On Dimensions,Standard Part Maps,and p-Adically Closed Fields*

Ningyuan Yao

Abstract. The aim of this paper is to study the dimensions and standard part maps between the field of p-adic numbers Qp and its elementary extension K in the language of rings Lr.We show that for any K-definable set X ?Km,dimK(X) ≥dimQp(X ∩).Let V ?K be convex hull of K over Qp,and st : V →Qp be the standard part map.We show that for any K-definable function f : Km →K,there is definable subset D ?such thatD has no interior,and for all x ∈D,either f(x) ∈V and st(f(st－1(x)))is constant,or f(st－1(x))∩V=?.We also prove that dimK(X)≥dimQp(st(X ∩V m))for every definable X ?Km.

1 Introduction

In[5],L.van den Dries consider a pair(R,V),where R is an o-minimal extension of a real closed field,and V is a convex hull of an elementary submodel M of R.Let μ ?R be the set infinitesimals over M and=V/μ be the reside field with residue class mapIf M is Dedekind complete in R,then= M and the residue class map coincide the standard part map st:RM.In this context,van den Dries showed the follows:

Theorem 1([5]).Let S ?Rnbe R-definable and

(i) S ∩Mnis definable in M and dimM(S ∩Mn)≤dimR(S)?

(ii)st(S)is definable in M and dimM(st(S))≤dimR(S).

Theorem 2([5]).Let f :RmR be an R-definable function.Then here is a finite partition P of Mminto definable sets,where each set in the partition is either open in Mmor lacks of interior.On each open set C ∈P we have:

(ii) or there is a continuous function g : CM,definable in M,such that f(x)∈V and st(f(x))=g(st(x)),for all x ∈Ch,

where Chis the hull of C defined by

Remark 1.1.For any topological space Y,and X ?Y,by Int(X)we mean the set of interiors in X.Namely,x ∈Int(X)iffthere is an open neighborhood B ?Y of x contained in X.

There are fairly good analogies between the field of reals R and the field of p-adic numbers Qp,in both model-theoretic and field-theoretic view.For example,both of them are complete and locally compact topological fields,are distal and dpminimal structures,have quantifier eliminations with adding the new predicates for n-th power,and have cell decompositions.

In this paper,we treat the p-adic analogue of above two Theorems,where M is replaced by Qp,and R is replaced by an arbitrary elementary extension K of Qp.In our case,the convex hull V is the set

and μ,the infinitesimals of K over Qp,is the set

By [12,Lemma 2.1],for every x ∈V,there is a unique element st(x)in Qpsuch that a-st(a) ∈μ,we call it the standard part of a and st : ast(a)the standard part map.It is easy to see that st:VQpis a surjective ring homomorphism and st－1(0)=μ.So=V/μ is isomorphic to Qpin our context.With the notations as above,we now highlight our main results.

Theorem 1.2.Let S ?Knbe K-definable.Then

(ii)st(S ∩Vn)is definable in Qpand dimQp(st(S ∩Vn))≤dimK(S).

Theorem 1.3.Let f : KmK be a K-definable function.Then here is a finite partition P of Qpinto definable sets,where each set in the partition is either open inor lacks of interior.On each open set C ∈P we have:

(ii) or there is a continuous function g : CQp,definable in Qp,such that f(x)∈V and st(f(x))=g(st(x)),for all x ∈Ch.

In the rest of this introduction we give more notations and model-theoretic approach.

1.1 Notations

Let p be a fixed prime number,Qpthe field the p-adic field,and v :Qp{0}Z the valuation map.Let K be a fixed elementary extension of Qp.Then valuation v extends to a valuation map from K{0} to ΓK,we also denote it by v,where(ΓK,+,＜,0)is the corresponding elementary extension of(Z,+,＜,0).

Fact 1.4.Let v :K{0}ΓKbe as above.Then we have

· v(xy)=v(x)+v(y)for all x,y ∈K?

· v(x+y) ≥min{v(x),v(y)},and v(x+y)= min{v(x),v(y)} if v(x)≠v(y)?

· For x ∈Qp,|x|= p－v(x) if x≠0 and |x|=0 if x=0 defines a non-archimedean metric on Qp.

· K is a non–archimedean topological field.

We will assume a basic knowledge of model theory.Good references are[13]and[10].We will be referring a lot to the comprehensive survey[1]for the basic model theory of the p-adics.A key point is Macintyre’s theorem[9]that Th(Qp,+,×,0,1)has quantifier elimination in the language L=Lr∪{Pn|n ∈N+},where Lris the language of rings,and the predicate Pnis interpreted as the n-th powers

for each n ∈N+.Note that Pnis definable in Lr.Moreover,the valuation is definable in Lras follows.

Fact 1.5([3]).Let f,g ∈K[x1,...,xm].Thenis definable.

Remark 1.6.It is easy to see from Fact 1.5 that {a ∈K|v(a)= γ} and {a ∈K|v(a)＜γ}are definable for any fixed γ ∈ΓK.

For A a subset of K,by an LA-formula we mean a formula with parameters from A.Bywe mean arbitrary n-variables and∈Kndenote n-tuples in Knwith n ∈N+.Bywe mean the length of the tupleWe say that X ?Kmis A-definable if there is a LA-formula φ(x1,...,xm)such that

We also denote X by φ(Km)and say that X is defined byWe say that X is definable in K if X ?Kmis K-definable.If X ∈is defined by some LQpformulaThen by X(K)we mean ψ(Km),namely,the realizations of ψ in K,which is a definable subset of Km.

For any subset A of K,by acl(A)we mean the algebraic closure of A.Namely,b ∈acl(A)if and only if there is a formula φ(x)with parameters from A such that b ∈φ(K)and φ(K)is finite.Let α=(α1,...,αn) ∈Km,we denote acl(A ∪{α1,...,αn})by acl(A,α).

By a saturated extension K of K,we mean that|K|is a sufficiently large cardinality,and every type over A ?K is realized in K whenever|A|＜|K|.

1.2 Preliminaries

The p-adic field Qpis a complete,locally compact topological field,with basis given by the sets

for a ∈Qpand n ∈Z.The elementary extension K ?Qpis also a topological field but not need to be complete or locally compact.Let X ?Km,we say that∈X is an interior if there is γ ∈Γksuch that

Let V={x ∈K |x=0 ∨?n ∈Z(v(x)＞n)}.We call V the convex hull of Qp.It is easy to see that for any a,b ∈K,if b ∈V and v(a)＞v(b),then a ∈V.As we said before,for every a ∈V,there is a unique a0∈Qpsuch that v(a-a0)＞n for all n ∈Z.This gives a map aa0from V onto Qp.We call this map the standard part map,denoted by st : VQp.For any=(a1,...,am) ∈Vm,by st()we mean(st(a1),...,st(am)).Let f()∈Kbe a polynomial with every coefficient contained in V.Then by st(f),we mean the polynomial over Qpobtained by replace each coefficient of f by its standard part.Let

which is the collection of all infinitesimals of K over Qp.It is easy to see that for any a ∈K{0},aViffa－1∈μ.

Any definable subset X ?Knhas a topological dimension which is defined as follows:

Definition 1.7.Let X ?Kn.By dimK(X),we mean the maximal k ≤n such that the image of the projection

has interiors,for suitable 1 ≤r1＜...＜rk≤n.We call dimK(X)the topological dimension of X.

Recall that Qpis a geometry structure(see[8,Def.2.1,Prop.2.11]),so any K |=Th(Qp)is a geometry structure.The fields has geometric structure are certain fields in which model-theoretic algebraic closure equals field-theoretic algebraic closure.

Every geometry structure is a pregeometry structure,which means that for any=(a1,...,an) ∈Knand A ?K,dim(/A)makes sense,which by definition is the maximal k such that ar1acl(A)and ari+1acl(A,ar1,...,ari)for some subtuple(ar1,...,ark)of.We call dim(/A)the algebraic dimension ofover A.

Fact 1.8([8]).Let A be a subset of K and X an A-definable subset of Km.

(ii) Let K ?K be a saturated model.Then dimK(X)=X(K)}.

(iii) Let φ(x1,...,xm,y1,...,yn)be any LA-formula and r ∈N.Then the set

is A-definable.

(iv) If X ?K is K-definable.Then X is infinite iff dimK(X)≥1.

(v) Let A0be a countable subset of Qp,and let Y be an A0-definable subset of

It is easy to see from Fact 1.8 that for any LK-formula φ(x1,...,xn)and K′?K,we have

We will write dimK(X)by dim(X)if there is no ambiguity.If the function f :XK is definable in K,and Y ?X ×K is the graph of f.Then we conclude directly that dim(X)=dim(Y)by Fact 1.8(ii).

For later use,we recall some well-known facts and terminology.

Hensel’s Lemma.Let Zp= {x ∈Qp|x=0 ∨v(x) ≥0}be the valuation ring of Qp.Let f(x)be a polynomial over Zpin one variable x,and let a ∈Zpsuch that v(f(a)) ＞2n+1 and v(f′(a)) ≤n,where f′denotes the derivative of f.Then there exists a uniquea ∈Zpsuch that

We say a field E is a Henselian field if Hensel’s Lemma holds in E.Note that to be a henselian field is a first-order property of a field in the language of rings.Namely,there is a Lr-sentence σ such that E |= σiffE is a henselian field.So any K ?Qpis henselian.

2 Main Results

2.1 Some properties of Henselian fields

Since Qpis complete and local compact,it is easy to see that:

Fact 2.1.Suppose that E is a finite(or algebraic)field extension of Qp.Then for any α ∈EQp,there is n ∈Z such that v(α-a) ＜n(|α-a| ＞p－n)for all a ∈Qp.Namely,Qpis closed in E.

We now show that Fact 2.1 holds for any K |=Th(Qp).

Lemma 2.2.Let K be a henselian field,R= {x ∈K | x=0 ∨v(x) ≥0}be the valuation ring of K,and f(x) ∈R[x]a polynomial,D ?ΓKa cofinal subset,and X={xd|d ∈D}?R.If

Then there exist a cofinal subset I ?D and a ∈K such that

ProofInduction on deg(f).Suppose that f has degree 1,say,f(x)=αx+β.Then for any γ ∈ΓK,there is d0∈D such that v(f(xd)) ＞γ for all d0＜d ∈D.Now v(αxd+β)＞γ implies that＞γ-v(α).So

and hence

as required.

Now suppose that deg(f)=n+1 ＞1.We see that the derivative f′has degree n.

If there are γ0∈ΓKand ε0∈D such that v(f′(xε)) ≤γ0for all ε0＜ε ∈D.Take ε0sufficiently large such that

for all ε0＜ε ∈D.Then,by Hensel’s Lemma,we see that for all ε ＞ε0,there issuch that

As f has at most finitely many roots,there is a cofinal subset I ?D and some∈K such that

for all i ∈I.Since v(f(xi))+∞,we see that+∞.Thus we have

as required.

Otherwise,if for every γ ∈ΓK,there is γ ＜dγ∈D such that v(f′(xdγ))＞γ.Then there is a cofinal subset I={dγ|γ ∈ΓK}?D such that

Then,by induction hypothesis,there exist a cofinal subset J ?I and b ∈K such that

Since f is continuous,limj∈J,j→+∞f(xj)= f(b).Now J is cofinal in I,and I is cofinal in D,we conclude that J is cofinal in D.This complete the proof. □

Proposition 2.3.If K is a henselian field,and E is a finite extension of K.Then for any α ∈EK,there is γ0∈ΓKsuch that v(α-a)＜γ0for all a ∈K.Namely,K is closed in E.

ProofBy[6,Lem.4.1.1],the valuation of K extends uniquely to E.For each β ∈E,Let g(x)= xn+an－1xn－1+...+a1x+a0be the minimal polynomial of β over K,then the valuation of β is exactly(See[7,Prop.5.3.4]).

Let α ∈EK,and d(x)be the minimal polynomial of α over K with degree k.Then d(x+a)is the minimal polynomial of(α-a)over K for any a ∈K.Since d(x+a)=xf(x)+d(a)for some f(x)∈K[x],we see that v(α-a)=We claim that there is γ0∈ΓKsuch that v(d(a))＜γ0for all a ∈K.Otherwise,we will find a sequence{aγ|γ ∈ΓK}such that v(d(aγ))＞γ.Replace d(x)by ?d(x)with some ? sufficiently close to 0,we may assume that d ∈R[x].Moreover,fix γ0∈Γ,if v(α-a)＞γ0,and v(α-b)＞2γ0,then v(a-b)≥γ0.So

for some δ0∈K,and hence

Let δ=kδ0.If δ ∈R,then,by Lemma 2.2,there is b ∈K such that d(b)=0.However d is minimal polynomial of degree ＞1,so has no roots in K.A contradiction.

Let

We see that h(x)∈R[x]and

Now we have

For γ ＞γ0,we have aγ∈δR.Therefore δ－1aγ∈R for all γ ＞γ0.Applying Lemma 2.2 to h(x),we can find c ∈K such that

So d(δc)=0.A contradiction. □

Now we assume that K is an elementary extension of Qpin the language of rings Lr.This follow result was proven by[14]in the case of K=Qp.

Lemma 2.4.LetThen there is a partition of

into finitely many definable subsets S,over each of which f has some fixed number k ≥1 of distinct roots in K with fixed multiplicities m1,...,mk.For any fixed∈S,let the roots of f(,y)be r1,...,rk,and e=max{v(ri-rj)|1 ≤i ＜j ≤k}.Thenhas a neighborhood N ?Km,γ ∈ΓK,and continuous,definable functions F1,...,Fk: S ∩NK such that for each∈S ∩N,are roots of f(,y)of multiplicities m1,...,mkand

ProofThe proof of[14,Lem 1.1]applies almost word for word to the present context.The only problem is that the authors used Fact 2.1 in their proof.But the Proposition 2.3 saying that we could replace Qpby arbitrary K |=Th(Qp)in our argument.□

Remark 2.5.Lemma 1.1 of[14]saying that definable functions F1,...,Fkare not only continuous but analytic.However we can’t prove it in arbitrary K |=Th(Qp)as K might not be complete as a topological field.

Similarly,Lem.1.3 in [14]could be generalized to arbitrary K |=Th(Qp)as follows:

Lemma 2.6.If A ?Kmand f :AK is definable.Then there is a definable set B ?A,open in Kmsuch that AB has no interior and f is continuous on B.

ProofThe proof of Lem.1.3 in [14]applies almost word for word to the present context. □

2.2 Dimensions

We now assume that K is an elementary extension of Qp.

Lemma 2.7.Suppose that A ?K,X,Y are A-definable in K,f : XY is an A-definable function.If f is a finite-to-one map,dim(X)=dim(f(X)).

ProofLet K be a saturated elementary extension of K.By Fact 1.8 (iii),there is r ∈N such that|f－1(y)|≤r for all y ∈Y(K).For any a ∈X(K),since

we see that a ∈acl(A,f(a)).So dim(a/A,f(a))=0.By Fact 1.8(i)we have

dim(a/A)=dim(a,f(a)/A)=dim(a/A,f(a))+dim(f(a)/A)=0+dim(f(a)/A).

So dim(a/A)=dim(f(a)/A).By Fact1.8(ii),we conclude that dim(X)=dim(Y).□

Lemma 2.8.Suppose that A ?K,f : XY is an A-definable function in K.Then

ProofGenerally,we have

dim(a/A)=dim(a,f(a)/A)=dim(a/A,f(a))+dim(f(a)/A)≥dim(f(a)/A).

By Fact1.8(ii),we conclude that dim(X)≥dim(Y). □

Corollary 2.9.Suppose that A ?K,f : XY is an A-definable bijection function in K.Then

Prooff－1is a definable function as f is bijection.So we conclude that

Lemma 2.10.Suppose that X,Y are A-definable in K.Then

ProofBy Fact1.8(ii). □

Lemma 2.11.Let X ?Kn.Then dim(X)is the minimal k ≤n such that there is definable Y ?X with dim(Y)=dim(X)and projection

is a finite-to-one map on Y,for suitable 1 ≤r1＜...＜rk≤n.

ProofLet k be as above and π : XKkbe a projection with π(x1,...,xn)=(xr1,...,xrk).If Y ?X such that the restriction π ?Y : YKkis a finite-toone map.Then by Lemma 2.7 we have dim(Y)=dim(π(Y))and hence

Now suppose that dim(X)= l ≤k.Without loss of generality,we assume that f : XKl?(x1,...,xn)(x1,...,xl)is a projection such that f(X)has nonempty interior.Then we prove a claim:

Claim.Let Z0={b ∈Kl|f－1(b)is finite}and Z1=Kl0.Then dim(Z1)＜l.Clearly,

is definable in K.If dim(Z1)=l.Then,there is β ∈Z1(K)such that dim(β/A)=l,where is K ?K is saturated.Since dim(f－1(β)) ≥1,by Fact 1.8 (ii),there is α ∈dim(f－1(β))such that dim(α/A,β)≥1.By Fact 1.8(i),we conclude that

dim(α/A)=dim(α,f(α)/A)=dim(a/A,f(α))+dim(f(α)/A)≥l+1.But dim(α/A)≤dim(X)=l.A contradiction.

Since dim(Z1) ＜l,by Lemma 2.10,dim(Z0)= l.The restriction of f on f－1(Z0)is a finite-to-one map,we conclude that

by Lemma 2.7.Now dim(f－1(Z0))=dim(X)and the restriction of f on f－1(Z0)is a finite-to-one map.So k ≤l as k is minimal.We conclude that k=l=dim(X)as required. □

Corollary 2.12.Let X ?Knbe definable with dim(X)= k.Then there exists a partition of X into finitely many K-definable subsets S such that whenever dim(S)=dim(X),there is a projection πS:SKkon k suitable coordinate axes which is finite-to-one.

ProofLet X0= X and[n]kbe the set of all subset of{1,...,n}of cardinality k.By Lemma 2.11,there exist D0= {r1,...,rk} ∈[n]k, S0?X with dim(S0)=dim(X0),such that the projection

is finite-to-one on S0and infinite-to-one on X0S0.If dim(X0S0)＜dim(X0),then the partition{X0S0,S0}meets our requirements.

Otherwise,let X1=X0S0,we could find D1∈[n]k{D0}and S1∈X1such that the projection on coordinate axes from D1is finite-to-one over S1.Repeating the above steps,we obtained sequences Xiand Sisuch that Xi+1=XiSi.As[n]kis finite,there is a minimal t ∈N such that dim(Xt)＜dim(X0)and dim(Si)=dim(X0)for all i ＜t.It is easy to see that{S0,...,St－1,Xt}meets our requirements. □

Recall that by[4],Th(Qp)admits definable Skolem functions.Namely,we have

Fact 2.13(([4])).Let A ?K andbe a LA-formula such that

Then there A-definable function f :KmK such that

With the above Fact,we could refine Corollary 2.12 as follows:

Corollary 2.14.Let X ?Knbe definable with dim(X)= k.Then there exists a partition of X into finitely many K-definable subsets S such that whenever dim(S)=dim(X),there is a projection πS:SKkon k suitable coordinate axes which is injective.

ProofLet X0= X.By Corollary 2.12,we may assume that the projection π :X0Kkgiven by(x1,...,xn)(x1,...,xk)is finite-to-one.By compactness,there is r ∈N such that

By our induction hypothesis,there is a partition of X1into finitely may definable subsets meets our requirements.This completes the proof. □

Theorem 2.15.Let B ?Kmbe definable in K.Then dimK(B)≥

ProofSuppose that dimK(B)= k.By Lemma 2.10 and Corollary 2.14,we may assume that π :BKkis injective.The restriction of π tois a injective projection fromBy Lemma 2.11,≤k. □

Note that Pn(K)= {a ∈K | a0 ∧?b ∈K(a= bn)} is an open subset of K whenever K is a hensilian field.For any polynomial f(x1,...,xm) ∈K[x1,...,xm],

is an open subset of Kmsince f is continuous.

2.3 Standard part map and definable functions

The following Facts will be used later.

Fact 2.16(([2])).Every complete n-type over Qpis definable.Equivalently,for any K ?Qp,any Lr-formula φ(x1,...,xn,y1,...,ym),and any∈Km,the set

is definable in Qp.

Fact 2.17(([11])).Let X ?Kmbe a Qp-definable open set,let Y ?X be a K-definable subset of X.Then either Y or XY contains a Qp-definable open set.

Fact 2.18(([11])).Let X ?Kmbe a K-definable set.Then st(X)∩st(KmX)has no interior.

Recall that μ is the collection of all infinitesimals of K over Qp,which induces a equivalence relation ∽μon K,which is defined by

Definition 2.19.be polynomials.By f ∽μg we mean that

Lemma 2.20.be polynomials over K with f ∽μg.=(b1,...,bn)are tuples from K with ai∽μbifor each i ≤n.

ProofWe see that α ∈μiffv(α) ＞Z.Since v(α+β) ≥min{v(α),v(β)}and v(αβ)=v(α)+v(β),we see that μ is closed under addition and multiplication.As polynomials are functions obtained by compositions of addition and multiplication,we conclude that□

Since V ?K is also closed under addition and multiplication.We conclude directly that:

Corollary 2.21.Let=(x1,...,xn),andbe a polynomial with every coefficient contained in V.If a=(a1,...,an)∈Vn,then st(f)(st(a))=st(f(a)).

Corollary 2.22.Let f(x)= anxn+...+a1x1+a0be a polynomial over K with every coefficient contained in V and anμ.If b ∈K such that f(b)=0.Then b ∈V.

ProofSuppose for a contradiction that bV.Then st(b－1)=0.Clearly,we have

Let

Then g(b－1)=0.By Corollary 2.21,we have st(g)(st(b－1))=0.As st(b－1)=0,we see that st(an)=0,which contradicts to anμ. □

Fact 2.23([11]).Let S ?Kmbe definable in K.Then st(S ∩Vm) ?is definable in Qp.

Lemma 2.24.Let f :KkK be definable in K.Then

(i) X∞=is definable in Qp.

ProofBy Fact 2.16,there is asuch that for all a ∈and b ∈Qp,we have

Hence

which shows that X∞is definable in Qp.Again by Fact 2.16,there isψ(x,y1,y2)such that for all a ∈

Therefore

Lemma 2.25.Let X ?be a clopen subset ofX,thenX(K).

ProofAs X is clopen andX,there is N ∈Z such that

Lemma 2.26.If X ?Kmand f : XK are definable in K,then there is a polynomial q(x1,...,xm,y)such that the graph of f is contained in the variety

ProofLet Y be the graph of f.Since Th(Qp)has quantifier elimination,Y is defined by a disjunctionwhere eachis a conjunction

where g’s and h’s belong to KNow eachdefines an open subset of Km+1.Since dim(Y)≤m,we see that for each i ≤s,there is f(i)≤lisuch thatThen

as required. □

Proposition 2.27.If f :KmK is definable in K.Let X=V}.Then

has no interiors.

ProofBy Lemma 2.26,there is a polynomial

such that the graph of f is contained in the variety of g.Without loss of generality,we may assume that each coefficient of g is in V,otherwise,we could replace g by g/c,where c is a coefficient of g with minimal valuation.Moreover,we could assume that at least one coefficient of g is not in μ.

Suppose for a contradiction that DXcontains a open subset ofShrink DXif necessary,we may assume that DX?X is a Qp-definable open set inBy Lemma 2.4,there is a partition P of DX(K) ?Kminto finitely many definable subsets S,over each of which g has some fixed number k ≥1 of distinct roots in K with fixed multiplicities m1,...,mk.For any fixed∈S,let the roots ofbe r1,...,rk,and e=max{v(ri-rj)|1 ≤i ＜j ≤k}.Thenhas a neighborhood N ?Km,γ ∈ΓK,and continuous,definable functions F1,...,Fk:S ∩NK such that for each ˉx ∈S ∩N,are roots ofof multiplicities m1,...,mkand

Since DX(K)is a Qp-definable open subset of X(K).By Fact 2.17,some S ∈P contains a Qp-definable open subset ψ(Km)of X(K).Where ψ is an LQpformula.Let A0=Then A0?A is an open subset ofand over A0(K)we have

Since the family of clopen subsets forms a base for topology onwe may assume that A0is clopen.We now claim that

Claim 1.For every

ProofOtherwise,by Corollary 2.21,we have st(gn)(st())=st(gn())=0.SoBy Lemma 2.25,we see thatA0(K).A contradiction. □

By Claim 1 and Corollary 2.22,we see that for every∈A0(K)and b ∈K,if=0,then b ∈V.By Corollary 2.21,we conclude the following claim

Claim 2.For every∈A0(K)and b ∈K,if=0,then b ∈V and

Now st(g)is a polynomial over Qp.Applying Lemma 2.4 to st(g)and Fact 2.17,and shrink A0if necessary,we may assume that

By Fact 2.18,each st(Di)∩st(Dj)has no interior.By Fact 2.17,A0has no interiors.A contradiction. □

ProofOtherwise,suppose that U ?Kmis open.Applying Proposition 2.27 to g(x)=(f(x))－1,we see that g(U) ?V,and for all∈U there are∈st－1(a)such that=0.A contradiction. □ProofLet X0=As we showed in Lemma 2.24,both X0and X1are Qp-definable sets.Let

be the K-definable function given by

Let Y ?Kk+1be the graph of f and Z ?Kk+1be the graph of g.For eachlet

Theorem 2.30.Let f : KmK be an K-definable function.Then here is a finite partition P of Qpinto definable sets,where each set in the partition is either open inor lacks of interior.On each open set C ∈P we have:

(ii) or there is a continuous function g : CQp,definable in Qp,such that f(x)∈V and st(f(x))=g(st(x)),for all x ∈Ch.

ProofLet X,X∞be as in Lemma 2.24, DXas in Proposition 2.27,and U as in Corollary 2.28,then DXand U have no interior,and by Lemma 2.29,they are definable.Now {DX,XDX,U,X∞U} is a partition ofClearly, {Int(X∞U),(X∞U) Int(X∞U)} is a partition of X∞U where Int(X∞U)is open and(X∞U)Int(X∞U)lacks of interior.

Let h : XDXQpbe a definable function defined by xst(f(x)).By Theorem 1.1 of [14],there is a finite partition P*of XDXinto definable sets,on each of which h is analytic.Each set in the partition is either open inor lacks of interior.

Clearly,the partition

satisfies our condition. □

We now prove our last result.

Lemma 2.31.Let Z ?Knbe definable in K of dimension k ＜n,and the projection

is injective on Z.Then dimQp(st(Z ∩Vn))≤k.

ProofAs π is injective on X,there is a definable function

such that

By Lemma 2.26,for each i ≤n,there is a polynomialsuch that the graph of fiis contained in the variety

of Fi.We assume that each coefficient belongs to V.It is easy to see that for each

we have fi(π(a1,...,an))=ai.So Fi(a1,...,ak,ai)=0.By Corollary 2.21,

So st(Z ∩Vn)is contained in the variety

Let A ?Qpbe the collection of all coefficients from st(Fi)’s.Then for each

we see that aiis a root of Fi(a1,...,ak,u),and hence ai∈acl(A,a1,...,ak),where i ≤n.This implies that

for all(a1,...,an)∈V(st(F1),...,st(Fn)).By Fact 1.8(v),we see that

So st(Z ∩Vn)≤k as required. □

Theorem 2.32.Let Z ? Knbe definable in K.Then dimQp(st(Z ∩Vn)) ≤dimK(Z).

ProofSince st(X∪Y)=st(X)∪st(Y)and dim(X∪Y)=max{dim(X),dim(Y)}hold for all definable X,Y ?Kn.Applying Corollary 2.14,we many assume that dim(Z)=k and π :(x1,...xn)(x1,...,xk)is injective on Z.If k=n,then

as st(Z ∩Vn)If k ＜n,then by Lemma 2.31,

as required. □