数理解析研究所講究録 1290 巻 2002 年 7-24

7On Hilbert’s 17th Problem and Real Nullstellensatz for
Definable Analytic Functions
京都大学

理学研究科

・

・

数学教室

雅人 (Masato Fujita)

藤田Department of Mathematics, Graduate School of Science,
Kyoto University
名古屋大学

・

多元数理科学研究科

塩田

昌弘 (Masahiro Shiota)

Graduate School of Mathematics,
Nagoya University

1

Introduction.

David Hilbert proposed the 23 famous problems in his address at the 1900 International Congress of Mathematics in Paris. The following is the 17th prolemof them.

Hilbert’s 17th Problem. Is every nonnegative polynomial function on
sum of squares of rational functions?
Let be acommutative ring. Remember that an ideal I ofif, for every sequence ,,of elements of with,,. On the other hand, we set$A$$a_{1}$$a_{1}$$\ldots$$\ldots$$A$$a_{p}$$A$$\mathbb{R}^{n}$a

is areal ideal, we have$a_{1}^{2}+\cdots+a_{p}^{2}\in I$$a_{p}\in I$$Z_{\mathrm{p}\mathrm{o}1\mathrm{y}}(I):=\{a\in \mathbb{R}^{n}; f(a)=0(\forall f\in I)\}$

and$\mathrm{I}_{\mathrm{p}\mathrm{o}1\mathrm{y}}(Z):=\{f\in R[x_{1}, \ldots, x_{n}];f(a)=0(\forall a\in Z)\}$

for every ideal I of$\mathbb{R}[x_{1}, \ldots, x_{n}]$and for any subset $Z$ of Rn.

Real Nullstellensatz. Let I be an ideal of
I if and only if I is areal ideal.

$\mathbb{R}[x_{1},$$x_{n}]$\ldots ,. Then$\mathrm{I}_{\mathrm{p}\mathrm{o}1\mathrm{y}}(Z_{\mathrm{p}\mathrm{o}1\mathrm{y}}(I))$

Hilbert’s 17th Problem was solved by Emil Artin in 1927 [A] and it is known
that above Real Nullstellensatz also holds true. Hilbert 17th Problem and Real
Nullstellensatz are problems on polynomial functions. Real algebraic geometerssolved the problems on Nash functions similiar to Hilbert’s 17th Problem andReal Nullstellensats.
Hilbert’s 17th Problem for Nash functions. Let
be areal closed field
and $M\subset R^{n}$ be asemialgebraically connected Nash submanifold. Thenevery nonnegative Nash function on $M$ is asum of squares in the field of
fractions of the ring $N(M)$ of Nash functions on $M$ .
$R$

Real Nullstellensatz for Nash functions. Let R be areal closed field andM
be aNash submanifold. We set$\subset R^{n}$$Z_{\mathrm{N}\mathrm{a}\mathrm{s}\mathrm{h}}(I):=\{a\in M;f(a)=0(\forall f\in I)\}$

and$\mathrm{I}_{\mathrm{N}\mathrm{a}\mathrm{s}\mathrm{h}}(Z):=\{f\in N(M);f(a)=0(\forall a\in Z)\}$8
for every ideal I of $N(M)$ and for any subset Z of M. Then an ideal I of
$N(M)$ is areal ideal if and only if$\mathrm{I}_{\mathrm{N}\mathrm{a}\mathrm{s}\mathrm{h}}(Z_{\mathrm{N}\mathrm{a}\mathrm{s}\mathrm{h}}(I))=I$

An -minimal expansion of the real field is acollection of subsets of Eu,
clidean spaces ,satisfying afew simple axioms which is satisfied for thecollection of all semialgebraic subsets. In other word, the notion of 0-minimal
structures is asimple model-theoritical generalization of Real algebraic geometry. Therefore, it is natural to propose the problems whether the ringof definable analytic functions on $M$ satisifies the following Hilbert’s 17th Problem and Real Nullstellensatz for definable analytic functions for each 0-minimal
expansion of the real field. Here $M$ denotes adefinable analytic submanifoldof some Euclidean space.
$\overline{\mathbb{R}}$$0$$\mathbb{R}^{1}$$\mathbb{R}^{2}$$\ldots$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$\tilde{\mathbb{R}}$Hilbert’s 17th Problem for definable analyic functions. Assume that $M$
is connected. Then every nonnegative definable analytic function on $M$ isasum of squares in the field of fractions of the ring.
$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$Real Nullstellensatz for definable analytic functions. We set$Z(I)=Z_{\mathrm{d}\mathrm{f}}(I):=\{a\in M;f(a)=0(\forall f\in I)\}$

and$\mathrm{I}(Z)=\mathrm{I}_{\mathrm{d}\mathrm{f}}(Z):=\{f\in C_{\mathrm{d}\mathrm{f}}^{\omega}’(M);f(a)=0(\forall a\in Z)\}$

for every ideal I ofand for any subsetof
is areal ideal if and only if$C_{\mathrm{d}\mathrm{f}}^{\mathrm{t}v}’(M)$$Z$

of $M$ . Then an ideal$\mathrm{I}_{\mathrm{d}\mathrm{f}}(Z_{\mathrm{d}\mathrm{f}}(I))=I$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$I$.

We solve these problems in the low dimensional cases in the present paper.
Precisely,
Theorem 1.1. Consider an -minimal expansionof the real field and$M$analytic
space.
some
Euclidean
Assume further
definable
submanifoldofthat one of the following conditions is satisfied.
$\overline{\mathbb{R}}$

$0$

$a$

1. $\dim(M)\leq 1$

2. All compact analytic subsets of Euclidean spaces are definable in$\tilde{\mathbb{R}}$and$\dim(M)=2$

3. The -minimal structureis the structure such that all compact analytic
subsets of Euclidean spaces are definable, it admits an analytic cylindricalare Noetherian for all
definable cell decomposition and the rings$N$
analytic
spaces
Euclidean
definable
submanifold ofof dimensions 3.
Assume further $\dim(M)=3$ .
$\tilde{\mathbb{R}}$$0$$C_{df}^{\omega}(N)$Then the folloing statements hold true.

Hilbert’s 17th Problem. Any nonnegative definable analytic function on $M$
is a sum of squares in the field of fractions of the ringwhen $M$ isconnected.
$C_{df}^{\omega}(M)$Real Nullstellensatz. An ideal I of.
$\mathrm{I}_{df}(Z_{df}(I))=I$$C_{df}^{\omega}(M)$is a real ideal if and only if

92Review: Real Algebra.
We review the notions concerning areal spectrum.
be aunitary commutative ring. The
Definition 2.1 (prime cone). Let
subset of is aprime cone of if asatisfies the following conditions.
$A$

$A$

$\alpha$

1.

2.

$A$

$\alpha+\alpha\subset\alpha$$\alpha\cdot$$\alpha\subset\alpha$for every3.

$a^{2}\in\alpha$4.

$-1\not\in \mathrm{c}\mathrm{r}$5. If$ab\in\alpha$

$a$

in

$A$

for , $b\in A$ , then$a$$a\in\alpha \mathrm{o}\mathrm{r}-b\in\alpha$.

Notation 2.2. The set
ce -ci is areal prime ideal of . Let. The prime cone induces
denote the field of fractions of. The ordering is defined as follows.
an orderingof the field$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)=$$A$$\cap$$A/\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)$$k(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha))$$\alpha$$k(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha))$$\leq_{\alpha}$$0\leq_{\alpha}\overline{a}\Leftrightarrow a\in\alpha$,. We simply denotefor every $a\in A$ , where denotes the class of in, $a(\alpha)<0$ and $a(\alpha)>0$ in the same
. We also define
if$a$$\overline{a}$$a(\alpha)\geq 0$$k(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha))$$\mathrm{a}(\mathrm{a})=0$$\overline{a}\geq_{\alpha}0$

way.

, is the. The real spectrum of , denoted
Definition 2.3
topological space whose points are the prime cones of , and whose topology is
given by the basis of open subsets$A$$(\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}A)$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}A$$A$$\mathcal{U}(a_{1}, \ldots, a_{n})=\{\alpha\in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}A;a_{1}(\alpha)>0, \ldots, a_{n}(\alpha)>0\}$

,

where
is any finite family of elements of . This topology is called
the spectral topology.
. If the conditionis satisfied, we say
Let , be two points ofthat is aspecialization of , or that is generalization of . We abbreviate.
this condition to$A$$\{a_{1}, \ldots, a_{n}\}$$\alpha$$\beta$$\alpha\subset\beta$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}A$$\beta$$\alpha$$\beta$$\alpha$$\alphaarrow\beta$

The following theorems are direct consequences of Real Algebra.

Theorem 2.4 (Hilbert’s 17th Problem for acommutative field). Let $K$
be a commutative field. Define
as the set of all finite sums of squares of.
elemets of K. Then$\sum K^{2}$$\sum K^{2}=\bigcap_{\alpha\in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}K}\alpha$

Proof [BCR, Corollary 1.1.11]$\blacksquare$

Theorem 2.5 (Real Going-down for Regular Homomorphisms). Letin: $Aarrow B$ be a homomorphism berween commutative rings. Let, and letbe such that$\betaarrow\alpha$$\phi$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}(A)$$\alpha’\in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}(B)$$\phi^{*}(\alpha’):=\{a\in A;\phi(a)\in\alpha’\}=\alpha$.

10
Suppose that the local domain
is excellent and the\rightarrow
B\otimes\^A
induced homomorphism
a is regular. Then there exists,
such that
and.
$A_{\beta\alpha}:=A_{\sup \mathrm{p}(\alpha)}/\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta)$$\beta’\in$$A_{\beta\alpha}$$\beta’arrow\alpha’$$\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}(B)$$ht(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta’))=ht(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta))$

Proof. [ABR, Theorem 7.1]$\phi^{*}(\beta’)=\beta$$\blacksquare$

Proposition 2.6 (Artin-Lang Property for convergent power series).
denote the ring of convergent power series. Let be a prmeLet
, , be a sequence of elements
cone
and $f1$ ,$f_{1}(\alpha)>0$$f_{p}(\alpha)>0$,, $g(\alpha)=0$ . We may regard ,,, , as realsuch thatanalytic functions on an open neighborhood at the or igin in the n-dimensional, $fp(x)>$Euclidean space. Then the germ of the set { $x\in U;f1(x)>0$ ,, $g(x)=0\}$ at the origin is not empty.
$\mathbb{R}\{x_{1}, \ldots, x_{n}\}$$\alpha$$of\mathbb{R}\{x_{1}, \ldots, x_{n}\}$$\ldots$$f_{p}$$of\mathbb{R}\{x_{1}, \ldots, x_{n}\}$$g$$f_{1}$$\ldots$$\ldots$$f_{p}$$g$$U$$\ldots$$0$

Proof. [ , Proposition 3.4] lm$\mathrm{R}$3One-dimensional Case.
We prove Theorem 1.1 when $\dim(M)=1$ in this section. We may assume thatis connected without loss of generality.
Let be aprime ideal ofwith $Z(p)\neq M$ . Fix
. For
,
all
denote the polynomial functions on
defined by $S_{a,x}(y)=$dist $(a, y)^{2}$ -dist $(a, x)^{2}$ . Here dist(a, ) denotes the distance between and .
suitably. Then we may assume the following.
Choose , , and$M\subset \mathbb{R}^{n}$$x\in \mathrm{Z}(\mathrm{p})$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$p$$a\in \mathbb{R}^{n}$$S_{a,x}$$\mathbb{R}^{n}$

$x$

$a$

$\bullet$

$b$

$c$

$a$

$x$

$d\in \mathbb{R}^{n}$

The zero set of.
set of$S_{a,x}$intersects with $M$ transversally and so does the zero

$S_{b,x}$

and $S_{b,x}^{-1}(\mathrm{O})\cap M=\{x,x’\}$$\bullet$$S_{a,x}^{-1}(\mathrm{O})\cap M=\{x, x’\}$$\bullet$$S_{a,x}^{-1}(0)\cap S_{b,x}^{-1}(\mathrm{O})\cap M=\{x\}$$\bullet$$S_{c,x}^{-1},(\mathrm{O})\cap M=\{x’\}$and $S_{d,x}^{-1},,(\mathrm{O})\cap M=\{x’\}$

From now on,
also denotes the restiction ofnotation will not confuse the readers.
$S_{a,x}$$S_{a,x}$

to $M$ . This abuse of

We first show that $p=m_{x}:=\{f\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M);f(x)=0\}$ and is finitely.
generated. We have only to show that is generated only byand$S_{c,x’}f$
$f\in
p$
,
However, for any
is divisible by
and $S_{d,x’}f$ is also divisible by. Especially $S(a, x)$ , $S(b, x)\in p$ because is prime. Remark that $\pm S_{c,x’}\pm$
if we choose the signs properly. Therefore, $f\in$
is an unit in$p$$S_{a,x}$$p$$S_{b,x}$$S_{a,x}$$S_{b,x}$$p$$S_{d,x^{ll}}$$(S_{a,x}, S_{b,x})$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$.

Proof of Real Nullstellesatz. It is obvious when I is aprime ideal because
all prime ideals are of the formor (0). It is also trivial that I is real if$m_{x}$$\mathrm{I}(Z(I))=I$Next consider the case when I is areal ideal. Remark that areal ideal is aradical ideal. Consider the irreducible primary decomposition $I=p_{1}\cap\cdots\cap p_{m}$ .

11
are prime. We show that all
Since I is radical, we may assume that all
are real. Let , , be definable analytic functions with $f_{1}^{2}+\cdots+f_{r}^{2}\in pj$ .
. Then $(hf_{1})^{2}+\cdots+(hf_{r})^{2}\in I$ . Since I is real, all
Choose. We havebecauseis in . Therefore,
are in . Particularly,
that
,
it
obviousisallfor
are real. Since
shown that all$p_{j}$$p_{j}$$f_{1}$$\ldots$$f_{r}$$hf_{k}$$h \in\bigcap_{i\neq j}p_{i}\backslash p_{j}$$I$$hf_{k}$$f_{k}\in p_{j}$$p_{j}$$j$$p_{j}=I_{\mathrm{d}\mathrm{f}}(Z(p_{j}))$$p_{j}$$I=I_{\mathrm{d}\mathrm{f}}(Z(I))$$h\not\in p_{j}$. II

Proof of Hlber ’s 17th Problem. Any nonnegative definable analytic function$f$

if $f(x)=0$ . Hence $f=u \prod_{i=1}^{m}(S_{a_{i},x:}^{2}+$is obviously divisible by, where is apositive function on $M$ . We have finished the proof. @$S_{a,x}^{2}+S_{b,x}^{2}$$S_{b.,x_{\mathrm{i}}}^{2})^{m_{\mathrm{t}}}$$u$and elements , , , $d\in A$ , the equation $(a^{2}+$$b^{2})(c^{2}+d^{2})=(ad-bc)^{2}+(ac+bd)^{2}$ holds true. Hence the above proof ofHilbert’s 17th Problem claims the stronger fact that any nonnegative definable
analytic function on $M$ is asum of two definable analytic functions on $M$ if

Remark 3.1. For any ring

$A$

$a$

$b$

$c$

$\dim(M)=1$ .

4Basic lemmas
We first fix an arbitrary -minimal expansion
submanifold $M$ of an Euclidean space.

$\tilde{\mathbb{R}}$$0$

Lemma 4.1. Any maximal ideal $m$ of$C_{df}^{\omega}’(M)$of the real field and adefinable
is of the form

$m_{x}=m_{x}^{M}:=\{f\in C_{df}^{\omega}(M);f(x)=0\}$,

where $x\in M$ .

Proof Any real analytic set is locally homeomorphic to an union of finite

cones at any point by [ , Theorem , p.96]. Remark that, if areal analytic set
is locally homeomorphic to some Euclidean space at some point, the germ of
this analytic set at the given point is analytically irreducible as agerm of real
analytic set. Since an -minimal structure admits finite definable cell decomp0sitions, any definable real analytic set particularly has only finite analytically$\mathrm{I}\mathrm{I}$$\mathrm{S}$$0$of definable analytic funcirreducible components. Whence, for any family$M$, $f_{k}\in\Psi$ with
tions on , there exists finite definable analytic functions $f1$ ,.
, $f_{k}\in m$ such that $\bigcap_{f\in m}Z(f)=$There particularly exist finite elements ,is invertible. Contrais empty. If the zero set of. It is obvious
diction. Choose apoint of the zero set of the function
because $m$ is maximal.
that$\Psi$$\ldots$$\bigcap_{f\in\Psi}f^{-1}(0)=\bigcap_{i=1}^{k}f_{i}^{-1}(0)$$f_{1}$$\ldots$$\sum_{i=1}^{k}f_{i}^{2}$$\sum_{i=1}^{k}f_{i}^{2}$$\bigcap_{\mathrm{i}\leq i\leq k}Z(f_{i})$$\sum_{i=1}^{k}f_{i}^{2}$$x$$m\subset m_{x}^{M}$

Lemma 4.2. Let$\blacksquare$$\alpha\in Spec_{r}(C_{df}^{\omega}’(M))$

be a prime cone. Then the set$\{x\in M;f_{1}(x)\geq 0, \ldots, f_{m}(x)\geq 0\}$is not empty for any finite family$f_{1}$,$\ldots$,$f_{m}\in\alpha$.

12
As was shown in thesuch that

Proof. We show it by the reduction to absurdity.

proof of Lemma 4.1, there exist finite elements. We set $f_{m+1}=- \sum_{i=1}^{k}h_{i}$ .
for any $i=$
Set $F_{j}=\{x\in M;fj\{x)\geq 0\}$ and $G_{j}=\{x\in M;fj\{x$ )$m+1$ . The intersection
is empty because$m+1$ } for all $j=1$ ,,$(\{x\in M;fj\{x)<0\})_{i=1,\ldots,m+1}$ is an open covering of $M$ by the definition.
as follows.
Define the definable continuous function $:$h_{1}$,$\ldots$,$h_{k}\in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)$$\bigcap_{h\in\sup \mathrm{p}(\alpha)}h^{-1}(0)=\bigcap_{1\leq i\leq k}h_{i}^{-1}(0)$$\geq f_{j}(x)$$1$$F_{j}\cap G_{j}$$\ldots$$\ldots$$Marrow \mathbb{R}$$\phi(x)=(m+2)\cross\frac{|\max_{i=1,\ldots,m+1}f_{i}(x)|}{|\min_{i=1,\ldots,m+1}f_{i}(x)|}$

We will construct positive definable analytic functions
isfying the following conditions for any .

$P_{1}$,$\ldots$, $P_{m+1}$ on $M$ sat-$j$

$P_{j}>\phi$$P_{j}<1$onon$G_{j}$$F_{j}$

Fix the number . Arranging (fj) in asuitable order, we may assume that$j=m+1$ . First set
$j$

{( ,

$N=\mathbb{R}^{m+1}\backslash$$x_{1}$$\ldots$,$x_{m}$,$y)\in \mathbb{R}^{m+1}$; $y\geq 0$ , $x_{i}\geq 0$ for all },$i$$F=$ $\{(x_{1}, \ldots, x_{m}, y)\in N;y\geq 0\}$ and$G=${$(x_{1}$,$\ldots$,$x_{m}$, $y)\in N;x_{i}\geq 0$ for all

$i$

}.

denote the distance function from $F$ and , respectively. We
andLet
on $N$ , where is anext define two kinds of semialgebraic functions andpositive number.
$d_{F}$$G$$d_{G}$$Q$$r$$\psi_{r}$$Q(x_{1}, \ldots, x_{m}, y)=\sqrt{y^{2}+(x_{1}+y)^{2}+\cdots+(x_{m}+y)^{2}}\mathrm{x}$$(m+2)$$\psi_{r}(x_{1}, \ldots, x_{m}, y)=\frac{rd_{G}}{d_{F}+d_{G}}Q-\frac{yd_{F}}{2(d_{F}+d_{G})}$

on $N$ if $r’>r$ . There exists
It is easy to see that $\psi_{r}>0$ on $N$ andaNash function $P’$ on $N$ with $|\emptyset_{2}-P’|<\psi_{2}$ by Efroymson’s Approximation
and $P’>Q$ on $F$ . Hence,
Theorem. Obviously, $P’>0$ on $N$ , $P’<|y|$ on$\psi_{r’}\geq\psi_{r}$$G$$Q/P’>(m+2) \cross\frac{|\max\{x_{1}+y,\ldots,x_{m}+y,y\}|}{|\min\{x_{1}+y,\ldots,x_{m}+y,y\}|}$

on

$G$

and $Q/P’<1$ on $F$ . The

function $P_{m+1}$ on $M$ defined by$P_{m+1}(x)=, \frac{Q(f_{1}(x)-f_{m+1}(x),\ldots,f_{m}(x)-f_{m+1}(x),f_{m+1}(x))}{P(f_{1}(x)-f_{m+1}(x),\ldots,f_{m}(x)-f_{m+1}(x),f_{m+1}(x))}$satisfies the required condition.
Any positive definable analytic function on $M$ is contained in because$g=(\sqrt{g})^{2}$ and. Henceis adefinable analytic function. Particularly,
is contained in ;nevertheless, it
the definable analytic function$M$$|P_{j}f_{j}|>(m+2)|\max f_{j}|$ whenreaches to the
becauseis negative on. Therefore it is not contained in . Conminimum value among
tradiction.
$\alpha$$g$$P_{j}\in\alpha$$\sqrt{g}$$\sum_{i=1}^{m+1}P_{i}f_{i}$$\alpha$$f_{j}$$(f_{i})_{i=1,\ldots,m+1}$$\blacksquare$$\alpha$13
We next consider an -minimal expansion of the ristricted analytic field.
member that the -minimal structure
is by definition the smallest-minimal expansion of the real field such that all compact analytic subsets of
Euclidean spaces are definable. Hence all compact analytic subsets of Euclidean
spaces are definable in any -miniaml expansion of.
$\tilde{\mathbb{R}}$$0$${\rm Re}$$\mathbb{R}_{\mathrm{a}\mathrm{n}}$$0$$\mathbb{R}_{\mathrm{a}\mathrm{n}}$$\mathrm{o}$$\tilde{\mathbb{R}}$$0$$\mathbb{R}_{\mathrm{a}\mathrm{n}}$Proposition 4.3. Consider an -minimal expansion. Let $M$ be
of
$N$
and
be a definable open subset of $M$ .
definable analytic submanifold of
Assume thatandare both noetherian. Then the localization
is an excellent and regular local ring for any prime ideal . Furthermore, the natural ring homomor phisms
and$M$
are regular homomorphisms, where is a point of .
$\tilde{\mathbb{R}}$

$0$

$a$

$\mathbb{R}_{an}$$\mathbb{R}^{n}$$C_{df}^{\omega}(M)$$C_{df}^{\omega}’(N)$$C_{df}^{\omega}(M)_{p}$$p$$C_{df}^{\omega}(M)arrow O_{x}$$C_{df}^{\omega}(M)arrow C_{df}^{\omega}(N)$$x$is more than or
Proof. It is obvious that the Krull dimension ofequal to $\dim(M)$ for $x\in M$ . We first show thatis aregular ring
whose Krull dimension $\dim(M)$ and that.
is generated by
Fix apoint $x\in M$ . We may assume without loss of generality that $x=$(0,
, 0, 1) and that $M$ is contained in the unit sphere $S^{n-1}$ . Choose alinear
function onsuch that $x\in f_{1}^{-1}(0)$ ,is transversal to $M$ at and theconnected component ofbecontaining is compact. Let
the ring of germs of analytic functions on . Since any connected componentof an analytic set is also an analytic set, there exists an analytic function onwithsuch thatis divisible byfor any analytic
function on $M$ vanishing on $X$ .
Let $mx$ be the ideal of analytic functions on $X$ vanishing at . Since $X$
is acoherent analytic set, one can easily show that $mx$ is generated by thecoordinate functions ,,[Cartan, Th\’eor\‘eme 3].
by Cartan’s Theorem
It is obvious that $\dim(M)-1$ of them, say, ,, $x_{\dim(M)-1}$ is nonsingular at
on $X$ and the other coordinate functions are generated by ,, $x_{\dim(M)-1}$$C^{\omega}(X)_{mx}$as elements of. Particularly,
is aregular local ring and, $f_{\dim(M)}’\in C^{\omega}(X)$ which generates the maximal ideal
there exist ,in. By Cartan’s Theorem , there exist analytic functions ,, $f_{\dim(M)}$$S^{n-1}$$X$onsuch that the restriction of
to aneighborhood of
coincides with
, $\dim(M)$ .
for any $j=2$ ,$f_{\dim(M)}$ .
We show thatFix adeis generated by ,$M$$f$finable analytic function onvanishing at . There exist analytic func, $g_{\dim(M)}’$ and $h’$ on $X$ such that $h’f= \sum_{i=2}^{\dim(M)}g_{i}’f_{i}$ on $X$ and,than$h’(x)\neq 0$ . Choose an analytic extension, $g_{\dim(M)}’$, $g_{\dim(M)}$ and of ,,and $h’$ on $S^{n-1}$ , respectively. It is possible by Cartan’s Theorem B. By theway of construction, $hf- \sum_{i=2}^{\dim(M)}g_{i}f_{i}$ is an analytic function on $M$ vanishing on $X$ . Therefore there exists an analytic function
on $M$ with $g_{1}f_{1}=$, that is,. All , , andare definable for $i=1$ , , $\dim(M)$ and for $j=2$ , , $\dim(M)$ because they are
analytic on $S^{n-1}$ or on . It is obvious that is definable. We have shown thatis aregular local ring whose Krull dimension is $\dim(M)$ . The ringis regular and its Krull dimension equals $\dim(M)$ by Lemma 4.1 and by$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{m_{x}}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{m_{x}^{M}}$$m_{x}^{M}$$m_{x}^{N}C_{\mathrm{d}\mathrm{f}}^{\omega}(N)_{m_{x}^{N}}$$\ldots$$f_{1}^{-1}(0)$$\mathbb{R}^{n}$$f_{1}$$f_{1}^{-1}(\mathrm{O})\cap M$$x$$C^{\omega}(\overline{M})$$x$$\overline{M}$$\tilde{h}(x)\neq 0$$\tilde{h}\in C^{\omega}(\overline{M})$$\tilde{h}f$$f_{1}$$f$$x$$x_{1}$$\ldots$$\mathrm{B}$$x_{n}$$x_{1}$$\ldots$$x$$x_{1}$$\ldots$$C^{\omega}(X)_{m_{X}}$$f_{2}’$$m_{X}$$\ldots$$C^{\omega}(X)_{m_{X}}$$\mathrm{B}$$f_{2}$$\ldots$$f_{j}$$f_{j}’$$\ldots$$f_{1}$$m_{x}^{M}C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{m_{x}^{M}}$$\ldots$$x$$\mathrm{s}$$g_{2}’$$\ldots$$g_{2}$$h$$\ldots$$g_{2}’$$\ldots$$g_{1}$$\tilde{h}hf-\sum_{i=2}^{\dim(M)}\tilde{h}g_{i}f_{i}$$\tilde{h}hf=g_{1}f_{1}+\sum_{i=2}^{\dim(M)}\tilde{h}g_{i}f_{i}$$\ldots$$\ldots$$\overline{M}$$g_{1}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{m_{x}^{M}}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$f_{i}$$g_{j}$$h$$\tilde{h}$14[Ml, Theorem 19.3]. We can show in the same way thatis excellent
102].
[M2,
We omit the details. Remark that ,using, $f_{\dim(M)}$
Theorem.
also generate the ideal
We next show that the natural ring homomorphisms
andare regular homomorphisms. The proof of thatisregular is similar to that of the regularity of the homomorphism. Therefore we only show the regularity of the homomorphism:. Fix $x\in N$ . Consider the following diagram.
$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{m_{x}^{\Lambda I}}$$f_{1}$$\ldots$$C_{\mathrm{d}\mathrm{f}}^{\omega}/(N)_{m_{x}^{N}}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)arrow C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)arrow O_{x}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)arrow O_{x}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)arrow$$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$\phi$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)arrow C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$C_{\mathrm{d}\mathrm{f}}^{\omega}/(M)_{m_{x}^{M}}-C_{\mathrm{d}\mathrm{f}}^{\omega}\overline{(M)}_{m_{x}^{M}}$$\downarrow$$\underline{\downarrow}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)_{m_{x}^{N}}$$-C_{\mathrm{d}\mathrm{f}}^{\omega}(N)_{m_{x}^{N}}$The second vertical arrow is an isomorphism because $\phi(m_{x}^{M})=m_{x}^{N}$ . Since bothare both excellent rings, the first horizontal arrows
andare regular and faithfully flat. The first vertical arrow is therefore aregularhomomorphism by [Ml, Theorem 32.1]. Summing up, the homomorphism isregular by Lemma 4.1. @$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)_{m_{x}^{N}}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{m_{x}^{M}}$$\phi$

Corollary 4.4. Let
cohomology classdomain.

$\tilde{\mathbb{R}}$and $M$ be the same as in Proposition 4.3. If the firstis zero, thenis an unique factorization$H^{1}(M, \mathbb{Z}_{2})$$C_{df}^{\omega}’(M)$Proof. The proof is exactly the same as [BE, Theorem 4.1]. Hence we omit

the proof.

$\blacksquare$

5TwO-dimensional Case.
in this section.
We fix an -minimal expansion of. We setFix aprime cone ofas the family of all closed definable
subsets of the form$\tilde{\mathbb{R}}$$0$$\mathbb{R}_{\mathrm{a}\mathrm{n}}$$\alpha$$\Gamma_{\alpha}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$S=\{y\in M;f_{1}(y)\geq 0, \ldots, f_{m}(y)\geq 0\}$,where $f1$ ,is afinite family of elements inofis not empty by Lemma 4.2.
$\ldots$$f_{m}$$\alpha$,. Remark that any element$\Gamma_{\alpha}$. Assume that the intersection
Lemma 5.1. Fix a
coneof
is not an empty set. Then the set
consists of only one point
and the prime cone has the specialization , where $\alpha_{x}=\{f\in C_{df}^{\omega}(M);f(x)\geq$$p7\dot{\tau}me$$\alpha$$\bigcap_{S\in\Gamma_{\alpha}}S$$\bigcap_{S\in\Gamma_{\alpha}}S$

$x$

$0\}$

$C_{df}^{\omega}(M)$$\alpha_{x}$.

Proof. Obvious.

$\blacksquare$

Theorem 5.2. Consider an -minimal $ex$ ansion ofand a 2-dimensi0nal$M$analytic
space.
Then
isdefinable
submanifold of some Euclidean
Noether $ian$ ring$\tilde{\mathbb{R}}$$0$$\mathbb{R}_{an}$$C_{df}^{\omega}(M)$

$a$

15
and we can also assume thatbe contained and bounded in$M$ is connected. Let
denote acomplexification of $M$ and its germandand of class
at $M$ , respectively. (A complexification is always considered in. ) Here note thatis unique but
is not so. We set.
and define
to be the sheaf of holomorphic functions on, let $Z(p)$ denote the common zero set of and be
For asubset of
the germ at $M$ of acomplex
called aglobal analytic set. We next define
analytic set of someas follows. The sheaf $pO$ of $O$ -ideals generated by isis extendable as acoherent sheaf
coherent. It is easychoosewhere
to be theofdenote such an extension. We define-ideals. Let. Remark. We regardandgerm at $M$ ofand $Z(p)=Z^{\mathbb{C}}(p)\cap M$ . The notation dimc $Z(p)$ denotes
acomplex. We call agerm of the form
the complex dimension ofanalytic set germ (at $M$ ). It is known that acomplex analytic set admits aunique decomposition into irreducible ones.
is finitely generated if $Z(p)$We first show that aprime ideal ofis compact. Remark thatis finitely generated. Hence there exists,
afinite family ,of elements of which generates $pC^{\omega}(M)$ . Chaging,,, $f_{m}+K( \sum_{j=1}^{m}f_{j}^{2})$ for some large $K$ , we,with
may assume thatare compact for all . Fix an arbitrary element $f$ of .
Remark that $(f_{1}, \ldots, f_{m})C^{\omega}(M)=H^{0}(M, (f_{1}, \ldots, f_{m})O)$ by Cartan’s Theorem
B. Since $Z(p)$ is compact, $f^{l}\in(f_{1}, \ldots, f_{m})C^{\omega}(M)$ by Hilbert Nullstellensatz.
-ideals with compact support
Let mean the trivial extension of asheaf Iof. By Cartan’s theorem
to one offor-ideals defined by, the sequences

Proof. Let$M$$\mathbb{R}^{n}$$\tilde{M}^{\mathbb{C}}$$M^{\mathbb{C}}$$\mathbb{C}^{n}$$O^{\mathbb{C}}=O\otimes_{\mathbb{R}}\mathbb{C}$$\tilde{M}^{\mathbb{C}}$$M^{\mathbb{C}}$$C^{\omega}$$\overline{M}^{\mathbb{C}}$$\tilde{O}^{\mathbb{C}}$$C^{\omega}(M)$

$p$

$p$

$Z^{\mathbb{C}}(p)$$\tilde{M}^{\mathbb{C}}$$p$$\tilde{M}^{\mathbb{C}}$$pO^{\mathbb{C}}$$\underline{\mathrm{t}\mathrm{o}}$$\tilde{O}^{\mathbb{C}}$$Z^{\mathbb{C}}(p)$$pO^{\mathbb{C}}$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\overline{O}^{\mathbb{C}}/\overline{pO^{\mathbb{C}}})$$Z^{\mathbb{C}}(p)\subset M^{\mathbb{C}}$$M\subset M^{\mathbb{C}}$$Z(p)=\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(O/pO)$$Z^{\mathbb{C}}(p)$$Z^{\mathbb{C}}(p)$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$p$$pC^{\omega},(M)$$f_{1}$$f_{1}$$\ldots$$\ldots$$f_{m}$$p$$f_{1}+K( \sum_{j=1}^{m}f_{j}^{2})$$f_{m}$$\ldots$$f_{i}^{-1}(0)$

$i$

$p$

$O_{M}$$\overline{\mathrm{I}}$$\overline{\mathrm{I}}_{x}=O_{x}$$O_{\mathbb{R}^{n}}$$x\in \mathbb{R}^{n}\backslash M$$\mathrm{B}$$0arrow H^{0}(M, f_{1}O)arrow H^{0}$(A#,$\mathrm{I}(A)$)$arrow H^{0}q$(A#,$\mathrm{I}(A)/f_{1}O$) $arrow 0$$H^{0}(\mathbb{R}^{n}, (f_{2}, \ldots, f_{k})O)arrow H^{0}\mathfrak{W}^{n}(\mathbb{R}^{n},\mathrm{I}(A)/\overline{f_{1}O})$

are exact, and we can identify $H^{0}(M,\mathrm{I}(A)/f_{1}O)$ with$H^{0}(\mathbb{R}^{n},\mathrm{I}_{\mathbb{R}^{n}}(A)/\overline{f_{1}O})$.

, and, there
Hence there issuch that$g_{1}\inC^{\omega}(M)$$f-h_{1}|_{M}=g_{1}f_{1}$$f-h_{1}|_{M}\in
H^{0}(M,f_{1}O)$
.
fore,
. Let
be such that
and, hence,Then
are definable. In the same way, we can construct
such that $h_{1}|_{M}-h_{2}|_{M}=g_{2}f_{2}$ .
and,satisfying the required
Repeating these arguments, we can construct ,property.
is finitely generated if $Z(p)$We next show that aprime ideal of
has no analytically irreducible
is not compact. We want to show that$C\capM$
cmponents such thatis compact. Assume the contrary.
$\dim_{\mathbb{C}}(C)=1$
on $M$ such,If
there exists adefinable analytic function$C$
is divisible by inthat $Z^{\mathbb{C}}(h)=C$ and any analytic function vanishing onof $C$ . Fix an
definition
as shown before. Remark that ! by thearbitrary element of . Then $f/h^{k}\in p$ vanishes on all analytically irreducible. Contradiction to the definition
exceptcomponents offor some$h_{1}\in H^{0}(\mathbb{R}^{n}, \overline{(f_{2},}\ldots, f_{k})O)$$h_{1}|_{M}$$q_{\mathbb{R}^{n}}(h_{1})=q(f)$$g_{1}$$h_{2}\in H^{0}(\mathbb{R}^{n},\overline{(f_{3},}\ldots, f_{k})O)$$g_{2}\in C_{\mathrm{d}\mathrm{f}}^{\omega}.(M)$$g_{3}$$\ldots$$g_{k}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$p$$Z^{\mathbb{C}}(p)$

$C$

$h$

$h$

$h$

$C^{\omega}(M)$$f$$Z^{\mathbb{C}}(p)$

$p$

$p$

$C$

$k\in \mathrm{N}$16
If $\dim_{\mathbb{C}}(C)=0$ , we can easily construct adefinable analytic function $f\in p$
containing $C$ satissuch that any analytically irreducible component offies the condition $\dim(Y\cap M)=0$ . Therefore, as in the case when $\dim_{\mathbb{C}}(C)=1$ ,has no anwe can construct adefinable analytic function $g\in p$ such that
$C$ except, wheredenotealytically irreducible components containing$C$ .
$Z(p)$
components
containingofContradiction
the analytically irreeducible
has no analytically irreducible
to the definition of $C$ . We have shown that$C\capM$$C$
is compact.
cmponents such that
. Set $X= \bigcup_{i}X_{i}$ . We
be the irreducible components ofLet. Assume the contrary. Remark. Clearly
want to see
is also principal
is principal for each $x\in M$ . Obviously
that$0\neq
g\in
p$
.
Then
for
and
choosebe its generator
by [Cain]. Let. This is a
is an element of , but does not vanish on some
some
contradiction.
We show that is finitely generated. Choose an analytically irreducible. There exists a
whose. Remark thatcomponent
just 1. It is easy to construct $g\in p$ such thatmultiplicity on$M$$g\geq0$
andon . Let $\{Yj\}$ be analytically irreducible components of. Choose one point $y_{j}\in M\cap Y_{j}$ . Theneach one of which is different from all$(yj)\neq
0$ and the multiplicity of $gj$ onsuch that
we can choose
be analytically irreducible components of
is 1where $g_{j}=f+c_{j}g$ . Let. Choosing points and real numbers, we can constructafinite family of definable analytic functions whose complex common zero set is. Continuing
and which have the multiplicity 1onstrictly smaller thanand,with multiplicity 1onin this way, we can find ,$X$
coincides with .
is principal for all an. Remark thatFix an arbitraryof complex dimension $=1$ . Hence,ofalytically irreducible component
is divisible by
we can construct definable analytic functions such that
$Y$

$Z^{\mathbb{C}}(f)$$Z^{\mathbb{C}}(g)$$\{Y_{i}\}$$\{Y_{i}\}$$Z^{\mathbb{C}}(p)$$Z^{\mathbb{C}}(p)$$\{X_{i}\}$$p\subset I_{\mathrm{d}\mathrm{f}}(X)$$p=I_{\mathrm{d}\mathrm{f}}(X)$$I_{\mathrm{d}\mathrm{f}}(X)^{2}$$I_{\mathrm{d}\mathrm{f}}(X)O_{x}$$g^{2}/f^{m}$$f$$m\in \mathrm{N}$$X_{i}$$p$$p$$I_{\mathrm{d}\mathrm{f}}(X_{1})=p$$X_{1}$$f\in I_{\mathrm{d}\mathrm{f}}(X)$$(g^{\mathbb{C}})^{-1}(\mathrm{O})=X$$X_{1}$$(f^{\mathbb{C}})^{-1}(0)$$X_{i}$$c_{j}\in \mathbb{R}$$X_{1}$$\mathit{9}j$$\{Z_{k}\}$$(f^{\mathbb{C}})^{-1}(0) \cap\bigcap_{j}(g_{j}^{\mathbb{C}})^{-1}(0)$$X_{1}$$\cup Z_{k}$$f_{1}$$\ldots$$f\iota$$X_{1}$$\in I_{\mathrm{d}\mathrm{f}}(X_{1})$$\bigcap_{t=1}^{l}(f_{t}^{\mathbb{C}})^{-1}(0)$$I_{\mathrm{d}\mathrm{f}}(Y)^{2}$$h\in I_{\mathrm{d}\mathrm{f}}(X_{1})$$Y$$M^{\mathbb{C}}$$hh_{t}$$h_{t}$

and$\mathrm{n}_{t=1}^{l}(h_{t}^{\mathbb{C}})^{-1}(0)=\emptyset$$I_{\mathrm{d}\mathrm{f}}(X_{1})$is generated by. Therefore$f_{t}$$h= \sum_{t=1}^{l}\frac{h_{t}^{2}h/f_{t}}{\sum_{t=1}^{l}h_{t}^{2}},$ $f_{t}$,$f_{t}$. We have shown that. @. Let $M$ be a 2Theorem 5.3. Consider an -minimal expansionofdimensional definable analytic manifold, and let be a prime cone of,with supp(\mbox{\boldmath $\alpha$}) $=(0)$ . Assume
for given $f1$ , , , $g\in$, $fi(y)>0$ for some $y\in M$ .
,. Then$\tilde{\mathbb{R}}$$\mathbb{R}_{an}$$\mathit{0}$$C_{df}^{\omega}(M)$$\alpha$$that/\mathrm{i}(\mathrm{a})>0$$C_{df}^{\omega}(M)$$\mathrm{f}\mathrm{i}(\mathrm{y})>0$$\ldots,$$/\mathrm{i}(\mathrm{a})>0$$\ldots$$f_{s}$$\ldots$

is noetherian for any open set $N$ of $M$ . We firstis not empty. The prime cone is ageneralizaconsider the case whenbe aby Lemma 5.1. Let
than of, wheredenotescanonical homomorphism. Remark that
aprime cone $\{f\in O_{x}; f(x)\geq 0\}$ . Therefore, there exists aprime cone of, $f_{p}(\beta)>0$ , $g(\beta)=0$ .
by Theorem 2.5. Hence $f1(\beta)>0$ ,such that
The set $\{x\in M;f_{1}(x)>0, \ldots, f_{p}(x)>0, g(x)=0\}$ is not empty by Propositio

Proof. Remark that

$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$\alpha$$\bigcap_{S\in\Gamma_{\alpha}}S$$\alpha_{x}:=\{f\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M);f(x)\geq 0\}$$\phi:C_{\mathrm{d}\mathrm{f}}^{\omega}(M)arrow O_{x}$$\phi^{*}(\alpha_{\mathrm{a}\mathrm{n}\mathrm{a},x})=\alpha_{x}$$\alpha_{\mathrm{a}\mathrm{n}\mathrm{a},x}$$\beta$$\phi^{*}(\beta)=\alpha$$O_{x}$$\ldots$$\mathrm{n}$

17
. If
We next consider the other case. Set $d( \alpha)=\min\{\dim(S);$ S$d(\alpha)=2$ , this theorem is trivial. It is also trivial in the case when g is not azero function because the set$\in\Gamma_{\alpha}\}$$\{y\in M;f_{1}(y)\geq 0, \ldots, f_{s}(y)\geq 0, g(y)=0\}$

is not enpty by Lemma 4.2. Therefore we only consider the case when
) $=1$ andon $M$ .
We lead the contradiction with the assumption that the set$(/)$$\bigcap_{S\in\Gamma_{\alpha}}S=$$g\equiv 0$$\mathrm{d}\{\mathrm{a}$$\{y\in M;f_{1}(y)>0, \ldots, f_{s}(y)>0\}$is empty. We first prove aclaim which is necessary for the proof of this theorem.
We denotefor any definable analytic function on $M$ if and only
, whereif
andis a1-dimensional analytically irreducible$M$
subset ofandis the ideal of germs of analytic functions on $M$ at a
isnonsingular point $y\in Y$ which vanish on . It is well known thatindependent of the choice out of the nonsingular point . Remark that there.
exists anatural extension mult :$f$$m=\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f)$$Y$$f\not\in p_{Y}^{m+1}$$f\in p_{Y}^{m}$$p_{Y}$$Y$$\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f)$$y$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{pY}arrow \mathbb{N}\cup\{0\}$Claim. Let denote a1-dimensional definable analytic subset of $M$ suchthatis aprime ideal. Thenis aprime ideal generated.
by adefinable function
with
$Y$

$p_{Y}=I_{\mathrm{d}\mathrm{f}}(Y)$$p_{Y}C_{\mathrm{d}\mathrm{f}}^{\omega}/(M)_{p_{Y}}$$h_{Y}$

Proof of Claim. The ring$\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(h_{Y})=1$

is aregular local ring. It is especiallyan unique factorization Noetherian domain by [Ml, Theorem 20.3]. Hence,
Since the natural mapping$W$ is aprincipal ideal by [Ml, Theorem 20.1].
is definable and analytic, there exists an$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{p_{Y}}$$T\mathbb{R}^{n}\supset\{(y, v)\in M\cross T_{y}\mathbb{R}^{n}\}arrow TM$-derivation $D$ :for any$\mathbb{R}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{\mathrm{P}Y}arrow C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{p_{Y}}$$f\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{p_{Y}}$withsuch that. In fact, we have only to take the$\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(D(f))=\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f)-1$$\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f)>0$

are real numbers,
such that it is not zero in $TyM$ at some nonsingular point $y\in X$ . Therefore the. Claim is proved.
ofgenerator
has multiplicity
derivation induced by the image of$h_{Y}$$\sum_{k=1}^{n}a_{k}\frac{\partial}{\partial x_{k}}$, where$a_{k}$$\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(h_{Y})=1$$p_{Y}C_{\mathrm{d}\mathrm{f}}^{\omega}(M)_{pY}$The analytic closure of the set $Z(F)=\{y\in M;g_{1}(y)\geq 0, \ldots, g_{p}(y)\geq 0\}$in
is a1-dimensional analytic set for any finite family $F=\{g_{1}, \ldots, g_{p}\}\subset\alpha$with $\dim Z(F)\neq 2$ . This analytic closure has only finite analytically irreducible
components as was shown in the proof of Lemma 4.1. At least one analyticallyirreducible analytic set is an analytically irreducible component of the analytic
with $\dim Z(F)\neq 2$ because any
closure of $Z(F)$ for all finite families
by the assumption.
element ofhas dimension$F=\{g_{1},\ldots,g_{p}\}$
Fix afamilywith $\dim Z(F)=1$ . Assume that there
existssuch that, for any $y\in S\cap Z(F)\cap Y$ , the relations, , whereis asufficiently small
andhold true for some $i=1$ ,$M$such that
open neighborhood of in . There exist ,without
by Claim. We may assume that , ,
$M$

$Y$

$F\subset\alpha$$\geq 1$$\Gamma_{\alpha}$$g_{i}|_{Y}\equiv 0$$S\in\Gamma_{\alpha}$$g_{i}|_{U_{y}}\leq 0$$\ldots$$y$$WiQi=z_{i}h_{Y}^{\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(_{\mathit{9}:})}$$p$$w_{i}$$U_{y}$$z_{i}\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M)\backslash I_{\mathrm{d}\mathrm{f}}(Y)$$w_{i}$$z_{i}$$h_{Y}\in\alpha$$1\mathrm{o}\mathrm{s}$18
is even.
becauseis not positive on
and . Therefore theby the definition of
Furthermore,as an analytically irreducible
does not haveanalytic closure ofcomponent, where $F’=F\cup\{w_{i}\cdot z_{\mathrm{i}};i=1, \ldots,p\}$ . Contradiction to the definitionof .
forConsider the case when $F=\{f1, \ldots, f_{s}\}$ . By Claim,$F’=\{f_{i},u_{i},
v_{i};
i=1,\ldots,
s\}$
.
.
As
was
Set
some ,
shown in the last paragraph, there exists apoint $y\in Y\cap Z(F’)$ such that, for any
and is not contained inopen neighborhood of in $M$ ,except for any
any 1-dimensional analytically irreducible component of$\phi\in F’$ . Fix such asmall neighborhoodof . The equation $f_{q}|_{Y}=f_{r}|_{Y}\equiv 0$for some and . We may assume
is negative onholds true and,withoutofloss
generality. Remark that $uq$ ,that multy
are positive on . These facts, however, contradicts to the equation
andbecause the number
is an even number.

of generality. Obviously,$w_{i}\cdot$$\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(g_{i})$$U_{y}$$z_{i}$$Y\not\subset w_{i}^{-1}(0)$ $\cup z_{i}^{-1}(0)$$w_{i}$$z_{i}$$Y$$Z(F’)\in\Gamma_{\alpha}$$Y$$vlfi=u_{i}h_{Y}^{\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f_{i})}$$v_{i}$$u_{i}\in\alpha\cap(C_{\mathrm{d}\mathrm{f}}^{\omega}(M)\backslash I_{\mathrm{d}\mathrm{f}}(Y))$$U$$U\cap\phi^{-1}((0, \infty))\neq\emptyset$

$y$

$y$

$Y$

$\phi^{-1}(0)$$U$$f_{q}\cdot$$y$$U\backslash Y$$f_{r}$$r$$q$$(fq)\geq \mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f_{r})$$v_{q}$$u_{r}$$U$$v_{r}$$u_{r}v_{q}f_{q}=u_{q}v_{r}f_{r}h_{Y}^{\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f_{q}\rangle-\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f_{r})}$$\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f_{q})-\mathrm{m}\mathrm{u}1\mathrm{t}_{Y}(f_{r})$$\blacksquare$

Proof of Theorem 1.1 when $\dim(M)=2$ . We first show Hilbert’s 17th
. Assume thatfor
Problem. Let $K$ be the quotient field of
$K$
$x\in
M$
$-f(\alpha)>0$
,
by
.
for
some
Since
$-f(x)>0$
some positive cone of. Theorem is proved
Theorem 5.3. Contradiction. Therefore,
2.4.
by Theorem
We next show Real Nullstellensatz. We have only to show as in the case when$\dim(M)=1$ that $\mathrm{I}(Z(p))=p$ if is areal prime ideal. Real Nullstellensatz. If $\dim(Z(p))=$is obvious when $\dim(Z(p))=1$ because
$x\in
Z(p)$
.
. Consider the homomorphism, choose apoint
Remark that there exists aprime cone with support if is areal prime ideal.
andofsuch that
Hence, there exists aprime coneby Theorem 2.5. Contradiction to Proposition 2.6. @$f\not\in\alpha$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$f \in\bigcap_{\alpha\in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}(K)}\alpha$$p$$\dim(C_{\mathrm{d}\mathrm{f}}^{\omega}(M))=2$$0$$C_{\mathrm{d}\mathrm{f}}^{\omega}/(M)arrow O_{x}$$p$$\alpha’$$O_{x}$$p$$\mathrm{h}\mathrm{t}(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha’))=1$$\dim(Z(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)))=0$6Th$\mathrm{r}\mathrm{e}\mathrm{e}$-dimensional Case.

such that it admits
denotes an -minimal expansion of
In this section,arean analytic cylindrical definable cell decompositon and the rings$N$of EuNoetherian rings for all 3-dimensional definable analytic submanifold
are Noetherian rings for all 3-dimensi0nal
clidean spaces. Remark that.
definable analytic submanifold $N$ of Euclidean spaces when
The purpose of this section is to finish the proof of Theorem 1.1.
$\tilde{\mathbb{R}}$$0$$\mathbb{R}_{\mathrm{a}\mathrm{n}}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$\tilde{\mathbb{R}}=\mathbb{R}_{\mathrm{a}\mathrm{n}}([\mathrm{F}\mathrm{S}])$

Lemma 6.1. Let $M$ be a bounded definable analytic submanifold of . Let
such thatbe a prime cone offor some 1-dimensional definable
set C. Then one of the following prime cones is a specialization of .
$\mathbb{R}^{n}$$C_{df}^{\omega}’(M)$$\alpha$$C\in\Gamma_{\alpha}$$\alpha$Points of $C\alpha_{x}=\{f\in C_{df}^{\omega}(M);f(x)\geq 0\}$ , where x $\in C$ .

Curve germ The$p7\dot{\tau}me$cone ac,x defined as follows, where x is a point in19$\overline{C}\backslash C$. The function f is an element of

the set$C_{/}\cap\{y\in M,\cdot \mathrm{f}(\mathrm{y})\geq 0\}$in$\mathbb{R}^{n}$$\alpha_{C,x}$

if and only if the closure ofcontains the point x.

contains apoint , the prime cone a. We consider the case when this intersection
obviously has aspecialization
is notis empty. Since the closureof $M$ is compact, the set. The set consists of only one point because
empty and contained in
any two points are separated by some definable analytic function. It is easy tosee by using Lemma 4.2 that the closure of $C\cap\{y\in M;f(y)\geq 0\}$ contains thepoint for any.

Proof. When the intersection$x$$\bigcap_{S\in\Gamma_{\alpha}}S$$\alpha_{x}$$\lrcorner l^{-}I$$T= \bigcap_{S\in\Gamma_{\alpha}}\overline{S}$$T$$\overline{C}\backslash C$$f\in\alpha$$x$

Lemma 6.2. Let$0,g(\alpha)=0$ for some
, $g(x)=0$ for some$\blacksquare$$\alpha\in Spec_{r}(C_{df}^{\omega}(\mathbb{R}))$

$0$

$x$

$f_{1}$,$x\in \mathbb{R}$Proof. Remark that$\ldots$,$f_{s}$,. Suppose that $f1(\alpha)>0$ ,. Then $f_{1}(x)>0$ ,$g\in C_{df}^{\omega}(\mathbb{R})$\ldots ,$\ldots$$f_{s}(\alpha)>$, $f_{s}(x)>$.
is amaximal ideal or (0). When
. Therefore, $f_{1}(x)>0$ ,for someis, $f_{s}(x)>$amaximal ideal,, $g(x)=0$ .
We consider the case whennext. Remark that $g\equiv 0$ in thiscase. One can easily show that any definable analytic function $f$ is factorized$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)$$\alpha=\alpha_{x}$$x\in \mathbb{R}$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)$$\ldots$$0$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)=(0)$as follows.
$f=u \prod_{i=1}^{n}(t-a_{i})$,where allare real numbers and is an unit. We may therefore assume thatthe analytic functions
are of the form $f_{i}=t-a_{i}$ . If there exist no points$f_{1}(x)>0$,
such that both
, $f_{s}(x)>0$ , there exists
such that$t-a$ and $a-t$ are in
by Lemma 4.2. Contradiction to the assumption that.
$u$$a_{i}$$f_{i}$$x\in \mathbb{R}$$a\in \mathbb{R}$$\ldots$$\alpha$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\alpha)=(0)$$\blacksquare$. We
Let and be (not necessarily analytic) definable functions onthen anifdenotefor some $r>0$ . The relationequivalence relation. We set
as the set of equivalence classes with respect
to this relation.
$f_{1}$$(0, \infty)$$f_{2}$$f_{1}\sim f_{2}$$\sim \mathrm{i}\mathrm{s}$$f_{1}|_{(0,r)}\equiv f_{2}|_{(0,r)}$$K(\tilde{\mathbb{R}})$is a real closed field and the prime cone
under theofdefined in Lemma 6.1 lies over the prime cone.
natural injection

Lemma 6.3. The set$K(\tilde{\mathbb{R}})$$\alpha(0,\infty),0$$K(\tilde{\mathbb{R}})^{2}$$C_{df}^{\omega}((0, \infty))$$C_{df}^{\omega}((0, \infty))arrow K(\tilde{\mathbb{R}})$

Proof.

It is obvious thatis afield and $K(\tilde{\mathbb{R}})^{2}=\{[f];f|(0,r)\geq$
for some }, where $[f]$ represents the equivalence class of adefinable function . Clearly,
is apositive cone. See [ABR] for the definition of apositive cone.
Fix apolynomial $P(t, X)\in K(\tilde{\mathbb{R}})[X]$ of odd degree. There exists apositiveis adefinable function
number such that $P(t, X)= \sum_{k=0}^{n}f_{k}(t)X^{k}$ , where$f_{n}(t)\neq
0$
$(0,
r)$
$t\in(0,r)$
closed field, there
.
areal
with
on
Since isfor any
exists aroot of $P(t, X)$ in for any $t\in(0, r)$ . Hence there exists adefinable. Namely, $P(t, X)$ has aroot infunction : $(0, r’)arrow \mathbb{R}$ with
$K(\tilde{\mathbb{R}})$

$0$

$r$

$K(\tilde{\mathbb{R}})^{2}$$f$$f_{k}(t)$$r$$\mathbb{R}$$\mathbb{R}$$g$$\mathrm{P}(\mathrm{t},g(t))\equiv 0$

20
is areal closed field by [BCR, Theorem 1.2.2]. The lastpart of this lemma is obvious by the definition.
$K(\tilde{\mathbb{R}})$. Therefore,$K(\tilde{\mathbb{R}})$$\blacksquare$

Theorem 6.4. Let $M$ be a 3-dimensional definable analytic submanifold ofwith support (0). Assume thatand letbe a prime cone of$f_{s}(\alpha)>0$$f1$, $f_{s}(y)>0$. Then $f_{1}(y)>0$ ,,,for given , ,$\mathbb{R}^{n}$$f_{1}(\alpha)>$$C_{df}^{\omega}’(M)$$\alpha$$0$$\ldots$$\ldots$$f_{s}\in C_{df}^{\omega}(M)$$\ldots$

for some $y\in M$ .

Proof We may assume thatis connected and bounded in. The proof.
of this theorem is the same as that of Theore 5.3 in the case whenis empty. Set $d( \alpha)=\min\{\dim(S);S\in$Hence we may assume that the set, $d(\alpha)\neq 0$ . This theorem is trivial in the case
. By the definition ofwhen $d(\alpha)=3$ . We can prove this theorem in the same way as Theorem 5.3.
when $d(\alpha)=2$ . Therefore we may also assume that has aspecialization
Here is a1-dimensional definable analytic submanifold of $M$ , is apoint in, and
is aprime cone ofdefined in Lemma
the boundary of , in$M$$\mathbb{R}^{n}$$\bigcap_{S\in\Gamma_{\alpha}}\neq\emptyset$$\bigcap_{S\in\Gamma_{\alpha}}$$d(\alpha)$$\Gamma_{\alpha}\}$$\alpha$$\alpha c,x$

$C$

$x$

$C$

$\mathbb{R}^{n}$$C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$\mathrm{a}\mathrm{c},\mathrm{x}$6.1.
Take adefinable analytic tubler neighborhood $N$ of $C$ in $M$ and the definable
be the prime cone ofsuch thatanalytic retraction : $Narrow C$ . Let$C\cap\{y\inN;h(y)\geq
0\}$
if and only if the closure of the set
containsofwhich is lying overthe point . Then there exists aprime coneby Lemma 6.1, Proposition 4.3 and Theorem 2.5. Take ashorter curve $C$ and asmaller tubler neighborhood $N$ . Then we can easily construct adefinable andanalytic diffeomorphism $(N, C)\simeq((0, \infty)\cross \mathbb{R}^{n-1},$ $(0, \infty)\cross\{0\})$ . Hence we may
assume $M=(0, \infty)\cross \mathbb{R}^{n-1}$ , $C=(0, \infty)\cross\{0\}$ and $x=(0$ , , 0 . Let be the, $x_{n-1}$ be the other coordinate functions.
first coordinate function and ,and that $x_{j}=f_{j}$ for $j=$belong to
We may further assume that all,,, $n-1$ without loss of generality. We will construct functions ,, $fx\{y)>0\}$such that the set { $y\in U;f_{1}(y)>0$ ,is some open
contains the set $\{y\in U;h_{1}(y)>0, \ldots, h_{s}(y)>0\}$ , where
neighborhood of $(0, r)$ $\cross\{0\}\subset C$ and is apositive real number. We firstare constructed infinish the proof with the assumption that such functions
advance.
conas the maximal positive number such thatSet
tains the cube {0} $\cross(-\rho(t), \rho(t))^{n-1}$ for any 0 $<t<r$ . Since the funcis analytic. Set $D=$tion : $(0, r)arrow \mathbb{R}$ is definable, the restriction$r’)\cross\mathbb{R}^{n-1};-\rho(t)<x_{i}<\rho(t)$, $x_{n-1})\in(0,$,for any }. Then there{(
lying over by Theorem 2.5, Lemma 6.1
exists aprime coneand Proposition 4.3. Consider the embedding : $D’=(0, r’)\cross(-1,1)^{n-1}arrow M$
given by $\iota(t, x_{1}, \ldots, x_{n-1})=(t, \rho(t)x_{1}$ , $\rho(t)x_{n-1})$ . This embedding obviously induces adefinable and analytic diffeomorphism between $D’$ and $D$ . The. The integral domain
on $D’$ are also infunctionsis contained in the real closed fieldby Lemma 6.3. The canonical homomorphism
is regular
Lemma 1] and [Ml, Theorem 7.2]. We show
and faithfully flat by [M2,is regular and excellent. Regularity follows fromthat$\alpha_{C,x}’$$\rho$$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$h\in\alpha_{C,x}’$$\alpha’$$x$$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$\alpha$$t$$)$$\ldots$$x_{1}$$\ldots$$\alpha$$x_{j}$$1$$h_{1}$$\ldots$$C_{\mathrm{d}\mathrm{f}}^{\omega}((0, \infty))[x_{1}, \ldots, x_{n-1}]$$\ldots$$\ldots$$U$$r$$h_{j}$$U\cap(\{t\}\cross \mathbb{R}^{n-1})$$\rho(t)$$\rho|(0,r’)$$\rho$$t,x_{1}$$i.$$\ldots$$\alpha’$$\tilde{\alpha}\in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}_{r}(C_{\mathrm{d}\mathrm{f}}^{\omega}(D))$$\iota$$\ldots$$C_{\mathrm{d}\mathrm{f}}^{\omega}((0, r’))[x_{1}, \ldots, x_{n}]$$h_{j}\circ\iota$$K(\tilde{\mathbb{R}})$$C_{\mathrm{d}\mathrm{f}}^{\omega}((0, r’))$$C_{\mathrm{d}\mathrm{f}}^{\omega}((0, r’))[x_{1}, \ldots, x_{n-1}]arrow K(\tilde{\mathbb{R}})[x_{1}, \ldots, x_{n-1}]$$33.\mathrm{B}$$C_{\mathrm{d}\mathrm{f}}^{\omega}((0, r’))[x_{1}, \ldots, x_{n-1}]$$h_{S}\in$21
is regular. Excellentness follows from
[Ml, Theorem 19.5] becauseof[M2, Theorem 73, 77]. Therefore some prime coneconsisting of all polynomials whose
lies overbecause the prime cone
. By Positivstellensatz, there exconstant terms are positive lies overists atuple $(x_{1}(t), \ldots, x_{n-1}(t))\in K(\tilde{\mathbb{R}})$ such that $h_{1}(x_{1}(t), \ldots, x_{n-1}(t))>$. The definable curve,, $h_{s}(x_{1}(t), \ldots, x_{n-1}(t))>0$ as elements of, $x_{n-1}(t))$ is well defined for any small enoughgiven by $\gamma(t)=(t, x_{1}(t)$ ,positive real number . Particularly, the set $\{y\in U;f1(y)>0, \ldots, f_{s}(y)>0\}$
is not empty.
We give some notations. Fix adefinable analytic function $f$ . Set$C_{\mathrm{d}\mathrm{f}}^{\omega}’((0, r’))$$K(\tilde{\mathbb{R}})[x_{1}, \ldots, x_{n-1}]$$\alpha_{\mathrm{p}\mathrm{o}1\mathrm{y}}$$\tilde{\alpha}$$\beta_{\mathrm{p}\mathrm{o}1\mathrm{y}}$$\alpha c,x$$K(\tilde{\mathbb{R}})$$0$$\ldots$$\gamma$$\ldots$$t$$\mathrm{C}(i_{1},\ldots,i_{n-1})=$$c_{(i_{1\prime}\ldots,i_{n-1})}^{f}= \frac{1}{i_{1}!\cdots i_{n-1}!}\frac{\partial^{i_{1}+\cdots+i_{n-1}}f}{\partial x_{1}^{i_{1}}\cdots\partial x_{n-1}^{i_{n-1}}}(t, 0, \ldots, 0)\in C_{\mathrm{d}\mathrm{f}}^{\omega}((0, \infty))$. For $k=$satisfying that is not. We also
in the ideal generated by
but in that generated bysatisfying thatdefine the number. Remark that, for any
for any
and$M$, there exist definable analytic functions
definable analytic function on,,
on A# with,,
. In fact, we have only to set
because$1$,$\ldots$, $n-1$ , define the natural number$f$$\kappa_{1}^{k}=\kappa_{1}^{k}(f)$$x_{k}^{\kappa_{1}^{k}(f)}$$x_{k}^{\kappa_{1}^{k}(f)+1}$$\kappa_{2}^{k}=\kappa_{2}^{k}(f)$$l<\kappa_{2}^{k}$$c_{(\kappa_{1}^{1},\ldots,\kappa_{1}^{k-1},l,\kappa_{1}^{k+1},\ldots,\kappa_{1}^{n-1})}^{f}\equiv 0$$c_{(\kappa_{1}^{1},..,\kappa_{1}^{k-1},\kappa_{2}^{k},\kappa_{1}^{k+1},\ldots,\kappa_{1}^{n-1})}\not\equiv 0$$g$$\mathrm{c}(\mathrm{t})$$\mathrm{d}\mathrm{i}(\mathrm{t}, \mathrm{x}\mathrm{i})$$\mathrm{d}\mathrm{i}(\mathrm{t}, x_{1}, \mathrm{x}\mathrm{i})$$\mathrm{d}$$7=c+ \sum_{i=1}^{n-1}d_{i}x_{i}$$\mathrm{i}(\mathrm{t}, x_{1}, \ldots, x_{n-1})$$\ldots$$M=(0, \infty)\cross \mathbb{R}$if$\frac{g(t,x_{1},\ldots,x_{j},0,\ldots,0)-c(t)-\sum_{i=1}^{j-1}x_{i}d_{i}(t,x_{1},\ldots,x_{i})}{x_{\mathrm{j}}}$$x_{\mathrm{j}}\neq 0$$d_{j}=\{$$\frac{\partial(g-c-\sum_{i=1}^{j-1}x_{i}d_{i})}{\partial x_{j}}$otherwise$(t, x_{1}, \ldots, x_{j-1}, 0, \ldots,0)$which is analytic because the numerator vanishes on $\{x_{i}=0\}$ . Set$\mathrm{J}(\mathrm{f}):=\{i=(i_{1}, \ldots, i_{n-1});0\leq i_{1}\leq\kappa_{2}^{1}(f), \ldots, 0\leq i_{n-1}\leq\kappa_{2}^{n-1}(f)\}$$J(f):=\{j=(j_{1}, \ldots,j_{n-1});0\leq j_{1}\leq\kappa_{2}^{1}(f)+1$$\exists k,j_{k}=\kappa_{2}^{k}(f)+1\}$

Then there exist,$\ldots$and, $0\leq j_{n-1}\leq\kappa_{2}^{n-1}(f)+1$ ,.

$d_{(j_{1},\ldots,j_{n-1})}=d_{(j_{1},\ldots,j_{n-1})}^{f}\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$with$f_{q}= \sum_{i\in I(f)}c_{i}x_{1}^{i_{1}}\cdots x_{n-1}^{i_{n-1}}+\sum_{j\in J(f)}d_{j}x_{1}^{j_{1}}\cdots x_{n-1}^{j_{n-1}}$.

definedcoincides with the prime cone
The prime coneandin Lemma 6.1 by Lemma 6.2. Therefore, for any nonzero
at is positive if and only if, the value
is positive by Lemma 4.2. Fix adefinable analytic function on $M$ with
$at\cap C_{\mathrm{d}\mathrm{f}}^{\omega}((0, \infty))$$\alpha_{(0,\infty),(0)}$$c\in C_{\mathrm{d}\mathrm{f}}^{\omega}((0, \infty))$$P_{j}\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$\mathrm{c}(\mathrm{a})+\sum_{j=1}^{n-1}x_{j}(\alpha)P_{j}(\alpha)$$\alpha$$f$$c(\alpha)$$f(\alpha)>0$

Claim.

.

$P(\alpha)x_{k}(\alpha)^{\kappa_{2}^{k}(f)+1}$

is smaller than$/(\mathrm{a})$any .

for any$P\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$and for$k$

Proof of Claim. Assume that

$P(\alpha)x_{1}(\alpha)^{\kappa_{2}^{1}(f)+1}$

is larger than$f(\alpha)$

for some

22. We construct adeConsider the case when,
finable analytic function satisfying
and thatis larger thanfor some. Set $Q=f$ ifaanndd sseett $Q=Px_{1}^{\kappa_{2}^{1}(f)+1}-f$ otherwise. By thedefinition, $Q(\alpha)>0$ . Furthermore, there exist definable analytic functions$P\in C_{\mathrm{d}\mathrm{f}}^{\omega}/(M)$.

$\kappa_{1}^{1}(f)\neq\kappa_{2}^{1}(f)$$\kappa_{2}^{1}(g)=\kappa_{2}^{1}(f)$$P(\alpha)x_{1}(\alpha)^{\kappa_{2}^{1}(g)+1}$$g(\alpha)$$\kappa_{1}^{1}(g)>\kappa_{1}^{1}(f)$$P\in C_{\mathrm{d}\mathrm{f}}^{\omega}(M)$$c_{\kappa_{2}^{1}(f),\kappa_{1}^{2}(f),\ldots,\kappa_{1}^{n-1}(f)}^{f}(\alpha)<0$$P_{0}(t, x_{2}, \ldots, x_{n-1})$,$\ldots$,$P_{\kappa_{2}^{1}(f)-1}(t, x_{2}, \ldots, x_{n-1})$, $R(t, x_{1}, \ldots, x_{n-1})$

on $M$ such thatand $R(\alpha)<0$ . Then $g=$
satisfies the required property. Therefore we have only to prove. When, the equation
this claim in the case when
$Q= \sum_{i=0}^{\kappa_{2}^{1}(f)-1}P_{i}x_{1}^{i}+x_{1}^{\kappa_{2}^{1}(f)}R$$Q-x_{1}^{\kappa_{2}^{1}(f)}R$$\kappa_{1}^{1}(f)=\kappa_{2}^{1}(f)$holds for any$\kappa_{1}^{1}(f)=\kappa_{2}^{1}(f)$by the definition.
(ci) $<0$ and set
otherwise. By
Set $Q=f$ if
the definition, $Q(\alpha)>0$ . The function , however, is of the form$\kappa_{1}^{k}(f)=\kappa_{2}^{k}(f)$$k$$Px_{1}^{\kappa_{2}^{1}(f)+1}-f$$c_{\kappa_{1}^{1}(f),\kappa_{1}^{2}(f),\ldots,\kappa_{1}^{n-1}(f)}^{f}$$Q$$Q= \prod_{i=1}^{n-1}x_{i}^{\kappa \mathrm{i}(f)}(c(t)+\sum_{j=1}^{n-1}x_{j}P_{j}’)$,where $c(t)$ is adefinable analytic function onwith $c(\alpha)<0$ andare$M$$Q(\alpha)<0$. Contradiction. We
definable analytic functions on . Particularly,
have shown the claim.
$(0, \infty)$$P_{j}’$, . Choose an arbitrary . Whenfrom $f1$ ,We construct ,$q\leq n-1$ , set $h_{q}=f_{q}=x_{q}$ . We construct
for $q\geq n$ . Any
is of, where ,is adefinable
the formandis adefinable analytic function on
analytic function on$M$ . Set$h_{1}$$\ldots$$h_{s}$$\ldots$$f_{s}$$f_{q}$$d_{(j_{1},\ldots,j_{n-1})}^{f_{q}}$$h_{q}$$C_{(j_{1},\ldots,j_{n-1})}^{q}+ \sum_{i=1}^{n-1}x_{i}P_{(j_{1},\ldots,j_{n-1}),i}^{q}$$(0, \infty)$$1\leq$$C_{(j_{1},\ldots,j_{n-1})}^{q}$$P_{(j_{1},\ldots,j_{n-1}),i}^{q}$$T_{(j_{1},\ldots,j_{n-1})}^{q}=-1- \sum_{i=1}^{n-1}x_{i}P_{(j_{1}}^{q},\ldots$, $j_{n-1})$ , .
$i$

Then

$T_{(j_{1},\ldots,j_{n-1})}^{q}(\alpha)<0$. Set$J_{q}=\{(j_{1}, \ldots,j_{n-1});0\leq j_{1}\leq\kappa_{2}^{1}(f_{q})+1$$\exists k,j_{k}=\kappa_{2}^{k}(f_{q})+1\}$,$\ldots$, $0\leq j_{n-1}\leq\kappa_{2}^{n-1}(f_{q})+1$ ,,$h_{q}=f_{q}+ \sum_{(j_{1},\ldots,j_{n-1})\in J}T_{(\mathrm{j}_{1,\ldots\prime}j_{n-1})}^{q}x_{1}^{\mathrm{j}_{1}}\cdots x_{n-1}^{\mathrm{j}_{n-1}}$and.

$U=\cap^{s}\cap\{y\in M;T_{(j_{1\prime}\ldots,j_{n-1})}^{q}(y)q=n(j_{1},\ldots,\mathrm{j}_{n-1})\in J_{q}<0\}$Thenby the way of construction.
is negative on ,Since
is an open neighborhood of $C$ . As wasshown befor\^e 0
for $n\leq q\leq s$ . Hence the constructedandsatisfy the required
properties.
$\bigcap_{q=1}^{s}h_{q}^{-1}((0, \infty))\cap U\subset\bigcap_{q=1}^{s}f_{q}^{-1}((0, \infty))\cap U$$T_{(j_{1},\ldots,j_{n-1})}^{q}$

$C$

$U$

$<- \sum_{(j_{1},\ldots,j_{n-1})\in J}T_{(j_{1},\ldots,j_{n-1})}^{q}(\alpha)x_{1}(\alpha)^{j_{1}}\cdots x_{n-1}(\alpha)^{j_{n-1}}<$$f_{q}(\alpha)$$h_{q}$$\blacksquare$

$U$

23
Proof of Theorem 1.1 in the case when $\dim(M)=3$ . The proof of Hilbert’s
17th Problem is the same as in the case when $\dim(M)=2$ . Hence we omit theproof of Hilbert’s 17th Problem
We will show Real Nullstellensatz. As shown in the proof of Real Nullstellensatz in lower dimensional case, we have only to prove thatwhen is areal prime ideal. We can prove Real Nullstellensatz in the same
way as in the lower dimensional case if
is compact or height of is 2.
Hence we have only to show that is not areal ideal whenand.
Assume the contrary. There exists aprime cone of C$f(M) whose support
coincides with . If the intersection
is not empty, ahas aspecialization
by Lemma 5.1. There exists aprime cone oflying over withby Proposition 4.3 and Theorem 2.5. Since the germ of zeros of
is of dimension, it contradicts to Proposition 2.6. We next consider
the case when
. The prime cone has aspecialization
suchthat $C$ is adefinable analytic submanifold of $M$ contained in $Z(p)$ by Lemma
6.1. Choose asufficiently short curve , then we may assume that there exists
atubler neighborhood $N$ of such that the pair $(N, C)$ is definably and analytically diffeomorphic to the pair
, $(0, \infty)$. As was mentioned
some times in the present paper, there exists aprime cone oflying overawith
by Proposition 4.3 and Theorem 2.5. It is obvious thatis of dimension $=1$ . Set, then is generated
by one definable analytic function on $N$ by Corollary 4.4 and [Ml, Theorem
20.1]. We show that and $-f$ are not asum of finite squares of elements of
the quotient field of. Assume the contrary, then there exist definable
analytic functions ,,,on $N$ with $Q^{2}f=P_{1}^{2}+\cdots+P_{m}^{2}$ . Since isreal, allare in . Hence,for some definable analytic,,’
functionsbecauseis adomain. There therefore exists a
definable analytic function $Q’$ on $N$ with $Q=fQ’$ and.
Continuing in this way, we obtain that
. The function
vanishes on $N$ by Krull intersection theorem [Ml, Theorem 8.10]. Contradiction.
Whence, changes the sign on some open set which intersects with $f^{-1}(0)$ by
Hilbert’s 17th Problem. However, the zero set $f^{-1}(0)$ is of codimension $>1$ .
They contradicts each other. We have shown Real Nullstellensatz in the casewhen I is aprime ideal.
$\mathrm{I}_{\mathrm{d}\mathrm{f}}(Z_{\mathrm{d}\mathrm{f}}(p))=p$$p$$Z_{\mathrm{d}\mathrm{f}}(p)$$p$$\dim Z_{\mathrm{d}\mathrm{f}}(p)=1$$p$$\mathrm{h}\mathrm{t}(p)=1$$\alpha$$p$$\bigcap_{S\in\Gamma_{\alpha}}S$$\beta$$\alpha_{x}$$O_{x}$$\alpha$$\mathrm{h}\mathrm{t}(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta))=1$$\leq 1$$\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta)$$\bigcap_{S\in\Gamma_{\alpha}}S=\emptyset$$\alpha$$\alpha c,x$$\mathrm{C}$$C$$((0, \infty)\cross \mathbb{R}^{2}$$\cross\{0\})$$\beta$$C_{\mathrm{d}\mathrm{f}}^{\omega}’(N)$$\mathrm{h}\mathrm{t}(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta))=\mathrm{h}\mathrm{t}(p)=1$$Z(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta))$$q=\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\beta)$

$q$

$f$

$f$

$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$P_{1}$$P_{j}$$P_{1}’$$\ldots$$P_{m}$$q$$Q^{2}=(P_{1}^{\prime 2}+\cdots+P_{m}^{\prime 2})f$$q$$\ldots$$Q\not\equiv \mathrm{O}$$P_{m}’$$C_{\mathrm{d}\mathrm{f}}^{\omega}(N)$$Q^{\prime 2}f=P_{1}^{\prime 2}+\cdots+P_{m}^{\prime 2}$$Q \in\bigcap_{n\in \mathrm{N}}q^{n}$

$Q$

$f$

$\blacksquare$

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