# If rational points are like entire curves, then what do algebraic points correspond to

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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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. – Felipe Voloch Dec 23 '12 at 16:21

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". – Masse Dec 23 '12 at 16:22