Analytic finiteness for preparation

Written by Claude Opus 5.5 (Anthropic), October 2026; self-checked by the writing AI. Original text: public domain (CC0). The theorems and the route of proof are those of Guillaume Valette, On subanalytic geometry, arXiv:2507.23622v1, Chapter 1 and §2.3. Global subanalytic sets are defined by the projective product convention of Jean-Marie Lion and Jean-Philippe Rolin, Théorème de préparation pour les fonctions logarithmico-exponentielles.

This reading derives the finiteness theorems of global subanalytic geometry from Weierstrass division. Division makes the ring of convergent real power series Noetherian, and the Artin–Rees lemma and Krull’s intersection theorem follow. A linear approximation theorem then shows that a real analytic function of $(u,z)$ is controlled, near a point, by finitely many of its Taylor coefficients in $z$:

$$\psi(u,z)=\sum_{i=0}^{d}c_i(u)(z-z_0)^iA_i(u,z), \qquad A_i(u_0,z_0)=1,$$

with analytic functions $A_i$. Dividing by the largest coefficient produces a family whose order in $z$ is bounded uniformly. This is the analytic input to the preparation theorem: on the cells of a finite decomposition, every function built from restricted analytic functions and rational powers becomes a base coefficient times a power of a distance $|z-\theta(x)|$ times an analytic unit. The cell decomposition theorem for globally subanalytic sets, Gabrielov’s complement theorem, convergent Puiseux expansions and the dimension theory of the last section are consequences.

Real analytic division and finite generation

Let $R_n=\mathbb R\{x_1,\ldots,x_n\}$ be the ring of germs at $0\in\mathbb R^n$ of real analytic functions, that is, of real power series with a positive radius of convergence, and let $\mathfrak m_n\subset R_n$ consist of the germs vanishing at $0$. Translation transfers every statement below to germs at an arbitrary point.

The ring $R_n$ is local with maximal ideal $\mathfrak m_n$. A germ with $f(0)\ne0$ has an analytic reciprocal near $0$, and a germ with $f(0)=0$ has no inverse. The ideal $\mathfrak m_n$ is generated by $x_1,\ldots,x_n$. To see this, give each monomial $x^\alpha\ne1$ of a series without constant term to the smallest index $i$ with $\alpha_i>0$, and write it as $x_i x^{\alpha-e_i}$. The monomials given to $i$ form $x_ig_i$, where $g_i$ is a subseries of an absolutely convergent series, so convergent on the same polydisc.

The complex preparation and division theorems are proved in Preparation by the zeros in one fibre and Division, finite generation and elementary algebra, Theorem 1 and Corollary 3 there. They descend to real germs because the factors are unique. Call $g\in R_n$ regular of order $s$ in $x_n$ if $g(0,\ldots,0,x_n)=x_n^s\,h(x_n)$ with $h(0)\ne0$. Regard $g$ as a complex germ and write $g=uP$, with $u$ a unit and $P=x_n^s+\sum_{k=1}^sa_k(x')x_n^{s-k}$ a Weierstrass polynomial. Replacing every coefficient of the power series of $u$ and $P$ by its complex conjugate gives a second such factorization of the real germ $g$. By uniqueness the two coincide, so $u$ and $P$ have real coefficients. In the same way, for real $f$ the unique division

$$f=qP+\sum_{j=0}^{s-1}r_j(x')x_n^j$$

has real $q$ and real $r_j\in R_{n-1}$. Thus a real germ regular of order $s$ in $x_n$ is a real unit times a real Weierstrass polynomial of degree $s$, and every $f\in R_n$ is uniquely $qP$ plus a polynomial of degree less than $s$ in $x_n$ with coefficients in $R_{n-1}$.

Regularity can be produced by a real linear change of coordinates. A nonzero real polynomial $F$ does not vanish on all of $\mathbb R^n$. Indeed, view $F$ as a polynomial in $x_n$ whose coefficients are polynomials in the other variables. By induction one of these coefficients is nonzero at some real point, and there the resulting nonzero polynomial in $x_n$ has a non-root. If $g\ne0$ has order $m$ and lowest homogeneous part $g_m$, choose $v\in\mathbb R^n$ with $g_m(v)\ne0$ and a linear isomorphism $L$ with $L(e_n)=v$. Then $g(L(0,\ldots,0,t))=g(tv)=t^mg_m(v)+O(t^{m+1})$, so $g\circ L$ is regular of order $m$ in $x_n$. Applying this to a product of finitely many germs makes all of them regular at once.

We use the module fact established in the proof of Theorem 6 of the division reading: over a Noetherian ring every submodule of a finite free module is finitely generated.

Noetherianity of real convergent germs. The ring $R_n$ is Noetherian.

Proof. We induct on $n$; the field $R_0=\mathbb R$ is Noetherian. Let $J\subset R_n$ be a nonzero ideal. A linear change of coordinates is a ring automorphism, so we may assume that $J$ contains a germ $g$ regular of some order $s$ in $x_n$. Write $g=uP$ as above. Then $P=u^{-1}g\in J$. If $s=0$, $P=1$ and $J=R_n$. Otherwise divide every $f\in J$ by this one polynomial: $f=qP+r(f)$. The remainder $r(f)=f-qP$ lies in $J$, and by uniqueness of division the map $f\mapsto r(f)$ is $R_{n-1}$-linear. Its image is an $R_{n-1}$-submodule of the polynomials of degree less than $s$, a free module of rank $s$. By induction and the module fact it is generated by finitely many remainders $r(f_1),\ldots,r(f_k)$. Every $f\in J$ is then $qP$ plus an $R_{n-1}$-combination of these, so $J=(P,r(f_1),\ldots,r(f_k))$. $\square$

Polynomial rings and stable filtrations

Hilbert basis theorem. If $A$ is a Noetherian ring, so is $A[T]$.

Proof. Let $J\subset A[T]$ be an ideal. The leading coefficients of the elements of $J$, together with $0$, form an ideal of $A$: multiplying by a power of $T$ aligns degrees before adding. Pick finitely many elements $f_1,\ldots,f_k$ of $J$ such that their leading coefficients generate that ideal, and let $D$ be the largest of their degrees. If $f\in J$ has degree $N\ge D$ and leading coefficient $\sum c_i\operatorname{lc}(f_i)$, then $f-\sum c_iT^{N-\deg f_i}f_i$ lies in $J$ and has smaller degree. Repeating, every element of $J$ is a combination of the $f_i$ plus an element of $J$ of degree less than $D$. Those elements form a submodule of $A^D$, finitely generated by the module fact. So $J$ is finitely generated. By induction, polynomial rings in finitely many variables over $A$ are Noetherian, and so are their quotients, since ideals of a quotient pull back to ideals. $\square$

Let $A$ be Noetherian, $I\subset A$ an ideal and $M$ a finitely generated $A$-module. A decreasing sequence of submodules $M=M_0\supset M_1\supset\cdots$ is an $I$-filtration if $IM_i\subset M_{i+1}$ for all $i$. It is stable if $IM_i=M_{i+1}$ for all large $i$. The Rees algebra and the Rees module of the filtration are

$$A^\ast=\bigoplus_{i\ge0}I^iT^i\subset A[T],\qquad M^\ast=\bigoplus_{i\ge0}M_iT^i,$$

with $I^0=A$. Since $I^jM_i\subset M_{i+j}$, $M^\ast$ is a module over $A^\ast$. If $a_1,\ldots,a_\ell$ generate $I$, then $Z_j\mapsto a_jT$ maps $A[Z_1,\ldots,Z_\ell]$ onto $A^\ast$. By the Hilbert basis theorem $A^\ast$ is Noetherian.

Stability criterion. The module $M^\ast$ is finitely generated over $A^\ast$ if and only if the filtration is stable.

Proof. Suppose $IM_i=M_{i+1}$ for $i\ge e$. Then $M_{e+j}=I^jM_e$ for all $j\ge0$. Each $M_i$ is a submodule of the finitely generated module $M$, hence finitely generated, and generators of $M_0,\ldots,M_e$, placed in their degrees, generate $M^\ast$. Conversely, a finite generating set may be replaced by the homogeneous components of its elements. Let these be $m_kT^{e_k}$ with $m_k\in M_{e_k}$, and $e=\max e_k$. For $i\ge e$, every element of $M_{i+1}$ is a sum $\sum_ka_km_k$ with $a_k\in I^{i+1-e_k}=I\cdot I^{i-e_k}$. Since $I^{i-e_k}m_k\subset M_i$, it lies in $IM_i$. So $M_{i+1}\subset IM_i$, and the reverse inclusion holds for every filtration. $\square$

Artin–Rees lemma. For every submodule $N\subset M$ there is an $e$ with

$$N\cap I^{i+1}M=I\,(N\cap I^iM)\qquad(i\ge e),$$

and consequently $N\cap I^iM=I^{i-e}(N\cap I^eM)$ for $i\ge e$.

Proof. The filtration $(I^iM)_i$ is stable, so its Rees module is finitely generated over the Noetherian ring $A^\ast$. The sets $N\cap I^iM$ form an $I$-filtration of $N$, whose Rees module is an $A^\ast$-submodule of the former. It is therefore finitely generated, and the stability criterion gives the assertion. $\square$

The lemma holds for every ideal of a Noetherian ring. Below, $I$ will be the maximal ideal of a local Noetherian ring.

Krull intersection for modules

Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak m$. Its non-units are exactly the elements of $\mathfrak m$.

Nakayama’s lemma. If a finitely generated module $L$ satisfies $L=\mathfrak mL$, then $L=0$.

Proof. Suppose $L\ne0$ and let $e_1,\ldots,e_p$, $p\ge1$, be a generating set of minimal size. Write $e_p=\sum_jb_je_j$ with all $b_j\in\mathfrak m$. Then $(1-b_p)e_p=\sum_{j<p}b_je_j$. Since $1-b_p\notin\mathfrak m$, it is a unit. So $e_p$ lies in the span of $e_1,\ldots,e_{p-1}$, contradicting minimality; for $p=1$ it gives $e_1=0$ and $L=0$. $\square$

Krull intersection theorem. If $M$ is finitely generated, then $\bigcap_{i\ge0}\mathfrak m^iM=0$.

Proof. The intersection $L$ of all $\mathfrak m^iM$ is a submodule of $M$. For $i$ large, the Artin–Rees lemma with $N=L$ gives $L=L\cap\mathfrak m^{i+1}M=\mathfrak m(L\cap\mathfrak m^iM)=\mathfrak mL$. Nakayama’s lemma gives $L=0$. $\square$

Applied to $M/N$, whose submodules $\mathfrak m^i(M/N)$ are the images of $\mathfrak m^iM$, the theorem says that every submodule of a finitely generated module is closed in the $\mathfrak m$-adic sense:

$$\bigcap_{i\ge0}\left(N+\mathfrak m^iM\right)=N.$$

For an ideal $J\subset A$ this reads $\bigcap_i(J+\mathfrak m^i)=J$. Locality cannot be dropped. In $\mathbb R[x]$, with $J=(x-1)$ and $I=(x)$, the identity $1=x^i-(x^i-1)$ shows $J+I^i=\mathbb R[x]$ for every $i\ge1$, although $J$ is a proper ideal.

Formal linear solutions have convergent replacements

Let $A=R_n$, let $\widehat A=\mathbb R[[x_1,\ldots,x_n]]$ be the ring of formal power series, and let $H$ be a $b\times a$ matrix with entries in $A$. It defines linear maps $A^a\to A^b$ and $\widehat A^a\to\widehat A^b$.

Linear approximation theorem. Let $v\in A^b$, and suppose $H\widehat f=v$ for some $\widehat f\in\widehat A^a$. Then $Hf=v$ for some $f\in A^a$. For every $q\ge0$ the solution $f$ can be chosen to agree with $\widehat f$ in all Taylor coefficients of degree less than $q$.

Proof. Let $p_k$ be the vector of polynomials obtained by truncating each component of $\widehat f$ after degree $k$. Then $v-Hp_k=H(\widehat f-p_k)$ has no terms of degree at most $k$, because multiplication by a power series cannot lower the order. It is also a vector of convergent series. A convergent series without terms of degree at most $k$ belongs to $\mathfrak m^{k+1}$. To see this, assign each of its monomials to one of the finitely many monomials of degree $k+1$ dividing it, and collect: the series becomes a finite sum of such monomials times convergent subseries. Hence $v\in H(A^a)+\mathfrak m^{k+1}A^b$ for every $k$, and the closedness of submodules gives $v\in H(A^a)$.

For the second statement let $q\ge1$ and let $p$ be the truncation of $\widehat f$ below degree $q$. Grouping as before, $\widehat f=p+\sum_{|\alpha|=q}x^\alpha\widehat g_\alpha$ with formal vectors $\widehat g_\alpha$. So the linear system $\sum_{|\alpha|=q}x^\alpha Hg_\alpha=v-Hp$, in the unknowns $g_\alpha\in A^a$, has a formal solution. By the first part it has a convergent solution $g_\alpha$, and $f=p+\sum x^\alpha g_\alpha$ solves $Hf=v$ with the required low-degree coefficients. $\square$

The theorem replaces a formal solution by a convergent one. It says nothing about convergence of the given formal solution, and it concerns linear systems only.

A finite Taylor family with analytic units

Finite Taylor family. Let $\psi$ be real analytic near $(u_0,z_0)\in\mathbb R^{n-1}\times\mathbb R$, and let $c_j(u)=\frac1{j!}\partial_z^j\psi(u,z_0)$ be its Taylor coefficients in $z$. There are $d\ge0$ and analytic germs $A_0,\ldots,A_d$ at $(u_0,z_0)$ with $A_i(u_0,z_0)=1$ such that, near $(u_0,z_0)$,

$$\psi(u,z)=\sum_{i=0}^{d}c_i(u)(z-z_0)^iA_i(u,z).$$

Proof. Translate so that $(u_0,z_0)=(0,0)$; then $\psi=\sum_{j\ge0}c_j(u)z^j$ with $c_j\in R_{n-1}$. The ideals $(c_0,\ldots,c_k)$ increase with $k$, so by Noetherianity they are constant from some $d$ on. For $j>d$ choose $b_{ij}\in R_{n-1}$ with $c_j=\sum_{i\le d}b_{ij}c_i$. In $\mathbb R[[u,z]]$ put $\widehat F_i=\sum_{j>d}b_{ij}(u)z^{j-i-1}$. A given monomial $u^\beta z^\gamma$ receives a contribution only from $j=\gamma+i+1$, so $\widehat F_i$ is a well-defined formal series, and

$$\sum_{i=0}^dc_iz^{i+1}\widehat F_i=\sum_{j>d}\Bigl(\sum_{i\le d}b_{ij}c_i\Bigr)z^j=\psi-\sum_{i=0}^dc_iz^i.$$

The linear equation $\sum_{i\le d}c_iz^{i+1}F_i=\psi-\sum_{i\le d}c_iz^i$ in the unknowns $F_0,\ldots,F_d\in R_n$ therefore has a formal solution. By the linear approximation theorem it has a convergent solution $G_0,\ldots,G_d$. Then $A_i=1+zG_i$ satisfies $A_i(0,0)=1$ and gives the identity. If $\psi=0$, take $d=0$ and $A_0=1$. $\square$

The infinitely many $b_{ij}$ need not admit a common convergence estimate, so the formal series $\widehat F_i$ may diverge. The linear approximation theorem is what produces analytic $A_i$. Near $(u_0,z_0)$ all $A_i$ are nonzero.

Signed normalization and a uniform order bound

Suppose that near $(u_0,z_0)$

$$\psi(u,z)=\sum_{i=0}^dc_i(u)(z-z_0)^iA_i(u,z),\qquad A_i(u_0,z_0)\ne0.$$

For $0\le m\le d$ let $K_m=\{t\in[-1,1]^{d+1}:t_m=1\}$ and

$$Q(t,u,z)=\sum_{i=0}^dt_i(z-z_0)^iA_i(u,z).$$

Uniform regularity. There are a neighbourhood $V$ of $(u_0,z_0)$ and $\delta>0$ such that

$$\max_{0\le k\le m}\bigl|\partial_z^kQ(t,u,z)\bigr|\ge\delta\qquad(t\in K_m,\ (u,z)\in V).$$

Consequently, for $t\in K_m$ and $(u,z')\in V$, the function $z\mapsto Q(t,u,z)$ has order at most $m$ at $z'$.

Proof. At $(u_0,z_0)$ the $k$-th $z$-derivative of $(z-z_0)^iA_i$ vanishes for $k<i$ and equals $i!A_i(u_0,z_0)\ne0$ for $k=i$. Hence $\partial_z^kQ(t,u_0,z_0)$, $k\le m$, depends only on $t_0,\ldots,t_k$, through a triangular linear system with nonzero diagonal. As $t_m=1$, these $m+1$ derivatives are not all zero. Their maximum is a positive continuous function on the compact set $K_m$, with a minimum $2\delta>0$. The derivatives are uniformly continuous on $K_m$ times a closed box around $(u_0,z_0)$, which gives a neighbourhood $V$ on which the maximum stays at least $\delta$. $\square$

Here is how the bound is used. Let $\phi$ be a map from a set $B$ to the parameter space and put $g_i=c_i\circ\phi$. On a part of $B$ where $g_m\ne0$ and $|g_i|\le|g_m|$ for every $i$,

$$\psi(\phi(x),z)=g_m(x)\,Q\Bigl(\frac{g_0(x)}{g_m(x)},\ldots,\frac{g_d(x)}{g_m(x)},\phi(x),z\Bigr),$$

and the normalized vector lies in $K_m$ whatever the signs of the $g_i$. Where all $g_i$ vanish, $\psi(\phi(x),z)=0$. Arranging the $g_i$ into pieces of this kind is part of the cell induction below and is not assumed here. If $\psi$ is analytic near a compact set of parameters, finitely many of the neighbourhoods above cover it, and the induction refines their preimages compatibly.

How this supplies the geometric argument

The analytic chain proved so far is: real division gives Noetherianity of convergent germs; the Rees algebra gives the Artin–Rees lemma; Artin–Rees and Nakayama give Krull’s theorem and the closedness of submodules; closedness gives convergent solutions of formal linear systems; these give the finite Taylor family, whose normalization has a uniformly bounded order. The remaining sections build the cell induction on this input. They prove the preparation of functions built from restricted analytic functions and rational powers, the cell theorem, the complement theorem and convergent Puiseux expansions under the projective product definition. The curve-selection and Łojasiewicz reading uses the cell theorem and the Puiseux expansions proved here.

A convergent split before restricting to a hyperbola

On the hyperbola $c=zw$ a monomial $z^iw^j$ can be written as a power of $c$ times a pure power of $z$ or of $w$, according to whether $i>j$ or $i\le j$. Summing these rewrites gives a decomposition of an analytic function of $(u,z,w)$ into two functions, each of which depends on only one of the coordinates $z,w$ besides $c$.

Splitting lemma. Let $\psi(u,z,w)$ be real analytic near $(a,0,0)$, with $u\in\mathbb R^k$. There are $R>0$ and functions $\psi_+(u,c,z)$, $\psi_-(u,c,w)$, analytic for $|u_\nu-a_\nu|<R$, $|c|<R^2$ and $|z|,|w|<R$, such that

$$\psi(u,z,w)=z\,\psi_+(u,zw,z)+\psi_-(u,zw,w) \qquad(|u_\nu-a_\nu|<R,\ |z|,|w|<R).$$

Proof. The power series of $\psi$ at $(a,0,0)$ defines a holomorphic function on a complex polydisc around that point. Shrinking, we may take a common radius $2R$ for that polydisc with $|\psi|\le M$ on the closed polydisc of radius $R$. Write

$$\psi(u,z,w)=\sum_{\alpha\in\mathbb N^k}\ \sum_{i,j\ge0}b_{\alpha ij}(u-a)^\alpha z^iw^j .$$

Cauchy’s estimate in each variable gives $|b_{\alpha ij}|\le MR^{-|\alpha|-i-j}$. Define

$$\psi_+(u,c,z)=\sum_\alpha\sum_{i>j}b_{\alpha ij}(u-a)^\alpha c^j z^{i-j-1},\qquad \psi_-(u,c,w)=\sum_\alpha\sum_{i\le j}b_{\alpha ij}(u-a)^\alpha c^i w^{j-i}.$$

In the first sum, $(\alpha,i,j)\mapsto(\alpha,j,i-j-1)$ is a bijection onto $\mathbb N^k\times\mathbb N\times\mathbb N$, and the absolute value of a term is at most

$$\frac MR\prod_\nu\Bigl(\frac{|u_\nu-a_\nu|}R\Bigr)^{\alpha_\nu}\Bigl(\frac{|c|}{R^2}\Bigr)^j\Bigl(\frac{|z|}R\Bigr)^{i-j-1}.$$

So the sum converges absolutely, uniformly on smaller closed polydiscs, and defines an analytic function in the stated range. The second sum is treated in the same way with the bijection $(\alpha,i,j)\mapsto(\alpha,i,j-i)$ and the bound $M\prod_\nu(|u_\nu-a_\nu|/R)^{\alpha_\nu}(|c|/R^2)^i(|w|/R)^{j-i}$. If $|z|,|w|<R$, then $c=zw$ satisfies $|c|<R^2$. Substituting it, the terms of $z\psi_+$ become $b_{\alpha ij}(u-a)^\alpha z^iw^j$ for $i>j$, and those of $\psi_-$ become the same expression for $i\le j$. All series converge absolutely, so they can be recombined, which gives $\psi$. $\square$

For example, $\psi=1+z^2+zw+w^2$ gives $\psi_+=z$ and $\psi_-=1+c+w^2$: the term $z^2$ has $i>j$, while $1$, $zw$ and $w^2$ have $i\le j$. The identity holds for all signs of $z$ and $w$, and also at $z=0$. Where $z\ne0$, substituting $w=c/z$ gives

$$\psi(u,z,c/z)=z\,\psi_+(u,c,z)+\psi_-(u,c,c/z).$$

Localize on the actual image, rather than a larger product

Throughout the cell arguments, points of $\mathbb R^n$ are written $(x,z)$, with $x\in\mathbb R^{n-1}$ the base variables and $z$ the last coordinate. Let $C$ be a set of points with $z\ne0$. Let $\phi$ and $c$ be functions of $x$, with values in $\mathbb R^k$ and $\mathbb R$, such that

$$\Phi(x,z)=\bigl(\phi(x),z,c(x)/z\bigr)$$

is bounded on $C$. Let $\psi$ be analytic on an open neighbourhood $U$ of the compact set $K=\overline{\Phi(C)}$. Write $w=c/z$, so that $c=zw$ on $C$.

Localization lemma. There are finitely many open sets $V_1,\ldots,V_N$ covering $K$ such that, on each $C\cap\Phi^{-1}(V_l)$, the function $\psi\circ\Phi$ has one of the following forms.

1. $F(\phi,c,z)$, where $F(u,c,z)=\psi(u,z,c/z)$ is analytic near a closed box containing all values of $(\phi,c,z)$ on the piece, and $|z|$ is bounded below by a positive constant there.
2. $F(\phi,c,w)$, where $F(u,c,w)=\psi(u,c/w,w)$ is analytic near a closed box containing all values of $(\phi,c,w)$ on the piece, and $|w|$ is bounded below by a positive constant there.
3. $z\,\psi_+(\phi,c,z)+\psi_-(\phi,c,w)$, with the functions of the splitting lemma at a point $(a,0,0)\in K$, all values of $(\phi,c,z)$ and $(\phi,c,w)$ lying in closed boxes inside their domains of analyticity.

Proof. Let $p=(a,z_0,w_0)\in K$. Suppose first $z_0\ne0$. The map $(u,c,z)\mapsto(u,z,c/z)$ is analytic where $z\ne0$ and sends $(a,z_0w_0,z_0)$ to $p\in U$, so $F(u,c,z)=\psi(u,z,c/z)$ is analytic near $(a,z_0w_0,z_0)$. Choose a closed box around this point inside the domain of $F$, on which $|z|\ge|z_0|/2$. The continuous map $(u,z,w)\mapsto(u,zw,z)$ sends $p$ into the interior of the box, and therefore also some neighbourhood $V_p$ of $p$. On $C\cap\Phi^{-1}(V_p)$ we have $\psi(\phi,z,c/z)=F(\phi,c,z)$, and form 1 holds. If $w_0\ne0$, the same argument with $(u,c,w)\mapsto(u,c/w,w)$ gives form 2. If $z_0=w_0=0$, then $p=(a,0,0)$ lies in $K\subset U$, so $\psi$ is analytic near $p$. The splitting lemma applies, and a small $V_p$ keeps $(u,zw,z)$ and $(u,zw,w)$ inside closed boxes in the domains of $\psi_+$ and $\psi_-$. This gives form 3. Compactness of $K$ selects finitely many $V_p$. $\square$

Only actual points of $K$ are used. The function $\psi$ need not be analytic at $(a,0,0)$ for every $a$ in the closure of $\phi(C)$, and no analytic extension of $\psi$ beyond $U$ is required.

The signed balanced-coordinate calculation

In the remaining case of the two-coordinate argument below, both summands in form 3 are comparable, and the coordinate $z$ is comparable to a function of the base. The following lemma turns the function into an analytic composition in a single new coordinate.

Balanced substitution lemma. Keep the notation of the localization lemma, and assume moreover $c\ne0$ on $C$. Let $b>0$ be a function of $x$ with $\kappa^{-1}b\le|z|\le\kappa b$ on $C$, for a constant $\kappa\ge1$. Put $q=z/b$ and $d=c/b$. Then the map $\Gamma=(\phi,b,d,q)$ is bounded on $C$ and $|q|\ge\kappa^{-1}$ there. The function

$$\widetilde\psi(u,b,d,q)=\psi(u,bq,d/q)$$

is analytic on a neighbourhood of $\overline{\Gamma(C)}$, and $\psi\circ\Phi=\widetilde\psi\circ\Gamma$ on $C$.

Proof. Since $z$ is bounded, so is $b\le\kappa|z|$. Next, $|q|\in[\kappa^{-1},\kappa]$ and $d=(c/z)q=wq$ is bounded. On the closure of $\Gamma(C)$ we still have $|q|\ge\kappa^{-1}$, so the map $\Lambda(u,b,d,q)=(u,bq,d/q)$ is analytic near that closure. It sends $\Gamma(C)$ onto $\Phi(C)$, because $bq=z$ and $d/q=c/z$. By continuity it sends $\overline{\Gamma(C)}$ into $K\subset U$. The open set $\Lambda^{-1}(U)$ therefore contains $\overline{\Gamma(C)}$, and $\widetilde\psi=\psi\circ\Lambda$ is analytic there. $\square$

No sign of $z$, $c$ or of the two summands has been assumed. The choice of $b$ is made in the proof of two-coordinate preparation below.

Analytic units and the lower-dimensional induction

The preparation theorem is proved for a class of functions closed under the operations it needs. A restricted analytic function is a function on a closed cube $Q\subset\mathbb R^m$ that extends analytically to a neighbourhood of $Q$. An $\mathcal L$-function on a set $E\subset\mathbb R^m$ is a function given on $E$ by a finite expression built from the coordinate functions and real constants with sums, products, rational powers $t\mapsto t^\lambda$ and restricted analytic functions, each operation applied only where it is defined. A non-integer power requires $t>0$, a negative integer power requires $t\ne0$, and a restricted analytic function requires its argument to lie in its cube.

$\mathcal L$-cells are defined inductively, one ambient dimension at a time. The only cell in $\mathbb R^0$ is the point. A cell in $\mathbb R^n$ is either a graph $\{(x,\zeta(x)):x\in B\}$ or a band $\{(x,z):x\in B,\ \zeta_1(x)<z<\zeta_2(x)\}$. Here $B\subset\mathbb R^{n-1}$ is a cell, the base; $\zeta$, $\zeta_1$ and $\zeta_2$ are analytic $\mathcal L$-functions on $B$ with $\zeta_1<\zeta_2$; and $\zeta_1=-\infty$ or $\zeta_2=+\infty$ is allowed. Cells are connected. A graph is homeomorphic to its base. A band is homeomorphic to $B\times\mathbb R$ by $(x,\tau)\mapsto(x,z)$, with $z=\zeta_1+(\zeta_2-\zeta_1)/(1+e^{-\tau})$, $z=\zeta_1+e^\tau$, $z=\zeta_2-e^{-\tau}$ or $z=\tau$, according to which endpoints are finite. Induction on $n$ proves connectedness.

Let $C$ be a band with base $B$. A translation for $C$ is an analytic $\mathcal L$-function $\theta$ on $B$ whose graph does not meet $C$. Then $y=|z-\theta(x)|$ is positive on $C$. An analytic $\mathcal L$-unit in $y$ is a function

$$U(x,z)=\Psi\bigl(\beta(x),\,u(x)y^{1/s},\,v(x)y^{-1/s}\bigr),$$

where $s\ge1$ is an integer, $\beta$ (vector valued), $u$ and $v$ are analytic $\mathcal L$-functions on $B$, the argument map is bounded on $C$, and $\Psi$ is analytic and nowhere zero on a neighbourhood of the compact closure of its image. A function $f$ on $C$ is reduced if

$$f(x,z)=a(x)\,y^r\,U(x,z),\qquad r\in\mathbb Q,$$

with an analytic $\mathcal L$-function $a$ on $B$, possibly zero, and a unit $U$ in the same $y$. On a graph cell, a function is reduced if it is an analytic $\mathcal L$-function of the base. A function is reducible on a cell if it is reduced on each cell of a finite refinement. A unit is bounded and bounded away from zero, by continuity of $\Psi$ on the compact closure, and it has constant sign on the connected cell. We use the analytic form of units, not only these bounds.

We prove the preparation theorem by induction on $n$. The induction hypothesis $(\mathrm H_{<n})$ is the following statement in every dimension $m<n$: given finitely many cells of $\mathbb R^m$ and finitely many $\mathcal L$-functions, each defined on one of these cells, there is a finite cylindrical decomposition of $\mathbb R^m$ into cells, compatible with the given cells, with the following property. On each new cell $D$, the given functions whose domain contains $D$ are analytic and of constant sign, and any two of them are equal on $D$ or strictly ordered there. The theorem for arbitrary subsets of $\mathbb R^n$ is not assumed.

A function on a graph cell $\{z=\zeta(x)\}$ is reduced in the lower dimension. Substituting $z=\zeta(x)$ into an $\mathcal L$-expression gives an $\mathcal L$-function of $x$, and $(\mathrm H_{<n})$ makes it analytic on the cells of a refinement of the base. This is how graph cells are handled in every argument below, so those arguments only treat bands.

Finite endpoint refinements

Endpoint refinement. Let finitely many cells of $\mathbb R^n$ be given, together with finitely many $\mathcal L$-functions on their bases. Assume $(\mathrm H_{<n})$. There is a finite decomposition of $\mathbb R^n$ into cells that is compatible with the given cells and, over each base cell on which a given function is defined, with the graph of that function.

Proof. Apply $(\mathrm H_{<n})$ in $\mathbb R^{n-1}$ to the bases of the given cells and to all given functions together with the finite endpoints of the given cells. On a resulting base cell $D$, the relevant functions are analytic and any two are equal or strictly ordered. List the distinct ones as $\zeta_1<\cdots<\zeta_p$. The graphs of the $\zeta_l$, the bands between consecutive ones, and the two bands below $\zeta_1$ and above $\zeta_p$ (all of $D\times\mathbb R$ if $p=0$) partition $D\times\mathbb R$ into cells. A given cell is a union of these, because its base is a union of base cells and its endpoints occur among the $\zeta_l$. $\square$

A decomposition of this shape is called cylindrical: the bases of its cells form a cylindrical decomposition of $\mathbb R^{n-1}$, and over each base cell its cells are the graphs and bands cut out by finitely many ordered analytic functions. Every point of a base cell lies under each cell above it, so projecting a cylindrical decomposition onto the first $m$ coordinates gives a cylindrical decomposition of $\mathbb R^m$. In particular finitely many finite cell decompositions have a common cylindrical refinement. This uses no cell theorem for arbitrary sets. Three conventions follow from the lemma and are used without comment below.

• A function of the base alone is reduced on a band after cutting at $z=0$: on the bands $z>0$ and $z<0$ use $\theta=0$, $r=0$ and $U=1$. The graph $z=0$ is a graph cell.
• If $f=ay^rU$ is reduced, refine the base so that $a$ has constant sign. Since $y>0$ and $U$ has constant sign, $f$ has constant sign on each new cell.
• A reduction restricts to a subcell with the same base data. Its translation still misses the subcell, and its argument image only shrinks.

Collecting bounded monomials into two coordinates

In this and the following sections $C$ is a band with base $B$, $\theta$ is a translation for $C$, $y=|z-\theta(x)|>0$, and $(\mathrm H_{<n})$ is assumed. Every refinement of the base is made with $(\mathrm H_{<n})$, and every refinement of a cylinder is an endpoint refinement.

Two-coordinate lemma. Let $M_i=b_i(x)y^{q_i}$, $1\le i\le N$, with $q_i\in\mathbb Q$ and $\mathcal L$-functions $b_i$, be bounded on $C$. After refining the base, there are an integer $s\ge1$ and analytic $\mathcal L$-functions $u,v\ge0$ on the base such that $Z=uy^{1/s}$ and $W=vy^{-1/s}$ are bounded on $C$, and each $M_i$ is a monomial in $Z$ or in $W$ times a bounded analytic base function, or a bounded base function if $q_i=0$.

Proof. Choose $s$ with all $k_i=sq_i$ integers, and refine the base so that every $b_i$ is analytic and either identically zero or of constant sign, and the finitely many functions introduced below are ordered. For the indices with $k_i>0$ and $b_i\ne0$, put $u_i=|b_i|^{1/k_i}$ and let $u$ be the largest of them on each base cell; put $u=0$ if there are no such indices. If $u>0$, then

$$M_i=\operatorname{sign}(b_i)\Bigl(\frac{u_i}{u}\Bigr)^{k_i}Z^{k_i},\qquad 0\le\frac{u_i}u\le1,$$

and $Z$ equals $|M_i|^{1/k_i}$ for the index where $u_i=u$, so $Z$ is bounded. Indices with $k_i<0$ are treated in the same way with $v_i=|b_i|^{1/|k_i|}$, their maximum $v$ and $W=vy^{-1/s}$. For $k_i=0$, $M_i=b_i$ is a base function, bounded because every base point has a nonempty fibre in $C$. Roots, ratios and maxima of $\mathcal L$-functions are $\mathcal L$-functions, and the refinement makes them analytic. $\square$

The recovery is polynomial. Let $\Gamma=(\eta,Z,W)$ collect the new bounded arguments, with $\eta$ the bounded base functions, and let $P$ be the polynomial map with $P\circ\Gamma=(M_1,\ldots,M_N)$ on $C$. Then

$$P\bigl(\overline{\Gamma(C)}\bigr)\subset\overline{P(\Gamma(C))},$$

because $P$ is continuous. Hence an analytic function $\Psi$ given near the compact closure of the old arguments composes with $P$ to an analytic function near $\overline{\Gamma(C)}$, and if $\Psi$ has no zeros near the old closure, $\Psi\circ P$ has none near the new one. This lets any finite set of bounded monomial arguments be replaced by the two coordinates $Z$, $W$ and bounded base functions.

Closure and substitution for analytic units

Units form a group. If $U_1,U_2$ are units in the same $y$, then $U_1U_2$ and $1/U_1$ are units in $y$, and so is $|U_1|^\lambda$ for every $\lambda\in\mathbb Q$.

Proof. Collect the arguments of both units, which are bounded monomials and bounded base functions, by the two-coordinate lemma. With the polynomial recovery map $P$, both outer functions become analytic and zero-free near the compact closure of the new argument image. So do their product and the reciprocal of the first. For a power, note that the image $\Gamma(C)$ is connected, since $C$ is, so its closure is connected and $\Psi_1\circ P$ has a constant sign $\epsilon$ there. Then $\epsilon\,\Psi_1\circ P$ is positive on a neighbourhood of that closure, where its $\lambda$-th power is analytic and positive. $\square$

Change of distance. Let $h>0$ be another distance on $C$, $h=|z-\vartheta(x)|$, and suppose $h=A(x)\,y^e\,V(x,z)$ with an analytic $\mathcal L$-function $A>0$, $e\in\mathbb Q$ and a positive unit $V$ in $y$. Then every unit in $h$ is a unit in $y$. Consequently a reduction $f=a\,h^r\,U$ in $h$ becomes the reduction

$$f=\bigl(aA^r\bigr)\,y^{er}\,\bigl(V^rU\bigr)$$

in $y$.

Proof. An argument of a unit in $h$ is a bounded monomial $b(x)h^q=b(x)A(x)^qy^{eq}\,V^q$ or a base function. Since $V$ is bounded above and below by positive constants, the monomial $bA^qy^{eq}$ is bounded. Collect these monomials and the arguments of $V$ by the two-coordinate lemma. Recover $V$ through its outer function, then $V^q$ by an analytic power, then the old arguments by multiplication. The resulting analytic map is defined near the compact closure of the new argument image, coincides with the old argument map on $C$, and so sends that closure into the closure of the old argument image. The old outer function composes with it to a zero-free analytic function there. The last formula follows, and $V^rU$ is a unit by the group property. $\square$

A common translation for finitely many reductions

Common translation lemma. Let $f_1,\ldots,f_k$ be reduced on $C$ with translations $\theta_1,\ldots,\theta_k$. After an endpoint refinement, on each new band some $\theta_j$ serves as a common translation. More precisely, each distance $|z-\theta_i|$ equals $A_iy^{e_i}V_i$ with $y=|z-\theta_j|$, $A_i>0$, $e_i\in\{0,1\}$ and $V_i$ a positive unit, so every $f_i$ is reduced with translation $\theta_j$.