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Nov 5, 2012 at 7:19 comment added Laurent Moret-Bailly @stefan: every torsor under an étale group scheme is étale as a scheme (or algebraic space), because étaleness is an fppf-local property. Hence it is trivialized by an étale covering (namely itself).
Nov 4, 2012 at 16:25 comment added Chandan Singh Dalawat If you have a surjective morphism of commutative groups $G\to H$ with kernel $N$, then every fibre is a torsor under $N$.
Nov 4, 2012 at 16:25 comment added Will Sawin Yes. Let $x$ be an element which is not an $n$th power in $k$. Then the set of $n$th roots of $x$ is a torsor for the $n$th roots of unity in an obvious way, and has no $k$-valued points, thus is nontrivial.
Nov 4, 2012 at 16:22 comment added stefan the torsor is supposed to be an etale torsor (i.e. trivialized by an etale covering)
Nov 4, 2012 at 16:20 comment added stefan Thank you very much! Do you have an example of a non-trivial torsor over k with group $\mu_n$ in this situation?
Nov 4, 2012 at 16:08 comment added Chandan Singh Dalawat For $k=\mathbf{Q}$, it works for every $n>2$.
Nov 4, 2012 at 15:59 comment added stefan yes, that's right, i was just thinking about this. in fact one could take k to be a finite field, then for large enough n, this should work. Thank you very much.
Nov 4, 2012 at 15:27 comment added Chandan Singh Dalawat How about $\mu_n$, the $n$-th roots of $1$, over a field $k$ in which $n$ is invertible but which does not contain a primitive $n$-root of $1$ ?
Nov 4, 2012 at 15:19 history asked stefan CC BY-SA 3.0