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Let $k$ be an algebraically closed field of char($k$)=p>0, $X$ a smooth projective variety over $k$, $F:X\rightarrow X^{(1)}$ the relative Frobenius morphism. Let $E$ be an ample vector bundle on $X$. Then Frobenius direct image $F_*(E)$ is also an ample vector bundle on $X^{(1)}$?

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up vote 6 down vote accepted

Definitely no. Take $X=\mathbb{P}^1$, $E=\mathcal{O}(1)$. Write $$F_*(E)= \bigoplus_{i=1}^p \mathcal{O}(a_i)$$ If $a_i\ge 0$, for all $i$, then $$2=h^0(E)=h^0(F_*(E)) \ge \sum (a_i+1) \ge p$$ so it's not even semipositive when $p>2$.

Note this works for any finite map $F:\mathbb{P}^1\to \mathbb{P}^1$, so in particular for the $p$th power map in characteristic $0$.

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Thank you very much Prof. Arapura –  Universe Apr 18 '11 at 14:54
    
You're welcome. –  Donu Arapura Apr 18 '11 at 15:41
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