Timeline for Homologue of the Inertia group and of the Frobenius theorem for the group of values of a valuation
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2015 at 9:02 | comment | added | MikeTeX | David : I have just realized that I even not voted for your answers (this is done now). I appologize : I was entirely new in Mathoverflow and not used with the customs here. | |
| Dec 11, 2014 at 19:52 | vote | accept | MikeTeX | ||
| Dec 11, 2014 at 18:24 | comment | added | David Lampert | @MikeTex: Any s in D preserves the local ring at v so it preserves valuations for all elements of value 0 (they are the units of the local ring). The value group of K is of finite index in the value group of the finite extension L, so if x is in L then there exist integer n and y in K such that v(x^n/y)=0. This implies that nv(x)-v(y)=v(x^n/y)=0=v(s(x^n/y))=v(s(x)^n/y)=nv(s(x))-v(y) and thus v(s(x))=v(x) because the value group is ordered and so torsion free. | |
| Dec 11, 2014 at 16:23 | comment | added | MikeTeX | To be more explicit, D is by definition the group of automorphisms such that $\sigma O_v = O_v$. If $\tau$ is an automorphism of the ordered group of values of v, then the valuation v' defined by $v'(x) = \tau v(x)$ has the same valuation ring as v, but is not exactly equal to v, only equivalent to it. So, D is the group that preserves v up to an equivalence. | |
| Dec 11, 2014 at 16:00 | comment | added | MikeTeX | I am not sure you have understood my question. D consists in the automorphisms of G that preserve de valuation v of L UP TO AN EQUIVALENCE, but I asked for the group of automorphisms of G that preserve exactly the valuation v (let us denote this group by J, by analogy with the inertia group I which is the set of automorphisms of G that preserves exactly a place F, not up to an equivalence). My second question is related to an analogue of Frobenius theorem. | |
| Dec 11, 2014 at 15:05 | history | answered | David Lampert | CC BY-SA 3.0 |