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This is a follow up from this question, in characteristic $p>0$. First, some background. It is known that in characteristic $0$, a vector bundle $E$ on say a projective curve $X$ is ample if and only if every quotient bundle has positive degree. This is known as Hartshorne's Theorem (see this and this paper) which is known (due to Serre, see the first linked paper) to fail in positive characteristic $p$. Serre's example deals with a smooth quartic $X/\mathbb{F}_3$ in $\mathbb{P}^2$, i.e. a curve of genus $3$ and the construction goes as follows:

Let $X$ be given by the equation $x^3y+y^3z+z^3x=0$. One proves that the action of Frobenius $F:X\to X$ on $H^1(X, \mathcal{O}_X)$ is trivial (i.e. the Hasse-Witt matrix is identically zero). One then constructs a rank $2$ bundle which is a non-trivial extension: $0 \to \mathcal{O}_X \to E \to \mathcal{O}_X(p) \to 0$ for $p\in X$ such that pulling back by Frobenius trivialises the extension $F^*E = \mathcal{O}_X \oplus \mathcal{O}_X(3p)$ (note $F^*\mathcal{O}(p) = \mathcal{O}(3p)$) and also every quotient line bundle of $E$ has positive degree. Then it's a lemma that $F^*E$ is ample iff $E$ is and we know $F^*E$ is not ample since it has $\mathcal{O}_X$ as a quotient which is not ample.

[Edit: As pointed out in the comments, in this example, E is not globally generated as it has a degree $1$ quotient] My question is whether there is a counterexample to Hartshorne's theorem which is globally generated or if one can hope to prove ampleness using Hartshorne's theorem for a bundle in characteristic $p$ assuming global generation and vanishing $H^1$.

Original motivation: I am studying curves $f:C\to X$ in characteristic $p$ such that $H^1(C, f^*T_X) = 0$ and $f^*T_X$ is globally generated (aka free curves à la Kollàr) and was wondering whether these two conditions are enough to prove that $f^*T_X$ is ample. In characteristic $0$, if one assumes that the genus of $C$ is $\geq2$ then only the cohomology vanishing is enough to deduce the result (as shown here). Can one show ampleness by making the extra assumption (or not!) of global generation in characteristic $p$?

Thank you.

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Just to clarify: in your first paragraph, should X already be a curve? –  Artie Prendergast-Smith Jul 8 '11 at 17:07
    
Apologies yes, fixed. –  Frank Jul 8 '11 at 17:41
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It is clear that $E$ as above is not globally generated, since $\mathcal{O}_X(p)$ being a degree one line bundle on genus 3 curve can not be globally generated. –  Mohan Jul 8 '11 at 20:18
    
Of course, thank you! I suppose this leaves the question whether global generation can play a role in the existence of a counter-example –  Frank Jul 10 '11 at 13:56
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