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arXiv:1704.02910v2 [math.LO] 9 May 2017
DEFINABLE VALUATIONS INDUCED BY
MULTIPLICATIVE SUBGROUPS AND NIP FIELDS
KATHARINA DUPONT∗, ASSAF HASSON∗∗, AND SALMA KUHLMANN∗∗∗
Abstract. We study the algebraic implications of the non-independence
property (NIP) and variants thereof (dp-minimality) on infinite fields,
motivated by the conjecture that all such fields which are neither real
closed nor separably closed admit a (definable) henselian valuation. Our
results mainly focus on Hahn fields and build up on Will Johnson’s
preprint ”dp-minimal fields”, arXiv: 1507.02745v1, July 2015.
Introduction
The classification of ω-stable fields [24, Theorem 3.1] and later of super-
stable fields [2] is a cornerstone in the development of the interactions be-
tween model theory, algebra and geometry. Ever since, the theme of clas-
sifying algebraic structures according to their model theoretic properties is
a recurring theme in model theory. Despite some success in the classifica-
tion of groups of finite rank (with respect to various notions of rank), e.g..
[6],[27, Section 4] (essentially, generalising results from the stable context),
and most notably in the o-minimal setting (e.g., [11] and many references
therein) little progress has been made in the classification of infinite stable
(let alone simple) fields. Indeed, most experts view the conjecture asserting
that (super) simple fields are bounded (perfect) PAC, and even the consid-
erably weaker conjecture that stable fields are separably closed to be out of
the reach of existing techniques.
In the last decade, or so, the increasing interest in theories without the in-
dependence property (NIP theories), associated usually with the solution
of Pillay’s conjecture [11] and with the study of algebraically closed valued
fields, led naturally to analogous classification problems in that context. In
its full generality, the problem of classifying NIP fields encompasses the clas-
sification of stable fields, and may, therefore, be too ambitious. In [26], as an
Date: May 11, 2017.
2010 Mathematics Subject Classification. 03C68.
Key words and phrases. NIP, strong NIP, definable valuations, henselian fields, Hahn
series, dp-minimal fields.
∗Partially supported by a Minerva fellowship and a post-doctoral fellowship from Ben
Gurion University.
∗∗ Partially supported by GIF grant 2165/2011 and by ISF grant 181/16.
∗∗∗ Partially supported by AFF grant from the University of Konstanz.
1
2K. DUPONT, A. HASSON, AND S. KUHLMANN
attempt to find the right analogue of super-stability in the context of NIP
theories, Shelah introduced the notion of strong NIP. As part of establishing
this analogy, Shelah showed [26, Claim 5.40] that the theory of a separa-
bly closed field that is not algebraically closed is not strongly NIP. In fact
Shelah’s proof actually shows that strongly NIP fields are perfect1. Shelah
conjectured [26, Conjecture 5.34] that (interpreting its somewhat vague for-
mulation) strongly NIP fields are real closed, algebraically closed or support
a definable non-trivial (henselian) valuation. Recently, this conjecture was
proved2by Johnson [16] in the special case of dp-minimal fields (and, inde-
pendently, assuming the definability of the valuation, henselianity is proved
in [15]).
The two main open problems in the field are the following two questions:
(1) Let Kbe an infinite (strongly) NIP field that is neither separably
closed nor real closed. Does Ksupport a non-trivial definable valu-
ation?
(2) Are all (strongly) NIP fields henselian (i.e., admit some non-trivial
henselian valuation) or, at least, t-henselian (i.e., elementarily equiv-
alent in the language of rings, to a henselian field)?
As a positive answer to Question (1) in its full generality may be out of
reach at present, addressing the question in the more restrictive strongly
dependent setting suggests a search for field theoretic division lines equiv-
alent – or at least implied – by the strongness assumptions. In the present
paper we show that perfection and boundedness – the conjectural division
lines for simple fields – are not valid in the NIP case. Building on previous
results of Delon [4], B´elair [1] and Jahnke-Simon [14] we study the model
theoretic properties of Hahn series to provide the following examples (see
Theorem 3.2): There are NIP fields with the following properties:
(1) A strongly NIP field that is not dp-minimal.
(2) A strongly NIP field Ksuch that [K×: (K×)q] = ∞for some prime
q.
(3) A perfect NIP field that is not strongly NIP.
(4) An unbounded strongly NIP field.
En passant we point out that our example of an unbounded strongly NIP
field shows that the result of [18, Corollary 3.13], stating that in a strongly
NIP field [K×: (K×)q] = ∞for at most finitely many qis best possible.
Building on results of Koenigsmann (reviewed and corrected in [5]) we show,
Proposition 3.13, that Hahn series which are NIP support a definable non-
trivial valuation with a t-henselian topology. In most cases we can show
that, in fact, the definable valuation is henselian.
1Shelah’s proof only uses the simple fact that if char(K) = p > 0 then either Kis
perfect or [K×: (K×)p] is infinite. See, e.g., [21, Remark 2.5]
2The existence of a definable valuation is implicit in Johnson’s work. See Remark 2.3.
DEFINABLE VALUATIONS IN NIP FIELDS 3
Section 2 is dedicated to proving that a finite extension of a dp-minimal
field is dp-minimal (see Theorem 2.4).
The proof builds heavily on Johnson’s classification of dp-minimal fields.
We also point out that dp-minimal fields which are neither real closed nor
separably closed also have a non-trivial definable valuation, allowing us to
conclude that, restricted to the dp-minimal case, the V-topology constructed
by Johnson, and the topology TGarising from Koenigsmann’s construction
(see Section 1) are equivalent.
Motivated by the above observation, the last part of the paper studies more
closely Koenigsmann’s construction of the definable valuation, in the aim
of suggesting a strategy for constructing a non-trivial definable valuation
on NIP fields. We give, provided the field Kis neither real closed nor
separably closed (without further model theoretic assumptions), an explicit
first order sentence ψKin the language of rings such that K|=ψKimplies
a positive answer to Question (1). Assuming a positive answer to Question
(2) (even in its weaker form), the sentence K|=ψKis, in fact, equivalent to
Ksupporting a definable non-trivial valuation (Corollary 1.3). In that case
it follows from Fact 1.1 that this definable valuation extends to a definable
henselian valuation on some finite extension L≥K.
Implicit in the work of Koenigsmann [19], a sentence with roughly the same
properties as ψKabove can certainly be extracted from [5]. However, the
sentence ψKobtained in Proposition 4.9 of this paper is simpler in quantifier
depth and in length. As a result the strategy proposed for tackling Question
(1) above can be summarised as follows:
0.1. Conjecture. Let Kbe an infinite field not separably closed. For any
prime q6= char(K) let Tq:= (K×)q+ 1. Assume that
(1) Tq6=K\{1}
(2) √−1∈K
(3) There exists ζq∈Ka primitive q-th root of unity.
and at least one of the following holds:
(1) K|= (∃a1, a2)({0}=a1Tq∩a2Tq)
(2) K|= (∀a1, a2∃b)(b /∈Tq∧b∈(a1Tq∩a2Tq)−(a1Tq∩a2Tq))
(3) K|= (∀a1, a2∃b)(b /∈Tq∧b∈(a1Tq∩a2Tq)·(a1Tq∩a2Tq))
(4) K|= (∀a1, a2∃x, y)(xy ∈a1Tq∩a2Tq∧x /∈a1Tq∩a2Tq∧y /∈a1Tq∩
a2Tq)
Then Khas IP.
Acknowledgements. We would like to thank I. Efrat, F. Jahnke and F.-V.
Kuhlmann for several ideas and suggestions.
4K. DUPONT, A. HASSON, AND S. KUHLMANN
1. Preliminaries
Let Kbe a field and let Gbe a multiplicative subgroup of K×with G6=K×.
We will be using the following notation and definitions: Owill denote a
valuation ring on Kwith maximal ideal M. Valuations on Kwill be denoted
by vand Ov:= {x∈K:v(x)≥0},Mvthe valuation ring associated with
vand its maximal ideal respectively. If Ldenotes a language and L(K) the
extension of the language Lby constants for all elements of K, we say that
a valuation von Kis L-definable if Ovis L(K)-definable. Throughout the
paper Lwill be the language of rings.
Given a group G≤K×we let TGbe the coarsest topology for which G
is open and linear transformations are continuous. As shown in [5, Theo-
rem 3.3] SG:= {a·G+b:a∈K×, b ∈K}is a subbase of TG. Hence
BG:= (n
\
i=1
(ai·G+bi)n∈N, a1,...,an∈K×, b1,...,bn∈K)
is a base for TG.
A simple calculation shows that
NG:= {U∈ BG|0∈U}=(n
\
i=1
ai·(−G+ 1) n∈N, ai∈K×)
is a base of neighbourhoods of zero for TG. If further −1∈Gthen
NG=(n
\
i=1
ai·(G+ 1) : n∈N, ai∈K×).
Throughout the paper U, V and W, possibly with indices, will always denote
elements of NG.
It follows from [5, Lemma 3.6 and Corollary 3.8] that, if TGis a basis for a
V-topology (see Fact 1.1 below) then already
(a1·G+b1)∩(a2·G+b2) : a1, a2∈K×, b1, b2∈K
is a base for TG. Hence, if −1∈G,
N′
G:= (a1·(G+ 1)) ∩(a2·(G+ 1)) : a1, a2∈K×
is a base of the neighbourhoods of zero for TG. As in most of the paper it
will be more convenient to work with arbitrary intersections, we will mostly
choose to work with NG. The advantage of the basis N′
Gis that it is a
definable basis of 0-neighbourhoods.
The starting point of this paper is [5, Corollary 5.16], suitably adapted to
the present, more restricted, setting:
DEFINABLE VALUATIONS IN NIP FIELDS 5
1.1. Fact. Let Kbe a field. Let char (K)6=qand if q= 2 assume Kis not
euclidean. Assume that there exists a primitive qth-root of unity ζq∈K.
Let G:= (K×)q6=K×. Assume that NGis a basis of 0-neighbourhoods
for a V-topology on K(see Definition 4.1). Then Kadmits a non-trivial
∅-definable valuation for which NGis a neighbourhod basis at 0.
In [5] to Gas above is associated a canonical valuation ring OG. Though OG
is not necessarily definable, it follows readily from the proof of [5][Corollary
5.17] that the definable valuation appearing in the above conclusion induces
the same topology as OG. Thus, under the assumptions of Fact 1.1, OGis
non-trivial if and only if NGis the basis of a V-topology, if and only if there
exists a definable non-trivial valuation inducing the same topology.
Let us now explain how the above fact will be applied. Let Kbe an NIP field.
We aim to find conditions for the existence of a definable non-trivial valua-
tion on K. By [25, Theorem II.4.11] ([26, Observation 1.4]) if Tis (strongly)
NIP then so is Teq. Thus any finite extension of Kis also (strongly) NIP.
It will suffice, therefore, to find a definable non-trivial valuation on some
finite extension L≥K(since if Ois a non-trivial valuation ring on Lthen
O ∩ Kis a non-trivial valuation ring in K). It is therefore, harmless to
assume that √−1∈K. By [17, Theorem 4.4] Kis Artin-Schreier closed.
So the same is true of any finite extension L≥K. This implies (e.g., [21,
Lemma 2.4]3) that either Kis separably closed, or there exists some finite
extension L≥Kand q6= char(K) such that (L×)q6=L×. Since √−1∈Lit
follows that, letting L(q) denote the q-closure of L, we have [L(q) : L] = ∞.
So extending La little more, there is no harm assuming that there exists
ζq∈L, a primitive qth root of unity. Thus, at the price of, possibly, losing
the ∅-definability of the resulting valuation, the basic assumptions of Fact
1.1 are easily met. So that the application of this result reduces to proving
that for Land qas above, NGis a 0-neighbourhood basis for a V-topology
on L. Thus, we get the following result (see also [5]):
1.2. Corollary. Let Kbe an NIP field that is neither separably closed nor
real closed. Assume that Kis t-henselian. Then for any finite field extension
L≥Kand any qsuch that (L×)q6=L×the group Gq(L) := (L×)qsatisfies
conditions (V1)-(V6) of Definition 4.1 below.
Proof: Assume first that Kis henselian, witnessed by a valuation v. Then,
by the above discussion, as Kis neither real closed nor algebraically closed,
there is some finite field extension L≥Kand prime qsuch that Gq(L) :=
(L×)qis a proper subgroup of L×. Since vis henselian, it extends to a
henselian valuation on Lwhich by abuse of notation we will also denote v.
By [5, Theorem 5.18] there exists a definable valuation won Linducing the
3Krupinski’s argument assumes that the field is perfect to conclude that it is alge-
braically closed. Descarding this additional assumption, and restricting to separable ex-
tensions the stronger result follows.
6K. DUPONT, A. HASSON, AND S. KUHLMANN
same topology as both vand OGq(L). In particular OGq(L)is non-trivial. So,
by the above discussion, NGq(L)is a basis for a V-topology, i.e., it satisfies
(V1)-(V6), as required.
Now let K ≻ Kbe ℵ1-saturated. Then Kis henselian. Let L≥Kbe a
finite separable extension such that Gq(L) is a proper subgroup. By the
primitive element theorem there exists α∈Lsuch that L=K(α). Let
L:= K(α). Then L ≻ Land Gq(L) is a proper subgroup. By what we have
already shown the group Gq(L) satisfies conditions (V1)-(V6). So N′
Gq(L)is
a also a basis for the topology, so it satisfies the corresponding statements
(V1)′-(V6)′. Since those are first order statements without parameters, they
are also satisfied by Gq(L), so Gq(L) also satisfies (V1)-(V6) as required.
We remind also that by a Theorem of Schmidt [7, Theorem 4.4.1] any two
henselian valuations on a non-separably closed field Kare dependent (i.e.,
generate the same V-topology). So we get:
1.3. Corollary. Let Kbe an NIP field that is neither real closed, nor sep-
arably closed. If Kis henselian, then Ksupports a definable non-trivial
valuation. Moreover, there exists a finite extension L≥Kand a prime q
such that Gq(L)∩Kgenerates the same Vtopology as any henselian valu-
ation on K.
Proof: There is no harm assuming that √−1∈K. As is well knonw (and
see the beginning of Section 5), if for all finite Land all qwe have Lq=L,
then as Kand all its finite extensions are Artin-Schreier closed, we get that
Kis separably closed, contradicting our assumption. So there are L≥K,
a finite extension, and qsuch that Lq6=L. Since √−1∈Kwe get that
L(ζq)q6=L(ζq) for any ζq, a primitive qth root of unity. So there is no harm
assuming ζq∈L. Since Kis henselian, so is L. By the previous corollary,
Gq(L) generates on Lthe same topology as any henselian valuation on L.
The corollary follows.
Replacing (V1)-(V6) with (V1)′-(V6)′as in the proof of Corollary 1.2 we get
that modulo the conjecture that all NIP fields are henselian4, the existence
of a definable non-trivial valuation (on a field that is neither real closed
nor algebraically closed) can be stated in a single first order sentence, ψK.
Without assuming that all strongly NIP valued fields are henselian, we still
get that ψKimplies the existence of a definable non-trivial valuation.
In Section 4 below we reduce ψKin both quantifier depth (reducing it to an
AE-sentence) and in length. This reduction is obtained without any model
theoretic assumptions on K.
4The usage of NIP in the above corollaries can be replaced with the somewhat less
natural assumption that the field and all of its separable extensions are Artin-Schreier
closed.
DEFINABLE VALUATIONS IN NIP FIELDS 7
Throughout the paper we will be using without further reference the facts
that strongly NIP fields are perfect, that NIP fields are Artin-Schreier closed,
and that NIP valued fields of characteristic p > 0 have a p-divisible value
group ([17, Proposition 5.4]).
2. dp-minimal fields
Dp-minimal fields are classified in Theorem 1.2 of [16]:
2.1. Theorem (Johnson).A sufficiently saturated field Kis dp-minimal if
and only if Kis perfect and there exists a valuation von Ksuch that:
(1) vis henselian.
(2) vis defectless (i.e., any finite extension of (L, v)over (K, v)is de-
fectless).
(3) The residue field Kv is either an algebraically closed field of charac-
teristic por elementarily equivalent to a local field of characteristic
0.
(4) The valuation group Γvis almost divisible, i.e., [Γv:nΓv]<∞for
all n.
(5) If char(Kv) = p6= char(K)then [−v(p), v(p)] ⊆pΓv.
Given a dp-minimal field Kthat is not strongly minimal, Johnson constructs
an (externally definable) topology [16, §3], which he then proves to be a V-
topology [16, §3 , §4]. Pushing these results further he proceeds to show
[16, Theorem 5.14] that Kadmits a henselian topology (not necessarily
definable). From this we immediately get:
2.2. Corollary. Any dp-minimal field is either real closed, algebraically
closed or admits a non-trivial definable henselian valuation. In particu-
lar, the V-topology constructed by Johnson is definable and coincides with
Koenigsmann’s topology, TG(L)∩K, for some finite extension L≥Kand
some (equivakently, any) G:= (L×)psuch that G6=L×.
Proof: Let Kbe a dp-minimal field that is neither real closed nor alge-
braically closed. By [16, Theorem 5.14] Kis henselian, and therefore so
is any finite extension of K. Let Lbe a finite extension of Ksuch that
Gq(L)6=L×and Lcontains a primitve q-th root of unity. Then by Fact 1.1
and Corollary 1.2 we get that Ladmits a non-trivial definable valuation. So
Kadmits a non-trivial definable valuation, and by [16, Theorem 5.14] all
definable valuations on Kare henselian.
Since Kis not separably closed it follows that Ksupports a unique non-
trivial t-henselian topology so the V-topology constructed by Johnson co-
incides with the topology associated with the definable henselian valuation,
and is therefore definable.
8K. DUPONT, A. HASSON, AND S. KUHLMANN
2.3. Remark. (1) The above corollary is implicit in Johnson’s work. By
inspecting his proof of Theorem 1.2 ([16, §6]) one sees that unless K
is real closed or algebraically closed the valuation ring O∞appear-
ing in the proof, the intersection of all definable valuation rings on
K, is non-trivial, implying that Ksupports a non-trivial definable
valuation.
(2) The same result can also be inferred from [15, §7]. In that paper it
is shown that a dp-minimal valued field which is neither real closed
nor algebraically closed supports a non-trivial henselian valuation
definable already in the pure field structure. By Johonson’s The-
orem 5.14 we know that Kadmits a henselian valuation, which is
externally definable. Since an expansion of a dp-minimal field by
externally definable sets is again dp-minimal, the result follows.
We note that the proof of the first part of the above corollary shows that
the same results remain true for finite extensions of dp-minimal fields. This
follows also from the following, somewhat surprising, corollary of Theorem
2.1:
2.4. Theorem. Let Kbe a dp-minimal field, La finite extension of K.
Then Lis dp-minimal.
Proof: Since dp-minimality is an elementary property, we may assume that
Kis saturated. Indeed, since Lis a finite extension of Kit is interpretable
in K, and if K′≻Kis saturated, the field L′interpreted in K′by the
same interpretation is a saturated elementary extension of L. Thus, it will
suffice to show that there exists a valuation von Lsatisfying conditions
(1)-(5) of Theorem 2.1. Since Kis saturated, there is such a valuation on
K, extending uniquely to L. By abuse of notation we will let vdenote also
this extension.
Conditions (1) and (2) of the theorem are automatic and condition (4) is
an immediate consequence of the fundamental inequality (e.g., Theorem
3.3.4[7]). Condition (3) is automatic if Kv is real closed or algebraically
closed. So it remains to check that if Kv is elementarily equivalent to a
finite extension of Qpthen so is Lv. This is probably known, but as we
could not find a reference, we give the details.
By Krasner’s Lemma any finite extension of Qpis of the form Qp(δ) for
some δalgebraic over Qand Qphas only finitely many extensions of degree
n(for any n). Denoting e(n) the number of extensions of Qpof degree n,
there are P1(x),...,Pe(n)(x)∈Qirreducible such that any finite extension
of Qpof degree nis generated by a root of one of P1(x),...,Pe(n)(x). As
this is clearly an elementary property, we get that the same remains true if
F≡Qp. Of course, all of the above remains true if we replace Qpby some
finite extension L≥Qp. So if L′≡Land F′≥L′is an extension of degree
nit must be that F′=L(δ) for some δrealising on of P1(x),...,Pe(n)(x),
DEFINABLE VALUATIONS IN NIP FIELDS 9
implying that F′is elementarily equivalent to F, the algebraic extension of
Lobtained by realising the same polynomial.
It remains to show that if (K, v) is of mixed characteristic then [−v(p), v(p)] ⊆
pΓ where p= charKv. By [16, Lemma 6.8] and the sentence following it, to
verify this condition it suffices to show that [−v(p), v(p)] is infinite. Towards
that end it will suffice to show that [−v(p), v(p)] ∩v(K) is infinite. Indeed,
by assumption (K, v) satisfies (5) of Theorem 2.1, so [−v(p), v(p)] ⊆pΓ. As
we are in the mixed characteristic case v(p)>0. Since v(p)⊆pΓ, there
is some g1∈Γ such that pg1=v(p) := g0. So 0 < g1< g0, and by in-
duction, for all nwe can find 0 < gn< gn−1< g0=v(p). This show that
[−v(p), v(p)] ∩v(K) is infinite, concluding the proof of the theorem.
As already mentioned in the beginning of this section, the V-topologies con-
structed by Johnson and Koenigsmann coincide in the dp-minimal case.
However, in order to start Koenigsmann’s construction we first need to
assure that Gq(K)6=K×, and for that we may have to pass to a finite
extension. Let us now point out that in the dp-minimal case this is not
needed:
2.5. Lemma. Let Kbe a dp-minimal field that is neither real closed nor
algebraically closed. Then Gq(K)6=K×for some q.
Proof: Let vbe as provided by Theorem 2.1. It will suffice to show that the
value group is not divisible. This is clear if the residue field is elementarily
equivalent to a finite extension of Qp. Indeed, any finite extension Lof Qp
is henselian with value group isomorphic to Z, which is not ndivisible for
any n > 1. So Gn(L)6=L×for any such n. As this is expressible by a first
order sentence with no parameters, it remains true in any L′≡L.
If Kv |=AC F0or Kv |=RC F , the value group cannot be divisible, as then
Kwould be algebraically closed (resp. real closed). If K v |=ACFpthen,
as vis henselian defectless (K, v) is algebraically maximal, in which case
divisibility of the value group would again imply that K|=AC F .
3. Hahn Series and related constructions
Little is known on the construction of simple fields. The situation is different
in the NIP setting where strong transfer principles for henselian valued fields
(see, e.g., [14] and references therein for the strongest such result to date)
allow the construction of many examples of NIP fields. In the present section
we sharpen some of these results and exploit them to construct various
examples.
For the sake of clarity we remind the definition of strong dependence (in the
formulation most convenient for our needs. See [26, §2] for more details):
10 K. DUPONT, A. HASSON, AND S. KUHLMANN
3.1. Definition. (1) A theory Tis strongly dependent if whenever I
is an infinite linear order, {at}t∈Ian indiscernible sequence (of α-
tuples, some α), and ais a singleton there is an equivalence relation
Eon Iwith finitely many convex classes such that for s∈Ithe
sequence {at:t∈s/E}is a-indiscernible.
(2) We will say that a type pis strongly dependent if any indiscernible
sequence {ai}i∈ωin psatisfies the above conclusion.
We show:
3.2. Theorem. There are NIP fields with the following properties:
(1) A strongly NIP field that is not dp-minimal.
(2) A strongly NIP field Ksuch that [K×: (K×)q] = ∞for some prime
q.
(3) A perfect NIP field that is not strongly NIP.
(4) An unbounded strongly NIP field.
Recall that a field is bounded5if for all n∈Nit has finitely many sep-
arable extensions of degree n. Super-simple fields are bounded,[23], and
conjecturally, so are all simple fields. As pointed out to us by F. Wag-
ner, it follows, e.g., from [24, Theorem 5.10] that bounded stable fields are
separably closed.
For the sake of completeness we give a different proof, essentially, due to
Krupinski, with a less stability-theoretic flavour: Let Kbe a bounded stable
field. Since stability implies NIP Kand all its finite extensions are, as
already mentioned, Artin-Schreier closed. By an easy strengthening of [21,
Lemma 2.4], it will suffice to show that Kq=Kfor all prime q6= char(K).
Boundedness6implies that were this not the case for some qwe would have
1<[K×,(K×)q]<∞. By [20, Proposition 4.8] this implies that Kis
unstable (in fact, that the formula ∃z(x−y=zq) has the order-property).
As we will see in the concluding section of the present paper, boundedness
may also have a role to play in the study of the two questions stated in the
Introduction. In view of the results of Theorem 3.2 it seems natural to look
for model theoretic division lines that will separate the bounded NIP fields.
3.3. Remark. (1) In [18, Corollary 3.13] it is shown that in a strongly
dependent field Kfor all but finitely many primes pwe have [K×:
(K×)p]<∞. Clause (2) of Theorem 3.2 shows that this result is
optimal.
(2) Clause (4) of Theorem 3.2 answers (partially) [12, Question 3.11].
5In the literature e.g., [23], [22] a slightly stronger condition is used. The restriction
to separable extensions seems, however, more natural and even implicitly implied in some
applications.
6In [21] Krupinski introduces the slightly weaker radical boundedness, which suffices for
the argument.
DEFINABLE VALUATIONS IN NIP FIELDS 11
We will use Hahn series to construct the desired examples. The basic facts
that we need are:
3.4. Fact. A henselian valued field (K, v)of equi-characteristic 0is (strongly)
NIP if and only if the value group and the residue field are (strongly) NIP.
If (K, v)is dp-minimal then so are the residue field and the value group.
The NIP case of the above fact is due to Delon [4] and the strongly NIP
case is due to Chernikov [3]. We get:
3.5. Lemma. Let kbe a field of characteristic 0,Γan ordered abelian group.
Then the Hahn series k((tΓ)) is NIP as a valued field if and only if kis NIP
as a pure field. It is strongly NIP if and only if kand Γare.
Proof: Hahn series are maximally complete, and therefore henselian. So
the result follows from the previous fact.
In order to prove clauses (1) and (3) of Theorem 3.2 it will suffice, therefore,
to find strongly NIP ordered abelian groups that are not dp-minimal and
ones that are not strongly NIP. We start with the latter:
3.6. Example. Consider Γ := ZNas an abelian group (with respect to
pointwise addition) with the lexicographic order. Then Γ is NIP but not
strongly NIP.
Proof: The group Γ is ordered abelian, and therefore NIP by [8]. But
[Γ : nΓ] = ∞for all n > 1, contradicting [18, Corollary 3.13].
3.7. Remark. In [26] Shelah considers a closely related example of an or-
dered abelian group that is not strongly dependent.
3.8. Example. Let Γ := ZN. If kis an NIP field of characteristic 0 then
K:= k((tΓ)) is NIP by the previous lemma. It is not strongly NIP because
Γ is not strongly NIP. It is unbounded, since by the fundamental inequality
it has infinitely many Kummer extensions of any prime degree q. Indeed,
for any natural number nlet {a1,...,an} ∈ Γ be pairwise non-equivalent
modulo qΓ. Let c1,...,cn∈Kbe such that v(ci) = aifor all i. Let
L≥Kbe the extension obtained by adjoining qth -roots for all ci. Let
∆ = v(L), where vis identified with its unique extension to L. Then Γ 6⊆ q∆.
Otherwise [q∆ : qΓ] = [q∆ : Γ][Γ : qΓ] = ∞, whereas [∆ : Γ] ≤[L:K] and
[q∆ : qΓ] ≤[∆ : Γ]. This is a contradiction. Since nwas arbitrary, this
shows that Khas infinitely many Kummer extensions of degree q.
Note that by [18, Corollary 3.13] and [15] if Gis an ordered abelian group
that is strongly dependent and not dp-minimal then there are finitely many
primes qsuch that [G:qG] = ∞. So the previous example with Greplacing
Γ will give an example for Theorem 3.2(1), (2) and (5).
The details of the following example can be found in [9]:
12 K. DUPONT, A. HASSON, AND S. KUHLMANN
3.9. Fact. Let (2)Zbe the localisation of Zat (2). Let Bbe a base for
Ras a vector space over Qand let hBibe the Z-module generated by B.
Let G:=(2) Z⊗ hBi. Viewed as an additive subgroup of Rthe group Gis
naturally ordered. It is strongly dependent but not dp-minimal.
In positive characteristic, the situation is slightly different. The basic result
is due to B´elair [1]:
3.10. Fact. Let (K, v)be an algebraically maximal Kaplansky field of char-
acteristic p > 0. Then Kis NIP as a valued field if and only if the residue
field kis NIP as a pure field.
We could not find a reference for the generalisation of Fact 3.9 to the strong
NIP case, so we give the details here. The proof follows closely the proof
of the main theorem of [14], so we will be brief. In [14, Theorem 2.3] the
authors prove that if Tis any theory of valued fields such that:
(SE): the value group and the residue field are stably embedded; and
(Im): if K≺ U |=Tand a∈ U is such that K(a)/K is an immediate
extension then tp(a/K) is isolated by an NIP type,
then Tis NIP. They then show that under a weaker assumption than that of
the field being Kaplansky and algebraically maximal the assumptions (SE)
and (Im) hold (Lemma 3.1 and Lemma 3.2, respectively). The proof of the
latter lemma actually shows that if Kis algebraically maximal Kaplansky,
and K(a)/K is immediate then tp(a/K) is isolated by quantifier-free for-
mulas (in fact, in the applications, those are formulas of the form v(x) = y
and xv =y). This implies, immediately, that if the value group and the
residue field are strongly dependent then tp(a/K) is isolated by a strongly
dependent type. So in order to prove Fact 3.10 in the strong NIP setting it
will suffice to show that if in assumption (Im), above we replace “NIP” with
“strong NIP” then we can make the same replacement in the conclusion. So
it will suffice to prove:
3.11. Proposition. Let Tbe any theory of algebraically maximal Kaplansky
fields. Assume that whenever K≺ U |=Tand a∈ U is such that K(a)/K
is an immediate extension then tp(a/K)is isolated by strongly NIP type.
Then Tis strongly NIP.
Proof: Assume, towards a contradiction that Tis not strongly dependent.
Then there exists an indiscernible sequence {ai}i∈ωand asuch that for
n∈ωthere is a formula (with no parameters) φn(x, y) such that φn(a, ai)
alternates more than ntimes on {ai}i∈ω.
It follows from (the proof of ) [14, Theorem 2.3] that we can find an in-
discernible sequence {Ni}i∈ωof models such that the type of the sequence
{Ni}i∈ωover aextends the type of {ai}i∈ωover aand N0(a)/N0is immedi-
ate.
DEFINABLE VALUATIONS IN NIP FIELDS 13
By assumption tp(a/N0) is isolated by a strongly dependent type. More pre-
cisely, tp(a/N0) is isolated by formulas of the form v(t(x)) = γand t(x)v=r
where t(x) is a rational function over N0, applying [3, Lemma 7.1] we get
strong dependence of the type. So there exists a convex equivalence rela-
tion Ewith finitely many classes such that {Ni}i∈j/E is an a-indiscernible
sequence for every j∈ω. But, if |ω/E|=nthen φn(a, ai) alternates more
than ntimes on ω, which is impossible.
Summing up the above discussion we get an extension of Fact 3.10 to the
strongly NIP case in analogy with Fact 3.4:
3.12. Corollary. Let kbe an infinite NIP field (of any characteristic) and
Γan ordered abelian group. Then k((tΓ)) is NIP provided that, if p=
char(k)>0, then Γis p-divisible. It is strongly NIP if and only if kand Γ
are.
It is natural to ask whether all NIP fields constructed as Hahn series satisfy
Shelah’s conjecture, namely, whether they all support a definable henselian
valuation. It follows immediately from Corollary 1.2 that:
3.13. Proposition. Let kbe an NIP field, Γan ordered abelian group which
is p-divisible if char(k) = p > 0. Then K:= k((tΓ)) is either algebraically
closed, or real closed or it supports a definable non-trivial valuation.
This answers Question (1) for Hahn fields. Whether NIP Hahn fields support
a definable henselian valuation is more delicate. In positive characteristic
this follows from [13, Corollary 3.18]. In some cases we can be even more
precise. E.g., Hong, [10] gives conditions on the value group implying the
definability of the canonical valuation on k((tΓ)):
3.14. Fact. Let (K, O)be a henselian field. If the value group contains a
convex p-regular subgroup that is not p-divisible, then Ois definable in the
language of rings.
In characteristic 0, to the best of our knowledge, the question is not fully
solved. Corollary 1.3 gives full control over the parameters needed for the
definition of the valuation. In particular, if Gq(K)6=K×and ζq∈Kthen
the valuation is definable without parameters, implying that it is henselian.
Thus, if Γ is not divisible, then the resulting definable valuation is henselian,
possibly after passing to K(ζ) for ζa root of unity. On the residue field side,
we get the exact same result if khas Kummer-extensions.
In all the examples discussed in the present section, the source of the com-
plexity of the field (unbounded, strongly dependent not dp-minimal etc.)
can be traced back to the value group of the canonical valuation. For ex-
ample, as shown in [15], an ordered abelian group Γ is dp-minimal if and
only if [Γ : pΓ] is finite for all primes p. By Theorem 2.1 dp-minimal fields
are henselian with dp-minimal value groups. We note that it also follows
14 K. DUPONT, A. HASSON, AND S. KUHLMANN
from the same theorem that dp-minimal fields are bounded. Indeed7, for
any Henselian field Kwe have that GK∼
=T⋊Gkwhere GK, Gkare the
respective absolute Galois groups of Kand k=K v, and Tis the inertia
group. If Kis dp-minimal then8T=Ql∈ΩZdimFlΓ/lΓ
lfor a certain set of
primes Ω. Since Γ := vK is dp-minimal, this implies that Tis small. Since
kis eaither real closed, algerbaically closed or elementarily equivalent to a
finite extension of Qp, also GKis small.
It seems, therefore, natural to ask whether the complexity of the value group
in the above examples can be recovered definably. Can any (model theoretic)
complexity of an NIP field be traced back to that of an ordered abelian
group:
3.15. Question. Let Kbe a non separably closed NIP field. Does Kin-
terpret a dp-minimal field? If Kis not strongly dependent (dp-minimal)
does Kintepret an ordered abelian group Gthat is not strongly dependent
(dp-minimal)?
4. The Axioms of V-Topologies for TG
We are now returning to that construction of V topologies from multiplica-
tive subgroups, as described in Section 1. Throughout this section no model
theoretic assumptions are made, unless explicitly stated otherwise.
For ease of reference, we remind the axioms of V topology:
4.1. Definition. A collection of subsets Nof a field Kis a basis of 0-
neighbourhoods for a V-topology on Kif is satisfies the following axioms:
(V 1): TN:= TU∈N U={0}and {0}/∈ N;
(V 2): ∀U, V ∃W W ⊆U∩V;
(V 3): ∀U∃V V −V⊆U;
(V 4): ∀U∀x, y ∈K∃V(x+V)·(y+V)⊆x·y+U;
(V 5): ∀U∀x∈K×∃V(x+V)−1⊆x−1+U;
(V 6): ∀U∃V∀x, y ∈K x ·y∈V−→ x∈U∨y∈U.
Notation: From now on Gwill denote a multiplicative subgroup of K×
with −1∈Gand T:= G+ 1. We let NG:= {Tn
i=1 ai·T:ai∈K×}, as
defined in the opening paragraphs of Section 1.
In this setting the first part of (V 1) is automatic, and (V 1) holds by
definition:
4.2. Lemma. (1) TNG={0}.
(2) ∀U, V ∃W W ⊆U∩V.
7This argument was sugegsted to us by I. Efrat. Any mistake is, of course, solely, ours.
8In equicharacteristic 0 use the fact that Kis elementarily equivalent to a Hahn field.
In positive residue characteristic use the fact that, with respect to the valuation provided
in Theorem 2.1, kis algebraically closed.
DEFINABLE VALUATIONS IN NIP FIELDS 15
Proof: For every x∈K×we have x6∈ x·T∈ NG. As −1∈Gfurther
0∈x·Tfor every x∈K×. Hence TNG={0}. This proves (1), item (2)
holds by the definition of NG.
We will come back to the second part of Axiom (V 1) later. Axiom (V 3)
is simplified as follows:
4.3. Lemma. The following are equivalent
(V 3): ∀U∃V V −V⊆U.
(V 3)′:∃V V −V⊆T.
(V 3)∗:∀U∃V V +V⊆U.
Proof: The first implication is obvious. By (V 3)′there exists V=Tn
j=1 bj·
T∈ NGsuch that V−V⊆T. Let U=Tm
i=1 ai·T∈ NG. For all
i∈ {1,...,m},j∈ {1,...,n}. Let V′:= Tm
i=1 Tn
j=1 (ai·bj·T). Then by
direct computation
V′−V′⊆
m
\
i=1
ai·(V−V)⊆
m
\
i=1
ai·T=U.
This shows that (V 3) follows from (V 3)′. Replacing Vwith V∩(−V)
(throughout) we may assume that V=−V, proving the equivalence with
(V 3)∗
In order to simplify Axiom (V 4) we need:
4.4. Lemma. If ∃V V ·V⊆Tthen ∀U∃V V ·V⊆U.
Proof: Let U=Tm
i=1 ai·T∈ NG. By assumption there exist
V=Tn
j=1 bj·T∈ NGsuch that V·V⊆T.
Let V′:= Tm
i=1 Tn
j=1 ai·bj·T∩Tn
j=1 bj·T∈ NG. Then by direct
computation
V′·V′=
m
\
i=1
ai·(V·V)⊆
m
\
i=1
ai·T=U.
This proves the claim.
Now we can prove:
4.5. Lemma. The axiom
(V 4) ∀U∀x, y ∈K∃V(x+V)·(y+V)⊆x·y+U
is equivalent to the conjunction of
(V 4)′∃V V ·V⊆Tand
(V 4)′′ ∀x∈K∃V(x+V)·(1 + V)⊆x+T.
16 K. DUPONT, A. HASSON, AND S. KUHLMANN
Proof: (V 4)′and (V 4)′′ are special cases of (V 4). So we prove the other
implication.
Let x, y ∈Kand U=Tm
i=1 ai·T∈ NG. The case x=y= 0 is Lemma 4.4.
So we assume that y6= 0. For every i∈ {1,...,m}we define eai:= ai·y−1
and xi:= x·ea−1
i. By (V 4)′′ there exists Vi∈ NGsuch that
(1) (xi+Vi)·(1 + Vi)⊆xi+T.
Let V:= Tm
i=1 Tni
j=1 eai·Vi∩Tni
j=1 (y·Vi)∈ NG. Then
(x+V)·(y+V)⊆
m
\
i=1 eai·y·(xi+Vi)·(1 + Vi)
⊆
m
\
i=1 eai·y·(xi+T)=x·y+U.
Where the last inclusion follows from Equation (1). This finishes the proof.
Assuming (V 3)′we can simplify further:
4.6. Lemma. The axioms (V 3)′and (V 4)′imply axion (V 4).
Proof: By the previous lemma it will suffice to prove the lemma for U=
Tand y= 1. The case x= 0 is automatic from the assumptions and
Lemma 4.3. So assume x∈K×. By Lemma 4.3 there exist V1, V2such that
V1+V1⊆T,V2+V2⊆V1. Further by Lemma 4.4 there exists V3with
V3·V3⊆V2. Define V:= x−1·V1∩V2∩V3∈ NG. Let v, w ∈V.
(x+v)·(1 + w)∈x+V+x·V+V·V⊆x+V2+x·x−1·V1+V3·V3
⊆x+V2+V1+V2⊆x+V1+V1⊆x+T.
Hence (x+V)·(1 + V)⊆x+T, as required.
The axiom (V 5) holds without further assumptions:
4.7. Lemma. Let Kbe a field. Let Gbe a multiplicative subgroup of Kwith
−1∈G. Then (V 5) ∀U∀x∈K×∃V(x+V)−1⊆x−1+Uholds.
Proof: We will first show
(2) ∀x∈K×∃V(x+V)−1⊆x−1+T.
For x=−1 let V:= T. We have (x+T)−1= (−1 + G+ 1)−1=G−1=
G=x−1+T.
If x∈K×\{−1}, let b1=−x2·(1 + x)−1,b2=−xand V:= b1·T∩b2·T=
b1·(G+ 1) ∩b2·(G+ 1). Let z∈(x+V)−1. Let g1, g2∈Gsuch that
z= (x+b1·g1+b1)−1= (x+b2·g2+b2)−1. We have
(3) z= (x+b2·g2+b2)−1= (x−x·g2−x)−1=−x−1·g−1
2.
DEFINABLE VALUATIONS IN NIP FIELDS 17
Further we have z−1=x+b1+b1·g1and therefore 1−b1·g1·z= (x+b1)·z.
This implies
z= (1 −b1·g1·z)·(x+b1)−1
=x−1+ 1 + x·g1·z
(3)
=x−1+ 1 −x·g1·x−1·g−1
2
=x−1+−g1·g−1
2+ 1 ∈x−1+G+ 1.
Hence (x+V)−1⊆x−1+T. This proves Equation (2).
Now let x∈K×and U=Tm
i=1 ai·T∈ NG. For every i∈ {1,...,m}let
xi:= ai·x. By Equation (2) there exists Visuch that
(4) (xi+Vi)−1⊆x−1
i+T.
For V:= Tm
i=1 a−1
i·Vi
(x+V)−1=
m
\
i=1
ai·(xi+Vi)−1(4)
⊆
m
\
i=1
ai·x−1
i+T=x−1+U.
Therefore (V 5) holds.
The axiom (V 6) can be reduced as follows:
4.8. Lemma. The following are equivalent
(V 6): ∀U∃V∀x, y ∈K(x·y∈V→x∈U∨y∈U)
(V 6)′:∃V∀x, y ∈K(x·y∈V→x∈T∨y∈T).
Proof: We assume (V 6)′and show (V 6). We will show by induction
on m, that for all a1,...,am∈K×, there exists V∈ NGsuch that for all
x, y ∈K, if x·y∈Vthen x∈Tm
i=1 aior y∈Tm
i=1 ai.
Let a1∈K×and U:= a1·T∈ NG. By (V 6)′there exists Vsuch that for
all x, y ∈K, if x·y∈Vthen x∈Tor y∈T. Define V′:= a2
1·V∈ NG.
For all x, y ∈Ksuch that x·y∈V′we have x·a−1
1·y·a−1
1∈a−2
1·V′=V
and therefore x·a−1
1∈Tor y·a−1
1∈Tand hence x∈Uor y∈U.
Now let a1, a2∈K×and U:= T2
i=1 ai·T∈ NG. By assumption there exists
Vsuch that for all x, y ∈Kif x·y∈Vthen x∈Tor y∈T. Define
V′=: a2
1·V∩a2
2·V∩a1·a2·V. Let x, y ∈Ksuch that x·y∈V′. Then
x·a−1
1·y·a−1
1∈a−2
1·V′⊆Vand therefore as above
(5) x∈a1·Tor y∈a1·T.
and
(6) x∈a2·Tor y∈a2·T.
If, by way of contradiction, x·a−1
1/∈Tand y·a−1
2/∈T, then x·a−1
1·y·a−1
2/∈V,
implying x·y /∈a1·a2·V⊇V′contradicting the choice of xand y. Therefore
(7) x∈a1·Tor y∈a2·T.
18 K. DUPONT, A. HASSON, AND S. KUHLMANN
and, similarly,
(8) y∈a1·Tor x∈a2·T.
A straightforward verification shows that equations (5)-(8) implies that if
x·y∈V′then either x∈Uor y∈U.
Now let m≥3. Assume that for all a1,...,am−1there exists Vsuch that
for all x, y ∈K, if x·y∈Vthen x∈Tm−1
i=1 ai·Tor y∈Tm−1
i=1 ai·T.
Let a1,...,am∈K×and U:= Tm
i=1 ai·T∈ NG. By induction hypothesis
for every j∈ {1,...,m}there exists V6=jsuch that for all x, y ∈K, if
x·y∈V6=jthen x∈
m
\
i=1
i6=j
ai·Tor y∈
m
\
i=1
i6=j
ai·T. Define V:= Tm
i=1 V6=i. Let
x, y ∈K×such that x·y∈V. If x∈ai·Tfor all i∈ {1,...,m}then
x∈Uand we are done. Otherwise let j∈ {1,...,m}with x /∈aj·T. Let
k, ℓ ∈ {1,...,m} \ {j}with k6=ℓ. We have x·y∈Tm
i=1 V6=i⊆V6=k. As
x /∈aj·T⊇
m
\
i=1
i6=k
ai·Twe have y∈
m
\
i=1
i6=k
ai·T. Analogous we show y∈
m
\
i=1
i6=ℓ
ai·T.
Therefore y∈
m
\
i=1
i6=k
ai·T∩
m
\
i=1
i6=ℓ
ai·T=U.
Hence for all Uthere exists Vsuch that for all x, y ∈K, if x·y∈Vthen
x∈Uor y∈U.
Summing up all the simplifications of the present section we obtain:
4.9. Proposition. Let Kbe a field. Let char (K)6=qand if q= 2 assume
Kis not euclidean. Assume that for the primitive qth-root of unity ζq∈K.
Let G:= (K×)q6=K×. Assume that
(V 1)′:{0}/∈ NG;
(V 3)′:∃V V −V⊆T
(V 4)′:∃V V ·V⊆T
(V 6)′:∃V∀x, y ∈K x ·y∈V→x∈T∨y∈T.
Then Kadmits a non-trivial definable valuation.
Proof: With Lemma 4.2, Lemma 4.2, Lemma 4.3, Lemma 4.6, Lemma 4.7
and Lemma 4.8 the result follows directly from Corollary 1.1.
5. Back to NIP fields
As already explained in the opening sections, our main motivation in the
present paper is to study the existence of definable valuations on (strongly)
NIP fields. We also hope that such a project may shed some light on the
DEFINABLE VALUATIONS IN NIP FIELDS 19
long standing open conjecture that stable fields are separably closed. We
have already explained that in the stable case this conjecture can be rather
easily settled under the further assumption that the field is bounded. It is
therefore natural to ask whether the same assumption can help settle the
questions stated in the Introduction. In the present section we show how
boundedness gives quite easily Axiom [(V 1)′] (stating that {0}/∈ NG).
If Kis an infinite NIP field, i.e. a field definable in a monster model sat-
isfying NIP, then by [20, Corollary 4.2] there is a definable additively and
multiplicatively invariant Keisler measure on K. In the whole section if not
stated differently let Kbe an infinite NIP field and µan additively and
multiplicatively invariant definable Keisler measure on K.
By [20, Proposition 4.5] for any definable subset Xof Kwith µ(X)>0 and
any a∈K, we have
(♣)µ((a+X)∩X) = µ(X).
5.1. Lemma. Let a1,...,am∈K×and G⊆K×a multiplicative subgroup
with −1∈Gand µ(G)>0. Then Tm
i=1 aiT){0}.
Proof: As −1∈Git follows that 0 ∈Tm
i=1 aiT.
This also implies that
(*) G+a−1={s: 1 ∈a(G+s)}
for any a∈K×.
By additivity of the measure (♣) applied to the left hand side of (∗) gives
µ(
m
\
i=1
(G+a−1
i∩G)) = µ(G)>0.
So by the right hand side of (∗) we have t0∈Tm
i=1{s∈G: 1 ∈ai(G+s)}. So
1∈ai(G+t0) for all 1 ≤i≤m, and as t0∈Gwe get t−1
0∈Tm
i=1 ai(G+1) =
Tm
i=1 aiT.
5.2. Corollary. Let Kbe an infinite NIP field with √−1∈K. Let G:=
(K×)q6=K×for some q6=char(K)prime with ζq∈K. Assume that
[K×:G]<∞and that for T:= G+ 1 we have:
(V 3)′:∃V V −V⊆T
(V 4)′:∃V V ·V⊆T
(V 6)′:∃V∀x, y ∈K x ·y∈V→x∈T∨y∈T.
Then Kadmits a non-trivial ∅-definable valuation.
Proof: By additivity and invariance of µwe get that µ(G) = [K×:G]−1.
The result now follows directly from Proposition 4.9 using Proposition 5.1
20 K. DUPONT, A. HASSON, AND S. KUHLMANN
As mentioned in Section 1,
N′
G:= (a1·(G+ 1)) ∩(a2·(G+ 1)) : a1, a2∈K×
is a base of the neighbourhoods of zero of TG. We obtain the following
corollary:
5.3. Corollary. Let Kbe an infinite NIP field with √−1∈K. Let G:=
(K×)q6=K×for some q6=char(K)prime with ζq∈K. Assume that
[K×:G]<∞. Then for T:= G+ 1 we have that
(V 3)′:∃V∈ NGV−V⊆T
(V 4)′:∃V∈ NGV·V⊆T
(V 6)′:∃V∈ NG∀x, y ∈K x ·y∈V→x∈T∨y∈T.
if and only if
(V 3)′
2:∃e
V∈ N′
Ge
V−e
V⊆T
(V 4)′
2:∃e
V∈ N′
Ge
V·e
V⊆T
(V 6)′
2:∃e
V∈ N′
G∀x, y ∈K x ·y∈e
V→x∈T∨y∈T.
Proof: As N′
G⊆ NGit is clear that if (V 3)′
2,(V 4)′
2and (V 6)′
2hold,
then so do (V 3)′,(V 4)′and (V 6)′.
On the otherhand if (V 3)′,(V 4)′and (V 6)′hold, then TGis a V-topology
and N′
Gis a 0-neighbourhood basis for TG. Therefore, for any V∈ NG
witnessing (V i) (i= 3,4,6), there exists e
V∈ NGsuch that e
V⊆V, and as
– for a fixed V– the axiom (V i) is universal, it is automatically satisfied
by ˜
V.
Note that (V 3)′
2,(V 4)′
2and (V 6)′
2are first order sentences in the language
of rings (appearing explicitly in the statement of Conjecture 0.1). Let us
denote their conjunction as ψK. Thus, if Kis a bounded9NIP field such
that K|=ψKthen Ksupports a definable valuation.
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DEFINABLE VALUATIONS IN NIP FIELDS 21
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∗FB Mathematik und Statistik, Universit¨
at Konstanz, 78457 Konstanz, Ger-
many
E-mail address:katharina.dupont@uni-konstanz.de
∗∗Department of mathematics, Ben Gurion University of the Negev, Be’er
Sehva, Israel
E-mail address:hassonas@math.bgu.ac.il
∗∗∗FB Mathematik und Statistik, Universit¨
at Konstanz, 78457 Konstanz, Ger-
many
E-mail address:salma.kuhlmann@uni-konstanz.de
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