# Fields interpretable in $P$-minimal fields

Yatir Halevi  Department of Mathematics, Ben Gurion University of the Negev,
Be’er-Sheva 84105, Israel<EMAIL_ADDRESS>,  Assaf Hasson  Department
of Mathematics, Ben Gurion University of the Negev, Be’er-Sheva 84105, Israel
<EMAIL_ADDRESS>and  Ya’acov Peterzil  Department of Mathematics,
University of Haifa, Haifa, Israel<EMAIL_ADDRESS>
###### Abstract.

We prove that an infinite field interpretable in a $p$-adically closed field
$K$ is definably isomorphic to a finite extension of $K$. The result remains
true in any $P$-minimal field where definable functions are generically
differentiable.

The first author would like to thanks the Israel Science Foundation for its
support of this research (grant No. 181/16) and the Kreitman foundation
fellowship. The third author was partially supported by Israel Science
Foundation grant number 290/19

In [5], interpretable fields in various expansions of real closed valued
fields were classified using analysis of one dimensional quotients which are
definably embedded in the field. Here we use similar methods to classify
interpretable fields in various expansions of $p$-adically closed fields.

In [8] Pillay proved that any field definable in $\mathbb{Q}_{p}$ is definably
isomorphic to a finite extension of $\mathbb{Q}_{p}$, and asked whether the
same result is true of interpretable fields. Our main theorem gives a positive
answer to Pillay’s question.

######  Theorem 1.

Let $K$ be a $P$-minimal field. If, additionally, definable functions in $K$
are generically differentiable, then any infinite field interpretable in $K$
is definably isomorphic to a finite extension of $K$. In particular, the
result holds for $p$-adically closed fields.

The proof does not use elimination of imaginaries (that is not known, in
general, for $P$-minimal fields). It uses implicitly the fact that $P$-minimal
fields are uniformities (see [13] for details). The results of Sections 4-5 do
not use $P$-minimality beyond those consequences that are treated
axiomatically in [13] (and even [2]). For that reason we expect that our
methods can also be applied outside of the $P$-minimal context.

Note that recently Walsberg showed, [15], that a positive answer to the above
question of Pillay’s would provide an example of an NIP theory not
interpreting an infinite field whose Shelah expansion does. We mention also
that a positive answer to Pillay’s question on interpretable fields in
$\mathbb{Q}_{p}$ was announced also by E. Alouf, A. Fornasiero and J. de la
Nuez Gonzalez.

Acknowledgements We thank Immanuel Halupczok and Dugald Macpherson for several
discussions during the preparation of the article.

##  1\. Background and preliminaries

The notion of $P$-minimal fields was introduced by Haskell and Macpherson in
[4]. Recall that a valued field $K$ is _$P$ -minimal_ if it is $p$-valued, its
value group is a ${\mathbb{Z}}$-group and in every structure elementary
equivalent to $K$ every definable subset of $K$ is quantifier-free definable
in the Macintyre language for valued fields. We denote its value group by
$\Gamma$ and its valuation ring by ${\mathcal{O}}$. It is one of the main
results of [4] that $P$-minimal fields are henselian, and therefore, as pure
valued fields $p$-adically closed.

There are various cell decomposition results for $P$-minimal fields, but we
will only explicitly use the fact, following from the very definition, that
definable subsets of $K$ itself can be partitioned into finite many cells, as
in the case of $p$-adically closed fields. I.e. every definable set is a
disjunction of sets of the form

$\\{x\in K:\gamma_{1}<v(x)<\gamma_{2}\wedge P_{n}(\lambda\cdot x)\\},$

where $P_{n}$ is the $n$-th power predicate, $\lambda\in K$ and
$\gamma_{1},\gamma_{2}\in\Gamma\cup\\{\infty,-\infty\\}$ and $n\in\mathbb{N}$.

In particular, every infinite definable subset of $K$ has non-empty interior
with respect to the valuation topology (this also follows from Simon’s [11]).
Since any $P$-minimal field is dp-minimal, this shows that $P$-minimal fields
satisfy the assumptions of Simon and Walsberg’s tameness conditions for
topological dp-minimal structures, [13]. Most of the topological properties of
$P$-minimal fields we will be using below follow from this work of Simon and
Walsberg, though – in most cases – specialised to the case of $P$-minimal
fields they were already known by earlier work.

In particular, this implies that $P$-minimal fields eliminate
$\exists^{\infty}$ (see [13, Lemma 2.1]) in the valued field sort. We will use
this fact in several places without explicit reference. We note however, that
this property does not transfer to imaginary sorts (such as the value group
and $K/{\mathcal{O}}$). We will also be using the fact that
$\operatorname{acl}$ (the model theoretic algebraic closure) satisfies the
Steiniz Exchange principle in $P$-minimal fields ([4, Corollary 6.2]). As a
consequence, in the valued field sort of $K$, dp-rank, topological dimension
and $\operatorname{acl}$-dimension coincide (see [13, Proposition 2.4]) and
dp-rank is additive. Most of the result we need from [13] also appear in the
the work of Dolich and Goodrick on viscerality, [2], and may be relevant for
possible generalisations of the present work into structures with IP.

##  2\. Strong internality to $K$

Our aim in this section is, given an interpretable field ${\mathcal{F}}$, to
find a dp-minimal $S\subseteq{\mathcal{F}}$ that is in definable bijection
with a subset of $K$. We proceed in three steps, we first reduce the problem
to subsets of ${\mathcal{F}}$ that are in definable bijection with an infinite
subset of $K/E$, where $E$ is some definable equivalence relation on $K$. We
then show that for any interpretable set $S$ of the form $K/E$, there exists a
finite-to-one (partial) function $f:S\to V$ with infinite domain, where $V$ is
either $K$, $\Gamma$ or $K/{\mathcal{O}}$. Finally we show that if
$S\subseteq{\mathcal{F}}$ then this set can only be $K$.

We begin with several general results about definable equivalence relations in
$P$-minimal fields.

###  2.1. Definable one-dimensional quotients in $P$-minimal fields

We assume that $K$ is a $P$-minimal field and $E$ a definable equivalence
relation on $K$ with infinitely many classes. Our aim is to find a definable
function with finite fibres from an infinite subset of $K/E$ into either
$\Gamma$, $K/{\mathcal{O}}$ or $K$ itself. We need some preperation. For
$\gamma\in\Gamma$ and $a\in K$, we let

$B_{\gamma}(a)=\\{x\in K:v(x-a)>\gamma\\},$

and call $\gamma$ the radius of the $\gamma$-ball $B_{\gamma}(a)$.

By using the fact that any $P$-minimal valued field is elementary equivalent
to a $p$-adic field, it is not hard to verify the following.

######  Lemma 2.1.

  1. (1)

For any $\gamma\in\Gamma$ and $n\in\mathbb{N}$, any $\gamma$-ball contains
only finitely many $\gamma+n$-balls.

  2. (2)

If a $\gamma$-ball is covered by finitely many balls with radii in
$\mathbb{Z}$ then $\gamma\in\mathbb{Z}$.

The following is the key lemma for this part of the argument. We thank D.
Macpherson for sketching the proof for us. We first recall [4, Lemma 2.3]:

######  Fact 2.2.

Let $K$ be a $p$-adically closed field and let $n\in\mathbb{N}$ with $n>1$ and
$x,y,a\in K$. Suppose that $v(y-x)>2v(n)+v(y-a)$. Then $x-a,y-a$ are in the
same coset of $P_{n}$.

######  Lemma 2.3.

Assume that $X\subseteq K$ is a definable set, $\gamma_{0}\in\Gamma$ and $X$
contains infinitely many $\gamma_{0}$-balls. Then for every $k\in\mathbb{N}$,
$X$ contains a ball of radius $\gamma_{0}-k$.

In particular, if $K$ is $\aleph_{0}$-saturated then $X$ contains a ball of
radius $\gamma$ satisfying $\gamma<\gamma_{0}-k$ for all $k\in\mathbb{N}.$

###### Proof.

We assume that $X$ contains infinitely many $\gamma_{0}$-balls.

Partitioning $X$ into cells, and, if needed, translating by an element of $K$,
we may assume that $X$ has the form

$\\{x\in K:\gamma_{1}<v(x)<\gamma_{2}\wedge P_{n}(\lambda\cdot x)\\},$

where $\lambda\in K$, $\gamma_{1}<\gamma_{2}\in\Gamma\cup\\{\infty,-\infty\\}$
and $n\in\mathbb{N}$. Multiplying $X$ by a scalar, we may assume that
$\gamma_{0}=0$.

By Lemma 2.1, $X$ is not contained in a ball $B_{-m}(0)$ for any
$m\in{\mathbb{N}}$. So for every $k\in\mathbb{N}$, there is some $y_{0}\in X$
such that $v(y_{0})<-k$. We can thus fix $y_{0}\in X$, such that
$v(y_{0})+2v(n)<-1$. We claim that $B_{-1}(y_{0})\subseteq X$.

Indeed, assume that $v(x-y_{0})>-1$. Then, since $v(y_{0})<-1$, we have
$v(x)=v(y_{0})$ and therefore $\gamma_{1}<v(x)=v(y_{0})<\gamma_{2}$. Thus it
is sufficient to see that $x$ and $y_{0}$ are in the same $P_{n}$-coset.

By our choice of $y_{0}$ we have $v(x-y_{0})>-1>v(y_{0})+2v(n)$, hence by Fact
2.2 (with $a=0$ there), $x$ and $y_{0}$ are in the same $P_{n}$-coset, so
$x\in X$. Thus $B_{-1}(y_{0})\subseteq X$.

We have thus shown that $X$ contains at least one ball of radius $-1$. After
removing from $X$ a single ball of radius $-1$ it still contains infinitely
many $0$-balls so we can find in $X$ a second ball of radius $-1$. Continuing
in this manner we find infinitely many balls in $X$ of radius $-1$.

It follows that if $X$ contains infinitely many balls of radius $\gamma_{0}$
then it also contains infinitely many balls of radius $\gamma_{0}-1$.
Repeating the process we obtain in $X$ infinitely many balls of radius
$\gamma_{0}-k$, for every $k\in\mathbb{N}$.

If $K$ is $\aleph_{0}$-saturated then the existence of a ball of radius
$\gamma$ with $\gamma<\gamma_{0}-k$ for all $k\in\mathbb{N}$ follows by
saturation. ∎

######  Lemma 2.4.

Let $S=\\{S_{t}:t\in T\\}$ be an infinite definable family of pairwise
distinct finite sets of $0$-balls. Assume that there exists an integer
$n\in\mathbb{N}$ with $|S_{t}|\leq n$ for all $t\in T$. Then there exists a
definable infinite subset $T^{\prime}\subseteq T$ and a definable finite-to-
one $f$ from $T^{\prime}$ into either $\Gamma$ or $K/{\mathcal{O}}$.

###### Proof.

We may assume that $K$ is $\aleph_{0}$-saturated, by passing to a large enough
elementary extension.

By possibly passing to a definable subset of $T$, we may assume that
$|S_{t}|=|S_{t^{\prime}}|=n$ for all $t,t^{\prime}\in T$, for some
$n\in\mathbb{N}$.

Given two $0$-balls, $b_{1}\neq b_{2}$ denote

$d(b_{1},b_{2})=\\{v(x_{1}-x_{2}):\mbox{ for any }x_{i}\in b_{i}\\},$

and $d(b_{1},b_{1})=1$ (notice that $d(b_{1},b_{2})$ is independent of the
choice of $x_{i}\in b_{i}$). And for $t\in T$, let
$d(t)=\min\\{d(b_{1},b_{2}):b_{i}\in S_{t}\\}$ and note that $d(t)\leq 1$.

Case 1 There exists $d\in\Gamma$ such that $\\{t\in T:d(t)=d\\}$ is infinite.

Reducing $T$, we may assume that for every $t\in T$, $d(t)=d$.

Case 1(a) $d\in\mathbb{Z}$.

Given a $0$-ball $b$, and $\gamma<0$ let $B_{\gamma}(b)$ be the unique ball of
radius $\gamma$ containing $b$. With this notation, for $t\in T$ and
$b,b^{\prime}\in S_{t}$ we get $B_{d-2}(b)=B_{d-2}(b^{\prime})$ (since
$d(b_{1},b_{2})=d$). In other words, for $t\in T$ the ball $B_{t}:=B_{d-2}(b)$
for some $b\in S_{t}$ is independent of the choice of $b\in S_{t}$.

Define $t_{1}\sim t_{2}$ if $B_{t_{1}}=B_{t_{2}}$. Because $d-2\in\mathbb{Z}$,
each $B_{t}$ contains only finitely many $0$-balls (Lemma 2.1). In fact each
$B_{t}$ contains at most $r:=p^{d-2}$-many $0$-balls. Thus each $\sim$-class
contains at most $r$-many elements. Therefore, the map $t\to B_{t}$ from $T$
into $K/p^{d-2}{\mathcal{O}}$ has fibres of cardinality at most $r$. Since
$K/r{\mathcal{O}}$ is definably isomorphic to $K/{\mathcal{O}}$, we are done.

Case 1(b) $d+k<0$ for all $k\in\mathbb{N}$.

For each $t\in T$, let $S^{\prime}_{t}:=\\{b\in S_{t}:(\exists b^{\prime}\in
S_{t})\,d(b,b^{\prime})=d\\}$. Replacing $S_{t}$ with $S^{\prime}_{t}$ we may
assume that every $b\in S_{t}$ has some $b^{\prime}\in S_{t}$ such that
$d(b,b^{\prime})=d$.

Let $\mathcal{B}$ be the collection of all balls of radius $d+1$ in $K$. For
any $B\in\mathcal{B}$ there is no $t\in T$ with $\bigcup S_{t}\subseteq B$,
for otherwise we could find $b,b^{\prime}\in S_{t}$ with $d(b,b^{\prime})=d$
and $b,b^{\prime}\subseteq B$, which is absurd. Let $X=\bigcup_{t\in T}S_{t}$.
By assumption, $X$ is a union of infinitely many $0$-balls. By Lemma 2.3 $X$
contains some ball $B_{\gamma}(x_{0})$ with $\gamma+k<0$ for all
$k\in\mathbb{N}$. By Lemma 2.1(2) $B_{\gamma}(x_{0})$ intersects infinitely
many of the $S_{t}$.

Consider the ball $B_{0}:=B_{d+1}(x_{0})\in\mathcal{B}$. If $d+1\leq\gamma$
then $B_{0}\supseteq B_{\gamma}(x_{0})$ and hence it clearly intersects
infinitely many of the $S_{t}$. If $d+1>\gamma$ then $B_{0}\subseteq
B_{\gamma}(x_{0})$, thus (since $d<\mathbb{Z}$) $B_{0}$ contains infinitely
many $0$-balls inside $B_{\gamma}(x_{0})$. Because every such ball belongs to
some $S_{t}$ it follows that $B_{0}$ intersects infinitely many of the $S_{t}$
as well.

In abuse of notation we write $B_{0}\cap S_{t}$ for the set of balls in
$S_{t}$ which are contained in $B_{0}$. If $|B_{0}\cap S_{t}|\leq 1$ for all
$t\in T$, then by passing to a definable subset of $T$ we may assume that
$|B_{0}\cap S_{t}|=1$ for all $t\in T$ and as a conclusion we get a definable
function (with finite fibers) from $T$ into $K/{\mathcal{O}}$. Otherwise, by
replacing $S$ with $\\{S_{t}\cap B_{0}:t\in T\\}$, we may finish the proof by
induction on $K$, since as we noted above, $|S_{t}\cap B_{0}|<n$.

Case 2 For every $d\in\Gamma$ the set $\\{t\in T:d(t)=d\\}$ is finite.

In this case the map $t\mapsto d(t)$ a finite-to-one function into $\Gamma$. ∎

We are now in position to obtain the second goal of our strategy:

######  Proposition 2.5.

Let $X=\\{X_{t}:t\in T\\}$ be a definable family of infinite pairwise disjoint
subsets of $K$. Then,

  1. (1)

There exists a definable set $D\subseteq K$ such that for each $t\in T$, the
set $D\cap X_{t}$ is the union of finitely many balls of equal radius.

  2. (2)

There exists a definable infinite subfamily $T^{\prime}\subseteq T$ and a
definable finite-to-one $f$ from $T^{\prime}$ to either $\Gamma$ or
$K/{\mathcal{O}}$.

###### Proof.

Recall that a definable set $D\subseteq K$ is bounded if it is contained in
some ball $B_{\gamma}(0)$. We first reduce to the case where all the $X_{t}$
are bounded, by replacing each $X_{t}$ with the set $B_{\gamma(t)}(0)\cap
X_{t},$ for $\gamma(t)\in\Gamma$ maximal satisfying
$\mathrm{Int}(B_{\gamma}(0)\cap X_{t})\neq\emptyset$ (since each $X_{t}$ is
infinite and $K$ has no definable infinite discrete sets, each $t$ has some
$\gamma\in\Gamma$ such that $B_{\gamma}(0)\cap X_{t}$ is infinite, so has non
empty interior). The family $\\{B_{\gamma(t)}(0)\cap X_{t}:t\in T\\}$ is
definable, so we may assume that each $X_{t}$ is bounded.

Consequently, for any $t\in T$ the set $\\{\gamma\in\Gamma:(\exists a\in
X_{t})(B_{\gamma}(a)\subseteq X_{t})\\}$ has a minimum. Let
$\gamma_{t}\in\Gamma$ be this minimum. By minimality of $\gamma_{t}$ and Lemma
2.3, for any $t\in T$ $X_{t}$ contains only finitely many balls of radius
$\gamma_{t}$. We may now choose, uniformly in $t$, the set $D_{t}\subseteq
X_{t}$, consisting of the union of those balls of radius $\gamma_{t}$ which
are contained in $X_{t}$. Let $D=\bigcup_{t}D_{t}$. This ends (1).

Consider the definable map $T\to\Gamma$ sending $t$ to $\gamma_{t}$.

If the map is finite-to-one, we are done. Otherwise, there exists
$\gamma\in\Gamma$ with $\\{t\in T:\gamma_{t}=\gamma\\}$ infinite. By passing
to a definable subfamily, we may assume that $\gamma_{t}=\gamma$ for all $t\in
T$. After rescailing, we may assume that $\gamma=0$. We are now in the
situation to apply Lemma 2.4. ∎

###  2.2. Interpretable fields

We now return to our problem of an interpretable field ${\mathcal{F}}$. We
start with the following general lemma. It uses the well known coding of
finite sets using symmetric functions:

######  Fact 2.6.

Let $(L,+,\cdot,\dots)$ be any field, possibly with additional structure. Then
$L$ eliminates bounded finite imaginaries. I.e. if $\\{X_{t}\subseteq
L^{n}:t\in T\\}$ is a definable family of finite sets uniformly bounded in
size then there exists a definable map $f:T\to L^{m}$, for some integer $m$,
satisfying that $f(t_{1})=f(t_{2})$ if and only if $X_{t_{1}}=X_{t_{2}}$.

######  Lemma 2.7.

Let ${\mathcal{F}}$ be a field interpretable in an $\aleph_{0}$-saturated
structure ${\mathcal{M}}$. Then there exists an infinite definable subset
$S\subseteq{\mathcal{F}}$ that is in definable bijection with $M/E$ for some
definable equivalence relation $E$ on $M$.

###### Proof.

For any infinite definable subset $S\subseteq{\mathcal{F}}$ there exist an
integer $n$, definable subset $X\subseteq M^{n}$ and a definable equivalence
relation $E$ such that $S$ is in definable bijection with $X/E$.

Choose $S$ so that $n$ is minimal possible and let $X$ and $E$ be the
corresponding definable set $X\subseteq K^{n}$ and definable equivalence
relation, respectively. Assume, for simplicity, that $S=X/E$. We claim that
$n=1$. Otherwise, let $\pi:X\to M^{n-1}$ be the projection onto the first
$n-1$ coordinates. Let $W=\pi(X)$. If for some $w\in W$ the set $X_{w}/E$ is
infinite, where $X_{w}:=X\cap\pi^{-1}(w)$ the set $X_{w}$ contradicts the
minimality of $n$ (since we can definably identify $X_{w}$ with a subset of
$M$).

So $|X_{w}/E|$ is finite for all $w\in W$. By saturation, there exists a
uniform bound $k$ on $|X_{w}/E|$ as $w$ ranges over $W$. Reducing $X$, we may
assume that $|X_{w}/E|=r$ (for some $r\leq k$) for all $w\in W$. This gives a
definable correspondence $C:W\to{\mathcal{F}}$ given by $w\mapsto X_{w}/E$.
Using Fact 2.6 we can replace $C$ with a function $c:W\to{\mathcal{F}}^{r}$
whose image is infinite. So there is a projection
$\tau:{\mathcal{F}}^{r}\to{\mathcal{F}}$ such that $\tau\circ c$ has infinite
image in ${\mathcal{F}}$. Define an equivalence relation $E^{\prime}$ on $W$
by $E^{\prime}(u,v)$ if $\tau\circ c(u)=\tau\circ c(v)$. This gives us a
bijection between an infinite subset of ${\mathcal{F}}$ and $W/E^{\prime}$.
Since $W\subseteq M^{n-1}$, this contradicts the minimality of $n$.

Hence $S$ is in definable bijection with $X/E$, where $X\subseteq M$ is a
definable subset. By possibly enlarging $S$ by one element, we may assume that
$X=M$. ∎

We now add our underlying assumption that $K$ is a $P$-minimal field. We first
note that neither $K/{\mathcal{O}}$ nor $\Gamma$ “eliminate
$\exists^{\infty}$” and this remains true for any infinite definable subset.

######  Lemma 2.8.

Let $V$ be either $\Gamma$ or $K/{\mathcal{O}}$ with the induced structure.
Let $X$ be an infinite definable subset of $V$. Then there exists a formula
$\varphi(x,y)$ satisfying that for every $n<\omega$ there exists $b_{n}$ such
that $\varphi(V,b_{n})\subseteq X$, $|\varphi(V,b_{n})|<\omega$ and
$\sup_{n}|\varphi(V,b_{n})|=\omega$.

###### Proof.

As usual, there is no harm assuming that $K$ is $\aleph_{0}$-saturated. Assume
that $V=\Gamma$. Since $X$ is infinite, by quantifier elimination, we may
assume without loss of generality that $X$ is of the form
$\\{x\in\Gamma:\gamma_{1}<x<\gamma_{2}\land P_{m}(x)\\}$ for some
$\gamma_{1},\gamma_{2}\in\Gamma\cup\\{\pm\infty\\}$. Fix some $a\in X$. Since
$X$ is infinite, at least one of $\gamma_{1}$ or $\gamma_{2}$ is not in the
same archimedean component as $a$. So the set $\varphi(x,n):=x\in
X\cap(a-n,a+n)$ satisfies the requirements.

Assume that $V=K/{\mathcal{O}}$. Let $\pi:K\to K/{\mathcal{O}}$ be the natural
projection and $Y=\pi^{-1}(X)$. Since $X$ is infinite, $Y$ contains infinitely
many balls of radius $0$. By Lemma 2.3, there exists $a\in Y$ such that
$B_{-k}(a)\subseteq Y$ for all $k\in\mathbb{N}$. By Lemma 2.1, for each
$k\in\mathbb{N}$, $B_{-k}(a)$ contains only finitely many, say $n_{k}$, balls
of radius $0$. In fact, $\sup_{k\in\mathbb{N}}n_{k}=\omega$ by Lemma 2.1(2).
By choosing elements $b_{k}\in K$ with $v(b_{k})=-k$, the definable sets
$\pi(B_{v(b_{k})}(a))\subseteq X$ satisfy the requirements. ∎

######  Proposition 2.9.

For $V=\Gamma$ or $V=K/{\mathcal{O}}$, there is no infinite definable subset
$I\subseteq{\mathcal{F}}$ and finite-to-one definable function $f:I\to V$.

###### Proof.

Assume towards a contradiction that such $I\subseteq{\mathcal{F}}$ and $f:I\to
V$ existed. Since $I$ is infinite and $f$ is finite-to-one, $f(I)\subseteq V$
is a definable infinite subset. Let $\varphi(x,y)$ and
$\\{b_{n}:n\in\mathbb{N}\\}$ be as provided by Lemma 2.8. Let $\psi(u,y)$ be
the formula $(\exists x)(u\in I\wedge f(u)=x\wedge\varphi(x,y))$. Since $f$ is
finite-to-one, for every $n\in\mathbb{N}$,
$n<|\psi(\mathcal{F},b_{n})|<\omega$. On the other hand, $\mathcal{F}$, with
its induced structure, is a field of finite $dp$-rank, and thus we get a
contradiction to [3, Corollary 2.2]. ∎

We are ready to show that no interpretable field contains an infinite
definable subset almost internal to either $\Gamma$ or $K/{\mathcal{O}}$:

######  Corollary 2.10.

There exists a definable dp-minimal $I\subseteq{\mathcal{F}}$, and a definable
injection $I\hookrightarrow K$.

###### Proof.

Let $X\subseteq K^{n}$ be a definable set $E$ a definable equivalence relation
on $X$ such that $X/E$ is the universe of ${\mathcal{F}}$. By Lemma 2.7 there
exists a definable infinite $S\subseteq{\mathcal{F}}$ and an equivalence
relation $E^{\prime}$ on $K$ such that $K/E^{\prime}$ is in definable
bijection with $S$.

We first claim that only finitely many $E^{\prime}$-classes could be infinite.
Indeed, if $E^{\prime}$ had infinitely many infinite classes then by
restricting to those classes with nonempty interior, we may assume that all
classes are infinite. By applying Proposition 2.5, we obtain a definable
injection of some infinite subset of ${\mathcal{F}}$ into either $\Gamma$ or
$K/{\mathcal{O}}$, contradicting Proposition 2.9.

We may, therefore, assume that all $E^{\prime}$-classes are finite, and by
uniform finiteness, can restrict to the case where all classes are of size $m$
for some $m\in\mathbb{N}$. The definable quotient map $\pi:K\to K/E^{\prime}$
gives rise to a definable family $\\{\pi^{-1}(e):e\in K/E^{\prime}\\}$ of
distinct finite sets of cardinality $m$. Hence by Fact 2.6, there exists a
definable injection $\widehat{f}:K/E^{\prime}\to K^{n}$ for some integer $n$.
Composing, we get a definable injection $f:S\to K^{n}$.

Since $\mathrm{dp\text{-}rk}(S)=1$ the set $f(S)\subseteq K^{n}$ is one-
dimensional and hence, by [13, Proposition 4.6], there exists a definable
infinite $V\subseteq f(S)$ and a projection map $\pi:K^{n}\to K$ such that
$\pi|V$ is a bijection with an open subset of $K$. ∎

##  3\. Subset of ${\mathcal{F}}$ strongly internal to $K$

Let $K$ be a $P$-minimal valued. The previous section was concluded with the
proof that there exists an infinite $I\subseteq\mathcal{F}$ “strongly
internal” to $K$, where we borrow the following terminology from [5]:

######  Definition 3.1.

A definable set $S\subseteq{\mathcal{F}}$ is strongly internal to $K$ over $A$
if there exists an $A$-definable injection $f:S\to K^{n}$ for some $n$. It is
called strongly internal to $K$ if it is strongly internal over some $A$.

Since, by [12, Theorem 0.3.$\oplus_{0}$], dp-rank is additive we may conclude:

######  Remark 3.2.

If $S$ is strongly internal to $K$ over $A$ then for any $a,b\in S$

$\mathrm{dp\text{-}rk}(a,b/A)=\mathrm{dp\text{-}rk}(a/bA)+\mathrm{dp\text{-}rk}(b/A).$

In the present section we study subsets of ${\mathcal{F}}$ strongly internal
to $K$ of maximal dp-rank. Our aim is to show that such sets, at least on some
generic subset, are not too far from being closed under the field operations.
The proof is built on the analogous statement from [5]. Specifically, if
$Y\subseteq{\mathcal{F}}$ is of maximal dp-rank among all the subsets of
${\mathcal{F}}$ strongly internal to $K$. We show, Lemma 3.5, that the
function $\langle x,y,z\rangle\mapsto xy-z$ maps a generic subset of
$(I+b)\times Y^{2}$ (some $b$) into $Y$.

######  Lemma 3.3.

Let $X\subseteq Z\subseteq K^{n}$ be $\emptyset$-definable sets with
$\mathrm{dp\text{-}rk}(X)=\mathrm{dp\text{-}rk}(Z)$. For any $d\in X$ with
$\mathrm{dp\text{-}rk}(d)=\mathrm{dp\text{-}rk}(X)$ there exists an open
neighborhood $U\subseteq K^{n}$ such that $U\cap X=U\cap Z$.

###### Proof.

The relative interior of $X$ in $Z$ is the set $D$ of $x\in X$ such that there
exists $U\ni x$ an open subset of $K^{n}$ such that $U\cap Z\subseteq X$. By
[13, Corollary 4.4], $\mathrm{dp\text{-}rk}(X\setminus
D)<\mathrm{dp\text{-}rk}(X)$ and hence $d\in D$ so there is an open set $U\ni
d$ such that $U\cap Z=U\cap X$. ∎

The following lemma plays an important role in our argument.

######  Lemma 3.4.

Let $V\subseteq K^{n}$ be an open set, $x\in K^{m}$ any element and $A$ an
arbitrary set of parameters. Then there exists $B\supseteq A$ and a
$B$-definable open subset $U\subseteq V$ such that
$\mathrm{dp\text{-}rk}(x/B)=\mathrm{dp\text{-}rk}(x/A)$. Moreover, if $x\in V$
then we can find such $U$ with $x\in U$.

###### Proof.

Since $V$ is open, and is not assumed to be defined over $A$, we may assume
that $V=B_{v(g_{1})}(a_{1})\times\dots\times B_{v(g_{n})}(a_{n})$ for some
$a_{1},\ldots,a_{n}\in K$ and $g_{1},\dots,g_{n}\in\Gamma$.

We prove the result by induction on $n$. For $n=1$, we use compactness to find
$r_{1}\in K$ with $v(r_{1})>v(g_{1})$ such that
$\mathrm{dp\text{-}rk}(r_{1}/xA)=1$ and then find $e_{1}\in
B_{v(r_{1})}(a_{1})$, such that $\mathrm{dp\text{-}rk}(e_{1}/r_{1}xA)=1$. We
have $U:=B_{v(r_{1})}(e_{1})\subseteq B_{\gamma_{1}}(a_{1})$ and by exchange,
$\mathrm{dp\text{-}rk}(x/e_{1}r_{1}A)=\mathrm{dp\text{-}rk}(x/A)$, so
$B=e_{1}r_{1}A$ satisfies the lemma. If $x\in V$ then we may start with
$a_{1}=x$, and then $x\in B_{v(r_{1})}(e_{1})$.

As for the general case, we first replace, by induction,
$B_{v(g_{1})}(a_{1})\times\dots\times B_{v(g_{n-1})}(a_{n-1})$ by an open
subset $U_{1}\subseteq K^{n-1}$ definable over $A_{1}\supseteq A$ such that
$\mathrm{dp\text{-}rk}(x/A_{1})=\mathrm{dp\text{-}rk}(x/A)$. Then we apply the
case $n=1$ to the last coordinate with $A_{1}$ replacing $A$. ∎

We can now prove the main lemma of this section that is the main technical
lemma of the paper:

######  Lemma 3.5.

Let $I\subseteq{\mathcal{F}}$ be as provided by Corollary 2.10 and
$Y\subseteq{\mathcal{F}}$ strongly internal to $K$ of maximal dp-rank. Assume
that both are defined over a parameter set $A\subseteq K$.

Then, there exists a definable $J\subseteq I$ and a definable $S\subseteq
Y^{2}$ such that $\mathrm{dp\text{-}rk}(J)=1$ and
$\mathrm{dp\text{-}rk}(S)=2\mathrm{dp\text{-}rk}(Y)$, and there exists $b\in
J$, such that for every $\langle x,y,z\rangle\in J\times S\subseteq J\times
Y^{2}$, we have $(x-b)y+z\in Y$.

Furthermore, if $\langle b,c,d\rangle\in I\times Y^{2}$ are such that
$\mathrm{dp\text{-}rk}(b,c,d/A)=2\mathrm{dp\text{-}rk}(Y)+1$ then we can
choose $J$ and $S\ni\langle c,d\rangle$, definable over some set $B\ni b$ such
that $\mathrm{dp\text{-}rk}(c,d/B)=2\mathrm{dp\text{-}rk}(Y)$.

###### Proof.

We fix $A$-definable injections $g_{1}:I\to K$ and $g_{2}:Y\to K^{l}$. For
simplicity of notation assume $A=\emptyset$. By assumption
$\mathrm{dp\text{-}rk}(I)=1$, and denote $\mathrm{dp\text{-}rk}(Y)=n$ for some
integer $n>0$ and fix some $\langle b,c,d\rangle\in I\times Y\times Y$ such
that $\mathrm{dp\text{-}rk}(b,c,d)=2n+1$. Note that this implies that

$2n+1=\mathrm{dp\text{-}rk}(b,c,d)\leq\mathrm{dp\text{-}rk}(b,d/c)+\mathrm{dp\text{-}rk}(c)\leq\mathrm{dp\text{-}rk}(b,d)+\mathrm{dp\text{-}rk}(c)\leq
2n+1$

and hence $\mathrm{dp\text{-}rk}(b,d/c)=n+1$, $\mathrm{dp\text{-}rk}(c)=n$.
Similarly, $\mathrm{dp\text{-}rk}(b,c)=n+1$ and
$\mathrm{dp\text{-}rk}(b/c)=1$.

For $\langle x,y,z\rangle\in I\times Y\times Y$ consider the function
$f_{y}(x,z)=xy-z$. Let $e=f_{c}(b,d)$.

######  Claim 3.5.1.

$b\notin\operatorname{acl}(c,e)$.

###### Proof.

Assume towards a contradiction that there existed an algebraic formula
$\varphi(x,e,c)$ isolating $\operatorname{tp}(b/c,e)$, in particular,
$\varphi(x,e,c)$ would imply that $x\in I$. By the definition of $f_{c}$,
$d\in\operatorname{dcl}(b,c,e)$, therefore $\varphi(x,e,c)\wedge f_{c}(x,y)=e$
is an algebraic formula isolating $\operatorname{tp}(b,d/c,e)$. Hence, there
is some integer $m$ with $\exists^{=m}(x,y)(\varphi(x,e,c)\wedge
f_{c}(x,y)=e)$. By compactness, there is a formula
$\psi(z,c)\in\operatorname{tp}(e/c)$ that implies
$\exists^{=m}(x,y)(\varphi(x,z,c)\wedge f_{c}(x,y)=z)$. In other words, for
$D:=\psi({\mathcal{F}},c)$, there is a 1-to-$m$ definable partial
$c$-definable correspondence $G$ from $D$ into $I\times Y$, sending $e$ to
$(b,d)$. Note that if $e_{1}\neq e_{2}\in D$ then $f_{c}^{-1}(e_{1})\cap
f_{c}^{-1}(e_{2})=\emptyset$, thus $G(e_{1})\cap G(e_{2})=\emptyset$.

The image of $G$ in $I\times Y$ is a $c$-definable set containing $\langle
b,d\rangle$ and since $\mathrm{dp\text{-}rk}(b,d/c)=n+1$, we have
$\mathrm{dp\text{-}rk}(\mathrm{Im}(G))=n+1$.

Recall that $(g_{1},g_{2}):I\times Y\to K\times K^{l}$ is a
$\emptyset$-definable injection, so $H=(g_{1},g_{2})\circ G:\mathrm{Dom}(G)\to
K^{1+l}$ is a one-to-$m$ correspondence, with disjoint images corresponding to
distinct $e_{1},e_{2}\in\mathrm{Dom}(G)$. By Fact 2.6, $H$ induces a definable
injection from $\mathrm{Dom}(G)$ into $K^{N}$, for some integer $N$, as
$\mathrm{dp\text{-}rk}(\mathrm{Im}(G))>\mathrm{dp\text{-}rk}(Y)$ this is a
contradiction to the choice of $Y$. ∎

We thus conclude that
$b\notin\operatorname{acl}(c,f_{c}(b,d))=\operatorname{acl}(c,e)$ and in
particular $f_{c}^{-1}(e)\subseteq I\times Y$ is infinite. Notice that by the
definition of $f_{c}$, for every $b^{\prime}\in I$, and $d_{1}\neq d_{2}\in
Y$, we have $f_{c}(b_{1},e_{1})\neq f_{c}(b_{1},e_{2})$, thus the projection
of $f_{c}^{-1}(e)$ on the first coordinate, call it $J$, is infinite. By
definition of $J$, for every $x\in J$ there is $z\in Y$ with $xc-z=bc-d=e$ and
since $\mathcal{F}$ is a field the map from $J$ to $Y$ mapping $x\in J$ to
$z=xc-e=(x-b)c+d$ is injective.

Because $g_{1}$ is injective, $g_{1}(J)$ is an infinite subset of $K$ and
hence has non-empty interior. By Lemma 3.4, we can find a $B$-definable open
subset $U\subseteq g_{1}(J)$, such that

$\mathrm{dp\text{-}rk}(g_{2}(c),g_{2}(d)/Bg_{1}(b))=\mathrm{dp\text{-}rk}(g_{2}(c),g_{2}(d)/g_{1}(b))=2n.$

Set $J^{\prime}=g_{1}^{-1}(U)$. It is an infinite $B$-definable subset of $J$,
so for every $x$ in $J^{\prime}$, we have $(x-b)c+d\in Y$. Let

$S=\\{\langle y,z\rangle\in Y^{2}:(\forall x\in J^{\prime})((x-b)y+z\in
Y)\\}.$

Note that $\langle c,d\rangle\in S$ and that $S$ is $Bb$-definable. Because
$(c,d,b)$ and $(g_{1}(b),g_{2}(c),g_{2}(d))$ are inter-definable over
$\emptyset$ it follows that $\mathrm{dp\text{-}rk}(c,d/Bb)=2n$, so the proof
is completed. ∎

######  Remark 3.6.

Note that in Lemma 3.5 we do not claim that the set $I\times Y\times Y$ is
mapped under $xy+z$, (or under $(x-b)y+z$) onto a subset of ${\mathcal{F}}$
whose dp-rank equals $\mathrm{dp\text{-}rk}(Y)$ (this will turn out to be true
once we complete the proof of the main theorem). Instead, at this stage, we
only found a subset $J\times S\subseteq I\times Y\times Y$ of full dp-rank
whose image has the same rank as $\mathrm{dp\text{-}rk}(Y)$.

##  4\. infinitesimal neighbourhoods and topology

In the present section we use methods similar to those in [5] in order to
construct a type definable, “infinitesimal” subgroup of $({\mathcal{F}},+)$
which is definably embedded into $K^{n}$ for some $n$. The field
${\mathcal{F}}$ itself will later be embedded into some $K^{m}$, using the
subgroup of infinitesimals.

We assume in this section that $K$ is $P$-minimal. We will repeatedly use the
fact (already mentioned in the intoduction) that in $P$-minimal
$\operatorname{acl}$ satisfies the Steiniz Exchange prilnciple. Throughout
$+$, $-$, $\cdot$ and $(\,)^{-1}$ denote the operations in ${\mathcal{F}}$.

######  Definition 4.1.

For any $Z\subseteq{\mathcal{F}}$ and a definable injective $g:Z\to K^{m}$,
let $\tau_{Z,g}$ be the topology on $Z$ given by $\\{g^{-1}(U):U\subseteq
K^{m}\text{ open}\\}$

We observe that because $K^{n}$ has a definable basis for its topology (given,
say, by the family of open balls), each $\tau_{Z,g}$ has a definable basis as
well. Also, if $Z_{1},Z_{2}$ are both strongly internal to $K$ via definable
injections $g_{1}$ and $g_{2}$, respectively, then $\tau_{Z_{1}\times
Z_{2},g_{1}\times g_{2}}$ is equal to the topology generated by
$\tau_{Z_{1},g_{1}}\times\tau_{Z_{2},g_{2}}$.

We will repeatedly use the following

######  Lemma 4.2.

Let $Z_{1},Z_{2},\subseteq{\mathcal{F}}$ be strongly internal to $K$,
witnessed by $A$-definable injections $g:Z_{1}\to K^{m}$ and $h:Z_{2}\to
K^{n}$. Let $f:Z_{1}\to Z_{2}$ be an $A$-definable function between the
topological spaces $(Z_{1},\tau_{Z_{1},g})$ and $(Z_{2},\tau_{Z_{2},h})$. Then
the set $C\subseteq Z_{1}$ of $\tau_{Z_{1},g}$-continuous points of $f$ is
$A$-definable, and $\mathrm{dp\text{-}rk}(Z_{1}\setminus
C)<\mathrm{dp\text{-}rk}(Z_{1})$.

###### Proof.

The definability of $C$ follows from the definability of a basis for
$\tau_{Z,g}$. The dimension statement follows from the analogous result for
definable functions on subsets of $K^{m}$, in $P$-minimal fields, [4, Theorem
5.1]. ∎

We thus have:

######  Lemma 4.3.

Let $Z\subseteq{\mathcal{F}}$ be definable and $g:Z\to K^{m}$, $h:Z\to K^{n}$
two $A$-definable injections. Then, $\tau_{Z,g}$ and $\tau_{Z,h}$ agree at
every $z\in Z$ with $\mathrm{dp\text{-}rk}(z/A)=\mathrm{dp\text{-}rk}(Z)$.
Namely, there is a common basis for the $\tau_{Z,g}$-neighbourhoods and the
$\tau_{Z,h}$-neighbourhoods of $z\in Z$.

###### Proof.

Apply Lemma 4.2 to $\operatorname{id}:Z\to Z$.∎

######  Definition 4.4.

For $Z\subseteq{\mathcal{F}}$ a definable set, $g:Z\to K^{m}$ a definable
injection, and $d\in Z$, let $\nu_{Z,g}(d)$ be the partial type given by all
definable $\tau_{Z,g}$-open sets containing $d$. We call it the infinitesimal
neighborhood of $d$ with respect to $\tau_{Z,g}$.

######  Remark 4.5.

If $Z_{1},Z_{2}$ are both strongly internal to $K$ over $A$ via injections
$g_{1},g_{2}$, respectively, and $\langle d_{1},d_{2}\rangle\in Z_{1}\times
Z_{2}$ with
$\mathrm{dp\text{-}rk}(d_{1},d_{2}/A)=\mathrm{dp\text{-}rk}(Z_{1})+\mathrm{dp\text{-}rk}(Z_{2})$
then $\nu_{Z_{1},g_{1}}(d_{1})\times\nu_{Z_{2},g_{2}}(d_{2})=\nu_{Z_{1}\times
Z_{2},g_{1}\times g_{2}}(d_{1},d_{2})$ (see above discussion on
$\tau_{Z_{1},g_{1}}\times\tau_{Z_{2},g_{2}}$).

By Lemma 4.3, we have:

######  Corollary 4.6.

If $Z\subseteq{\mathcal{F}}$ is strongly internal to $K$ over $A$ and $d\in Z$
is such that $\mathrm{dp\text{-}rk}(d/A)=\mathrm{dp\text{-}rk}(Z)$ then the
infinitesimal neighborhood of $d$ in $Z$ does not depend on any particular
$A$-definable injection of $Z$ into some $K^{m}$. We denote it by
$\nu_{Z}(d)$.

We also have:

######  Lemma 4.7.

Assume that $Z\subseteq{\mathcal{F}}$ is strongly internal to $K$ over $A$ as
witnessed by $g$, and $Z_{1}\subseteq Z$ is $A$-definable with
$\mathrm{dp\text{-}rk}(Z_{1})=\mathrm{dp\text{-}rk}(Z)$. If $d\in Z_{1}$ is
such that $\mathrm{dp\text{-}rk}(d/A)=\mathrm{dp\text{-}rk}(Z)$ then
$\nu_{Z_{1}}(d)=\nu_{Z}(d)$.

###### Proof.

By Lemma 3.3, the topologies $\tau_{Z,g}$ and $\tau_{Z_{1},g}$ agree on a
neighborhood of $d$, thus $\nu_{Z}(d)=\nu_{Z_{1}}(d)$.∎

######  Lemma 4.8.

Let $Y_{1},Y_{2}\subseteq{\mathcal{F}}$ be strongly internal to $K$ over $A$.
If $f:Y_{1}\to Y_{2}$ is an $A$-definable partial function, and
$a\in\operatorname{dom}(f)$ with
$\mathrm{dp\text{-}rk}(a/A)=\mathrm{dp\text{-}rk}(Y_{1})$ then $f$ takes the
partial type $\nu_{Y_{1}}(a)$ into $\nu_{Y_{2}}(f(a))$. If $f$ is injective
then $f(\nu_{Y_{1}}(a))=\nu_{Y_{2}}(f(a))$.

###### Proof.

By Lemma 4.2, $f$ is continuous at $a$ with respect to $\tau_{Y_{1},g}$ and
$\tau_{Y_{2},h}$ (for any $A$-definable $g,h$ witnessing the strong
internality of $Y_{1}$, $Y_{2}$, respectively). It is now easy to conclude
that $f$ maps $\nu_{Y_{1}}(a_{1})$ into $\nu_{Y_{2}}(f(a))$.∎

The main result in this section is the following.

######  Lemma 4.9.

Assume that $Y_{1},Y_{2},Y_{3}\subseteq{\mathcal{F}}$ are strongly internal to
$K$ over $A$, with
$\mathrm{dp\text{-}rk}(Y_{1})=\mathrm{dp\text{-}rk}(Y_{2})=\mathrm{dp\text{-}rk}(Y_{3})=k$,
and assume that $Y_{1}+Y_{2}\subseteq Y_{3}$. Then

  1. (1)

For every $c\in Y_{1}$ and $d\in Y_{2}$ such that
$\mathrm{dp\text{-}rk}(c,d/A)=2k$, we have
$\nu_{Y_{1}}(c)-c=\nu_{Y_{2}}(d)-d$, and for every $d^{\prime}\in Y_{2}$ with
$\mathrm{dp\text{-}rk}(d^{\prime}/A)=k$ we have
$\nu_{Y_{2}}(d)-d=\nu_{Y_{2}}(d^{\prime})-d^{\prime}$.

  2. (2)

For every $d\in Y_{2}$ such that $\mathrm{dp\text{-}rk}(d/A)=k$, the partial
type $\nu_{Y_{2}}(d)-d$ is a type definable subgroup of $({\mathcal{F}},+)$

###### Proof.

It is convenient to work in a $|K|^{+}$-saturated elementary extension
${\mathcal{K}}$ of $K$ and work with the realizations in ${\mathcal{K}}$ of
the infinitesimal types. We first make some general observations.

Consider the function $F:Y_{1}\times Y_{2}\to Y_{3}$,
$F(y_{1},y_{2})=y_{1}+y_{2}$. Applying Lemma 4.8 to $F$ (and using Remark
4.5), we see that
$F(\nu_{Y_{1}}(c)\times\nu_{Y_{2}}(d))\subseteq\nu_{Y_{3}}(c+d)$. Consider
also the function $H:Y_{1}\times Y_{3}\to Y_{2}$,
$H(y_{1},y_{3})=y_{3}-y_{1}$. By Lemma 4.8 (and Remark 4.5), it sends
$\nu_{Y_{1}}(c)\times\nu_{Y_{3}}(c+d)$ into $\nu_{Y_{2}}(d)$. It follows that
for every $c_{1}\in\nu_{Y_{1}}(c)$, the function $y\mapsto c_{1}+y$ is a
bijection between $\nu_{Y_{2}}(d)$ and $\nu_{Y_{3}}(c+d)$. Indeed,
$F(c_{1},-)$ is a function from $\nu_{Y_{2}}(d)$ into $\nu_{Y_{3}}(c+d)$,
whose inverse is $H(c_{1},-)$.

By the same observations we show that for every $d_{1}\in\nu_{Y_{2}}(d)$, the
function $x\mapsto x+d_{1}$ is a bijection of $\nu_{Y_{1}}(c)$ and
$\nu_{Y_{3}}(c+d)$.

In order to prove (1) we show that the sets of realizations in ${\mathcal{K}}$
of the partial types $\nu_{Y_{1}}(c)-c$ and $\nu_{Y_{2}}(d)-d$ are equal.

Let $c_{1}\in\nu_{Y_{1}}(c)$. By our above discussion, there is
$d_{1}\in\nu_{Y_{2}}(d)$ such that $c_{1}+d_{1}=c+d$. Thus,
$c_{1}-c\in\nu_{Y_{2}}(d)-d$, so $\nu_{Y_{1}}(c)-c\subseteq\nu_{Y_{2}}(d)-d$.
The other inclusion is proved similarly, hence
$\nu_{Y_{1}}(c)-d=\nu_{Y_{2}}(d)-d$.

Assume next that $d^{\prime}\in Y_{2}$ is such that
$\mathrm{dp\text{-}rk}(d^{\prime}/A)=k$. We fix $c\in Y_{1}$ such that
$\mathrm{dp\text{-}rk}(c/d,d^{\prime}A)=k$. It follows, by Remark 3.2, that
$\mathrm{dp\text{-}rk}(c,d/A)=2k$ and
$\mathrm{dp\text{-}rk}(c,d^{\prime}/A)=2k$. And then, by what we just saw,
$\nu_{Y_{2}}(d)-d=\nu_{Y_{1}}(c)-c=\nu_{Y_{2}}(d^{\prime})-d^{\prime}$.

(2) We claim that $\nu_{Y_{2}}(d)-d$ is a subgroup of $({\mathcal{F}},+)$.
Given $d_{1},d_{2}\in\nu_{Y_{2}}(d)$, we need to show that
$(d_{1}-d)-(d_{2}-d)=d_{1}-d_{2}$ is also in $\nu_{Y_{2}}(d)-d$.

By our above observation, there is $c_{1}\in\nu_{Y_{1}}(c)$ such that
$c+d_{1}=c_{1}+d_{2}$. It follows that
$d_{1}-d_{2}=c_{1}-c\in\nu_{Y_{1}}(c)-c$. By what we just showed,
$\nu_{Y_{1}}(c)-c=\nu_{Y_{2}}(d)-d$, hence $d_{1}-d_{2}\in\nu_{Y_{2}}(d)-d$.∎

We are now ready to construct the maximal infinitesimal subgroup of
$({\mathcal{F}},+)$.

######  Proposition 4.10.

Let $Y\subseteq{\mathcal{F}}$ be strongly internal to $K$ over $A$, of maximal
dp-rank $n$.

  1. (1)

Let $d\in Y$ be with $\mathrm{dp\text{-}rk}(d/A)=n$. Then $\nu_{Y}(d)-d$ is a
type definable subgroup of $({\mathcal{F}},+)$. Moreover, it is independent of
the choice of $d$, and we denote it $\nu_{Y}$.

  2. (2)

For any strongly internal $Y_{0}\subset{\mathcal{F}}$, with
$\mathrm{dp\text{-}rk}(Y_{0})=n$, $\nu_{Y}=\nu_{Y_{0}}$.

###### Proof.

(1) We return to our one dimensional $I\subseteq{\mathcal{F}}$ which is
strongly internal to $K$ (Corollary 2.10). We fix any $b\in I$ and $c\in Y$
such that $\mathrm{dp\text{-}rk}(b,c,d/A)=2n+1$ and apply Lemma 3.5.

We obtain $A\subseteq B$-definable $J\subseteq I$ and $S\subseteq Y\times Y$
containing $\langle c,d\rangle$, with $\mathrm{dp\text{-}rk}(J)=1$,
$\mathrm{dp\text{-}rk}(S)=2n$, and $B\ni b$, such that the map $(x-b)y+z$
sends $I\times S$ into $Y$ and such that $\mathrm{dp\text{-}rk}(c,d/B)=2n$.
Because $\langle c,d\rangle\in S\subseteq Y\times Y$ has maximal dp-rank, it
follows by Lemma 3.4 that there exist $B\subseteq B_{0}$-definable
$\tau_{Y}$-open definable sets $Y_{1},Y_{2}\subseteq Y$, neighborhoods of $c$
and $d$, respectively, such that $Y_{1}\times Y_{2}\subseteq S$ and
$\mathrm{dp\text{-}rk}(c,d/B_{0})=\mathrm{dp\text{-}rk}(c,d/B)=2n$.

Since $\mathrm{dp\text{-}rk}(J\times S)=2n+1$, we can find $b_{1}\in J$ with
$\mathrm{dp\text{-}rk}(b_{1},c,d/B_{0})=2n+1$. Let
$Y_{1}^{\prime}=(b_{1}-b)Y_{1}$ and $c^{\prime}=(b-b_{1})c$. By our
assumptions, $Y_{1}^{\prime}+Y_{2}\subseteq Y$ and
$\mathrm{dp\text{-}rk}(Y_{1}^{\prime})=\mathrm{dp\text{-}rk}(Y_{2})=\mathrm{dp\text{-}rk}(Y)$.
Therefore, the assumptions of Lemma 4.9 are satisfied, with
$\mathrm{dp\text{-}rk}(c^{\prime},d/b_{1}B_{0})=2n$. Thus, $\nu_{Y_{2}}(d)-d$
is a subgroup of $({\mathcal{F}},+)$, equal to
$\nu_{Y_{1}^{\prime}}(c^{\prime})-c^{\prime}$. By Lemma 4.7,
$\nu_{Y_{2}}(d)=\nu_{Y}(d)$ and
$\nu_{Y_{1}^{\prime}}(c)-c=\nu_{(b-b_{1})Y}(c^{\prime})-c^{\prime}$. Thus,
$\nu_{Y}(d)-d=\nu_{(b-b_{1})Y}(c^{\prime})-c^{\prime}$ is a subgroup of
$({\mathcal{F}},+)$.

If we now have $d^{\prime}\in Y$ such that
$\mathrm{dp\text{-}rk}(d^{\prime}/A)=n$, then we choose $c\in Y$ with
$\mathrm{dp\text{-}rk}(c,d/A)=\mathrm{dp\text{-}rk}(c,d^{\prime}/A)=2n$, and
then $b_{1}\in J$ such that
$\mathrm{dp\text{-}rk}(b_{1},c,d/B_{0})=\mathrm{dp\text{-}rk}(b_{1},c,d^{\prime}/B_{0})=2n+1$.
Repeating the above argument we conclude that

$\nu_{Y}(d)-d=\nu_{(b-b_{1})Y}(c^{\prime})-c^{\prime}=\nu_{Y}(d^{\prime})-d^{\prime}.$

(2) Let $Y_{0}\subseteq{\mathcal{F}}$ be any set strongly internal to $K$ with
$\mathrm{dp\text{-}rk}(Y_{0})=n$ and let $Y^{\prime}=Y\cup Y_{0}$. It follows
from Lemma 4.7 that $\nu_{Y}=\nu_{Y^{\prime}}=\nu_{Y_{0}}$. ∎

######  Corollary 4.11.

The partial type $\nu$ is invariant under multiplication by scalars from
${\mathcal{F}}$.

###### Proof.

Let $c\in{\mathcal{F}}$ and let $Y\subseteq{\mathcal{F}}$ be any definable
subset strongly internal to $K$ over $\emptyset$, of maximal dp-rank $n$. Let
$d\in Y$ be such that $\mathrm{dp\text{-}rk}(d/c)=n$, so also
$\mathrm{dp\text{-}rk}(c\cdot d)=n$. The function $x\mapsto cx$ sends $Y$ to
$cY$, and by lemma 4.8, it sends $\nu_{Y}(d)$ onto $\nu_{cY}(cd)$. Hence,

$c\nu=c(\nu_{Y}(d)-d)=c\nu_{Y}(d)-cd=\nu_{cY}(cd)-cd=\nu.$

∎

##  5\. Embedding the field ${\mathcal{F}}$ into $M_{n}(K)$

Assume that $K$ is a P-minimal field. In the present section we prove that
${\mathcal{F}}$ is in definable isomorphism with a finite extension of $K$. We
do so by identifying ${\mathcal{F}}$ with a subfield of $M_{n}(K)$. Using dp-
minimality of $K$, we show that ${\mathcal{F}}$ must be a finite extension of
the canonical embedding of $K$ into $M_{n}(K)$.

In order to embed ${\mathcal{F}}$ into the ring of matrices we need to endow
the additive subgroup $\nu$ introduced in the previous section, with a
${\mathcal{D}}$-structure with respect to $K$. That is, we will see that $\nu$
is an open set, and that group operations are differentiable. Recall,

######  Definition 5.1.

Given $U\subseteq K^{n}$ open, a map $f:U\to K^{m}$, is _differentiable_ at
$x_{0}\in U$ if there exists a linear map $D_{x_{0}}f:K^{n}\to K^{m}$ such
that

$\lim_{x\to x_{0}}\frac{f(x)-f(x_{0})-D_{x_{0}}f(x-x_{0})}{x-x_{0}}=0.$

If $f$ is differentiable at a point $x_{0}$ we say that $x_{0}$ is a
${\mathcal{D}}^{1}$-point of $f$ and that $f$ is ${\mathcal{D}}^{1}$ at
$x_{0}$. It is easy to see that if $f$ is definable then set of $x_{0}$’s at
which $f$ is ${\mathcal{D}}^{1}$ is definable as well. For the purposes of the
present section we assume:

(A) For every definable open $D\subseteq K^{n}$, and definable $f:D\to K$,
there exists an open subset $U\subseteq K^{n}$ such that $f\restriction U\cap
D$ is differentiable.

######  Remark 5.2.

It is standard to see that (A) implies that for any such $\emptyset$-definable
$f:D\to K$, any $d\in D$ with
$\mathrm{dp\text{-}rk}(D)=\mathrm{dp\text{-}rk}(d)$ is a
${\mathcal{D}}^{1}$-point of $f$.

By [13, Proposition 4.6] if $S\subseteq K^{n}$ has dimension $d$ then there
exists an open set $U\subseteq K^{n}$ and a projection $\pi:K^{n}\to K^{d}$
such that $\pi(U)$ is open and $\pi:(U\cap S)\to K^{d}$ is a homeomorphism
onto its image. Thus, if $Y\subseteq{\mathcal{F}}$ is strongly internal to $K$
with $\mathrm{dp\text{-}rk}(Y)=n$ we can find $Y_{0}\subseteq Y$ and a
definable injection $g_{0}:Y_{0}\to K^{n}$ with open image. This is the
setting for the following proof, due to Marikova [6], who proved it for
continuity in the o-minimal context. The exact same proof goes through for any
property that is generically true for all definable functions.

######  Proposition 5.3.

Let $Y\subseteq{\mathcal{F}}$ be a definable subset with dp-rank $n$, $g:Y\to
K^{n}$ an injective $A$-definable map with open image, and let ${\mathcal{K}}$
be a $|K|^{+}$-saturated elementary extension of $K$. Then there exists $a\in
Y$, with $\mathrm{dp\text{-}rk}(a/A)=\mathrm{dp\text{-}rk}(Y)$, such that,

  1. (1)

The map $x\mapsto g(x+a)$ induces on $\nu({\mathcal{K}})$ a structure of a
${\mathcal{D}}^{1}$-group, namely, the group operations are
${\mathcal{D}}^{1}$ when read via $g$.

  2. (2)

Any $K$-definable endomorphism
$\alpha:({\mathcal{F}}({\mathcal{K}}),+)\to({\mathcal{F}}({\mathcal{K}}),+)$
sending $\nu({\mathcal{K}})$ to itself is a ${\mathcal{D}}^{1}$-map with
respect to the above differential structure on $\nu$.

###### Proof.

All group operations appearing in the proof are the restriction to $Y$ of
${\mathcal{F}}$ addition (and subtraction).

Let $d\in Y$ be such that
$\mathrm{dp\text{-}rk}(d/A)=\mathrm{dp\text{-}rk}(Y)$. By replacing $Y$ with
$Y-d$, we may assume that $\nu\vdash Y$. Now absorb $A$ and $d$ into the
language. In order to keep notation simple we identify $Y$ with its image
under $g$ (so $g=\operatorname{id}$).

(1) Note that $\nu=\\{U\subseteq K^{n}:U\text{ is $K$-open and
}0_{\mathcal{F}}\in U\\}$ and thus $\nu({\mathcal{K}})$ is open in
${\mathcal{K}}^{n}$.

Since $\nu$ is a group, type definable over $K$, there are, by compactness,
$K$-definable open sets $V_{1},V_{0}$, such that $\nu\vdash V_{1}\subseteq
V_{0}\subseteq K^{n}$ and

$\varphi_{4}:\langle x,y\rangle\mapsto x+y:V_{1}^{2}\to V_{0}.$

Similarly, we find $V_{2}\subseteq V_{1}$ such that

$\varphi_{3}:\langle x,y,z\rangle\mapsto\langle-z,x+y\rangle:V_{2}^{2}\to
V_{1}^{2}.$

We also find $V_{3}\subseteq V_{2}$ such that

$\varphi_{2}:\langle x,y,z\rangle\mapsto\langle x+y,z\rangle:V_{3}^{3}\to
V_{2}^{2},$

and $V_{4}\subseteq V_{3}$ such that

$\varphi_{1}:\langle x,y,z,w\rangle\mapsto\langle w+x,-y,z\rangle:V_{4}^{3}\to
V_{3}^{3}.$

We may assume that all the above are $\emptyset$-definable. Let $a,b\in V_{4}$
with
$\mathrm{dp\text{-}rk}(a,b)=2\mathrm{dp\text{-}rk}(Y)=2\mathrm{dp\text{-}rk}(V_{4})$.
By assumption (A), $\varphi_{1}(x,y,z,b)$ is ${\mathcal{D}}^{1}$ at $(a,a,a)$,
$\varphi_{2}$ is ${\mathcal{D}}^{1}$ at $(b+a,-a,a)$, $\varphi_{3}(x,y,b)$ at
$(b,a)$ and $\varphi_{4}$ at $(b^{-1},b+a)$. Composing, we obtain
$\varphi_{4}\circ\varphi_{3}\circ\varphi_{2}\circ\varphi_{1}(x,y,z)=x-y+z$, so
we have shown that the map $\langle x,y,z\rangle\mapsto x-y+z$ is
${\mathcal{D}}^{1}$ at $(a,a,a)$. In fact, the proof actually provides an open
set, $U\subseteq V_{4}$ which, by compactness, we may take to be
$K$-definable, with $a\in U$, such that $\langle x,y,z\rangle\mapsto
x-y+z:U^{3}\to V_{0}$ is ${\mathcal{D}}^{1}$.

Our goal is to show that the push-forward of $+\restriction\nu^{2}$ and
$-()\restriction\nu$ under the map $x\mapsto x+a$ are ${\mathcal{D}}^{1}$ in
the sense of $K$. Namely, we need to prove that the functions $(x-a)+(y-a)+a$
and $-(x-a)+a$ on $\nu_{Y}(a)^{2}$ and $\nu_{Y}(a)$, respectively, are
${\mathcal{D}}^{1}$. Both follow immediately from our choice of $U$.

(2) Let $e\in\nu({\mathcal{K}})$ with
$\mathrm{dp\text{-}rk}(e/K)=\mathrm{dp\text{-}rk}(\nu)=n$. By assumption
$(A)$, $e$ is a ${\mathcal{D}}^{1}$-point of $\alpha$. Since $\alpha$ is a
homomorphism and $\nu$ is a ${\mathcal{D}}^{1}$-group, $\alpha$ is a
${\mathcal{D}}^{1}$-function on $\nu$. ∎

We need the following easy and well known fact:

######  Remark 5.4.

If $K$ is a dp-minimal field, then $K$ has no definable infinite subfields.
Indeed, if $L$ were such a field then $L$ itself is dp-minimal. And if we had
some $v\in K\setminus L$ we could define $T:L^{2}\to K$ by $(a,b)\mapsto
a+bv$. Since $v$ is $L$-linearly independent of $1$, we get that $T$ is a
linear injection, so $\mathrm{dp\text{-}rk}(T(L^{2}))=2$ which is impossible.

######  Proposition 5.5.

The field ${\mathcal{F}}$ is definably isomorphic to a finite extension of
$K$.

###### Proof.

Let $\nu$ be as before, let $Y\subseteq{\mathcal{F}}$ be strongly internal of
maximal dp-rank $n$ and assume that $g:Y\to K^{n}$ is the corresponding
injective map.

By Corollary 4.11, for any $z\in{\mathcal{F}}$ the function
$\lambda_{z}:x\mapsto zx$ leaves the type $\nu$ invariant. By Proposition 5.3,
$\lambda_{z}$ is a ${\mathcal{D}}^{1}$ map on $\nu({\mathcal{K}})$. Thus, for
every $z\in{\mathcal{F}}$ there exists a definable neighborhood $V\ni
0_{{\mathcal{F}}}$ such that $\lambda_{z}$ is ${\mathcal{D}}^{1}$ on $V$. This
is a first order property which holds in ${\mathcal{K}}$ as well.
Consequently, to each $z\in{\mathcal{F}}$ may associate, definably, the
Jacobian matrix $J_{z}\in M_{n}(K)$ of $\lambda_{z}$ at $0$.

As in the proof of [7, Lemma 4.3], an application of the chain rule [9, Remark
4.1.ii], implies that the map $z\mapsto J_{z}$ is a ring homomorphism sending
$1\in{\mathcal{F}}$ to the identity matrix $I_{n}$. Since ${\mathcal{F}}$ is a
field, the map is injective so we may embed ${\mathcal{F}}$ into a definable
subring of $M_{n}(K)$.

We may now view ${\mathcal{F}}$ as a definable subfield of $M_{n}(K)$. Let
$K_{0}=\\{xI_{n}:x\in K\\}$, where now we take the usual scalar multiplication
in the algebra of matrices. Thus $K_{0}\cap F$ is an infinite definable
subfield of $K$ (as they both contain $I_{n}$ and are of characteristic $0$).
Since $K_{0}\cong K$ is dp-minimal, it has no infinite definable subfield, so
$K_{0}\cap{\mathcal{F}}=K_{0}$ i.e. $K_{0}\subseteq{\mathcal{F}}$. Thus
${\mathcal{F}}$ is a finite extension of $K_{0}$. ∎

Summing all the above we get:

######  Theorem 5.6.

Let $K$ be a $P$-minimal field $K$. If for every definable $f:D\to K$,
$D\subseteq K^{n}$ with non-empty interior, there exists an open subset
$U\subseteq K^{n}$ such that $f\restriction U\cap D$ is differentiable then
every infinite interpretable field is definably isomorphic to a finite
extension of $K$.

Let ${\mathcal{L}}^{an}$ be the subanalytic language for the p-adics, see [1].
For any prime $p$, let $\mathbb{Q}_{p}^{an}$ be the p-adic field in the
subanalytic language.

######  Corollary 5.7.

Let $(K,v)$ be a valued field. Assume that either

  1. (1)

$K$ is elementary equivalent to $\mathbb{Q}_{p}^{an}$ in the subanalytic
language ${\mathcal{L}}^{an}$ or

  2. (2)

$K$ is p-adically closed.

Then every infinite field interpretable in $K$ is definably isomorphic to a
finite extension of $K$.

###### Proof.

In case of (1), $K$ is $P$-minimal by [14, Theorem B]). As a result it is
sufficient to verify assumption (A). Since having non-empty interior is
definable in families and being differentiable is definable (using the
parameters needed to define the function), it is enough to check assumption
(A) for $\emptyset$-definable functions whose domain has non-empty interior
and assume that either $f$ is a definable function in $\mathbb{Q}_{p}^{an}$ or
in a finite extension of $\mathbb{Q}_{p}$ (in the valued field language).
Since every analytic map is differentiable [9, Proposition 6.1], the result
follows readily from [1, Proposition 3.29] in case (1) and [10, Theorem 1.1]
in case (2)111See, also, the discussion in [10, Section 5]. ∎

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