})$", finiteness ought to follow for dimension reasons (since when $g \geq 2$ we have $1 = \dim_{\mathbb{C}} C < \dim_{\mathbb{C}} J = g$). He also mentions that to date (the notes seem to have been updated last in 2008) there is no proof of Faltings' theorem that follows this strategy.

My question is to ask whether a refinement of Mordell's conjecture, following the heuristic above, is expected to hold. Indeed one can view $J(K)$ as a finitely generated subgroup, or $\mathbb{Z}$-module, inside $J(\mathbb{C})$ that in some sense is independent of $C(\mathbb{C})$. Thus let $\Gamma$ be a finitely generated subgroup of $J(\mathbb{C})$. Is it then expected that $\Gamma$ will intersect $C(\mathbb{C})$ inside $J(\mathbb{C})$ finitely many times? In other words, is Mordell's conjecture really about the geometry of finitely generated subgroups inside $J(\mathbb{C})$? Or is there some subtle arithmetic significance to the group of rational points over a number field?

$\endgroup$
  • 5
    $\begingroup$ You seem to be asking about exactly the Mordell-Lang conjecture. Yes, Faltings proved this too. $\endgroup$ – Daniel Loughran Mar 14 at 12:45
  • 1
    $\begingroup$ Decades prior to Faltings's proof this had already been known, by a specialization argument from a finitely generated field to a number field, to be equivalent to Mordell's original conjecture on $C(K)$. $\endgroup$ – Vesselin Dimitrov Mar 14 at 18:48
4
$\begingroup$

In fact, you don't need $C$ to be defined over a number field: if $C$ is curve of genus $\geq 2$ defined over $\mathbb{C}$, and $\Gamma$ is a finitely generated subgroup of $J(\mathbb{C})$, then $C \cap \Gamma$ is finite. In fact, you can even take a subgroup $\Gamma$ of finite rank, i.e. $\Gamma \otimes \mathbb{Q}$ is finite dimensional over $\mathbb{Q}$. So for example, you can throw in $\Gamma$ all the torsion points of $J$. There is also a generalization for an algebraic variety $X$ sitting inside a semi-abelian variety $G$. The conclusion is that $X \cap \Gamma$ is finite unless $X$ contains a translate of an algebraic subgroup of $G$ of dimension $\geq 1$.

The history of this theorem is long, see for example Hindry's survey and Silverman's review of Zannier's book Some problems of unlikely intersections in arithmetic and geometry. I'm not in the domain, but there is currently much research about unlikely intersections.

$\endgroup$
  • 2
    $\begingroup$ Still the original form of the question, with the finitely generated group instead of the generalization to finite rank groups (allowing for $C / \mathbb{C}$), is almost trivially equivalent to the original Mordell $\#C(K) < \infty$ conjecture for $C/K$ over number fields. $\endgroup$ – Vesselin Dimitrov Mar 14 at 22:49

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.