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As I said previously, I have some problems in the theory of valuations and places.

Let L/K be a finite (say) Galois extension, F a place of L, and v a valuation of L. I denote by l and k the residue field of F and of the restriction of F to K resp. It is known that the map from the decomposition group D of F to the group of automorphisms of the residual field extension l/k is onto. This means that for every k-automorphism t of l/k, there exists an automorphism s of Gal(L/K) such that tF = Fs.
Also the inertia group of F is exactly the set of automorphisms s of Gal(L/K) such that Fs = F. But what is known about the set of automorphisms s of Gal(L/K) such that vs = v ? and is there a theorem according to which the map from the decomposition group D of v to the group of automorphisms t of the Abelian ordered group of values of v that fixes the group of values of the restriction of v to K, is onto ? (it is clear that an automorphism s of D induces such an automorphism via the formula tv(x) = v(sx), but is this homomorphism surjective) ? I have found not hint in Fried and jarden, nor in Bourbaki.

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D consists of the automorphisms which preserve the valuation v of L, by definition of decomposition group. In the case of a discrete valuation, D has a filtration by "ramification groups" given by valuation of (sx/x - 1) for s in D (where x is a uniformizer for v). See Serre, "Local Fields".

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  • $\begingroup$ 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. $\endgroup$ – MikeTeX Dec 11 '14 at 16:00
  • $\begingroup$ 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. $\endgroup$ – MikeTeX Dec 11 '14 at 16:23
  • $\begingroup$ @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. $\endgroup$ – David Lampert Dec 11 '14 at 18:24
  • $\begingroup$ 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. $\endgroup$ – MikeTeX Jan 26 '15 at 9:02

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