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Let $X$ be a compact connected Riemann surface of genus $g>0$.

Suppose that $X$ can be defined over a number field (as an algebraic curve). Then, is it clear that each Weierstrass point of $X$ is algebraic?

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    $\begingroup$ Yes it is clear. $\endgroup$ – Felipe Voloch Oct 9 '11 at 13:36
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Edit: as pointed out by Felipe below, what used to be here was sort of wrong-headed, but I can't delete this answer since it has already been accepted. Anyway you should do what he says: the divisor of all Weierstrass points counted with weights is defined by the vanishing of the Wronskian determinant of a basis of the space of $1$-forms on $X$. This divisor is then manifestly defined over $k$ and so all the Weierstrass points are defined over a finite extension.

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  • $\begingroup$ I've deleted my earlier comment, since it is no longer relevant, as Dan has edited his answer. Hopefully this won't start an infinite recursion. $\endgroup$ – Felipe Voloch Oct 10 '11 at 15:23

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