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I read somewhere that if $X$ is a projective variety of general type over a number field $K$, then rational points are an analogue of entire curves $\mathbf{C}\to X^{an}$ (with $X^{an}$ the analytification of $X_{\mathbf{C}}$ for some $K\to \mathbf C$).

Rational points are algebraic points of degree $1$ on $X$ and they "correspond" to entire curves.

Naive question:

Let $m\geq 2$. What do algebraic points of degree $m$ on $X$ "correspond" to?

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  • $\begingroup$ In the analogy between Nevanlinna theory and Diophantine Geometry suggested by Vojta, an entire curve corresponds to an infinite set of rational points, not to just a point. See Vojta's book LNM 1239 Ch 3. Vojta discusses algebraic points of bounded degree but does not explicitly state an analogy. $\endgroup$
    – Felipe Voloch
    Commented Dec 23, 2012 at 16:21

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Rational points are (kind of) like maps from a fixed curve (say P^1) to X.

Algebraic points of degree m are like curves endowed with a map to X and a degree-m map to P^1.

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  • $\begingroup$ That seems reasonable. Just to convince myself a bit more. Let $X$ be curve of genus at least $two$ over a number field $K$ of big gonality. Then the set of quadratic points on $X$ is finite by Faltings-Frey. In particular, this should correspond to the fact that there are only finitely many hyperelliptic curves dominating $X$. And that of course follows from the big gonality of $X$. (Actually there are no hyp ell curves dominating $X$.) So ok, I'm convinced that this "works". $\endgroup$
    – Masse
    Commented Dec 23, 2012 at 16:22
  • $\begingroup$ @Masse: Actually, for quadratic points, gonality is sharp. In other words, if X is not hyperelliptic, then it has only finitely many quadratic points. My recollection is that the same is true for cubics (but I may be misremembering), and in any case, it's not true in general for higher degree. For references, there are various papers by Joe Harris, Dan Abramovich, and me. (I think that the first one was mine and Joe Harris's that did the quadratic points case.) I'm not quite sure which result of Faltings-Frey you mean, but of course all these results rely on Faltings' Annals paper. $\endgroup$
    – Joe Silverman
    Commented Dec 24, 2012 at 2:51

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