Invariant valuations and solutions of l.d.e.
2.1. Group actions on the Riemann-Zariski variety. Let K → F denote
an arbitrary field extension of finite type and G ⊂ Aut(F/K) be a subgroup of the
automorphism group of the field extension. We define a natural action of G on
= S(F/K) by permutation of the valuations in the following way:
(σ, ν) −→ σ · ν = ˜ ν
where, for all x ∈ F ,
(1) ˜(x) ν =
The induced action on the valuations rings is given by:
= σ(Rν) and m˜
Hence every σ in G induces a K-isomorphism on the residue fields:
(3) ¯ σ : kν −→ k˜
(This last arrow determines the action of G on the corresponding places.)
Remark 7. For any such G-action, two conjugate valuations ν and ˜ ν share the
same value groups by (1); they have K-isomorphic valuation rings by (2), and K-
isomorphic residue fields by (3). Since ν and ˜ ν have the same value group, they have
the same rank ( p 10). As a consequence, the rank, rational rank, transcendence
degree of the residue fields are invariant for this action. It is, therefore, impossible
to have a transitive action on
as soon as tr . deg(F/K) 2.
Definition 8. With the notations above, we would say that ν is G-invariant
if and only if the valuation ring Rν is mapped onto itself by all σ in G. Similarly,
ν is strongly invariant if and only if ν(σ(x)) = ν(x) for all x in F and all σ in G.
Strong invariance implies invariance; but the converse is false as shown in the
following example suggested by M.Spivakovsky: let F = K((t )) be the field of a
formal power series expansion with rational exponents and well-ordered support.
Let ν = νt be the t-adic valuation. Then Γν = . Let g be the F/K-automorphism
defined by: g(ϕ(t)) =
for all ϕ ∈ F . The element g generates a group G
isomorphic to and the following two relations hold: σ(Rν) = Rν for all σ in G,
and ν(g(ϕ)) = 3ν(ϕ) for all ϕ in F . Therefore, this valuation is G-invariant but
not strongly invariant.
The following definition and proposition describe a context where these two
notions of invariance are going to coincide.