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Let's fix a scheme $S$ and consider some Grothendieck topology $\tau\in\{Zariski,\acute{e}tale,fppf\}$ on the category of schemes over $S$. Define a $\tau$ fiber bundle with fiber type $F$ to be a map $f:X\to Y$ of schemes over $S$ that $\tau$-locally on $Y$ looks like a product with $F$.

Given $\tau,\tau'\in\{Zariski,\acute{e}tale,fppf\}$ with $\tau'$ coarser than $\tau$, are there conditions on $F$ that imply that every $\tau$ fiber bundle with fiber type $F$ is also a $\tau'$ fiber bundle?

I've seen examples of etale fiber bundles that aren't Zariski bundles, but it's my understanding that for $F=\mathbb{P}^1$ this can't happen (I'm not positive about this), and I'm wondering about other $F$'s. I don't know if there are fppf fiber bundles that aren't etale bundles, and if there are I'd love to see an example.

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In characteristic $p$, the Frobenius morphism $F: \mathbb{P}^1\to \mathbb{P}^1$, defined by $[X,Y]\mapsto [X^p,Y^p]$, is fppf locally trivial with fiber $\text{Spec}k[\epsilon]/\langle \epsilon^p \rangle$, yet it is neither Zariski locally trivial nor 'etale locally trivial. –  Jason Starr Sep 23 '13 at 18:47
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If $Y$ is the spectrum of a field $k$ and $X$ is a $k$-conic without rational points, then $X$ is an étale $\mathbb{P}^1$-bundle which is not Zariski-locally trivial. –  Laurent Moret-Bailly Sep 23 '13 at 20:29
    
@LaurentMoret-Bailly Good point. Might it be true that etale $\mathbb{P}^1$ bundles are Zariski locally trivial when $S=\mathrm{Spec}(\overline{k})$? –  Julian Rosen Sep 23 '13 at 20:46
    
No, since there are conic bundles without generic sections: take $Y=\mathbb{G}_m\times \mathbb{G}_m$ with coordinates $t$, $u$, and $X\subset Y\times \mathbb{P}^2_k$ given by $x^2+ty^2+uz^2=0$. It is an exercise to check that this equation has no nontrivial solutions with $x$, $y$, $z$ in $k(t,u)$. –  Laurent Moret-Bailly Sep 24 '13 at 18:53

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