I'm reading about the general position theorem in Arabello, Cornalba, Griffiths, and Harris' Geometry of Algebraic Curves volume 1. Everything is over $\Bbb C$, and for these authors a curve is a reduced complete scheme of finite type over $\Bbb C$. Here's the statement of the theorem and the proof:
Theorem: Let $C\subset \Bbb P^r$, $r\geq 2$, be an irreducible, non-degenerated, possibly singular, curve of degree $d$. Then a general hyperplane meets $C$ in $d$ points, any $r$ of which are linearly independent.
Proof: i) For a general point $p\in C$, the projection $\pi_p: C \to \Bbb P^{r-1}$ is birational onto its image. By the previous lemma (same assumptions as the theorem, conclusion is that a general hyperplane meets $C$ in $d$ points, no three of which are collinear), we may find a pair of points $p,q\in C$ so that $\overline{pq}\cap C$ is just $\{p,q\}$, and then $\pi_p$ is one-to-one over $\pi(q)$.
ii) Next, we let $U\subset(\Bbb P^r)^*$ be the open set of hyperplanes transverse to $C$, and consider the incidence correspondence $I\subset C\times U$ consisting of pairs $(p,H)$ where $p\in H\cap C$. We claim that if the statement of the theorem is false, then for a general pair $(p,H)\in I$ there is a dependent set of points $p=p_1,p_2,\dots,p_r\in H\cap C$. Indeed, we first observe that $I$ is irreducible of dimension $r$. Now we consider $I_0\subset I$ the subvariety consisting of pairs $(p,H)$ where $p\in H\cap C$ is part of a dependent set of points. It is clear that if the general position theorem is false, then $\dim I_0=r$. By irreducibility, we must have $I_0=I$.
iii) Finally, for $p\in C$ a general point we consider the projection $\pi_p:C\to C'\subset\Bbb P^{r-1}$. By (i), $\pi_p$ is birational onto $C'$. By (ii), if the general position theorem fails for $C\subset \Bbb P^r$, then it also fails for $C'\subset\Bbb P^{r-1}$. But the lemma is equivalent to the general position theorem in the case $r=3$, and this contradiction shows that the lemma implies the general position theorem.
Question: Does the same proof work in positive characteristic? I was talking to a friend who solved Hartshorne exercise IV.3.10 which asks one to prove the same statement with the additional assumption that the characteristic of the base field is zero, and we couldn't figure out whether that was actually necessary. Parts (ii) and (iii) seem characteristic-independent - it seems like the only way something bad could happen here is in (i), but we're not sure.
Guess: We think we've got an idea where characteristic zero matters, and it's indeed in part (i). The projection described in (i) is one-to-one on points near $\pi(q)$, but this doesn't guarantee birationality except when the characteristic is zero because the projection could come from a purely inseparable extension: consider the projection of the plane curve $y=x^p$ onto the $y$-axis, which is one-to-one everywhere and which gives the map $k(y)\to k(x)$ by $y\mapsto x^p$, a purely inseparable extension.
To prove the claim in characteristic zero, we would use the fact that any field extension in characteristic zero is separable, so the remark here shows that generically the cardinality of the fiber must be equal to the degree of the morphism and the lemma shows that this is generically one. Can anyone confirm this?