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Aug 6, 2013 at 11:43 comment added Jason Starr @Chris: It is not the opposite. The same argument as in the exercise applies to any (tame) multiplicative group $T$. For every smooth, connected, linear algebraic group $G$ (over a field in that exercise, but the same argument works relatively), every $T$-torsor over $G$ is induced from a unique extension of group schemes of $G$ by $T$. Now take $G$ to be $\mathbb{G}_m$ and take $T$ to be a finite Abelian group of order prime to $p$.
Aug 6, 2013 at 9:13 comment added Chris Schommer-Pries I looked up exercise 7.1 in Dolgachev's "Lectures on Invariant Theory". It has to do with $\mathbb{G}_m$ bundles over a connected algebraic group G. This is opposite to what I was asking. I want G-bundles not $\mathbb{G}_m$-bundles. Moreover in my case G is a finite group, so not connected.
Aug 6, 2013 at 1:22 history edited Jason Starr CC BY-SA 3.0
Added tame hypothesis -- still considering the wild case.
S Aug 6, 2013 at 1:09 history answered Jason Starr CC BY-SA 3.0
S Aug 6, 2013 at 1:09 history made wiki Post Made Community Wiki by Jason Starr