Dp-finite fields I: infinitesimals and positive characteristic
Will Johnson

Diversity with Cooperation: Ensemble Methods for Few-Shot Classification
    Nikita Dvornik,  Cordelia Schmid,  Julien Mairal



Univ. Grenoble Alpes, Inria, CNRS, Grenoble INP, LJK, 38000
Grenoble, France

firstname.lastname@inria.fr

    December 30, 2023
==================================================================================================================================================================


  2010 Mathematical Subject Classification: 03C45

  Key words and phrases: dp-rank, NIP fields, modular lattices



  We prove that NIP valued fields of positive characteristic are
  henselian. Furthermore, we partially generalize the known results on
  dp-minimal fields to dp-finite fields. We prove a dichotomy: if K
  is a sufficiently saturated dp-finite expansion of a field, then
  either K has finite Morley rank or K has a non-trivial
  (K/A)-invariant valuation ring for a small set A. In the
  positive characteristic case, we can even demand that the valuation
  ring is henselian. Using this, we classify the positive
  characteristic dp-finite pure fields.




§ INTRODUCTION

The two main conjectures for NIP fields are


  * The henselianity conjecture: any NIP valued field
  (K,𝒪) is henselian.

  * The Shelah conjecture: any NIP field K is algebraically
  closed, real closed, finite, or admits a non-trivial henselian
  valuation.

By generalizing the arguments used for dp-minimal fields (for example
in Chapter 9 of <cit.>), we prove the henselianity conjecture
in positive characteristic, and the Shelah conjecture for positive
characteristic dp-finite fields.  This yields the
positive-characteristic part of the expected classification of
dp-finite fields.

We also make partial progress on dp-finite fields of characteristic
zero.  Let (K,+,·,…) be a sufficiently saturated dp-finite
field, possibly with extra structure.  Then either


  * K has finite Morley rank, or

  * There is an (K/A)-invariant non-trivial valuation ring on
  K for some small set A.

Unfortunately, we can only prove henselianity of this valuation ring
in positive characteristic.

Following the approach used for dp-minimal fields, there are three
main steps to the proof:


  * Construct a type-definable group of infinitesimals.

  * Construct a valuation ring from the infinitesimals.

  * Prove henselianity.

We discuss each of these steps, explaining the difficulties that arise
when generalizing from rank 1 to rank n.


 §.§ Constructing the infinitesimals

Mimicking the case of dp-minimal fields, we would like to define the
group I_M of M-infinitesimals as

    ⋂_X “big” and M-definable{δ∈ K  |  X
      ∩ (X + δ)  is “big”}

for some notion of “big.”  In the dp-minimal case, “big” was
“infinite.”  By analyzing the proof for dp-minimal fields, one can
enumerate a list of desiderata for bigness:


  * Non-big sets should form an ideal.

  * Bigness should be preserved by affine transformations.

  *  Bigness should vary definably in
  families.

  * The universe K should be big.

  *  If X, Y are big, the set {δ | 
  X ∩ (Y + δ)  is big} should be big.

  * Bigness should be coherent on externally definable sets, to the
  extent that:
  
  
  * If X is M-definable and big for a small model M ≼
    K, if Y is K-definable, and if X(M) ⊆ Y, then Y
    is big.
  
  * If X is big and X ⊆ Y_1 ∪⋯∪ Y_n for
    externally definable sets Y_1, …, Y_n, then there is a big
    definable subset X' ⊆ X such that X' ⊆ Y_i for
    some i.
  

The intuitive guess is that for a field of dp-rank n, “big” should
mean “rank n.”  But there is no obvious proof of the definability
condition (<ref>), as noted in 3.2 of
Sinclair's thesis <cit.>.  An alternative, silly guess is
that “big” should mean “infinite.” This fails to work in some of
the simplest examples, such as (,+,·,).  However, the
silly guess nearly works; the only requirement that can fail is
(<ref>).

The key insight that led to the present paper was the realization that
in rank 2, any failure of (<ref>) for the silly
option (“big”=“infinite”) fixes (<ref>) for
the intuitive option (“big”=“rank 2”).  Indeed, if X, Y are
infinite sets but X ∩ (Y + δ) is finite for almost all
δ, then


  * By counting ranks, X, Y must be dp-minimal.

  * The map X × Y → X - Y is almost finite-to-one.

  * By a theorem of Pierre Simon <cit.>, “rank 2” is
  definable on X × Y.[He proves this under the
    assumption that the entire theory is dp-minimal, but the proof
    generalizes to products of definable sets of dp-rank 1.]  This
  ensures that “rank 2” is definable on X - Y.

  * A definable set D ⊆ K has rank 2 if and only if some
  translate of D has rank 2 intersection with X - Y.

So, in the rank-2 setting, one can first try “big”=“infinite,” and
if that fails, take “big”=“rank 2.”


  A prototype of this idea appears in Peter Sinclair's thesis
  <cit.>.  In his 3.3, he observes that the machinery of
  infinitesimals goes through when “rank n”=“infinite,” and
  conjectures that this always holds in the pure field reduct.


By extending this line of thinking to higher ranks, we obtain a notion
of heavy sets satisfying the desired properties.[In
  later work, it has been shown that the heavy sets are exactly the
  sets of full rank (<cit.>, Theorem 5.9.2).  However, the proof
  relies on the construction of the infinitesimals, and only works in
  hindsight.] See <ref> for details; the technique
is reminiscent of Zilber indecomposability in groups of finite Morley
rank.  Once heavy and light sets are defined, the construction of
infinitesimals is carried out in <ref> via a
direct generalization of the argument for dp-minimal fields.  For
certain non-triviality properties, we need to assume that K does not
have finite Morley rank.  The relevant dichotomy is proven in
<ref>; it is closely related to Sinclair's Large Sets
Property (Definition 3.0.3 in <cit.>), but with heaviness
replacing full dp-rank.

For ranks greater than 2, we need to slightly upgrade Simon's results
in <cit.>.  We do this in <ref>.  Say that
an infinite definable set Q is quasi-minimal if (D) ∈{0, (Q)} for every definable subset D ⊆ Q.  The main
result is the following:

  Let M be an NIP structure eliminating ∃^∞.  Let Q_1,
  …, Q_n be quasi-minimal sets, and m = (Q_1 ×⋯× Q_n).  Then “rank m” is definable in families of
  definable subsets of Q_1 ×⋯× Q_n.

This is a variant of Corollary 3.12 in <cit.>.  Note that
dp-minimal sets are quasi-minimal, and quasi-minimal sets are
guaranteed to exist in dp-finite structures.



 §.§ Getting a valuation

Say that a subring R ⊆ K is a good Bezout domain if
R is a Bezout domain with finitely many maximal ideals, and
(R) = K.  This implies that R is a finite intersection of
valuation rings on K.  Ideally, we could prove the following

  The M-infinitesimals I_M are an ideal in a good Bezout domain R
  ⊆ K.

Assuming the Conjecture, one can tweak R and arrange for I_M to be
the Jacobson radical of R.  This probably implies that


  * The ring R is ∨-definable, and so are the associated
  valuation rings.

  * The canonical topology is a field topology, and has a definable
  basis of opens.

This would put us in a setting where we could generalize the
infinitesimal-based henselianity proofs from the dp-minimal case,
modulo some technical difficulties in characteristic zero.  The
strategy would be to prove that R is the intersection of just one
valuation ring; this rules out the possibility of K carrying two
definable valuations, leading easily to proofs of the henselianity and
Shelah conjectures.[The details of this strategy have been
  verified in <cit.>, which proves the classification of
  dp-finite fields contingent on the above Conjecture.  The
  “technical difficulties in characteristic zero” involve showing
  that 1 + I_M is exactly the group of “multiplicative
  infinitesimals.”  This is <cit.>, Proposition 5.12.]

As it is, the Conjecture is still unknown, even in positive
characteristic.  Nevertheless, we can prove that a non-trivial
(K/A)-invariant good Bezout domain R exists for some small set
A.  Unfortunately, the connection between R and I_M is too weak
for the henselianity arguments, and we end up using a different
approach, detailed in the next section.

How does one obtain a good Bezout domain?  It is here that we diverge
most drastically from the dp-minimal case.  Say that a type-definable
group G is 00-connected if G = G^00.  The group I_M of
M-infinitesimals is 00-connected because it is the minimal
M-invariant heavy subgroup of (K,+).  In the dp-minimal case (rank
1), the 00-connected subgroups of (K,+) are totally ordered by
inclusion.  Once one has any non-trivial 00-connected proper
subgroup J < (K,+), the set {x ∈ K  |  x · J ⊆ J}
is trivially a valuation ring.

In higher rank, the lattice 𝒫 of 00-connected subgroups is
no longer totally ordered.  Nevertheless, dp-rank bounds the
complexity of the lattice.  Specifically, it induces a rank function
_dp(A/B) for A ≥ B ∈𝒫
probably[One must verify that the basic properties of
  dp-rank, including subadditivity, go through for hyperimaginaries.]
satisfying the following axioms:


  * _dp(A/B) is a nonnegative integer, equal to zero iff A =
  B.

  * If A ≥ B ≥ C, then _dp(A/C) ≤_dp(A/B) +
  _dp(B/C).

  * If A ≥ A' ≥ B, then _dp(A/B) ≥_dp(A'/B).

  * If A ≥ B' ≥ B, then _dp(A/B) ≥_dp(A/B').

  * For any A, B
  
    _dp(A ∨ B/B)     = _dp(A/A ∧ B) 
    _dp(A ∨ B/A ∧ B)    = _dp(A/A ∧ B) + _dp(B/A ∧ B).


This rank constrains the lattice substantially.  For example, if n =
(K) there can be no “(n+1)-dimensional cubes” in
𝒫, i.e., no injective homomorphisms of unbounded lattices
from ([n]) to 𝒫.

Somehow, the rank needs to be leveraged to recover a valuation.  First we need some lattice-theoretic way to recover valuations.
Recall that in a lower-bounded modular lattice (P,∧,∨,),
there is a modular pregeometry on the set of atoms.  More generally,
if x is an element in an unbounded modular lattice
(P,∧,∨), there is a modular pregeometry on the set of
“relative atoms” over x, i.e., elements y ≥ x such that the
closed interval [x,y] has size two.  If
(K,𝒪,𝔪) is a model of ACVF and P_n is the
lattice of definable additive subgroups of K^n, then the pregeometry
of relative atoms over 𝔪^n ∈ P_n is projective
(n-1)-space over the residue field k, because the relative atoms
all lie in the interval [𝔪^n,𝒪^n] whose quotient
𝒪^n/𝔪^n is isomorphic to k^n.  Furthermore,
the specialization map on Grassmannians

    (ℓ, K^n) →(ℓ, k^n)

is essentially the map sending V to the set of relative atoms below
V + 𝔪^n.

For dp-finite fields, we do something formally similar: we consider
the lattice 𝒫_n of 00-connected type-definable subgroups
of K^n, we consider the pregeometry 𝒢_n of relative
atoms over (I_M)^n, and we extract a map f_n from the lattice of
K-linear subspaces of K^n to the lattice of closed sets in
𝒢_n.  Morally, this leads to a system of maps

    f_n, ℓ : (ℓ, K^n) →(r ·ℓ, k^n)

where r = (K) and k is some skew field.  The maps f_n,
  ℓ satisfy some compatibility across n and ℓ.  Call this
sort of configuration an r-fold specialization.  The hope was
that r-fold specializations would be classified by valuations (or
good Bezout domains).  Unfortunately, this turns out to be far from
true.  All we can say is that there is a way of “mutating” r-fold
specializations such that, in the limit, they give rise to Bezout
domains.  See Remark <ref> and Section <ref>
for further details.[Luckily, we can avoid the machinery of
  r-fold specializations in the present paper.  We defer a proper
  treatment of r-fold specializations until a later paper,
  <cit.>.]

The above picture is complicated by three technical issues.  First and
most glaringly, the modular lattices we are considering don't really
have enough atoms.  For example, there are usually no minimal non-zero
type-definable subgroups of K.  So instead of using atoms, we use
equivalence classes of quasi-atoms.  In a lower-bounded modular
lattice (P,∧,∨,), say that an element x > is a
quasi-atom if the interval (,x] is a sublattice, and two
quasi-atoms x, y are equivalent if x ∧ y >.
There is always a modular geometry on equivalence classes of
quasi-atoms.  In the presence of a subadditive rank such as
_dp(-/-), there are always “enough” quasi-atoms, and the
modular geometry has bounded rank.  We verify these facts in
<ref>, in case they are not yet known.

Second of all, equation (<ref>) says that an
ℓ-dimensional subspace V ⊆ K^n must map to a closed set
in 𝒢_n of rank exactly r ·ℓ, where r =
(K).  This only works if the pregeometry of relative atoms over
I_M has the same rank as the dp-rank of the field.  This turns out
to be false in general, necessitating two modifications:


  * Rather than using dp-rank, we use the minimum subadditive rank
  on the lattice.  We call this rank “reduced rank,” and verify its
  properties in <ref>.  In equation
  (<ref>), r should be the reduced rank rather than the
  dp-rank.

  * Rather than using I_M, we use some other group J which
  maximizes the rank of the associated geometry.  We call such J
  special, and consider their properties in
  <ref>.

With these changes, equation (<ref>) can be recovered; see
Lemma <ref>.

Lastly, there is an issue with the lattice meet operation.  Note that
the lattice operations on the lattice of 00-connected type-definable
subgroups are

    A ∨ B     := A + B 
    
      A ∧ B     := (A ∩ B)^00.

The 00 in the definition of A ∧ B is a major
annoyance[Specifically, it would derail the proof of
  Proposition <ref>.<ref>, among other
  things.]; it would be nicer if A ∧ B were simply A ∩ B.
The following trick clears up this headache:

  There is a small submodel M_0 ≼ K such that J = J^00 for
  every type-definable M_0-linear subspace J ≤ K.  Consequently,
  if 𝒫' ⊆𝒫 is the sublattice of
  00-connected M_0-linear subspaces of K, then the lattice
  operations on 𝒫' are given by
  
    A ∨ B     = A + B 
    
        A ∧ B     = A ∩ B.


This is a corollary of the following uniform bounding principle for
dp-finite abelian groups:

  If H is a type-definable subgroup of a dp-finite abelian group
  G, then |H/H^00| is bounded by a cardinal κ(G)
  depending only on G.

We prove both facts in <ref>.



 §.§ Henselianity

There is comparatively little to say about henselianity.  In the
dp-minimal case, henselianity of definable valuations was first proven
by Jahnke, Simon, and Walsberg <cit.>.  Independently around the same time,
the author proved henselianity using the machinery of infinitesimals.
The outline of the proof is


  * Assume non-henselianity

  * Pass to a finite extension and get multiple incomparable
  valuations.

  * Pass to a coarsening and get multiple independent valuations.

  * Consider the multiplicative homomorphism f(x) = x^2 in
  characteristic 2 or the additive homomorphism f(x) = x^2 -
  x.

  * Use strong approximation to produce an element x such that x
  ∉ I'_M but f(x) ∈ I'_M.

  * Argue that f(I_M') is strictly smaller than I_M',
  contradicting the minimality of the group of infinitesimals among
  M-definable infinite type-definable groups.

In the last two steps, I_M' denotes the multiplicative
infinitesimals 1 + I_M or the additive infinitesimals I_M as
appropriate.

The “strong approximation” step requires the infinitesimals to be an
intersection of valuation ideals.  As noted in the previous section,
this property is unknown for higher rank, so we are not able to prove
much.

But then a miracle occurs.  In the additive Artin-Schreier case, the
only real properties of I_M being used are that I_M is the
Jacobson radical of a good Bezout domain, and I_M = I_M^00.  If
𝒪_1 and 𝒪_2 are two incomparable valuation
rings on an NIP field, then R = 𝒪_1 ∩𝒪_2 is
a good Bezout domain whose Jacobson radical is easily seen to be
00-connected, provided that the residue fields are infinite.  This
yields an extremely short proof of the henselianity conjecture for NIP
fields in positive characteristic.  See <ref>,
specifically Lemma <ref>.

This argument does not directly apply to the A-invariant valuation
rings constructed using modular lattices.  Nevertheless, a variant can
be made to work leveraging the infinitesimals, utilizing the existence
of G^000 for type-definable G.  See <ref>
for details.  Putting everything together, we find that if K is a
sufficiently saturated dp-finite field of positive characteristic,
then either K has finite Morley rank, or K admits a non-trivial
henselian valuation.

This in turn yields a proof of the Shelah conjecture for dp-finite
positive characteristic fields, as well as the expected
classification.  See <ref>.



 §.§ Outline

In <ref> we prove the henselianity conjecture in
positive characteristic.  In <ref> we generalize
Pierre Simon's results <cit.> about definability of dp-rank
in dp-minimal theories eliminating ∃^∞.  In
<ref>, we apply this to define a notion of “heavy”
sets which take the place of infinite sets in the construction of the
group of infinitesimals in <ref>.  Section
<ref> verifies the additional property of heavy sets that
holds when the field is not of finite Morley rank.  Section
<ref> applies the infinitesimals to the problem of
proving henselianity of A-invariant valuation rings in positive
characteristic.  In <ref> we prove the technical fact
that |H/H^00| is uniformly bounded as H ranges over
type-definable subgroups of a dp-finite abelian group.  Section
<ref> is a collection of abstract facts about modular
lattices, probably already known to experts.  Specifically, we verify
that “independence” and “cubes” make sense
(<ref>-<ref>), that there is a minimum
subadditive rank (<ref>), and that there is a
modular pregeometry on quasi-atoms (<ref>).  Then, in
<ref>, we apply the abstract theory to construct a
valuation ring.  Finally, in <ref> we verify the Shelah
conjecture for positive characteristic dp-finite fields, and enumerate
the consequences.





§ HENSELIANITY IN POSITIVE CHARACTERISTIC
 
In this section, we prove that definable NIP valuations in positive
characteristic must be henselian.  We also consider the situation of
(/A)-invariant valuation rings.

  
  Let 𝒪_1 and 𝒪_2 be two incomparable
  valuation rings on a field K.  For i = 1, 2 let 𝔪_i
  be the maximal ideal of 𝒪_i.  Consider the sets
  
    R    = 𝒪_1 ∩𝒪_2 
    
        I_1    = 𝔪_1 ∩𝒪_2 
    
        I_2    = 𝒪_1 ∩𝔪_2 
    
        J    = 𝔪_1 ∩𝔪_2.

  Then R is a Bezout domain with exactly two maximal ideals I_1,
  I_2 and Jacobson radical J.  The quotient R/I_i is isomorphic
  to the residue field 𝒪_i/𝔪_i.  Moreover,
  
    (a + 𝔪_1) ∩ (b + 𝔪_2) ∅

  for any a ∈𝒪_1 and b ∈𝒪_2.

(This is mostly or entirely well-known.)

  Let _i : K^×→Γ_i denote the valuation associated
  to 𝒪_i.  Note that x|y in R if and only if
  _1(x) ≤_1(y) and _2(x) ≤_2(y).
   
    For any x, y ∈ R, the ideal (x,y) is generated by x or y
    or x - y.
  
  
    The following cases are exhaustive:
    
    
  * If _1(x) ≤_1(y) and _2(x) ≤_2(y)
      then (x,y) is generated by x.
    
  * If _1(x) ≥_1(y) and _2(x) ≥_2(y),
      then (x,y) is generated by y.
    
  * If _1(x) < _1(y) and _2(x) > _2(y), then
      
    _1(x - y ) =    _1(x) < _1(y) 
    _2(x - y) =    _2(y) < _2(x)

      so (x,y) is generated by x- y.
    
  * If _1(x) > _1(y) and _2(x) < _2(y), then
      similarly (x,y) is generated by x - y. 
  
  By the claim, R is a Bezout domain.
  
    
    There is an element u ∈𝔪_1 ∩ (1 +
    𝔪_2).
  
  
    By incomparability, we can find a ∈𝒪_1 ∖𝒪_2 and b ∈𝒪_2 ∖𝒪_1.
    Then _1(a) ≥ 0 > _1(b) and _2(b) ≥ 0 >
    _2(a), so a/b ∈𝔪_1 and b/a ∈𝔪_2.  Let u = a/(a + b).  Then
    
    u     = a/b/a/b + 1∈𝔪_1 
    
          1 - u     = b/a/b/a + 1∈𝔪_2

  
  If u is as in Claim <ref>, then u ∈ I_1 but u ∉
  I_2, so I_1  I_2.
   
    For any a ∈𝒪_1, (a + 𝔪_1) ∩𝒪_2 ∅.
  
  
    If a ∈𝔪_1 then 0 is in the intersection.  If a
    ∈𝒪_2 then a is in the intersection.  So we may
    assume a ∈𝒪_1^×∖𝒪_2.  Then
    for u as in Claim <ref>,
    
    (a^-1 + u)^-1∈ (a + 𝔪_1) ∩𝒪_2.

    Indeed, a^-1 and a^-1 + u have the same residue class
    modulo 𝔪_1, so their inverses have the same residue
    class.  And
    
    a ∉𝒪_2  a^-1∈𝔪_2
           a^-1 + u ∈ 1 + 𝔪_2  (a^-1 +
          u)^-1∈ 1 + 𝔪_2 ⊆𝒪_2.

  
  Claim <ref> says that the natural inclusion
  
    R/I_1 = (𝒪_1 ∩𝒪_2)/(𝔪_1 ∩𝒪_2) ↪𝒪_1/𝔪_1

  is onto, hence an isomorphism.  Therefore R/I_1 is a field and
  I_1 is a maximal ideal.  Similarly I_2 is a maximal ideal and
  R/I_2 ≅𝒪_2/𝔪_2.
  
    There cannot exist a third maximal ideal I_3.
  
  
    Otherwise use the Chinese remainder theorem to find x ∈ (1 +
    I_1) ∩ (1 + I_2) ∩ I_3 and y ∈ I_1 ∩ (1 + I_2) ∩
    (1 + I_3).  Then
    
    (x) ⊆    I_3 ⊉(x,y) 
    
          (y) ⊆    I_1 ⊉(x,y) 
    
          (x - y) ⊆    I_2 ⊉(x,y)

    contradicting Claim <ref>.
  
  Consider the composition
  
    R/(I_1 ∩ I_2) ↠ (R/I_1) × (R/I_2)
        ∼→ (𝒪_1/𝔪_1) ×
        (𝒪_2/𝔪_2).

  The first map is surjective by the Chinese remainder theorem (as
  I_1 and I_2 are distinct maximal ideals).  The second map was
  shown to be an isomorphism earlier.  Therefore the composition is
  surjective, which exactly means that for any a ∈𝒪_1
  and b ∈𝒪_2, the intersection (a + 𝔪_1)
  ∩ (b + 𝔪_2) is non-empty.




  
  Let K be a field and 𝒪_i be a valuation ring on K
  for i = 1,2.  If the structure (K,𝒪_1,𝒪_2)
  is NIP and K has characteristic p > 0, then 𝒪_1 and
  𝒪_2 are comparable.


  Otherwise, let 𝔪_1, 𝔪_2, R, I_1, I_2, J be
  as in Lemma <ref>.  The incomparability of the
  𝒪_i implies that K is infinite, hence Artin-Schreier
  closed (by the Kaplan-Scanlon-Wagner theorem <cit.>).  Consider the
  homomorphism
  
    f : (J,+) → (/p,+)

  defined as follows: given x ∈ J = 𝔪_1 ∩𝔪_2, take an Artin-Schreier root y ∈ K such that
  y^p - y = x.  Then y ∈ (a + 𝔪_1) ∩ (b +
  𝔪_2) for unique a, b ∈/p.  Define f(x) = a -
  b.  This is independent of the choice of the root y, and the
  homomorphism f is definable.  By Lemma <ref> there
  is some y such that y ∈ (1 + 𝔪_1) ∩𝔪_2; then f(y^p - y) = 1.  Therefore f is onto and
  f is a definable subgroup of J of index p.  Recall that
  J^00 exists in NIP theories.  It follows that J^00⊊
  J.  Now for any a ∈ R, we have
  
    a · J^00 = (a · J)^00⊆ J^00,

  and so J^00 is an ideal in R.  Choose b ∈ J ∖
  J^00 and let I = {x ∈ R | xb ∈ J^00}.  Then I is a
  proper ideal of R, so I ≤ I_i for i = 1 or i = 2.  The
  maps
  
    (R · b)/(R · b ∩ J^00)    ≅ R/I ↠ R/I_i ≅𝒪_i/𝔪_i 
    
        (R · b)/(R · b ∩ J^00)    ≅ (R · b +
        J^00)/J^00↪ J/J^00

  together show that |J/J^00| ≥ |𝒪_i/𝔪_i|.  But
  the fact that K is Artin-Schreier closed forces the residue field
  𝒪_i/𝔪_i to be Artin-Schreier closed, hence
  infinite, hence unbounded in elementary extensions.  This
  contradicts the definition of J^00.



  Let (K,𝒪) be a valued field and L/K be a finite normal
  extension.  Every extension of 𝒪 to L is definable
  (identifying L with K^d for d = [L : K]).


  This essentially follows from Beth implicit definability.  Naming parameters, we may assume that 𝒪 is 0-definable
  and L is 0-interpretable.  Recall that (L/K) acts
  transitively on the set of extensions (essentially because valued
  fields can be amalgamated, i.e., ACVF has quantifier elimination).
  So there are only finitely many extensions to L, and it suffices
  to show that at least one extension is definable.  For any formula
  ϕ(x;y), the condition “ϕ(L;a) is a valuation ring on L
  extending 𝒪” is expressed by a formula ψ(a), so we
  may replace the original (K,𝒪) with an elementary
  extension.  Fix some 𝒪_L on L extending K, and pass
  to an elementary extension if necessary to ensure that
  
  
  * M^+ := (L,K,𝒪_L,𝒪) is
    ℵ_0-saturated.
  
  * M^- := (L,K,𝒪) is ℵ_0-homogeneous and
    ℵ_0-saturated.
  
  Because there are only finitely many extensions of 𝒪 to
  𝒪_L we can find a_1, …, a_n, b_1, …, b_m ∈
  L such that 𝒪_L is the unique extension of
  𝒪 to L containing A = {a_1,…,a_n} and
  disjoint from B = {b_1, …, b_m}.  Now if σ∈(M^-/AB), then σ preserves 𝒪 setwise, and
  therefore moves 𝒪_L to some extension
  𝒪_L' of 𝒪.  But σ fixes A and B
  so 𝒪_L' must still contain A and be disjoint from B.
  Thus 𝒪_L' = 𝒪_L, so (M^-/AB) fixes
  𝒪_L setwise.  By ℵ_0-homogeneity, we see that
  
    _M^-(c/AB) = _M^-(c'/AB)  (c ∈𝒪_L
         c ∈𝒪_L')

  for c, c' ∈ L.  Then by considering the map of type spaces
  
    S_M^+^L(AB) → S_M^-^L(AB)

  the clopen set in S_M^+^L(AB) corresponding to 𝒪_L
  must be the preimage of some subset (necessarily clopen) in
  S_M^-^L(AB), implying that 𝒪_L is AB-definable in
  M^-.  Here we are using ℵ_0-saturation to ensure that every
  M^+ and M^- type is realized in L.



 
  Let (K,𝒪) be an NIP valued field of positive
  characteristic.  Then (K,𝒪) is henselian.


  Otherwise there is some finite normal extension L/K such that
  𝒪 has multiple extensions to L.  If 𝒪_1,
  𝒪_2 are two distinct extensions, then
  (L,𝒪_1,𝒪_2) is an NIP 2-valued field.  By
  Lemma <ref>, 𝒪_1 and 𝒪_2 must be
  comparable.  But (L/K) acts transitively on the set of
  extensions, and a finite group cannot act transitively on a poset
  unless all elements are incomparable.




  The same proof should work when 𝒪 is ∨-definable,
  or equivalently, 𝔪 is type-definable.

We will also need a variant of the above results for invariant
valuation rings.  Here and in what follows, A-invariant means
(/A)-invariant, and we will often use “invariant” to mean
“A-invariant for some small A ⊆”.


  
  A valuation ring 𝒪⊆ K is spanned by a
  set S if
  
  
  * Every element of S has positive valuation.
  
  * For every x ∈ K of positive valuation, there is y ∈ S
    such that (y) ≤(x).
  
  In other words, (S) is downwards-cofinal in the interval
  (0,+∞) ⊆Γ.  Equivalently, S is a set of
  generators for the maximal ideal of 𝒪.


Recall that G^000 exists for type-definable abelian G in NIP
theories, by a theorem of Shelah (<cit.>, Theorem 1.12).

  
  Let (K,…) be a monster NIP field of positive characteristic,
  and let 𝒪_1, 𝒪_2 be two invariant valuation
  rings.  Suppose that 𝒪_1 and 𝒪_2 are both
  spanned by some type-definable subgroup (I,+) ≤ (K,+) with I =
  I^000.  Then 𝒪_1 and 𝒪_2 are comparable.


  Assume not.  Take a small set A such that 𝒪_1,
  𝒪_2 are A-invariant and I is type-definable over A.
  Let 𝔪_1, 𝔪_2, R, I_1, I_2, J be as in
  Lemma <ref>.  Define f : (J,+) → (/p,+) as in
  the proof of Lemma <ref>, and let J' be the kernel of
  f.  As in the proof of Lemma <ref> f is a surjective
  homomorphism, and so J' is an index p subgroup of J.  Moreover,
  J' is A-invariant (though not necessarily definable).  We claim
  that R · I = J.  Indeed, if x ∈ J, then _1(x) > 0
  and _2(x) > 0, so we may find y, z ∈ I such that
  _1(y) ≤_1(x) and _2(z) ≤_2(x).  As R is
  a Bezout domain, there is w ∈ R such that (w) = (y,z).  Then
  
    w|y    _1(w) ≤_1(y) _1(w) ≤_1(x) 
    
        w|z    _2(w) ≤_2(z) _2(w) ≤_2(x)

  so w|x.  This in turn implies that (x) ⊆ (w) = (y,z)
  ⊆ R · I.  Thus J ⊆ R · I, and the
  converse holds because I ⊆𝔪_i for i = 1, 2.

  For any r ∈ R, the group homomorphism
  
    I x ↦ r · x⟶ r · I
        ↪ J ↠ J/J' ≅/p

  is rA-invariant.  The kernel of this map is rA-invariant of
  finite index in I, so the kernel must be all of I because I =
  I^000.  Therefore r · I ⊆ J' for any r ∈ R.  It
  follows that R · I ⊆ J', contradicting the fact that
  R · I = J ⊋ J'.



  
  Let (K,…) be a monster NIP field of positive characteristic,
  and 𝒪 be an invariant valuation ring, spanned by some
  type-definable group I ≤ (K,+) with I = I^000.  Then
  𝒪 is henselian.


  Let Γ be the value group of 𝒪.  Then Γ is
  p-divisible (because K is Artin-Schreier closed or finite).  For
  any positive γ∈Γ and positive integer n, we can
  find ε∈ I such that
  
    (ε) ≤γ/p^n < γ/n,

  because γ/p^n ∈Γ_> 0.  If henselianity fails, there
  is a finite normal extension L/K such that 𝒪 has at
  least two (and at most [L : K]) extensions 𝒪_1 and
  𝒪_2 to L.  Let A be a set such that 𝒪
  and I are A-invariant and L is interpretable over A.  Any
  σ∈(K/A) fixes 𝒪 setwise, hence permutes
  the finitely many extensions of 𝒪 to L.  After adding
  finitely many parameters to A, we may assume that (K/A)
  fixes 𝒪_1 and 𝒪_2.  Thus 𝒪_1 and
  𝒪_2 are A-invariant.  If Γ, Γ_1, Γ_2
  are the value groups of 𝒪, 𝒪_1, 𝒪_2
  respectively, then |Γ_i/Γ| ≤ [L : K] < ℵ_0 for i
  = 1, 2.  Thus Γ_i ⊗ = Γ⊗.  By
  (<ref>) it follows that 𝒪_1 and
  𝒪_2 are both spanned by I.  By
  Lemma <ref> the valuation rings 𝒪_1
  and 𝒪_2 must be comparable, which is absurd.




§ BROAD AND NARROW SETS
 
Almost everything in this section is a straightforward generalization
of the second half of <cit.>.


 §.§ The general setting

Let T be a theory, not assumed to eliminate imaginaries.  Work in a
monster model .

  Let X_1, …, X_n be definable sets and Y ⊆ X_1 ×⋯× X_n be type-definable.  Then Y is broad if
  there exist a_i,j∈ X_i for 1 ≤ i ≤ n and j ∈
  such that the following conditions hold:
  
  
  * For fixed i, the a_i,j are pairwise distinct.
  
  * For any function η : [n] →,
    
    (a_1,η(1),…,a_n,η(n)) ∈ Y.

  
  Otherwise, we say that Y is narrow.

 
  If n = 1, then a type-definable set Y ⊆ X_1 is broad if
  and only if it is infinite.

 
  By compactness and saturation, Y is broad if and only if the
  following holds: for every m ∈ there exist a_i,j∈ X_i
  for 1 ≤ i ≤ n and 1 ≤ j ≤ m such that
  
  
  * For fixed i, the a_i,j are pairwise distinct.
  
  * For any function η : [n] → [m],
    
    (a_1,η(1),…,a_n,η(n)) ∈ Y.

  
  An equivalent condition is that for every m ∈ there exist
  subsets S_1, …, S_n with S_i ⊆ X_i, |S_i| = m,
  and S_1 ×⋯× S_n ⊆ Y.


  
  Fix a product of definable sets ∏_i = 1^n X_i.
  
  
  * A partial type Σ(x⃗) on ∏_i = 1^n X_i is
    broad if and only if every finite subtype is broad. 
  
  * If {D_b}_b ∈ Y is a definable family of definable
    subsets of ∏_i = 1^n X_i, then the set of b such that
    D_b is broad is type-definable. 
  
  * Let A be some small set of parameters and D be a definable
    set.  The set of tuples (a_1, …, a_n, b) ∈ X_1 ×⋯× X_n × D such that (a⃗/bA) is broad is
    type-definable. 
  


  
  
  * This follows immediately by compactness.
  
  * The statement that D_b is broad is equivalent to the small
    conjunction
    
    ⋀_m ∈∃ a_1,1, …, a_n,m(
          ⋀_i ∈ [n], 1 ≤ j < j' ≤ m a_i,j a_i,j'∧⋀_η : [n] → [m]
          (a_1,η(1),…,a_n,η(m)) ∈ D_b ).

  
  * The statement that (a⃗/bA) is broad is equivalent to
    the type-definable condition
    
    ⋀_m ∈∃ a_1,1, …, a_n,m(    ⋀_i ∈ [n], 1 ≤ j < j' ≤ m a_i,j a_i,j'
       ∧⋀_η : [n] → [m]
          (a_1,η(1),…,a_n,η(m)) ≡_bA
          (a_1,…,a_n) ).

    Note that a⃗≡_Ac⃗b⃗ is a type-definable
    condition on (a⃗,b⃗,c⃗) for a fixed small set
    A, and that in a monster model the type-definable conditions are
    closed under quantification and small conjunctions. 


  Let A be a small set of parameters.  Suppose X_1, …, X_n
  are A-definable sets and Y ⊆ X_1 ×⋯×
  X_n is type-definable over A.  Then Y is broad if and only if
  there exists a mutually A-indiscernible array ⟨ a_ij⟩_i ∈ [n], j ∈ such that for fixed i the a_ij
  are pairwise distinct elements of X_i, and such that for any η
  : [n] →, the tuple
  (a_1,η(1),a_2,η(2),…,a_n,η(n)) is an element
  of Y.  In other words, the witnesses of broadness can be chosen to
  be mutually A-indiscernible.


  This follows from the fact that we can extract mutually
  indiscernible arrays.

 
  Let X_1, …, X_n be infinite definable sets, and let Y, Z be
  type-definable subsets.
  
  
  * If Y ⊆ Z and Y is broad, then Z is broad.
  
  *  If Y ∪ Z is broad, then Y is broad
    or Z is broad.
  
  Equivalently, narrow sets form an ideal:
  
  
  * If Y ⊆ Z and Z is narrow, then Y is narrow.
  
  * If Y and Z are narrow, then Y ∪ Z is narrow.
  


  If Y is broad then Z is broad because the witnesses of broadness of
  Y show broadness of Z.  If Y ∪ Z is broad, we can take a small
  set A over which the X_i, Y, Z are defined and then find a
  mutually A-indiscernible array ⟨ a_i,j⟩_i ∈
    [n], j ∈ such that the ith row is a sequence of distinct
  elements, and such that for every η : [n] →,
  
    (a_1,η(1),…,a_n,η(n)) ∈ Y ∪ Z.

  Choose some fixed η_0 : [n] →.  Without loss of
  generality (interchanging Y and Z),
  
    (a_1,η_0(1),…,a_n,η_0(n)) ∈ Y.

  By mutual A-indiscernibility, it follows that
  
    (a_1,η(1),…,a_n,η(n)) ∈ Y

  for all η, and so Y is broad.

 
  Let A be a small set of parameters, let X_1,…,X_n be
  A-definable, and let Y ⊆ X_1 ×⋯× X_n be
  type-definable over A.  Then Y is broad if and only if
  (a⃗/A) is broad for some a⃗∈ Y.


  If (a⃗/A) is broad then Y is broad as it contains (the set
  of realizations of) (a⃗/A).  Conversely, suppose Y is
  broad.  Choose a mutually A-indiscernible array {α_i,j}_i
    ∈ [n], j ∈ witnessing broadness.  Let a⃗ =
  (α_1,1,α_2,1,…,α_n,1).  By mutual
  indiscernibility, for any η : [n] → the n-tuple
  (α_1,η(1),…,α_n,η(n)) realizes
  (a⃗/A), so (a⃗/A) is broad.

The trick in the following proof is taken from Theorem 3.10 in
<cit.>.
 
  Assume NIP.  Let X_1, …, X_n be infinite definable sets, and
  let Y ⊆ X_1 ×⋯× X_n be definable,
  not just type-definable.  Assume Y is broad and n ≥ 2.
  
  
  *  There exists some b ∈ X_n such that the slice
    
    {(a_1,…,a_n-1) ∈ X_1 ×⋯× X_n-1 | 
          (a_1,…,a_n-1,b) ∈ Y}

    is broad as a subset of X_1 ×⋯× X_n-1.
  
  *  There exists a broad definable subset D_< n⊆ X_1 ×⋯× X_n-1 and an infinite
    definable subset D_n ⊆ X_n such that
    
    (D_< n× D_n) ∖ Y

    is a “hyperplane,” in the sense that for every b ∈ D_n, the
    definable set
    
    {(a_1,…,a_n-1) ∈ D_< n |  (a_1,…,a_n-1,b)
          ∉ Y}

    is narrow in X_1 ×⋯× X_n-1.
  


  Choose a small set A over which the X_i and Y are definable.
  Choose a mutually indiscernible array {a_i,j}_i ∈ [n], j ∈ witnessing broadness of Y.  Let b_j = a_n,j be the
  elements of the bottom row.  Note that the top n-1 rows
  {a_i,j}_i ∈ [n-1], j ∈ form a mutually indiscernbile
  array over Ab⃗.  In particular, if e⃗ denotes the
  first column
  
    e⃗ = (e_1,…,e_n-1) := (a_1,1,a_2,1,…,a_n-1,1),

  then (e⃗/Ab⃗) is broad (within X_1 ×⋯× X_n-1).  Note that the n-tuple e⃗b_j lies in Y
  for any j ∈.  In particular, e⃗ lies in the slice
  over b_1, so the slice over b_1 is broad, proving the first point.

  For the second point, consider sequences c_1, c_2, …, c_m
  satisfying the following constraints:
  
  
  * The sequence I = b_1 c_1 b_2 c_2 ⋯ b_m c_m b_m+1 b_m+2
    b_m+3⋯ is A-indiscernible.
  
  * The type (e⃗/AI) is broad.
  
  * The tuple e⃗c_j is not in Y for j = 1,
    …, m.
  
  There is at least one such sequence, namely, the empty sequence.
  Because of NIP, the first and third conditions imply some absolute
  bound on m, so we can find such a sequence with m maximal.  Fix
  such a sequence
  
    I = b_1 c_1 b_2 c_2 ⋯ b_m c_m b_m+1 b_m+2 b_m+3⋯

  and choose c_m+1 such that b_1 c_1 ⋯ b_m
  c_m b_m+1 c_m+1 b_m+2 b_m+3⋯ is A-indiscernible.
  
    If α⃗∈ X_1 ×⋯× X_n-1 and β∈ X_n satisfy
    
    α⃗≡_AIe⃗, β≡_AI c_m+1, and  
          (α⃗,β) ∉ Y

    then (α⃗/AIβ) is narrow.
  
  
    Applying an automorphism over AI, we may assume α⃗ =
    e⃗.  If (e⃗/AIβ) is broad, then
    
    
  * The sequence I' = b_1 c_1 b_2 c_2 ⋯ b_m c_m b_m+1β b_m+2 b_m+3⋯ is A-indiscernible, because
      β≡_AI c_m+1.
    
  * The type (e⃗/AI') is broad.
    
  * The tuple e⃗β is not in Y, and neither are
      e⃗c_j for j = 1,…,m.
    
    This contradicts the maximality of m.
  
  By Remark <ref>.<ref> and compactness, there
  must be formulas φ(x⃗) ∈(e⃗/AI) and ψ(y)
  ∈(c_m+1/AI) such that
  
    φ(α⃗) ∧ψ(β) ∧((α⃗,β) ∉ Y)

  implies narrowness of (α⃗/AIβ).  Let D_<n be
  the subset of X_1 ×⋯× X_n-1 cut out by
  φ(x⃗), and let D_n be the subset of X_n cut out by
  ψ(y).  Because the sequence
  
    b_1 c_1 b_2 c_2 ⋯ b_m c_m b_m+1 c_m+1 b_m+2 b_m+3 b_m+4⋯

  is A-indiscernible and non-constant (as the b_i are distinct),
  it follows that no term is in the algebraic closure of the other
  terms.  In particular
  
    c_m+1∉(Ab_1c_1 … b_mc_m b_m+1 b_m+2⋯) = (AI).

  Therefore D_n is infinite.  And D_<n is a broad subset of X_1
  ×⋯× X_n-1 because (e⃗/AI) is broad (by
  choice of c_1,…,c_m).  It remains to show that if β∈ D_n, then the set
  
    {α⃗∈ D_< n |  (α⃗,β) ∉ Y}

  is narrow as a subset of X_1 ×⋯× X_n-1.  Indeed,
  the set in question is AIβ definable, and if α⃗
  belongs to the set then (α⃗/AIβ) is narrow by
  choice of φ(x⃗) and ψ(y).  By
  Proposition <ref> the set in question is narrow.


 
  Assume NIP.  Let X_1, …, X_n be definable and Y be a
  definable subset of X_1 ×⋯× X_n.  Then Y is broad
  if and only if there exist infinite subsets D_i ⊆ X_i such
  that (D_1 ×⋯× D_n) ∖ Y is a
  “hyperplane,” in the sense that for every b ∈ D_n the set
  
    {(a_1,…,a_n-1) ∈ D_1 ×⋯× D_n-1 | 
        (a_1,…,a_n-1,b) ∉ Y}

  is narrow as a subset of X_1 ×⋯× X_n-1.


  For the “if” direction, first note that D_1 ×⋯×
  D_n is broad: for each i ∈ [n] we can choose a_i,1, a_i,2,
  … to be an arbitrary infinite sequence of distinct elements in
  D_i; the array {a_i,j}_i ∈ [n], j ∈ then witnesses
  broadness of D_1 ×⋯× D_n.  Consider the sets
  
    Y'    := (D_1 ×⋯× D_n) ∩ Y 
    
        H    := (D_1 ×⋯× D_n) ∖ Y.

  Then D_1 ×⋯× D_n is the (disjoint) union of Y'
  and H, so at least one of Y' and H must be broad.  If H were
  broad, by Lemma <ref>.<ref> it could not be a
  hyperplane.  Therefore H is narrow and Y' is broad, and so Y is
  broad, proving the “if” direction.

  We prove the “only if” direction by induction on n.  For the
  base case n = 1, we can take D_1 = Y by Remark <ref>.
  Next suppose n > 1.  By Lemma <ref>.<ref> there
  exist definable sets D_<n⊆ X_1 ×⋯×
  X_n-1 and D_n ⊆ X_n with D_< n broad, D_n infinite,
  and (D_<n× D_n) ∖ Y a hyperplane.  By induction
  there exist infinite definable sets D_i ⊆ X_i for 1 ≤ i
  < n such that the set H' = (D_1 ×⋯× D_n-1)
  ∖ D_<n is a hyperplane.  By
  Lemma <ref>.<ref> the set H' is narrow.  For any
  b ∈ D_n, the two sets
  
    {a⃗∈ D_1 ×⋯× D_n-1 | a⃗∉ D_< n} = H' 
       {a⃗∈ D_< n |  (a⃗,b) ∉ Y}

  are both narrow (the latter because (D_< n× D_n) ∖
  Y is a hyperplane).  The union of these two sets contains
  
    {a⃗∈ D_1 ×⋯× D_n-1 |  (a⃗,b) ∉
        Y}

  which must therefore be narrow.  Therefore (D_1 ×⋯×
  D_n) ∖ Y is a hyperplane, completing the proof.



  Assume that T is NIP and eliminates ∃^∞.  Then
  “broadness is definable in families” on the product X_1 ×⋯× X_n.  In other words, if {Y_b}_b ∈ Z is a
  definable family of definable subsets of X_1 ×⋯×
  X_n, then the set {b ∈ Z  |  Y_b  is broad} is
  definable.

In fact, we only need T to eliminate ∃^∞ on the sets
X_1,…,X_n.

  We proceed by induction on n.  The base case n = 1 is equivalent
  to elimination of ∃^∞ by Remark <ref>.  Suppose
  n > 1 and {Y_b}_b ∈ Z is a definable family of definable
  subsets of X_1 ×⋯× X_n.  Let A be a set of
  parameters over which everything is defined.  The set {b ∈ Z  | 
  Y_b  is broad} is type-definable by
  Remark <ref>.<ref>.  It
  remains to show that the set is also ∨-definable: if Y_b_0
  is broad, then there is some A-definable neighborhood N of b_0
  such that Y_b is broad for b ∈ N.  Indeed, by
  Theorem <ref> there exist formulas
  ϕ_i(x,z) and elements c_1,…,c_n such that
  
    ⋀_i = 1^n ∃^∞ x ∈ X_i : ϕ_i(x,c_i)

  and
  
    ∀ y ∈ X_n : ϕ_n(x,c_n) ∃^broad   
        (x_1,…,x_n-1) ∈ X_1 ×⋯× X_n-1 : 
       ((x_1,…,x_n-1,y) ∉ Y_b_0) ∧⋀_i = 1^n-1ϕ_i(x_i,c_i)

  where ∃^broadx⃗ : P(x⃗) means that the set of
  x⃗ such that P(x⃗) holds is broad.  (This quantifier is
  eliminated, by induction.)  We take N to be the A-definable set of
  b such that
  
    ∃ c_1, …, c_n :    ( ⋀_i = 1^n
        ∃^∞ x ∈ X_i : ϕ_i(x,c_i) )  
       ∧( ∀ y ∈ X_n : ϕ_n(x,c_n) ∃^broad (x_1,…,x_n-1) ∈ X_1 ×⋯×
        X_n-1 : 
       ((x_1,…,x_n-1,y) ∉ Y_b) ∧⋀_i =
          1^n-1ϕ_i(x_i,c_i)).

  If b ∈ N, then the sets ϕ_i(;c_i) show that Y_b is broad
  by Theorem <ref>.  This proves
  ∨-definability of the set of b such that Y_b is broad,
  completing the inductive step and the proof.


  
  Assume T is NIP and eliminates ∃^∞.  Let X_1,
  …, X_n be definable sets and {D_b}_b ∈ Y be a
  definable family of subsets of X_1 ×⋯× X_n.  Then
  there is some constant m depending on the family such that for any
  b ∈ Y, the set D_b is broad if and only if there exist
  {a_i,j}_i ∈ [n], j ∈ [m] such that
  
  
  * For fixed i, the a_i,j are pairwise distinct elements of
    X_i.
  
  * For any η : [n] → [m], the tuple
    (a_1,η(1),…,a_n,η(n)) belongs to D_b.
  
  Equivalently, there is an m such that for every b, D_b is
  broad if and only if there exist finite subsets S_i ⊆ X_i
  of cardinality m such that S_1 ×⋯× S_n ⊆
  D_b.


  This follows by compactness and Remark <ref>, once we know
  that “broad” is a definable condition.




 §.§ Externally definable sets
 
It is a theorem of Chernikov and Simon that in NIP theories
eliminating ∃^∞, any infinite externally definable set
contains an infinite internally definable set (Corollary 1.12 in
<cit.>).  As a consequence, if an infinite (internally)
definable set Y is covered by finitely many externally definable
sets D_1, …, D_ℓ, then one of the D_i contains an infinite
internally definable set.

The following lemma is an analogue for broad sets.

  Assume T is NIP and eliminates ∃^∞.  Let M ≼ be a small model.  Let X_1, …, X_n be M-definable
  infinite sets and Y ⊆ X_1 ×⋯× X_n be an
  M-definable broad set.  Let D_1, …, D_ℓ be
  -definable subsets of X_1 ×⋯× X_n such that
  
    Y(M) ⊆⋃_k = 1^ℓ D_k.

  Then there exists some k and some M-definable broad set Y'
  ⊆ Y such that Y'(M) ⊆ D_k.


  For every m ∈ we can find {a_i,j}_i ∈ [n], j ∈
    [m] satisfying the following conditions:
  
  
  * a_i,j∈ X_i
  
  * a_i,j a_i,j' for j  j'.
  
  * For any η : [n] → [m] the tuple
    (a_1,η(1),…,a_n,η(n)) lies in Y.
  
  Because X_i and Y are M-definable, we can even choose the
  a_i,j to lie in M.  By the Ramsey-theoretic statement
  underlying Proposition <ref>.<ref>, there is some
  fixed k ∈ [ℓ] such that for every m ∈ we can find
  {a_i,j}_i ∈ [n], j ∈ [m] satisfying the following
  conditions:
  
  
  * a_i,j∈ X_i(M)
  
  * a_i,j a_i,j' for j  j'.
  
  * For any η : [n] → [m] the tuple
    (a_1,η(1),…,a_n,η(n)) lies in Y ∩ D_k.
  
  By honest definitions (<cit.>, Remark 3.14), the externally
  definable set Y(M) ∩ D_k can be approximated by internally
  definable sets in the following sense: there is an M-definable
  family {F_b} such that for every finite subset S ⊆ Y(M)
  ∩ D_k there is a b ∈(M) such that
  
    S ⊆ F_b(M) ⊆ Y(M) ∩ D_k.

  Take m as in Corollary <ref> for the family {F_b}.
  Take a_i,j for i ∈ [n] and j ∈ [m] such that
  
  
  * a_i,j∈ X_i(M)
  
  * a_i,j a_i,j' for j  j'.
  
  * For any η : [n] → [m] the tuple
    (a_1,η(1),…,a_n,η(n)) lies in Y ∩ D_k (hence
    in Y(M) ∩ D_k).
  
  Let S be the finite subset of tuples of the form
  (a_1,η(1),…,a_n,η(n)).  Take b ∈(M) such
  that S ⊆ F_b(M) ⊆ Y(M) ∩ D_k.  The a_i,j
  show that F_b is broad, by choice of m.  Let Y' = F_b.  The
  fact that Y'(M) ⊆ Y(M) implies that Y' ⊆ Y, as
  Y and Y' are both M-definable and M ≼.



  
  Let M ≼ be a small model and X_1, …, X_n be infinite
  M-definable sets.  Suppose that Y ⊆ X_1 ×⋯× X_n is broad and M-definable, and that W ⊆ X_1
  ×⋯× X_n is -definable.  If Y(M) ⊆
  W, then W is broad.


  For every m, we can find {a_i,j}_i ∈ [n], j ∈ [m]
  satisfying the following conditions:
  
  
  * a_i,j∈ X_i.
  
  * a_i,j a_i,j' for j  j'.
  
  * For any η : [n] → [m] the tuple
    (a_1,η(1),…,a_n,η(n)) lies in Y.
  
  The requirements on the a_i,j are M-definable, and [n] ×
  [m] is finite.  Therefore, for any m we can choose the a_i,j
  to lie in M.  Having done so,
  
    (a_1,η(1),…,a_n,η(n)) ∈ Y(M) ⊆ W

  for any η.  The existence of the a_i,j for all m imply
  that W is broad.




 §.§ The finite rank setting

We now turn to proving Theorem <ref>, which relates
broadness and narrowness to dp-rank, under certain assumptions.

  
  Let X_1, …, X_n be A-definable sets, and let
  (a_1,…,a_n) be a tuple in X_1 ×⋯× X_n.
  Suppose that there exists a sequence b_1, b_2, …∈ X_n of
  pairwise distinct elements such that
  
    (a_1,…,a_n-1,b_i) ≡_A (a_1,…,a_n-1,a_n)

  for every i, and such that (a_1,…,a_n-1/Ab⃗) is
  broad.  Then (a_1,…,a_n/A) is broad.


  Let {e_i,j}_i ∈ [n-1], j ∈ witness broadness of
  (a_1,…,a_n-1/Ab⃗).  For any η : [n-1] →,
  
    (e_1,η(1),…,e_n-1,η(n-1)) ≡_A b⃗ (a_1,…,a_n-1).

  Thus, for any η : [n-1] → and any j,
  
    (e_1,η(1),…,e_n-1,η(n-1),b_j)    ≡_A
        (a_1,…,a_n-1,b_j) 
       ≡_A (a_1,…,a_n-1,a_n).

  If we set e_n,j := b_j, then the e_i,j show that
  (a⃗/A) is broad.


 
  Let X_1, X_2, …, X_n be A-definable sets of finite dp-ranks
  r_1, …, r_n.  Let (a_1,a_2,…,a_n) be a tuple in X_1
  ×⋯× X_n.  Then
  
    (a_1,…,a_n/A) ≤ r_1 + ⋯ + r_n

  by subadditivity of dp-rank, and if equality holds then
  
    (a_i/Aa_1a_2 ⋯ a_i-1 a_i+1⋯ a_n) = r_i

  for every i.  Indeed, otherwise
  
    (a_i/Aa_1a_2 ⋯ a_i-1 a_i+1⋯ a_n) < r_i

  and so
 
    r_1 + ⋯ + r_n    ≤(a_i/Aa_1⋯ a_i-1a_i+1⋯
       a_n) + (a_1,…,a_i-1, a_i+1, …, a_n/A) 
        < r_i + (r_1
       + ⋯ + r_i-1 + r_i+1 + ⋯ + r_n)

  which is absurd.


The technique for the next proof comes from Proposition 3.4 in
<cit.>.

  
  Let X, Y be infinite A-definable sets of finite dp-rank n and
  m, respectively.  Let (a,b) be a tuple in X × Y with
  (a,b/A) = n + m.  Then there exist pairwise distinct b_1,
  b_2, … such that ab_i ≡_A ab and such that (a/Ab_1
  …, b_ℓ) = n for every ℓ.


  Take an ict-pattern of depth n + m in (ab/A), extract a
  mutually A-indiscernible ict-pattern, and extend the pattern to
  have columns indexed by × (with lexicographic order).
  This yields formula ϕ_i(x,y;z) for 1 ≤ i ≤ n + m and
  elements c_i,j,k for 1 ≤ i ≤ n + m and j, k ∈ such
  that c_i;j,k is a mutually A-indiscernible array, and such
  that
  
    ⋀_i = 1^n + m( ϕ_i(x,y;c_i;0,0) ∧⋀_(j,k)  (0,0)ϕ_i(x,y;c_i;j,k) )

  is consistent with (a,b/A).  Moving the c's by an
  automorphism over A, we may assume that (a,b) realizes this
  type, so that ϕ_i(a,b;c_i;j,k) holds if and only if (j,k) =
  (0,0).  By the proof of subadditivity of dp-rank, there exist m
  rows which form a mutually Aa-indiscernible array.  Without loss
  of generality, these are the rows i = n + 1, …, n + m.  For
  j ∈ let d_j be an enumeration of { c_i;j,k}_n < i
    ≤ n+m,  k ∈.  The d_j form an indiscernible sequence
  over Aa, so we may choose b_j such that
  
    b_jd_j ≡_Aa bd_0

  and such that b_0 = b.  In particular, ab_j ≡_A ab for any
  j.
  
    For any ℓ > 0 the tuple (a,b_0,b_1,…,b_ℓ - 1) has
    dp-rank n + ℓ m over A.
  
  
    The dp-rank is at most n + ℓ m by subadditivity of dp-rank
    (as b_j ≡_A b  b_j ∈ Y).  It thus suffices to
    exhibit (n + ℓ m)-many mutually A-indiscernible sequences,
    none of which are indiscernible over A a b_0 b_1 … b_ℓ -
      1:
    
    
  * For 1 ≤ i ≤ n, the sequence
      
    c_i,0,0, c_i,0,1, c_i,0,2, …,

      which fails to be indiscernible over ab_0 = ab on account of
      the fact that
      
    ϕ_i(a,b,c_i,0,k)  k = 0.

    
  * For n < i ≤ n + m and for 0 ≤ j < ℓ, the sequence
      
    c_i,j,0, c_i,j,1, c_i,j,2, …

      which fails to be indiscernible over ab_j on account of the
      fact that
      
    ϕ_i(a,b_j,c_i,j,k) ϕ_i(a,b,c_i,0,k)  k = 0.

    
    These sequences are indeed mutually A-indiscernible, because we
    can split the mutually indiscernible array c_i;j,k into a
    mutually indiscernible array c_i,j;k.
  
  Now by Remark <ref>, it follows that
  
    (a/Ab_0 b_1 ⋯ b_ℓ - 1) = n

  for each ℓ.  Moreover,
  
    (b_ℓ/Aa b_0 b_1 ⋯ b_ℓ - 1) = m > 0,

  implying that b_ℓ∉(A a b_0 ⋯ b_ℓ - 1).  In
  particular, b_ℓ b_j for j < ℓ, and the b_j are
  pairwise distinct.



 
  For i = 1, …, n let X_i be a definable set of finite
  dp-rank r_i > 0.  Let Y ⊆ X_1 ×⋯× X_n
  be type-definable.  If (Y) = r_1 + ⋯ + r_n then Y is
  broad.


  We proceed by induction on n.  Assume (Y) = r_1 + ⋯ +
  r_n.  For the base case n = 1, we see that (Y) = r_1 > 0,
  so Y is infinite.  By Remark <ref>, Y is broad.  Suppose n
  > 1.  Let A be a small set of parameters such that the X_i are
  definable over A and Y is type-definable over A.  Take
  (a_1,…,a_n) ∈ Y such that (a_1,…,a_n/A) = r_1 +
  ⋯ + r_n.  By Lemma <ref> applied to
  (a_1,…,a_n-1;a_n), there exist pairwise distinct b_1, b_2,
  … such that
  
    (a_1,…,a_n-1,a_n) ≡_A (a_1,…,a_n-1,b_j)

  for every j and such that
  
    (a_1,…,a_n-1/Ab_1… b_j) = r_1 + ⋯ + r_n-1

  for every j.  By induction, (a_1,…,a_n-1/Ab_1 …
  b_j) is broad for every j.  By
  Remark <ref>.<ref>, it follows that
  (a_1,…,a_n-1/Ab_1 b_2 ⋯) is broad.  By
  Lemma <ref>, it follows that (a_1,…,a_n/A)
  is broad, and so Y is broad.



  A definable set D is quasi-minimal if D has finite
  dp-rank n > 0, and every definable subset D' ⊆ D has
  dp-rank 0 or n.  Equivalently, every infinite definable subset
  of D has the same (finite) dp-rank as D.


  If D is a definable set of dp-rank 1, then D is quasi-minimal.

 
  If D is an infinite definable set of finite dp-rank, then D contains a
  quasi-minimal definable subset.






 
  Assume NIP.  Let X_1, …, X_n be quasi-minimal definable sets
  of rank r_1, …, r_n, and let Y ⊆ X_1 ×⋯× X_n be definable.  Then Y is broad if and only if (Y) =
  r_1 + ⋯ + r_n.


  Note that the r_i > 0 by definition of quasi-minimal.  The “if”
  direction then follows by Proposition <ref>.  We prove
  the “only if” direction by induction on n.  For n = 1, Y is
  broad if and only if Y is infinite (by Remark <ref>) if and
  only if (Y) = r_1 (by definition of quasi-minimal).  Assume n
  > 1.  Let Y be broad.  By Theorem <ref>,
  there are infinite definable subsets D_1 ×⋯× D_n
  such that for every b ∈ D_n, the set
  
    H_b := {(a_1,…,a_n-1) ∈ D_1 ×⋯× D_n-1 | 
        (a_1,…,a_n-1,b) ∉ Y}

  is narrow (as a subset of X_1 ×⋯× X_n-1).  Note
  that (D_i) = r_i by quasi-minimality.  Let
  
    H    = (D_1 ×⋯× D_n) ∖ Y = ∐_b ∈ D_n H_b 
    
        Y'    = (D_1 ×⋯× D_n) ∩ Y.

  By induction, (H_b) < r_1 + ⋯ + r_n-1 for every b ∈
  D_n.  By subadditivity of the dp-rank, it follows that
  
    (H) < r_1 + ⋯ + r_n-1 + (D_n) = r_1 + ⋯ + r_n.

  Now D_1 ×⋯× D_n = H ∪ Y', so
  
    r_1 + ⋯ + r_n = (H ∪ Y') = max((H),(Y')).

  Given the low rank of H, this forces Y' to have rank r_1 +
  ⋯ + r_n.  Then
  
    r_1 + ⋯ + r_n = (Y') ≤(Y) ≤(X_1 ×⋯× X_n) = r_1 + ⋯ + r_n

  completing the inductive step.

Combining Theorem <ref> with Corollary <ref>
yields the following:
 
  Assume T is NIP and eliminates ∃^∞, and let X_1,
  …, X_n be quasi-minimal sets of finite dp-rank.  Let r =
  (X_1 ×⋯× X_n).  Given a definable family
  {D_b}_b ∈ Y of subsets of X_1 ×⋯× X_n,
  the set of b such that (D_b) = r is definable.  In fact,
  there is some m depending on the family such that (D_b) = r
  if and only if there exist finite subsets S_i ⊆ X_i of
  cardinality m such that S_1 ×⋯× S_n ⊆
  D_b.




§ HEAVY AND LIGHT SETS
 
In this section, we assume that  is a dp-finite field, possibly
with additional structure.  The goal is to define a notion of
“heavy” and “light” sets, and prove Theorem <ref>,
which verifies that heaviness satisfies the properties needed for the
construction of infinitesimals.

The heavy sets turn out to be exactly the definable sets of full rank,
but this will not be proven until a later paper (<cit.>,
Theorem 5.9.2).

By the assumption of finite dp-rank, T is NIP.  Moreover, T
eliminates ∃^∞ by a theorem of Dolich and Goodrick
(<cit.>, Corollary 2.2).  Therefore, the
results of the previous section apply.



 §.§ Coordinate configurations


  A coordinate configuration is a tuple (X_1,…,X_n,P)
  where X_i ⊆ are quasi-minimal definable sets, where P
  ⊆ X_1 ×⋯× X_n is a broad definable set,
  and where the map
  
    P    →
    
        (x_1,…,x_n)    ↦ x_1 + ⋯ + x_n

  has finite fibers.  The target of the coordinate
  configuration is the image of this map.  The rank of the
  coordinate configuration is the sum ∑_i = 1^n (X_i).


  If (X_1,…,X_n,P) is a coordinate configuration with target
  Y and rank r, then
  
    (Y) = (P) = (X_1 ×⋯× X_n) = r

  by Theorem <ref> and the subadditivity of dp-rank.



  
  Let (X_1,…,X_n,P) be a coordinate configuration with target
  Y.
  
  
  *  Full dp-rank is definable on Y: if {D_b} is
    a definable family of subsets of Y, the set of b such that
    (D_b) = (Y) is definable. 
  
  * If Y' is a definable subset of Y and (Y') = (Y),
    then Y' is the target of some coordinate configuration.
  


  Let r be the rank of the coordinate configuration, or
  equivalently, r = (Y).  Let π : P → Y be the map
  π(x_1,…,x_n) = ∑_i = 1^n x_i.  Because π is
  surjective with finite fibers, a subset D_b ⊆ Y has rank
  r if and only if π^-1(D_b) has rank r.  Therefore, the
  definability of rank r on Y follows from the definability of
  rank r on X_1 ×⋯× X_n, which is
  Corollary <ref>.  If Y' is a full-rank subset of
  Y, then π^-1(Y) is a broad subset of X_1 ×⋯× X_n, and so (X_1,…,X_n,π^-1(Y)) is a coordinate
  configuration with target Y'.

 
  Let X_1, …, X_n be quasi-minimal and A-definable.  Let
  (a_1,…,a_n) be an element of X_1 ×⋯× X_n
  such that (a_1,…,a_n/A) is broad.  Let s = a_1 + ⋯ +
  a_n.  If a⃗∈(sA), then there is a broad set P
  ⊆ X_1 ×⋯× X_n containing a⃗ such
  that (X_1,X_2,…,X_n,P) is a coordinate configuration with
  target containing s.


  We can find some formula ϕ(x_1,…,x_n,y) (with suppressed
  parameters from A) such that ϕ(a_1,…,a_n,s) holds, and
  ϕ(,s) is finite of size k.  Strengthening ϕ, we may
  assume that ϕ(,s') has size at most k for any s'.  We
  may also assume that ϕ(x_1,…,x_n,y)  y = x_1 +
  ⋯ + x_n.  Let P be the definable set
  
    P = {(x_1,…,x_n) ∈ X_1 ×⋯× X_n  | ϕ(x_1,…,x_n,x_1 + ⋯ + x_n)}.

  Then P is A-definable, and broad by virtue of containing
  a⃗.  The map
  
    P    →
    
        (x_1,…,x_n)    ↦ x_1 + ⋯ + x_n

  has fibers of size at most k.  Therefore (X_1,…,X_n,P) is a
  coordinate configuration.



 
  Let X_1,…,X_n be quasi-minimal A-definable sets.  For each
  i let p_i be a global A-invariant type in X_i.  Assume p_i
  is not realized (i.e., not an algebraic type).  Then (p_1 ⊗⋯⊗ p_n) | A is broad.


  Because p_i is not algebraic, any realization of p_i^⊗ω is an indiscernible sequence of pairwise distinct
  elements.  Let {a_i,j}_i ∈ [n], j ∈ be a realization
  of p_1^⊗ω⊗⋯⊗ p_n^⊗ω over A.  Then for every η : [n] →, the tuple
  
    (a_1,η(1),…,a_n,η(n))

  is a realization of p_1 ⊗⋯⊗ p_n over A, and
  the rows in this array are sequences of distinct elements.


  
  Let (X_1,…,X_n,P) be a coordinate configuration.  There
  exists a small set A and non-algebraic global A-invariant types
  p_i on X_i such that if (a_1,…,a_n)  p_1 ⊗⋯⊗ p_n|A then (a_1,…,a_n) ∈ P.


  By Lemma <ref>.<ref> and
  Theorem <ref>, we can find infinite definable
  sets D_1, D_2, …, D_n such that (D_1 ×⋯×
  D_n) ∖ P is a hyperplane, hence narrow.  Choose a model M
  defining the X_i's, D_i's, and P, and let p_i be any
  M-invariant non-algebraic global type in D_i, such as a coheir
  of a non-algebraic type in D_i over M.  If (a_1,…,a_n)
  ⊗_i = 1^n p_i|M, then (a_1,…,a_n/M) is
  broad, hence cannot live in the hyperplane.  But a_i ∈ D_i, so
  a⃗∈ P.




 §.§ Critical coordinate configurations

In this section, we carry out something similar to the Zilber
indecomposability technique in groups of finite Morley rank.  We
consider ways to expand coordinate configurations to higher rank, and
deduce consequences of being a maximal rank coordinate configuration.

  The critical rank ρ is the maximum rank of any
  coordinate configuration.  A coordinate configuration is
  critical if its rank is the critical rank ρ.  A set Y
  is a critical set if Y is the target of some critical
  coordinate configuration.


  
  If Y is a critical set, then (Y) = ρ.  If Y' ⊆
  Y is definable of rank ρ, then Y' is critical.


  Indeed, Y' is the target of some coordinate configuration by
  Proposition <ref>.<ref>. This coordinate
  configuration has rank ρ and is thus critical.


  
  If Y is a critical set and α∈, the translate α
  + Y is critical.


  If (X_1,…,X_n,P) is a coordinate configuration with target
  Y, then (X_1 + α, X_2, …, X_n, P') is a coordinate
  configuration with target Y', where
  
    P' := {(a_1 + α, a_2, a_3, …, a_n)  |  (a_1,…,a_n) ∈ P}.

  Because (Y) = (Y + α), Y + α is critical.



  
  Let (X_1,…,X_n,P) be a critical coordinate configuration and
  Q be a quasi-minimal set.  Let A be a small set of parameters
  over which the X_i,P, and Q are defined.  Let
  (a_1,…,a_n,b) be a tuple in X_1 ×⋯× X_n
  × Q such that (a_1,…,a_n,b/A) is broad and
  (a_1,…,a_n) ∈ P.  Let s = a_1 + ⋯ + a_n + b.  Then
  b ∉(sA).


  Suppose otherwise.  Let s' = a_1 + ⋯ + a_n = s - b.  If b
  ∈(sA), then s' ∈(sA).  Because a⃗∈ P, we
  have a⃗∈(s'A) by definition of coordinate
  configuration.  It follows that (a⃗,b) ∈(sA).  By
  Lemma <ref>, there is a coordinate configuration
  (X_1,…,X_n,Q,P').  This coordinate configuration has higher
  rank than (X_1,…,X_n,P), contradicting criticality.



  
  Let A be a small set.  Suppose b ∉(A), and suppose a
   p|Ab for some global A-invariant type p.  Then there is
  a small model M containing A and a global M-invariant
  non-constant type r such that (a,b)  p ⊗ r |
  M, i.e., b  r | M and a  p|Mb.


  Let b_1, b_2, … be an A-indiscernible sequence of pairwise
  distinct realizations of (b/A).  Let M_0 be a small model
  containing A.  Let b_1', b_2', … be an M_0-indiscernible
  sequence extracted from b_1, b_2, ….  Thus each b_i'
  realizes (b_i/A) and the b'_i are pairwise distinct.  Let
  σ∈(/A) move b'_1 to b, and let M_1 =
  σ(M_0).  Then σ(b'_1), σ(b'_2), … is an
  M_1-indiscernible sequence whose first entry is b and whose
  entries are pairwise distinct.  Therefore M_1 contains A but not
  b.  Let r_0 be a global coheir of tp(b/M_1); note that r_0
  is non-constant as b ∉(M_1).  Let a' realize p|M_1b;
  thus (a',b) realizes p ⊗ r_0 | M_1.  Now a' and a both
  realize p|Ab, because A ⊆ M_1.  Therefore, there is an
  automorphism τ∈(/Ab) such that τ(a') = a.  Let
  M = τ(M_1) and r = τ(r_0).  Then (a,b)  p ⊗
  r|M because (a',b)  p ⊗ r_0|M_1.



  
  Let Y be a critical set, let Q be quasi-minimal, and let t ≥
  1 be an integer.  There exist pairwise distinct q_1, q_2, …,
  q_t ∈ Q such that
  
    (⋂_i = 1^t (Y + q_i)) = (Y).

  In particular, by Remarks <ref> and <ref>, the
  intersection ⋂_i = 1^t (Y + q_i) is a critical set.


  Suppose otherwise.  Then for any distinct q_1, …, q_t in Q,
  the intersection ⋂_i = 1^t (Y + q_i) has rank at most
  ρ - 1.  Choose a critical coordinate configuration
  (X_1,…,X_n,P) with target Y.  Note that ∑_i (X_i) =
  ρ.  By Lemma <ref>, we may find a small set A
  and global A-invariant non-algebraic types p_i on X_i and
  p_0 on Q such that
  
  
  * The sets X_i, P, Q are A-definable.
  
  * The type p_1 ⊗⋯⊗ p_n lives on P.
  
  
    For any k, let Ω_k be the set of
    (q_0,a_1,1,…,a_1,n,…,a_k,1,…,a_k,n) ∈ Q
    × (X_1 ×⋯× X_n)^k such that
    
    
  * For each i ∈ [k], the tuple (a_i,1,…,a_i,n)
      lies on P.
    
  * There are infinitely many q ∈ Q such that
      
    ⋀_i = 1^k ( q_0 + ∑_j = 1^n a_i,j∈ Y
            + q ).

    
    Then Ω_k is narrow as a subset of Q × (X_1 ×⋯× X_n)^k for sufficiently large k.
  
  
    Let h = (Q) > 0 and take k so large that th + k(ρ - 1)
    < h + kρ.  The set Ω_k is definable (because
    ∃^∞ is eliminated).  If Ω_k were broad, it
    would have dp-rank h + kρ by Theorem <ref>.
    Therefore Ω_k would contain a specific tuple
    (q_0,a_1,1,…,a_k,n) of dp-rank h + kρ over A.
    Let s_i = a_i,1 + ⋯ + a_i,n for i ∈ [k].  Because
    (a_i,1,…,a_i,n) ∈ P, we have s_i ∈ Y.  Moreover,
    by definition of coordinate configuration,
    
    (a_i,1,…,a_i,n) ∈(s_iA).

    By definition of Ω_k there are infinitely many q such
    that
    
    {q_0 + s_1, q_0 + s_2, …, q_0 + s_k}⊆ Y + q.

    Choose q_1, …, q_t-1 distinct such q, not equal to
    q_0.  Then
    
    q_0 + s_i ∈ (Y + q_0) ∩ (Y + q_1) ∩⋯∩ (Y +
          q_t-1)

    for every i.  Therefore
    
    (a_i,1,…,a_i,n/Aq_0q_1 ⋯ q_t-1) =
          (s_i/Aq_0q_1 ⋯ q_t-1) ≤(⋂_i =
            0^t-1(Y + q_i)) < ρ.

    By subadditivity of dp-rank,
    
    (a_1,1,…,a_k,n,q_0,q_1,…,q_t-1/A) ≤ k(ρ - 1) + th < kρ + h

    contradicting the fact that
    
    (a_1,1,…,a_i,n,q_0/A) = kρ + h.

  
  Fix k as in the Claim.  Choose
  
    (a_k,1,…,a_k,n,…,a_1,1,…,a_1,n,q_0)

  realizing (p_1 ⊗⋯⊗ p_n)^⊗ k⊗ p_0
  over A.  Let s_i = a_i,1 + ⋯ + a_i,n.  By choice of
  p_1 ⊗⋯⊗ p_n, each (a_i,1,…,a_i,n)
  lives on P, and therefore s_i ∈ Y.  By
  Lemma <ref>, the type of (a⃗,q_0) over A is
  broad in (X_1 ×⋯× X_n)^k × Q.  Therefore the
  tuple (q_0,a⃗) cannot lie in the narrow set Ω_k.
  Consequently, there are only finitely many q ∈ Q such that
  
    {s_1 + q_0, …, s_k + q_0}⊆ Y + q.

  As s_i ∈ Y, one such q is q_0.  Therefore,
  
    q_0 ∈(A,s_1 + q_0,…, s_k + q_0).

  Choose j minimal such that
  
    q_0 ∈(A,s_1 + q_0, …, s_j + q_0).

  Note that j ≥ 1 because (q_0/A) is the non-algebraic type
  p_0.  Let A' = A ∪{s_1 + q_0, …, s_j-1 + q_0}.  By
  choice of j, q_0 ∉(A').  Note that
  (a_j,1,…,a_j,n) realizes the A-invariant type p_1
  ⊗⋯⊗ p_n over A'q_0 ⊆(q_0,a_1,1,…,a_j-1,nA).  Applying
  Lemma <ref>, we find a small model M ⊇ A'
  and a non-constant global M-invariant type r such that
  
    (a_j,1,…,a_j,n,q_0)  p_1 ⊗⋯⊗
        p_n ⊗ r|M.

  The type r extends (q_0/M) and therefore lives in the M-definable set Q.
  By Lemma <ref>, it follows that
  
    (a_j,1,…,a_j,n,q_0/M)

  is broad, as a type in X_1 ×⋯× X_n × Q.
  This contradicts Lemma <ref>: the X_i, P, Q are all
  M-definable, the type of (a_j,1,…,a_j,n,q_0) over M
  is broad, the tuple (a_j,1,…,a_j,n) lives on P, and
  for
  
    s = a_j,1 + ⋯ + a_j,n + q_0 = s_j + q_0,

  we have
  
    q_0 ∈(A',s_j + q_0) ⊆(sM).




  Let Y be a critical set and Q_1, …, Q_n be quasi-minimal.
  Then for every m there exist {q_i,j}_i ∈ [n], j ∈ [m]
  such that
  
  
  * For fixed i ∈ [n], the sequence q_i,1,…,q_i,m
    consists of m distinct elements of Q_i.
  
  * The intersection
    
    ⋂_η : [n] → [m](Y + ∑_i = 1^n q_i,η(i))

    is critical.
  


  We prove this by induction on n; the base case n = 1 is
  Lemma <ref>.  If n > 1, by induction there are
  q_1,1,…,q_n-1,m in the correct places, such that
  
    Y' = ⋂_η : [n-1] → [m](Y + ∑_i = 1^n
        q_i,η(i))

  is critical.  By Lemma <ref> there are
  q_n,1,…,q_n,m∈ Q_n, pairwise distinct, such that
  
    Y” = (Y' + q_n,1) ∩⋯∩ (Y' + q_n,m)

  is critical.  But x ∈ Y” if and only if for every j ∈ [m]
  and every η : [n-1] → [m] the element x - q_n,j -
  q_n-1,η(n-1) - ⋯ - q_1,η(1) lies in Y.  Thus
  
    Y” = ⋂_η : [n] → [m](Y + ∑_i = 1^n
        q_i,η(i)).




  
  Let Y be a critical set and Q_1,…,Q_n be quasi-minimal.
  There exists a δ∈ such that
  
    { (x_1,…,x_n) ∈ Q_1 ×⋯× Q_n  |  x_1 +
        ⋯ + x_n ∈ Y + δ}

  is broad, as a subset of Q_1 ×⋯× Q_n.


  For any m ≥ 1, we can find {q_i,j}_i ∈ [n], j ∈ [m]
    such that
  
  
  * q_i,j∈ Q_i
  
  * q_i,j q_i,j' for j  j'.
  
  * The intersection
    
    ⋂_η : [n] → [m](Y - ∑_i = 1^n q_i,η(i))

    is non-empty.
  
  Indeed, this follows from Proposition <ref> applied to the
  quasi-minimal sets -Q_i.  If -δ is an element of the
  intersection, then for any η : [n] → [m] we have
  
    -δ∈ Y - ∑_i = 1^n q_i,η(i)

  or equivalently,
  
    ∑_i = 1^n q_i, η(i)∈ Y + δ.

  Thus, for every m we can find {q_i,j}_i ∈ [n], j ∈ [m]
  and δ∈ such that
  
  
  * q_i,j∈ Q_i.
  
  * q_i,j q_i,j' for j  j'.
  
  * For any η : [n] → [m] the sum ∑_i = 1^n
    q_i,η(i) lies in Y + δ.
  
  By compactness, we can find {q_i,j}_i ∈ [n], j ∈ and
  δ satisfying these same properties.  This means that
  
    {(q_1,…,q_n) ∈ Q_1 ×⋯× Q_n  |  q_1 + ⋯
        + q_n ∈ Y + δ}

  is broad as a subset of Q_1 ×⋯× Q_n.




 §.§ Heavy sets

We are nearly ready to define heavy and light sets.

  Let Y be a critical set.  Say that a definable set X ⊆ is Y-heavy if there is δ∈ such that
  (Y ∩ (X + δ)) = (Y).

 
  Let Y be a critical set and {D_b} be a definable family of
  subsets of .  Let S be the set of b such that D_b is
  Y-heavy.  Then the set S is definable.  Indeed, this follows
  almost immediately from
  Proposition <ref>.<ref>.


  Let Y, Y' be two critical sets and X ⊆ be definable.
  Then X is Y-heavy if and only if X is Y'-heavy.


  Suppose X is Y-heavy.  Let X' := Y ∩ (X + δ_0) have
  full rank in Y.  Then X' is critical by Remark <ref>.  We
  claim that X' is Y'-heavy.  Let (A_1,…,A_n,P) be a
  coordinate configuration for Y'.  The set P is broad in A_1
  ×⋯× A_n, so there exist infinite definable subsets
  Q_i ⊆ A_i such that (Q_1 ×⋯× Q_n)
  ∖ P is a hyperplane, and in particular is narrow.  By
  Corollary <ref>, there is a δ_1 ∈ such that
  
    {(q_1,…,q_n) ∈ Q_1 ×⋯× Q_n  |  q_1 + ⋯
        + q_n ∈ X' + δ_1}

  is broad as a subset of Q_1 ×⋯× Q_n, hence broad
  as a subset of A_1 ×⋯× A_n.  Because (Q_1 ×⋯× Q_n) ∖ P is narrow, it follows that
  
    {(q_1,…,q_n) ∈ P  |  q_1 + ⋯ + q_n ∈ X' + δ_1}

  is broad as a subset of A_1 ×⋯× A_n.  Because the
  map (q_1,…,q_n) ↦ q_1 + ⋯ + q_n has finite fibers
  on P, it follows that
  
    {q_1 + ⋯ + q_n  | q⃗∈ P  and  q_1 + ⋯ + q_n ∈
        X' + δ_1} = Y' ∩ (X' + δ_1)

  has full rank.  Therefore, X' is Y'-heavy.  Now X' ⊆ (X +
  δ_0), and so Y' ∩ (X + δ_0 + δ_1) has full rank
  in Y', implying that X is Y'-heavy.  The converse follows by
  symmetry.


  A definable subset X ⊆ is heavy if it is Y-heavy
  for any/every critical set Y ⊆, and light
  otherwise.


 
  Let X, Y be definable subsets of .
  
  
  * (Assuming  is infinite) If X is finite, then X is
    light.
  
  *  If X, Y are light, then X ∪ Y is
    light.
  
  * If Y is light and X ⊆ Y, then X is light.
  
  * If {D_b} is a definable family of subsets of , then
    
    {b  |  D_b  is light}

    is definable.
  
  *  is heavy.
  
  * If X is heavy, then α· X is heavy for α∈^×.
  
  *  If X is heavy, then α + X is
    heavy for α∈.
  
  * If X and Y are heavy, the set
    
    X  Y := {δ∈ |  X ∩ (Y + δ)  is heavy}

    is heavy.
  


  
  
  * Indeed, if X is heavy then some translate of X contains a
    critical set, and so X has rank at least ρ.  The critical
    rank ρ cannot be 0 because there is at least one
    quasi-minimal set Q_i (by Remark <ref>), and
    (Q_i,Q_i) is a coordinate configuration with target Q_i.
    Therefore heavy sets are infinite.
  
  * We show the contrapositive: if X ∪ Y is heavy, then X is
    heavy or Y is heavy.  Fix some critical set Z.  The fact that X
    ∪ Y is Z-heavy implies that there is a δ such that Z
    ∩ ((X ∪ Y) + δ) has full rank.  But
    
    Z ∩ ((X ∪ Y) + δ) = (Z ∩ (X + δ)) ∪ (Z
          ∩ (Y + δ))

    so at least one of Z ∩ (X + δ) and Z ∩ (Y + δ)
    has full rank ρ.
  
  * If X ⊆ Y and X is Z-heavy because Z ∩ (X +
    δ) has full rank ρ, then certainly Z ∩ (Y +
    δ) has full rank ρ.
  
  * This is Remark <ref>.
  
  * Critical sets exist, and if Y is a critical set, then
    (Y ∩) = (Y).
  
  * Heaviness was defined without any reference to multiplication,
    so it is preserved by definable automorphisms of the additive
    group (,+).
  
  * Heaviness is translation-invariant by definition.
  
  * Shrinking X and Y we may assume that X and Y are
    translates of critical sets–or merely critical sets by
    Remark <ref>.  Let (A_1,…,A_n,P) be a coordinate
    configuration for Y.  By Theorem <ref> we
    can find infinite definable subsets Q_i ⊆ A_i such that
    (Q_1 ×⋯× Q_n) ∖ P is narrow.  By
    Corollary <ref>, there is a δ∈ such that
    the set
    
    {(q_1,…,q_n,q'_1,…,q'_n) ∈ (Q_1 ×⋯× Q_n)^2  |  q_1 + ⋯ + q_n + q'_1 + ⋯ + q'_n ∈ X
          + δ}

    is broad as a subset of Q_1 ×⋯× Q_n × Q_1
    ×⋯× Q_n.  In particular, we can find q_i,j
    and q'_i,j for i ∈ [n] and j ∈ such that
    
    
  * q_i,j and q'_i,j are in Q_i.
    
  * q_i,j q_i,j' and q'_i,j q'_i,j' for j
       j'.
    
  * For any η : [n] → and η' : [n] →, we
      have
      
    ∑_i = 1^n (q_i,η(i) + q'_i,η'(i)) ∈ X +
            δ.

    
    Let s_η denote the sum ∑_i = 1^n q_i,η(i).  Note
    that for any η, the q'_i,j show that the set
    
    {(q'_1,…,q'_n) ∈ Q_1 ×⋯× Q_n  |  s_η +
          q'_1 + ⋯ + q'_n ∈ X + δ}

    is broad as a subset of Q_1 ×⋯× Q_n.  It follows
    that
    
    {(q'_1,…,q'_n) ∈ P  |  s_η +
          q'_1 + ⋯ + q'_n ∈ X + δ}

    has full rank ρ, and so (Y + s_η) ∩ (X + δ) has
    full rank ρ for any η.  The translate X ∩ (Y +
    s_η - δ) has full rank and is trivially X-heavy.
    Therefore s_η - δ∈ X  Y.  The fact that this
    holds for all s_η implies that
    
    {(q_1,…,q_n) ∈ Q_1 ×⋯× Q_n  |  q_1 +
          ⋯ + q_n + δ∈ X  Y}

    is broad as a subset of Q_1 ×⋯× Q_n.  Again, this
    implies that (Y + δ) ∩ (X  Y) has full rank
    ρ, and so X  Y is heavy. 




 §.§ Externally definable sets

We prove the analogues of <ref> for heavy and light
sets.

  
  Let M ≼ be a small model defining a critical coordinate
  configuration.  Let Z be an M-definable heavy set.  Let D_1,
  …, D_m be -definable sets such that
  
    Z(M) ⊆ D_1 ∪⋯∪ D_m.

  Then there is a j ∈ [m] and a M-definable heavy set Z'
  ⊆ Z such that Z'(M) ⊆ D_j.


  Let (X_1,…,X_n,P) be an M-definable critical coordinate
  configuration, with target Y.  Because Z is Y-heavy, there is
  some δ such that Y ∩ (Z + δ) has full rank in Y.
  Full rank in Y is a definable condition, clearly M-invariant, so
  we may take δ∈ M.  Replacing Z and the D_i's with Z +
  δ and D_i + δ, we may assume δ = 0.  The fact
  that Y ∩ Z has full rank implies that the set
  
    R := {(x_1,…,x_n) ∈ P  |  x_1 + ⋯ + x_n ∈ Z}

  is broad.  Furthermore, R(M) is a union of the
  externally-definable sets
  
    {(x_1,…,x_n) ∈ P(M)  |  x_1 + ⋯ + x_n ∈ Z ∩ D_j.}

  By Lemma <ref>, there is an M-definable R' ⊆
  R and a j ∈ [m] such that R' is broad and
  
    (x_1,…,x_n) ∈ R'(M)  x_1 + ⋯ + x_n ∈ D_j.

  Let Z' be the M-definable set
  
    {x_1 + ⋯ + x_n  |  (x_1,…,x_n) ∈ R' }.

  Because R' ⊆ R ⊆ P, the set Z' ⊆ Y ∩
  Z.  The set of (x_1,…,x_n) ∈ P such that x_1 + ⋯ +
  x_n ∈ Z' contains the broad set R', and is therefore broad.  It
  follows that Z' has full rank in Y, and so Z' is Y-heavy.
  Lastly, we claim that
  
    Z'(M) ⊆ D_j.

  Indeed, suppose y ∈ Z'(M).  Then there exists an n-tuple
  (x_1, …, x_n) ∈ R' such that y = x_1 + ⋯ + x_n.  The
  fact that M ≼ implies that we can take x⃗∈(M).  Then by choice of R', it follows that
  
    y = x_1 + ⋯ + x_n ∈ D_j.

  Therefore Z'(M) ⊆ D_j.


  Let M be a small model defining a critical coordinate
  configuration.  Let Z be an M-definable heavy set, and let W
  be an -definable set.  If Z(M) ⊆ W, then W is
  heavy.


  Take an M-definable coordinate configuration (X_1,…,X_n,P)
  with target Y.  As in the proof of Lemma <ref> we may
  assume that Z ∩ Y has full rank in Y, and so the set
  
    {(x_1,…,x_n) ∈ P  |  x_1 + ⋯ + x_n ∈ Z}

  is broad in X_1 ×⋯× X_n.  This set is
  M-definable, and its M-definable points lie in the set
  
    {(x_1,…,x_n) ∈ P  |  x_1 + ⋯ + x_n ∈ W},

  which is therefore broad by Lemma <ref>.  This
  implies that W ∩ Y has full rank, and so W is (Y-)heavy.




§ THE EDGE CASE: FINITE MORLEY RANK
 
If our end goal is to construct non-trivial valuation rings on
dp-finite fields, we had better quarantine the stable fields at some
point.  In other words, we need to find some positive consequence of
the hypothesis “not stable.”  We prove the relevant dichotomy in
this section.

  Let X_1, …, X_n be infinite definable sets.  If the number of
  broad global types in X_1 ×⋯× X_n is finite, then
  every X_i has Morley rank 1.


  Suppose, say, X_1 has Morley rank greater than 1.  Then we can
  find a sequence D_1, D_2, D_3, … of length ω where
  each D_i is an infinite definable subset of X_1 and the D_i
  are pairwise disjoint.  Let P_i = D_1 × X_2 ×⋯× X_n.  Each P_i is broad and the P_i are pairwise
  disjoint.  By Proposition <ref>, each P_i is inhabited
  by a broad type over , so there are at least ℵ_0 broad
  global types.  If there are a finite number of broad global types,
  it follows that every X_i has Morley rank 1.



  At least one of the following holds:
  
  
  * The field  has finite Morley rank.
  
  * There are two disjoint heavy sets.
  


  Let (X_1,…,X_n,P) be a critical coordinate configuration.  We
  may find infinite definable subsets Q_i ⊆ X_i such that for
  
    H    := (Q_1 ×⋯× Q_n) ∖ P 
    
        P'    := (Q_1 ×⋯× Q_n) ∩ P

  the set H is narrow in Q_1 ×⋯× Q_n, and
  therefore P' is broad.  Then (Q_1,…,Q_n,P') is a coordinate
  configuration of the same rank as (X_1,…,X_n,P).  In
  particular (Q_1,…,Q_n,P') is critical and its target Y is a
  critical set.

  First suppose that at least one Q_i has Morley rank greater than 1
  (possibly ∞).  By Lemma <ref>, there are
  infinitely many broad global types in Q_1 ×⋯×
  Q_n.  Let p_1, p_2, p_3, … be pairwise distinct broad global
  types.  Because H is narrow, each p_i must live in P.  Let m
  bound the fiber size of the map Σ : P' → Y.  There is a
  pushforward map Σ_* from the space of -types in P' to
  the space of -types in Y, and this map has fibers of size no
  greater than m.  Consequently, we can find i, j such that
  Σ_* p_i Σ_* p_j.  Without loss of generality
  Σ_* p_1 Σ_* p_2.  We can then write Y as a
  disjoint union Y = Z_1 ∪ Z_2 where Σ_* p_i lives on
  Z_i.  For i = 1, 2, the definable set
  
    {(x_1,…,x_n) ∈ P'  |  x_1 + ⋯ + x_n ∈ Z_i}

  is broad as a subset of Q_1 ×⋯× Q_n because it
  contains the broad type p_i.  Therefore this set has dp-rank
  ρ and so (Z_i) = ρ = (Y).  So the two sets Z_1,
  Z_2 are full-rank subsets of Y, and are therefore Y-heavy.  We
  have constructed two disjoint heavy sets.

  Next suppose that every Q_i has Morley rank 1.  Let
  
    W = Q_1 + ⋯ + Q_n.

  The set W contains the critical set Y, so W is heavy.  If W -
  W =, then there is a surjective map
  
    Q_1 ×⋯× Q_n × Q_1 ×⋯× Q_n    →
    
        (x_1,…,x_n,x'_1,…,x'_n)    ↦ x_1 + ⋯ + x_n - x'_1 - ⋯ - x'_n.

  The set (Q_1 ×⋯× Q_n)^2 has finite Morley
  rank,[See Proposition 1.5.16 in <cit.> or the
    proof of Corollary 2.14 in <cit.>.], and so  must
  have finite Morley rank.  Otherwise, take x ∉ W - W.  Then
  the sets W and W + x are disjoint, and both are heavy.



  Theorem <ref> is closely related to Sinclair's
  Large Sets Property (LSP).  A field  of dp-rank n has
  the LSP if there do not exist two disjoint subsets of  of
  dp-rank n (<cit.>, Definition 3.0.3).  Sinclair
  conjectures that fields with LSP are stable, hence algebraically
  closed.  Theorem <ref> says something similar, with
  “heavy” in place of dp-rank n.




§ INFINITESIMALS

In this section, we carry out the construction of infinitesimals.  The
arguments are copied directly from the dp-minimal setting (<cit.>, Chapter 9).


 §.§ Basic neighborhoods


  If X, Y are definable sets, X  Y denotes the set
  
    {δ∈ |  X ∩ (Y + δ)  is heavy}.


 
  Let X, Y be definable sets.
  
  
  * X  Y is definable.
  
  * X  Y ⊆ X - Y, where
    
    X - Y = {x - y  |  x ∈ X,   y ∈ Y}.

  
  * If X or Y is light, then X  Y is empty.
  
  * If X and Y are heavy, then X  Y is heavy.
  
  * For any δ_1, δ_2 ∈,
    
    (X + δ_1)  (Y + δ_2) = (X  Y) +
          (δ_1 - δ_2).

  
  * For any α∈^×,
    
    (α· X)  (α· Y) = α· (X
           Y).

  
  * If X' ⊆ X and Y' ⊆ Y then
    
    X'  Y' ⊆ X  Y.

  
  Indeed, all of these properties follow immediately from
  Theorem <ref>.


  A basic neighborhood is a set of the form X  X,
  where X is heavy.


  If M is a small model, then every M-definable basic
  neighborhood is of the form X  X with X heavy and
  M-definable.


  Let U = X  X be a basic neighborhood which is
  M-definable.  Write X as ϕ(;b).  Because heaviness is
  definable in families (by Theorem <ref>), the set of
  b' such that ϕ(;b') ϕ(;b') = U is definable.
  This set is M-invariant, and non-empty because of b, so it
  intersects M.  Therefore, we can find M-definable X' =
  ϕ(;b') such that U = X'  X'.


   
  
  
  *  If U is a basic neighborhood, then U is
    heavy.
  
  *  If U is a basic neighborhood, then 0 ∈
    U.
  
  *  If U is a basic neighborhood and α∈^×, then α· U is a basic neighborhood.
  
  *  If U_1, U_2 are basic neighborhoods,
    there is a basic neighborhood U_3 ⊆ U_1 ∩ U_2.  If
    U_1, U_2 are M-definable, we can choose U_3 to be
    M-definable.
  
  *  If  is not of finite Morley rank, then
    for every α 0 there is a basic neighborhood U ∌α.  If α is M-definable, we can choose U to be
    M-definable.
  


  
  
  * This is part of Remark <ref>.
  
  * If X is heavy, then 0 ∈ X  X because X ∩ (X +
    0) is the heavy set X.
  
  * If U = X  X, then α· U = (α· X)
     (α· X) by Remark <ref>, and
    α· X is heavy by Theorem <ref>.
  
  * Let U_i = X_i  X_i.  By Theorem <ref> the
    set X_1  X_2 is heavy, hence non-empty.  Take δ∈
    X_1  X_2.  Then X_3 := X_1 ∩ (X_2 + δ) is heavy.
    By Remark <ref>
    
    X_3  X_3    ⊆ X_1  X_1 = U_1 
    
          X_3  X_3    ⊆ (X_2 + δ)  (X_2 +
          δ) = X_2  X_2 = U_2.

    If U_1 and U_2 are M-definable and X_3 = ϕ(;b_0),
    the set of b such that
    
    ∅ϕ(;b) ϕ(;b) ⊆ U_1 ∩ U_2

    is M-definable and non-empty.  Therefore we can move X_3 to be
    M-definable.
  
  * If  is not of finite morley rank, we can find two
    disjoint heavy sets X, Y.  By Theorem <ref>, X
     Y ∅.  Take δ∈ X  Y.  Then
    the sets
    
    X'     := X ∩ (Y + δ) 
    
          Y'     := (X - δ) ∩ Y

    are heavy—X' is heavy by choice of δ and Y' is a
    translate of X'.  Also,
    
    Y' ∩ (Y' + δ) = Y' ∩ X' ⊆ Y ∩ X = ∅,

    so δ∉ Y'  Y'.  Because Y' is heavy, there is
    a basic neighborhood U_0 = Y'  Y' such that δ∉ U_0.  By part 2, δ 0.  If α is a
    non-zero element of , then we can find β∈^×
    such that α = β·δ.  Then α = β·δ is not in U = β· U_0, which is a basic
    neighborhood by part 3.

    Finally, suppose that α is a non-zero element of a small
    model M.  We know that there is some heavy set ϕ(;b)
    such that
    
    α∉ϕ(;b) ϕ(;b).

    As before, the set of b with this property is M-definable,
    hence we can pull b into M, producing an M-definable basic
    neighborhood that avoids α. 




 §.§ The set of infinitesimals


  Let M be a small model.  An element ε∈ is an
  M-infinitesimal if ε lies in every M-definable
  basic neighborhood.  Equivalently, for every M-definable heavy set
  X, the intersection X ∩ (X + ε) is heavy.  We let
  I_M denote the set of M-infinitesimals.


  Note that I_M is type-definable over M, as it is the small
  intersection
  
    I_M = ⋂_X  heavy and M-definable X  X.

  Moreover, this intersection is directed, by
  Proposition <ref>.


  
  If M' ≽ M, then I_M'⊆ I_M because there are
  more neighborhoods in the intersection.


  Let M be a small model.
  
  
  *  If X is a definable set containing
    I_M, then X is heavy.  In particular, if  is infinite,
    then I_M  0.
  
  * 0 ∈ I_M.
  
  *  I_M · M ⊆
    I_M.
  
  * If  is not of finite Morley rank, then I_M ∩ M =
    {0}.
  


  
  
  * The fact that X ⊇ I_M implies by compactness that
    there exist M-definable basic neighborhoods U_1, …, U_n
    such that
    
    X ⊇ U_1 ∩⋯∩ U_n.

    However, the fact that the intersection is directed (i.e.,
    Proposition <ref>.<ref>) implies that
    we can take n = 1.  Then X ⊇ U_1 and U_1 is heavy
    (by Proposition <ref>.<ref>), so X is
    heavy.
  
  * Every basic neighborhood contains 0 by
    Proposition <ref>.<ref>.
  
  * Suppose a ∈ M.  We claim that a · I_M ⊆ I_M.
    If a = 0 this is the previous point.  Otherwise, the two
    collections
    
    {U     |  U  an M-definable basic neighborhood}
    {a · U     |  U  an M-definable basic neighborhood}

    are equal by Proposition <ref>.<ref>, so they
    have the same intersection.  But the intersection of the first is
    I_M and the intersection of the second is a · I_M.  Thus
    a · I_M = I_M.
  
  * This follows directly from
    Proposition <ref>.<ref>. 




 §.§ The group of infinitesimals


  Let M be a small model and X ⊆ be M-definable.  An
  element δ∈ is said to M-displace the set X
  if
  
    x ∈ X ∩ x ∈ M  x + δ∉ X.



  Let M ≼ M' be an inclusion of small models, and ε,
  ε' be elements of .  Suppose that (ε'/M')
  is an heir of (ε/M).
  
  
  * If ε is an M-infinitesimal, then ε' is an
    M'-infinitesimal.
  
  * If X ⊆ is M-definable and M-displaced by
    ε, then X is M'-displaced by ε'.
  


  The assumptions imply that ε' ≡_M ε, and the
  statements about ε are M-invariant, so we may assume
  ε' = ε.
  
  
  * Suppose ε fails to be M'-infinitesimal.  Then there
    is a tuple b ∈(M') such that ϕ(;b) is heavy and
    ε∉ϕ(;b) ϕ(;b).  These
    conditions on b are Mε-definable.  Because
    (b/Mε) is finitely satisfiable in M, we can find
    such a b in M.  Then ε fails to be M-infinitesimal
    because it lies outside ϕ(;b) ϕ(;b).
  
  * Suppose ε fails to M'-displace X.  Then there
    exists b ∈ X(M') such that b + ε∈ X = X().
    These conditions on b are Mε-definable, so we can find
    such a b in M.  Then ε fails to M-displace X. 


  
  Let M be a small model defining a critical coordinate
  configuration.  Let ε be an M-infinitesimal, and X
  ⊆ be an M-definable set that is M-displaced by
  ε.  Then X is light.


  Build a sequence ε_0, ε_1, … and M = M_0
  ≼ M_1 ≼ M_2 ≼⋯ so that for each i,
  
  
  * M_i+1∋ε_i
  
  * (ε_i/M_i) is an heir of (ε/M).
  
  By Lemma <ref>,
  
  
  * ε_i is M_i-infinitesimal.
  
  * The set X is M_i-displaced by ε_i.
  
  For α a string in {0,1}^< ω, define X_α
  recursively as follows:
  
  
  * X_{} = X.
  
  * If α has length n, then X_α 0 = {x ∈
    X_α |  x + ε_n ∉ X}.
  
  * If α has length n, then X_α 1 = {x ∈
    X_α |  x + ε_n ∈ X}.
  
  For example
  
    X_0 = {x ∈ X  |    x + ε_0 ∉ X}
    
        X_1 = {x ∈ X  |    x + ε_0 ∈ X}
    
        X_011 = {x ∈ X  |    x + ε_0 ∉ X,
       x + ε_1
        ∈ X,
       x + ε_2 ∈ X}.

  Note that some of the X_α must be empty, by NIP.  Note also
  that if α has length n, then X_α is M_n-definable.
  
    If X_α is heavy, then X_α 1 is heavy.
  
  
    Let α have length n.  Then X_α is heavy and
    M_n-definable, and ε_n is M_n-infinitesimal.
    Consequently X_α∩ (X_α - ε_n) is heavy.
    But
    
    X_α 1 = X_α∩ (X - ε_n) ⊇ X_α∩ (X_α - ε_n),

    and so X_α 1 must be heavy.
  
  
    If X_α is heavy, then X_α 0 is heavy.
  
  
    Let α have length n.  The set X_α is
    M_n-definable.  Note that
    
    x ∈ X_α(M_n)  x ∈ X(M_n)  x +
          ε_n ∉ X(M_n)

    because X is M_n-displaced by ε_n.  Therefore
    
    X_α(M_n) ⊆ X_α 0.

    By Lemma <ref> it follows that X_α 0
    is heavy.
  
  If X = X_{} is heavy, then the two claims imply that every
  X_α is heavy, hence non-empty, for every α.  This
  contradicts NIP.



  
  Let M be a model defining a critical coordinate configuration.
  Let ε_1, ε_2 be two M-infinitesimals.  Then
  ε_1 - ε_2 is an M-infinitesimal.


  Let X be an M-definable heavy set; we will show that X ∩ (X
  + ε_1 - ε_2) is heavy.  Note that X is covered by
  the union of the following three sets:
  
    D_0    := {x ∈ X  |  x + ε_1 ∈ X,  x + ε_2 ∈ X}
    
        D_1     := {x ∈ |  x + ε_1 ∉ X }
    
        D_2     := {x ∈ |  x + ε_2 ∉ X }.

  By Lemma <ref>, there is a j ∈{0,1,2} and an
  M-definable heavy set X' ⊆ X such that X'(M) ⊆
  D_j.  If j > 0, then
  
    x ∈ X'(M)  x ∈ D_j  x + ε_j ∉ X
         x + ε_j ∉ X'.

  In other words, X' is M-displaced by ε_j.  But X' is
  heavy and ε_j is an M-infinitesimal, so this would
  contradict Lemma <ref>.  Therefore j = 0.  The fact that
  X'(M) ⊆ D_0 implies that D_0 is heavy, by
  Lemma <ref>.  By definition of D_0,
  
    D_0 + ε_1    ⊆ X 
    
        D_0 + ε_2    ⊆ X 
    
        D_0 + ε_1    ⊆ X + ε_1 - ε_2 
    
        D_0 + ε_1    ⊆ X ∩ (X + ε_1 - ε_2).

  Heaviness of D_0 then implies heaviness of X ∩ (X + ε_1
  - ε_2).


 
  Let U be an M-definable basic neighborhood.  Then there is an
  M-definable basic neighborhood V such that V - V ⊆ U.


  Let M' ≽ M be a larger model that defines a critical
  coordinate configuration.  By Lemma <ref> and Remark <ref>,
  
    I_M' - I_M'⊆ I_M'⊆ I_M ⊆ U.

  The M'-definable partial type asserting that
  
    x    ∈ I_M'
    
        y    ∈ I_M'
    
        x - y    ∉ U

  is inconsistent.  Because I_M' is the directed intersection of
  M'-definable basic neighborhoods, there is an M'-definable basic
  neighborhood V' such that 
  
    x ∈ V' ∧ y ∈ V'  x - y ∈ U,

  i.e., V' - V' ⊆ U.  In particular, there is a formula
  ϕ(x;z) and c ∈(M') such that ϕ(;c) is heavy
  and
  
    (ϕ(;c) ϕ(;c)) - (ϕ(;c) ϕ(;c)) ⊆ U.

  But heaviness of ϕ(;c) and condition (<ref>) are both
  M-definable constraints on c, so we may find c_0 ∈(M)
  satisfying the same conditions.  Then
  
    V := ϕ(;c_0) ϕ(;c_0)

  is an M-definable basic neighborhood such that V - V ⊆
  U.



  
  If M is any model, the set I_M of M-infinitesimals is a
  subgroup of the additive group (,+).


  The fact that 0 ∈ I_M follows by <ref>, and the fact
  that I_M - I_M ⊆ I_M is a consequence of
  Lemma <ref>.




 §.§ Consequences

We mention a few corollaries of Theorem <ref>.
 
  If M is a field of finite dp-rank but not finite Morley rank, then
  there is a (unique) Hausdorff non-discrete group topology on (M,+)
  such that
  
  
  * The basic neighborhoods X  X form a basis of
    neighborhoods of 0.
  
  * For any α∈ M, the map x ↦ x ·α is
    continuous.
  
  Indeed, this follows formally from the fact that basic neighborhoods
  are filtered with intersection I_M, and
  
    I_M - I_M    ⊆ I_M 
    
        0    ∈ I_M 
    {0}    I_M 
    
        I_M ∩ M     = {0}
    
        M · I_M    ⊆ I_M.

  On the other hand, we have not yet shown that there is a definable
  basis of opens, or that multiplication is continuous in general
  (which would require I_M · I_M ⊆ I_M).

Later (Corollary <ref>), we will see that multiplication
is continuous.

 
  Let J be a subgroup of (,+), type-definable over a small
  model M.  Suppose that every M-definable set containing J is
  heavy.  Then J ⊇ I_M.


  Let D be any M-definable set containing J.  We claim D
  contains I_M.  Because J - J ⊆ J ⊆ D there is,
  by compactness, an M-definable set D' such that D' ⊇ J
  and D' - D' ⊆ D.  By assumption, D' is heavy.  Then D'
   D' is an M-definable basic neighborhood, and so
  
    I_M ⊆ D'  D' ⊆ D' - D' ⊆ D.

  As this holds for all M-definable D ⊇ J, we see that
  I_M ⊆ J.


Because we are in an NIP setting, G^00 and G^000 exist for any
type-definable group G, by <cit.>, Theorem 1.12 in the abelian case and <cit.> in general.
 
  For any small model M, I_M = I_M^00 = I_M^000.


  Let M' ⊇ M be a model containing a representative from
  each coset of I_M^000 in I_M.  Take a ∈ I_M such that a
  does not lie in any M'-definable light set; this is possible by
  Theorem <ref>.<ref> and
  Proposition <ref>.<ref>.  If δ is the
  M'-definable coset representative for a + I_M^000, then a' :=
  a - δ lies in I_M^000, but not in any M'-definable light
  set (by Theorem <ref>.<ref>).  Let Y
  be the set of realizations of (a'/M).  By M-invariance of
  I_M^000 we have Y ⊆ I_M^000.  On the other hand,
  every M-definable neighborhood D ⊇ Y is heavy, so I_M
  ⊆ D  D ⊆ D - D.  As Y is the directed
  intersection of M-definable D ⊇ Y, it follows that I_M
  ⊆ Y - Y ⊆ I_M^000.




§ FURTHER COMMENTS ON HENSELIANITY
 
Now that we have a type-definable group of infinitesimals, we can
reduce the Shelah conjecture for dp-finite fields of positive
characteristic to the construction of invariant valuation rings.


  
  Let X be a definable set of full dp-rank.  Then X is heavy.


  Let n be ().  Let (Q_1,…,Q_m,P) be a critical
  coordinate configuration of rank ρ.  Shrinking Q_i, we may
  assume that (Q_1 ×⋯× Q_m) ∖ P is narrow.
  Let M be a small model over which X, Q_1, …, Q_m, P are
  defined.  Take a tuple
  
    (a,b_1,…,b_m) ∈ X × Q_1 ×⋯× Q_m

  such that
  
    (a,b_1,…,b_m/M) = (X) + (Q_1) + ⋯ +
        (Q_m) = n + ρ.

  By Remark <ref>, (b⃗/M) = ρ.  Let
  δ = a - (b_1 + ⋯ + b_m).  Then
  
    n + ρ = (a, b⃗/M) = (δ,b⃗/M) ≤(b⃗/δ M) + (δ/M).

  As (δ/M) ≤() = n, it follows that
  
    ρ≤(b⃗/δ M) ≤(b⃗/M) = ρ.

  By Proposition <ref>, (b⃗/δ M) is
  broad.  It follows that the set
  
    {(b_1,…,b_m) ∈ Q_1 ×⋯× Q_m  |  b_1 + ⋯
        + b_m + δ∈ X}

  is broad.  As we arranged (Q_1 ×⋯× Q_m) ∖
  P to be narrow, the set
  
    {(b_1,…,b_m) ∈ P  |  b_1 + ⋯ + b_m + δ∈ X}

  is also broad, and therefore has dp-rank ρ.  If Y is the
  target of (Q_1,…,Q_n,P), then
  
    {y ∈ Y  |  y + δ∈ X}

  has dp-rank ρ, as the map P → Y has finite fibers.
  Therefore X is Y-heavy.

In fact, the converse to Lemma <ref> is true.  See Theorem 5.9.2 in <cit.>.

  
  Let M be a small model and 𝒪 be a non-trivial
  M-invariant valuation ring.  Then the maximal ideal 𝔪
  contains I_M, and the subideal of 𝒪 generated by I_M
  is prime.


  Let n = ().  Take an element a of positive valuation.
  Take an element b ∈ with (b/aM) = n.  Replacing b
  with a/b if necessary, we may assume b ∈𝔪.  Let
  Y be the type-definable set of realizations of (b/M).  Note
  (Y) ≥(b/aM) = n, so (Y) = n.  Also note that Y
  ⊆𝔪 because 𝔪 is M-invariant.
  Every M-definable set D ⊇ Y is heavy, by
  Lemma <ref>.  Therefore, I_M ⊆ D - D.  As D
  is arbitrary,
  
    I_M ⊆ Y - Y ⊆𝔪 - 𝔪⊆𝔪.

  Now let J be the ideal of 𝒪 generated by I_M.
  Evidently J ≤𝔪 and J is M-invariant.  Suppose
  J is not prime.  By the characterization of ideals in valuation
  rings in terms of cuts in the value group, non-primality of J
  implies that the ideal J' := J · J is a strictly smaller
  non-zero ideal in 𝒪, and in particular J' ⊉I_M (by choice of J).  On the other hand, J' is
  M-invariant.  Because J' is non-zero, there exists c ∈^× such that c ·𝔪⊆ J'.  Let d
  be an element of Y such that (d/cM) = n.  Then (d ·
  c / cM) = (d / cM) = n, so (d · c / M) = n.  Let Z
  be the set of realizations of (d · c / M).  As d · c
  ∈ c · Y ⊆ c ·𝔪⊆ J', the
  usual argument shows that I_M ⊆ Z - Z ⊆ J' - J' =
  J', a contradiction.  Therefore J is prime.



  
  Let M be a small model and 𝒪 be a non-trivial
  M-invariant valuation ring.  Then there is an M-invariant
  coarsening 𝒪' ⊇𝒪 such that I_M
  spans 𝒪', in the sense of
  Definition <ref>.


  It is a general fact that if 𝔭 is a prime ideal in a
  valuation ring 𝒪, then 𝔭 is the maximal
  ideal of some (unique) coarsening 𝒪' ⊇𝒪.  In our specific case,
  
    I_M    ⊆𝔭 = 𝒪· I_M 
     I_M    ⊆𝔭⊆𝒪' ·
        I_M ⊆𝒪' ·𝔭 = 𝔭
     I_M    ⊆𝔭 = 𝒪' · I_M

  so I_M spans 𝒪'.



  
  If  has characteristic p > 0 and if at least one non-trivial
  invariant valuation ring exists, then  admits a non-trivial
  henselian valuation.


  By Proposition <ref> there is a small model M and
  an M-invariant valuation ring 𝒪 spanned by I_M.  The
  fact that I_M  0 prevents 𝒪 from being the trivial
  valuation ring.  Moreover, I_M = I_M^000 by
  Corollary <ref>.  By
  Proposition <ref> it follows that 𝒪
  is henselian.




§ BOUNDS ON CONNECTED COMPONENTS
 
In this section, (G,+,…) is a monster-model abelian group,
possibly with additional structure, of finite dp-rank n.

 
  Let G_0, …, G_n be type-definable subgroups of G.  There is
  some 0 ≤ k ≤ n such that
  
    ( ⋂_i = 0^n G_i )^00 = ( ⋂_i 
          k G_i )^00.


This follows by the proof of Proposition 4.5.2 in <cit.>.




































  
  Let H be a type-definable subgroup of G.  There is a cardinal
  κ depending only on H and G such that if H < H' < G for
  some type-definable subgroup H', and if H'/H is bounded, then
  H'/H has size at most κ.  This κ continues to work
  in arbitrary elementary extensions.


  Naming parameters, we may assume that H (but not H') is
  type-definable over ∅.  By Morley-Erdős-Rado there is
  some cardinal κ with the following property: for any sequence
  {a_α}_α < κ of elements of G, there is some
  ∅-indiscernible sequence {b_i}_i ∈ such that
  for any i_1 < ⋯ < i_n there is α_1 < ⋯ < α_n
  such that
  
    a_α_1⋯ a_α_n≡_∅ b_i_1⋯
        b_i_n.

  Let H' be a subgroup of G, containing H, type-definable over
  some small set A.  Suppose that |H'/H| ≥κ.  We claim
  that H'/H is unbounded.  Suppose for the sake of contradiction
  that |H'/H| < λ in all elementary extensions.  Take a
  sequence {a_α}_α < κ of elements of H' lying
  in pairwise distinct cosets of H.  Let {b_i}_i ∈ be an
  ∅-indiscernible sequence extracted from the a_α by
  Morley-Erdős-Rado.  Because the a_α live in pairwise
  distinct cosets of H and H is ∅-definable, the b_i
  live in pairwise distinct cosets of H.  By indiscernibility, there
  is a 0-definable set D ⊇ H such that b_i - b_j ∉ D
  for i  j.  Consider the ∗-type over A in variables
  {x_α}_α < λ asserting that
  
  
  * x_α∈ H' for every α < λ
  
  * If α_1 < ⋯ < α_n, then
    
    x_α_1⋯ x_α_n≡_∅ b_1 ⋯ b_n.

  
  This type is consistent.  Indeed, if
  Σ_α_1,…,α_n(x⃗) is the sub-type
  asserting that
  
    x_α_1, …, x_α_n   ∈ H' 
    
          x_α_1⋯ x_α_n   ≡_∅ b_1 ⋯ b_n

  then Σ_α_1,…,α_n(x⃗) is satisfied by
  (a_β_1,…,a_β_n) for some well chosen β_i,
  by virtue of how the b_i were extracted.  Moreover, the full type
  is a filtered union of Σ_α⃗(x⃗)'s, so it is
  consistent.  Let {c_α}_α < λ be a set of
  realizations.  Then every c_α lies in H', but
  
    c_α - c_α'∉ D ⊇ H

  for αα'.  Therefore, the c_α lie in pairwise
  distinct cosets of H, and |H'/H| ≥λ, a contradiction.



  
  For any cardinal κ there is a cardinal τ(κ) with
  the following property: given any family {H_α}_α <
    τ(κ) of type-definable subgroups of G, there exist
  subsets S_1, S_2 ⊆τ(κ) such that S_1 is finite,
  |S_2| = κ, and
  
    ( ⋂_α∈ S_1 H_α)^00⊆⋂_α∈ S_2 H_α.



  Without loss of generality κ≥ℵ_0.  By the
  Erdős-Rado theorem (or something weaker), we can choose
  τ(κ) such that any coloring of the n+1-element subsets
  of τ(κ) with n+1 colors contains a homogeneous subset of
  cardinality κ^+.  Now suppose we are given H_α for
  α < τ(κ).  Given α_1 < ⋯ < α_n+1,
  color the set {α_1, …, α_n+1} with the smallest
  k ∈{1, …, n+1} such that
  
    ( ⋂_i = 1^n+1 H_α_i)^00 = (
        ⋂_i = 1^k-1 H_α_i∩⋂_i = k + 1^n+1
        H_α_i)^00.

  This is possible by Fact <ref>.  Passing to a
  homogeneous subset and re-indexing, we get {H_α}_α <
    κ^+ such that every (n+1)-element set has color k for
  some fixed k.  In particular, for any α_1 < ⋯ <
  α_n+1 < κ^+, we have
  
    H_α_k⊇( ⋂_i = 1^n+1 H_α_i)^00 = ( ⋂_i = 1^k - 1 H_α_i∩⋂_i = k + 1^n+1 H_α_i)^00.

  Thus, for any β_1 < β_2 < ⋯ < β_2n+1 < κ^+
  we have
  
    H_β_n+1⊇( ⋂_i = n - k + 2^n
        H_β_i∩⋂_i = n + 2^2n - k + 2 H_β_i)^00⊇ ( ⋂_i = 1^n H_β_i∩⋂_i = n+ 2^2n + 1 H_β_i)^00

  by taking
  
    (α_1, …, α_n+1) =
        (β_n-k+2,…,β_2n-k+2).

  Then, for any β∈ [n+1,κ],
  
    H_β⊇( H_1 ∩⋯∩ H_n ∩ H_κ +
          1∩ H_κ + n)^00,

  so we may take S_1 = {1, …, n, κ + 1, …, κ +
  n} and S_2 = [n+1,κ].


 
  There is a cardinal κ, depending only on the ambient group
  G, such that for any type-definable subgroup H < G, the index of
  H^00 in H is less than κ.  This κ continues to
  work in arbitrary elementary extensions.


  Say that a subgroup K ⊆ G is ω-definable if
  it is type-definable over a countable set.  Note that if K is
  ω-definable, so is K^00.  Moreover, if K_1, K_2 are
  ω-definable, then so are K_1 ∩ K_2 and K_1 + K_2.
  Also note that if H is any type-definable group, then H is a
  small filtered intersection of ω-definable groups.

  Up to automorphism, there are only a bounded number of
  ω-definable subgroups of G, so by Lemma <ref>
  there is some cardinal κ_0 with the following property: if
  K is an ω-definable group and if K' is a bigger
  type-definable group, then either |K'/K| < κ_0 or |K'/K| is
  unbounded.
   
    If H is a type-definable group and K is an ω-definable
    group containing H^00, then |H/(H ∩ K)| < κ_0.
  
  
    Note that
    
    H^00⊆ H ∩ K ⊆ H,

    so H/(H ∩ K) is bounded.  On the other hand, H/(H ∩ K)
    is isomorphic to (H + K)/K, which must then have cardinality
    less than κ_0.
  
  Let κ_1 = τ((2^κ_0)^+) where τ(-) is as in
  Lemma <ref>.
  
    If H is a type-definable subgroup of G, then there are fewer
    than κ_1 subgroups of the form H ∩ K where K is
    ω-definable and K ⊇ H^00.
  
  
    Otherwise, choose {K_α}_α∈κ_1 such that
    K_α is ω-definable, K_α⊇ H^00, and
    
    H ∩ K_α H ∩ K_α'

    for α < α' < κ_1.  By Lemma <ref>,
    there are subsets S_1, S_2 ⊆κ_1 such that |S_1| <
    ℵ_0, |S_2| = (2^κ_0)^+, and
    
    ( ⋂_α∈ S_1 K_α)^00⊆⋂_α∈ S_2 K_α.

    Let J be the left-hand side.  Then J is an ω-definable
    group containing H^00, so |H/(H ∩ J)| < κ_0 by
    Claim <ref>.  Now for any α∈ S_2,
    
    J ⊆ K_α H ∩ J ⊆ H ∩ K_α⊆ H.

    There are at most 2^|H/(H ∩ J)|≤ 2^κ_0 groups
    between H ∩ J and J, so there are at most 2^κ_0
    possibilities for H ∩ K_α, contradicting the fact that
    |S_2| > 2^κ_0 and the H ∩ K_α are pairwise
    distinct for distinct α.
  
  Now given the claim, we see that the index of H^00 in H can be
  at most κ_0^κ_1.  Indeed, let 𝒮 be the
  collection of ω-definable groups K such that K ⊇
  H^00, and let 𝒮' be a subcollection containing a
  representative K for every possibility of H ∩ K.  By the
  second claim, |𝒮'| < κ_1.  Every type-definable
  group is an intersection of ω-definable groups, so
  
    H^00 = ⋂_K ∈𝒮 K = ⋂_K ∈𝒮 (H ∩ K) = ⋂_K ∈𝒮' (H ∩ K).

  Then there is an injective map
  
    H/H^00↪∏_K ∈𝒮' H/(H ∩ K),

  and the right hand size has cardinality at most
  κ_0^κ_1.  But κ_0^κ_1 is independent of
  H.


  
  Let  be a field of finite dp-rank.  There is a cardinal
  κ with the following property: if M ≼ is any
  small model of cardinality at least κ, and if J is a
  type-definable M-linear subspace of , then J = J^00.
  More generally, if J is a type-definable M-linear subspace of
  ^k, then J = J^00.

Note that we are not assuming J is type-definable over M.

  Take κ as in the Theorem, M a small model of size at least
  κ, and J a type-definable M-linear subspace of .
  For any α∈^×, we have (α· J)^00 =
  α· J^00.  Restricting to α∈ M^×, we see
  that α· J^00 = J^00.  In other words, J^00 is an
  M-linear subspace itself.  The quotient J/J^00 naturally has
  the structure of a vector space over M.  If it is non-trivial, it
  has cardinality at least κ, contradicting the choice of
  κ.  Therefore, J/J^00 is the trivial vector space, and
  J^00 = J.  For the “more generally” claim, apply
  Theorem <ref> to the groups ^k and take the
  supremum of the resulting κ.




§ GENERALITIES ON MODULAR LATTICES
 
Recall that a a lattice is modular if the identity

    (x ∨ a) ∧ b = (x ∧ b) ∨ a

holds whenever a ≤ b.
Modularity is equivalent to the statement that for any a, b, the
interval [a ∧ b, a] is isomorphic as a poset to [b, a ∨ b]
via the maps

    [a ∧ b, a]    → [b, a ∨ b] 
    
      x    ↦ x ∨ b

and

    [b, a ∨ b]    ↦ [a ∧ b, a] 
    
      x    ↦ x ∧ a.



 §.§ Independence
 
Let (P,<) be a modular lattice with bottom element .
 
  A finite sequence a_1, …, a_n of elements of P is
  independent if a_k ∧⋁_i = 1^k-1 a_i =
  for 2 ≤ k ≤ n.

Note that if a and b are independent (i.e., a ∧ b =),
then there is an isomorphism between the poset [, a] and [b, a
  ∨ b] as above.

  If the sequence {a, b, c} is independent, then the sequence
  {a, c, b} is independent.


  By assumption,
  
    b ∧ a    = 
    
        c ∧ (a ∨ b)    = .

  First note that
  
    c ∧ a ≤ c ∧ (a ∨ b) = ,

  so {a,c} is certainly independent.  Let x = b ∧ (a ∨
  c).  We must show that x =.  Otherwise,
  
    < x ≤ b.

  Applying the isomorphism [, b] ≅ [b, a ∨ b],
  
    a < x ∨ a ≤ a ∨ b.

  Applying the isomorphism [, a ∨ b] ≅ [c, a ∨ b ∨
    c],
  
    a ∨ c < x ∨ a ∨ c ≤ a ∨ b ∨ c.

  But x ∨ (a ∨ c) = a ∨ c, as x ≤ (a ∨ c).

 
  Independence is permutation invariant: if {a_1,…,a_n} is
  independent and π is a permutation of [n], then
  {a_π(1),…,a_π(n)} is independent.


  It suffices to consider the case where π is the transposition of
  k and k+1.  Let b = a_1 ∨⋯∨ a_k-1.  Then we know
  
    a_k ∧ b    = 
    
        a_k+1∧ (b ∨ a_k)    =

  and we must show
  
    a_k+1∧ b    ?=
    
        a_k ∧ (b ∨ a_k+1)    ?=.

  This is exactly Lemma <ref>.

 
  Let c_1, …, c_n and a_1, …, a_m be two sequences.  Set
  a_0 = c_1 ∨⋯∨ c_n.  Then the following are
  equivalent:
  
  
  * The sequence c_1, …, c_n is independent and the
    sequence a_0, a_1, …, a_m is independent.
  
  * The sequence c_1, …, c_n, a_1, …, a_m is
    independent.
  


  Trivial after unrolling the definition.


  
  Let a_1, …, a_n be an independent sequence and S be a
  subset of [n].  Let b = ⋁_i ∈ S a_i (understood as
   when S = ∅).  Then {b}∪{a_i  |  i ∉
  S} is an independent sequence.


  Modulo Proposition <ref>, this is the (2)  (1)
  direction of Proposition <ref>.


  
  Let a_1, …, a_n be an independent sequence.  Let S_1,
  …, S_m be pairwise disjoint subsets of [n].  For j ∈ [m]
  let b_j = ⋁_i ∈ S_j a_i, or  if S_j is empty.
  Then b_1, …, b_m is an independent sequence.


  Iterate Remark <ref>.


  
  If a, b, c is an independent sequence, then (a ∨ c) ∧ (b
  ∨ c) = c.


  As c ≤ b ∨ c,
  
    (a ∨ c) ∧ (b ∨ c) = (a ∧ (b ∨ c)) ∨ c = ∨ c = c,

  where the first equality is by modularity, and the second equality
  is by independence of ⟨ b,c,a ⟩.


  
  Let a_1, …, a_n be an independent sequence.  For S ⊆
  [n], let a_S = ⋁_i ∈ S a_i.  Then the following facts
  hold:
  
  
  * a_S ∪ S' = a_S ∨ a_S'.
  
  * a_∅ =.
  
  * a_S ∩ S' = a_S ∧ a_S'.
  
  In other words, S ↦ a_S is a homomorphism of lower-bounded
  lattices from ([n]) to P.


  The first two points are clear by definition.  For the third point, let
  
    S_1    = S ∖ S' 
    
        S_2    = S' ∖ S 
    
        S_3    = S ∩ S'.

  By Lemma <ref>, the sequence a_S_1, a_S_2, a_S_3 is
  independent.  By Lemma <ref>,
  
    a_S ∩ S' = a_S_3!= (a_S_1∨ a_S_3)
        ∧ (a_S_2∨ a_S_3) = a_S_1 ∪ S_3∧ a_S_2
          ∪ S_3 = a_S ∧ a_S'.



  
  In Proposition <ref>, suppose in addition that a_i > for 1 ≤ i ≤ n.  Then the map S ↦ a_S is an isomorphism onto its image:
  
    S ⊆ S'  a_S ≤ a_S'.



  The  direction follows formally from the fact that S
  ↦ a_S is a lattice morphism.  Conversely, suppose S
  ⊈S' but a_S ≤ a_S'.  Take i ∈ S
  ∖ S'.  Then a_i ≤ a_S ≤ a_S', so
  
    a_i = a_i ∧ a_S' = a_S' ∩{i} = a_∅ = ,

  contradicting the assumption.


  
  Let (P,≤) and (P',≤) be two bounded-below modular lattices,
  and f : P → P' be a map satisfying the following conditions:
  
    f(x ∧ y)    = f(x) ∧ f(y) 
    
        f(_P)    = _P'.

  If a_1, …, a_n is an independent sequence in P then
  f(a_1), …, f(a_n) is an independent sequence in P'.


  Note that
  
    x ≤ y  x = x ∧ y  f(x) = f(x) ∧ f(y) 
        f(x) ≤ f(y),

  so f is weakly order-preserving.  In particular, for any x, y
  
    f(x)    ≤ f(x ∨ y) 
    
        f(y)    ≤ f(x ∨ y) 
    
        f(x) ∨ f(y)    ≤ f(x ∨ y).

  Therefore for 2 ≤ k ≤ n we have
  
    f(a_k) ∧⋁_i < k f(a_i) ≤ f(a_k) ∧ f(
        ⋁_i < k a_i ) = f(a_k ∧⋁_i < k
        a_i ) = f(_P) = _P'.





 §.§ Cubes and relative independence
 
We continue to work in a modular lattice (P,≤), but drop the
assumption that a bottom element exists.

  Let b be an element of P.
  
  
  * A sequence a_1, …, a_n of elements of P is
    independent over b if it is an independent sequence in
    the bounded-below modular lattice {x ∈ P  |  x ≥ b}.
    Equivalently, a_1, …, a_n is independent over b if a_i
    ≥ b for each i, and
    
    a_k ∧⋁_i < k a_i = b

    for 2 ≤ k ≤ n.
  
  * A sequence a_1, …, a_n of elements of P is
    co-independent under b if a_i ≤ b for each i, and
    
    a_k ∨⋀_i < k a_i = b

    for 2 ≤ k ≤ n.
  


  
  Let b and c be elements of P.
  
  
  * If a_1, …, a_n is an independent sequence over b,
    then a_1 ∧ c, a_2 ∧ c, …, a_n ∧ c is an
    independent sequence over b ∧ c.
  
  * If a_1, …, a_n is a co-independent sequence under b,
    then a_1 ∨ c, a_2 ∨ c, …, a_n ∨ c is a
    co-independent sequence under b ∨ c.
  


  Part 1 follows from Lemma <ref> applied to the
  function
  
    f : {x ∈ P  |  x ≥ b}   →{x ∈ P  |  x ≥ b ∧ c}
    
        x    ↦ x ∧ c,

  and part 2 follows by duality.

In what follows, we require “lattice homomorphisms” to preserve
∨ and ∧, but not necessarily ⊤ and  when they
exist.

  An n-cube in P is a family {a_S}_S ⊆ [n] of
  elements of P such that S ↦ a_S is a lattice homomorphism
  from ([n]) to P.  A strict n-cube is an n-cube
  such that this homomorphism is injective.  The top and
  bottom of an n-cube are the elements a_∅ and
  a_[n], respectively.

 
  Lattice homomorphisms are weakly order-preserving, and injective
  lattice homomorphisms are strictly order-preserving.

 
  Let b be an element of P.
  
  
  * If {a_S}_S ⊆ [n] is an n-cube with bottom b,
    then a_1, …, a_n is an independent sequence over b.
  
  * This establishes a bijection from the collection of n-cubes
    with bottom b to the collection of independent sequences over
    b of length n.
  
  * If a_1, …, a_n is an independent sequence over b, the
    corresponding n-cube is strict if and only if every a_i is
    strictly greater than b.
  


  
  
  * By definition of n-cube and Remark <ref>,
    
    a_i    ≥ a_∅ = b 
    
          a_i ∧⋁_j < i a_j     = a_i ∧ a_[i-1] = a_i
            ∩ [i-1] = a_∅ = b.

  
  * Injectivity: if {a_S}_S ⊆ [n] and {a'_S}_S
    ⊆ [n] are two n-cubes such that a_S = a'_S when |S|
    ≤ 1, then
    
    a_S = ⋁_i ∈ S a_i = ⋁_i ∈ S a'_i = a'_S

    when |S| > 1.

    Surjectivity: let a_1, …, a_n be a sequence independent
    over b.  By Proposition <ref> applied to the
    sublattice {x ∈ P  |  x ≥ b}, there is an n-cube with
    bottom b.
  
  * Let {a_S}_S ⊆ [n] be the corresponding n-cube.
    If a_i > b for i ∈ [n], then the cube is strict by
    Proposition <ref>.  Conversely, if the cube is
    strict, then a_i > a_∅ = b. 

Dually,

  
  Let b be an element of P.
  
  
  * Let {a_S}_S ⊆ [n] be an n-cube with top b.
    For i ∈ [n] let c_i = a_[n] ∖{i}.  Then c_1,
    …, c_n is a co-independent sequence under b.
  
  * This establishes a bijection from the collection of n-cubes
    with top b to the collection of co-independent sequences under
    b of length n.
  
  * If c_1, …, c_n is a co-independent sequence under b,
    the corresponding n-cube is strict if and only if every c_i is
    strictly less than b.
  




 §.§ Subadditive ranks
 

  The reduced rank of a modular lattice (P,≤) is the
  supremum of n ∈ such that a strict n-cube exists in P,
  or ∞ if there is no supremum.  If a ≥ b are two elements
  of P, the reduced rank (a/b) is the reduced rank of
  the sublattice [b,a] := {x ∈ P  |  b ≤ x ≤ a}.


  
    
  
  
  * Let P_1 and P_2 be two modular lattices.  If the reduced
    rank of P_1 is at least n and the reduced rank of P_2 is at
    least m, then the reduced rank of P_1 × P_2 is at least
    n + m.
  
  *  If P_1 is a modular lattice and P_2 is a
    sublattice, the reduced rank of P_1 is at least the reduced rank
    of P_2.
  
  * If a, b are two elements of a modular lattice (P,≤),
    then there is an injective lattice homomorphism
    
    [a ∧ b, a] × [a ∧ b, b]    → [a ∧ b, a ∨ b] 
    
          (x,y)    ↦ x ∨ y.

  
  *  If a, b are two elements of a modular
    lattice (P,≤) such that (a/a ∧ b) ≥ n and
    (b/ a ∧ b) ≥ m, then (a ∨ b/a ∧ b)
    ≥ n + m.
  


  
  
  * Let {a_S}_S ⊆[n] and {b_S}_S ⊆ [m]
    be a strict n-cube in P_1 and a strict m-cube in P_2.
    Then
    
    (S,T) ↦ (a_S,b_T)

    is an injective lattice homomorphism ([n]) ×([m]) →
    P_1 × P_2.  But ([n]) ×([m]) is isomorphic to
    ([n+m]).
  
  * Any n-cube in P_2 is already an n-cube in P_1.
  
  * The map is clearly a homomorphism of ∨-semilattices.
    
      If (x, y) ∈ [a ∧ b, a] × [a ∧ b, b], then x
      ∨ y = (x ∨ b) ∧ (y ∨ a).
    
    
      By modularity,
      
    (b ∨ x) ∧ a = (b ∧ a) ∨ x.

      Simplifying, we see that a ∧ (x ∨ b) = x.  Because y
      ≤ b, we have x ∨ y ≤ x ∨ b.  By modularity,
      
    (a ∨ (x ∨ y)) ∧ (x ∨ b) = (a ∧ (x ∨ b))
            ∨ (x ∨ y).

      Simplifying, we see that
      
    (y ∨ a) ∧ (x ∨ b) = x ∨ x ∨ y = x ∨ y.

    
    In light of the claim, the map (x,y) ↦ x ∨ y is the
    composition
    
    [a ∧ b, a] × [a ∧ b, b] ∼→ [b,
            a ∨ b] × [a, a ∨ b] → [a ∧ b, a ∨ b]

    where the first map is the product of the poset isomorphisms
    
    [a ∧ b, a]    ∼→ [b, a ∨ b] 
    
          [a ∧ b, b]    ∼→ [a, a ∨ b]

    and the second map is the ∧-semilattice homomorphism
    (x',y') ↦ x' ∧ y'.  As a composition of a poset
    isomorphism and a ∧-semilattice homomorphism, the original
    map in question is a ∧-semilattice homomorphism.  Finally,
    for injectivity, note that the composition of (x,y) ↦ x
    ∨ y with z ↦ (z ∨ b, z ∨ a) is
    
    (x,y) ↦ (x ∨ b, y ∨ a)

    which is the aforementioned isomorphism
    
    [a ∧ b, a] × [a ∧ b, b] ∼→ [b,
            a ∨ b] × [a, a ∨ b].

    Therefore (x,y) ↦ x ∨ y is injective on [a ∧ b,
      a] × [a ∧ b, b].
  
  * This follows by combining the previous three points. 

Work in a modular lattice (P,≤).

  
  Let x ≤ y ≤ z be three elements of P.  If there is a strict
  n-cube {a_S}_S ⊆ [n] in [x,z], then we can write
  n = m + ℓ and find a strict m-cube in [x,y] and a strict
  ℓ-cube in [y,z].


  Passing to the sublattice [x,z], we may assume that , ⊤
  exist and equal x, z respectively.
   
    For any strict inequality b < c in P, at least one of
    the following strict inequalities holds:
    
    b ∧ y    < c ∧ y 
    
          b ∨ y    < c ∨ y.

  
  
    Otherwise, b ∧ y = c ∧ y and b ∨ y = c ∨ y.
    Then
    
    c = (y ∨ c) ∧ c = (y ∨ b) ∧ c = (y ∧ c)
          ∨ b = (y ∧ b) ∨ b = b

    where the middle equality is the modular law.
  
  Take S_0 ⊆ [n] maximal such that a_S_0∧ y =
  a_∅∧ y.  Applying a permutation, we may assume S_0 =
  [ℓ] for some ℓ≤ n.  Let m = n - ℓ.  Note that there
  are injective lattice homomorphisms
  
    ([ℓ])    → P 
    
        S    ↦ a_S 
    ([n] ∖ [ℓ])    → P 
    
        S    ↦ a_S ∪ [ℓ]

  obtained by restricting the original lattice homomorphism S ↦
  a_S.  By Proposition <ref> applied to the
  first of these two cubes, there is a co-independent sequence b_1,
  …, b_ℓ under a_[ℓ] given by
  
    b_i = a_[ℓ] ∖{i},

  and b_i < a_[ℓ] for i ∈ [ℓ].  Similarly, by
  Proposition <ref> applied to the second of
  these two cubes, there is an independent sequence c_1, …, c_m
  over a_[ℓ] given by
  
    c_i = a_[ℓ] ∪{ℓ + i},

  and c_i > a_[ℓ] for i ∈ [m].  Note that
  
    a_∅∧ y ≤ b_i ∧ y ≤ a_[ℓ]∧ y ≤
        c_j ∧ y

  for i ∈ [ℓ] and j ∈ [m].  By choice of S_0, we in fact
  have
  
    a_∅∧ y = b_i ∧ y = a_[ℓ]∧ y < c_j ∧ y.

  Then b_i ∧ y = a_[ℓ]∧ y and b_i < a_[ℓ]
  imply b_i ∨ y < a_[ℓ]∨ y by
  Claim <ref>.  Thus
  
    b_i ∨ y     < a_[ℓ]∨ y 
    
        a_[ℓ]∧ y     < c_j ∧ y

  for all i ∈ [ℓ] and j ∈ [m].  By
  Lemma <ref>, the sequence {b_i ∨ y} is
  co-independent under {a_[ℓ]∨ y}, yielding a strict
  ℓ-cube in [y,⊤].  Similarly, the sequence {c_j ∧
  y} is independent over {a_[ℓ]∧ y}, yielding a
  strict m-cube in [,y].


  
  Let a ≥ b ≥ c be three elements of a modular lattice
  (P,≤).  If (a/b) < ∞ and (b/c) < ∞,
  then
  
    (a/c) ≤(a/b) + (b/c) < ∞.



  
  A subadditive rank on a modular lattice is a function
  assigning a non-negative integer (a/b) to every pair a ≥ b,
  satisfying the following axioms
  
  
  *  (a/b) = 0 if and only if a = b.
  
  * If a ≥ b ≥ c, then
    
    (a/c)    ≤(a/b) + (b/c) 
    (a/c)    ≥(a/b) 
    (a/c)    ≥(b/c).

  
  * If a, b are arbitrary, then
    
    (a/a ∧ b)    = (a ∨ b/b) 
    (b/a ∧ b)    = (a ∨ b/a) 
    (a ∨ b / a ∧ b)     = (a/a ∧ b) + (b/a ∧ b).

  
  A weak subadditive rank is a function satisfying the axioms
  other than (<ref>).

 
  We could also define the notion of a subadditive rank on an abelian
  category.  This should be a function (-) from objects to
  nonnegative integers satisfying the axioms:
  
  
  * (A) = 0 if and only if A = 0.
  
  * If f : A → B is epic (resp. monic) then (A) ≥(B) (resp. (A) ≤(B)).
  
  * In a short exactly sequence
    
    0 → A → B → C → 0,

    we have (B) ≤(A) + (C) with equality if the sequence
    is split.
  
  A weak subadditive rank would satisfy all the axioms except the
  first.  With these definitions, a weak or strong subadditive rank on
  an abelian category 𝒞 should induce a weak or strong
  subadditive rank on the subobject lattice of any A ∈𝒞.[These ideas are developed further in
    <cit.>, Appendix E.]


  
  If (-/-) is a subadditive rank and a_1, a_2, …, a_n is
  relatively independent over b, then
  
    (a_1 ∨⋯∨ a_n/b) = ∑_i = 1^n (a_i/b),

  by induction on n.

 
  Let K be a κ-saturated field of dp-rank n < ∞.  Let
  P be the lattice of subgroups of (K,+) that are type-definable
  over sets of size less than κ.  There is probably a weak
  subadditive rank on P such that _dp(A/B) is the dp-rank of
  the hyper-definable set A/B.  To verify this, we would merely need
  to double-check that subadditivity of dp-rank continues to hold on
  hyperimaginaries.  Although this example motivates the notion of
  “subadditive rank,” we will never specifically need it.


  
  
  * Let P be a modular lattice with a weak subadditive rank.
    Define x ≈ y if (x/x ∧ y) = (y/x ∧ y) =
    0.  Then ≈ is probably a lattice congruence, so we can
    form the quotient lattice P/≈.  The weak rank on P
    should induce a (non-weak) subadditive rank on P/≈.
  
  * In the example of Remark <ref>, G ≈ H
    if and only if G^00 = H^00, and P/≈ is presumably
    isomorphic to the subposet
    
    P' := {G ∈ P  |  G = G^00},

    which is a lattice under the operations
    
    G ∨ H     := G + H 
    
          G ∧ H     := (G ∩ H)^00.

    The fact that dp-rank is weak and P' is not a sublattice of P
    is an annoyance that motivates <ref>.  By taking a
    large enough small model K_0 ≼ K and restricting to the
    sublattice
    
    P” := {G ∈ P  |  G  is a K_0-linear subspace of  K}

    the equation G = G^00 is automatic, and dp-rank presumably
    induces a (non-weak) subadditive rank on P”.
  

 
  Let (P,≤) be a modular lattice.
  
  
  * If the reduced rank (a/b) is finite for all pairs a
    ≥ b, then  is a (non-weak) subadditive rank.
  
  * If there is a subadditive rank  on P, then (a/b)
    ≤(a/b) < ∞ for all a ≥ b.
  


  
  
  * Assume that (a/b) is finite for all pairs a ≥ b.
    We verify the definition of subadditive rank.
    
    
  * Note that a strict 1-cube is just a pair (x,y) with x <
      y.  Therefore, (a/b) ≥ 1 if and only if there exist
      x, y ∈ [b,a] such that x < y, if and only if a > b.
      Conversely, (a/b) = 0 if and only if a = b.
    
  * Suppose a ≥ b ≥ c.  The inequalities
      
    (a/c)    ≥(a/b) 
    (a/c)    ≥(b/c)

      follow by Lemma <ref>.<ref> applied to
      the inclusions [c,b] ⊆ [c,a] and [b,a] ⊆
      [c,a].  The requirement that (a/c) ≤(a/b) +
      (b/c) is Corollary <ref>.
    
  * Let a, b be arbitrary.  By modularity, we have lattice
      isomorphisms
      
    [a ∧ b, a]    ≅ [b, a ∨ b] 
    
            [a ∧ b, b]    ≅ [a, a ∨ b],

      so certainly
      
    (a/a ∧ b)    = (a ∨ b/b) 
    (b/a ∧ b)    = (a ∨ b/a).

      By subadditivity, it remains to show that
      
    (a ∨ b / a ∧ b) ≥(a / a ∧ b) +
            (b / a ∧ b).

      This is Lemma <ref>.<ref>.
    
  
  * Suppose (a/b) ≥ n; we will show (a/b) ≥ n.
    By definition of reduced rank, there is a strict n-cube
    {c_S}_S ⊆ [n] in the interval [b, a].  By
    definition of strict n-cube, we have
    
    c_i ∧ c_[i-1]   = c_∅
    
          c_i ∨ c_[i-1]   = c_[i]
    
          c_i    > c_∅

    for 0 < i ≤ n.  By the axioms of subadditive rank,
    
    (c_[i]/c_∅) = (c_[i-1]/c_∅) +
          (c_i/c_∅) > (c_[i-1]/c_∅).

    By induction on i, we see that (c_[i]/c_∅) ≥ i.
    In particular,
    
    (a/b) ≥(c_[n]/c_∅) ≥ n.

  


  A modular lattice (P,≤) admits a subadditive rank if and only if
  (a/b) < ∞ for all pairs a ≥ b, in which case
  (a/b) is the unique minimum subadditive rank.


  
  Let K be a κ-saturated field of dp-rank n, and P be the
  lattice of subgroups of (K,+) that are type-definable over sets of
  size less than κ.
  
  
  * P need not have finite reduced rank.  For example, if K
    has positive characteristic, there are arbitrarily big cubes made
    of finite subgroups of (K,+).
  
  * Let P' be the subposet of G ∈ P such that G = G^00.
    Then P' should be a modular lattice with lattice operations
    given by G ∧ H = (G ∩ H)^00 and G ∨ H = G + H.
    The map G ↦ G^00 should be a surjective lattice
    homomorphism from P to P'.  The lattice P' should have
    reduced rank at most n by Fact <ref>.
  
  Moreover, the same facts should hold when K is a field of finite
  burden n.  However, G^00 can fail to exist in this
  case, and we can only realize P' as the quotient of P modulo the
  00-commensurability equivalence relation where G ≈ H iff
  G/(G ∩ H) and H/(G ∩ H) are bounded.[These claims
    have now been verified in <cit.>, 6.4.]  In particular,
  we get a weak subadditive rank on the lattice P and a subadditive
  rank on the lattice P' without needing the conjectural
  subadditivity of burden.

We will not directly use Remark <ref>.  Instead,
we will take the following variant approach which avoids all G^00
issues:
 
  Let  be a field, possibly with additional structure.  Assume
   is a monster model, and () = n < ∞.  Let M_0
  be a small submodel as in Corollary <ref>.  Let P
  be the modular lattice of M_0-linear subspaces of ,
  type-definable over small[Here and in what follows, we will
    be a bit sloppy with what exactly “small” means, but this can
    probably be fixed.] parameter sets.  Then P has reduced rank at
  most n.


  Otherwise, there is an (n+1)-cube {H_S}_S ⊆ [n+1] of
  type-definable subgroups which happen to be M_0-linear subspaces
  of .  For i = 1, …, n+1, let G_i = H_[n+1] ∖{i}.  Note that
  
    ⋂_i = 1^n+1 G_i    = H_∅
    ⋂_i  k G_i    = H_k,

  for any k ∈ [n+1], by definition of (n+1)-cube.  Because the
  cube is strict, we have H_∅⊊ H_k, so
  
    ⋂_i = 1^n+1 G_i ⊊⋂_i  k G_i

  for each k.  Both sides are M_0-linear type-definable subspaces.
  By choice of M_0 (i.e., by Corollary <ref>), for
  every k we have
  
    ( ⋂_i = 1^n+1 G_i)^00 = ⋂_i = 1^n+1 G_i
        ⊊⋂_i  k G_i = ( ⋂_i  k G_i
        )^00

  contradicting Fact <ref>.




 §.§ The modular pregeometry on quasi-atoms
 
Work in a modular lattice (P,≤) with bottom element .

  An element a > is a quasi-atom if the set
  
    {x ∈ P  |  < x ≤ a}

  is downwards directed.  Equivalently, if < x ≤ a and < y ≤ a, then < x ∧ y.


  
   
  
  
  * If a is a quasi-atom and < a' < a then a' is a
    quasi-atom.
  
  * Every atom is a quasi-atom.
  
  *  Non-independence is an equivalence relation
    on quasi-atoms.  Two quasi-atoms a, a' are equivalent if a
    ∧ a' >.  This equivalence relation is the transitive
    symmetric closure of ≥ on the set set of quasi-atoms.
  
  *  Suppose a subadditive rank exists.  For every a >,
    there is a quasi-atom a' ≤ a.
  


  
  
  * If < x, y and x,y ≤ a' then x,y ≤ a so x
    ∧ y >.
  
  * If a is an atom, then {x ∈ P  |  < x ≤ a} = {a}
    which is trivially directed.
  
  * By definition, two arbitrary elements a, b are
    non-independent exactly if a ∧ b >.  Now restrict to
    the case of quasi-atoms.  The relation a ∧ a' > is
    clearly symmetric.  It is reflexive because a > is part of
    the definition of quasi-atom.  For transitivity, suppose that a,
    a', a” are quasi-atoms, and a ∧ a' >  < a' ∧
    a”.  Then
    
    a ∧ a”≥ (a ∧ a') ∧ (a”∧ a') >

    where the strict inequality holds because a' is a quasi-atom and
    
    < a ∧ a' ≤ a' 
       < a”∧ a' ≤ a'.

    Thus transitivity holds.  This equivalence relation contains the
    restriction of ≥ to quasi-atoms, because if a and a' are
    quasi-atoms such that a ≥ a', then a ∧ a' = a' >.
    Finally, it is contained in the transitive symmetric closure of
    ≥ because if a ∧ a' >, then a” := a ∧ a'
    is a quasi-atom by part 1, and a ≥ a”≤ a.
  
  * Among the elements of the set {x ∈ P  |  < x ≤ a},
    choose an element a' such that (a'/) is minimal.  We
    claim that a' is a quasi-atom.  Otherwise, there exist x, y ≤
    a' such that x, y >  = x ∧ y.  Then
    
    (a'/) ≥(x ∨ y / ) = (x/) + (y/).

    As y >, it follows that (y/) > 0 and so
    (x/) < (a'/), contradicting the choice of a'. 


  In particular, if K is a field of finite dp-rank and P is the
  lattice of type-definable subgroups G ≤ (K,+) such that G =
  G^00, then every non-zero element G ∈ P has a quasi-atomic
  subgroup G' < G.  The same fact might hold in strongly dependent
  fields, even if a finite subadditive rank is lacking.  Otherwise,
  one can take a counterexample G and split off an infinite
  independent sequence H_1, H_2, … of subgroups of G.  This
  sequence might violate strong dependence.

Say that two quasi-atoms a, a' are equivalent if a ∧ a'
>, as in Proposition <ref>.<ref>.

  
  Let a, a' be equivalent quasi-atoms, and b be an arbitrary
  element.  Then {a,b} is independent if and only if {a',b} is
  independent.


  By Proposition <ref>.<ref>, we may
  assume a ≤ a'.  Then
  
    a' ∧ b =  a ∧ b =

  trivially.  Conversely, suppose a ∧ b = but a' ∧ b
  >.  Note that
  
    {a, a' ∧ b}⊆{x ∈ P  |  < x ≤ a'}.

  As a' is a quasi-atom, we have
  
    = a ∧ b = (a ∧ a') ∧ b = a ∧ (a' ∧
        b) !>.


 
  Let a_1, …, a_n be an independent sequence.  Suppose that
  a_k is a quasi-atom for some k.  Let a'_k be an equivalent
  quasi-atom.  Then a_1, …, a_k-1, a'_k, a_k+1, …, a_n
  is an independent sequence.


  By Proposition <ref> we may assume k = n.  Let b = a_1
  ∨⋯∨ a_n-1.  Then {b, a_n} is independent and it
  suffices to show that {b, a'_n} is independent.  This is
  Lemma <ref>.



In the next few lemmas, we adopt the following notation.  The set of
quasi-atoms will be denoted by Q.  If x ∈ P, then V(x) will
denote the set of a ∈ Q such that a ∧ x >.  Thus Q
∖ V(x) is the set of quasi-atoms independent from x.
 
  V(x) ∩ V(y) = V(x ∧ y) for any x, y ∈ P.


  Let a be a quasi-atom.  Note that
  
    (a ∧ x) ∧ (a ∧ y) > ( (a ∧ x
        > )  and  (a ∧ y > ));

  the left-to-right implication holds generally and the
  right-to-left implication holds because a is a quasi-atom.  The
  left hand side is equivalent to a ∈ V(x ∧ y) and the
  right hand side is equivalent to a ∈ V(x) ∩ V(y).


  
  Let S be a finite subset of Q.
  
  
  * There is a closure operation on S whose closed sets are
    exactly the sets of the form V(x) ∩ S.
  
  *  Let a_1, …, a_n be a sequence of elements
    of S, independent in the sense of Definition <ref>.
    The following are equivalent for b ∈ S:
    
    
  *  b lies in the closure of
      {a_1,…,a_n}.
    
  *  b ∈ V(a_1 ∨⋯∨ a_n).
    
  *  The sequence a_1, …, a_n, b is
      not independent.
    
  
  * If a_1, …, a_n is a sequence in S, there is an
    subsequence b_1, …, b_m having the same closure, and
    independent in the sense of Definition <ref>.
  
  * The closure operation on S satisfies exchange, i.e., it is a
    pregeometry.
  
  * A sequence a_1, …, a_n in S is independent with
    respect to the pregeometry if and only if it is independent in the
    sense of Definition <ref>.
  


  
  
  * Let 𝒞 be the collection of sets of the form V(x)
    ∩ S for x ∈ P.  Then 𝒞 is closed under 2-ary
    intersections by Lemma <ref>.  It is closed under 0-ary
    intersections because if {a_1,…,a_n} is an enumeration of
    S, then every a_i lies in V(a_1 ∨⋯∨ a_n).
    Therefore, 𝒞 is the set of closed sets with respect to
    some closure operation on S.
  
  * The set V(a_1 ∨⋯∨ a_n) ∩ S is closed and
    contains each a_i, so it contains the closure of
    {a_1,…,a_n}.  This shows that
    (<ref>)(<ref>).  Since a_1, …,
    a_n is independent, the sequence a_1, …, a_n, b is
    non-independent if and only if b ∧ (a_1 ∨⋯∨
    a_n) >, if and only if b ∈ V(a_1 ∨⋯∨ a_n).
    Thus (<ref>)(<ref>).  Finally, suppose
    (<ref>) holds.  Let x be an element of P such that
    V(x) ∩ S equals the closure of {a_1,…,a_n}.  Let
    a_i' = a_i ∧ x.  By definition of V(x) these elements are
    non-zero, and a_i' is a quasi-atom equivalent to a_i by
    Proposition <ref>.  By
    Lemma <ref>, the sequence a_1', a_2', …,
    a_n', b is not independent, but a_1', a_2', …, a_n'
    is independent.  Therefore,
    
    < b ∧ (a_1' ∨⋯∨ a_n') ≤ b ∧ x

    because a_i' ≤ x.  Then b ∈ V(x) ∩ S, so
    (<ref>) holds.
  
  * Let b_1, …, b_m be the subsequence consisting of those
    a_i which do not lie in the closure of the preceding a_i's.
    This subsequence has the same closure as the original sequence.
    The sequence {b_1,…,b_j} is independent for j ≤ m, by
    induction on j, using the equivalence
    (<ref>)(<ref>) in the previous point.
  
  * Let a_1, …, a_n, b, c be elements of S.  Suppose that
    b is not in the closure of {a_1, …, a_n}, and
    c is not in the closure of {a_1,…,a_n,b}.  We
    must show that b is not in the closure of
    {a_1,…,a_n,c}.  By the previous point, we may assume that
    the a_i's are independent.  In this case, the equivalence
    (<ref>)(<ref>) in (<ref>) immediately
    implies that the sequence {a_1,…,a_n,b} is independent,
    and then that {a_1,…,a_n,b,c} is independent.  By
    permutation invariance, it follows that {a_1,…,a_n,c} and
    {a_1,…,a_n,c,b} are independent.  By the equivalence,
    this means that b is not in the closure of
    {a_1,…,a_n,c}.
  
  * We proceed by induction on n.  For n = 1, note that
    V() ∩ S = ∅ and so every sequence of length 1 is
    independent with respect to the pregeometry (and vacuously
    independent with respect to the lattice).  Suppose n > 1.  Let
    a_1, …, a_n-1, a_n be a sequence.  We may assume that
    a_1, …, a_n-1, a_n is independent with respect to the
    pregeometry or the lattice.  Either way, a_1, …, a_n-1 is
    independent with respect to the pregeometry or the lattice, hence
    with respect to the lattice or the pregeometry (by induction).
    Then a_1, …, a_n is independent with respect to the
    pregeometry if and only if a_n is not in the closure of a_1,
    …, a_n-1.  By (<ref>)'s equivalence, this is the same
    as a_1, …, a_n being independent with respect to the
    lattice. 


  
  Let Q be the set of quasi-atoms in (P,≤).
  
  
  * There is a finitary pregeometry on the set Q of quasi-atoms,
    characterized by the fact that a finite set {a_1,…,a_n}
    of quasi-atoms is independent with respect to the pregeometry if
    and only if it is independent in the lattice-theoretic sense of
    Definition <ref>.
  
  * If x is any element of P, then the set V(x) = {a ∈ Q  | 
    a ∧ x > } is a closed subset of Q.
  
  * Two quasi-atoms a, b are parallel (i.e., have the same
    closure) if and only if they are equivalent in the sense of
    Proposition <ref>.  Consequently, there is an
    induced geometry on the equivalence classes of quasi-atoms.
  


  To specify a finitary closure operation on an infinite set Q, it
  suffices to give a closure operation on each finite subset S
  ⊆ Q, subject to the compatibility requirements that when
  S' ⊆ S, the closed sets on S' are exactly the sets of
  the form V ∩ S' for V a closed set on S.  We clearly have
  this compatibility.  Furthermore, the induced finitary closure
  operation on Q satisfies exchange if and only if the closure
  operation on each finite S ⊆ Q satisfies exchange.  When
  this holds, a finite set I is independent with respect to the
  pregeometry on Q if and only if it is independent with respect to
  the pregeometry on S, for any/every finite S containing I.
  Moreover, a set V ⊆ Q is closed if and only if V ∩ S
  is a closed subset of S for every finite S ⊆ Q.
  Therefore, the sets V(x) are certainly closed.  For the final
  point, it follows on general pregeometry grounds that two
  non-degenerate elements a, b are parallel if and only if {a,b}
  is independent.  Because independence in the pregeometry agrees with
  lattice-theoretic independence, we see that a, b are parallel if
  and only if they are equivalent in the sense of
  Proposition <ref>.


  
  Let x, a, b be three elements of P, such that the sets {x,
  a}, {x, b}, {a,b} are independent, but {x, a, b} is not
  independent.  Suppose that a is a quasi-atom.  Then there is a
  quasi-atom w ≤ x such that {w, a, b} is not independent (but
  {w, a}, {w, b}, {a, b} are independent).


  Let w = (a ∨ b) ∧ x.  As {a, b} is independent but
  {a, b, x} is not, we must have w >.  As w ≤ x it
  follows that {w, b} and {w, a} are independent.  From the
  independence of {b, w}, there is an isomorphism of posets
  
    (,w] ≅ (b, b ∨ w].

  Now b ∨ w ≤ a ∨ b because w ≤ a ∨ b (by definition)
  and b ≤ a ∨ b.  Therefore (b, b ∨ w] ⊆ [b, a ∨
      b].  By the independence of a and b, there is a poset
    isomorphism
  
    [b, a ∨ b]    ∼→ [, a] 
    
        x    ↦ x ∧ a.

  This isomorphism induces an isomorphism of the subposets
  
    (b, b ∨ w] ≅ (, a ∧ (b ∨ w)].

  In particular, there is a chain of poset isomorphisms
  
    (,w] ≅ (b, b ∨ w] ≅ (, a ∧ (b ∨ w)].

  The left hand side is non-empty, because w >.  Therefore, the
  right hand side is non-empty, and so a' := a ∧ (b ∨ w) is
  greater than .  On the other hand, a' is less than or equal
  to the quasi-atom a, so a' is a quasi-atom.  Therefore, the
  right hand side is a downwards directed poset.  The same must hold
  for the left hand side, implying that w is a quasi-atom.  Note
  that w ∧ (a ∨ b) = w >, so the sequence {a, b, w}
  is not independent.


 
  The pregeometry of Corollary <ref> is
  modular.


  
    
    Let a, b, c_1, …, c_n be elements of Q, with n > 0.
    Suppose that a ∈{b, c_1, …, c_n}.  Then there is
    c' ∈{c_1, …, c_n} such that a ∈{b, c'}.
  
  
    We may assume that the c_i are independent, passing to an
    independent subsequence otherwise[This preserves n > 0
      because no quasi-atom is in the closure of the empty set, as
      sequences of length 1 are always independent].  We may assume
    that a ∉{b}, or else take c' = c_1.  We may assume
    that a ∉{c_1,…,c_n}, or else take c' = a.
    Then b ∉{c_1,…,c_n}.  It follows that the
    sequences
    
    
  * {a, b}
    
  * {c_1,…,c_n,a}
    
  * {c_1,…,c_n,b}
    
    are independent, but the sequence {c_1, …, c_n, a, b} is
    not independent.  Let x = c_1 ∨⋯∨ c_n.  Then
    
    a ∧ b    = 
    
          a ∧ x    = 
    
          b ∧ x    =

    but {a, b, x} is not independent.  By
    Lemma <ref>, there is a quasi-atom w ≤ x such
    that {a, w}, {b, w}, {a, b} are independent but {a, b,
    w} is not. Now w ≤ x = c_1 ∨⋯∨ c_n, so the
    sequence {c_1, …, c_n, w} is not independent.  Therefore
    w ∈{c_1,…,c_n}.  The fact that {b, w} is
    independent but {a, b, w} is not implies that a ∈{b,
    w}.  Take c' = w.
  
  Given the claim, modularity follows on abstract grounds.  First, one
  upgrades the claim to the following:
  
    If X, Y are two non-empty closed subsets of Q and a ∈(X
    ∪ Y), then there is x ∈ X and y ∈ Y such that a ∈{x,y}.
  
  
    Because closure is finitary, there exist b_1, …, b_n ∈ X
    and c_1, …, c_m ∈ Y such that a ∈{b_1,…,b_n,c_1,…,c_m}.  By repeated applications
    of the previous claim one can find
    
    
  * p_1 ∈{b_2, …, b_n, c_1, …, c_m} such
      that a ∈{b_1,p_1}.
    
  * p_2 ∈{b_3, …, b_n, c_1, …, c_m} such
      that p_1 ∈{b_2,p_2}.
    
  * …
  * p_k ∈{b_k+1,…,b_n,c_1,…,c_m} such that
      p_k-1∈{b_k,p_k}.
    
  * …
  * p_n ∈{c_1,…,c_m} such that p_n-1∈{b_n,p_n}.
    
    Then {b_1, b_2, …, b_n, p_n} contains p_n-1, hence
    also p_n-2, p_n-3, …, p_2, p_1, and a.  Thus a ∈{b_1,b_2,…,b_n,p_n}.  By one final application of
    Claim <ref>, there is q ∈{b_1,b_2,…,b_n} such that a ∈{p_n,q}.  Take
    x = q and y = p_n.
  
  Finally, let V_1, V_2, V_3 be closed subsets of Q with V_1
  ⊆ V_2.  Write + for the lattice join on the lattice of
  closed sets.  We must show one direction of the modular equation:
  
    (V_1 + V_3) ∩ V_2 ⊆ V_1 + (V_3 ∩ V_2)

  as the ⊇ direction holds in any lattice.  If V_1 or
  V_3 is empty, the equality is clear, so we may assume both are
  non-empty.  Take x ∈ (V_1 + V_3) ∩ V_2.  By the previous
  claim, there are (y,z) ∈ V_1 × V_3 such that x ∈{y,z}.  If x ∈{y}, then x ∈ V_1 so certainly x
  ∈ V_1 + (V_3 ∩ V_2).  Otherwise, x ∈{y,z}∖{y}, so z ∈{x,y}.  Now x ∈ V_2 by choice of x,
  and y ∈ V_1 ⊆ V_2.  Therefore, z ∈{x,y}⊆ V_2.  The element z is also in V_3, so z ∈ (V_3
  ∩ V_2).  Thus x ∈{y,z} where y ∈ V_1 and z ∈
  V_3 ∩ V_2.  It follows that x ∈ V_1 + (V_3 ∩ V_2),
  completing the proof of modularity.

In a later paper (<cit.>), we will see that equivalence classes
of quasi-atoms in M correspond to atoms in the category M
obtained by formally adding filtered infima to M.  The category
M is itself a modular lattice, and the modular pregeometry
constructed in Corollary <ref> and
Proposition <ref> comes from the usual modular
geometry on atoms.  See 7.1-7.2 of <cit.> for details.


  Let P be the modular lattice of type-definable subgroups of
  (K,+) in a dp-finite field K.  Or let P be an interval inside
  that lattice.  In these cases, the modular geometry on quasi-atoms
  in P should consist of finitely many pieces, each of which is
  canonically a projective space over a division ring.  In particular,
  
  
  * Exotic non-Desarguesian projective planes shouldn't appear.
  
  * Even if a component has rank 1, it should still carry the
    structure of ^1(D) for some division ring D.
  
  These nice properties should hold because of the fact that we can
  embed P into a larger lattice P_n of type-definable subgroups of
  K^n.  The existence of the family of P_n's for n ≥ 1 should
  ensure that all the connected components of the modular geometry can
  be embedded into connected modular geometries of arbitrarily high
  rank, forcing Desargues' theorem to hold.  Similar statements should
  hold in the abstract setting of
  Remark <ref>.[All these claims
    have now been verified in <cit.> 2.2 and 7.4.]




 §.§ Miscellaneous facts


 
  Let (P,≤) be a modular lattice with top and bottom elements
  ⊤ and .  Suppose that P admits a subadditive rank.
  
  
  * The rank of the modular pregeometry on quasi-atoms is at most
    the reduced rank (⊤/).  In particular, it is
    finite.
  
  * Every closed set is of the form V(x) for some x.
  
  * Let x be any element of P, and y_1, …, y_n be a
    basis for V(x).  Let z_1, …, z_m be a sequence of
    quasi-atoms.  Then {x, z_1, …, z_m} is independent if and
    only if 
    {y_1, …, y_n, z_1, …, z_m}
 is independent.
  
  *  If x is any element of P, there is a pregeometry basis of
    the form y_1, …, y_n, z_1, …, z_m where
    
    
  * Each y_i ≤ x.
    
  * The set {y_1,…,y_n} is a basis for V(x).
    
  * The set {x, z_1, …, z_m} is independent.
    
  


  
  
  * If the pregeometry rank is at least n, then there is a
    sequence of independent quasi-atoms a_1, …, a_n.  Each
    a_i is greater than , so this determines a strict n-cube
    by Proposition <ref>.
  
  * Let V be a closed set.  Because the pregeometry rank is
    finite, we can find a basis a_1, …, a_n for V.  Note that
    {a_1,…,a_n} is an independent set.  Let x = a_1 ∨⋯∨ a_n.  If b is any element of Q, then the
    following statements are equivalent
    
    
  * b is in V.
    
  * b is in the closure of {a_1,…,a_n}.
    
  * The set {a_1,…,a_n,b} is not independent.
    
  * b ∧ x >.
    
  * b ∈ V(x).
    
    Therefore V = V(x).
  
  * Let y'_i = y_i ∧ x.  By definition of V(x), each
    y'_i >, and so y'_i is a quasi-atom equivalent to y_i.
    By Lemma <ref> we may replace y_i with
    y'_i and assume that each y_i ≤ x.  Let x' = y_1 ∨⋯∨ y_n ≤ x.  If {x, z_1, …, z_m} is
    independent, then so is {x', z_1, …, z_m}.  As the y_i
    are independent, it follows by Proposition <ref> that
    {y_1, …, y_n, z_1, …, z_m} is independent.
    Conversely, suppose that {y_1,…,y_n,z_1,…,z_m} is
    independent, but {x,z_1,…,z_m} is not.  Choose m
    minimal such that this occurs.  Let b = z_1 ∨⋯∨
    z_m-1.  Then (x ∨ b) ∧ z_m >, but
    
    x ∧ b    = 
    
          b ∧ z_m    = 
    
          x ∧ z_m    =

    respectively: by choice of m; because z⃗ is independent;
    and because z_m is independent from y_1,…,y_n hence not
    in {y_1,…,y_n} = V(x).  By Lemma <ref>
    there is a quasi-atom w ≤ x such that {w, b, z_m} is not
    independent.  But w ∈ V(x), so w ∈{y_1,…,y_n}.
    The fact that {y_1,…,y_n,z_1,…,z_m-1,z_m} is
    independent then implies on pregeometry-theoretic grounds that
    {w,z_1,…,z_m-1,z_m} is independent, which implies on
    lattice-theoretic grounds that {w, b, z_m} is independent, a
    contradiction.
  
  * Take {y_1, …, y_n} a basis for V(x) and extend it
    to a basis {y_1,…,y_n,z_1,…,z_m} for the entire
    pregeometry.  As in the proof of the previous point, we may
    replace y_i with y_i ∧ x, and thus assume that y_i ≤
    x.  By the previous point, independence of
    {y_1,…,y_n,z_1,…,z_m} implies independence of
    {x,z_1,…,z_m}. 


 
  Let (P,≤) be a modular lattice with bottom element  and
  reduced rank n < ∞.  The pregeometry on quasi-atoms has rank
  at most n.  If a_1, …, a_m is a basis, then the following
  facts hold:
  
  
  * Any independent sequence {b_1, …, b_ℓ} of
    non- elements has size ℓ≤ m.
  
  * If ℓ = m then each b_i is a quasi-atom.
  


  
  
  * For each b_i, we may find a quasi-atom b'_i ≤ b_i, by Proposition <ref>.<ref>.  The
    sequence b'_1, …, b'_ℓ is pregeometry-independent, so
    ℓ≤ m.
  
  * Suppose ℓ = m.  If, say, b_ℓ is not a quasi-atom, we
    can find x, y ∈ (,b_ℓ] such that x ∧ y =.
      Replacing b_ℓ with x ∨ y ≤ b_ℓ, and then applying
      Proposition <ref>, we see that b_1, …, b_ℓ
        - 1, x, y is an independent sequence contradicting part 1. 



  If (P,≤) is a modular lattice and a ≥ b are elements, we
  define (a/b) to be the supremum of n such that there is a
  strict n-cube in [b,a] with bottom b.

Equivalently (by Proposition <ref>),
(a/b) is the supremum over n such that there exist c_1,
…, c_n ∈ (b,a] relatively independent over b.  The rank
  (-) is not a subadditive rank in the sense of
  Definition <ref>.
 
  If a ≥ b, then
  
  
  *  (a/b) ≤(a/b)
  
  * If (a/b) < ∞, there exists b' ∈ [b,a] such
    that (a/b') = (a/b').  Indeed, take a strict
    n-cube in [b,a] for maximal n and let b' be the bottom of
    the cube.
  
  * If (a/b) < ∞, then (a/b) is the rank of
    the pregeometry of quasi-atoms in [b,a], by
    Remark <ref>.
  

 
  For any a, b
  
    (a ∨ b / a ∧ b) ≥(a/a ∧ b) + (b/a ∧ b).

  Also, (a ∨ b/a) = (b/a ∧ b).


  If c_1, …, c_n is a strict independent sequence in [a ∧
    b, a] and d_1, …, d_m is a strict independent sequence in
  [a ∧ b, b], then c_1, …, c_n, d_1, …, d_m is a
  strict independent sequence in [a ∧ b, a ∨ b] by a couple
  applications of Proposition <ref>.  The “also” clause
  is unrelated and follows immediately from the isomorphism of
  lattices between [a, a ∨ b] and [a ∧ b, b].







§ INVARIANT VALUATION RINGS
 
Let (,+,·,…) be a monster-model finite dp-rank expansion
of a field.  Assume that  is not of finite Morley rank.  Fix a
small model M_0 large enough for Corollary <ref> to
apply.  Thus, for any type-definable M_0-linear subspace J ≤^n, we have J = J^00.

Let 𝒫_n be the poset of type-definable M_0-linear
subspaces of ^n, let 𝒫 = 𝒫_1, and let
𝒫^+ be the poset of non-zero elements of 𝒫.

We collect the basic facts about these posets in the following
proposition:

  
   
  
  
  * For each n, 𝒫_n is a bounded lattice.
  
  *  For any small model M ⊇ M_0,
    the group I_M is an element of 𝒫.  In particular,
    𝒫 contains an element other than = 0 and ⊤
    =.
  
  * If J ∈𝒫 is non-zero, every definable set D
    containing J is heavy.
  
  *  If J ∈𝒫^+ is
    M-definable for some small model M ⊇ M_0, then J
    ⊇ I_M.
  
  * If J ∈𝒫_n, then J = J^00.
  
  *  𝒫^+ is a sublattice of
    𝒫, i.e., it is closed under intersection.  Thus,
    𝒫^+ is a bounded-above lattice.
  
  *  𝒫 has reduced rank r for
    some 0 < r ≤().  The reduced rank of 𝒫^+
    is also r, and the reduced rank of 𝒫^n is rn.
  


  
  
  * Clear—the lattice operations are given by
    
    G ∨ H    = G + H 
    
          G ∧ H    = G ∩ H 
       = 0 
    ⊤   = ^n.

  
  * By Remark <ref> and Theorem <ref>, I_M is
    a non-zero M_0-linear subspace of , distinct from 0 and
    .
  
  * Replacing J with a · J for some a ∈^×, we
    may assume 1 ∈ J.  Then M_0 = M_0 · 1 ⊆ J by
    M_0-linearity.  Let D be any definable set containing J.  By
    Lemma <ref> with Z = and W = D, the
    set D is heavy.
  
  * Corollary <ref>.
  
  * By choice of M_0.
  
  * Let J_1, J_2 be two non-zero elements of 𝒫.  Let
    M be a small model containing M_0, over which both J_1 and
    J_2 are type-definable.  Then J_1 ∩ J_2 ≥ I_M >.
    Therefore 𝒫^+ is closed under intersection.
  
  * Proposition <ref> gives the bound r ≤
    n.  Then
    
    0 < (𝒫^+) ≤(𝒫) = r ≤ n,

    where the left inequality is sharp because 𝒫 has at
    least three elements by part <ref>.  If r >
    (𝒫^+), there is a strict r-cube in
    𝒫 which does not lie in 𝒫^+.  The bottom
    of this cube must be , the only element of 𝒫∖𝒫^+.  Then r ≤(𝒫).
    However, part <ref> says (𝒫)
    ≤ 1, so r ≤ 1 ≤(𝒫^+), a contradiction.
    Therefore (𝒫^+) = r = (𝒫).
    Finally, in 𝒫_n, if we let J_i = 0^⊕ (i - 1)⊕⊕ 0^⊕ (n - i) for i = 1, …, n, then
    the sequence J_1, …, J_n is independent and J_1 ∨⋯∨ J_n = ^n.  Thus
    
    (^n/0) = ∑_i = 1^n (J_i/0)

    by Remark <ref>.  However, (J_i/0)
    = r because of the isomorphism of lattices
    
    𝒫   → [0,J_i] 
    
          X    ↦ 0^⊕(i-1)⊕ X ⊕ 0^⊕(n-i).

  


In what follows, we will let r be (/0).

  If r = 1, then 𝒫 is totally ordered and we can reuse
  the arguments for dp-minimal fields to immediately see that I_M is
  a valuation ideal.  Usually we are not so lucky.




 §.§ Special groups
 

  An element J ∈𝒫_n is special if
  (^n/J) = (^n/J) = rn.

Equivalently, J ∈𝒫_n is special if (^n/J) ≥
rn.  This follows by Proposition <ref>.<ref> and Remark <ref>.<ref>.

  
   
  
  
  * There is at least one non-zero special J ∈𝒫 =
    𝒫_1.
  
  *  Let J ∈𝒫 be special.  Let A_1,
    …, A_r be a basis of quasi-atoms over J.  Let G ∈𝒫 be arbitrary.  If G ∩ A_i ⊈J for
    each i, then G ⊇ J.
  
  *  If J ∈𝒫 is special
    and nonzero and type-definable over a small model M ⊇
    M_0, then
    
    I_M · J ⊆ I_M ⊆ J

  
  *  If I ∈𝒫_n and J ∈𝒫_m are special, then I ⊕ J ∈𝒫_n+m is special.
  
  *  If I ∈𝒫_n is special and
    α∈^×, then α· I is special.
  


  
  
  * By Proposition <ref>.<ref> the
    reduced rank of 𝒫^+ is exactly r, so we can find a
    strict r-cube in 𝒫^+.  The base of such a cube is a
    non-zero special element of 𝒫.
  
  * For i = 1, …, r take a_i ∈ (G ∩ A_i) ∖
    J.  Then for each i, we have
    
    a_i ∈ G ∩ A_i ⊆ (G + J) ∩ A_i ⊇ J ∩ A_i = J 
        a_i ∉ J 
    
          ()     (G + J) ∩ A_i ⊋ J.

    Consequently,
    
    A_1 ∩ (G + J), A_2 ∩ (G + J), …, A_r ∩ (G + J)

    is a strict independent sequence over J.  It follows that
    
    (G/(G ∩ J)) = ((G+J)/J) ≥((G+J)/J) ≥ r.

    On the other hand
    
    (G/(G ∩ J)) + (J/(G ∩ J)) = ((G+J)/(G ∩ J)) ≤ r,

    and so (J/(G ∩ J)) = 0.  This forces J = G ∩ J, so
    J ⊆ G.
  
  * The inclusion I_M ⊆ J is
    Proposition <ref>.<ref>. Let
    A_1, …, A_r be a basis of quasi-atoms over J as in part
    <ref>.  For each A_i choose an element a_i ∈
    A_i∖ J.  Let M' be a small model containing M and the
    a_i's.  We first claim that I_M'· J ⊆ I_M'.
    Let ε be a non-zero element of I_M'.  As I_M' is
    closed under multiplication by (M')^×, we have M'
    ⊆ε^-1· I_M'.  In particular, a_i ∈ε^-1· I_M' for each i.  Then
    
    (ε^-1· I_M') ∩ A_i ⊈J

    for each i, so by part <ref> we have
    
    ε^-1· I_M'⊇ J.

    In other words, ε· J ⊆ I_M'.  As
    ε was an arbitrary non-zero element of I_M', it
    follows that I_M'· J ⊆ I_M'.  Now suppose that
    D is an M-definable basic neighborhood.  Then D is an
    M'-definable basic neighborhood.  By the above and compactness,
    there is an M'-definable basic neighborhood X  X and a
    definable set D_2 ⊇ J such that (X  X) · D_2
    ⊆ D.  Furthermore, D_2 can be taken to be
    M-definable, because J is a directed intersection of
    M-definable sets.  Having done this, we can then pull the
    parameters defining X into M, and assume that X is
    M-definable.  (This uses the fact that heaviness is definable in
    families).  Then we have an M-definable basic neighborhood X
     X and an M-definable set D_2 ⊇ J such that
    (X  X) · D_2 ⊆ D.  As D was arbitrary, it
    follows that
    
    I_M · J ⊆ I_M.

  
  * The interval [I,^n] in 𝒫_n is isomorphic to
    [I ⊕ J, ^n ⊕ J] in 𝒫_n+m, so
    
    rn = (^n/I)    = ((^n ⊕ J)/(I ⊕ J)) 
    
          rm = (^m/J)    = ((I ⊕^m)/(I ⊕ J)),

    where the second line is true for similar reasons.  By
    Lemma <ref>,
    
    ((^n ⊕^m)/(I ⊕ J)) ≥ rn + rm.

    On the other hand
    
    ((^n ⊕^m)/(I ⊕ J)) ≤((^n
          ⊕^m)/(I ⊕ J)) ≤ r(n+m)

    so equality holds and I ⊕ J is special.
  
  * For any α∈^×, the map X ↦α· X is an automorphism of 𝒫_n. 

 
  For any model M, I_M · I_M ⊆ I_M.  The topology in
  Remark <ref> is a ring topology, not just a group
  topology.


  First suppose M ⊇ M_0.  Let J be a non-zero special
  element of 𝒫_1.  Then I_M ⊆ J, and so
  
    I_M · I_M ⊆ I_M · J ⊆ I_M

  by Proposition <ref>.<ref>.
  Then we can shrink M using the technique of the proof of
  Proposition <ref>.<ref>.

With much more work, one can show that the canonical topology is a
field topology.  See Corollary 5.15 in <cit.>.

  Say that J ∈𝒫_1 is bounded if J ≤ J' for
  some special J'.  Based on the argument in
  Proposition <ref>.<ref>-<ref>,
  it seems that J is bounded if and only if α· J
  ⊆ I_M for some α∈^× and some small model
  M.  Bounded elements should form a sublattice of
  𝒫_1.[These ideas have been developed in 8 of
    <cit.>.]





  Let I ∈𝒫_n be special and D ∈𝒫_n be
  arbitrary.  Then D dominates I if D ≥ I and
  (D/I) = nr.


  
  Let J ∈𝒫_n be special, let A_1, …, A_nr be a
  basis of quasi-atoms in [J,^n], and D be arbitrary.  Then D
  dominates J if and only if D ∩ A_i ⊋ J for each i.
  In particular, this condition doesn't depend on the choice of the
  basis {A_1,…,A_nr}.


  Suppose D dominates J.  Let B_1, …, B_nr be a basis of
  quasi-atoms in [J,D].  The B_i are independent quasi-atoms in
  the larger interval [J,^n], so {B_1,…,B_nr} is
  another basis of quasi-atoms in [J,^n].  Therefore, for every
  i the sequence {B_1,…,B_nr,A_i} is not
  independent over J.  Consequently
  
    D ∩ A_i ⊇ (B_1 + ⋯ + B_nr) ∩ A_i ⊋
        J.

  Conversely, suppose D ∩ A_i ⊋ J for each i.  Then
  certainly D ⊇ J, and it remains to show (D/J) ≥
  nr.  Let A'_i := D ∩ A_i.  Then each A'_i is a quasi-atom
  over J, equivalent to A_i, and so {A'_1,…,A'_nr} is
  another basis of quasi-atoms over J.  As each A'_i lies in
  [J,D], it follows that (D/J) ≥ nr.




  
  Let I ∈𝒫_n be special, and V be a k-dimensional
  -linear subspace of ^n.  Then
  
    ((V + I)/I)    = ((V + I)/I) = kr 
    (V/(V ∩ I))    = (V/(V ∩ I)) = kr 
    (^n/(V + I))    = (^n/(V + I)) = (n-k)r.

  Moreover, there exist A_1, …, A_kr, B_1, …, B_(n-k)r∈𝒫_n such that the following conditions hold:
   
  
  * The set {A_1, …, A_kr, B_1, …, B_(n-k)r} is
    a basis of quasi-atoms in [I, ^n].
  
  * Let Ã_i = A_i ∩ V.  Then {Ã_1, …,
    Ã_kr} is a basis of quasi-atoms in [V ∩ I, V].
  
  * Let B̃_i = B_i + V.  Then {B̃_1, …,
    B̃_(n-k)r} is a basis of quasi-atoms in [V + I,
      ^n].
  
  Given a D ∈𝒫_n dominating I, we may choose the
  A_i and B_i to lie in [I,D].


  Let W be a complementary (n-k)-dimensional -linear
  subspace, so that V + W = ^n.  Let V' = V + I and W' = W+I.
  Then
  
    nr = (^n/I)    ≤(^n/V') + (V'/I) 
        = ((V'+W')/V') + (V'/I) 
        = (W'/(W' ∩ V')) + (V'/I) 
       ≤(W'/I) + (V'/I) 
        = (W/(W ∩ I)) + (V/(V ∩ I)) 
       ≤(W/0) + (V/0).

  Now any -linear isomorphism ϕ : ^k ∼→
  V induces an isomorphism of posets from 𝒫_k to [0,V]
  ⊆𝒫_n, so
  
    (V/0)    = (𝒫_k) = kr 
    (W/0)    = (𝒫_n-k) = (n-k)r,

  where the second line follows similarly.  Therefore the inequalities
  above are all equalities, and
  
    ((V+I)/I)    = (V/(V ∩ I)) = kr 
    (^n/(V+I))    = (^n/V') = (n-k)r.

  By Proposition <ref>.<ref>, there
  is a basis {A_1, …, A_m, B_1, …, B_nr-m} of
  independent quasi-atoms in [I, ^n] such that
  
  
  * Each A_i ⊆ V+I.
  
  * The sequence (V+I), B_1, …, B_nr-m is independent
    over I.
  
  Given a bound D dominating I, we may replace each A_i with
  A_i ∩ D and B_i with B_i ∩ D, and assume henceforth that
  A_i, B_i ⊆ D.  By Remark <ref>,
  
    nr = (^n/I)    ≥((V+I)/I) + (B_1/I) + ⋯
        + (B_nr-m/I) 
        = kr + (B_1/I) + ⋯ + (B_nr-m/I).

  Each B_i is strictly greater than I, so
  
    nr ≥ kr + nr - m,

  and thus m ≥ kr.  On the other hand, the set {A_1, …,
  A_m} is a set of independent quasi-atoms in [I, V + I], so
  
    m ≤((V+I)/I) ≤((V+I)/I) = kr.

  Thus equality holds, m = kr, and the set {A_1, …, A_m} is
  a basis of quasi-atoms in [I, V + I].  Applying the isomorphism
  
    [I, V + I]    ∼→ [V ∩ I, V] 
    
        X    ↦ X ∩ V

  shows that the Ã_i form a basis of quasi-atoms in [V ∩
    I, V].  Next, let Q = B_1 ∨⋯∨ B_(n-k)r.  (Note
  that nr - m = (n-k)r.)  The fact that (V+I), B_1, …,
  B_(n-k)r is independent over I implies that (V+I) ∩ Q = I.
  Therefore, there is an isomorphism
  
    [I,Q]    ∼→ [V+I,V+I+Q] 
    
        X    ↦ X + (V + I) = X + V.

  The elements {B_1, …, B_(n-k)r} are independent
  quasi-atoms in [I,Q], and therefore the B̃_i are a set of
  independent quasi-atoms in [V+I,V+I+Q] or even in [V+I,^n].
  It follows that
  
    (n-k)r ≤(^n/(V+I)) ≤(^n/(V+I)) = (n-k)r,

  so equality holds and the B̃_i are a basis of quasi-atoms
  in [V+I,^n].


 
  Fix a special element I of 𝒫_1.  For every n, I^n
  is a special element of 𝒫_n.  Let 𝒢_n be the
  modular geometry associated to the pregeometry of quasi-atoms over
  I^n in 𝒫_n.  By utilizing the embeddings between the
  𝒢_n, one should be able to prove the following: there is
  an object T in a semi-simple abelian category 𝒞 such
  that T has length r, and for every n the geometry
  𝒢_n is the geometry of atoms in the lattice of
  subobjects of T^n.  The category 𝒞 should be enriched
  over M_0-vector spaces.  For every n there is a map f_n from
  the lattice of -linear subspaces of ^n to the lattice of
  subobjects of T^n, defined in such a way that f_n(V) corresponds
  to the closed set cut out by V + I^n in the pregeometry of
  quasi-atoms in [I^n,^n].  These maps f_n should have the
  following properties:
  
  
  * V ≤ W should imply f_n(V) ≤ f_n(W).
  
  * The length of f_n(V) should be r times the dimension of
    V, essentially by Lemma <ref>.  In particular,
    f_1(0) = 0 and f_1() = T.
  
  * For each n, the map f_n should be GL_n(M_0)-equivariant.
  
  * f_n(V) ⊕ f_m(W) should equal f_n+m(V ⊕ W).
  
  When r = 1, this should secretly amount to a valuation on .
  Indeed, T is then a simple object so we may assume that
  𝒞 is the category of finite-dimensional D-modules for
  some skew field D extending M_0, and T is just D^1.  Then
  the above configuration should necessarily come from a map →
  k → D where → k is a valuation/place, and k → D is an
  inclusion.  The hope was that similar configurations for r > 1
  should come from multi-valued fields, but several things go wrong.
  
  
  * The idea is to recover a valuation on  where the residue
    of a is α∈ End(T) if
    
    f_2({(x,ax)  |  x ∈}) = {(x, α x)  |  x ∈ T}.

    This line of thinking is what underlies the ring R_J and ideal
    I_J in Proposition <ref> below–R_J is the
    set of a such that the left hand side is the graph of any
    endomorphism of T and I_J is the set of a such that the left
    hand side is the graph of 0.
  
  * Ideally, we would get R_J equal to a Bezout domain with
    finitely many maximal ideals.  This amounts to the requirement
    that for any α∈ and any distinct q_0, …, q_n
    ∈ M_0, at least one of 1/(α - q_i) lies in R_J.
  
  * However, this fails when the system of maps looks like
    
    (V ≤ ((√(2)))^n) ↦ (V + V^†, V ∩ V^†),

    where the right hand side should be thought of as an element of
    FinVec() × FinVec(), and V^† is the image of V
    under the non-trivial element of ((√(2))/).
  
  * There is a certain way to mutate f that improves the
    situation, namely
    
    f'_n(V) := f_2n({(x⃗,a ·x⃗)  | x⃗∈ V})

    for well-chosen a, e.g., a = √(2).  This mutation is what
    underlies Lemma <ref> below.
  
  * Originally, I had hoped that finitely many mutations would
    reduce to the “good” Bezout case, but there are genuine
    type-definable J in dp-finite fields for which this fails.  For
    example, if Γ is the group (1/3) and K is
    ((Γ)), we can make K have dp-rank 2 by adding a
    predicate for the subfield ((2 ·Γ)).  Any Hahn
    series x ∈ K can be decomposed as x_odd + x_even where
    x_odd has the terms with odd exponents and x_even has the
    terms with even exponents.  The decomposition is definable.
    Taking J to be the set of x such that (x_even) > 0 and
    (x_odd) ≥ 1 (or something similar), one arrives at a
    situation where R_J and I_J cannot be fixed via any finite
    sequence of mutations.
  
  * Nevertheless, one can still recover a good Bezout domain in
    the limit, which is Theorem <ref>.  Unfortunately,
    the resulting Bezout domain is not closely tied to the
    infinitesimals—not enough to run the henselianity machine from
    the dp-minimal case.
  
  For details on the above construction, see <cit.>.

We now return to actual proofs.

  
  Let I, J ∈𝒫_n be special.  Then I + J and I ∩
  J are special.  Furthermore, there exists
  
  
  * a basis of quasi-atoms Â_1, …, Â_n in [I
    ∩ J, ^n],
  
  * a basis of quasi-atoms B̂_1, …, B̂_n in [I
    + J, ^n], and
  
  * a basis of quasi-atoms A_1, …, A_n, B_1, …, B_n
    in [I ⊕ J, ^2n]
  
  related as follows:
  
    Â_i     = {x⃗∈^n  |  (x⃗,x⃗) ∈ A_i}
    B̂_i     = {x⃗ - y⃗ |  (x⃗,y⃗) ∈ B_i}.

  Given D ∈𝒫_2n dominating I ⊕ J, we may
  choose the A_i and B_i to lie in [I ⊕ J, D].


  For any J ∈𝒫_n, define
  
    Δ(C)    = {(x⃗,x⃗)  | x⃗∈ C}∈𝒫_2n
    ∇(C)    = {(x⃗, x⃗ + y⃗)  | x⃗∈^n,  y⃗∈ C}∈𝒫_2n.

  Let V = Δ(^n) = ∇(0).  The maps Δ(-),
  ∇(-) yield isomorphisms
  
    Δ : 𝒫_n    ∼→ [0,V] ⊆𝒫_2n
    ∇ : 𝒫_n    ∼→ [V,^2n]
        ⊆𝒫_2n.

  Indeed, the inverses are given by
  
    Δ^-1 : [0, V]    ∼→ [0, ^n] 
    
        C    ↦{x⃗ |  (x⃗,x⃗) ∈ C}
    ∇^-1 : [V, ^2n]    ∼→ [0, ^n] 
    
        C    ↦{x⃗ - y⃗ |  (x⃗,y⃗) ∈ C}.

  Note that Δ^-1(V ∩ (I ⊕ J)) = I ∩ J and
  ∇^-1(V + (I ⊕ J)) = I + J.  Therefore, Δ^-1
  and ∇^-1 restrict to isomorphisms
  
    Δ^-1 : [V ∩ (I ⊕ J), V]    ∼→ [I
          ∩ J, ^n] 
    ∇^-1 : [V + (I ⊕ J), ^2n]    ∼→
              [I + J, ^n].

  It follows that
  
    (^n/(I ∩ J))    = (V/(V ∩ (I ⊕ J))) =
        ((V + (I ⊕ J))/V) 
    (^n/(I + J))    = (^2n/(V + (I ⊕ J))/V).

  Now I ⊕ J is special in 𝒫_2n by
  Proposition <ref>.<ref>, and V is an
  n-dimensional -linear subspace of ^2n, so by
  Lemma <ref>,
  
    (^n/(I ∩ J))    = ((V + (I ⊕ J))/V) = rn
        
    (^n/(I + J))    = (^2n/(V + (I ⊕ J))) = rn.

  Therefore I ∩ J and I + J are special.  Furthermore, by
  Lemma <ref> there exists a basis
  {A_1,…,A_rn,B_1,…,B_rn} of quasi-atoms over I
  ⊕ J such that
  
  
  * The elements Ã_i := A_i ∩ V form a basis of
    quasi-atoms in [V ∩ (I ⊕ J), V].
  
  * The elements B̃_i := B_i + V form a basis of
    quasi-atoms in [V + (I ⊕ J), ^2n].
  
  (Additionally, the A_i and B_i can be chosen below any given D
  dominating I ⊕ J.)  Applying Δ^-1 and ∇^-1
  we see that the elements
  
    Â_i = Δ^-1(A_i ∩ V)     = {x⃗ | 
        (x⃗,x⃗) ∈ A_i ∩ V}
        = {x⃗ |  (x⃗,x⃗) ∈ A_i}
    B̂_i = ∇^-1(A_i + V)     = {x⃗ - y⃗ | 
        (x⃗,y⃗) ∈ B_i + V}
        = {x⃗ - y⃗ |  (x⃗,y⃗) ∈ B_i}

  form bases of quasi-atoms for [I ∩ J, ^n] and [I + J,
    ^n], respectively.


  By Lemma <ref> special elements of 𝒫_n
  form a sublattice.  Can this be proven directly (lattice
  theoretically) within 𝒫_n without using the larger
  lattice 𝒫_2n?




 §.§ The associated rings and ideals


  Let J ∈𝒫_n be special, and a ∈^×.  Say
  that a contracts J if a = 0 or J dominates a ·
  J (i.e., (J/a · J) = nr).

Note that when a  0, J dominates a · J if and only if
a^-1· J dominates J.

  
   
  
  
  *  Let A_1, …, A_nr be a basis of
    quasi-atoms over J ∈𝒫_n and a be an element of
    .  If a · A_i ⊆ J for all i, then a
    contracts J.  Conversely, suppose a contracts J.  Then there
    exists A'_i ∈ (J,A_i] such that A'_1, …, A'_nr is a
      basis of quasi-atoms over J and a · A'_i ⊆ J for
      each i.
  
  *  If a contracts J and b ∈^×, then
    a contracts b · J.
  
  *  If a, b contract J then a + b contracts J.
  
  *  If a contracts J and b · J ⊆ J,
    then a · b contracts J.
  
  * If a contracts both I ∈𝒫_n and J ∈𝒫_m, then a contracts I ⊕ J ∈𝒫_n+m.
  
  *  If a contracts I, J ∈𝒫_n, then
    a contracts I ∩ J and I + J.
  


  
  
  * First suppose a · A_i ⊆ J.  If a = 0 then a
    contracts J by definition, so suppose a  0.  Then
    
    (a^-1· J) ∩ A_i = A_i ⊋ J

    for each i, so by Lemma <ref> the
    group a^-1· J dominates J, or equivalently, J
    dominates a · J.  Thus a contracts J.  Conversely,
    suppose that a contracts J.  If a = 0 then a · A_i
    ⊆ J so we may take A'_i = A_i.  Otherwise, note that
    a^-1· J dominates J, so by
    Lemma <ref>,
    
    A'_i := (a^-1· J) ∩ A_i ⊋ J.

    Then A'_i is a quasi-atom over J, equivalent to A_i, so
    {A'_1, …, A'_nr} is another basis of quasi-atoms over
    J.  Furthermore A'_i ⊆ a^-1· J, so a ·
    A'_i ⊆ J.
  
  * Multiplication by b induces an automorphism of
    𝒫_n sending the interval [a · J, J] to [a ·
    (b · J), b · J], so (J/a · J) = (b ·
    J / (ab) · J).
  
  * Take a basis of quasi-atoms A_1, …, A_rn over J.
    By part 1, we may shrink the A_i and assume that a · A_i
    ⊆ J.  Shrinking again, we may assume b · A_i
    ⊆ J.  Then
    
    (a + b) · A_i ⊆ a · A_i + b · A_i ⊆ J + J = J

    so by part 1, a+b contracts J.
  
  * Suppose a contracts J and b · J ⊆ J.  Then
    
    (J/a · b · J) ≥(b · J/ a · b
          · J) = (J/a · J) = nr.

  
  * Let A_1, …, A_rn be a basis of quasi-atoms in
    [I,^n], and B_1, …, B_rm be a basis of quasi-atoms
    in [J,^m].  Shrinking the A_i and B_i, we may assume a
    · A_i ⊆ I and a · B_i ⊆ J.  Note that
    the sequence
    
    A_1 ⊕ J, A_2 ⊕ J, …, A_rn⊕ J, I ⊕
          B_1, I ⊕ B_2, …, I ⊕ B_rm

    is a basis of quasi-atoms in [I ⊕ J, ^n+m].
    Multiplication by a collapses each of these quasi-atoms into I
    ⊕ J (using the fact that a · I ⊆ I and a
    · J ⊆ J).  Therefore a contracts I ⊕ J.
  
  * We may assume a  0.  By the previous point, a^-1·
    (I ⊕ J) dominates I ⊕ J.  By
    Lemma <ref>, I + J and I ∩ J are special.
    Moreover, there is a basis of quasi-atoms A_1, …, A_n, B_1,
    …, B_n in [I ⊕ J, ^2n] such that for
    
    Â_i    = {x⃗∈^n  |  (x⃗,x⃗) ∈ A_i}
    B̂_i    = {x⃗ - y⃗ |  (x⃗,y⃗) ∈ B_i}

    the set {Â_1, …, Â_n} is a basis of
    quasi-atoms over I ∩ J and the set {B̂_1, …,
    B̂_n} is a basis of quasi-atoms over I + J.  Furthermore
    Lemma <ref> ensures that the A_i and B_i can
    be chosen in [I ⊕ J, a^-1· (I ⊕ J)].  Thus a
    · A_i ⊆ I ⊕ J and a · B_i ⊆ I
    ⊕ J.  Then
    
    x⃗∈Â_i  (x⃗,x⃗) ∈ A_i  (a
          ·x⃗, a ·x⃗) ∈ I ⊕ J  a ·x⃗∈ I ∩ J,

    so a ·Â_i ⊆ I ∩ J.  As the Â_i
    form a basis of quasi-atoms over I ∩ J, it follows that a
    contracts I ∩ J.  Similarly,
    
    (x⃗,y⃗) ∈ B_i  (a ·x⃗, a ·y⃗) ∈ I ⊕ J  a · (x⃗ - y⃗) ∈
          I + J

    so a ·B̂_i ⊆ I + J.  Thus a contracts I +
    J. 


  
  For any special J ∈𝒫 = 𝒫_1, let R_J be
  the set of a ∈ such that a · J ⊆ J, and let
  I_J be the set of a ∈ that contract J.
  
  
  * R_J is a subring of , containing M_0.
  
  * I_J is an ideal in R_J.
  
  *  If b ∈^× then R_J = R_b ·
    J and I_J = I_b · J.
  
  * If J is type-definable over M ⊇ M_0, then R_J
    and I_J are M-invariant.
  
  *  If J is non-zero and type-definable over M
    ⊇ M_0 then I_M ⊆ I_J.
  
  *  If J_1 and J_2 are special, then
    
    R_J_1∩ R_J_2   ⊆ R_J_1 ∩ J_2
    
          I_J_1∩ I_J_2   ⊆ I_J_1 ∩ J_2

  
  *  (1 + I_J) ⊆ R_J^×.
    Consequently, I_J lies inside the Jacobson radical of R_J.
  


  
  
  * Straightforward.
  
  * The set I_J is a subset of R_J.  The fact that I_J 
    R_J is exactly Lemma <ref>.<ref>-<ref>.
  
  * For I_J this is Lemma <ref>.<ref>.  For
    R_J this is clear:
    
    a · J ⊆ J  (ab) · J ⊆ b · J.

  
  * The definitions are (/M)-invariant.
  
  * Let A_1, …, A_r be a basis of quasi-atoms over J.
    For each i let a_i be an element of A_i∖ J.  Let
    M' be a small model containing M and the a_i's.
    
      Any ε∈ I_M' contracts J.
    
    
      We may assume ε 0.  Let D = ε^-1· I_M'.  By
      Remark <ref>.<ref>, a_i
      ·ε∈ I_M', or equivalently a_i ∈ D.
      Thus
      
    D ∩ A_i ⊈J

      By Proposition <ref>.<ref>, D
      ⊇ J.  Then
      
    D ∩ A_i ⊋ J

      so by Lemma <ref>, D dominates
      J.  By
      Proposition <ref>.<ref>,
      ε^-1· J ⊇ε^-1· I_M' =
      D.  Thus ε^-1· J dominates J.
    
    Let ε be a realization of the partial type over M'
    asserting that ε∈ I_M' and ε∉ X for
    any light M'-definable set X.  This type is consistent because
    M'-definable basic neighborhoods are heavy
    (Proposition <ref>.<ref>) and no heavy set
    is contained in a finite union of light sets
    (Theorem <ref>).  Then ε∈ I_M'⊆
    I_J.  As I_J is M-invariant, every realization of
    (ε/M) is in I_J.  Let Y be the M-definable set
    of realizations of (ε/M).  For any M-definable X
    ⊇ Y we have
    
    I_M ⊆ X  X ⊆ X - X.

    Therefore I_M ⊆ Y- Y.  But Y - Y ⊆ I_J - I_J = I_J.
  
  * If a ∈ R_J_1 and a ∈ R_J_2, then
    
    a · (J_1 ∩ J_2) = (a · J_1) ∩ (a · J_2)
          ⊆ J_1 ∩ J_2

    so a ∈ R_J_1 ∩ J_2.  The inclusion I_J_1∩ I_J_2⊆ I_J_1 ∩ J_2 is Lemma <ref>.<ref>.
  
  * First note that 1 does not contract J.  Indeed,
    (J/J) = 0  r.  Thus 1 ∉ I_J.  As I_J is an
    ideal, it follows that -1 ∉ I_J.
    
      If ε∈ I_J then ε/(1 + ε) ∈ I_J.
    
    
      We may assume ε 0.  Using
      Lemma <ref>.<ref> choose a basis
      {A_1,…,A_r} of quasi-atoms over J such that ε· A_i ⊆ J.  For each i choose a_i ∈ A_i
      ∖ J.  Then ε· a_i ∈ J, so (1 +
      ε) · a_i ∈ A_i ∖ J.  Let β = (1 +
      ε)/ε.  Then
      
    β· (ε· a_i)    ∈β· J 
    
             (β·ε) · a_i = (1 + ε) · a_i    ∈ A_i ∖ J.

      In particular
      
    (β· J) ∩ A_i ⊈J,

      for every i, so β· J ⊇ J by
      Proposition <ref>.<ref>.  Then
      (β· J) ∩ A_i ⊋ J for every i, so β· J dominates J by
      Lemma <ref>.  This means that
      β^-1 = ε/(1 + ε) lies in I_J.
    
    Now if ε∈ I_J, then
    
    1/1 + ε = 1 - ε/1 + ε∈ 1
          + I_J ⊆ R_J.

  


  Proposition <ref>.<ref> also holds for
  R_J_1 + J_2 and I_J_1 + J_2.


  In Proposition <ref>.<ref>, not only is I_M
  a subset of I_J, it is a subideal in the ring R_J.  One
  can probably prove this by first increasing M to contain a
  non-zero element j_0 of J.  Then for any ε∈ I_M
  and a ∈ R_J, we have
  
    ε· a · j_0 ∈ I_M · R_J · J ⊆ I_M · J ⊆ I_M,

  so ε· a ∈ j_0^-1 I_M = I_M.  Thus I_M · R_J
  ⊆ I_M.  Then one can probably shrink M back to the
  original model by the usual methods.

  Now, the important thing is that if R_J were already a good Bezout
  domain, then there would be some valuations _i : →Γ_i and cuts Ξ_i in Γ_i such that
  
    I_M = {x ∈ | _1(x) > Ξ_1 ∧_2(x) > Ξ_2 ∧⋯}.

  By replacing Ξ_i with 2 Ξ_i or Ξ_i/2 we could obtain
  alternate M-invariant subgroups smaller than I_M, contradicting
  the analogue of Corollary <ref> for invariant
  (not type-definable) subgroups.  Unless, of course, the Ξ_i lie
  on the edges of convex subgroups.  Then, by coarsening, one could
  essentially arrive at a situation where I_M is the Jacobson
  radical of an M-invariant good Bezout domain.  This would allow
  some of the henselianity arguments used in the dp-minimal case to
  directly generalize, though there are some additional complications
  in characteristic 0.[The details are worked out in
    <cit.>.]




 §.§ Mutation and the limiting ring
 
The next two lemmas provide a way to “mutate” a special group J
and obtain a better special group J' for which R_J' is closer
than R_J to being a good Bezout domain.

  
  Let J ∈𝒫 be special and non-zero.  Let a_1, …,
  a_n be elements of ^×.  Let J' = J ∩ a_1 · J ∩
  a_2 · J ∩⋯∩ a_n · J.  Then J' is special and
  non-zero, R_J ⊆ R_J', and I_J ⊆ I_J'.


  By Proposition <ref>.<ref>, each
  a_i · J is special, so the intersection J' is special by
  Lemma <ref>.  It is nonzero by
  Proposition <ref>.<ref>.  By
  Proposition <ref>.<ref> we have R_J =
  R_a_i · J and I_J = I_a_i · J for each i.  Then the
  inclusions R_J ⊆ R_J' and I_J ⊆ I_J' follow
  by an interated application of
  Proposition <ref>.<ref>.


Recall that r is the reduced rank of 𝒫.

  
  Let J ∈𝒫 be special and non-zero.  Let α∈^× be arbitrary.  Let J' = J ∩ (α· J) ∩⋯∩ (α^r-1· J).  Let q_0, q_1, …, q_r be
  r+1 distinct elements of M_0.  Then there is at least one i
  such that α q_i and
  
    1/α - q_i∈ R_J'.



  For each 0 ≤ i ≤ r let
  
    α_i     := α - q_i 
    
        G_i     := {x ∈ | α_ix ∈ J ∧α_i^2 x ∈
        J∧⋯∧α_i^rx ∈ J}
    
        H_i     := J ∩ G_i = {x ∈ |  x ∈ J ∧α_ix
        ∈ J ∧⋯∧α_i^rx ∈ J}.

  Also let
  
    H = {x ∈ |  x ∈ J ∧α x ∈ J ∧⋯∧α^r x ∈ J}.

   
    H_i = H for any i.
  
  
    Note α = α_i + q_i.  If x ∈ H_i then
    
    α^n x = (α_i + q_i)^n x = ∑_k = 0^n nkα_i^k q_i^n-k x ∈ J

    for 0 ≤ n ≤ r, because α_i^k x ∈ J, q_i^n-k∈
    M_0, and J is an M_0-vector space.  Thus H_i ⊆ H;
    the reverse inclusion follows by symmetry.
  
  Because the q_i are distinct, the (r+1) × (r+1) Vandermonde
  matrix built from the q_i is invertible.  Let f : ^r+1→^r+1 be the -linear map sending (1,q_i,…,q_i^r) to
  the ith basis vector.  Let g : →^r+1 be the map
  
    g(x) = (x, α x, …, α^r x).

  
    
    The composition
    
    g→^r+1f→^r+1↠ (/J)^r+1

    has kernel H, and maps G_i into 0^i ⊕ (/J) ⊕
    0^r-i.
  
  
    The invertible matrix defining f has coefficients in M_0, and
    J is closed under multiplication by M_0, so f maps J^r+1
    isomorphically to J^r+1.  Therefore,
    
    f(g(x)) ∈ J^r+1 g(x) ∈ J^r+1 x ∈ H,

    where the second  is the definition of H.  Now suppose x
    ∈ G_i.  Then g(x) - (x, q_ix, …, q_i^r x) ∈ J^r+1.
    Indeed, for any 0 ≤ n ≤ r we have
    
    α^n x = (α_i + q_i)^n x = q_i^n x + ∑_k = 1^n nk q_i^n-k (α_i^k x),

    and the sum is an element of J by definition of G_i.  As f
    preserves J^r+1, it follows that
    
    f(g(x)) ≡ f(x, q_ix, …, q_i^r x) = x · e_i 
          J^r+1,

    where e_i is the ith basis vector.
  
  
    
    If (x_0, x_1, …, x_r) ∈ G_0 ×⋯× G_r has
    x_0 + ⋯ + x_r ∈ H, then each x_i ∈ H.
  
  
    For 0 ≤ i ≤ r let p_i : ^r+1→/J be the
    composition of the ith projection and the quotient map →/J.  Claim <ref> implies that
    
    x ∈ H     p_i(f(g(x))) = 0 
    
          x ∈ G_j     p_i(f(g(x))) = 0     if  i  j.

    Thus
    
    0 = p_i(f(g(x_0 + ⋯ + x_r))) = p_i(f(g(x_i))).

    As p_j(f(g(x_i))) = 0 for j  i, it follows that
    p_j(f(g(x_i))) = 0 for all j.  In other words, f(g(x_i)) ∈
    J^r+1.  By Claim <ref>, x_i ∈ H.
  
  Now Claim <ref> implies that the map
  
    (G_0/H) ×⋯× (G_r/H)    →/H 
    
        (x_0,…,x_r)    ↦ x_0 + ⋯ + x_r

  is injective.  The image is D/H for some type-definable D ∈𝒫, namely D = G_0 + ⋯ + G_r.  Then the interval
  [H^r+1,G_0 ⊕⋯⊕ G_r] in 𝒫_r+1 is
  isomorphic to the interval [H,D] in 𝒫_1.  Thus
  
    r ≥(D/H) = (G_0/H) + ⋯ + (G_r/H).

  Therefore G_i = H = H_i for at least one i.  By definition of
  G_i and H_i, this means that
  
    α_i x ∈ J ∧⋯∧α_i^r x ∈ J 
        x ∈ J

  for any x ∈.  As J  0, this implies α_i  0.
  Then (<ref>) can be rephrased as
  
    α_i^-1· J ∩⋯∩α_i^-r· J
        ⊆ J.

  Define
  
    J”    := J ∩α_i^-1 J ∩⋯∩α_i^-(r-1) J 
        = J ∩α^-1 J ∩⋯∩α^-(r-1) J,

  where the second equality follows by the proof of
  Claim <ref>.  By (<ref>),
  
    α_i^-1· J” = α_i^-1 J ∩⋯∩α_i^-r!⊆ J ∩α_i^-1 J ∩⋯∩α_i^-(r-1) J =
        J”.

  Therefore α_i^-1∈ R_J”.  But
  
    J' = J ∩⋯∩α^r-1 J = α^r-1· (J
        ∩⋯∩α^-(r-1) J) = α^r-1 J”.

  Thus, by Proposition <ref>.<ref>
  
    α_i^-1∈ R_J” = R_J'.




  
  Let J ∈𝒫_1 be special, non-zero, and type-definable
  over M ⊇ M_0.  Then there is an M-invariant ring
  R^∞_J and ideal I^∞_J  R^∞_J satisfying the
  following properties:
  
  
  * R^∞_J and I^∞_J are M-invariant.
  
  * (1 + I^∞_J) ⊆ (R^∞_J)^×, so
    I^∞_J is a subideal of the Jacobson radical of
    R^∞_J.
  
  * The M-infinitesimals I_M are a subgroup of I^∞_J
    (and therefore of the Jacobson radical).
  
  * M_0 ⊆ R^∞_J.
  
  * R^∞_J is a Bezout domain with at most r maximal
    ideals.
  
  * The field of fractions of R^∞_J is .
  


  Let P be the set of finite S ⊆^× such that 1
  ∈ S.  Then P is a commutative monoid with respect to the
  product S · S' = {x · y  |  x ∈ S, y ∈ S'}.  For any
  S ∈ P and G ∈𝒫_1, define
  
    G^S := ⋂_s ∈ S s · G.

  Note that (G^S)^S' = G^S · S'.  If G is special and
  non-zero then by Lemma <ref> G^S is special and non-zero,
  and there are inclusions R_G ⊆ R_G^S and I_G ⊆
  I_G^S.  Define sets
  
    R_J^∞    := ⋃_S ∈ P R_J^S
    
        I_J^∞    := ⋃_S ∈ P I_J^S.

  These sets are clearly M-invariant.  Moreover, the unions are
  directed: given any S and S' we have
  
    R_J^S∪ R_J^S'   ⊆ R_J^S · S'
    
        I_J^S∪ I_J^S'   ⊆ I_J^S · S'.

  Therefore R^∞_J is a ring and I^∞_J is an ideal.  The
  fact that (1 + I^∞_J) ⊆ (R^∞_J)^× also
  follows (using Proposition <ref>.<ref>).
  Taking S = {1}, we see that I_J ⊆ I^∞_J.
  Proposition <ref>.<ref> says I_M ⊆
  I_J, so I_M ⊆ I^∞_J as desired.  Similarly, M_0
  ⊆ R_J ⊆ R^∞_J.
   
    If q_0, q_1, …, q_r are distinct elements of M_0 and
    α∈^×, then at least one of 1/(α - q_i)
    is in R^∞_J.
  
  
    By Lemma <ref>, at least one of 1/(α - q_i)
    lies in R_J^S for S = {1, α, …, α^r-1}.
  
  It follows formally that R^∞_J is a Bezout domain with no
  more than r maximal ideals.  Let a, b be two elements of
  R^∞_J.  We claim that the ideal (a,b) is principal.  This
  is clear if a = 0 or b = 0.  Otherwise, let α = a/b.  As
  M_0 is infinite, Claim <ref> implies that
  
    b/a - q b = 1/a/b - q∈ R^∞_J

  for some q ∈ M_0.  Then the principal ideal (a - qb) 
  R^∞_J contains b, hence qb and thus a.  Therefore (a -
  qb) = (a, b).

  Next, we show that R^∞_J has at most r maximal ideals.
  Suppose for the sake of contradiction that there were distinct
  maximal ideals 𝔪_0, …, 𝔪_r in
  R^∞_J.  As R^∞_J is an M_0-algebra, each quotient
  R^∞_J/𝔪_i is a field extending M_0.  Take
  distinct q_0, …, q_r ∈ M_0, and find an element x ∈
  R^∞_J such that x ≡ q_i 𝔪_i for each
  i, by the Chinese remainder theorem.  Then x - q_i ∈𝔪_i ⊆ R^∞_J ∖ (R^∞_J)^×
  for each i.  So 1/(x - q_i) does not lie in R^∞_J for any
  0 ≤ i ≤ r, contrary to Claim <ref>.

  Lastly, note that if x is any element of ^×, then 1/(x
  - q) ∈ R^∞_J for some q ∈ M_0, q  x.  As q ∈ M_0
  ⊆ R^∞_J, the field of fractions of R^∞_J
  contains x.  So the field of fractions must be all of .




 §.§ From Bezout domains to valuation rings

We double-check some basic facts about Bezout domains.

  
  Let R be a Bezout domain.
  
  
  * For each maximal ideal 𝔪, the localization
    R_𝔪 is a valuation ring on the field of fractions
    of R.
  
  * R is the intersection of the valuation rings
    R_𝔪.
  


  
  
  * Given non-zero a, b ∈ R, we must show that a/b or b/a
    lies in R_𝔪.  If c is such that (c) = (a,b),
    then we may replace a and b with a/c and b/c, and assume
    that (a,b) = (1) = R.  Both of a and b cannot lie in
    𝔪, and so either a/b or b/a is in
    R_𝔪.
  
  * Given a ∈⋂_𝔪 R_𝔪, we will
    show a ∈ R.  Let I = {x ∈ R  |  ax ∈ R}.  This is an
    ideal in R.  If I is proper, then I ≤𝔪 for
    some 𝔪.  As a ∈ R_𝔪 we can write a
    = b/s where b ∈ R and s ∈ R ∖𝔪.  Then
    s ∈ R and as = b ∈ R, so s ∈ I ≤𝔪, a
    contradiction.  Therefore I is an improper ideal, so 1 ∈ I,
    and a = a · 1 ∈ R. 



  
  Let  be a sufficiently saturated dp-finite field, possibly with
  extra structure.  Suppose  is not of finite Morley rank.  Then
  there is a small set A ⊆ and a non-trivial
  A-invariant valuation ring.


  Take M_0 as usual in this section.  By
  Proposition <ref> there is a non-zero special J
  ∈𝒫_1.  The group J is type-definable over some small
  M ⊇ M_0.  Let R be the R^∞_J of
  Theorem <ref>.  Then R is an M-invariant Bezout
  domain with at most r maximal ideals, the Jacobson radical of R
  is non-zero (because it contains I_M), and (R) =.  Let
  𝔪_1, …, 𝔪_k enumerate the maximal
  ideals of R.  Let 𝒪_i be the localization
  R_𝔪_i.  By Remark <ref>, each
  𝒪_i is a valuation ring on , and
  
    R = 𝒪_1 ∩⋯∩𝒪_k.

  At least one 𝒪_i is non-trivial; otherwise R = and
  has Jacobson radical 0.[Tracing through the proof, here is
    what explicitly happens. If ε∈ I_M then
    -1/ε cannot be in R^∞_J, or else ε∈ I_M ⊆ I_J ⊆ I_J^S R_J^S and
    -1/ε∈ R_J^S for large enough S, so that -1 ∈
    I_J^S, contradicting
    Proposition <ref>.<ref>.]  Without loss
  of generality 𝒪_1 is non-trivial.  By the Chinese
  remainder theorem, choose a ∈ R such that a ≡ 1 𝔪_1 and a ≡ 0 𝔪_i for i 
  1.  We claim that 𝒪_1 is (/aM)-invariant.  If
  σ∈(/aM), then σ∈(/M) so σ
  preserves R setwise.  It therefore permutes the finite set of
  maximal ideals.  As 𝔪_1 is the unique maximal ideal not
  containing a, it must be preserved (setwise).  Therefore σ
  preserves the localization 𝒪_1 setwise.


  Stable fields do not admit non-trivial invariant valuation rings
  (<cit.>, Lemma 2.1).  Consequently,
  Theorem <ref> can be used to give an extremely
  roundabout proof of Halevi and Palacín's theorem that stable
  dp-finite fields have finite Morley rank (<cit.>,
  Proposition 7.2).




§ SHELAH CONJECTURE AND CLASSIFICATION
 
 
  Let K be a sufficiently saturated dp-finite field of positive
  characteristic.  Then one of the following holds:
  
  
  * K has finite Morley rank (and is therefore finite or
    algebraically closed).
  
  * K admits a non-trivial henselian valuation.
  


  This is Theorem <ref> and
  <ref>.


 
  Let K be a sufficiently saturated dp-finite field of positive
  characteristic.  Assume K is infinite.  Let 𝒪_∞
  be the intersection of all K-definable valuation rings on K.
  Then 𝒪_∞ is a henselian valuation ring on K whose
  residue field is algebraically closed.


  The proof for dp-minimal fields (Theorem 9.5.7 in <cit.>)
  goes through without changes, using Lemma <ref>,
  Theorem <ref>, and
  Proposition <ref>.  Additionally, we must rule out the
  possibility that the residue field is real closed or finite.  The
  first cannot happen because we are in positive characteristic.  The
  second cannot happen because K is Artin-Schreier closed, a
  property which transfers to the residue field.


 
  Let K be a sufficiently saturated infinite dp-finite field of
  positive characteristic.  If every definable valuation on K is
  trivial, then K is algebraically closed.


 
  Let K be a dp-finite field of positive characteristic.  Then one
  of the following holds:
  
  
  * K is finite.
  
  * K is algebraically closed.
  
  * K admits a non-trivial definable henselian valuation.
  


  Suppose K is neither finite nor algebraically closed.  Let K'
  ≽ K be a sufficiently saturated elementary extension.  Then
  K' is neither finite nor algebraically closed.  By
  Corollary <ref> there is a non-trivial definable valuation
  𝒪 = ϕ(K',a) on K'.  The statement that ϕ(x;a)
  cuts out a valuation ring is expressed by a 0-definable condition on
  a, so we can take a ∈(K).  Then ϕ(K,a) is a
  non-trivial valuation ring on K, henselian by
  Theorem <ref>.

So the Shelah conjecture holds for dp-finite fields of positive
characteristic.

By Proposition 3.9, Remark 3.10, and Theorem 3.11 in
<cit.>, this implies the following classification
of dp-finite fields of positive characteristic: up to elementary
equivalence, they are exactly the Hahn series fields 𝔽_p((Γ))
where Γ is a dp-finite p-divisible group.  Dp-finite ordered
abelian groups have been algebraically characterized and are the same
thing as strongly dependent ordered abelian groups
(<cit.>, <cit.>,
<cit.>).


  The author would like to thank


  * Meng Chen, for hosting the author at Fudan University, where
  this research was carried out.

  * Jan Dobrowolski, who provided some helpful references.

  * John Goodrick, for suggesting that the ideal of finite sets
  could be replaced with other ideals in the construction of
  infinitesimals.

  * Franziska Jahnke, for convincing the author that the proof in
  <ref> was valid.

  * Shichang Song, who invited the author to give a talk on
  dp-minimal fields.  The ideas for
  <ref>-<ref> arose while preparing
  that talk.

  * The UCLA model theorists, who read parts of this paper and
  pointed out typos.

This material is based upon work supported by the National Science
Foundation under Award No. DMS-1803120.  Any opinions, findings, and
conclusions or recommendations expressed in this material are those of
the author and do not necessarily reflect the views of the National
Science Foundation.


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