Return to Revisions

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.