By Guillaume Duval

During this paper, valuation idea is used to examine infinitesimal behaviour of suggestions of linear differential equations. For any Picard-Vessiot extension $(F / ok, \partial)$ with differential Galois staff $G$, the writer appears on the valuations of $F$ that are left invariant via $G$. the most reason behind this is often the next: If a given invariant valuation $\nu$ measures infinitesimal behaviour of features belonging to $F$, then conjugate components of $F$ will percentage an identical infinitesimal behaviour with admire to $\nu$. This memoir is split into seven sections

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Extra resources for Valuations and differential Galois groups

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By induction on the degree, it is enough to prove the ﬁrst step for E/K. More precisely if we take H to be a maximal normal subgroup properly included in G, we are reduced to the particular case where Gal(E/K) = G/H is a simple group. Let’s assume that F/K is a simple Galois extension. By ([34], Th 21 p. 69), kν /kν0 is a normal extension. Since it is of characteristic zero, it is Galois, and one gets the following exact sequence of groups: É É É É É É É 1 → GI →Gal(F/K) → Gal(kν /kν0 ) → 1, where GI = {σ ∈ Gal(F/K)|σ(x) − x ∈ mν for all x ∈ Rν } is the inertia group.

An important family of analytic valued ﬁelds comes from real analysis thanks to the following 4. CONTINUITY OF DERIVATIONS, GEOMETRY AND EXAMPLES 27 Proposition 35. For any Hardy ﬁeld H with its natural valuation v given by the place that computes limits of functions when the real time t tends to +∞, the is a diﬀerential ﬁeld endowed with a new valuation vˆ that ring H = H ⊗Ê d still satisﬁes the l’Hopital rule extends v and for which Γvˆ = Γv . t. vˆ. Furthermore, when rat . rank(v) = dimÉ (Γv ⊗ ) is ﬁnite, H and H are analytic valued ﬁelds.

Let’s denote by f (X) the monic minimal polynomial of α in K[X]. We are going ﬁrst to prove that (25) deg f (X) = n and F = K(α). Since ν is a G-invariant valuation, all the conjugates of α are units of Rν . Factoring f (X) as a product of linear terms, we obtain: f (X) ∈ Rν [X] ∩ K[X] = R0 [X]. ¯ hence: Its reduction f¯ ∈ kν0 [X] is a monic polynomial vanishing at X = α, deg(F/K) = n deg f (X) = deg f¯(X) deg(α) ¯ = n. This proves (25). The polynomial ∂ f¯ ∂X (X) is a nonzero element of kν0 [X] of degree n − 1.