Maximal immediate extensions of valued differential fields

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by  Matthias Aschenbrenner

MAXIMAL IMMEDIATE EXTENSIONS OF VALUED
DIFFERENTIAL FIELDS
7
1
MATTHIASASCHENBRENNER,LOUVANDENDRIES,ANDJORISVANDERHOEVEN
0
2
n Abstract. We show that every valued differential field has an immediate
a strict extension that is spherically complete. We also discuss the issue of
J uniqueness uptoisomorphismofsuchanextension.

4
2

]
C Introduction
A In this paper a valued differential field is a valued field K of equicharacteristic
. zero, equipped with a derivation ¶ : K →K that is continuous with respect to the
h
t valuation topology on the field. (The difference with [1] and [2] is that there the
a
definition did not include the continuity requirement.)
m
LetK beavalueddifferentialfield. Unlessspecifiedotherwise,¶ isthederivation
[ ofK,andweletv: K× →Γ=v(K×)bethevaluation,withvaluationringO=Ov
1 and maximal ideal O = Ov of O; we use the subscript K, as in ¶ K, vK, ΓK, OK,
v OK, if we wish to indicate the dependence of ¶ , v, Γ, O, O on K. We denote the
1 residue field O/O of K by res(K). When the ambient K is clear from the context
9
we often write a′ instead of ¶ (a) for a∈K, and set a† :=a′/a for a∈K×.
6
By [1, Section 4.4], the continuity requirement on ¶ amounts to the existence of
6
0 a φ ∈ K× such that ¶ O ⊆ φO; the derivation of K is said to be small if this holds
1. for φ = 1, that is, ¶ O ⊆ O. By an extension of K we mean a valued differential
0 field extension of K. An extension L of K is said to be strict if for every φ∈K×,
7
1 ¶ O ⊆ φO ⇒ ¶ LOL ⊆ φOL, ¶ O ⊆ φO ⇒ ¶ LOL ⊆ φOL.
:
v With these conventions in place, our goal is to establish the following:
i
X Theorem. Every valued differential field has an immediate strict extension that is
r spherically complete.
a
WeconsiderthisasadifferentialanalogueofKrull’swell-knowntheoremin[4,§13]
thateveryvaluedfieldhasasphericallycompleteimmediatevaluedfieldextension.
(Recall that for a valued field the geometric condition of spherical completeness is
equivalentto the algebraicconditionof being maximalin the sense of nothaving a
proper immediate valued field extension.)
Throughout this paper K is a valued differential field. For the sake of brevity
wesaythatK hasthe Krull property if K has asphericallycomplete immediate
strict extension. Let us first consider two trivial cases:
Case Γ = {0}. Then K itself is a spherically complete immediate strict extension
of K, and thus K has the Krull property.
Date:January2017.
1
2 ASCHENBRENNER,VANDENDRIES,ANDVANDERHOEVEN
Case ¶ = 0. Take a spherically complete immediate valued field extension L of
the valued field K. Then L with the trivial derivation is a spherically complete
immediate strict extension of K, so K has the Krull property.
Thus towards proving our main theorem we can assume Γ 6= {0} and ¶ 6= 0 when
convenient. We shall freely use facts (with detailed references) from Sections 3.4,
4.1, 4.2, 4.3, 4.4, 4.5, 5.7, 6.1, 6.2, 6.3, 6.5, 6.6, 6.9, 9.1, 9.2, 10.5, and 11.1 in [1].
Specialcasesofthemaintheoremare[1,Corollaries6.9.5,11.4.10]. Whatisnew
comparedtotheproofsofthesespecialcases? Mainlythenotionofstrictextension,
the invariant convex subgroup S(¶ ) of Γ, the flexibility condition on K, and the
lemmasaboutthese(related)concepts;seeSections1,3,4,6. Wealsogeneralizein
Section2thenotionofNewton degree from[1,11.1,11.2]tooursetting. This gives
us the tools to adapt in Section 5 the proofs of these special cases to deriving our
main theorem for K such that Γ> has no least element and S(¶ )={0}. Section 6
shows how that case extends to arbitrary K using coarsening by S(¶ ).
Our focus in [1] was on asymptotic fields—a special kind of valued differential
field—and in this connection we note that by Lemma 1.12 below any immediate
strictextensionofanasymptoticfieldisagainasymptotic. Takingthisintoaccount,
andrelatedresultssuchas[1,Lemmas9.1.2,10.5.8],wenotethatourtheoremalso
answersquestions about Hardy fields and H-fields that havebeen aroundfor some
time. For example, it gives a positive answer to Question 2 in Matusinski [5].
(However, in [5] the notion of H-field is construed too narrowly.) See also the
remark at the end of Section 3.
Uniqueness. By Kaplansky [3], a valued field F of equicharacteristic zero has
up to isomorphism over F a unique spherically complete immediate valued field
extension. In Section 7 we indicate some classes of valued differential fields where
this uniqueness holds in the setting of valued differential fields, where of course we
only allow strict extensions. We also discuss a conjecture from [1] about this.
In Section 8 we give an example of an H-field where such uniqueness fails. Here
we use some basic facts related to transseries from Sections 10.4, 10.5, 13.9, and
Appendix A in [1].
Notations and conventions. We borrow these notational conventions from [1].
For the reader’s convenience we repeat what is most needed in this paper. We set
N:={0,1,2,...} and let m, n range over N.
A valuation (tacitly, on a field) takes values in an ordered (additively written)
abelian group Γ, where “ordered” here means “totally ordered”, and for such Γ,
Γ< := {γ ∈Γ: γ <0}, Γ6 := {γ ∈Γ: γ 60},
and likewise we define the subsets Γ> and Γ> of Γ.
For a field E we set E× := E \{0}. Let E be a valued field with valuation
v: E× →ΓE =v(E×), valuation ring OE and maximal ideal OE of OE. When the
ambient valued field E is clear from the context, then for a,b∈E we set
a≍b :⇔ va=vb, a4b :⇔ va>vb, a≺b :⇔ va>vb,
a<b :⇔ b4a, a≻b :⇔ b≺a, a∼b :⇔ a−b≺a.
Itiseasyto checkthatifa∼b,thena,b6=0,andthat∼is anequivalencerelation
on E×; let a∼ be the equivalence class of an element a ∈ E× with respect to ∼.
We use pc-sequence to abbreviate pseudocauchy sequence; see [1, Sections 2.2, 3.2].
MAXIMAL IMMEDIATE EXTENSIONS OF VALUED DIFFERENTIAL FIELDS 3
Let also a valued field extension F of E be given. Then we identify in the usual
way res(E) with a subfield of res(F), and Γ with an ordered subgroup of Γ .
E F
Next, let E be a differential field of characteristic 0 (so the field E is equipped
with a single derivation ¶ : E → E, as in [1]). Then we have the differential ring
E{Y}=E[Y,Y′,Y′′,...]ofdifferentialpolynomialsinanindeterminateY,andwe
set E{Y}6= :=E{Y}\{0}. Let P =P(Y)∈E{Y} have order at most r ∈N, that
is,P ∈E[Y,Y′,...,Y(r)]. ThenP = P Yi,asin[1,Section4.2],withiranging
i i
over tuples (i0,...,ir)∈N1+r, Yi :=Yi0(Y′)i1···(Y(r))ir, and the coefficients Pi
P
are in E, and P 6=0 for only finitely many i. For such i we set
i
|i| := i +i +···+i , kik := i +2i +···+ri ,
0 1 r 1 2 r
the degree of P 6= 0 is degP := max |i| : P 6= 0 ∈ N, and the weight of P is
i
wtP :=max kik: P 6=0 ∈N. WealsousethedecompositionP = P Y[σ];
i (cid:8) (cid:9) σ [σ]
here σ range(cid:8)s over words(cid:9)σ = σ1···σd ∈ {0,...,r}∗, Y[σ] := YP(σ1)···Y(σd),
all P ∈ E and P 6= 0 for only finitely many σ, and P = P for all
[σ] [σ] [σ] [π(σ)]
σ = σ ···σ and permutations π of {1,...,d}, with π(σ) = σ ···σ . We
1 d π(1) π(d)
set kσk := σ +···+σ for σ = σ ···σ , so kik = kσk whenever Yi = Y[σ].
1 d 1 d
We also use for a ∈ E the additive conjugate P := P(a+Y) ∈ E{Y} and the
+a
multiplicative conjugate P :=P(aY)∈E{Y}.
×a
For a valued differential field K we construe the differential fraction field KhYi
of K{Y} as a valued differential field extension of K by extending v: K× → Γ to
the valuation KhYi× →Γ by requiring vP =minvP for P ∈K{Y}6=.
i
1. Preliminaries
We recall some basics about valued differential fields, mainly from Section 4.4 and
Chapter 6 of [1], and add further material on compositional conjugation, strict
extensions, the set Γ(¶ ) ⊆ Γ, the convex subgroup S(¶ ) of Γ, and coarsening. We
finishthispreliminarysectionwithfactsaboutthedominantdegreeofadifferential
polynomial as needed in the next section. In this section φ ranges over K×.
Compositional conjugation. The compositional conjugate Kφ of K is the val-
ued differential field that has the same underlying valued field as K, but with
derivation φ−1¶ . Let L be an extension of K. Then Lφ extends Kφ, and
L strictly extends K ⇐⇒ Lφ strictly extends Kφ.
Therefore,K hastheKrullpropertyiffKφhastheKrullproperty: Lisaspherically
completeimmediatestrictextensionofK iffLφ isasphericallycompleteimmediate
strict extension of Kφ. Moreover,
¶ O⊆φO ⇐⇒ the derivation φ−1¶ of Kφ is small.
Thus for the purpose of showing that K has the Krull property it suffices to deal
with the case that its derivation ¶ is small.
Strict extensions. Suppose ¶ is small. Then ¶ O ⊆ O by [1, Lemma 4.4.2], so ¶
induces a derivation
a+O7→(a+O)′ :=a′+O
on the residue field res(K); the residue field of K with this derivation is called
the differential residue field of K and is denoted by res(K) as well. Note that the
derivation of res(K) is trivial iff ¶ O⊆O.
4 ASCHENBRENNER,VANDENDRIES,ANDVANDERHOEVEN
The field C((t)) of Laurent series with derivation ¶ =d/dt and the usual valuation,
where O =C[[t]] and O =tC[[t]], is a valued differential field, since ¶ O=O=t−1O.
It is an example of a valued differential field with ¶ O ⊆ O, but ¶ O 6⊆ O. On the
other hand, under a mild assumption on Γ we do have ¶ O⊆O⇒¶ O ⊆O:
Lemma 1.1. Suppose ¶ O ⊆O and Γ> has no least element. Then ¶ O⊆O.
Proof. For f ∈O we have f =gh with g,h∈O, so f′ =g′h+gh′ ∈O. (cid:3)
Lemma 1.2. Suppose ¶ O ⊆O and ¶ O6⊆O. Then for all φ: ¶ O⊆φO ⇔ φ<1.
Proof. From ¶ O ⊆φO, we get φ−1¶ O ⊆O, so the derivation φ−1¶ is small, and thus
φ−1¶ O ⊆ O, hence ¶ O ⊆ φO, which in view of ¶ O 6⊆ O gives φ < 1. For the
converse,note that if φ<1, then O ⊆φO. (cid:3)
This leads easily to:
Lemma 1.3. Suppose ¶ is small and the extension L of K has small derivation.
Then the differential residue field res(L) of L is an extension of the differential
residue field res(K) of K. If in addition ¶ O 6⊆O, then L is a strict extension of K.
Lemma 1.4. Let L be an algebraic extension of K. Then L strictly extends K.
Proof. If ¶ O ⊆φO, then φ−1¶ O ⊆O, hence φ−1¶ LOL ⊆OL by [1, Proposition 6.2.1],
so¶ LOL ⊆φOL. Next, assume¶ O⊆φO. Thenφ−1¶ issmallandinduces the trivial
derivationonres(K). Henceφ−1¶ issmall,andthederivationitinducesonres(L)
L
extends the trivial derivation on res(K), so is itself trivial, as res(L) is algebraic
over res(K). Thus φ−1¶ LOL ⊆OL, that is, ¶ LOL ⊆φOL. (cid:3)
For immediate extensions, strictness reduces to a simpler condition:
Lemma 1.5. Let L bean immediate extension of K suchthat for all φ, if ¶ O⊆φO,
then ¶ LOL ⊆φOL. Then L is a strict extension of K.
Proof. Suppose ¶ O ⊆ φO. Given f ∈ OL we have f = g(1+ε) with g ∈ O and
ε∈OL, hence f′ =g′(1+ε)+gε′ ∈φOL. (cid:3)
The following related fact will also be useful:
Lemma 1.6. Suppose ¶ is small and L is an immediate extension of K such that
¶ LOL ⊆OL. Then ¶ L is small.
Proof. Ifa∈OL,thena=b(1+ε)withb∈O,ε∈OL,soa′ =b′(1+ε)+bε′ ∈OL. (cid:3)
Let us record the following observations on extensions M ⊇L⊇K:
(1) If M ⊇K is strict, then L⊇K is strict.
(2) If M ⊇L and L⊇K are strict, then so is M ⊇K.
(3) If L is an elementary extension of K, then L⊇K is strict.
(4) Any divergent pc-sequence in K pseudoconverges in some strict extension
ofK; this is aneasyconsequenceof(3), cf.[1,Remarkafter Lemma 2.2.5].
MAXIMAL IMMEDIATE EXTENSIONS OF VALUED DIFFERENTIAL FIELDS 5
The set Γ(¶ ). Note that if a,b ∈ K×, a 4 b, and O ⊆ aO, then O ⊆ bO. The set
Γ(¶ )⊆Γ, denoted also by Γ (¶ ) if we need to specify K, is defined as follows:
K
Γ(¶ ) := {vφ: ¶ O ⊆φO}.
ThisisanonemptydownwardclosedsubsetofΓ,withanupperboundinΓif¶ 6=0.
Moreover,Γ(¶ )<v(y′) whenever y ∈O. Lemma 1.2 has a reformulation:
Corollary 1.7. If ¶ O⊆O and ¶ O 6⊆O, then Γ(¶ )=Γ6.
Lemma 1.8. If vφ∈Γ(¶ ) is not maximal in Γ(¶ ), then ¶ O⊆φO.
Proof. Leta∈K×besuchthatvφ<va∈Γ(¶ ). Thena−1¶ issmall,soa−1¶ O⊆O,
and thus ¶ O⊆aO⊆φO. (cid:3)
Corollary 1.9. Suppose Γ (¶ ) has no largest element, L extends K, and for all φ,
K
if ¶ O ⊆φO, then ¶ LOL ⊆φOL. Then L strictly extends K.
Proof. If vφ ∈ Γ (¶ ), then vφ ∈ Γ (¶ ), but vφ is not maximal in Γ (¶ ), and thus
K L L
¶ OL ⊆φOL by Lemma 1.8. (cid:3)
Lemma 1.10. If L strictly extends K with Γ =Γ, then Γ (¶ )=Γ (¶ ).
L L K
The case of asymptotic fields. In this subsection we assume familiarity with
Sections6.5,9.1,andtheearlypartsofSection9.2in[1]. Recallfrom[1,Section9.1]
that by definition K is asymptotic iff for all f,g ∈ K× with f,g ≺ 1 we have
f ≺ g ⇐⇒ f′ ≺ g′. In that case we put Ψ := v(f†) : f ∈ K×,f 6≍ 1 ⊆ Γ; if we
need to make the dependence on K explicit we denote Ψ by Ψ .
K
(cid:8) (cid:9)
Lemma 1.11. Suppose K and L are asymptotic fields, and L is an immediate
extension of K. Then L is a strict extension of K.
Proof. Assume ¶ O ⊆ O; we show that ¶ OL ⊆ OL. (Using Lemma 1.5, apply this
to φ−1¶ in the role of ¶ , for vφ∈Γ(¶ ).) Now ¶ O⊆O means that there is no γ ∈Γ<
such that Ψ 6 γ, by [1, Lemma 9.2.9]. Since Γ = Γ, we have Ψ = Ψ, and so
L L
there is no γ ∈Γ<L such that ΨL 6γ, which gives ¶ OL ⊆OL. (cid:3)
Here is a partial converse:
Lemma 1.12. Suppose K is asymptotic and L strictly extends K with Γ = Γ.
L
If K is ungrounded or res(K)=res(L), then L is asymptotic.
Proof. Let a ∈ L×, a 6≍ 1. Then a = bu with b ∈ K× and a ≍ b, so u ≍ 1 and
a† =b†+u†. By [1, Proposition 9.1.3] it is enough to show that a† ≍b†, which in
turn will follow from u′ ≺b† in view of u′ ≍u†.
Suppose that Ψ has no largest element. Take φ with v(b†) < v(φ) ∈ Ψ. Then
¶ O⊆φO, so ¶ OL ⊆φOL, hence ¶ OL ⊆φOL, and thus u′ 4φ≺b†.
Next, suppose that res(K) = res(L). Then in the above we could have taken
u=1+ε with ε≺1. Then ¶ O⊆b†O, so ¶ OL ⊆b†OL, hence u′ =ε′ ≺b†. (cid:3)
For an example of a non-strict extension L ⊇ K of asymptotic fields, take K = R
with the trivial valuation and trivial derivation, and L = R((t)) with the natural
valuation f 7→order(f): L× →Z given by order(f)=k for
f = f tk+f tk+1+···
k k+1
with f 6= 0 and all coefficients f ∈ R, and derivation ¶ = d/dt given by
k k+n
¶ (f)= kf tk−1 for f = f tk. Note that Γ (¶ )=Z<, so 0∈/ Γ (¶ ).
k6=0 k k k L L
P P
6 ASCHENBRENNER,VANDENDRIES,ANDVANDERHOEVEN
The following isn’t needed, but is somehow missing in [1].
Lemma 1.13. If K is asymptotic and Γ> has a least element, then K is grounded.
Proof. SupposeΓ>hasleastelement1. Aftershiftingwecanassumethatψ(1)=0.
There is no γ ∈Γ with 0<γ <1=1+ψ(1), so 0=maxΨ. (cid:3)
Another class of valued differential fields considered more closely in [1] is the class
of monotone fields: by definition, K is monotone iff f† 4 1 for all f ∈ K×.
If K is monotone, then so is any strict extension L of K with Γ = Γ, by [1,
L
Corollary6.3.6]. Note that K has a monotonecompositionalconjugate ifffor some
φ ∈ K× we have f† 4 φ for all f ∈ K×. If K has a monotone compositional
conjugate, then clearly any strict extension L of K with Γ =Γ has as well.
L
The stabilizer of Γ(¶ ). By this we mean the convex subgroup
S (¶ ) := γ ∈Γ: Γ(¶ )+γ =Γ(¶ )
K
of Γ, also denoted by S(¶ ) if K is(cid:8)clear from the context. T(cid:9)hus
S(¶ )> = γ ∈Γ> : Γ(¶ )+γ ⊆Γ(¶ ) .
Note that Γ(¶ ) is a union of c(cid:8)osets of S(¶ ). We have Γ(cid:9)(a¶ ) = Γ(¶ ) + va and
S(¶ )=S(a¶ )foralla∈K×. SoS(¶ )isinvariantundercompositionalconjugation.
Normalizing ¶ so that 0∈Γ(¶ ) has the effect that S(¶ )⊆Γ(¶ ).
If K is asymptotic,thenΓ(¶ )= γ ∈Γ: γ <(Γ>)′ ,andthus S(¶ )={0}. The
following is also easy to verify.
(cid:8) (cid:9)
Lemma 1.14. If Γ(¶ ) has a supremum in Γ, then S(¶ )={0}.
In particular, if Γ(¶ ) has a maximum, then S(¶ ) = {0}. If ¶ 6= 0 and Γ is
archimedean, then clearly also S(¶ )={0}. In Section 3 we need the following:
Lemma 1.15. Suppose S(¶ ) = {0}. Then for any ε ∈ Γ> there are γ ∈ Γ(¶ ) and
δ ∈Γ\Γ(¶ ) such that δ−γ 6ε.
Proof. Let ε∈Γ>. Then ε∈/ S(¶ ), so we get γ ∈Γ(¶ ) with δ :=γ+ε∈/ Γ(¶ ). (cid:3)
For cases where S(¶ ) 6= {0}, let k be a field of characteristic 0 with a valuation
w: k→∆=w(k×), and let K =k((t)) be the field of Laurent series over k. Then
we have the valuation f 7→ order(f): K× → Z, where order(f) = k means that
f = f tk +f tk+1+··· with f 6= 0 and all coefficients f ∈ k. We combine
k k+1 k k+n
these two valuations into a single valuation v: K× →Γ extending the valuation w
onk,withΓhaving∆andZasorderedsubgroups,∆convexinΓ,andΓ=∆+Z;
it is given by v(f) = w(f )+k, with k = order(f). Next, we equip K with the
k
derivation ¶ = t·d/dt given by ¶ (f) = kf tk for f = f tk. Then K with
k k k k
the valuation v and the derivation ¶ is a (monotone) valued differential field, with
¶ O= f ∈K : order(f)>1 . It followsPeasily that P
(cid:8) Γ(¶ ) =(cid:9){γ ∈Γ: γ 6δ for some δ ∈∆},
and thus S(¶ )=∆.
MAXIMAL IMMEDIATE EXTENSIONS OF VALUED DIFFERENTIAL FIELDS 7
Coarsening. Webeginwithremindersaboutcoarseningfrom[1,Sections3.4,4.4].
Let ∆ be a convex subgroup of Γ. This yields the ordered abelian quotient group
Γ˙ =Γ/∆ of Γ, with the coarsened valuation
v˙ =v : K× →Γ˙, v˙(a):=v(a)+∆
∆
on the underlying field of K. The ∆-coarsening K of K is the valued differential
∆
field with the same underlying differential field as K, but with valuation v˙. Its
valuation ring is O˙ = {a ∈ K : va > δ for some δ ∈ ∆} ⊇ O, with maximal ideal
O˙ = {a ∈ K : va > ∆} ⊆ O. The residue field K˙ := res(K∆) = O˙/O˙ is itself a
valued field with valuation v: K˙× → ∆ given by v(a+O˙) := v(a) for a ∈ O˙ \O˙,
and with valuation ring {a+O˙ : a ∈ O}. We identify res(K) with res(K˙) via
res(a) 7→ res(a+O˙) for a ∈ O. Suppose the derivation ¶ of K is small. Then ¶ is
also small as a derivation of K , by [1, Corollary 4.4.4], and the derivation on K˙
∆
induced by the derivation of K is small as well. This derivation on K˙ induces
∆
the same derivation on res(K˙) as ¶ on K induces on res(K). The operation of
coarsening commutes with compositional conjugation: (Kφ) and (K )φ are the
∆ ∆
same valued differential field, to be denoted by Kφ.
∆
Thenextlemma describesthe downwardclosedsubsetΓ˙(¶ )ofΓ˙ almostcompletely
in terms of Γ(¶ ) and the canonical map π :Γ→Γ˙. Let α range over Γ.
Lemma 1.16. If ¶ O˙ ⊆ φO˙, then vφ ∈ α>∆Γ(¶ )+α. As a consequence we have
either Γ˙(¶ )=πΓ(¶ ), or Γ˙(¶ )=πΓ(¶ )∪{µ˙} with µ˙ =maxΓ˙(¶ ).
T
Proof. Suppose ¶ O˙ ⊆ φO˙. Then φ−1¶ O˙ ⊆ O˙, so φ−1¶ O˙ ⊆ O˙, hence ¶ O˙ ⊆ φO˙. For
a ∈ K× with α = va > ∆ we have aO˙ ⊆ O, so ¶ O˙ ⊆ φO˙ ⊆ φa−1O, and thus
¶ O⊆φa−1O, which gives vφ−α∈Γ(¶ ). We conclude that
vφ∈ Γ(¶ )+α.
α>∆
\
Itfollowsfrom[1, Corollary4.4.4]thatπΓ(¶ )⊆Γ˙(¶ ). Suppose thatvφ>Γ(¶ )+∆.
Thenby the above,v˙φ−α˙ ∈πΓ(¶ ) forall α˙ ∈Γ˙>. If πΓ(¶ ) has nolargestelement,
then we get v˙φ = supπΓ(¶ ). If πΓ(¶ ) has a largest element, then v˙φ−maxπΓ(¶ )
must be the least positive element of Γ˙>, and Γ˙(¶ )=πΓ(¶ )∪{v˙φ}. (cid:3)
The following will be needed in deriving Proposition 7.2:
Lemma 1.17. Let L be an immediate strict extension of K such that res(L ) =
∆
res(K ). Then L is a strict extension of K .
∆ ∆ ∆
Proof. NotethatL∆isanimmediateextensionofK∆. Suppose¶ O˙ ⊆ φO˙;applying
Lemmas 1.5 and 1.6 to the extension L of K , it suffices to derive from this
∆ ∆
assumption that ¶ O˙L ⊆ φO˙L. The proof of Lemma 1.16 gives ¶ O ⊆ φa−1O for
every a ∈ K× with va > ∆, and so ¶ OL ⊆ φa−1OL for such a. Thus for f ∈ OL
we have v(f′) > vφ−α for all α > ∆, so v(f′/φ) > −α for all α > ∆, that is,
f′/φ∈O˙L, so f′ ∈φO˙L. This shows ¶ O˙L ⊆¶ OL ⊆φO˙L, as desired. (cid:3)
Dominant degree. We summarize herefrom[1,Section6.6]whatweneed about
thedominantpartanddominantdegreeofadifferentialpolynomialanditsbehavior
under additive and multiplicative conjugation. We give the definitions, but refer
to[1,Section6.6]fortheproofs. Inthis subsectionwe assumethederivation ¶ of K
8 ASCHENBRENNER,VANDENDRIES,ANDVANDERHOEVEN
is small, and we choose for every P ∈K{Y}6= an element d ∈K× with d ≍P,
P P
such that d =d whenever P ∼Q, P,Q∈K{Y}6=. Let P ∈K{Y}6=.
P Q
We have d−1P ≍ 1, in particular, d−1P ∈ O{Y}, and we define the dominant
P P
part D ∈res(K){Y}6= to be the imageofd−1P under the naturaldifferentialring
P P
morphism O{Y}→res(K){Y}. Note that degD 6degP. The dominant degree
P
of P is defined to be the natural number ddegP :=degD ; unlike D it does not
P P
depend on the choice of the elements d ∈ K×. Given also Q ∈ K{Y}6= we have
P
ddegPQ=ddegP+ddegQ. Iff 41inanextensionLofK withsmallderivation
satisfies P(f)=0, then DP(f +OL)=0 and thus ddegP >1.
Lemma 1.18. If a∈K and a41, then ddegP =ddegP.
+a
Lemma 1.19. Let a,b∈K, g ∈K× be such that a−b4g. Then
ddegP = ddegP .
+a,×g +b,×g
Lemma 1.20. If g,h∈K× and g 4h, then ddegP 6ddegP .
×g ×h
2. Eventual Behavior
In this section Γ6={0}. We let φ range over K×, and σ, τ over N∗. We also fix a
differential polynomial P ∈ K{Y}6=. Here we generalize parts of [1, Sections 11.1,
11.2]by dropping the assumption there that K is asymptotic. The condition vφ<
(Γ>)′ there becomes the condition vφ∈Γ(¶ ) here.
Behavior of vFn(φ). The differential polynomials Fn(X) ∈ Q{X} ⊆ K{X}
k k
(0 6 k 6 n) were introduced in [1, Section 5.7] in connection with compositional
conjugation. We also recall from there that for τ =τ ···τ >σ =σ ···σ ,
1 d 1 d
Fτ := Fτ1···Fτd.
σ σ1 σd
Inordertobetterunderstandv(Pφ)asafunctionofφweusefrom[1,Lemma5.7.4]
the identity
(Pφ) = Fτ(φ)P . (1)
[σ] σ [τ]
τ>σ
X
The next two lemmas have the same proof as [1, Lemmas 11.1.1, 11.1.2].
Lemma 2.1. If ¶ O ⊆O and φ41, then v(Pφ)>v(P), with equality if φ≍1.
We set d =φ−1¶ in the next two results.
Lemma 2.2. Suppose that d O⊆O, and let 06k 6n.
(i) If φ† 4φ, then Fn(φ)4φn and Fn(φ)=φn.
k n
(ii) If φ† ≺φ and k <n, then Fn(φ)≺φn.
k
Corollary 2.3. Suppose that d O⊆O and φ† 4φ, and τ >σ. Then Fστ(φ)4φkτk
and Fτ(φ)=φkτk. If φ† ≺φ and τ >σ, then Fτ(φ)≺φkτk.
τ σ
Let P have order 6 r, so P = P Yi with i ranging over N1+r. We define the
i i
dominant degree ddegP ∈N and dominant weight dwtP ∈N by
P
ddegP =max |i|: P ≍P , dwtP =max kik: P ≍P .
i i
This agrees with the do(cid:8)minant degre(cid:9)e and dominant weig(cid:8)ht from [1, Se(cid:9)ctions 4.5,
6.6]forK withsmallderivation,althoughthedefinitionthereissomewhatdifferent.
Lemma 2.4. Suppose ¶ O ⊆O and φ≍1. Then ddegPφ =ddegP.
MAXIMAL IMMEDIATE EXTENSIONS OF VALUED DIFFERENTIAL FIELDS 9
Proof. Set d:=ddegP, I :={i: P ≍P, |i| =d}, I :={i: P ≍ P, |i| < d}.
d i <d i
Then P =Q+R+S with Q:= P Yi, R:= P Yi, and so
i∈Id i i∈I<d i
Pφ =Qφ+Rφ+Sφ, Pφ ≍PP, Qφ ≍Q≍P,PRφ ≍R, Sφ ≍S ≺P,
by Lemma 2.1, and R ≍ P if R 6= 0. Also degQφ = degQ = d and degRφ =
degR<d by [1, Corollary 5.7.5], and thus ddegPφ =d. (cid:3)
It is convenient to introduce two operators D,W: K{Y}6= →K{Y}6=:
D(P) := P Yi, I := {i: P ≍P},
i i
i∈I
X
W(P) := P Yi, J := i∈I : kik=dwtP .
i
i∈J
X (cid:8) (cid:9)
Thus D(P) and W(P) are of degree ddegP, and every monomial Yi occurring
in W(P) has weight kik=dwtP. Note that P ≍D(P)≍W(P).
Lemma 2.5. Suppose Γ> has no smallest element and ¶ O ⊆ O. Then there exists
an α∈Γ< such that for w :=dwtP we have
D(Pφ) ∼ φwW(P)
for all φ with α<vφ<0, so ddegPφ =ddegP and dwtPφ =dwtP for such φ.
Proof. For any monomial Yi = Y[τ] we have (Y[τ])φ = Fτ(φ)Y[σ] by the
σ6τ σ
first part of the proof of [1, Lemma 5.7.4]. Now let φ ≻ 1. Then φ† ≺ φ: this
P
is clear if φ′ 4 φ, and follows from [1, Lemma 6.4.1(iii)] when φ′ ≻ φ. Thus by
Corollary 2.3 and using kik=kτk we have
(Yi)φ ∼ φkikYi.
Now P = W(P)+Q with Q ∈ K{Y}, and for each monomial Yi, either Q ≺ P,
i
or Q =P ≍P and kik<dwtP. Then
i i
Pφ = W(P)φ+Qφ, W(P)φ ∼φwDw(P) for w :=dwtP.
ByconsideringtheindividualmonomialsinQweobtaintheexistenceofanα∈Γ<
suchthatQφ ≺φwW(P)wheneverα<vφ<0. Anysuchαwitnessestheproperty
stated in the lemma. (cid:3)
Corollary 2.6. If Γ> has no least element, φ ∈K× and v(φ )∈Γ(¶ ), then there
0 0
exists α<v(φ0) such that ddegPφ0 =ddegPφ whenever α<v(φ)<v(φ0).
Proof. Apply Lemma 2.5 to Kφ0 and Pφ0 in the role of K and P. (cid:3)
Newton degree. In this subsection we assume Γ> has no least element. Let
P ∈K{Y}6= have order 6r ∈N. For d6degP we define
Γ(P,d) := γ ∈Γ(¶ ): ddegPφ =d for some φ with vφ=γ .
Note that in this defi(cid:8)nition of Γ(P,d) we can replace “some” by “a(cid:9)ll” in view
of Lemma 2.4, and hence the nonempty sets among the Γ(P,d) with d 6 degP
partition Γ(¶ ). Note also that if γ ∈ Γ(P,d), then (γ −α,γ] ⊆ Γ(P,d) for some
α∈Γ> by Corollary 2.6, so each convex component of Γ(P,d) in Γ is infinite.
Lemma 2.7. The set Γ(P,d) has only finitely many convex components in Γ.
10 ASCHENBRENNER,VANDENDRIES,ANDVANDERHOEVEN
Proof. Letirangeoverthetuples(i ,...,i )∈N1+r with|i|6degP,andlikewise
0 r
for j. Let N be the number of pairs (i,j) with i 6= j. We claim that for ev-
eryφ ∈K× withvφ ∈Γ(¶ )thesetΓ(P,d) hasatmostN+1convexcomponents
0 0
with an element 6vφ . (It follows easily from this claim that Γ(P,d) has at most
0
N +1 convex components.) By renaming Kφ0 and Pφ0 as K and P it suffices to
provetheclaimforφ0 =1. Soweassumethat¶ O ⊆OandhavetoshowthatΓ(P,d)
has at most N +1 components with an element 60. We now restrict i further by
the requirement that P 6=0, and likewise for j. By the proof of Lemma 2.5,
i
ddegPφ = max |i|: vP +kikvφ=minvP +kjkvφ for vφ<0.
i j
j
(cid:26) (cid:27)
For each i we have the function f : QΓ → QΓ given by f (γ) = vP +kikγ. For
i i i
any i, j, either f = f or we have a unique γ = γ ∈ QΓ with f (γ) = f (γ).
i j i,j i j
Let γ < ···< γ with M 6N be the distinct values of γ < 0 obtained in this
1 M i,j
way, and set γ := −∞ and γ := 0. Then on each interval (γ ,γ ) with
0 M+1 m m+1
0 6 m 6 M, the functions f −f have constant sign: −, 0, or +. In view of the
i j
above identity for ddegPφ it follows easily that for each m with 0 6 m 6 M the
value of ddegPφ is constantas vφ rangesover(γ ,γ )∩Γ. Thus Γ(P,d) has at
m m+1
most M +1 convex components. (cid:3)
ItfollowsfromLemmas2.5and2.7thatthereexistsd6degP andaφ ∈K× such
0
that vφ ∈Γ(¶ ), vφ is not maximal in Γ(¶ ), and ddegPφ = d for all φ 4φ with
0 0 0
vφ ∈ Γ(¶ ). We now define the Newton degree ndegP of P to be this eventual
value d∈N of ddegPφ. Note that if Γ(¶ ) does have a maximal element vφ, then
ndegP = ddegPφ.
Also, for f ∈K× and Q∈K{Y}6= we have
ndegPf = ndegP, ndegPQ = ndegP +ndegQ.
Newton degree and multiplicative conjugation. In this subsection Γ> has no
least element. Here we consider the behavior of ndegP as a function of g ∈K×.
×g
Indeed, ndegP > 1 is a useful necessary condition for the existence of a zero
×g
f 4 g of P in a strict extension of K, as stated in the following generalization of
[1, Lemma 11.2.1]:
Lemma 2.8. Let g ∈ K× and suppose some f 4 g in a strict extension of K
satisfies P(f)=0. Then ndegP >1.
×g
Proof. For such f we have f = ag with a 4 1, and Q(a) = 0 for Q := P . So
×g
Qφ(a) = 0 for all φ with vφ ∈ Γ(¶ ), hence ddegQφ > 1 for those φ, and thus
ndegQ>1. (cid:3)
Next some results on Newton degree that follow easily from corresponding facts at
the endofSection1ondominant degree, usingalsothatcompositionalconjugation
commutes with additive and multiplicative conjugation by [1, Lemma 5.7.1].
Lemma 2.9. If a∈K and a41, then ndegP =ndegP.
+a
Lemma 2.10. Let a,b∈K, g ∈K× be such that a−b4g. Then
ndegP = ndegP .
+a,×g +b,×g
Lemma 2.11. If g,h∈K× and g 4h, then ndegP 6ndegP .
×g ×h