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Let $X$ be a "nice" scheme and $Z \hookrightarrow X$ closed of codimension $\geq 2$. Let $Y$ over $X \setminus Z$ be a torsor for a finite flat group scheme $G/X$.

Does $Y$ spread out to a $G$-torsor over the whole of $X$?

For $G/X$ finite étale one can use Zariski-Nagata purity to spread $Y$ out to a scheme and then [Szamuely, Galois Groups and Fundamental Groups], p. 171, Lemma 5.3.13 to show that the extension is in fact a $G$-torsor.

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You may find the Appendix of Gille, Pianzola "Isotriviality and etale cohomology of Laurent polynomial rings" relevant (though I don't think it addresses your question precisely). – Kestutis Cesnavicius Feb 4 '13 at 19:35
up vote 5 down vote accepted

It is true if $X$ is regular. This is stated (as Lemme 2) in my CRAS 1985 note on purity for families of curves:
where, unfortumately, the proof is missing (I only claim that it extends Auslander's proof of the étale case).

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Thank you very much! – Timo Keller Feb 5 '13 at 19:46

Here is a counterexample for $G =\alpha_p$. Suppose that $k$ is a field of characteristic $p > 0$, and set $X = \mathbb A^2_k$, $Z= \{0\}$. Set $U := X \smallsetminus Z$. Then $\mathrm H^1(U, \mathbb{G}_{\rm a}) = \mathrm{H}^1(U, \mathcal{O})$ is an infinite dimensional vector space over $k$; hence it is $p$-torsion, and from the long exact sequence associated with the exact sequence $$ 0 \longrightarrow \alpha_p\longrightarrow \mathbb{G}_{\rm a}\longrightarrow \mathbb{G}_{\rm a}\longrightarrow 0 $$ we see that $\mathrm H^1(U, \alpha_p)$ surjects onto $\mathrm H^1(U, \mathbb{G}_{\rm a})$. If we take any class in $\mathrm H^1(U, \alpha_p)$ such that its image in $\mathrm H^1(U, \mathbb{G}_{\rm a})$ is $\neq 0$, this represents a torsor that does not extend to $X$, since $\mathrm H^1(X, \mathbb{G}_{\rm a}) = 0$.

[Edit:] Anon is right, my construction is nonsense. I apologize.

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Doesn't the $p$-power map $\mathbb{G}_a \to \mathbb{G}_a$ induce the Frobenius on $H^1(U,\mathbb{G}_a)$ (which is injective)? Also, aren't the categories of finite flat schemes over $U$ and $X$ equivalent in this example (via Hartog's)? – anon Feb 4 '13 at 17:26

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