2. INVARIANT VALUATIONS AND SOLUTIONS OF L.D.E. 15
iii. The induced action on S(F/K)∗ is locally-finite and each G◦(C)-invariant val-
uation ν of F/K remains G◦(K)-invariant.
This theorem illustrates two notions of stability. On the one hand, thanks to
Proposition 10, every invariant valuation of F/C remains strongly invariant. On the
other hand, G◦(K) is in general a much bigger group than G◦(C) and invariance for
the small group implies invariance for the bigger one, when the valuation is trivial
on the ground field K. This latter fact is a powerful tool to compute the invariant
valuations of F/K, since it reduces the problem to a purely algebraic statement
in a way that we can completely forget the Picard-Vessiot structure of the field
extension. This idea will be properly explained in section 7 of this paper.
The following is a finiteness property whose proof is left to the reader:
Lemma 18. Let ν be a non-trivial valuation of the field extension F/C and
{z1,...,zd} be d elements of F . Set:
E = ν(
d
i=1
cizi), c = (c1,...,cd)
Cd\{0}}
Γν {∞}.
Then card(E) d.
Proof of Theorem 17.
(i) Let z T (F/K) and d = ordK (z). Let {z = z1,...,zd} be a C-basis of
V (z) = VectC {σ(z)|σ G(C)} = SolF {Lz(y) = 0} .
Then:
{ν(σ(z))|σ G(C)} E = ν(
d
i=1
cizi), c = (c1,...,cd)
Cd\{0}
.
Therefore Lemma 18 gives us the inequality of (i). Since F/K is a Picard-Vessiot
field extension, F coincides with the field of fractions of T (F/K). So every x F

can be written as a fraction x =
z1
z2
with z1,z2 in T
(F/K)∗.
By setting
Ei = {ν(σ(zi))|σ G(C)},i = 1, 2
Δ = {ν(σ(x))|σ G(C)}.
We get Δ E1 E2 where, E1 E2 stands for the Minkowski difference of the
two subsets E1 and E2 of Γν. Therefore
card(Δ) card(E1) · card(E2) ordK (z1) · ordK (z2) ∞.
This concludes the proof of (i).
(ii) With the previous notations, for all z T (F/K) with ordK (z) = d, the
locally finite action of G(C) on T (F/K) can be described by the following relations
(RC ) :σ(z) =
d
i=1
ai(σ)zi, ∀σ G(C).
Each coefficient ai(σ) belongs to the ring Γ(G(C)) of regular functions on G(C).
Since Γ(G(K)) = K ⊗C Γ(G(C)) the ai are regular functions onG(K). So, if it is
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