2. INVARIANT VALUATIONS AND SOLUTIONS OF L.D.E. 9
have a strictly increasing orbit in Γν, which is therefore infinite. As a consequence,
Aut+(Γν)
is a torsion free group. This is why, if G has a locally-finite action, φν
must be trivial.
Thanks to this observation, other examples, where invariance implies strong-
invariance, can be given. This occurs when we know that
Aut+(Γν)
is itself trivial.
This is the case when Γν r with the lexicographic order.
This naturally leads to the study of
Aut+(Γν)
for more sophisticated examples.
Let’s assume for instance that Γν = Γ is a subgroup of with its natural
ordering induced by . In this case, we claim that
Aut+(Γ)
is Abelian since each
ϕ
Aut+(Γ)
acts on Γ by scalar multiplication. Indeed, for all (x, y) Γ × Γ and
(p, q) × ,
ϕ(px qy) × (px qy) = (pϕ(x) qϕ(y)) × (px qy) 0.
Therefore, the discriminant Δ of this two-form in (p, q) must be 0. But
Δ = (ϕ(x)y ϕ(y)x)2, hence, Δ = 0. This proves the claim, by the existence
of a unique positive multiplicator m such that ϕ(x) = mx for all x Γ. Since
1 Γ,m = ϕ(1) Γ. Conversely, since 1 Im(ϕ), we must have also
1
m
Γ.
As a consequence if Γ = Γα = + α for some α \ . If
Aut+(Γα)
is
non-trivial, then α must be quadratic over .
Conversely, one can exhibit a lot of finite type -modules Γ with
Aut+(Γ)
as big as wanted. For any real number field of degree r 2 over , let Γ be its
associated ring of algebraic integers. Γ is a free -module of rank r and the latter
argument shows that
Aut+(Γ)
is isomorphic to the multiplicative group of positive
units of Γ. By the Dirichlet-Chevalley-Hasse Unit theorem (see [2], Th 8.1, p. 211),
Aut+(Γ)
is a free Abelian group of rank r 1.
The next proposition shows how from a G-strongly invariant valuation of rank
2, we can obtain another strongly G-invariant valuation by localisation. This
property is very natural since by localising a valuation ring we lose information.
Proposition 12. Let F/Kbe a field extension and G a subgroup of Aut(F/K).
Let R be a strongly G-invariant valuation ring of F/K.
Then every ideal of R is mapped onto itself by the G-action.
For every Q in Spec(R), RQ is a strongly G-invariant valuation ring.
Proof. Let I = {0} be an ideal of R. For all x I∗ for all σ G, σ(x) =
σ(x)
x
· x
belongs to I because
σ(x)
x
U(R) R. Hence σ(I) I for all σ G. Since G acts
by automorphism in R, we have σ(I) = I for all σ G.
Now if I = Q is a prime ideal, the above equality yields σ(RQ) = RQ for all
σ G. Hence RQis a G-invariant valuation ring. Since U(R) U(RQ), Proposition
10(ii) implies that RQ is a strongly G-invariant valuation ring.
Now the inverse problem will be addressed. Let ν0 be a G-invariant, (re-
spectively strongly G-invariant) valuation of F/K. How can the set of valuations
composed with ν0, which are G-invariant, (respectively strongly invariant) be de-
scribed ? The following description of the valuations composed with ν0 is needed.
Consider the place P0 : F −→ kν0 {∞} defined by ν0. Let
¯:
P kν0 −→ E {∞}
be another place where E/K is a field extension. The resulting composition map:
P =
¯
P P0 gives rise to a K-place of F , hence to a valuation, which is said to be
composed with ν0. Let’s denote by ν and ¯ ν the valuations respectively associated
to P and
¯.
P It is well known (see [34] chap VI pages 35, 43 or [33]) that
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