Simultaneous Monomialization

TL;DR

This paper presents a new proof of the simultaneous embedded local uniformization theorem in zero characteristic, utilizing a novel approach called simultaneous monomialization, which advances the understanding of resolution of singularities.

Contribution

It introduces a new proof technique for uniformization using simultaneous monomialization, extending existing methods in algebraic geometry.

Findings

New proof of uniformization theorem in zero characteristic

Development of the key elements theory for monomialization

Enhanced understanding of resolution of singularities

Abstract

We give a new proof of the simultaneous embedded local uniformization Theorem in zero characteristic for essentially of finite type rings and for quasi excellent rings. The results are a consequence of the simultaneaous monomialization presented here. The methods develop the key elements theory that is a more subtle notion than the notion of key polynomials.

Equations

$$R_{\nu }:=\left\{x\in K\text{ such that }\nu \left(x\right)\ge 0\right\},$$

$$\left(R,u\right)\to \left(R^{\prime },u^{\prime }=\left(u_1^{\prime },\dots u_n^{\prime }\right)\right)$$

$$\left(R,u\right)\to \cdots \to \left(R_i,u^{\left(i\right)}\right)\to \cdots$$

$${ϵ}_{\nu }\left(P\right):=\underset{b\in {\mathbb{N}}^{\ast }}{\max }\left\{\frac{\nu \left(P\right)-\nu \left({\partial }_{b}P\right)}{b}\right\}.$$

$${ϵ}_{\nu }\left(P\right)\ge {ϵ}_{\nu }(Q)\Rightarrow \mathrm{de}{\mathrm{g}}_{\mathrm{X}}(P)\ge \mathrm{de}{\mathrm{g}}_{\mathrm{X}}(Q).$$

$$\left(R,u\right)\to \cdots \to \left(R^{\prime },u^{\prime }\right)$$

$$\left(R,u\right)\to \cdots \to \left(T,t\right)$$

$$\left(R,u\right)\to \left(R_1,u^{(1)}\right)\to \cdots \to \left(R_i,u^{(i)}\right)\to \cdots$$

$$\forall f\in R\text{, }\exists i\text{ such that }\nu \left(f\right)={\nu }_{Q_i}\left(f\right).$$

$$\{\begin{array}{l}\forall i\text{, }{Q}_i<Q_{i+1}\text{ or }{Q}_i{<}_{\lim }{Q}_{i+1},\\ \forall Q\in \Lambda \exists i\text{ such that }ϵ(Q_i)\ge ϵ(Q).\end{array}$$

$$\left(R,u\right)\to \cdots \to \left(R_j,u^{(j)}\right)\to \cdots$$

$$\forall i\text{, }\exists j,k\text{ such that }{Q}_i^{\mathrm{strict},j}=u_k^{(j)},$$

$$\left(R,u\right)\to \cdots \to \left(R_s,u^{(s)}\right)\to \cdots$$

$$\left(R,u\right)\to \cdots \to \left(R_l,u^{\left(l\right)}\right)\to \dots$$

$$H:=\bigcap _{\beta \in \nu \left(R\setminus \left\{0\right\}\right)}{P}_{\beta }\hat{R}$$

$$\nu :R\to {\Gamma }_{\infty }$$

$$K\hookrightarrow K(X)$$

$${ϵ}_{\nu }\left(P\right):=\underset{b\in {\mathbb{N}}^{\ast }}{\max }\left\{\frac{\nu \left(P\right)-\nu \left({\partial }_{b}P\right)}{b}\right\}.$$

$$ϵ\left(z\right)=\underset{b\in {\mathbb{N}}^{\ast }}{\max }\left\{\frac{\nu \left(z\right)-\nu \left({\partial }_{b}z\right)}{b}\right\}=\frac{\nu \left(z\right)-\nu \left(\partial z\right)}{1}=\nu \left(z\right)=\left(0,1+\pi \right).$$

$${ϵ}_{\nu }\left(P\right)\ge {ϵ}_{\nu }(Q)\Rightarrow \mathrm{de}{\mathrm{g}}_{\mathrm{X}}(P)\ge \mathrm{de}{\mathrm{g}}_{\mathrm{X}}(Q).$$

$$b_{\nu }\left(P\right):=\min I(P)$$

$$I(P):=\left\{b\in {\mathbb{N}}^{\ast }\text{ such that }\frac{\nu \left(P\right)-\nu \left({\partial }_{b}P\right)}{b}={ϵ}_{\nu }\left(P\right)\right\}.$$

$$S_Q\left(P\right):=\left\{j\in \left\{0,\dots ,n\right\}\text{ such that }\nu \left(p_j{Q}^j\right)={\nu }_Q\left(P\right)\right\}$$

$${\delta }_Q\left(P\right):=\max \left\{S_Q\left(P\right)\right\}.$$

$${\tilde{P}}_{\nu ,Q}:=\sum _{j\in S_Q\left(P\right)}{p}_j{Q}^j.$$

$$\nu \left(r\right)=\nu \left(\prod _{i=1}^t{P}_i\right)<\nu \left(qQ\right).$$

$$\begin{array}[]{ccc}\nu(r)&=&\nu\left(P_{1}P_{2}-qQ\right)\\ &=&\nu\left(P_{1}P_{2}\right)\\ &<&\nu\left(qQ\right)\end{array}$$

$$\begin{array}[]{ccc}\deg_{X}\left(P_{1}P_{2}\right)&=&\deg_{X}\left(P_{1}\right)+\deg_{X}\left(P_{2}\right)\\ &<&2\deg_{X}\left(Q\right).\end{array}$$

$$\nu \left({\partial }_{b(Q)}\left(qQ\right)\right)=\nu \left(\sum _{j=0}^{b(Q)}\left({\partial }_{b(Q)-j}\left(Q\right){\partial }_j\left(q\right)\right)\right).$$

$$\begin{array}[]{ccc}\nu\left(q\partial_{b(Q)}\left(Q\right)\right)&=&\nu\left(q\right)+\nu\left(\partial_{b(Q)}\left(Q\right)\right)\\ &=&\nu\left(q\right)+\nu\left(Q\right)-b\left(Q\right)\epsilon\left(Q\right)\end{array}$$

Full text

1. I Introduction.

2. II Key polynomials.

3. 1 Key polynomials of Spivakovsky et al.

4. 2 Immediate successors.

5. III Simultaneous local uniformization in the case of rings essentially of finite type over a field.

6. 3 Preliminaries.

7. 3.1 Non-degenerate elements.

8. 3.2 Framed and monomial blow-up.

9. 3.3 Key elements.

10. 4 Monomialization in the non-degenerate case.

11. 4.1 Construction of a stricly decreasing numerical character.

12. 4.2 Divisibility and change of variables.

13. 4.3 Monomialization of non-degenerate elements.

14. 5 Non-degeneracy and key polynomials.

15. 6 Monomialization of the key polynomials.

16. 6.1 Generalities.

17. 6.2 Puiseux packages.

18. 6.3 Generalization.

19. 6.4 Divisibility.

20. 7 Conclusion.

21. IV Simultaneous local uniformization in the case of quasi-excellent rings for valuations of rank less than or equal to $2$.

22. 8 Preliminaries.

23. 8.1 Quasi-excellent rings and the implicit prime ideal.

24. 8.2 Numerical characters associated to a singular local noetherian ring.

25. 9 Implicit ideal.

26. 10 Monomialization of key polynomials.

27. 11 Reduction.

28. 12 Conclusion.

Part I Introduction.

The resolution of singularities can be formulated in the following way.

Let $V$ be a singular variety. The variety $V$ admits a resolution of singularities if there exists a smooth variety $W$ and a proper birational morphism $W\to V$.

This problem has been solved in many cases but remains an open problem in others. In characteristic zero Hironaka proved resolution of singularities in all dimensions ([33]) in 1964. Much work has been done since 1964 to simplify and better understand resolution of singularities in characteristic zero. We mention [7], [8], [9], [10], [12], [13], [14], [15], [19], [20], [17], [18], [26], [32], [44], [45], [46], [50], and [52]. The problem remains open in positive characteristic. The first proof for surfaces is due to S. Abhyankar in 1956 [1] with subsequent strengthenings by H. Hironaka [34] and J. Lipman [37] to the case of more general 2-dimensional schemes, with Lipman giving necessary and sufficient condition for a 2-dimensional scheme to admit a resolution of singularities. See also [25]. Still, Abhyankar’s proof is extremely technical and difficult and comprises a total of 508 pages ([2], [3], [4], [5], [6]). For a more recent and more palatable proof we refer the reader to [27]. It was not until much later that V. Cossart and O. Piltant settled the problem of resolution of threefolds in complete generality (their theorem holds for arbitrary quasi-excellent noetherian schemes of dimension three, including the arithmetic case) in a series of three long papers spanning the years 2008 to 2019 [21], [22] and [23].

To try to solve the problem of resolution of singularities numerous methods were introduced, in particular Zariski and Abhyankar used the local uniformization. But it does not allow at the moment to solve completely the problem.

We are interested in a stronger problem than the local uniformization: the monomialization problem. In this work we solve the monomialization problem in characteristic zero. We hope that these methods, applicable in positive characteristic, may help to attack the global problem of resolution of singularities on a different point of view.

One of the essential tools to handle the monomialization or the local uniformization is a valuation. Let us look on an example how valuations naturally fit into the problem.

Let $V$ be a singular variety and $Z$ be an irreducible closed set of $V$.

If we knew how to resolve the singularities of $V$, we would have a smooth variety $W$ and a proper birational morphism $W\to V$. In $W$, we can consider a irreducible set $Z^{\prime}$ whose image is $Z$. And so the regular local ring $\mathcal{O}_{W,Z^{\prime}}$ dominates the non regular local ring $\mathcal{O}_{V,Z}$. It means that we have an inclusion $\mathcal{O}_{V,Z}\subseteq\mathcal{O}_{W,Z^{\prime}}$ and the maximal ideal of $\mathcal{O}_{V,Z}$ is the intersection of those of $\mathcal{O}_{W,Z^{\prime}}$ with $\mathcal{O}_{V,Z}$. Up to a blow-up $Z^{\prime}$ is a hypersurface and so $\mathcal{O}_{W,Z^{\prime}}$ is dominated by a discrete valuation ring. In this case the valuation is the order of vanishing along the hypersurface.

Before stating the local uniformization Theorem, we need a classical notion that will be very important: the center of a valuation. For details, we can read ([54]) or ([47, sections 2 and 3]).

Let $K$ be a field and $\nu$ be a valuation defined over $K$. We set

[TABLE]

the valuation ring of $\nu$, and $\mathfrak{m}_{\nu}$ its maximal ideal.

We consider a subring $A$ of $K$ such that $A\subset R_{\nu}$. Then the center of $\nu$ in $A$ is the ideal $\mathfrak{p}$ of $A$ such that $\mathfrak{p}=A\cap\mathfrak{m}_{\nu}$.

Now we consider an algebraic variety $V$ over a field $k$ and $K$ its fractions field. Assume $V$ is an affine variety. Then $V=\mathrm{Spec}\left(A\right)$ where $A$ is a finite type integral $k$-algebra with $A\subseteq K$. If $A\subseteq R_{\nu}$, then the center of $\nu$ over $V$ is the point $\zeta$ of $V$ which corresponds to the prime ideal $A\cap\mathfrak{m}_{\nu}$ of $A$.

The irreducible closed sub-scheme $Z$ of $V$ defined by $A\cap\mathfrak{m}_{\nu}$ (it means the image of the morphism $\mathrm{Spec}\left(\frac{A}{A\cap\mathfrak{m_{\nu}}}\right)\to\mathrm{Spec}\left(A\right)$) has a generic point $\xi$. Equivalently $\zeta$ is the point associated to the zero ideal. We say that $Z$ is the center of $\nu$ over $V$. .

Now let us state the local uniformization Theorem. It has been proved in characteristic zero but it is always a conjecture in positive characteristic.

Theorem** (Zariski [54]).**

Let $X=\mathrm{Spec}\left(A\right)$ be an affine variety of fractions field $K$ over a field $k$. We consider $\nu$ a valuation over $K$ of valuation ring $R_{\nu}$.

Then $A$ can be embedded in a regular local sub-ring $A^{\prime}$ essentially of finite type over $k$ and dominated by $R_{\nu}$.

In this work we prove a stronger result: the simultaneous monomialization Theorem. We are going to explain what is the monomialization and what are the objects that we handle.

Let $k$ be a field of characteristic zero and $f\in k[u_{1},\dots,u_{n}]$ be a polynomial in $n$ variables, irreducible over $k$. We denote by $V\left(f\right)$ the hypersurface defined by $f$ and we assume that it has a singularity at the origin. Then we set $R:=k\left[u_{1},\dots,u_{n}\right]_{\left(u_{1},\dots,u_{n}\right)}$. This is a regular local ring that is essentially of finite type over the field $k$. The vector $u=\left(u_{1},\dots,u_{n}\right)$ is a regular system of parameters of $R$. We use the notation $\left(R,u\right)$ to express the fact that $u$ is a regular system of parameters of the regular local ring $R$.

Definition** (4.9).**

The element $f$ is monomializable if there exists a map

[TABLE]

that is a sequence of blow-ups such that the total transform of $f$ is a monomial. It means that in $R^{\prime}$, the total transform of $f$ is $v\prod\limits_{i=1}^{n}\left(u^{\prime}_{i}\right)^{\alpha_{i}}$, with $v$ a unit of $R^{\prime}$.

Now we can give a simplified version of one of the main theorems of this work.

Theorem** (7.1).**

Let $\left(R,u\right)$ be a regular local ring that is essentially of finite type over a field $k$ of characteristic zero.

Then there exists a countable sequence of blow-ups

[TABLE]

that monomializes simultaneously all the elements of $R$.

Equivalently, it means that for each element $f$ in $R$, there exists an index $i$ such that in $R_{i}$, $f$ is one monomial.

If $f$ is an irreducible polynomial of $k\left[u_{1},\dots,u_{n}\right]$, then $A:=\frac{R}{\left(f\right)}$ is a local domain. We can find a valuation $\nu$ over $\mathrm{Frac}\left(A\right)$ centered in $R$. One consequence of Theorem 7.1 is that the total transform of $f$ in one of the $R_{i}$ is $v\prod\limits_{j=1}^{n}\left(u_{j}^{\left(i\right)}\right)^{\alpha_{j}}$ . By the irreducibility of $f$ its strict transform is exactly $u_{n}^{\left(i\right)}$.

Hence there exists an embedding of $A$ into the ring $A^{\prime}=\frac{R_{i}}{\left(u_{n}^{\left(i\right)}\right)}$ which is dominated by $R_{\nu}$. So a consequence of Theorem 7.1 is the Local Uniformization Theorem as announced.

And we obtain a stronger result here: the total transform is a normal crossing divisor. We call this result the embedded local uniformization. We will give a new proof of this theorem in this work.

Let us explain why simultaneous monomialization is a stronger result than the embedded local uniformization Theorem. First we monomialize all the elements of $R$ with the same sequence of blow-ups. Secondly, this sequence is effective and at each step of the process we can express the $u^{(i+1)}$ in terms of the $u^{(i)}$. Indeed, we consider an essentially of finite type regular local ring $R$, and a valuation centered in $R$. Thanks to this valuation we construct an effective sequence of blow-ups that monomializes all the elements of $R$. One more advantage of the proof we give here is that in the essentially of finite type case, we prove the simultaneous embedded local uniformization whatever is the valuation. In particular we do not need any hypothesis on the rank of the valuation.

One of the most important ingredient in the proof of this theorem is the notion of key polynomial. We give here a new definition of key polynomial, introduced by Spivakovsky and appearing for the first time in ([28] and [41]). Let $K$ be a field, $\nu$ be a valuation over $K$ and we denote by $\partial_{b}:=\frac{1}{b!}\frac{\partial^{b}}{\partial X^{b}}$ the formal derivative of the order $b$ on $K\left[X\right]$. For every polynomial $P\in K[X]$, we set

[TABLE]

Definition** (1.7).**

Let $Q\in K[X]$ be a monic polynomial. The polynomial $Q$ is a *key polynomial for $\nu$ *if for every polynomial $P\in K[X]$:

[TABLE]

One of the interests of this new definition is the following notion:

Definition** (2.1).**

Let $Q_{1}$ and $Q_{2}$ be two key polynomials. We say that $Q_{2}$ is an immediate successor of $Q_{1}$ if $\epsilon(Q_{1})<\epsilon(Q_{2})$ and if $Q_{2}$ is of minimal degree for this property. We denote this by $Q_{1}<Q_{2}$.

We denote by $M_{Q_{1}}$ the set of immediate successors of $Q_{1}$. We assume that they all have the same degree as $Q_{1}$ and that $\epsilon\left(M_{Q_{1}}\right)$ does not have any maximal element.

Definition** (2.10).**

We assume that there exists a key polynomial $Q^{\prime}$ such that $\epsilon(Q^{\prime})>\epsilon(M_{Q_{1}})$. We call *immediate limit successor *of $Q_{1}$ every polynomial $Q_{2}$ of minimal degree satisfying $\epsilon(Q_{2})>\epsilon(M_{Q_{1}})$, and we denote this by $Q_{1}<_{\lim}Q_{2}$.

Let $Q_{1}$ and $Q_{2}$ be two key polynomials. Let us write $Q_{2}$ according to the powers of $Q_{1}$, $Q_{2}=\sum\limits_{i=0}^{s}q_{i}Q_{1}^{i}$ where the $q_{i}$ are polynomials of degree strictly less than $Q_{1}$. We call this expression the* $Q_{1}$-expansion of $Q_{2}$.*

An important result in this work, and the only one for which we need the characteristic zero hypothesis, is the following Theorem.

Theorem** (2.17).**

Let $Q_{2}$ be an immediate limit successor of $Q_{1}$. Then the terms of the $Q_{1}$-expansion of $Q_{2}$ that minimize the valuation are exactly those of degrees [math] and $1$.

Then the hypothesis of characteristic zero is necessary also for the results that follow from this theorem.

Here we give an idea of our proof of Theorem 7.1. Let us consider a regular local ring $R$ essentially of finite type over a field $k$ of characteristic zero. We fix $u=\left(u_{1},\dots,u_{n}\right)$ a regular system of parameters of $R$.

The first ingredient in the proof is the notion of non degeneration.

Definition** (3.1).**

We say that an element $f$ of $R$ is non degenerated with respect to $u$ if there exists an ideal $N$ of $R$, generated by monomials in $u$, such that $\nu\left(f\right)=\min\limits_{x\in N}\left\{\nu\left(x\right)\right\}$.

The first step is to monomialize all the elements that are non degenerated with respect to a regular system of parameters of $R$. So let $f$ be an element of $R$ that is non degenerated with respect to $u$. We construct a sequence of blow-ups

[TABLE]

such that the strict transform of $f$ in $R^{\prime}$ is a monomial in $u^{\prime}$.

There exist elements $f$ of $R$ that are not non degenerated with respect to $u$. So we wonder if we could find a sequence of blow-ups

[TABLE]

such that $f$ is non degenerated with respect to $t$. If we can, after a new sequence of blow-ups, we monomialize $f$. Doing this for all the elements of $R$ would be too complicated. So we would want to find a sequence of blow-ups $\left(R,u\right)\to\cdots\to\left(R^{\prime},u^{\prime}\right)$ such that all the elements of $R$ are non degenerated with respect to $u^{\prime}$. It is a little optimistic and we need to do something more subtle. We will find an infinite sequence of blow-ups

[TABLE]

such that for each element $f$ of $R$, there exists $i$ such that $f$ is non degenerated with respect to $u^{(i)}$.

For this, we need the second main ingredient: the key polynomials.

We construct a sequence of key polynomials $\left(Q_{i}\right)_{i}$ such that each element $f$ of $R$ is non degenerated with respect to some $Q_{i}$. It means that:

[TABLE]

We construct the sequence $\left(Q_{i}\right)_{i}$ step by step. We require the following properties for this sequence: for every index $i$, the polynomial $Q_{i+1}$ is an (eventually limit) immediate successor of $Q_{i}$. Furthermore the sequence $\left(\epsilon\left(Q_{i}\right)\right)_{i}$ is cofinal in $\epsilon\left(\Lambda\right)$ where $\Lambda$ is the set of key polynomials of the extension $k\left(u_{1},\dots,u_{n-1}\right)\left(u_{n}\right)$.

Equivalently it means:

[TABLE]

Assume now that we can construct a sequence of blow-ups

[TABLE]

such that all the $Q_{i}$ belong to a regular system of parameters. It means that

[TABLE]

where $Q_{i}^{\mathrm{strict},j}$ is the strict transform of $Q_{i}$ in $R_{j}$. Then every element $f$ of $R$ which is non degenerated with respect to $Q_{i}$ is non degenerated with respect to $u^{(j)}$. Thus it is monomializable. So the next step is to monomialize all the $Q_{i}$.

In order to do this once again we have to be subtle. The notion of key polynomial is not stable by blow-up, so we need a better notion: the notion of key element. Let $\left(Q_{i},Q_{i+1}\right)$ a couple of (eventually limit) immediate successors of our sequence. We consider $Q_{i+1}=\sum\limits_{j=0}^{s}q_{j}Q_{i}^{j}$ the $Q_{i}$-expansion of $Q_{i+1}$. Then we associate to $Q_{i+1}$ a key element $Q^{\prime}_{i+1}$ defined as follows.

Definition** (3.11).**

An element $Q^{\prime}_{i+1}=\sum\limits_{j=0}^{s}a_{j}q_{j}Q_{i}^{j}$ where the $a_{j}$ are units is called a key element associated to $Q_{i+1}$.

In fact we also have a notion of (eventually limit) immediate successors in this case.

Definition** (3.13 and 3.14).**

Let $P_{1}^{\prime}$ and $P_{2}^{\prime}$ be two key elements. We say that $P_{1}^{\prime}$ and $P_{2}^{\prime}$ are (eventually limit) immediate successors key elements if their respective associated key polynomials $P_{1}$ and $P_{2}$ are such that $P_{1}<P_{2}$ (eventually $P_{1}<_{\mathrm{lim}}P_{2}$).

After some blow ups we prove that (eventually limit) immediate successors become (eventually limit) immediate successors key elements. So we monomialize these key elements. For this we construct a sequence of blow-ups

[TABLE]

that monomializes all the key polynomials $Q_{i}$. More precisely, for every index $i$ there exists an index $s_{i}$ such that in $R_{s_{i}}$, $Q_{i}$ is a monomial in $u^{\left(s_{i}\right)}$ up to a unit of $R_{s_{i}}$.

So in the case of essentially of finite type regular local rings, no matter the rank of the valuation is, we prove the embedded local uniformization Theorem. And we do this using only a sequence of blow-ups for all the elements of the ring, and in an effective way. It means that every blow-up is effective and we know how to express all the systems of coordinates.

Then we want to prove the same kind of result over more general rings, even if it means adding conditions on the valuation. We work with quasi excellent rings. Indeed, Grothendieck and Nagata showed that there is no resolution of singularities for rings that are not quasi excellent.

The second main result of this paper can be express in the following simplified form.

Theorem** (12.3).**

Let $R$ be a noetherian quasi excellent complete regular local ring and $\nu$ be a valuation centered in $R$.

Assume that $\nu$ is of rank $1$, or of rank $2$ but composed with a discrete valuation, and that $\mathrm{car}\left(k_{\nu}\right)=0$.

There exists a countable sequence of blow-ups

[TABLE]

that monomializes all the element of $R$.

So let $R$ be a quasi excellent local domain. This time $R$ is not assume to be of finite type, so we cannot repeat what we did before. We need to introduce one more ingredient: the implicit prime ideal.

Let $\nu$ be a valuation of the fractions field of $R$ centered in $R$. We call implicit prime ideal of $R$ associated to $\nu$ the ideal of the completion $\widehat{R}$ of $R$ defined by:

[TABLE]

where $P_{\beta}:=\left\{f\in R\text{ such that }\nu\left(f\right)\geq\beta\right\}$.

One can show that in this case desingularizing $R$ means desingularizing $\widehat{R}$. In the last part of this work we also prove that to desingularize $\widehat{R}$, we only need to desingularize $\widehat{R}_{H}$ and (up to one more sequence of blow-ups) $\frac{\widehat{R}}{H}$. We prove that the implicit prime ideal satisfies the property that $\widehat{R}_{H}$ is regular. So we only have to desingularize $\frac{\widehat{R}}{H}$ and this is done by Theorem 11.2.

Acknowledgments.

The author is really grateful to her PHD advisor Mark Spivakovsky for all the helpful discussions.

Part II Key polynomials.

The notion of key polynomials was first introduced by Saunders Mac Lane in 1936, in the case of discrete valuations of rank 1. The first motivation to introduce this notion was to describe all the extensions of a valuation to a field extension. Let $K\to L$ be an extension of field and $\nu$ a valuation on $K$. We consider a valuation $\mu$ that extends $\nu$ to $L$. In the case where $\nu$ is of rank $1$ and where $L$ is a simple algebraic extension of $K$, Mac Lane created the notion of key polynomial for $\mu$. He also created the notion of augmented valuations. Given a valuation $\mu$ and $Q$ a key polynomial of Mac Lane, we write $f=\sum\limits_{i=0}^{r}f_{j}Q^{j}$ the $Q$-expansion of an element $f\in K\left[X\right]$. An augmented valuation $\mu^{\prime}$ of $\mu$ is the one defined by $\mu^{\prime}\left(f\right)=\min\limits_{0\leq j\leq r}\left\{\mu\left(f_{j}\right)+j\delta\right\}$ where $\delta>\mu\left(Q\right)$. He proved that $\mu$ is the limit of a family of augmented valuations over the ring $K[x]$. Michel Vaquié extended this definition to arbitrary valued field $K$, without assuming that $\nu$ is discrete. The most important difference between these notions is the fact that those of Vaquié involves limit key polynomials while those of Mac Lane not.

More recently, the notion of key polynomials has been used by Spivakovsky to study the local uniformization problem, and to do this he created a new notion of key polynomials. It is the one we use here.

1. Key polynomials of Spivakovsky et al.

For some results of this part, we refer the reader to [28], but we recall the definitions and properties used in this work to have a selfcontained manuscript.

First, recall the definition of a valuation.

Definition 1.1**.**

Let $R$ be a commutative domain with a unit element, $K$ be a commutative field and $\Gamma$ be a totally ordered abelian group. We set $\Gamma_{\infty}:=\Gamma\cup\left\{+\infty\right\}$.

A valuation of $R$ is a map

[TABLE]

such that:

1. (1)

$\forall x\in R$, $\nu(x)=+\infty\Leftrightarrow x=0$, 2. (2)

$\forall\left(x,y\right)\in R^{2}$, $\nu\left(xy\right)=\nu\left(x\right)+\nu\left(y\right)$, 3. (3)

$\forall\left(x,y\right)\in R^{2}$, $\nu\left(x+y\right)\geq\min\left\{\nu\left(x\right),\nu\left(y\right)\right\}$.

Let us give three examples of valuations.

Example 1.2**.**

The map $\nu_{1}:\mathbb{C}\left[x\right]\to\mathbb{Z}\cup\left\{+\infty\right\}$ which sends a polynomial $P=\sum\limits_{i=0}^{d}p_{i}x^{i}$ to $\min\left\{i\text{ such that }p_{i}\neq 0\right\}$ is a valuation.

Example 1.3**.**

We want to define a valuation $\nu_{2}$ on $\mathbb{C}\left(x,y,z\right)$. The value of a quotient $\frac{P}{Q}$ is $\nu_{2}\left(P\right)-\nu_{2}\left(Q\right)$.

And we define the value of a polynomial $P=\sum\limits_{i}p_{i}x^{i_{1}}y^{i_{2}}z^{i_{3}}$ as the minimal of the values of $p_{i}x^{i_{1}}y^{i_{2}}z^{i_{3}}$.

Then we only have to define the values of the generators $x$, $y$ and $z$.

Hence the map $\nu_{2}:\mathbb{C}\left(x,y,z\right)\to\mathbb{R}_{\infty}$ which sends $x$ to $1$, $y$ to $2\pi$ and $z$ to $1+\pi$ is a valuation.

Example 1.4**.**

Let us set $Q=z^{2}-x^{2}y$. Every polynomial $P\in\mathbb{C}\left[x,y,z\right]$ can be written according to the powers of $Q$. We write $P=\sum\limits_{i}p_{i}Q^{i}$ with the $p_{i}\in\mathbb{C}\left[x,y\right]\left[z\right]$ of degree in $z$ strictly less than $\deg_{z}\left(Q\right)=2$. Assume that the first non zero $p_{i}$ is $p_{n}$.

Then the map $\nu_{3}:\mathbb{C}\left(x,y,z\right)\to\left(\mathbb{R}^{2},\mathrm{lex}\right)$ which sends $P$ to $\left(n,\nu_{2}\left(p_{n}\right)\right)$ defines a valuation, with $\nu_{2}$ the valuation defined in Example 1.3.

Let $K$ be a field equipped with a valuation $\nu$ and consider a simple transcendental extension

[TABLE]

with a valuation $\nu$ that extends $\mu$ to $K(X)$. We still denote by $\nu$ the restriction of $\nu$ to $K[X]$.

For every non zero integer $b,$ we set $\partial_{b}:=\frac{1}{b!}\frac{\partial^{b}}{\partial X^{b}}$. This is called the formal derivative of order $b$.

For every polynomial $P\in K[X]$, we set

[TABLE]

Remark 1.5*.*

Most of the time we will note $\epsilon\left(P\right):=\epsilon_{\nu}\left(P\right)$.

Example 1.6**.**

We consider $\mathbb{C}\left(x,y\right)\left[z\right]$ and the valuation $\nu:=\nu_{3}$ defined in Example 1.4.

We have $\nu\left(z\right)=\left(0,1+\pi\right)$ and $\nu\left(\partial z\right)=\nu\left(1\right)=\left(0,0\right)$. So

[TABLE]

Also we have $\nu\left(x\right)=\left(0,1\right)$ and $\nu\left(\partial x\right)=\nu\left(0\right)=\left(+\infty,+\infty\right)$ so $\epsilon\left(x\right)=\left(-\infty,-\infty\right)$. Furthermore $\epsilon\left(y\right)=\left(-\infty,-\infty\right)$.

Finally, let us compute $\epsilon\left(Q\right)$. Recall that $Q=z^{2}-x^{2}y$. We have $\nu\left(Q\right)=\left(1,0\right)$, $\nu\left(\partial Q\right)=\nu\left(2z\right)=\left(0,1+\pi\right)$ and $\nu\left(\partial_{2}Q\right)=\nu\left(2\right)=\left(0,0\right)$.

So $\epsilon\left(Q\right)=\max\left\{\frac{\nu\left(Q\right)-\nu\left(\partial Q\right)}{1},\frac{\nu\left(Q\right)-\nu\left(\partial_{2}Q\right)}{2}\right\}=\max\left\{\frac{\left(1,0\right)-\left(0,1+\pi\right)}{1},\frac{\left(1,0\right)-\left(0,0\right)}{2}\right\}=\left(1,-1-\pi\right)$.

Definition 1.7**.**

Let $Q\in K[X]$ be a monic polynomial. We say that $Q$ is a key polynomial for $\nu$ if for every polynomial $P\in K[X]$, we have:

[TABLE]

Example 1.8**.**

We consider the same example as in example 1.6.

Let us show that $z$ is a key polynomial. We do a proof by contrapositive. Let $P$ be a polynomial of degree in $z$ strictly less than $\deg_{z}\left(z\right)=1$. So $P$ does not depend on $z$. Then we saw that $\epsilon\left(P\right)=\left(-\infty,-\infty\right)$. So $\epsilon\left(P\right)<\epsilon\left(z\right)$ and $z$ is a key polynomial.

Now, let us show that $Q=z^{2}-x^{2}z$ is a key polynomial. So we consider a polynomial $P$ such that $\epsilon\left(P\right)\geq\epsilon\left(Q\right)=\left(1,-1-\pi\right)$.

Then $\epsilon\left(P\right)=\left(n,\ast\right)$ where $n\geq 1$ and $\ast$ is a scalar. So $\nu\left(P\right)=\left(m,\ast\right)$ where $m\geq 1$. Hence $Q^{m}\mid P$ and so $\deg_{z}\left(P\right)\geq\deg_{z}\left(Q\right)$. We proved that $Q$ is a key polynomial.

We have two key polynomials $z$ and $Q$ and we have $\epsilon\left(z\right)<\epsilon\left(Q\right)$. One can show that $Q$ is of minimal degree for this property. In this situation we will say that $Q$ is an immediate successor of $z$.

For every polynomial $P\in K[X]$, we set

[TABLE]

where

[TABLE]

Again, if there is no confusion, we will omit the index $\nu$.

Let $P$ and $Q$ be two polynomials such that $Q$ is monic. Then $P$ can be written $\sum\limits_{j=1}^{n}p_{j}Q^{j}$ with $p_{j}$ polynomials of degree strictly less than the degree of $Q$. This expression is unique and is called the $Q$-expansion of $P$.

Definition 1.9**.**

Let $\left(P,Q\right)\in K[X]^{2}$ such that $Q$ is monic, and we consider $P=\sum\limits_{j=1}^{n}p_{j}Q^{j}$ the $Q$-expansion of the polynomial $P$. Then we set $\nu_{Q}\left(P\right):=\min\limits_{0\leq j\leq n}\nu\left(p_{j}Q^{j}\right)$. The map $\nu_{Q}$ is called the $Q$-truncation of $\nu$.

Also we set

[TABLE]

and

[TABLE]

Now, we set

[TABLE]

Remark 1.10*.*

In the general case, $\nu_{Q}$ is not a valuation. But if $Q$ is a key polynomial, we are going to show that $\nu_{Q}$ is a valuation.

In order to do that, we need the next result, which will also be needed for a proof of the fundamental theorem 2.17.

Lemma 1.11**.**

Let $t\in\mathbb{N}_{>1}$ and $Q$ be a key polynomial. We consider $P_{1},\dots,P_{t}$ some polynomials of $K[X]$ all of degree strictly less than $\deg\left(Q\right)$ and we set $\prod\limits_{i=1}^{t}P_{i}:=qQ+r$ the Euclidean division of $\prod\limits_{i=1}^{t}P_{i}$ by $Q$ in $K[X]$. Then:

[TABLE]

Proof.

We use induction on $t$.

Base of the induction: $t=2$. So we want to show that $\nu\left(P_{1}P_{2}\right)<\nu\left(qQ\right)$.

Indeed, if $\nu\left(P_{1}P_{2}\right)<\nu\left(qQ\right)$, then

[TABLE]

and we have the result.

Assume, aiming for contradiction, that $\nu\left(P_{1}P_{2}\right)\geq\nu\left(qQ\right)$ and so $\nu\left(r\right)\geq\nu\left(qQ\right)$. Since $Q$ is a key polynomial, every polynomial $P$ of degree strictly less than $\deg\left(Q\right)$ satisfies $\epsilon\left(P\right)<\epsilon\left(Q\right)$. In particular, for every non-zero integer $j$, we have $\nu\left(P\right)-\nu\left(\partial_{j}P\right)<j\epsilon\left(Q\right)$. So it is the case for $P_{1}$, $P_{2}$ and $r$. Since $P_{1}$ and $P_{2}$ are of degree strictly less than $\deg\left(Q\right)$, we have

[TABLE]

However, $\deg_{X}\left(P_{1}P_{2}\right)=\deg_{X}\left(qQ\right)=\deg_{X}\left(q\right)+\deg_{X}\left(Q\right)$. So $q$ is of degree strictly less than $\deg\left(Q\right)$ too, and then $q$ satisfies, for every non-zero integer $j$: $\nu\left(q\right)-\nu\left(\partial_{j}q\right)<j\epsilon\left(Q\right)$. We are going to compute $\nu\left(\partial_{b(Q)}\left(qQ\right)\right)$ in two different ways to get a contradiction.

First,

[TABLE]

Look at the first term of the sum: $q\partial_{b(Q)}\left(Q\right)$, and compute its value $\nu\left(q\partial_{b(Q)}\left(Q\right)\right)$. We are going to show that its value is the smallest of the sum.

We have

[TABLE]

by definition of $b\left(Q\right)$. But we know that for every non-zero integer $j$, we have $\nu\left(q\right)<j\epsilon\left(Q\right)+\nu\left(\partial_{j}q\right)$, so

[TABLE]

Then $q\partial_{b(Q)}\left(Q\right)$ is the term of smallest value in the sum. In particular,

[TABLE]

Now we compute this value in a different way. We have:

[TABLE]

But also:

[TABLE]

If $j\neq 0$, we have $\nu\left(P_{1}\right)<j\epsilon\left(Q\right)+\nu\left(\partial_{j}\left(P_{1}\right)\right)$ and so

[TABLE]

since $\deg_{X}\left(P_{1}\right)<\deg_{X}\left(Q\right)$. If $0\leq j<b\left(Q\right)$, we also have

[TABLE]

So if $0<j<b\left(Q\right)$, we have

[TABLE]

This inequality stays true if $j=0$ and $j=b\left(Q\right)$, so:

[TABLE]

By hypothesis, $\nu\left(P_{1}P_{2}\right)\geq\nu\left(qQ\right)$, so

[TABLE]

But since $r$ is of degree strictly less than $\deg\left(Q\right)$, we know that $\nu\left(\partial_{b\left(Q\right)}\left(r\right)\right)>\nu\left(r\right)-b\left(Q\right)\epsilon\left(Q\right)$, and by hypothesis $\nu\left(r\right)\geq\nu\left(qQ\right)$. Then $\nu\left(\partial_{b\left(Q\right)}\left(r\right)\right)>\nu\left(qQ\right)-b\left(Q\right)\epsilon\left(Q\right)$.

So

[TABLE]

which contradicts (1.1). So we do have $\nu\left(r\right)=\nu\left(P_{1}P_{2}\right)<\nu\left(qQ\right)$, and this completes the proof of the base of the induction.

We now assume the result true for $t-1\geq 2$ and we are going to show it for $t$. We set $P:=\prod\limits_{i=1}^{t-1}P_{i}$.

Let

[TABLE]

be the Euclidean division of $P$ by $Q$ and

[TABLE]

be that of $r_{1}P_{t}$ by $Q$. Since $PP_{t}=qQ+r$, we have $r=r_{2}$ and $q=q_{1}P_{t}+q_{2}$.

By the induction hypothesis, $\nu\left(r_{1}\right)=\nu\left(P\right)<\nu\left(q_{1}Q\right)$. In particular,

[TABLE]

Since the polynomials $r_{1}$ and $P_{t}$ are both of degree strictly less than $\deg\left(Q\right)$, we can apply the base of the induction and so

[TABLE]

So $\nu\left(r\right)=\nu\left(r_{2}\right)=\nu\left(r_{1}P_{t}\right)=\nu\left(\prod\limits_{i=1}^{t}P_{i}\right)$ and furthermore this value is strictly less than both $\nu\left(q_{1}P_{t}Q\right)$ and than $\nu\left(q_{2}Q\right)$. So it is strictly less than the minimum, which is less then or equal to $\nu\left(q_{1}P_{t}Q+q_{2}Q\right)$ by definition of a valuation. So

[TABLE]

which completes the proof. ∎

Theorem 1.12**.**

Let $Q$ be a key polynomial. The map $\nu_{Q}$ is a valuation.

Proof.

The only thing we have to prove is that for every $\left(P_{1},P_{2}\right)\in K[X]^{2}$, we have

[TABLE]

First case: $P_{1}$ and $P_{2}$ are both of degree strictly less than $\deg\left(Q\right)$. Then $\nu_{Q}\left(P_{1}\right)=\nu\left(P_{1}\right)$ and $\nu_{Q}\left(P_{2}\right)=\nu\left(P_{2}\right)$. Since $\nu$ is a valuation, we have $\nu\left(P_{1}P_{2}\right)=\nu\left(P_{1}\right)+\nu\left(P_{2}\right)$.

Then, $\nu\left(P_{1}P_{2}\right)=\nu_{Q}\left(P_{1}\right)+\nu_{Q}\left(P_{2}\right)$. Since $P_{1}$ and $P_{2}$ are both of degree strictly less than $\deg\left(Q\right)$, by previous Lemma, we have $\nu_{Q}\left(P_{1}P_{2}\right)=\nu\left(P_{1}P_{2}\right)$ and we are done.

Second case: $P_{1}=p_{i}^{(1)}Q^{i}$ and $P_{2}=p_{j}^{(2)}Q^{j}$, with $p_{i}^{(1)}$ and $p_{j}^{(2)}$ both of degree strictly less than $\deg\left(Q\right)$.

Let $p_{i}^{(1)}p_{j}^{(2)}=qQ+r$ be the Euclidean division of $p_{i}^{(1)}p_{j}^{(2)}$ by $Q$. Since $\deg_{X}\left(p_{i}^{(1)}p_{j}^{(2)}\right)<2\deg_{X}\left(Q\right)$, we know that $\deg_{X}\left(q\right)<\deg_{X}\left(Q\right)$, and by definition of the Euclidean division, we have $\deg_{X}\left(r\right)<\deg_{X}\left(Q\right)$. So $P_{1}P_{2}=qQ^{i+j+1}+rQ^{i+j}$ is the $Q$-expansion of $P_{1}P_{2}$.

We are going to prove that in this case we still have

[TABLE]

and since $\nu$ is a valuation, we will have the result. We have:

[TABLE]

However, we can apply thee previous Lemma to the product

[TABLE]

and conclude that $\nu\left(r\right)=\nu\left(p_{i}^{(1)}p_{j}^{(2)}\right)<\nu\left(qQ\right)$.

Then

[TABLE]

and we have the result.

Last case: general case. Since we only look at the terms of smallest value, we can replace $P_{1}$ by

[TABLE]

and $P_{2}$ by

[TABLE]

We know that

[TABLE]

and

[TABLE]

So

[TABLE]

However

[TABLE]

and

[TABLE]

So $\nu_{Q}\left(P_{1}P_{2}\right)\geq\nu_{Q}\left(P_{1}\right)+\nu_{Q}\left(P_{2}\right)$, and we only have to show that it is an equality. In order to do that, it is enough to find a term in the $Q$-expansion of $P_{1}P_{2}$ whose value is exactly $\nu_{Q}\left(P_{1}\right)+\nu_{Q}\left(P_{2}\right)$. Let us consider the term of smallest value in each $Q$-expansion, so let us consider $p_{n_{1}}^{(1)}Q^{n_{1}}$ and $p_{m_{2}}^{(2)}Q^{m_{2}}$, where $n_{1}=\min S_{Q}\left(P_{1}\right)$ and $m_{2}=\min S_{Q}\left(P_{2}\right)$.

Let $p_{n_{1}}^{(1)}p_{m_{2}}^{(2)}=qQ+r$ be the Euclidean division of $p_{n_{1}}^{(1)}p_{m_{2}}^{(2)}$ by $Q$, which is its $Q$-expansion too.

By Lemma 1.11, we have $\nu(r)=\nu\left(p_{n_{1}}^{(1)}p_{m_{2}}^{(2)}\right)$. In fact, in the $Q$-expansion of $P_{1}P_{2}$, there is the term $rQ^{n_{1}+m_{2}}$, and we have:

[TABLE]

This completes the proof. ∎

Remark 1.13*.*

For every polynomial $P\in K[X]$, we have

[TABLE]

It will be very important to be able to determine when this inequality is an equality.

A key polynomial $P$ which satisfies the strict inequality and which is of minimal degree for this property will be called an immediate successor of $Q$ (Definition 2.1). We will study these polynomials in more details in this work. First, let us concentrate on the equality case.

Definition 1.14**.**

Let $Q$ be a key polynomial and $P$ be a polynomial such that $\nu_{Q}\left(P\right)=\nu\left(P\right)$. We say that $P$ is non-degenerate with respect to $Q$.

Another very important thing is to be able to compare the $\epsilon$ of key polynomials. Indeed, if I have two key polynomials $Q_{1}$ and $Q_{2}$, do I have $\epsilon\left(Q_{1}\right)<\epsilon\left(Q_{2}\right)$, or do I have $\epsilon\left(Q_{1}\right)=\epsilon\left(Q_{2}\right)$ ? Being able to answer will be crucial. The next four results can be found in [28] but we recall them for more clarity.

Lemma 1.15**.**

For every polynomial $P\in K[X]$ and every stricly positive integer $d$, we have :

[TABLE]

Proof.

We consider the $Q$-expansion $P=\sum\limits_{i=0}^{n}p_{i}Q^{i}$ of $P$.

Assume we have the result for $p_{i}Q^{i}$. It means that

[TABLE]

for every index $i$. Then:

[TABLE]

and the proof is finished.

So we just have to prove the result for $P=p_{i}Q^{i}$.

First, we know that $\nu_{Q}\left(\partial_{d}Q\right)\geq\nu_{Q}\left(Q\right)-d\epsilon\left(Q\right)$. Now we will prove that we have the result for $P=p_{i}$. Then we will conclude by showing that if we have the result for two polynomials, we have the result for the product.

So let us prove the result for $P=p_{i}$.

Since $\deg_{X}\left(p_{i}\right)<\deg_{X}\left(Q\right)$ and since $Q$ is a key polynomial, we have $\epsilon\left(p_{i}\right)<\epsilon\left(Q\right)$. So, for every strictly positive integer $d$, we have:

[TABLE]

Now, it just remains to prove that if we have the result for two polynomials $P$ and $S$, then we have it for $PS$. Assume the result proven for $P$ and $S$. Then:

[TABLE]

This completes the proof. ∎

Proposition 1.16**.**

Let $Q$ be a key polynomial and $P\in K[X]$ a polynomial such that $S_{Q}\left(P\right)\neq\left\{0\right\}$.

Then there exists a strictly positive integer $b$ such that

[TABLE]

Proof.

First, by Lemma 1.15, we can replace $P$ by $\widetilde{P}_{\nu,Q}=\sum\limits_{i\in S_{Q}\left(P\right)}p_{i}Q^{i}$.

We want to show the existence of a strictly positive integer $b$ such that $\nu_{Q}\left(P\right)-\nu_{Q}\left(\partial_{b}P\right)=b\epsilon\left(Q\right)$.

Since $S_{Q}\left(P\right)\neq\left\{0\right\}$, we can consider the smallest non-zero element $l$ of $S_{Q}\left(P\right)$. We write $l=p^{e}u$, with $p\nmid u$.

We are going to prove that we have the desired equality for the integer $b:=p^{e}b\left(Q\right)>0$. To do this, we need to compute $\partial_{b}\left(P\right)$, it is the objective of the following technical lemma.

Lemma 1.17**.**

We have $\partial_{b}\left(P\right)=urQ^{l-p^{e}}+Q^{l-p^{e}+1}R+S$, where:

1. (1)

The polynomial $r$ is the remainder of the Euclidean division of $p_{l}\left(\partial_{b\left(Q\right)}Q\right)^{p^{e}}$ by $Q$, 2. (2)

The polynomials $R$ and $S$ satisfy

[TABLE]

Proof.

First let us show that the Lemma is true for $P=p_{l}Q^{l}$ and that for every $j\in S_{Q}\left(P\right)\setminus\left\{l\right\}$, we have

[TABLE]

where $R_{j}$ and $S_{j}$ are two polynomials, and where $\nu_{Q}\left(S_{j}\right)>\nu_{Q}\left(P\right)-b\epsilon\left(Q\right)$.

So we consider $j\in S_{Q}\left(P\right)$. We set

[TABLE]

The generalized Leibniz rule tells us that:

[TABLE]

where

[TABLE]

with $C\left(B_{s}\right)$ some elements of $K$ whose exact value can be found in [35]. We set

[TABLE]

Recall that $I\left(Q\right)=\left\{d\in\mathbb{N}^{\ast}\text{ such that }\frac{\nu\left(Q\right)-\nu\left(\partial_{d}Q\right)}{d}=\epsilon\left(Q\right)\right\}$. We set

[TABLE]

[TABLE]

and finally we set

[TABLE]

If $j=l$, the term $T\left(\alpha\right)$ appears $\binom{l}{p^{e}}=u$ times in $\partial_{b}\left(p_{l}Q^{l}\right)$. Equivalently, $C\left(\alpha\right)=u$ and so

[TABLE]

where $qQ+r$ is the Euclidean division of $p_{l}\left(\partial_{b\left(Q\right)}Q\right)^{p^{e}}$ by $Q$.

In other words

[TABLE]

So if $j\neq l$, then $\partial_{b}\left(p_{j}Q^{j}\right)=Q^{l-p^{e}+1}R_{j}+S_{j}$. It remains to prove that $\nu_{Q}\left(S_{j}\right)>\nu_{Q}\left(p_{j}Q^{j}\right)-b\epsilon\left(Q\right)$.

But:

[TABLE]

Since $B_{s}\in N_{j}$, we have two options. The first is $b_{0}=0$ and $\left\{b_{1},\dots,b_{s}\right\}\nsubseteq I\left(Q\right)$. In other words for every $i\in\left\{1,\dots,s\right\}$ we have $\nu\left(\partial_{b_{i}}\left(Q\right)\right)\geq\nu\left(Q\right)-b_{i}\epsilon\left(Q\right)$. And then the inequality is strict for at least one index $i\in\left\{1,\dots,s\right\}$. The second option is $b_{0}>0$ and then

[TABLE]

because $\deg_{X}\left(p_{j}\right)<\deg_{X}\left(Q\right)$ and $Q$ is a key polynomial. Equivalently,

[TABLE]

So if $b_{0}=0$ and $\left\{b_{1},\dots,b_{s}\right\}\nsubseteq I\left(Q\right)$, we have

[TABLE]

And if $b_{0}>0$, then

[TABLE]

So:

[TABLE]

as desired.

If $j=l$, then

[TABLE]

hand using the same argument as before, $\nu_{Q}\left(S_{l}\right)>\nu_{Q}\left(P\right)-b\epsilon\left(Q\right)$.

It remains to prove the general case. We have:

[TABLE]

Then:

[TABLE]

where

[TABLE]

and

[TABLE]

We have

[TABLE]

This completes the proof of the Lemma. ∎

Recall that we want to prove that

[TABLE]

We just saw that the $Q$-expansion of $\partial_{b}P$ contains the term $urQ^{l-p^{e}}$, some terms divisible by $Q^{l-p^{e}+1}$ and others of value strictly higher than $\nu_{Q}\left(P\right)-b\epsilon\left(Q\right)$. It is sufficient now to show that

[TABLE]

and that

[TABLE]

Let us compute $\nu_{Q}\left(urQ^{l-p^{e}}\right)$.

Recall that $p_{l}\left(\partial_{b\left(Q\right)}Q\right)^{p^{e}}=qQ+r$. By Lemma 1.11, we have $\nu\left(r\right)=\nu\left(p_{l}\left(\partial_{b\left(Q\right)}Q\right)^{p^{e}}\right)$.

So:

[TABLE]

The result now follows from Lemma 1.15. ∎

Remark 1.18*.*

One can show that the implication of the proposition is, in fact, an equivalence.

Proposition 1.19**.**

Let $Q$ be a key polynomial and $P$ a polynomial such that there exists a strictly positive integer $b$ such that

[TABLE]

and

[TABLE]

Then $\epsilon\left(P\right)\geq\epsilon\left(Q\right)$.

If, in addition, $\nu\left(P\right)>\nu_{Q}\left(P\right)$, then $\epsilon\left(P\right)>\epsilon\left(Q\right)$.

Proof.

We have

[TABLE]

We know that for every polynomial $P$, we have $\nu\left(P\right)\geq\nu_{Q}\left(P\right)$ , so $\epsilon\left(P\right)\geq\epsilon\left(Q\right)$. And if $\nu\left(P\right)>\nu_{Q}\left(P\right)$, we have the strict inequality $\epsilon\left(P\right)>\epsilon\left(Q\right)$. ∎

Proposition 1.20**.**

Let $Q_{1}$ and $Q_{2}$ be two key polynomials such that

[TABLE]

and let $P\in K[X]$ be a polynomial.

Then $\nu_{Q_{1}}\left(P\right)\leq\nu_{Q_{2}}\left(P\right)$.

Furthermore, if $\nu_{Q_{1}}\left(P\right)=\nu\left(P\right)$, then $\nu_{Q_{2}}\left(P\right)=\nu\left(P\right)$.

Proof.

First, we show that $\nu_{Q_{2}}\left(Q_{1}\right)=\nu\left(Q_{1}\right)$. If $\deg_{X}\left(Q_{1}\right)<\deg_{X}\left(Q_{2}\right)$, we do have this equality. Otherwise we have $\deg_{X}\left(Q_{1}\right)=\deg_{X}\left(Q_{2}\right)$ since $\epsilon\left(Q_{1}\right)\leq\epsilon\left(Q_{2}\right)$ and since $Q_{1}$ is a key polynomial.

Assume, aiming for contradiction, that $\nu_{Q_{2}}\left(Q_{1}\right)<\nu\left(Q_{1}\right)$.

So $S_{Q_{2}}\left(Q_{1}\right)\neq\left\{0\right\}$ and by Proposition 1.16, there exists a non-zero integer $b$ such that $\nu_{Q_{2}}\left(Q_{1}\right)-\nu_{Q_{2}}\left(\partial_{b}Q_{1}\right)=b\epsilon\left(Q_{2}\right)$. However $\deg_{X}\left(\partial_{b}Q_{1}\right)<\deg_{X}\left(Q_{2}\right)$, so $\nu_{Q_{2}}\left(\partial_{b}Q_{1}\right)=\nu\left(\partial_{b}Q_{1}\right)$ and by Proposition 1.19, we have $\epsilon\left(Q_{1}\right)>\epsilon\left(Q_{2}\right)$. This is a contradiction. So we do have $\nu_{Q_{2}}\left(Q_{1}\right)=\nu\left(Q_{1}\right)$.

Let $P=\sum\limits_{i=0}^{n}p_{i}Q_{1}^{i}$ be the $Q_{1}$-expansion of $P$.

For every $i\in\left\{0,\dots,n\right\}$, we have:

[TABLE]

But $\deg_{X}\left(p_{i}\right)<\deg_{X}\left(Q_{1}\right)\leq\deg_{X}\left(Q_{2}\right)$, so $\nu_{Q_{2}}\left(p_{i}\right)=\nu\left(p_{i}\right)$ and $\nu_{Q_{2}}\left(p_{i}Q_{1}^{i}\right)=\nu\left(p_{i}Q_{1}^{i}\right)$.

Then

[TABLE]

Assume that, in addition, $\nu_{Q_{1}}\left(P\right)=\nu\left(P\right)$. Then $\nu\left(P\right)\leq\nu_{Q_{2}}\left(P\right)$. By definition of $\nu_{Q_{2}}$, we have $\nu_{Q_{2}}\left(P\right)\leq\nu\left(P\right)$, and hence the equality. ∎

Proposition 1.21**.**

Let $P_{1},\dots,P_{n}\in K[X]$ be polynomials and set $d:=\max\limits_{1\leq i\leq n}\left\{\deg_{X}\left(P_{i}\right)\right\}$.

There exists a key polynomial $Q$ of degree less than or equal to $d$ such that all the $P_{i}$ are non-degenerate with respect to $Q$. In other words, there exists a key polynomial $Q$ such that for every $i$, we have $\nu_{Q}\left(P_{i}\right)=\nu\left(P_{i}\right)$.

Proof.

Assume the result for only one polynomial and let $n>1$.

So we have $Q_{1},\dots,Q_{n}$ some key polynomials of degrees less than or equal to $d$ such that for every $i\in\left\{1,\dots,n\right\}$, the polynomial $P_{i}$ is non-degenerate with respect to $Q_{i}$. In other words $\nu_{Q_{i}}\left(P_{i}\right)=\nu\left(P_{i}\right)$.

We can assume

[TABLE]

By Proposition 1.20, for every $i\in\left\{1,\dots,n\right\}$, we have $\nu_{Q_{i}}\left(P_{i}\right)=\nu\left(P_{i}\right)=\nu_{Q_{n}}\left(P_{i}\right)$. So all the $P_{i}$ are non-degenerate with respect to $Q_{n}$. This completes the proof.

It remains to show the result for $n=1$. We give a proof by contradiction. Assume the existence of a polynomial $P$ such that for every key polynomial $Q$ of degree less than or equal to $d$, we have $\nu_{Q}\left(P\right)<\nu\left(P\right)$. We choose $P$ of minimal degree for this property.

Let us show that there exists a key polynomial $Q$, of degree less than or equal to $d=\deg_{X}\left(P\right)$, such that for every $b>0$, we have $\nu_{Q}\left(\partial_{b}P\right)=\nu\left(\partial_{b}P\right)$.

First, for every $b>d$, we have $\partial_{b}P=0$. Then, by minimality of the degree of $P$, for every $b\in\left\{1,\dots,d\right\}$, there exists a key polynomial $Q_{b}$ such that $\nu_{Q_{b}}\left(\partial_{b}P\right)=\nu\left(\partial_{b}P\right)$.

Take an element $Q\in\left\{Q_{1},\dots,Q_{d}\right\}$ such that $\epsilon\left(Q\right)=\max\limits_{1\leq b\leq d}\left\{\epsilon\left(Q_{b}\right)\right\}$. By Proposition 1.20, we have $\nu_{Q}\left(\partial_{b}P\right)=\nu\left(\partial_{b}P\right)$, for every $b>0$.

So we have $\nu_{Q}\left(P\right)<\nu\left(P\right)$. In particular, $S_{Q}\left(P\right)\neq\left\{0\right\}$, and $\nu_{Q}\left(\partial_{b}P\right)=\nu\left(\partial_{b}P\right)$ for every $b>0$. By Proposition 1.16 and Corollary 1.19, we conclude that $\epsilon\left(P\right)>\epsilon\left(Q\right)$.

Let us show that this last inequality is true for every key polynomial of degree less than or equal than $\deg(P)$. Let $Q_{0}$ be such a key polynomial.

First case: $\epsilon\left(Q_{0}\right)\leq\epsilon\left(Q\right)$. Then $\epsilon\left(Q_{0}\right)<\epsilon\left(P\right)$ since $\epsilon\left(Q\right)<\epsilon\left(P\right)$.

Last case: $\epsilon\left(Q_{0}\right)>\epsilon\left(Q\right)$. By Proposition 1.20, we have $\nu\left(\partial_{b}P\right)=\nu_{Q}\left(\partial_{b}P\right)=\nu_{Q_{0}}\left(\partial_{b}P\right)$ for every $b>0$. By hypothesis we know that $\nu_{Q_{0}}\left(P\right)<\nu\left(P\right)$. So by Proposition 1.16 and Corollary 1.19, we have $\epsilon\left(P\right)>\epsilon\left(Q_{0}\right)$ as desired.

So we know that for every key polynomial of degree less than or equan than those of $P$, we have $\epsilon\left(P\right)>\epsilon\left(Q\right)$. But by definition of key polynomials, there exists a key polynomial $\widetilde{Q}$ of degree less than or equal than those of $P$ and such that $\epsilon\left(P\right)\leq\epsilon\left(\widetilde{Q}\right)$. Contradiction. This completes the proof. ∎

2. Immediate successors.

Definition 2.1**.**

Let $Q_{1}$ and $Q_{2}$ be two key polynomials. We say that $Q_{2}$ is an immediate successor of $Q_{1}$ and we write $Q_{1}<Q_{2}$ if $\epsilon(Q_{1})<\epsilon(Q_{2})$ and if $Q_{2}$ is of minimal degree for this property.

Remark 2.2*.*

We keep the hypotheses of Example 1.8. Then we have $z<z^{2}-x^{2}y$.

Definition 2.3**.**

It will be useful to have simpler ways to check if a key polynomial is an immediate successor of another key polynomial. This is why we give these two results.

Proposition 2.4**.**

Let $Q_{1}$ and $Q_{2}$ be two key polynomials. The following are equivalent.

1. (1)

The polynomials $Q_{1}$ and $Q_{2}$ satisfy $Q_{1}<Q_{2}$. 2. (2)

We have $\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})$ and $Q_{2}$ is of minimal degree for this property.

Proof.

First let us show that

[TABLE]

We set $b:=b(Q_{2})=\min\left\{b\in\mathbb{N}^{\ast}\text{ such that }\frac{\nu(Q_{2})-\nu(\partial_{b}Q_{2})}{b}=\epsilon(Q_{2})\right\}$.

We have

[TABLE]

because for every polynomial $g$, we have $\nu_{Q_{1}}(g)\leq\nu(g)$.

But by Lemma 1.15, $\nu_{Q_{1}}(Q_{2})-\nu_{Q_{1}}(\partial_{b}Q_{2})\leq b\epsilon(Q_{1})$, so

[TABLE]

Then $\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})$.

Now let us show that $\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})\Rightarrow\epsilon\left(Q_{1}\right)<\epsilon\left(Q_{2}\right)$. Assume, aiming for contradiction, that $\epsilon(Q_{1})\geq\epsilon(Q_{2})$. Then $\deg(Q_{1})\geq\deg(Q_{2}).$

If we have $\deg(Q_{1})>\deg(Q_{2})$, then $\nu_{Q_{1}}(Q_{2})=\nu(Q_{2})$ and this is a contradiction. Hence we assume that $Q_{1}$ and $Q_{2}$ have same degree.

Let $Q_{2}=Q_{1}+(Q_{2}-Q_{1})$ be the $Q_{1}$-expansion of $Q_{2}$.

If $\nu(Q_{1})\neq\nu(Q_{2}-Q_{1})$, then

[TABLE]

and again we have a contradiction.

So $\nu(Q_{1})=\nu(Q_{2}-Q_{1})=\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})$.

But $\nu(Q_{2})=\nu_{Q_{2}}(Q_{2})\leq\nu_{Q_{1}}(Q_{2})$ by Proposition 1.20. Again, this is a contradiction.

So we showed that $\epsilon(Q_{1})<\epsilon(Q_{2})\Leftrightarrow\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})$.

Let $Q_{2}$ be of minimal degree for the first property.

Assume the existence of $Q_{3}$ of degree strictly less than $Q_{2}$ such that $\nu_{Q_{1}}(Q_{3})<\nu(Q_{3})$. So $\epsilon(Q_{1})<\epsilon(Q_{3})$, which is in contradiction with the minimality of the degree of $Q_{2}$ for this property.

So we have $Q_{1}<Q_{2}\Rightarrow\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})$ and $Q_{2}$ is of minimal degree for this property.

Take $Q_{2}$ such that $\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})$ and $Q_{2}$ is of minimal degree for this property. Assume the existence of $Q_{3}$ of degree strictly less than $\deg\left(Q_{2}\right)$ and such that $\epsilon(Q_{1})<\epsilon(Q_{3})$. By this last property, we have $\nu_{Q_{1}}(Q_{3})<\nu(Q_{3})$, which is in contradiction with the minimality of the degree of $Q_{2}$ for this property.

This completes the proof. ∎

Proposition 2.5**.**

Let $Q_{1}$ and $Q_{2}$ be two key polynomials, and let

[TABLE]

be the $Q_{1}$-expansion of $Q_{2}$ .

The following are equivalent:

1. (1)

The polynomials $Q_{1}$ and $Q_{2}$ satisfy $Q_{1}<Q_{2}$. 2. (2)

We have $\sum\limits_{j\in S_{Q_{1}}\left(Q_{2}\right)}\mathrm{in}_{\nu}\left(q_{j}Q_{1}^{j}\right)=0$ with $Q_{2}$ of minimal degree for this property.

Proof.

First, let us show that

[TABLE]

Assume $Q_{1}<Q_{2}$. By Proposition 2.4, we know that $\nu_{Q_{1}}(Q_{2})<\nu(Q_{2})$. So by definition

[TABLE]

Furthermore, if $Q_{1}<Q_{2}$, we have that $Q_{2}$ is of minimal degree for this property by definition of immediate successor.

Now let us show that if $\sum\limits_{j\in S_{Q_{1}}\left(Q_{2}\right)}\mathrm{in}_{\nu}\left(q_{j}Q_{1}^{j}\right)=0$ with $Q_{2}$ of minimal degree for this property, then $Q_{1}<Q_{2}$.

Assume $\sum\limits_{j\in S_{Q_{1}}\left(Q_{2}\right)}\mathrm{in}_{\nu}\left(q_{j}Q_{1}^{j}\right)=0$. Then

[TABLE]

and so $Q_{2}>Q_{1}$ by Proposition 2.4. ∎

Remark 2.6*.*

Let $Q_{1}$ and $Q_{2}$ be key polynomials such that $Q_{2}$ is an immediate successor of $Q_{1}$ and let $Q_{2}=\sum\limits_{j\in\Theta}q_{j}Q_{1}^{j}$ be the $Q_{1}$-expansion of $Q_{2}$ . We set

[TABLE]

We will show that $\widetilde{Q}_{2}$ is an immediate successor of $Q_{1}$. Then we will always consider “optimal” immediate successor key polynomials. This means, by definition, that all the terms in their expansion in the powers of the previous key polynomial are of same value.

Proposition 2.7**.**

Let $Q_{1}$ and $Q_{2}$ be key polynomials such that $Q_{2}$ is an immediate successor of $Q_{1}$ and let $Q_{2}=\sum\limits_{j\in\Theta}q_{j}Q_{1}^{j}$ be the $Q_{1}$-expansion of $Q_{2}$ . We set

[TABLE]

Then $\widetilde{Q}_{2}$ is an immediate successor of $Q_{1}$.

Proof.

First, by definition of $\widetilde{Q}_{2}$, we have $\deg\left(\widetilde{Q}_{2}\right)\leq\deg\left(Q_{2}\right)$. We are going to show that this inequality is, in fact, an equality.

We have $\sum\limits_{j\in S_{Q_{1}}\left(Q_{2}\right)}\mathrm{in}_{\nu}\left(q_{j}Q_{1}^{j}\right)=\sum\limits_{j\in S_{Q_{1}}\left(\tilde{Q}_{2}\right)}\mathrm{in}_{\nu}\left(q_{j}Q_{1}^{j}\right)=0$. Since $Q_{2}$ is of minimal degree for this property, we know that its term of greatest degree appears in this sum. So $\deg_{X}\left(\widetilde{Q}_{2}\right)=\deg_{X}\left(Q_{2}\right)$.

Now let us show that $\epsilon\left(\widetilde{Q}_{2}\right)>\epsilon\left(Q_{1}\right)$.

Since $\sum\limits_{j\in S_{Q_{1}}\left(\widetilde{Q}_{2}\right)}\mathrm{in}_{\nu}\left(q_{j}Q_{1}^{j}\right)=0$, we have $\nu_{Q_{1}}\left(\widetilde{Q}_{2}\right)<\nu\left(\widetilde{Q}_{2}\right)$, and $\widetilde{Q}_{2}$ is still of minimal degree for this property. Then $S_{Q_{1}}\left(\widetilde{Q}_{2}\right)\neq\left\{0\right\}$ and for every non-zero integer $b$, we have $\nu_{Q_{1}}\left(\partial_{b}\widetilde{Q}_{2}\right)=\nu\left(\partial_{b}\widetilde{Q}_{2}\right)$. By Proposition 1.16, there exists a strictly positive integer $b$ such that $\nu_{Q}\left(P\right)-\nu_{Q}\left(\partial_{b}P\right)=b\epsilon\left(Q\right)$. So we can use Corollary 1.19 to conclude that

[TABLE]

Assume that we already know that $\widetilde{Q}_{2}$ is a key polynomial. Since $\deg\left(\widetilde{Q}_{2}\right)=\deg\left(Q_{2}\right)$, we have that $\widetilde{Q}_{2}$ is of minimal degree for the property $\epsilon\left(\widetilde{Q}_{2}\right)>\epsilon\left(Q_{1}\right)$, and so $Q_{1}<\widetilde{Q}_{2}$.

It remains to prove that $\widetilde{Q}_{2}$ is a key polynomial.

Assume, aiming for contradiction, that $\widetilde{Q}_{2}$ is not a key polynomial. Then there exists a polynomial $P\in K[X]$ such that

[TABLE]

and

[TABLE]

We take $P$ of minimal degree for this property. We can also assume that $P$ is monic. Let us show that $P$ is a key polynomial.

Let $S\in K[X]$ be a polynomial such that $\epsilon\left(S\right)\geq\epsilon\left(P\right)$. Then $\epsilon\left(S\right)\geq\epsilon\left(\tilde{Q}_{2}\right)$. If $\deg_{X}\left(S\right)\geq\deg_{X}\left(\widetilde{Q}_{2}\right)$, then $\deg_{X}\left(S\right)>\deg_{X}\left(P\right)$ and the proof is finished. So let us assume that $\deg_{X}\left(S\right)<\deg_{X}\left(\widetilde{Q}_{2}\right)$.

We have $\epsilon\left(S\right)\geq\epsilon\left(\widetilde{Q}_{2}\right)$ and $\deg_{X}\left(S\right)<\deg_{X}\left(\widetilde{Q}_{2}\right)$. By minimality of the degree of $P$ for this property, we have $\deg_{X}\left(S\right)\geq\deg_{X}\left(P\right)$, and hence $P$ is a key polynomial.

So there exists a key polynomial $P$ such that

[TABLE]

and

[TABLE]

Since $\epsilon\left(\widetilde{Q}_{2}\right)>\epsilon\left(Q_{1}\right)$, we also have $\epsilon\left(P\right)>\epsilon\left(Q_{1}\right)$. By minimality of the degree of $Q_{2}$ among the key polynomials satisfying this inequality, we have $\deg_{X}\left(Q_{2}\right)\leq\deg_{X}\left(P\right)<\deg_{X}\left(\widetilde{Q}_{2}\right)$ which is a contradiction by the equality of the degrees of $Q_{2}$ and $\widetilde{Q}_{2}$. Hence the polynomial $\widetilde{Q}_{2}$ is a key polynomial. ∎

Definition 2.8**.**

Let $Q_{1}$ and $Q_{2}$ be two key polynomials such that $Q_{1}<Q_{2}$. We say that $Q_{2}$ is an optimal immediate successor of $Q_{1}$ if all the terms of its $Q_{1}$-expansion have same value.

Remark 2.9*.*

Proposition 2.7 shows how to associate to every immediate successor $Q_{2}$ of $Q_{1}$ an optimal immediate successor $\widetilde{Q}_{2}$.

Hence, if $Q_{1}$ is not maximal in the set of the key polynomials $\Lambda$, it admits an optimal immediate successor.

Let $Q\in\Lambda$ be a key polynomial. We note

[TABLE]

Definition 2.10**.**

We assume that $M_{Q}$ does not have a maximal element and that for every element $P\in M_{Q}$ we have $\deg_{X}\left(P\right)=\deg_{X}\left(Q\right)$.

We also assume that there exists a key polynomial $Q^{\prime}\in\Lambda$ such that $\epsilon(Q^{\prime})>\epsilon(M_{Q})$.

We call a limit immediat successor of $Q$ every polynomial $Q^{\prime}$ of minimal degree which has this property, and we write $Q<_{\lim}Q^{\prime}$.

Proposition 2.11**.**

Let $Q$ and $Q^{\prime}$ be two key polynomials such that $\epsilon\left(Q\right)<\epsilon\left(Q^{\prime}\right)$. Then there exists a sequence $Q_{1}=Q,\dots,Q_{h}=Q^{\prime}$ where for every index $i$, the polynomial $Q_{i+1}$ is either an immediate successor of $Q_{i}$ or a limit immediate successor of $Q_{i}$.

Proof.

If $Q^{\prime}$ is an immediate successor of $Q,$ we are done, so we assume that $Q^{\prime}$ is not an immediate successor of $Q$, and we write this $Q\nless Q^{\prime}$.

Let us first look at $M_{Q}=M_{Q_{1}}$. If this set has a maximum, we denote this maximum by $Q_{2}$. We have:

[TABLE]

and by minimality of the degree of $Q_{2}$ we know that $\deg_{X}\left(Q_{2}\right)<\deg_{X}\left(Q^{\prime}\right)$. But $Q^{\prime}$ is a key polynomial, so $\epsilon\left(Q_{2}\right)<\epsilon\left(Q^{\prime}\right)$.

Then we have

[TABLE]

and since $Q<Q_{2}$, we know that $\deg_{X}\left(Q\right)\leq\deg_{Q}\left(Q_{2}\right)$.

We iterate the process as long as $M_{Q_{i}}$ has a maximum.

Assume that there exists an index $i$ such that $M_{Q_{i}}$ does not have a maximum.

Assume that $\epsilon\left(M_{Q_{i}}\right)\nless\epsilon\left(Q^{\prime}\right)$. So there exists $g_{i}\in M_{Q_{i}}$ such that $\epsilon\left(g_{i}\right)\geq\epsilon\left(Q^{\prime}\right)$. Since $Q^{\prime}$ is a key polynomial, we know that $\deg_{X}\left(g_{i}\right)\geq\deg_{X}\left(Q^{\prime}\right)$.

We have:

[TABLE]

By definition of immediate successors, we have $Q_{i}<Q^{\prime}$ and we set $Q_{i+1}=Q^{\prime}$. This completes the proof.

Now assume that $\epsilon\left(Q^{\prime}\right)>\epsilon\left(M_{Q_{i}}\right)$.

Since $\deg_{X}\left(Q\right)\leq\deg_{X}\left(Q_{i}\right)<\deg_{Q}\left(Q^{\prime}\right)$ for every index $i$, there exists a natural number $N$ such that for every index $j\geq N$ we have

[TABLE]

Let $P\in M_{Q_{N}}$. By construction, $\epsilon\left(P\right)\leq\epsilon\left(Q_{N+1}\right)<\epsilon\left(Q^{\prime}\right)$. If $Q^{\prime}$ is not of minimal degree for this property, then there exists a key polynomial $P^{\prime}$ limit immediate successor of $Q_{N}$, of degree strictly less than the degree of $Q^{\prime}$. So

[TABLE]

Then we replace $Q_{N+1}$ by $P^{\prime}$ and iterate the process, which ends because the sequence of the degrees increase strictly.

Otherwise, $Q^{\prime}$ is of minimal degree among all the key polynomials such that $\epsilon\left(M_{Q_{N}}\right)<\epsilon\left(Q^{\prime}\right)$, so $Q^{\prime}$ is a limit immediate successor of $Q_{N}$ and the process ends at $Q_{N+1}=Q^{\prime}$.

In each case, we construct a family of key polynomials which begins at $Q$, ends at $Q^{\prime}$ and such that for every index $i$, the polynomial $Q_{i+1}$ is either an immediate successor of $Q_{i}$, or a limit immediate successor of $Q_{i}$. This completes the proof. ∎

Proposition 2.12**.**

Let $Q$ and $Q^{\prime}$ be two key polynomials such that $\epsilon\left(Q\right)<\epsilon\left(Q^{\prime}\right)$. Then there exists a sequence $Q_{1}=Q,\dots,Q_{h}=Q^{\prime}$ where for every index $i$, the polynomial $Q_{i+1}$ is either an optimal immediate successor of $Q_{i}$ or a limit immediate successor of $Q_{i}$.

Proof.

Let $Q_{2}$ be an optimal immediate successor of $Q$. We look at $M_{Q}=M_{Q_{1}}$. If this set has a maximum, we denote this maximum by $P$.

If $\epsilon\left(Q_{2}\right)=\epsilon\left(P\right)$, we set $P=Q_{2}$. Otherwise, $\epsilon\left(Q_{2}\right)<\epsilon\left(P\right)$. Since $P$ and $Q_{2}$ are both immediate successors of $Q$, they have same degree.

Hence $P$ is an immediate successor of $Q_{2}$, of the same degree as $Q_{2}$. The polynomial $P$ is then an optimal immediate successor of $Q_{2}$.

So we set $Q_{3}=P$.

In fact, we have a finite sequence of optimal immediate successors which begins at $Q$ and ends at $P=\max\left\{M_{Q}\right\}$.

We iterate the process as long as $M_{Q_{i}}$ has a maximum. Assume that there exists an index $i$ such that $M_{Q_{i}}$ does not have a maximum.

Then we do exactly the same thing that we did in the proof ofcProposition 2.11 and this completes the proof. ∎

Lemma 2.13**.**

Let $Q$ and $Q^{\prime}$ be two key polynomials such that $Q<Q^{\prime}$ and we denote by $Q^{\prime}=\sum\limits_{j=0}^{m}q_{j}Q^{j}$ the $Q$-expansion of $Q^{\prime}$. Then $q_{m}=1$.

Proof.

Since $\epsilon\left(Q\right)<\epsilon\left(Q^{\prime}\right)$, we know by Proposition 2.5 that $\sum\limits_{j=0}^{m}\mathrm{in}_{\nu}\left(q_{j}Q^{j}\right)=0$. In fact we have

[TABLE]

Then, since $\mathrm{in}_{\nu}\left(q_{m}\right)\neq 0$, we have

[TABLE]

We set $a:=\deg_{X}\left(Q\right)$ and we consider $G_{<a}$ the subalgebra of $\mathrm{gr}_{\nu}\left(K\left[X\right]\right)$ generated by the initial forms of all the polynomials of degree strictly less than $a$.

Hence $G_{<a}$ is a saturated algebra, and all the coefficients of the form $\frac{\mathrm{in}_{\nu}\left(q_{i}\right)}{\mathrm{in}_{\nu}\left(q_{m}\right)}$ of the equation $\left(\ref{eq:premier coefficient egal a 1}\right)$ can be represented by polynomials. We denote by $h_{i}$ some liftings, of degrees strictly less than $a$.

The element $\mathrm{in}_{\nu}\left(Q\right)$ is hence a solution of a homogeneous monic equation with coefficients in $G_{<a}$ and whose leading coefficient is $1$.

We consider the polynomial $\widetilde{Q}=Q^{m}+\sum\limits_{j=0}^{m-1}h_{j}Q^{j}$, with, by hypothesis, $\deg_{X}\left(\widetilde{Q}\right)\leq\deg_{X}\left(Q^{\prime}\right)$. By construction we have

[TABLE]

and by the proof of the proposition 2.5, we have $\epsilon\left(\widetilde{Q}\right)>\epsilon\left(Q\right)$.

By minimality of the degree of $Q^{\prime}$ for this property, if we can show that $\widetilde{Q}$ is a key polynomial, then we would have $\deg_{X}\left(Q^{\prime}\right)=\deg_{X}\left(\widetilde{Q}\right)$ and so $q_{m}=1$.

Let us show that $\widetilde{Q}$ is a key polynomial.

Assume, aiming for contradiction, that it is not. Then there exists a polynomial $P$ such that $\epsilon\left(P\right)\geq\epsilon\left(\widetilde{Q}\right)$ and $\deg_{X}\left(P\right)<\deg_{X}\left(\widetilde{Q}\right)$. We choose $P$ monic and of minimal degree for this property. Let us show that $P$ is a key polynomial.

Let $S$ be a polynomial such that $\epsilon\left(S\right)\geq\epsilon\left(P\right)$. Then $\epsilon\left(S\right)\geq\epsilon\left(\widetilde{Q}\right)$.

If $\deg_{X}\left(S\right)\geq\deg_{X}\left(\widetilde{Q}\right)$, then, since $\deg_{X}\left(P\right)<\deg_{X}\left(\widetilde{Q}\right)$, the proof is finished.

So let us assume that $\deg_{X}\left(S\right)<\deg_{X}\left(\widetilde{Q}\right)$. Then $\epsilon\left(S\right)\geq\epsilon\left(\widetilde{Q}\right)$ and $\deg_{X}\left(S\right)<\deg_{X}\left(\widetilde{Q}\right)$. By minimality of the degree of $P$ for that property, $\deg_{X}\left(S\right)\geq\deg_{X}\left(P\right)$ and the proof is finished.

So there exists a key polynomial $P$ such that $\epsilon\left(P\right)\geq\epsilon\left(\widetilde{Q}\right)$ and $\deg_{X}\left(P\right)<\deg_{X}\left(\widetilde{Q}\right)$.

Since $\epsilon\left(\widetilde{Q}\right)>\epsilon\left(Q\right)$, we have $\epsilon\left(P\right)>\epsilon\left(Q\right)$.

So we have a key polynomial $P$ such that $\epsilon\left(P\right)>\epsilon\left(Q\right)$. By minimality of the degree of $Q^{\prime}$ for this property, we know that $\deg_{X}\left(Q^{\prime}\right)\leq\deg_{X}\left(P\right)$. But $\deg_{X}\left(P\right)<\deg_{X}\left(\widetilde{Q}\right)$, and this implies that $\deg_{X}\left(Q^{\prime}\right)<\deg_{X}\left(\widetilde{Q}\right)$, which is a contradiction.

Thus $\widetilde{Q}$ is a key polynomial. ∎

Proposition 2.14**.**

Let $Q$ and $Q^{\prime}$ be two key polynomials such that

[TABLE]

Let $c$ and $c^{\prime}$ be two polynomials of degrees strictly less than $\deg_{X}Q^{\prime}$ and let $j$ and $j^{\prime}$ be two integers such that :

[TABLE]

Then:

[TABLE]

If, in addition, either $j<j^{\prime}$ or $\nu_{Q}\left(c\left(Q^{\prime}\right)^{j}\right)<\nu_{Q}\left(c^{\prime}\left(Q^{\prime}\right)^{j^{\prime}}\right)$, then

[TABLE]

Proof.

We know that $\nu_{Q}\left(Q^{\prime}\right)\leq\nu\left(Q^{\prime}\right)$, hence

[TABLE]

Since we assumed that $j\leq j^{\prime}$, we have

[TABLE]

Furthermore, we know that $\nu_{Q}\left(c\left(Q^{\prime}\right)^{j}\right)\leq\nu_{Q}\left(c^{\prime}\left(Q^{\prime}\right)^{j^{\prime}}\right)$, hence

[TABLE]

So we have the inequality

[TABLE]

Equivalently, $\nu_{Q}\left(c\right)+j\nu\left(Q^{\prime}\right)\leq\nu_{Q}\left(c^{\prime}\right)+j^{\prime}\nu\left(Q^{\prime}\right)$.

But $\nu_{Q}\left(c\right)=\nu\left(c\right)$ and $\nu_{Q}\left(c^{\prime}\right)=\nu\left(c^{\prime}\right)$, so $\nu\left(c\left(Q^{\prime}\right)^{j}\right)\leq\nu\left(c^{\prime}\left(Q^{\prime}\right)^{j^{\prime}}\right)$.

If, in addition, either $j<j^{\prime}$ or $\nu_{Q}\left(c\left(Q^{\prime}\right)^{j}\right)<\nu_{Q}\left(c^{\prime}\left(Q^{\prime}\right)^{j^{\prime}}\right)$, then we have $\nu\left(c\left(Q^{\prime}\right)^{j}\right)<\nu\left(c^{\prime}\left(Q^{\prime}\right)^{j^{\prime}}\right)$. ∎

Lemma 2.15**.**

Let $Q$ and $Q^{\prime}$ be two polynomials such that

[TABLE]

and let $f\in K[X]$ be a polynomial whose $Q^{\prime}$-expansion is $f=\sum\limits_{j=0}^{r}f_{j}\left(Q^{\prime}\right)^{j}$. Then

[TABLE]

If we set

[TABLE]

then we have

[TABLE]

Proof.

Only for the purposes of this proof, we will write

[TABLE]

and

[TABLE]

Let us show that $\nu_{Q}\left(f\right)=\nu^{\prime}\left(f\right)$.

First, we have

[TABLE]

Set $b^{\prime}=\max T^{\prime}\left(f\right)$ and $b=\delta_{Q}\left(f_{b^{\prime}}\right)$. In other words

[TABLE]

where $f_{b^{\prime}}=\sum\limits_{j=0}^{n}a_{j}Q^{j}$. Hence, the expression $\sum\limits_{j\in T^{\prime}\left(f\right)}f_{j}\left(Q^{\prime}\right)^{j}$ contains the term

[TABLE]

Then for every $j\in\left\{0,\dots,r\right\}$ such that $f_{j}\neq 0$, we have:

[TABLE]

for every index $i\in T^{\prime}\left(f\right)$. So in particular,

[TABLE]

with strict inequality if $j\notin T^{\prime}\left(f\right)$.

So

[TABLE]

and

[TABLE]

By maximality of $b$ and $b^{\prime}$, the term $a_{b}c_{\deg_{Q}Q^{\prime}}Q^{b+b^{\prime}\deg_{Q}Q^{\prime}}$ cannot be cancelled and so $\nu_{Q}\left(f\right)=\nu\left(a_{b}c_{\deg_{Q}Q^{\prime}}Q^{b+b^{\prime}\deg_{Q}Q^{\prime}}\right)=\nu^{\prime}\left(f\right)$. In other words $\nu_{Q}\left(f\right)=\min\limits_{0\leq j\leq r}\left\{\nu_{Q}\left(f_{j}\left(Q^{\prime}\right)^{j}\right)\right\}$. So we also have

[TABLE]

Then $\sum\limits_{j\in T^{\prime}\left(f\right)}\mathrm{in}_{\nu_{Q}}\left(f_{j}\left(Q^{\prime}\right)^{j}\right)$ is a non-zero element of $G_{\nu_{Q}}$, equal to $\mathrm{in}_{\nu_{Q}}\left(f\right)$. This completes the proof. ∎

Corollary 2.16**.**

Let $Q$ and $Q^{\prime}$ be two key polynomials such that

[TABLE]

and let

[TABLE]

be the $Q^{\prime}$ and $Q$-expansions of an element $f\in K\left[X\right]$. We set

[TABLE]

and we assume that $\nu_{Q}\left(f_{\delta_{Q^{\prime}}\left(f\right)}\right)=\nu\left(f_{\delta_{Q^{\prime}}\left(f\right)}\right)$ and that $\nu_{Q}\left(f_{\theta}\right)=\nu\left(f_{\theta}\right)$.

Then:

1. (1)

$\delta_{Q^{\prime}}\left(f\right)\deg_{Q}Q^{\prime}\leq\delta_{Q}\left(f\right)$, and so $\delta_{Q^{\prime}}\left(f\right)\leq\delta_{Q}\left(f\right)$. 2. (2)

If $\delta_{Q}\left(f\right)=\delta_{Q^{\prime}}\left(f\right)$, we set $\delta:=\delta_{Q}\left(f\right)$ and then

[TABLE]

[TABLE]

and

[TABLE]

Proof.

First let us show the point $1$.

By the proof of the previous Lemma, we know that

[TABLE]

Furthermore,

[TABLE]

[TABLE]

By definition of $\delta=\delta_{Q^{\prime}}\left(f\right)$, we have $\nu\left(f_{\delta}\left(Q^{\prime}\right)^{\delta}\right)\leq\nu\left(f_{\theta}\left(Q^{\prime}\right)^{\theta}\right)$. We know by Lemma 2.15 that $\nu_{Q}\left(f\right)=\min\limits_{0\leq j\leq r}\left\{\nu_{Q}\left(f_{j}\left(Q^{\prime}\right)^{j}\right)\right\}$. Since $\theta=\min T_{Q,Q^{\prime}}\left(f\right)$, we have

[TABLE]

Hence $\nu_{Q}\left(f_{\theta}\left(Q^{\prime}\right)^{\theta}\right)\leq\nu_{Q}\left(f_{\delta}\left(Q^{\prime}\right)^{\delta}\right)$.

Then, since $\nu_{Q}\left(f_{\theta}\right)=\nu\left(f_{\theta}\right)$ and $\nu_{Q}\left(f_{\delta}\right)=\nu\left(f_{\delta}\right)$:

[TABLE]

Assume we have equality on $\nu$, it means that $\nu\left(f_{\theta}\left(Q^{\prime}\right)^{\theta}\right)=\nu\left(f_{\delta}\left(Q^{\prime}\right)^{\delta}\right)$. So $\nu\left(f_{\theta}\right)=\nu\left(f_{\delta}\right)+\delta\nu\left(Q^{\prime}\right)-\theta\nu\left(Q^{\prime}\right)$ and

[TABLE]

Since we know that $\epsilon\left(Q\right)<\epsilon\left(Q^{\prime}\right)$, by the proof of Proposition 2.4, we know that $\nu_{Q}\left(Q^{\prime}\right)<\nu\left(Q^{\prime}\right)$ and then $\delta-\theta\leq 0$, that is $\delta\leq\theta$.

Otherwise we have $\nu\left(f_{\delta}\left(Q^{\prime}\right)^{\delta}\right)<\nu\left(f_{\theta}\left(Q^{\prime}\right)^{\theta}\right)$.

Then the following four conditions hold:

[TABLE]

By the contrapositive of Proposition 2.14, we deduce that $\delta<\theta$.

In each case, we have $\delta\leq\theta$. Then since $\theta\deg_{Q}Q^{\prime}\leq\delta_{Q}\left(f\right)$, we know that $\delta\deg_{Q}Q^{\prime}\leq\delta_{Q}\left(f\right)$. So in particular $\delta_{Q^{\prime}}\left(f\right)\leq\delta_{Q}\left(f\right)$.

Now let us show the point $2$.

Assume $\delta_{Q^{\prime}}\left(f\right)=\delta_{Q}\left(f\right)=\delta$. We just saw that $\delta_{Q^{\prime}}\left(f\right)\deg_{Q}Q^{\prime}\leq\delta_{Q}\left(f\right)$, so we have $\deg_{Q}Q^{\prime}=1$. Then $Q^{\prime}=Q+b$ with $b$ a polynomial of degree strictly less than the degree of $Q$.

We know by the proof of point $1$ that $\delta\leq\theta$. Furthermore, we know that $\theta\deg_{Q}Q^{\prime}\leq\delta_{Q}\left(f\right)=\delta$, in other words $\theta\leq\delta$ since $\deg_{Q}Q^{\prime}=1$.

Hence $\delta\leq\theta\leq\delta$, hence $\theta=\delta=\min T_{Q,Q^{\prime}}\left(f\right)$. We now have to prove that for every index $j>\delta$, we have $j\notin T_{Q,Q^{\prime}}\left(f\right)$. Equivalently, that:

[TABLE]

And then we will have $T_{Q,Q^{\prime}}\left(f\right)=\left\{\delta\right\}$.

So let $j>\delta$. By definition of $\delta_{Q}\left(f\right)$ and $\delta_{Q^{\prime}}\left(f\right)$, we know that $\nu\left(f_{j}\left(Q^{\prime}\right)^{j}\right)>\nu_{Q^{\prime}}\left(f\right)$ and $\nu\left(a_{j}Q^{j}\right)>\nu_{Q}\left(f\right)$.

Furthermore, since $\delta\in T_{Q,Q^{\prime}}\left(f\right)$, we have $\nu_{Q}\left(f_{\delta}\left(Q^{\prime}\right)^{\delta}\right)=\nu_{Q}\left(f\right)$. We want to prove that $\nu_{Q}\left(f_{j}\left(Q^{\prime}\right)^{j}\right)>\nu_{Q}\left(f_{\delta}\left(Q^{\prime}\right)^{\delta}\right)$ for every index $j\in\left\{\delta+1,\dots,r\right\}$.

We know that:

[TABLE]

By the contrapositive of Proposition 2.14, we have

[TABLE]

So we have $T_{Q,Q^{\prime}}\left(f\right)=\left\{\delta\right\}$.

By Lemma 2.15, we have

[TABLE]

∎

Theorem 2.17**.**

Let $Q$ and $Q^{\prime}$ be two key polynomials such that

[TABLE]

We recall that $\mathrm{char}\left(k_{\nu}\right)=0$. If $Q^{\prime}$ is a limit immediate successor of $Q$, then $\delta_{Q}\left(Q^{\prime}\right)=1$.

Proof.

We give a proof by contradiction. Assume that $\delta_{Q}\left(Q^{\prime}\right)>1$. Among all the couples $\left(Q,Q^{\prime}\right)$ such that $Q^{\prime}$ is a limit immediat successor of $Q$ and such that $\delta_{Q}\left(Q^{\prime}\right)>1$, we choose $Q$ and $Q^{\prime}$ such that $\deg\left(Q^{\prime}\right)-\deg\left(Q\right)$ is minimal.

By definition of a limit immediate successor, for every sequence of immediate successors $\left(Q_{i}\right)_{i\in\mathbb{N}^{\ast}}$ with $Q_{1}=Q$, we have $Q_{i}\neq Q^{\prime}$ for every non-zero index $i$. By definition of limit key polynomials and by hypothesis, we know that $\deg\left(Q^{\prime}\right)-\deg\left(Q\right)$ is minimal for this property.

If we find a polynomial $\widetilde{Q}$ such that

[TABLE]

and $\deg\left(Q\right)<\deg\left(\widetilde{Q}\right)<\deg\left(Q^{\prime}\right)$, then by minimality of $\deg\left(Q^{\prime}\right)-\deg\left(Q\right)$, we know that there exists a finite sequence of immediate successors between $Q$ and $\widetilde{Q}$ and that there exists a finite sequence of immediate successors between $\widetilde{Q}$ and $Q^{\prime}$. Then we have a finite sequence of immediate successors between $Q$ and $Q^{\prime}$, which is a contradiction.

Hence there exists a key polynomial $\widetilde{Q}$ such that

[TABLE]

and $\deg\left(\widetilde{Q}\right)<\deg\left(Q^{\prime}\right)$, and so $\deg\left(Q\right)=\deg\left(\widetilde{Q}\right)$.

Let $\widetilde{Q}$ be a such key polynomial. We have $\widetilde{Q}:=Q-a$ where $a$ is a polynomial of degree strictly less than the degree of $Q$.

Since $\epsilon\left(Q\right)<\epsilon\left(\widetilde{Q}\right)$, by Proposition 2.5, we know that $\mathrm{in}_{\nu}\left(Q\right)=\mathrm{in}_{\nu}\left(a\right)$.

Consider the $Q$-expansion $\sum\limits_{j=0}^{n}a_{j}Q^{j}$ of $Q^{\prime}$. We may assume that $\delta_{Q}\left(Q^{\prime}\right)=\delta_{\widetilde{Q}}\left(Q^{\prime}\right)$ and we set $\delta:=\delta_{Q}\left(Q^{\prime}\right)$.

By Corollary 2.16, we know that $\mathrm{in}_{\nu_{Q}}\left(Q^{\prime}\right)=\mathrm{in}_{\nu_{Q}}\left(a_{\delta}\right)\mathrm{in}_{\nu_{Q}}\left(\widetilde{Q}\right)^{\delta}$. In other words $\mathrm{in}_{\nu_{Q}}\left(Q^{\prime}\right)=\mathrm{in}_{\nu_{Q}}\left(a_{\delta}\right)\mathrm{in}_{\nu_{Q}}\left(Q-a\right)^{\delta}$.

Furthermore, $\partial Q^{\prime}=\sum\limits_{j=0}^{n}\left[\partial\left(a_{j}\right)Q^{j}+a_{j}jQ^{j-1}\partial Q\right]$.

We first show that the terms $\partial\left(a_{j}\right)Q^{j}$ do not appear in $\mathrm{in}_{\nu}\left(\partial Q^{\prime}\right)$. So let $j\in\left\{0,\dots,n\right\}$.

We have

[TABLE]

But $Q$ is a key polynomial and $a_{j}$ is of degree strictly less than the degree of $Q$ since it is a coefficient of a $Q$-expansion. Then $\epsilon\left(a_{j}\right)<\epsilon\left(Q\right)$.

So

[TABLE]

By the proof of Proposition 1.16, we know that, since we are in characteristic zero,

[TABLE]

Then $\nu_{Q}\left(\partial a_{j}\right)>\nu_{Q}\left(a_{j}\right)-\nu_{Q}\left(Q\right)+\nu_{Q}\left(\partial Q\right)$. In fact,

[TABLE]

It means that $\nu_{Q}\left(Q\partial a_{j}\right)>\nu_{Q}\left(a_{j}\partial Q\right)$, and adding $\nu_{Q}\left(Q^{j-1}\right)$ to each side, we obtain:

[TABLE]

So

[TABLE]

Even though the expression $\sum\limits_{j=1}^{n}\left[ja_{j}Q^{j-1}\partial Q\right]$ need not be a $Q$-expansion, since $a_{j}$ and $\partial Q$ are of degrees strictly less than the degree of $Q$ in characteristic zero, by Lemma 1.11, the $\nu_{Q}$-initial form of $a_{j}\partial Q$ is equal to the initial form of its remainder after the Euclidean division by $Q$. So we conserve this expression and consider it a substitute of a $Q$-expansion.

Now let us prove that $\delta_{Q}\left(\partial Q^{\prime}\right)=\delta-1$.

Replacing $Q$ by $\widetilde{Q}$ in the computation of the initial form of $Q^{\prime}$ with respect to $Q$ (respectively $\widetilde{Q}$) does not change the problem, and we assume that $\delta$ stabilizes starting with $Q$. Then, if $\delta_{Q}\left(\partial Q^{\prime}\right)=\delta-1$, we would also have $\delta_{\widetilde{Q}}\left(\partial Q^{\prime}\right)=\delta-1$.

Let $j>\delta$. Let us first show that

[TABLE]

It is enough to show that

[TABLE]

But by definition of $\delta$, we have $\nu_{Q}\left(a_{j}Q^{j}\right)>\nu_{Q}\left(a_{\delta}Q^{\delta}\right)$. So

[TABLE]

hence $\nu_{Q}\left(ja_{j}Q^{j-1}\right)>\nu_{Q}\left(\delta a_{\delta}Q^{\delta-1}\right)$.

We now have to prove that the value of the term $\delta-1$ is minimal.

Let $j<\delta$. We know that $\nu_{Q}\left(a_{j}Q^{j}\right)=\nu_{Q}\left(a_{\delta}Q^{\delta}\right)$, and hence

[TABLE]

So $\nu_{Q}\left(ja_{j}Q^{j-1}\partial Q\right)=\nu_{Q}\left(\delta a_{\delta}Q^{\delta-1}\partial Q\right)$ since we are in characteristic zero.

So we do have $\delta_{Q}\left(\partial Q^{\prime}\right)=\delta_{\widetilde{Q}}\left(\partial Q^{\prime}\right)=\delta-1$. By Corollary 2.16, we have:

[TABLE]

In other words

[TABLE]

We know that $\nu_{Q}\left(Q-a\right)<\nu\left(Q-a\right)$. Then, since $\delta>1$,

[TABLE]

It means that the image by $\varphi\colon\mathrm{gr}_{\nu_{Q}}K\left[x\right]\to\mathrm{gr}_{\nu}K\left[x\right]$ of

[TABLE]

is zero. Then, the image by $\varphi$ of $\mathrm{in}_{\nu_{Q}}\left(\partial Q^{\prime}\right)$ is zero, and so

[TABLE]

By the proof of Proposition 2.4, we have $\epsilon\left(Q\right)<\epsilon\left(\partial Q^{\prime}\right)$. But we know that $\deg\left(\partial Q^{\prime}\right)<\deg\left(Q^{\prime}\right)$, and since $Q^{\prime}$ is a key polynomial, we have $\epsilon\left(\partial Q^{\prime}\right)<\epsilon\left(Q^{\prime}\right)$.

More generally, the above argument holds if we replace $Q$ by any key polynomial $\widetilde{Q}$ of the same degree as $Q$.

So for every key polynomial $\widetilde{Q}$ of the same degree as $\deg\left(Q\right)$, we have $\epsilon\left(\widetilde{Q}\right)<\epsilon\left(\partial Q^{\prime}\right)$.

In fact, $\epsilon\left(Q\right)<\epsilon\left(\partial Q^{\prime}\right)<\epsilon\left(Q^{\prime}\right)$ and $\deg\left(\partial Q^{\prime}\right)<\deg\left(Q^{\prime}\right)$. So if we show that $\partial Q^{\prime}$ is a key polynomial, we will have

[TABLE]

Let us show that $\partial Q^{\prime}$ is a key polynomial. We assume, aiming for contradiction, that it is not. There exists a polynomial $P$ such that $\epsilon\left(P\right)\geq\epsilon\left(\partial Q^{\prime}\right)$ and $\deg\left(P\right)<\deg\left(\partial Q^{\prime}\right)$. We choose $P$ of minimal degree for this property. Using the same idea as before, we can show that $P$ is a key polynomial.

We have $\deg\left(P\right)<\deg\left(\partial Q^{\prime}\right)$, hence $\deg\left(P\right)<\deg\left(Q^{\prime}\right)$ and since $Q^{\prime}$ is a key polynomial, we have $\epsilon\left(P\right)<\epsilon\left(Q^{\prime}\right)$.

Since $\epsilon\left(P\right)\geq\epsilon\left(\partial Q^{\prime}\right)$, we have $\epsilon\left(P\right)>\epsilon\left(Q\right)$.

Thus we have another key polynomial $P$ such that $\epsilon\left(Q\right)<\epsilon\left(P\right)<\epsilon\left(Q^{\prime}\right)$ and $\deg\left(P\right)<\deg\left(Q^{\prime}\right)$. Then $\deg\left(P\right)=\deg\left(Q\right)$. Hence the polynomial $P$ is a key polynomial of same degree as $Q$, and so $\epsilon\left(P\right)<\epsilon\left(\partial Q^{\prime}\right)$, which is a contradiction.

We have proved that $\partial Q^{\prime}$ is a key polynomial. Then $\deg\left(Q\right)=\deg\left(\partial Q^{\prime}\right)$. But then $\epsilon\left(\partial Q^{\prime}\right)<\epsilon\left(\partial Q^{\prime}\right)$ and this in a contradiction. This completes the proof. ∎

Part III Simultaneous local uniformization in the case of rings essentially

of finite type over a field.

The objective of this part is to give a proof of the local uniformization in the case of rings essentially of finite type over a field of zero characteristic without any restriction on the rank of the valuation. The proof of the local uniformization is well known in characteristic zero. It has been proved for the first time by Zariski in 1940 ([54]) in every dimension. The benefit of our proof is to present a universal construction which works for all the elements of the regular ring we start with, and in which the strict transforms of key polynomials become coordinates after blowing up. Thus we will have an infinite sequence of blow-ups given explicitly, together with regular systems of parameters of the local rings appearing in the sequence, and which eventually monomializes every element of our algebra essentially of finite type.

To do this, we will proceed in several steps. Let us give the idea.

Let $k$ be a field of characteristic zero, $R$ a regular local $k$-algebra essentially of finite type, with residual field $k$. Let $u=(u_{1},\ldots,u_{n})$ be a regular system of parameters of $R$, $\nu$ a valuation centered in $R$, $\Gamma$ the value group of $\nu$ and $K=k(u_{1},\dots,u_{n-1})$. We assume that $k=k_{\nu}$. This property is preserved under blowings-up. Thus every ring that will appear in our local blowing-up sequence along the valuation $\nu$ will have the same residue field: $k$.

We will construct a single sequence of blowings-up which monomializes every element of $R$ provided we look far enough in the sequence. To do this, we will construct a particular sequence of (possibly limit) immediate successors. We will show that every element $f$ of $R$ will be non-degenerate with respect to a key polynomial $Q$ of this sequence, in other words, that we will have $\nu_{Q}\left(f\right)=\nu\left(f\right)$. Furthermore, all the polynomials of this sequence will be monomializable. At this point we will have proved that every element of $R$ is non-degenerate with respect to a regular system of parameters of a suitable regular local ring $R_{i}$. Then we will just have to see that every element non-degenerate with respect to a regular system of parameters is monomializable by our sequence of blow-ups.

We will begin this part by some preliminaries, where we define non-degeneracy and framed and monomial blowing-up.

Then, we will see that every element non-degenerate with respect to a regular system of paramaters is monomializable. And then it will be sufficient to prove that it is the case of all the elements of $R$.

So, after that, we construct sequence of (possibly limit) immediate successors such that every element $f$ of $R$ is non-degenerate with respect to one of these key polynomials.

In sections $6$ and $7$ we prove that all the key polynomials of this sequence are monomializable, and that we have proven the simultaneous local uniformization. To do this we will need a new notion: the one of key element. Indeed, modified by the blow-ups, the key polynomials of the above mentioned sequence have no reason to still be polynomials. So we will give a new definition, this one of key element. This notion has the benefit to be conserved by blow-ups. We will monomialize the key elements and not the key polynomials, and the proof will be complete by induction.

3. Preliminaries.

Let $k$ be a field of characteristic zero and $R$ a regular local $k$-algebra which is essentially of finite type over $k$. We consider $u=(u_{1},\ldots,u_{n})$ a regular system of parameters of $R$ and $\nu$ a valuation centered on $R$ whose group of values is denoted by $\Gamma$. We write $\beta_{i}=\nu(u_{i})$ for every integer $i\in\left\{1,\dots,n\right\}$, and $K=k(u_{1},\dots,u_{n-1})$.

3.1. Non-degenerate elements.

Definition 3.1**.**

Let $f\in R$. We say that $f$ is non-degenerate with respect to $\nu$ and $u$ if we have $\nu_{u}(f)=\nu(f)$, where $\nu_{u}$ is the monomial valuation with respect to $u$.

We need a more convenient way of knowing whether an element is non-degenerate with respect to a regular system of parameters. It is the objective of the following Proposition.

Proposition 3.2**.**

Let $f\in R$. The element $f$ is non-degenerate with respect to $\nu$ and $u$ if and only if there exists an ideal $N$ of $R$ which contains $f$, monomial with respect to $u$ and such that

[TABLE]

Proof.

Let us show that if there exists an ideal $N$ of $R$ which contains $f$, monomial with respect to $u$ and such that

[TABLE]

then $\nu_{u}\left(f\right)=\nu\left(f\right)$. Let $N$ be such an ideal. As $N$ is monomial with respect to $u$, we have $\nu_{u}(N)=\nu(N)$ and $\nu_{u}(N)\leq\nu_{u}(f)$ since $f\in N$.

So $\nu(f)=\nu(N)\leq\nu_{u}(f)$, which give us the equality.

Now let us show that if $\nu_{u}\left(f\right)=\nu\left(f\right)$, then there exists an ideal $N$ of $R$ which contains $f$, monomial with respect to $u$ and such that $\nu(f)=\nu(N)=\min\limits_{x\in N}\left\{\nu(x)\right\}$.

Let us assume that $\nu_{u}\left(f\right)=\nu\left(f\right)$. Let $N$ be the smallest ideal of $R$ generated by monomials in $u$ containing $f$. So $\nu(N)=\nu_{u}(N)=\nu_{u}(f)$ and since $\nu_{u}(f)=\nu(f)$, we have $\nu(N)=\nu(f)$. ∎

3.2. Framed and monomial blow-up.

Let $J_{1}\subset\left\{1,\dots,n\right\}$, $A_{1}=\left\{1,\dots,n\right\}\setminus J_{1}$ and $j_{1}\in J_{1}$.

We write

[TABLE]

and we let $R_{1}$ be a localisation of $R^{\prime}=R\left[u^{\prime}_{J_{1}\setminus\left\{j_{1}\right\}}\right]$ by a prime ideal, say $R_{1}=R^{\prime}_{\mathfrak{m}^{\prime}}$ of maximal ideal $\mathfrak{m}_{1}=\mathfrak{\mathfrak{m}^{{}^{\prime}}}R_{1}$. Since $R$ is regular, $R^{\prime}$ and $R_{1}$ are regular. Let $u^{(1)}=\left(u_{1}^{(1)},\dots,u_{n_{1}}^{(1)}\right)$ be a regular system of parameters of $\mathfrak{m}_{1}$.

We write

[TABLE]

and

[TABLE]

Since $u$ is a regular system of parameters of $R$ , we have the disjoint union

[TABLE]

Let $\pi\colon R\to R_{1}$ be the natural map. Without loss of generality, we may assume that

[TABLE]

Definition 3.3**.**

We say that $\pi\colon\left(R,u\right)\to\left(R_{1},u^{(1)}\right)$ is a framed blow-up of $\left(R,u\right)$ along $\left(u_{J_{1}}\right)$ with respect to $\nu$ if there exists $D_{1}\subset\left\{1,\dots,n_{1}\right\}$ such that

[TABLE]

and if $\mathfrak{m}^{\prime}=\left\{x\in R^{\prime}\text{ such that }\nu(x)>0\right\}$.

Remark 3.4*.*

A blow-up $\pi$ is framed if among the given generators of the maximal ideal $\mathfrak{m}_{1}$ of $R_{1}$, we have all the elements of $u^{\prime}$, except, possibly, those that are in $u^{\prime}_{C_{1}}$. In other words, except, possibly, those that are invertibles in $R_{1}$.

It is framed with respect to $\nu$ if we localized in the center of $\nu$.

Let $\pi$ be such a blow-up.

Definition 3.5**.**

We say that $\pi$ is monomial if $B_{1}=J_{1}\setminus\{j_{1}\}$.

Remark 3.6*.*

Let $\pi$ be a monomial blow-up.

Then $n_{1}=n$ and $D_{1}=\left\{1,\dots,n\right\}$.

Definition 3.7**.**

Let $\pi\colon\left(R,u\right)\to\left(R_{1},u^{(1)}\right)$ be a framed blow-up and $T\subset\left\{1,\dots,n\right\}$.

We say that $\pi$ is independent of $u_{T}$ if $T\cap J_{1}=\emptyset$, in other words if $T\subset A_{1}$.

Remark 3.8*.*

Since we look at blow-ups with respect to a valuation $\nu$, we have blow-ups such that $\nu(R_{1})\geq 0$. Since $u^{\prime}_{q}\in R_{1}$ for every $q\in J_{1}$, we want $\nu\left(\frac{u_{q}}{u_{j_{1}}}\right)\geq 0$, so $\nu\left(u_{q}\right)\geq\nu\left(u_{j_{1}}\right)$ for every $q\in J_{1}\setminus\left\{j_{1}\right\}$ . So we can set $j_{1}$ to be an element of $J_{1}$ such that $\beta_{j_{1}}=\min\limits_{q\in J_{1}}\{\beta_{q}\}$.

We have :

[TABLE]

And $C_{1}=\left\{q\in J_{1}\setminus\left\{j_{1}\right\}\text{ such that }\beta_{q}=\beta_{j_{1}}\right\}$.

Let $k_{1}$ be the residue field of $R_{1}$ and $t_{k_{1}}$ the transcendence degree of $k\hookrightarrow k_{1}$. Let us show that $t_{k_{1}}\leq\sharp C$.

We write $\bar{R}=\frac{R^{\prime}}{\mathfrak{m}R^{\prime}}$. We denote by $\bar{u}_{q}$ the image of $u^{\prime}_{q}$ in $\bar{R}$ for every $q\in J_{1}\setminus\left\{j_{1}\right\}$. So $\bar{R}=k\left[\bar{u}_{B_{1}},\bar{u}_{C_{1}}^{\pm 1}\right]$. We have $R\to R^{\prime}\to R_{1}\to k_{1}$, which induces homomorphisms $k\to\bar{R}\to\frac{R_{1}}{\mathfrak{m}R_{1}}\to k_{1}$.

We have $\mathfrak{m}=\mathfrak{m}_{1}\cap R=\mathfrak{m}^{\prime}R_{1}\cap R=\mathfrak{m}^{\prime}\cap R$ . Let $\bar{\mathfrak{m}}=\frac{\mathfrak{m}^{\prime}}{\mathfrak{m}R^{\prime}}$. We have

[TABLE]

in other words

[TABLE]

Since $u^{\prime}_{A_{1}\cup B_{1}\cup\left\{j_{1}\right\}}\subset\mathfrak{m}^{\prime}$, for every $q\in A_{1}\cup B_{1}\cup\left\{j_{1}\right\}$, the image of $u^{\prime}_{q}$ in $k_{1}$ is zero. So $k_{1}$ is generated over $k$ by the images of the $u^{\prime}_{q}$ with $q\in C_{1}$. Hence $t_{k_{1}}\leq\sharp C_{1}$.

But we have $C_{1}:=J_{1}\setminus\left(B_{1}\cup\left\{j_{1}\right\}\right)$. So $\sharp C_{1}+\sharp B_{1}+1=\sharp J_{1}=h$, and:

[TABLE]

We will often set $J_{1}\subset\left\{1,\dots,r,n\right\}$ where $r$ is the dimension of $\sum\limits_{i=1}^{n}\mathbb{Q}\nu(u_{i})$ in $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}$. If $J_{1}\subset\left\{1,\dots,r\right\}$, the family $\beta_{J_{1}}$ is a family of $\mathbb{Q}$-linearly independent elements, and so $B_{1}=J_{1}\setminus\left\{j_{1}\right\}$.

Otherwise $n\in J_{1}$. Then we have $B_{1}=J_{1}\setminus\left\{j_{1}\right\}$ or $B_{1}=J_{1}\setminus\left\{j_{1},q_{1}\right\}$ where $q_{1}\in J_{1}\setminus\left\{j_{1}\right\}$. The interesting cases are those where $h-2\leq\sharp B_{1}$, in other words, those where $h-1\leq\sharp B_{1}+1$.

Since (3.2), we have $h-1+t_{k_{1}}\leq\sharp B_{1}+1+t_{k_{1}}\leq h$.

Then we have three cases.

The first one, $\sharp B_{1}+1=h$ and $t_{k_{1}}=0$, it occurs when the blow-up is monomial.

The second one, $\sharp B_{1}+1=h-1$ and $t_{k_{1}}=1$.

The last one, $\sharp B_{1}+1=h-1$ and $t_{k_{1}}=0$.

Fact 3.9**.**

In the cases $1$ and $3$, we have $n_{1}=n$ and in the case $2$ we have $n_{1}=n-1$.

Remark 3.10*.*

In the rest of the chapter, we will assume that the valuation ring has $k$ as residue field. So $k_{1}=k$ and $t_{k_{1}}=0$. Hence we will have $n_{1}=n$.

Since $k_{1}\simeq\frac{k\left[Z\right]}{\left(\lambda\left(Z\right)\right)}$, we know that $\lambda\left(Z\right)$ is a polynomial of degree $1$ over $k$.

3.3. Key elements.

We need a more general notion than the one of key polynomials. Indeed, after several blow-ups, a key polynomial might not be a polynomial anymore.

For example, we can have $\frac{1}{u_{n}+1}u_{n-1}$, which is not a polynomial.

Definition 3.11**.**

Let $P_{1},P_{2}$ be two key polynomials for the field extension $k\left(u_{1}^{(l)},\dots,u_{n-1}^{(l)}\right)\left(u_{n}^{(l)}\right)$ with $P_{2}$ and immediate successor of $P_{1}$. Let $P_{2}=\sum\limits_{j\in S_{P_{1}}\left(P_{2}\right)}a_{j}P_{1}^{j}$ be the $P_{1}$-expansion of $P_{2}$.

We call key element every element $P_{2}^{\prime}$ of the form

[TABLE]

where $b_{j}$ are units of $R_{l}=k\left(u_{1}^{(l)},\dots,u_{n}^{(l)}\right)_{\left(u_{1}^{(l)},\dots,u_{n}^{(l)}\right)}$. The polynomial $P_{2}$ is the key polynomial associated to the key element $P_{2}^{\prime}$.

Remark 3.12*.*

A key element is not necessarily a polynomial. Indeed, for example, $\frac{1}{1+u_{n_{l}}^{(l)}}$ is a unit of $R_{l}$.

Definition 3.13**.**

Let $P_{1}^{\prime}$ and $P_{2}^{\prime}$ be two key elements. We say that $P^{\prime}_{2}$ is an immediate successor of $P^{\prime}_{1}$, and we write , $P_{1}^{\prime}\ll P_{2}^{\prime}$, if their associated key polynomials are immediate successors of each other.

Now we define limit immediate successors key elements.

Definition 3.14**.**

Let $P_{1}^{\prime}$ and $P_{2}^{\prime}$ be two key elements. We say that $P_{2}^{\prime}$ is a *limit immediate successor * of $P^{\prime}_{1}$, and we write $P_{1}^{\prime}\ll_{\lim}P_{2}^{\prime}$, if their associated key polynomials $P_{1}$ and $P_{2}$ are such that $P_{2}$ is a limit immediate successor of $P_{1}$.

4. Monomialization in the non-degenerate case.

In this section, we will monomialize all the elements which are non-degenerate with respect to a system of parameters.

Let $\alpha$ and $\gamma$ be two elements of $\mathbb{Z}^{n}$, and let $\delta=\left(\min\left\{\alpha_{j},\gamma_{j}\right\}\right)_{1\leq j\leq n}$. We say that $u^{\alpha}\mid u^{\gamma}$ if for every integer $i$, $\alpha_{i}$ is less than or equal to $\gamma_{i}$ , in other words if $\alpha$ is componentwise less than or equal to $\beta$.

Let us set

[TABLE]

The objective is to build a sequence of blow-ups $(R,u)\to\cdots\to\left(R^{\prime},u^{\prime}\right)$ such that in $R^{\prime}$, we have $u^{\alpha}\mid u^{\gamma}$.

Definition 4.1**.**

We say that $\alpha\preceq\gamma$ if for every index $i$, we have $\alpha_{i}\leq\gamma_{i}$.

We assume that $\gamma\npreceq\alpha$ and that $\alpha\npreceq\gamma$ . So we may assume that $|\widetilde{\alpha}|\neq 0$, and $\widetilde{\alpha}_{i}>0$ for every integer $i\in\left\{1,\dots,a\right\}$.

Similarly, we set

[TABLE]

Interchanging $\alpha$ and $\gamma$, if necessary, we may assume that $0<|\widetilde{\alpha}|\leq|\widetilde{\gamma}|$.

4.1. Construction of a stricly decreasing numerical character.

Definition 4.2**.**

Let $\tau\colon\mathbb{Z}^{n}\times\mathbb{Z}^{n}\to\mathbb{N}^{2}$ be the map such that

[TABLE]

Let $J$ be a minimal subset of $\left\{1,\dots n\right\}$ such that $\{1,\dots,a\}\subset J$ and $\sum\limits_{q\in J}\widetilde{\gamma}_{q}\geq|\widetilde{\alpha}|$.

Let $\pi\colon\left(R,u\right)\to\left(R_{1},u^{(1)}\right)$ be a framed blow-up along $(u_{J})$. Let $j\in J$ be such that $R_{1}$ is a localization of $R\left[\frac{u_{J}}{u_{j}}\right]$.

If $q\in J\setminus\{j\}$, we recall that $u^{\prime}_{q}=\frac{u_{q}}{u_{j}}$, and $u^{\prime}_{q}=u_{q}$ otherwise.

We now define $\widetilde{\alpha}^{\prime}_{q}=\widetilde{\alpha}_{q}$ for $q\neq j$, and $\widetilde{\alpha}^{\prime}_{q}=0$ otherwise. We set $\widetilde{\gamma}^{\prime}_{q}=\widetilde{\gamma}_{q}$ if $q\neq j$, $\widetilde{\gamma}^{\prime}_{q}=\sum\limits_{q\in J}\widetilde{\gamma}_{q}-|\widetilde{\alpha}|$ otherwise.

And finally we define

[TABLE]

So we have:

[TABLE]

But for every $l\in J\setminus\left\{j\right\}$, we have $u_{l}=u_{l}^{\prime}\times u_{j}$ and for $l\notin J\setminus\left\{j\right\}$, we have $u_{l}=u_{l}^{\prime}$. Hence

[TABLE]

Let us isolate the term $u_{j}$. We obtain:

[TABLE]

and since $\tilde{\alpha}=\alpha-\delta$, we have $\alpha=\tilde{\alpha}+\delta$ and then

[TABLE]

But $\widetilde{\alpha}^{\prime}_{q}=\widetilde{\alpha}_{q}$ for $q\neq j$ and $\delta^{\prime}=(\delta_{1},\dots,\delta_{j-1},\sum\limits_{q\in J}\delta_{q}+|\widetilde{\alpha}|,\delta_{j+1},\dots,\delta_{n})$, so

[TABLE]

We include another time the term $l=j$ in the product, and then:

[TABLE]

But we have

[TABLE]

So $u^{\alpha}=\left(u^{\prime}\right)^{{}^{\widetilde{\alpha}^{\prime}+\delta^{\prime}}}$ , and similarly $u^{\gamma}=\left(u^{\prime}\right)^{{}^{\widetilde{\gamma}^{\prime}+\delta^{\prime}}}$ .

We set $\alpha^{\prime}=\delta^{\prime}+\widetilde{\alpha}^{\prime}$ and $\gamma^{\prime}=\delta^{\prime}+\widetilde{\gamma}^{\prime}$.

Proposition 4.3**.**

We have $\tau(\alpha^{\prime},\gamma^{\prime})<\tau(\alpha,\gamma)$.

Proof.

First case: $j\in\left\{1,\dots,a\right\}$. Then

[TABLE]

Second case: $j\in\left\{a+1,\dots,n\right\}$. Then $|\widetilde{\alpha}^{\prime}|=|\widetilde{\alpha}|$. Let us show that $|\widetilde{\gamma}^{\prime}|<|\widetilde{\gamma}|$.

We have

[TABLE]

By the minimality of $J$, we have $\sum\limits_{q\in J\setminus\left\{j\right\}}\widetilde{\gamma}_{q}-|\widetilde{\alpha}|<0$, and so

[TABLE]

In every case, we have $\left(|\widetilde{\alpha}^{\prime}|,|\widetilde{\gamma}^{\prime}|\right)<\left(|\widetilde{\alpha}|,|\widetilde{\gamma}|\right)=\tau(\alpha,\gamma)$.

If $|\widetilde{\alpha}^{\prime}|\leq|\widetilde{\gamma}^{\prime}|$, then $\tau(\alpha^{\prime},\gamma^{\prime})=\left(|\widetilde{\alpha}^{\prime}|,|\widetilde{\gamma}^{\prime}|\right)$ and this completes the proof.

Otherwise, $|\widetilde{\alpha}^{\prime}|>|\widetilde{\gamma}^{\prime}|$, so

[TABLE]

and the proof is complete. ∎

Renumbering the $u^{\prime}_{q}$, if necessary, we may assume that$u^{\prime}_{q}\notin R_{1}^{\times}$ for every $q\in\left\{1,\dots,s\right\}$ and $u^{\prime}_{q}\in R_{1}^{\times}$ otherwise. Since $\pi$ is a framed blow-up, we have $\left\{u^{\prime}_{1},\dots,u^{\prime}_{s}\right\}\subset u^{(1)}$, so renumbering again, if necessary, we may assume that $u^{\prime}_{q}=u_{q}^{(1)}$ for every $q\in\left\{1,\dots,s\right\}$. We set

[TABLE]

and

[TABLE]

We have $\tau\left(\alpha^{(1)},\gamma^{(1)}\right)\leq\tau(\alpha^{\prime},\gamma^{\prime})$. By Proposition 4.3, we have

[TABLE]

4.2. Divisibility and change of variables.

Let $s\in\left\{1,\dots,n\right\}$. We write $u=\left(w,v\right)$ where

[TABLE]

and

[TABLE]

Let $\alpha$ and $\gamma$ be two elements of $\mathbb{Z}^{s}$.

Proposition 4.4**.**

There exists a framed local sequence

[TABLE]

with respect to $\nu$, independent of $v$, such that in $R_{l}$, we have $w^{\alpha}\mid w^{\gamma}$ or $w^{\gamma}\mid w^{\alpha}$.

Proof.

Unless $\gamma\preceq\alpha$, or $\alpha\preceq\gamma$, we can iterate the above construction, choosing blow-up with respect to $\nu$ and independent of $v$. Since $\tau$ is a vector in $\mathbb{N}^{2}$ and is stricly decreasing, after a finite number of steps, the process stops. After these steps, we have $w^{\alpha}=U\times\left(u^{(l)}\right)^{\alpha^{(l)}}$, $w^{\gamma}=U\times\left(u^{(l)}\right)^{\gamma^{(l)}}$, with $U\in R_{l}^{\times}$ and with $\gamma^{(l)}\preceq\alpha^{(l)}$, or $\alpha^{(l)}\preceq\gamma^{(l)}$. So we do have $w^{\alpha}\mid w^{\gamma}$ or $w^{\gamma}\mid w^{\alpha}$ in $R_{l}$. ∎

Let us now study the change of variables we do at each blow-up. We consider $i$ and $i^{\prime}$ some indexes of the framed local sequence

[TABLE]

Proposition 4.5**.**

Let us consider $0\leq i<i^{\prime}\leq l$. We let $m$ be an element of $\{1,\dots,n_{i}\}$ and $m^{\prime}$ one of $\{1,\dots,n_{i^{\prime}}\}$. Then:

1. (1)

There exists a vector $\delta_{m}^{(i^{\prime},i)}\in\mathbb{N}^{\sharp D_{i}}$ such that

[TABLE] 2. (2)

If, in addition, the local sequence (4.1) is independent of $u_{T}$, with $T\subset\{1,\dots,n\}$ ; and if we assume that $u_{m}^{(i)}\notin u_{T}$, then $\left(u_{D_{i^{\prime}}}^{(i^{\prime})}\right)^{\delta_{m}^{(i^{\prime},i)}}$ is monomial in $u_{D_{i^{\prime}}}^{(i^{\prime})}\setminus u_{T}$. 3. (3)

We assume that $i^{\prime\prime}>0$ such that $i\leq i^{\prime\prime}<i^{\prime}$. We have $D_{i"}=\{1,\dots,n_{i^{\prime\prime}}\}$, and we assume that $m^{\prime}\in D_{i^{\prime}}$. Then there exists a vector $\gamma_{m^{\prime}}^{(i,i^{\prime})}$ of $\mathbb{Z}^{n_{i}}$ such that

[TABLE] 4. (4)

If, in addition, the local sequence (4.1) is independent of $u_{T}$ and if we assume that $u_{m^{\prime}}^{(i^{\prime})}\notin u_{T}$, then $u_{m^{\prime}}^{(i^{\prime})}$ is monomial in $u^{(i)}\setminus u_{T}$.

Proof.

We only consider the case $i^{\prime}=i+1$, the general case can be proved by induction on $i-i^{\prime}$. We can also assume that $i=0$.

Let us show $(1)$. By Definition 3.3, we have $u^{\prime}_{A_{1}\cup B_{1}\cup\left\{j_{1}\right\}}=u_{D_{1}}^{(1)}$.

We denote by $D_{1}=D_{1}^{A_{1}}\cup D_{1}^{B_{1}}$ where

[TABLE]

and

[TABLE]

If $m\in A_{1}\cup\{j_{1}\}$, so $u_{m}=u^{\prime}_{m}$ and the proof is finished. If $m\in B_{1}$ then $u_{m}=u_{j_{1}}u^{\prime}_{m}=u^{\prime}_{j_{1}}u^{\prime}_{m}$ and the proof is finished.

If $m\in C_{1}$, so $u_{m}=u^{\prime}_{j_{1}}u^{\prime}_{m}$ and by definition, $u^{\prime}_{m}\in R_{1}^{\times}$, which gives us the result.

Let us show $(3)$. We have $m^{\prime}\in D_{1}=D_{1}^{A_{1}}\cup D_{1}^{B_{1}}$ and $u^{\prime}_{A_{1}\cup B_{1}\cup\left\{j_{1}\right\}}=u_{D_{1}}^{(1)}$. If $m^{\prime}\in D_{1}^{A_{1}}$ then by definition $u_{m^{\prime}}^{(1)}\in u^{\prime}_{A_{1}}=u_{A_{1}}$ and we have the result. Otherwise $m^{\prime}\in D_{1}^{B_{1}}$. So

[TABLE]

This completes the proof of $(3)$.

Now let us assume that the sequence is independent of $u_{T}$. By definition we have $u_{J_{1}}\cap u_{T}=\emptyset$ and also

[TABLE]

Let us show $(2)$. Assume that $u_{m}\notin u_{T}$.

If $m\in A_{1}$, then $u_{m}=u^{\prime}_{m}\in u_{D_{1}^{A_{1}}}^{(1)}$ and $u_{m}\notin u_{T}$ and the proof is finished. Otherwise $m\in J_{1}$. We saw in the proof of $(1)$ that $m$ was monomial in $u_{D_{1}^{B_{1}}}^{(1)}$, and since $u_{D_{1}^{B_{1}}}^{(1)}\cap u_{T}=\emptyset$, this completes the proof of $(2)$.

It remains to prove $(4)$. We assume that $u_{m^{\prime}}^{(1)}\notin u_{T}$, with $m^{\prime}\in D_{1}=D_{1}^{A_{1}}\cup D_{1}^{B_{1}}$.

If $m^{\prime}\in D_{1}^{A_{1}}$, then $u_{m^{\prime}}^{(1)}\in u^{\prime}_{A_{1}}=u_{A_{1}}$. Since $u_{m^{\prime}}^{(1)}\notin u_{T}$, we have $u_{m^{\prime}}^{(1)}\in u\setminus u_{T}$.

Otherwise $m^{\prime}\in D_{1}^{B_{1}}$ and we saw that $u_{m^{\prime}}^{(1)}$ is monomial in $u_{B_{1}\cup[j_{1}\}}\subset u_{J}$. Since $u_{J}\cap u_{T}=\emptyset$, we are done. ∎

Remark 4.6*.*

Let $T\subset A$, be a set of cardinality $t$, and $s:=n-t$. We set

[TABLE]

and

[TABLE]

In this Remark, we only consider monomial blow-ups.

We have $u^{\prime}=(v,w^{\prime})$ where $w^{\prime}=(w^{\prime}_{1},\dots,w^{\prime}_{s})=(w^{\gamma(1)},\dots,w^{\gamma(s)})$ with $\gamma(i)\in\mathbb{Z}^{s}$, by Proposition 4.5. By the proof of this Proposition, the matrix $F_{s}=\left[\gamma(1)\dots\gamma(s)\right]$ is a unimodular matrix. For every $\delta\in\mathbb{Z}^{s}$, we have $w^{\prime\delta}=w^{\delta F_{s}}$. In the same vein $w_{i}=w^{\prime\delta(i)}$ and the $s$-vectors $\delta(1),\dots,\delta(s)$ form a unimodular matrix equal to the inverse of $F_{s}$. Then we have $w^{\prime\gamma}=w^{\gamma F_{s}^{-1}}$, for every $\gamma\in\mathbb{Z}^{s}$.

Proposition 4.7**.**

We have:

[TABLE]

Proof.

We have $u^{(l)}=\left(w_{1}^{(l)},\dots,w_{r_{l}}^{(l)},v\right)$.

By Proposition 4.5, there exists $\alpha^{(l)},\gamma^{(l)}\in\mathbb{N}^{r_{l}}$ and $y,z\in R_{l}^{\times}$ such that $w^{\alpha}=y\left(w^{(l)}\right)^{\alpha^{(l)}}$ and $w^{\gamma}=z\left(w^{(l)}\right)^{\gamma^{(l)}}$.

For every $i\in\left\{1,\dots,r_{l}\right\}$, we have $\nu(w_{i}^{(l)})\geq 0$ since the blow-up is with respect to $\nu$, so centered in $R_{l}$. By construction of $R_{l}$, we have that $\gamma^{(l)}\preceq\alpha^{(l)}$ or $\alpha^{(l)}\preceq\gamma^{(l)}$.

So

[TABLE]

hence

[TABLE]

∎

4.3. Monomialization of non-degenerate elements.

Let $N$ be an ideal of $R$ generated by monomials in $w$. We choose $w^{\epsilon_{0}},\dots,w^{\epsilon_{b}}$ to be a minimal set of generators of $N$, with $\nu(w^{\epsilon_{0}})\leq\nu(w^{\epsilon_{i}})$ for every $i$.

Proposition 4.8**.**

There exists a local framed sequence

[TABLE]

with respect to $\nu$, independent of $v$ and such that $NR_{l}=(w^{\epsilon_{0}})R_{l}$.

Proof.

Let

[TABLE]

Assume $b\neq 0$.

We let $\left(w^{\epsilon_{i_{0}}},w^{\epsilon_{j_{0}}}\right)$ be a pair for which the minimum

[TABLE]

is attained. By Proposition 4.3, $\tau(N,w)$ is strictly decreasing at each blow-up.

Since the process stops, $NR_{l}$ is generated by a unique element as an ideal of $R_{l}$. By Proposition 4.7, this element is $w^{\epsilon_{0}}$ (which has the minimal value), which divides the others. Then $NR_{l}=(w^{\epsilon_{0}})R_{l}$. ∎

Definition 4.9**.**

An element $f$ of $R$ is monomializable if there exists a sequence of blow-ups

[TABLE]

such that the total transformed of $f$ is a monomial. It means that in $R^{\prime}$, the total transform of $f$ is $v\prod\limits_{i=1}^{n}\left(u^{\prime}_{i}\right)^{\alpha_{i}}$, with $v$ a unit of $R^{\prime}$.

Theorem 4.10**.**

Let $f$ be a non-degenerate element with respect to $u=(w,v)$, and let $N$ be the ideal which satisfies the conclusion of the Proposition 3.2, generated by monomials in $w$.

Then there exists a local framed sequence, independent of $v$,

[TABLE]

such that $f$ is a monomial in $u^{\prime}$ multiplied by a unit of $R^{\prime}$. Equivalently, $f$ is monomializable.

Proof.

Let $\left(R,u\right)\to\left(R^{\prime},u^{\prime}\right)$ be the local framed sequence of the Proposition 4.8. We have $NR^{\prime}=w^{\epsilon_{o}}R^{\prime}$. Since $f\in N$, by the proof of the Proposition 3.2, there exists an element $z\in R^{\prime}$ such that $f=w^{\epsilon_{0}}z$. Since $\nu$ is centered in $R^{\prime}$, to show that $z$ is a unit of $R^{\prime}$, we will show that $\nu(z)=0$.

But $\nu(z)=\nu(f)-\nu(w^{\epsilon_{0}})=\nu(N)-\nu(w^{\epsilon_{0}})$ by Proposition 3.2.

Since $NR^{\prime}=w^{\epsilon_{o}}R^{\prime}$, we have $\nu(N)=\nu(w^{\epsilon_{0}})$, and so $\nu(z)=0$, and this completes the proof. ∎

5. Non-degeneracy and key polynomials.

Now that we monomialized every non-degenerate element with respect to the generators of the maximal ideal of our local ring, we are going to show that every element is non-degenerate with respect to a particular sequence of immediate successors. We denote by $\Lambda$ the set of key polynomials and

[TABLE]

Proposition 5.1**.**

We consider $\nu$ an archimedean valuation centered in a noetherian local domain $\left(R,\mathfrak{m},k\right)$. We denote by $\Gamma$ the value group of $\nu$ and we set $\Phi:=\nu\left(R\setminus\left(0\right)\right)$.

The set $\Phi$ does not contain an infinite bounded strictly increasing sequence.

Proof.

Assume, aiming for contradiction, that we have an infinite sequence

[TABLE]

of elements of $\Phi$ bounded by an element $\beta\in\Phi$.

Then we have an infinite decreasing sequence $\cdots\subseteq P_{\alpha_{2}}\subseteq P_{\alpha_{1}}$ such that for every index $i$, we have $P_{\beta}\subseteq P_{\alpha_{i}}$. And so we have an infinite decreasing sequence of ideals of $\frac{R}{P_{\beta}}$.

We set

[TABLE]

Since $\nu$ is archimedean, we know that there exists a non-zero integer $n$ such that $\beta\leq n\delta$, and so such that $\mathfrak{m}^{n}\subseteq P_{\beta}$. This way, we construct an epimorphism of rings $\frac{R}{\mathfrak{m}^{n}}\twoheadrightarrow\frac{R}{P_{\beta}}$. Since the ring $R$ is noetherian, $\frac{R}{\mathfrak{m}^{n}}$ is artinian, and so is $\frac{R}{P_{\beta}}$. This contradicts the existence of the infinite decreasing sequence of ideals of $\frac{R}{P_{\beta}}$. ∎

Definition 5.2**.**

Assume that the set $M_{\alpha}$ is non-empty and does not have an maximal element. Assume also that there exists a key polynomial $Q\in\Lambda$ such that $\epsilon(Q)>\epsilon(M_{\alpha})$. We call a limit key polynomial** every polynomial of minimal degree which has this property.

Definition 5.3**.**

Let $\left(Q_{i}\right)_{i\in\mathbb{N}}$ be a sequence of key polynomials. We say that it is a sequence of immediate successors if for every integer $i$, we have $Q_{i}<Q_{i+1}$.

Proposition 5.4**.**

If there are no limit key polynomials then there exists a finite or infinite sequence of immediate successors $Q_{1}<\ldots<Q_{i}<\ldots$ such that the sequence $\left\{\epsilon(Q_{i})\right\}$ is cofinal in $\epsilon(\Lambda)$. Equivalently, such that

[TABLE]

Proof.

We do the proof by contrapositive.

Assume that for every finite or infinite sequence of immediate successors key polynomials $(Q_{i})$, the sequence $\left\{\epsilon(Q_{i})\right\}$ is not cofinal in $\epsilon(\Lambda)$. Let us show that there exists a limit key polynomial.

First let assume that for every $\alpha\in\Omega=\left\{\beta\text{ such that }M_{\beta}\neq\emptyset\right\}$, $M_{\alpha}$ has a maximal element. It means that

[TABLE]

We set $M:=\left\{R_{\alpha}\right\}_{\alpha\in\Omega}$. All elements in $M$ are of distinct degree, so they are strictly ordered by their degrees. So if $\alpha<\alpha^{\prime}$, then $\deg(R_{\alpha})<\deg(R_{\alpha^{\prime}})$. Since $R_{\alpha^{\prime}}$ is a key polynomial, by definition, we have $\epsilon(R_{\alpha})<\epsilon(R_{\alpha^{\prime}})$ as soon as $\alpha<\alpha^{\prime}$. Then in $M$ the elements are strictly ordered by their values of$\epsilon$.

Let us show that they are immediate successors. Let $R_{\alpha}$ and $R_{\alpha^{\prime}}$ be two consecutive elements of $M$. We know that

[TABLE]

and $\epsilon(R_{\alpha})<\epsilon(R_{\alpha^{\prime}})$. We want to show that $R_{\alpha^{\prime}}$ is of minimal degree for the property. So let us set $R\in\Lambda$ such that $\epsilon(R_{\alpha})<\epsilon(R)$ and $\deg(R)\leq\deg(R_{\alpha^{\prime}})$. Let us show that $\deg(R)=\deg(R_{\alpha^{\prime}})=\alpha^{\prime}$. Since $\epsilon(R_{\alpha})<\epsilon(R)$ and since $R_{\alpha}$ is a key polynomial, by definition,

[TABLE]

Since $R$ is a key polynomial, if we had $\deg(R)=\deg(R_{\alpha})$, then we should have $\epsilon(R_{\alpha})\geq\epsilon(R)$, which is a contradiction. Let us set $\lambda:=\deg(R),$ so we have $\alpha<\lambda\leq\alpha^{\prime}$, $R\in M_{\lambda}$ and $R_{\lambda}\in M$. Since the polynomials in $M$ are strictly ordered by their degrees and that $R_{\alpha}$ and $R_{\alpha^{\prime}}$ are consecutive, then we have $\lambda=\alpha^{\prime}$, and so $R_{\alpha}<R_{\alpha^{\prime}}$.

So the set $M$ is a sequence of immediate successors. By hypothesis, the sequence $\epsilon(M)$ is not cofinal, so there exists $R\in\Lambda$ such that $\epsilon(R)>\epsilon(M)$. But then there exists $\alpha$ such that $R\in M_{\alpha}$ and then $\epsilon(R_{\alpha})\geq\epsilon(R)>\epsilon(R_{\alpha})$. It is a contradiction.

So there exists $\alpha\in\Omega$ such that $M_{\alpha}$ does not have any maximal ideal. Then we have a sequence:

[TABLE]

where $Q_{i}$ is an element of $M_{\alpha}$ for every integer $i$.

Let us show that the $Q_{i}$ are immediate successors. Let $R\in\Lambda$ such that $\epsilon(Q_{i})<\epsilon(R)$ and $\deg(R)\leq\deg(Q_{i+1})=\alpha$. Since $Q_{i}$ is a key polynomial, by definition, $\deg(R)\geq\deg(Q_{i})=\alpha$. So $\deg(R)=\deg(Q_{i+1})=\alpha$, and $Q_{i+1}$ is of minimal degree for the property. Then for every integer $i$, we have $Q_{i}<Q_{i+1}$.

By hypothesis, the sequence of the $Q_{i}$ is a sequence of immediate successors, so the sequence $\left(\epsilon(Q_{i})\right)_{i}$ is not cofinal. So there exists a key polynomial $Q\in\Lambda$ such that $\epsilon(Q)>\epsilon(Q_{i})$ for every integer $i$. Let $R\in M_{\alpha}$, since $M_{\alpha}$ does not have a maximal element, there exists $i$ such that $\epsilon(R)<\epsilon(Q_{i})<\epsilon(Q)$. So there exists a key polynomial $Q\in\Lambda$ such that $\epsilon(Q)>\epsilon(M_{\alpha})$. Then the polynomial $Q$ is a limit key polynomial. ∎

Theorem 5.5**.**

There exists a finite or infinite sequence $(Q_{i})_{i\geq 1}$ of key polynomials such that for each $i$ the polynomial $Q_{i+1}$ is either an optimal or a limit immediate successor of $Q_{i}$ and such that the sequence $\left\{\epsilon(Q_{i})\right\}$ is cofinal in $\epsilon(\Lambda)$ where $\Lambda$ is the set of key polynomials.

Proof.

We know that $x$ is a key polynomial. If for every key polynomial $Q\in\Lambda$, we have $\epsilon\left(x\right)\geq\epsilon\left(Q\right)$, then the sequence $\left\{\epsilon(x)\right\}$ is cofinal in $\epsilon(\Lambda)$ and it is done. Otherwise, it exists a key polynomial $Q\in\Lambda$ such that $\epsilon\left(x\right)<\epsilon\left(Q\right)$. If it exists a maximal element among the key polynomials of same degree than $Q$, then we exchange $Q$ by this element. By Proposition 2.12, it exists a finite sequence $Q_{1}=x<\dots<Q_{p}=Q$ of optimal (possibly limit) immediate successors which begins at $x$ and ends at $Q$.

If for every key polynomial $Q^{\prime}\in\Lambda$, there exists a key polynomial of this sequence $Q_{i}$ such that $\epsilon\left(Q_{i}\right)\geq\epsilon\left(Q^{\prime}\right)$, then the sequence $\left\{\epsilon(Q_{i})\right\}$ is cofinal in $\epsilon(\Lambda)$ and it is over.

Otherwise there exists a polynomial $Q^{\prime}\in\Lambda$ such that for every integer $i\in\left\{1,\dots,p\right\}$, we have $\epsilon\left(Q_{i}\right)<\epsilon\left(Q^{\prime}\right)$. So $\epsilon\left(Q_{p}\right)<\epsilon\left(Q^{\prime}\right)$ and we use Proposition 2.12 again to construct a sequence of optimal (possibly limit) immediate successors which begins at $Q_{p}$ and ends at $Q^{\prime}$. So we have a sequence $Q_{1}=x,\dots,Q_{r}=Q^{\prime}$ of optimal (possibly limit) immediate successors which begins at $x$ and ends at $Q^{\prime}$.

We iterate the process until the sequence $\left\{\epsilon(Q_{i})\right\}$ is cofinal in $\epsilon(\Lambda)$. If $Q_{i}$ is maximal among the set of key polynomials of degree $\deg_{X}\left(Q_{i}\right)$, then $\deg_{X}\left(Q_{i}\right)<\deg_{X}\left(Q_{i+1}\right)$. If $Q_{i}<_{\lim}Q_{i+1}$, we have again $\deg_{X}\left(Q_{i}\right)<\deg_{X}\left(Q_{i+1}\right)$. In fact, the degree of the polynomials of the sequence stricly increase at least each two steps, so the process stops. ∎

Proposition 5.6**.**

Assume that $k=k_{\nu}$. There exists a finite or infinite sequence $(Q_{i})_{i\geq 1}$ of key polynomials such that for each $i$ the polynomial $Q_{i+1}$ is either an optimal or a limit immediate successor of $Q_{i}$ and such that the sequence $\left\{\epsilon(Q_{i})\right\}$ is cofinal in $\epsilon(\Lambda)$ where $\Lambda$ is the set of key polynomials.

And this sequence is such that: if $Q_{i}<Q_{i+1}$, then the $Q_{i}$-expansion of $Q_{i+1}$ has exactly two terms.

Proof.

We have $Q_{1}=x$, and we assume that $Q_{1},Q_{2},\dots,Q_{i}$ have been constructed. We note $a:=\deg_{x}\left(Q_{i}\right)$ and recall that

[TABLE]

If $Q_{i}$ is maximal in $\Lambda$, we stop. Otherwise, $Q_{i}$ is not maximal and so it has an immediate successor.

We set $\alpha:=\min\left\{h\in\mathbb{N}^{\ast}\text{ such that }h\nu\left(Q_{i}\right)\in\Delta_{<a}\right\}$ where $\Delta_{<a}$ is the subgroup of $\Gamma$ generated by the values of the elements of $G_{<a}$.

In fact, there exists a polynomial $f$ of degree strictly less than $a$ such that $\alpha\nu\left(Q_{i}\right)=\nu\left(Q_{i}^{\alpha}\right)=\nu\left(f\right)\neq 0$.

Then, since $k_{\nu}=k$, there exists $c\in k^{\ast}$ such that $\mathrm{in}_{\nu}\left(Q_{i}^{\alpha}\right)=\mathrm{in}_{\nu}\left(cf\right)$.

We set $Q=Q_{i}^{\alpha}-cf$. By the proof of Proposition 2.5, we have $\epsilon\left(Q_{i}\right)<\epsilon\left(Q\right)$.

Let us show that $Q_{i}<Q$. We only have to show that $Q$ is of minimal degree. So let us set $P$ a key polynomial such that $\epsilon\left(Q_{i}\right)<\epsilon\left(P\right)$.

Assume by contradiction that $\deg\left(P\right)<a\alpha$. We set $P=\sum\limits_{j=0}^{\alpha-1}p_{j}Q_{i}^{j}$ the $Q_{i}$-expansion of $P$. Then by the proof of Proposition 2.5, we have $\sum\limits_{j=0}^{\alpha-1}\mathrm{in}_{\nu}\left(p_{j}\right)\mathrm{in}_{\nu}\left(Q_{i}\right)^{j}=0$, which contradicts the minimality of $\alpha$.

Then $Q$ is of minimal degree and $Q_{i}<Q$. Since it has just two terms in his $Q_{i}$-expansion, it is an optimal immediate successor of $Q_{i}$.

First case: $\alpha>1$. Then we set $Q_{i+1}:=Q$ and we iterate.

Second case: $\alpha=1$. Then all the elements of $M_{Q_{i}}$ have same degree than $Q_{i}$. If $M_{Q_{i}}$ does not have a maximal element, then we do the same thing than in the proof of Proposition 2.12 and we set $Q_{i+1}$ a limit immediate successor of $Q_{i}$.

Otherwise, $M_{Q_{i}}$ has a maximal element $Q_{i+1}$. This element has same degree as $Q_{i}$, so we have $Q_{i+1}=Q_{i}-h$ with $h$ of degree strictly less than the degree of $Q_{i}$. Then it is an immediate successor of $Q_{i}$ which $Q_{i}$-expansion admits uniquely two terms. So it is optimal, and this completes the proof. ∎

We now assume $k=k_{\nu}$ and consider $\mathcal{Q}:=\left(Q_{i}\right)_{i}$ a sequence of optimal (possibly limit) immediate successors such that $\left(\epsilon\left(Q_{i}\right)\right)_{i}$ is cofinal in $\epsilon\left(\Lambda\right)$ and such that if $Q_{i}<Q_{i+1}$, then the $Q_{i}$-expansion of $Q_{i+1}$ admits exactly two terms.

Remark 5.7*.*

We keep the same hypothesis as in Example 1.8. Then $\mathcal{Q}=\left\{z,Q\right\}$.

Corollary 5.8**.**

For every polynomial $f$, there exists an index $i$ such that $\nu_{Q_{i}}(f)=\nu(f)$.

Proof.

By Proposition 1.21, there exists a key polynomial$Q$ such that $\nu_{Q}(f)=\nu(f)$.

The sequence $\left\{\epsilon(Q_{i})\right\}$ being cofinal, there exists an index $i$ such that

[TABLE]

By Proposition 1.20, $\nu_{Q}(f)\leq\nu_{Q_{i}}(f)$ and since $\nu_{Q}(f)=\nu(f)$, we have $\nu_{Q_{i}}(f)=\nu(f)$. ∎

Remark 5.9*.*

So, for every polynomial $f$, there exists a key polynomial $Q_{i}$ of the sequence $\mathcal{Q}$ such that $f$ is non-degenerate with respect to $Q_{i}$.

Remark 5.10*.*

Let $Q_{i}\in\mathcal{Q}$. We don’t assume here $k=k_{\nu}$.

We set $a_{i}:=\deg_{x}\left(Q_{i}\right)$ and $\Gamma_{<a_{i}}$ the group $\nu\left(G_{<a_{i}}\setminus\left\{0\right\}\right)$.

If $\nu\left(Q_{i}\right)\notin\Gamma_{<a_{i}}\otimes_{\mathbb{Z}}\mathbb{Q}$, then $\epsilon\left(Q_{i}\right)$ is maximal in $\epsilon\left(\Lambda\right)$ and the sequence $\mathcal{Q}$ stops at $Q_{i}$.

6. Monomialization of

the key polynomials.

We set $K:=k\left(u_{1},\dots,u_{n-1}\right)$ and we consider the extension $K(u_{n})$. We consider also a sequence of key polynomials $\mathcal{Q}$ as in the section 5.

In other words, $\mathcal{Q}=\left(Q_{i}\right)_{i}$ is a sequence of optimal (possibly limit) immediate successors such that $\left(\epsilon\left(Q_{i}\right)\right)_{i}$ is cofinal in $\epsilon\left(\Lambda\right)$.

Let $f$ be an element of $R$. We know that this element is non-degenerate with respect to a key polynomial of the sequence $\mathcal{Q}$. We also know that every element non-degenerate with respect to a regular system of parameters is monomializable.

Then, to monomialize $f$, it is enough to monomialize the set of key polynomials of this sequence. We assume in this part that the residue field is $k$.

6.1. Generalities.

Let $r:=r\left(R,u,\nu\right)$ be the dimension of

[TABLE]

in $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}$. Renumbering, if necessary, we can assume that $\nu\left(u_{1}\right),\dots,\nu\left(u_{r}\right)$ are rationally independent and we consider $\Delta$ the subgroup of $\Gamma$ generated by $\nu(u_{1}),\ldots,\nu(u_{r})$.

Remark 6.1*.*

Let $\left(R,u\right)\to\left(R_{1},u^{\left(1\right)}\right)$ be a framed blow-up. Then $r\leq r_{1}:=r\left(R_{1},u^{\left(1\right)},\nu\right)$.

Remark 6.2*.*

We will consider the framed local blow-ups

[TABLE]

Then we write $r_{i}:=r\left(R_{i},u^{\left(i\right)},\nu\right)$.

We set $E:=\left\{1,\dots,r,n\right\}$ and $\overline{\alpha^{(0)}}:=\min\limits_{h\in\mathbb{N}^{\ast}}\left\{h\text{ such that }h\nu(u_{n})\in\Delta\right\}.$

So $\overline{\alpha^{(0)}}\nu(u_{n})=\sum\limits_{j=1}^{r}\alpha_{j}^{(0)}\nu(u_{j})$ with, renumbering the $\alpha_{i}^{(0)}$ if necessary,

[TABLE]

and

[TABLE]

We set

[TABLE]

and

[TABLE]

with $t=n-r-1$.

We set $x_{i}=\mathrm{in}_{\nu}u_{i}$ , and we have that $x_{1},\ldots,x_{r}$ are algebraically independent over $k$ in $G_{\nu}$. Let $\lambda_{0}$ be the minimal polynomial of $x_{n}$ over $k\left(x_{1},\ldots,x_{r}\right)$, of degree $\alpha$.

We set:

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

We have

[TABLE]

where $c_{0}\in k$ , and $Z+c_{0}$ is the minimal polynomial $\lambda_{z}$ of $z$ over $\mathrm{gr}_{\nu}k\left(x_{1},\dots,x_{r}\right)$.

Indeed, $k_{\nu}\simeq k\simeq\frac{k\left[Z\right]}{\left(\lambda_{z}\right)}$ so $\lambda_{z}$ is of degree $1$ in $Z$. Then $\lambda_{0}$ is of degree $\overline{\alpha^{(0)}}$, and so $\alpha=\overline{\alpha^{\left(0\right)}}$.

Definition 6.3**.**

We say that $Q_{i}$ is monomializable if there exists a sequence of blow-ups $\left(R,u\right)\to\left(R_{l},u^{\left(l\right)}\right)$ such that in $R_{l}$, $Q_{i}$ can be written as $u_{n}^{\left(l\right)}$ multiplied by a monomial in $\left(u_{1}^{\left(l\right)},\dots,u_{r_{l}}^{\left(l\right)}\right)$ up to a unit of $R_{l}$, where $r_{l}:=r\left(R_{l},u^{\left(l\right)},\nu\right)$.

We are going to show that there exists a local framed sequence that monomializes all the $Q_{i}$.

We have $Q_{1}=u_{n}$, it is a monomial. By the blow-ups, $Q_{1}$ stays a monomial. So we have to begin monomializing $Q_{2}$.

Since we want to monomialize the key polynomials $Q_{i}$ of the sequence $\mathcal{Q}$ constructed earlier by induction on $i$, we are going to do something more general here: we consider an immediate successors (possibly limit) key element $Q_{2}$ of $Q_{1}$ instead of immediate successor (possibly limit) key polynomial of $Q_{1}$.

First, let us consider

[TABLE]

where $b_{0}\in R$ such that $b_{\text{0}}\equiv c_{0}$ modulo $\mathfrak{m}$ and $a_{0}\in R^{\times}$.

A priori, $Q$ is not a key polynomial but we are going to prove that we can reduce this case to the case $Q_{2}=Q$ by a local framed sequence independent of $u_{n}$.

6.2. Puiseux packages.

Let

[TABLE]

and

[TABLE]

We have

[TABLE]

and

[TABLE]

So $\frac{w^{\delta}}{w^{\gamma}}=\frac{w_{n}^{\alpha}}{\prod\limits_{j=1}^{r}w_{j}^{\alpha_{j}^{(0)}}}=\overline{z}$.

Let us compute the value of $w^{\delta}$.

[TABLE]

Theorem 6.4**.**

There exists a local framed sequence

[TABLE]

with respect to $\nu$, independent of $v$, and that has the next properties:

For every integer $i\in\{1,\dots,l\}$, we write $u^{(i)}:=\left(u_{1}^{(i)},\dots,u_{n}^{(i)}\right)$ and we recall that $k$ is the residue field of $R_{i}$.

1. (1)

The blow-ups $\pi_{0},\dots,\pi_{l-2}$ are monomials. 2. (2)

We have $\overline{z}\in R_{l}^{\times}$. 3. (3)

We set $u^{\left(l\right)}:=\left(w_{1}^{(l)},\dots,w_{r}^{(l)},v,w_{n}^{(l)}\right)$. So for every integer $j\in\left\{1,\dots,r,n\right\}$, $w_{j}$ is a monomial in $w_{1}^{(l)},\dots,w_{r}^{(l)}$ multiplied by an element of $R_{l}^{\times}$. And for every integer $j\in\left\{1,\dots,r\right\}$, $w_{j}^{(l)}=w^{\eta}$ where $\eta\in\mathbb{Z}^{r+1}$. 4. (4)

We have $Q=w_{n}^{(l)}\times\overline{y}$.

Proof.

We apply Proposition 4.4 to $(w^{\delta},w^{\gamma})$ and so we obtain a local framed sequence for $\nu$, independent of $v$ and such that $w^{\gamma}\mid w^{\delta}$ in $R_{l}$.

By Proposition 4.7 and the fact that $w^{\delta}$ and $w^{\gamma}$ have same value, we have that $w^{\delta}\mid w^{\gamma}$ in $R_{l}$. In fact $\overline{z},\overline{z}^{-1}\in R_{l}^{\times}.$ So we have $(2)$.

We choose the local sequence to be minimal, in other words the sequence made by $\pi_{0},\ldots,\pi_{l-2}$ does not satisfy the conclusion of the Proposition 4.4 for $(w^{\delta},w^{\gamma})$. Now we are going to prove that this sequence satisfies the five properties of Theorem 6.4. Let $i\in\left\{0,\dots,l\right\}$. We write $w^{(i)}=\left(w_{1}^{(i)},\dots,w_{r}^{(i)},w_{n}^{(i)}\right)$, with $r=n-t-1$ and define $J_{i}$, $A_{i}$, $B_{i}$, $j_{i}$ and $D_{i}$ similarly that we defined $J$, $A$, $B$, $j$ and $D_{1}$, considering the $i$-th blow-up.

Since $D_{i}\subset\left\{1,\dots,n\right\}$, we have $\sharp D_{i}\leq n$. Hence $\sharp(A_{i}\cup(B_{i}\cup\left\{j_{i}\right\}))\leq n$, so $\sharp A_{i}+\sharp B_{i}+1\leq n$. As the sequence is independent of $v$, this implies that $T\subset A_{i}$, and so $\sharp T\leq\sharp A_{i}$. Then $\sharp T+1+\sharp B_{i}\leq n$, so $t+1\leq n$, and so $r\geq 0$. By the minimality of the sequence, we know that if $i<l$, $w^{\delta}\nmid w^{{}^{\gamma}}$ in $R_{i}$, and so $\sharp B_{i}\neq 0$, hence $r>0$.

For every integers $i\in\left\{1,\dots,l\right\}$ and $j\in\left\{1,\dots,n\right\}$, we set $\beta_{j}^{(i)}=\nu\left(u_{j}^{(i)}\right)$. For each $i<l$, $\pi_{i}$ is a blow-up along an ideal of the form $\left(u_{J_{i}}^{(i)}\right)$. Renumbering if necessary, we may assume that $1\in J_{i}$ and that $R_{i+1}$ is a localisation of $R_{i}\left[\frac{u_{J_{i}}^{(i)}}{u_{1}^{(i)}}\right]$. So we have $\beta_{1}^{(i)}=\min\limits_{j\in J_{i}}\left\{\beta_{j}^{(i)}\right\}$.

Fact 6.5**.**

Let $X=\left(x_{1},\dots,x_{n}\right)\in\mathbb{Z}^{n}$ be a vector whose elements are relatively prime. Then there exists a matrix $A\in\mathrm{SL}_{n}(\mathbb{Z})$ of determinant $1$ such that $X$ is the first line of $A$.

Proof.

This proof is made by induction on $n$ and using Bezout theorem. ∎

Lemma 6.6**.**

Let $i\in\left\{0,\dots,l-1\right\}$. We assume that the sequence $\pi_{0},\dots,\pi_{i-1}$ of 6.1 is monomial.

We set $w^{\gamma}=\left(w^{(i)}\right)^{\gamma^{(i)}}$ and $w^{\delta}=\left(w^{(i)}\right)^{\delta^{(i)}}$. Then:

1. (1)

[TABLE] 2. (2)

$\mathrm{pgcd}\left(\gamma_{1}^{(i)}-\delta_{1}^{(i)},\dots,\gamma_{r}^{(i)}-\delta_{r}^{(i)},\gamma_{n}^{(i)}-\delta_{n}^{(i)}\right)=1,$ 3. (3)

Every $\mathbb{Z}$-linear dependence relation between $\beta_{1}^{(i)},\dots,\beta_{r}^{(i)},\beta_{n}^{(i)}$ is an integer multiple of (6.2).

Proof.

1. (1)

We have $\nu\left(w^{\gamma}\right)=\nu\left(w^{\delta}\right)$, hence $\nu\left(\left(w^{(i)}\right)^{\gamma^{(i)}}\right)=\nu\left(\left(w^{(i)}\right)^{\delta^{(i)}}\right)$. So, since $w^{(i)}=\left(w_{1}^{(i)},\dots,w_{r}^{(i)},w_{n}^{(i)}\right)$:

[TABLE]

in other words

[TABLE]

By definition of $w^{(i)}$, for every integer $j\in\left\{1,\dots,r,n\right\}$, we have $w_{j}^{(i)}=u_{j}^{(i)}$, so $\nu(w_{j}^{(i)})=\beta_{j}^{(i)}$. Then:

[TABLE]

Then $\sum\limits_{j\in\left\{1,\dots,r,n\right\}}\left(\gamma_{j}^{(i)}-\delta_{j}^{(i)}\right)\beta_{j}^{(i)}=0$. 2. (2)

We do an induction. Case $i=0$.

We have

[TABLE]

By definition

[TABLE]

and

[TABLE]

So $\mathrm{pgcd}\left(\alpha_{1}^{(0)},\dots,\alpha_{s}^{(0)},\alpha_{s+1}^{(0)},\dots,\alpha_{r}^{(0)},-\alpha\right)=1$.

Case $i>0$. We assume the result shown at the previous rank. We have $\gamma^{(i)}=\gamma^{(i-1)}G^{(i)}$, $\delta^{(i)}=\delta^{(i-1)}G^{(i)}$ and $\beta^{(i)}=\beta^{(i-1)}F^{(i)}$ where $F^{(i)}=\left(G^{(i)}\right)^{-1}$ and $G^{(i)}\in\mathrm{SL}_{r+1}(\mathbb{Z})$ such that

[TABLE]

So $\left(\gamma^{(i)}-\delta^{(i)}\right)=\left(\gamma^{(i-1)}-\delta^{(i-1)}\right)G^{(i)}=\left(\gamma-\delta\right)G$ where $G$ is a product of unimodular matrixes, and so $G$ is unimodular.

By the case $i=0$, $\left(\gamma-\delta\right)$ is a vector whose elements are relatively prime. By 6.5 this vecteur can be complete as a base of $\mathbb{Z}^{r+1}$, which, by a unimodular matrix, stay a base of $\mathbb{Z}^{r+1}$. The vector $\left(\gamma^{(i)}-\delta^{(i)}\right)$ is then a vector of this base, so its elements are relatively prime. 3. (3)

Case $i=0$ is the fact that $\beta_{1},\dots,\beta_{r},\beta_{n}$ generate a vector space of dimension $r$.

Let

[TABLE]

But $\alpha\beta_{n}=\sum\limits_{j=1}^{r}\alpha_{j}^{(0)}\beta_{j}$, so:

[TABLE]

Since $\beta_{1},\dots,\beta_{r}$ are $\mathbb{Q}$-linearly independent elements, we have that $Z$ is a free $\mathbb{Z}$-module of rank $1$, so it is generated by a unique vector. By point $\left(1\right)$, the vector $(\gamma-\delta)$ is in $Z$, and by point $(2)$, it is composed of relatively prime elements. This vector generates the free $\mathbb{Z}$-module of rank $1$.

Let $i>0$. We already know that $\beta^{(i)}=\beta^{(i-1)}F^{(i)}=\beta F$ where $F$ is a unimodular matrix, so an automorphism of $\mathbb{Z}^{r}$.

Let

[TABLE]

So

[TABLE]

then

[TABLE]

Then the set $Z^{(i)}$ is a free $\mathbb{Z}$-module of rank $1$ by the case $i=0$. And we know by (3) that the vector $\left(\gamma^{(i)}-\delta^{(i)}\right)$ is a vector of $Z^{(i)}$ composed of relatively prime elements, so it generates $Z^{(i)}$. This completes the proof.

∎

Lemma 6.7**.**

The sequence (6.1) is not monomial.

Proof.

Assume, aiming for contradiction, that it is. By induction on $i$, we have $r_{i}=r$ for every $i\in\left\{0,\dots,l\right\}$. We know that $w^{(l)}$ is a regular system of parameters of $R_{l}$ and that $w^{\delta}$ and $w^{\gamma}$ divide each other in $R_{l}$.

We saw that

[TABLE]

and

[TABLE]

So $\delta^{(l)}=\gamma^{(l)}$ .

But $\left(\gamma^{(l)}-\delta^{(l)}\right)=\left(\gamma-\delta\right)G$ where $G$ is a unimodular matrix, hence $\gamma=\delta$, which is a contradiction. ∎

Lemma 6.8**.**

Let $i\in\{0,\dots,l-1\}$ and assume $\pi_{0},\dots,\pi_{i-1}$ are all monomials. Then the following assertions are equivalent:

1. (1)

The blow-up $\pi_{i}$ is not monomial. 2. (2)

There exists a unique index $q\in J_{i}\setminus\left\{1\right\}$ such that $\beta_{q}^{(i)}=\beta_{1}^{(i)}$. 3. (3)

We have $i=l-1$.

Proof.

$(3)\Rightarrow(1)$ by Lemma 6.7.

$(1)\Rightarrow(2)$ First, we prove the existence. We have$\beta_{1}^{(i)}=\min\limits_{j\in J_{i}}\left\{\beta_{j}^{(i)}\right\}$. So $\pi_{i}$ monomial $\Leftrightarrow B_{i}=J_{i}\setminus\left\{1\right\}\Leftrightarrow\beta_{q}^{(i)}>\beta_{1}^{(i)}$ for every $q\in J_{i}\setminus\left\{1\right\}$.

Since the blow-up is not monomial by hypothesis, there exists $q\in J_{i}\setminus\left\{1\right\}$ such that $\beta_{q}^{(i)}=\beta_{1}^{(i)}$.

Now let us show the unicity. Assume, aiming for contradiction, that there exist two different indexes $q$ and $q^{\prime}$ in $J_{i}\setminus\left\{1\right\}$ such that $\beta_{q}^{(i)}-\beta_{1}^{(i)}=0$ and $\beta_{q^{\prime}}^{(i)}-\beta_{1}^{(i)}=0$ .

Then we have two linear dependence relations between $\beta_{1}^{(i)},\dots,\beta_{r}^{(i)}$ and the element $\beta_{n}^{(i)}$, which are not linearly dependent. It is a contradiction by point $(4)$ of Lemma 6.6.

$(2)\Rightarrow(3)$

By Remark 4.6, we write $w_{1}^{(i)}=w^{\epsilon}$ and $w_{q}^{(i)}=w^{\mu}$ where $\epsilon$ and $\mu$ are two colons of an unimodular matrix. Then $\epsilon-\mu$ is unimodular, so its total pgcd is one.

So

[TABLE]

and

[TABLE]

By hypothesis, $\beta_{q}^{(i)}=\beta_{1}^{(i)}$. Then $\sum\limits_{s\in E}\left(\mu_{s}-\epsilon_{s}\right)\beta_{s}=0$ and by points $(3)$ and $(4)$ of Lemma 6.6, and the fact that the total pgcd of $\mu-\epsilon$ is one, we have $\mu-\epsilon=\pm(\gamma-\delta)$.

So $\frac{w_{q}^{(i)}}{w_{1}^{(i)}}=w^{\epsilon-\mu}=w^{\pm(\gamma-\delta)}=\overline{z}^{\pm 1}$, then either $\overline{z}\in R_{i+1}$ or $\overline{z}^{-1}\in R_{i+1}$.

To show that $i=l-1$, we are going to show that $i+1=l$ . And to do this, we are going to use the fact that $l$ has been chosen minimal such that $\overline{z}\in R_{l}^{\times}$. So let us show that $\overline{z}\in R_{i+1}^{\times}$.

Since $\overline{z}\in R_{i+1}$ or $\overline{z}^{-1}\in R_{i+1}$, we know that $w^{\delta}\mid w^{\gamma}$ in $R_{i+1}$ or the converse. By Proposition 4.7 and the fact that $w^{\delta}$ and $w^{\gamma}$ have same value, we have $w^{\delta}\mid w^{\gamma}$ in $R_{i+1}$ if and only if the converse is true. So $\overline{z}\in R_{i+1}^{\times}$, and the proof is complete. ∎

Doing an induction on $i$ and using Lemma 6.8, we conclude that $\pi_{0},\dots,\pi_{l-2}$ are monomials. So we have the first point of Theorem 6.4.

It remains to show the points $(3)$ and $(4)$.

By Lemma 6.8 there exists a unique element $q\in J_{l-1}\setminus\left\{j_{l-1}\right\}$ such that $\beta_{q}^{(l-1)}=\beta_{1}^{(l-1)}$, so we are in the case $\sharp B_{l-1}+1=\sharp J_{l-1}-1$. Now we have to see if we are in the case $t_{k_{l-1}}=0$ or in the case $t_{k_{l-1}}=1$.

We recall that $w_{1}^{(l-1)}=w^{\epsilon}$ and $w_{q}^{(l-1)}=w^{\mu}$ where $\epsilon$ and $\mu$ are two colons of a unimodular matrix such that $\mu-\epsilon=\pm(\gamma-\delta)$. So we have $x_{1}^{(l-1)}=x^{\epsilon}$ and $x_{q}^{(l-1)}=x^{\mu}$, then

[TABLE]

In other words

[TABLE]

Replacing $x_{1}^{(l-1)}$ and $x_{q}^{(l-1)}$ if necessary, we may assume $\frac{x_{q}^{(l-1)}}{x_{1}^{(l-1)}}=z$ .

Since $\beta_{1}^{(l-1)},\dots,\beta_{r}^{(l-1)}$ are linearly independent, we have $q=n$.

We recall that $\lambda_{0}=X^{\alpha}+c_{0}y$ where $c_{0}\in k$ , and $Z+c_{0}$ is the minimal polynomial $\lambda_{z}$ of $z$ on $\mathrm{gr}_{\nu}k\left(x_{1},\dots,x_{r}\right)$. By 3.9, we have

[TABLE]

Remark 6.9*.*

We know that $\overline{\lambda_{0}}\left(\overline{z}\right)=\overline{z}+b_{0}g_{0}$ where $g_{0}$ is a unit and$b_{0}\in R$ such that $b_{\text{0}}\equiv c_{0}$ modulo $\mathfrak{m}$. Then we choose $g_{0}=a_{0}$.

But $\overline{z}=\frac{w_{n}^{\alpha}}{\overline{y}}$, so

[TABLE]

as desired in point $(4)$.

Let us show the point $(3)$. We apply Proposition 4.5 at $i=0$ and $i^{\prime}=l$. By the monomiality of $\pi_{0},\dots,\pi_{l-2}$, we know that $D_{i}=\{1,\dots,n\}$ for each $i\in\{1,\dots,l-1\}$, and we know that $D_{l}=\{1,\dots,n\}$. We set $u_{T}=v$.

For every $j\in\left\{1,\dots,r,n\right\}$, the fact that $w_{j}=u_{j}$ is a monomial in $w_{1}^{(l)},\dots,w_{r}^{(l)}$, in other words in $u_{1}^{(l)},\dots,u_{r}^{(l)}$, multiplied by an element of $R_{l}^{\times}$ is a consequence of Proposition 4.5.

The fact that for every integer $j\in\{1,\dots,r\}$, we have $w_{j}^{(l)}=w^{\eta}$ is a consequence of the same Proposition. This completes the proof. ∎

Remark 6.10*.*

In the case $Q_{2}=Q$, we monomialized $Q_{2}$ as desired.

Definition 6.11**.**

[24] A local framed sequence that satisfies Theorem 6.4 is called a* $n$-*Puiseux package.

Let $j\in\left\{r+1,\dots,n\right\}$. A *$j$-Puiseux *package is a $n$-Puiseux package replacing $n$ by $j$ in Theorem 6.4.

Lemma 6.12**.**

Let $P=u_{n}^{\alpha}+c_{0}$ be the $u_{n}$-expansion of an immediate successor key element of $u_{n}$.

There exists a local framed sequence $\left(R,u\right)\to\left(R_{l},u^{(l)}\right)$, independant of $u_{n}$, that transforms $c_{0}$ in a monomial in $\left(u_{1}^{(l)},\dots,u_{r}^{(l)}\right)$, multiplied by a unit of $R_{l}$.

In particular, after this local framed sequence, the element $P$ is of the form $w_{n}^{\alpha}+a_{0}b_{0}\overline{y}$.

Proof.

We will prove this Lemma in a more general version in Lemma 6.16. ∎

Corollary 6.13**.**

Let $P$ be an immediate successor key element of $u_{n}$. Then $P$ is monomializable.

Proof.

If $u_{n}\ll P$, we use Lemma 6.12 to reduce to the case $P=w_{n}^{\alpha}+a_{0}b_{0}\overline{y}.$ By Theorem 6.4, we can monomialize $P$. ∎

Let $G$ be a local ring essentially of finite type over $k$ of dimension strictly less than $n$ that is equipped with a valuation centered on $G$.

Theorem 6.14**.**

Assume that for every ring $G$ as above, every element of $G$ is monomializable.

We recall that $\mathrm{car}\left(k_{\nu}\right)=0$. If $u_{n}\ll_{\lim}P$, then $P$ is monomializable.

Proof.

We write $P=\sum\limits_{j=0}^{N}b_{j}a_{j}u_{n}^{j}$ the $u_{n}$-expansion of $P$, with $a_{j}\in R^{\times}$ and $Q=\sum\limits_{j=0}^{N}b_{j}u_{n}^{j}$ a limite immediate successor of $u_{n}$.

By Theorem 2.17, we have $\delta_{u_{n}}\left(Q\right)=1$. Then:

[TABLE]

for every $j>1$.

The elements $a_{i}$ are units of $R$, so for every $j>1$ we have:

[TABLE]

In fact, $\nu\left(a_{1}b_{1}\right)<\nu\left(a_{0}b_{0}\right)$ and by hypothesis, after a sequence of blow-ups independent of $u_{n}$, we can monomialize $a_{j}b_{j}$ for every index $j$, and assume that $a_{1}b_{1}\mid a_{0}b_{0}$ by Proposition 4.7.

Then

[TABLE]

So $\nu\left(b_{0}\right)<\left(b_{j}\right)+j\left(\nu\left(b_{0}\right)-\nu\left(b_{1}\right)\right).$

In fact, $\nu\left(b_{1}^{j}\right)<\nu\left(b_{j}b_{0}^{j-1}\right)$. So after a sequence of blow-ups independent of $u_{n}$, we have $b_{1}^{j}\mid b_{j}b_{0}^{j-1}$. After a $n$-Puiseux package $\left(\ast\right)$ $\left(R,u\right)\to\dots\to\left(R^{\prime},u^{\prime}\right)$ in the special case $\alpha=1$, we obtain $P=\sum\limits_{j=0}^{N}b^{\prime}_{j}\left(u^{\prime}_{n}\right)^{j}$ with $b^{\prime}_{1}\mid b^{\prime}_{j}$ for every index $j$ with $u^{\prime}_{n}=\frac{b_{1}u_{n}}{b_{0}}+1$.

In fact, $\frac{P}{b^{\prime}_{1}}=u^{\prime}_{n}+\varphi$ with $\varphi\in\left(u^{\prime}_{1},\dots,u^{\prime}_{n-1}\right)$. So $u^{\prime\prime}:=\left(u^{\prime}_{1},\dots,u^{\prime}_{n-1},\frac{P}{b^{\prime}_{1}}\right)$ is a regular system of parameters of $R^{\prime}$. Then, the sequence $\left(R,u\right)\to\dots\to\left(R^{\prime},u"\right)$ given by $\left(\ast\right)$ changing uniquely the last parameter $u^{\prime}_{n}$ after the last blow-up is still a local framed sequence. So $P$ is monomializable. ∎

Remark 6.15*.*

Since $Q_{2}$ is an immediate successor (possibly limit) of $u_{n}$, this is in particular an immediate successor (possibly limit) key element of $u_{n}$. By Corollary 6.13, or Theorem 6.14, it is monomializable modulo Lemma 6.12.

6.3. Generalization.

Now we monomialized $Q_{2}$, but we want to monomialize every key polynomial of the sequence $\mathcal{Q}$. Here the key elements will be useful. Indeed, modified by the blow-ups which monomialized $Q_{2}$, we cannot know if $Q_{3}$ is still a key polynomial.

To be more general, we will show that if $Q_{i}\in\mathcal{Q}$ is monomializable, then $Q_{i+1}$ is monomializable.

Assume that the polynomial $Q_{i}$ is monomializable after a sequence of blow-ups $\left(R,u\right)\to\left(R_{l},u^{(l)}\right)$.

Let $\Delta_{l}$ be the group $\nu\left(k\left(u_{1}^{\left(l\right)},\dots,u_{n-1}^{\left(l\right)}\right)\setminus\left\{0\right\}\right)$. We set

[TABLE]

We set $X_{j}=\mathrm{in}_{\nu}\left(u_{j}^{(l)}\right)$, $W_{j}=w_{j}^{(l)}$ and $\lambda_{l}$ the minimal polynomial of $X_{n}$ over $\mathrm{gr}_{\nu}k\left(u_{1}^{\left(l\right)},\dots,u_{n-1}^{\left(l\right)}\right)$ of degree $\alpha_{l}$.

Since $k=k_{\nu}$, there exists $c_{0}\in\mathrm{gr}_{\nu}k\left(u_{1}^{\left(l\right)},\dots,u_{n-1}^{\left(l\right)}\right)$ such that

[TABLE]

Furthermore, we have $Q_{i}=\overline{\omega}w_{n}^{(l)}$ with $\overline{\omega}$ a monomial in $W_{1},\dots,W_{r_{l}}$ multiplied by a unit. We set $\omega:=\mathrm{in}_{\nu}\left(\overline{\omega}\right)$.

We know that $Q_{i+1}$ is an optimal immediate successor of $Q_{i}$, so we denote by

[TABLE]

the $Q_{i}$-expansion of $Q_{i+1}$ in $k\left(u_{1},\dots,u_{n-1}\right)\left[u_{n}\right]$ by Proposition 5.6 with $c_{0}=\mathrm{in}_{\nu}\left(b_{0}\right)$.

Since $Q_{i}=\overline{\omega}W_{n}$ and $Q_{i+1}=Q_{i}^{\alpha_{l}}+b_{0}$, we have

[TABLE]

We know that both terms of the $Q_{i}$-expansion of $Q_{i+1}$ have same value. So these two terms are divisible by the same power of $\overline{\omega}$ after a suitable sequence of blow-ups $(\ast_{i})$ independent of $u_{n}^{(l)}$.

We denote by $\widetilde{Q}_{i+1}$ the strict transform of $Q_{i+1}$ by the composition of $\left(\ast_{i}\right)$ with the sequence of blow-ups $\left(\ast^{\prime}_{i}\right)$ that monomialize $Q_{i}$. We denote this composition by $\left(c_{i}\right)$. We write $\left(R,u\right)\overset{\left(c_{i}\right)}{\to}\left(R_{l},u^{(l)}\right)$.

We know that $\widetilde{Q}_{i}$, the strict transform of $Q_{i}$ by $\left(c_{i}\right)$, is a regular parameter of $R_{l}$. Indeed, by Proposition 4.5, we know that every $u_{j}$ of $R$ can be written as a monomial in $w_{1}^{(l)},\dots,w_{r_{l}}^{(l)}$. In fact, the reduced exceptional divisor of this sequence of blow-ups is exactly $V\left(\overline{\omega}\right)_{\mathrm{red}}$. Then, since $Q_{i}=W_{n}\overline{\omega}$, we have that the strict transform of $Q_{i}$ is $\widetilde{Q}_{i}=W_{n}=w_{n}^{(l)}=u_{n}^{(l)}$. So it is a key polynomial in the extension $k\left(u_{1}^{(l)},\dots,u_{n-1}^{(l)}\right)\left(u_{n}^{(l)}\right)$.

Let us show that $\widetilde{Q}_{i+1}=\frac{Q_{i+1}}{\overline{\omega}^{\alpha_{l}}}$.

We have

[TABLE]

and also $u_{n}^{(l)}\nmid\overline{\omega}$. Thus $\overline{\omega}^{\alpha_{l}}$ divides $Q_{i}^{\alpha_{l}}$ and all the non-zero terms of the $Q_{i}$-expansion of $Q_{i+1}$. Furthermore, it is the greatest power of $\overline{\omega}$ that divides all the terms, so $\frac{Q_{i+1}}{\overline{\omega}^{\alpha_{l}}}$ is $\widetilde{Q}_{i+1}$, the strict transform of $Q_{i+1}$ by the sequence of blow-ups.

Let $G$ be a local ring essentially of finite type over $k$ of dimension strictly less than $n$ equipped with a valuation centered in $G$ whose residue field is $k$.

Lemma 6.16**.**

Assume that for every ring $G$ as above, every element of $G$ is monomializable.

Assume that $Q_{i}<Q_{i+1}$ in $\mathcal{Q}$.

There exists a local framed sequence $\left(R_{l},u^{(l)}\right)\to\left(R_{e},u^{(e)}\right)$ such that in $R_{e}$, the strict transform of $Q_{i+1}$ is of the form $\left(u_{n}^{\left(e\right)}\right)^{\alpha_{l}}+\tau_{0}\eta$, where $\tau_{0}\in R_{e}^{\times}$ and $\eta$ is a monomial in $u_{1}^{\left(e\right)},\dots,u_{r_{e}}^{\left(e\right)}$.

Proof.

By hypothesis, after a sequence of blow-ups independent of $u_{n}^{(l)}$, we can monomialize $b_{0}$ and assume that it is a monomial in $\left(u_{1}^{(l)},\dots,u_{n-1}^{(l)}\right)$ multiplied by a unit of $R_{l}$.

For every $g\in\left\{r_{l}+1,\dots,n-1\right\}$, we do a $g$-Puiseux package, and then we have a sequence

[TABLE]

such that every $u_{g}^{(l)}$ is a monomial in $\left(u_{1}^{(t)},\dots,u_{r_{t}}^{(t)}\right)$.

In fact, we can assume that $b_{0}$ is a monomial in $\left(u_{1}^{(l)},\dots,u_{r_{l}}^{(l)}\right)$ multiplied by a unit of $R_{l}$.

Since the strict transform $\widetilde{Q}_{i+1}=\left(u_{n}^{(l)}\right)^{\alpha_{l}}+\frac{b_{0}}{\overline{\omega}^{\alpha_{l}}}$ is an immediate successor key element of $\widetilde{Q}_{i}$. This completes the proof. ∎

Remark 6.17*.*

Lemma 6.12 is a special case of Lemma 6.16.

Let $G$ be a local ring essentially of finite type over $k$ of dimension strictly less than $n$ equipped with a valuation centered in $G$ whose residue field is $k$.

Theorem 6.18**.**

Assume that for every ring $G$ as above, every element of $G$ is monomializable.

We recall that $\mathrm{car}\left(k_{\nu}\right)=0$. If $Q_{i}$ is monomializable, there exists a local framed sequence

[TABLE]

that monomializes $Q_{i+1}$.

Proof.

There are two cases.

First: $Q_{i}<Q_{i+1}$. Then we just saw that the strict transform $\widetilde{Q}_{i+1}$ of $Q_{i+1}$ by the sequence $\left(R,u\right)\to\left(R_{l},u^{(l)}\right)$ that monomializes $Q_{i}$ is an immediate successor key element of $\widetilde{Q}_{i}=u_{n}^{(l)}$, and that we can reduce the problem to the hypotheses of Theorem 6.4 by Lemma 6.16. So we use Theorem 6.4 replacing $Q_{1}$ by $\widetilde{Q}_{i}$ and $Q_{2}$ by $\widetilde{Q}_{i+1}$.

Then we have constructed a local framed sequence (6.3) that monomializes $\widetilde{Q}_{i+1}$.

Second case: $Q_{i}<_{\lim}Q_{i+1}$.

Then we saw that the strict transform $\widetilde{Q}_{i+1}$ of $Q_{i+1}$ by the sequence $\left(R,u\right)\to\left(R_{l},u^{(l)}\right)$ that monomialize $Q_{i}$ is a limit immediate successor key element of $\widetilde{Q}_{i}=u_{n}^{(l)}$. Then we apply Theorem 6.14 replacing $Q_{1}$ by $\widetilde{Q}_{i}$ and $Q_{2}$ by $\widetilde{Q}_{i+1}$.

We have constructed a local framed sequence (6.3) that monomializes $\widetilde{Q}_{i+1}$. ∎

Theorem 6.19**.**

There exists a local sequence

[TABLE]

that monomializes all the key polynomials of $\mathcal{Q}$.

More precisely, for every index $i$, there exists an index $s_{i}$ such that in $R_{s_{i}}$, $Q_{i}$ is a monomial in $u^{\left(s_{i}\right)}$ multiplied by a unit of $R_{s_{i}}$.

Proof.

Induction on the dimension $n$ and on the index $i$ and we iterate the previous process. ∎

6.4. Divisibility.

We consider, for every integer $j$, the countable sets

[TABLE]

and

[TABLE]

with the convention that for every $i\in\left\{1,\dots,n\right\}$, $u_{i}^{(0)}=u_{i}$.

The set $\widetilde{\mathscr{S}}_{j}$ being countable for every integer $j$, we can number its elements, and then we write $\widetilde{\mathscr{S}}_{j}:=\left\{s_{m}^{(j)}\right\}_{m\in\mathbb{N}}$. We consider now the finite set

[TABLE]

Then $\bigcup\limits_{j\in\mathbb{N}}\left(\mathscr{S}_{j}\times\mathscr{S}_{j}\right)=\bigcup\limits_{j\in\mathbb{N}}\widetilde{\mathscr{S}}_{j}=\bigcup\limits_{j\in\mathbb{N}}\mathscr{S}^{\prime}_{j}$ is a countable union of finite sets.

Now we fix a local framed sequence

[TABLE]

Theorem 6.20**.**

There exists a finite local framed sequence

[TABLE]

such that for every integer $j\leq i$ and for every element $s$ of $\mathscr{S}^{\prime}_{j}$ , the first coordinate of $s$ divides its second coordinate in $R_{i+q_{i}}$.

Proof.

Consider an integer $j\leq i$ and an element $s=\left(s_{1},s_{2}\right)\in\mathscr{S}^{\prime}_{j}$ . We want to construct a sequence of blow-ups such that at the end we have $s_{1}\mid s_{2}$.

We know that $s\in\widetilde{\mathscr{S}}_{m}$ with $m\leq j$. All cases being similar, we may assume $s\in\widetilde{\mathscr{S}}_{j}$ and then we have

[TABLE]

and

[TABLE]

By Proposition 4.4 applied to $R_{i}$ instead of $R$, there exists a sequence $\left(R_{i},u^{(i)}\right)\to\cdots\to\left(R_{i+l},u^{(i+l)}\right)$ such that in $R_{i+l}$, $s_{1}\mid s_{2}$ or $s_{2}\mid s_{1}$. By definition $\nu(s_{1})\leq\nu(s_{2})$, so we have $s_{1}\mid s_{2}$ by Proposition 4.7.

By point $4$ of Theorem 6.4, we know that $\mathscr{S}_{j}\subseteq R_{i+l}^{\times}\mathscr{S}_{i+l}$. In other words every element of $\mathscr{S}_{j}$ can be written $z_{i+l}s_{i+l}$ with $z_{i+l}\in R_{i+l}^{\times}$ and $s_{i+l}\in\mathscr{S}_{i+l}$.

Let $\left(s_{3},s_{4}\right)\in\mathscr{S}^{\prime}_{j}$, be another pair of $\mathscr{S}^{\prime}_{j}$, let us say that it is still in $\widetilde{\mathscr{S}}_{j}$. We just saw that $s_{3},s_{4}\in R_{i+l}^{\times}\mathscr{S}_{i+l}$. Units don’t have an effect on divisibility, so we can only consider the part of $s_{3}$ and $s_{4}$ which is in $\mathscr{S}_{i+l}$. Hence we can iterate the Proposition 4.4 applying it to $\left(R_{i+l},u^{(i+l)}\right)$. So we constructed an other sequence of blow-ups

[TABLE]

such that $R_{i+h}$ we have $s_{3}\mid s_{4}$ or $s_{4}\mid s_{3}$. Since $\nu(s_{3})\leq\nu(s_{4})$, we know that $s_{3}$ divides $s_{4}$.

We iterate the process for all the pairs of $\mathscr{S}^{\prime}_{j}$, and for every $j\leq i$ . This is a finite number of times since $\mathscr{S}^{\prime}_{j}$ has a finite number of elements for every $j$ and since we consider a finite number of such sets. Then we obtain a finite sequence of blow-ups

[TABLE]

such that for every integer $j\leq i$ and every $s$ in $\mathscr{S}^{\prime}_{j}$ , the first coordinate of $s$ divides the second coordinate in $R_{i+q_{i}}$. ∎

The goal of the next theorem is to construct an infinite local framed sequence

[TABLE]

that monomializes all the key elements, as well as other elements specified below, and to ensure countably many divisibility conditions, also specified below. We will use the notation

[TABLE]

Theorem 6.21**.**

We recall that $\mathrm{car}\left(k_{\nu}\right)=0$. There exists an infinite sequence of blow-ups

[TABLE]

that monomializes all the key polynomials, all the elements of $B_{i}$ for every index $i$ and that has the following property:

[TABLE]

Proof.

The first key polynomial is a monomial, so for it we do not need to do anything. For $j=0$, the elements of $\mathscr{S}^{\prime}_{j}=\mathscr{S}^{\prime}_{0}$ are just pairs of monomials in $u$. Let us consider $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{0}$ and apply Proposition 4.4. We construct a sequence $p_{0}\colon R\to R_{q_{0}}$ such that in $R_{q_{0}}$, we have $s_{1}\mid s_{2}$ or $s_{2}\mid s_{1}$. Since $\nu\left(s_{1}\right)\leq\nu\left(s_{2}\right)$, we have $s_{1}\mid s_{2}$. We do the same for all the elements of $\mathscr{S}^{\prime}_{0}$ (recall that the set $\mathscr{S}^{\prime}_{0}$ is finite), and by abuse of notation we still denote by $p_{0}\colon R\to R_{q_{0}}$ the sequence obtained at the end. Now we have a sequence of blow-ups $p_{0}\colon R\to R_{q_{0}}$ such that the first key polynomial is a monomial and such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{0}$, we have $s_{1}\mid s_{2}$ in $R_{q_{0}}$.

We denote by $\left(P_{j}^{(i)}\right)_{j\in\mathbb{N}}$ the sequence of the generators of the $\nu$-ideals of the $B_{i}$. For the moment we only monomialize $P_{0}^{(0)}$ and still denote by $p_{0}\colon R\to R_{q_{0}}$ the sequence of blow-ups that monomializes the first key polynomial $P_{0}^{(0)}$ and such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{0}$, we have $s_{1}\mid s_{2}$ in $R_{q_{0}}$.

Arguing exactly as in the proof of Theorem 6.19, we show that there exists a sequence $\pi^{(2)}\colon R_{q_{0}}\to...\to R_{1}$ that monomializes the second key polynomial.

We have a sequence $\pi^{(2)}\circ p_{0}\colon R\to R_{q_{0}}\to R_{1}$ that monomializes the first two key polynomials, the element $P_{0}^{(0)}$, and such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{0}$, we have $s_{1}\mid s_{2}$ in $R_{q_{0}}$. Now, again by Proposition 4.4, we construct a sequence $p_{1}\colon R_{1}\to R_{q_{1}}$ such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{1}$, we have $s_{1}\mid s_{2}$ in $R_{q_{1}}$.

Now we monomialize all the $P_{j}^{(i)}$ for $i,j\leq 1$ and still denote, by abuse of notation, by $p_{1}\colon R_{1}\to R_{q_{1}}$ the sequence of blow-ups that monomializes these $P_{j}^{(i)}$ and such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{1}$, we have $s_{1}\mid s_{2}$ in $R_{q_{1}}$.

Arguing exactly as in the proof of Theorem 6.19, we show that there exists a sequence of blow-ups $\pi^{(3)}\colon R_{q_{1}}\to...\to R_{2}$ that monomializes the third key polynomial.

So we have a sequence $\pi^{(3)}\circ p_{1}\circ\pi^{(2)}\circ p_{0}\colon R\to R_{q_{0}}\to R_{q_{1}}\to R_{2}$ that monomializes the first three key polynomials, the elements $P_{j}^{(i)}$ for $i,j\leq 1$, and such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{0}$ or $\mathscr{S}^{\prime}_{1}$, we have $s_{1}\mid s_{2}$ in $R_{q_{0}}$ or in $R_{q_{1}}$. Now, again by Proposition 4.4, we construct a sequence $p_{2}\colon R_{2}\to R_{q_{2}}$ such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{2}$, we have $s_{1}\mid s_{2}$ in $R_{q_{2}}$.

Now we monomialize all the $P_{j}^{(i)}$ for $i,j\leq 2$ and still denote, by abuse of notation, by $p_{2}\colon R_{2}\to R_{q_{2}}$ the sequence of blow-ups that monomializes these $P_{j}^{(i)}$ and such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{2}$, we have $s_{1}\mid s_{2}$ in $R_{q_{2}}$.

Then we have a sequence $p_{2}\circ\pi^{(3)}\circ p_{1}\circ\pi^{(2)}\circ p_{0}$ that monomializes the first three key polynomials, the elements $P_{j}^{(i)}$ for $i,j\leq 2$, and such that for every $s=(s_{1},s_{2})\in\mathscr{S}^{\prime}_{i}$ for $i\in\left\{0,1,2\right\}$ we have $s_{1}\mid s_{2}$ in $R_{q_{i}}$. We iterate this process an infinite number of times. Hence we construct a sequence of blow-ups $(R,u)\to\cdots\to(R_{m},u^{(m)})\to\cdots$ that monomializes all the key polynomials, all the generators $P_{j}^{(i)}$ (and so all the elements of the $B_{i}$) and that has the last property of the statement of the Theorem. ∎

7. Conclusion.

Now we can prove the main result of this chapter, namely, simultaneaous embedded local uniformization for the local rings essentially of finite type over a field of characteristic zero.

A local algebra $K$ essentially of finite type over a field $k$ that has $k$ as residue field is an étale extension of

[TABLE]

Let $f\in K$ be an irreducible element over $k$ and

[TABLE]

The ideal $I$ is a prime ideal of height $1$, so $I$ principal. We consider a generator $\widetilde{f}$ of $I$. Then $\frac{K^{\prime}}{\left(\widetilde{f}\right)}\hookrightarrow\frac{K}{\left(f\right)}$ and each local sequence in $\frac{K^{\prime}}{\left(\widetilde{f}\right)}$ induced a local sequence in $\frac{K}{\left(f\right)}$.

So it is enough to prove local uniformization in the case of the rings $k\left[u_{1},\dots,u_{n}\right]_{\left(u_{1},\dots,u_{n}\right)}$ to prove it in the general case of algebras essentially of finite type over a field $k$.

Theorem 7.1**.**

Let us consider the sequence

[TABLE]

of Theorem 6.21.

Then for every element $f$ of $R$, there exists $i$ such that in $R_{i}$, $f$ is a monomial multiplied by a unit.

Proof.

Let $f\in R$. By Theorem 5.5, there exists a finite or infinite sequence $\left(Q_{i}\right)_{i}$ of key polynomials of the extension $K\left(u_{n}\right)$, optimal (possibly limit) immediate successors, such that $\left(\epsilon(Q_{i})\right)_{i}$ is cofinal in $\epsilon(\Lambda)$ where $\Lambda$ is the set of key polynomials.

Then by Remark 5.9, $f$ is non-degenerate with respect to one of these polynomials $Q_{i}$. But we saw in Theorem 6.21 that there exists an index $l$ such that in $R_{l}$, all the $Q_{j}$ with $j\leq i$ are monomials, hence $f$ is non-degenerate with respect to a regular system of parameters of $R_{l}$.

Let $N=\left(w_{1},\dots,w_{s}\right)$ be a monomial ideal in $u^{\left(l\right)}$ such that $\nu\left(N\right)=\nu\left(f\right)$ with $w_{j}$ monomials in $u^{\left(l\right)}$ such that $\nu\left(w_{1}\right)=\min\left\{\nu\left(w_{j}\right)\right\}$. By construction of the local framed sequence, there exists $l^{\prime}\geq l$ such that in $R_{l^{\prime}}$, $w_{1}\mid w_{j}$ for all $j$. So in $R_{l^{\prime}}$, $f$ is equal to $w_{1}$ multiplied by a unit of $R_{l^{\prime}}$. ∎

Theorem 7.2** (Embedded local uniformization).**

Let $k$ be a zero characteristic field and $f=\left(f_{1},\dots,f_{l}\right)\in k\left[u_{1},\dots,u_{n}\right]^{l}$ be a set of $l$ polynomials in $n$ variables, that are irreducible over $k$. We set $R:=k\left[u_{1},\dots,u_{n}\right]_{\left(u_{1},\dots,u_{n}\right)}$ and $\nu$ a valuation centered in $R$ such that $k=k_{\nu}$.

We consider the sequence $\left(R,u\right)\overset{}{\to}\cdots\overset{}{\to}\left(R_{m},u^{(m)}\right)\overset{}{\to}\cdots$ of Theorem 6.21.

Then there exists an index $j$ such that the subscheme of $\mathrm{Spec}\left(R_{j}\right)$ defined by the ideal $\left(f_{1},\dots,f_{l}\right)$ is a normal crossing divisor.

Proof.

Renumbering, if necessary, we may assume

[TABLE]

By Theorem 6.21 there exists an index $j_{1}$ such that in $R_{j_{1}}$, the total transform of $f_{1}$ is a monomial in $u^{(j_{1})}$, and so defines a normal crossing divisor.

Now we look at the equation $f_{2}$ in $R_{j_{1}}$. By Theorem 6.21, there exists an index $j_{2}$ such that in $R_{j_{2}}$, the total transform of $f_{2}$ defines a normal crossing divisor.

In $R_{2}$, the total transforms of $f_{1}$ and $f_{2}$ define normal crossing divisors.

We iterate the process until the total transforms of $f_{1},\dots,f_{l}$ define normal crossing divisors in $R_{j_{l}}$.

By construction of the local framed sequence $\left(R,u\right)\overset{}{\to}\cdots\overset{}{\to}\left(R_{m},u^{(m)}\right)\overset{}{\to}\cdots$, there exists $j\geq j_{l}$ such that in $R_{j}$, we have $f_{1}\mid f_{i}$ for every index $i$. ∎

Corollary 7.3**.**

We keep the same notation and hypotheses as in the previous Theorem.

Then $R_{\nu}=\lim\limits_{\rightarrow}R_{i}$.

Part IV Simultaneous local uniformization in the case of quasi-excellent

rings for valuations of rank less than or equal to $2$.

8. Preliminaries.

Let $R$ be a local noetherian domain of equicharacteristic zero and $\nu$ a valuation of $\mathrm{Frac}\left(R\right)$ of rank $1$, centered in $R$ and of value group $\Gamma_{1}$. We are going to define the implicit prime ideal $H$ of $R$ for the valuation $\nu$, which is a key object in local uniformization. Indeed, this ideal will be the ideal we have to desingularize. We are going to prove in this part that to regularize $R$, hence to construct a local uniformization, we only have to regularize $\widehat{R}_{H}$ and $\frac{\widehat{R}}{H}$. At this point, the hypothesis of quasi excellence is very important: if $R$ is quasi excellent, the ring $\widehat{R}_{H}$ is regular. So we will only have to monomialize the elements of $\frac{\widehat{R}}{H}$.

8.1. Quasi-excellent rings and the implicit prime ideal.

Definition 8.1**.**

Let $R$ be a domain. We say that $R$ is a *G-ring *if for every prime ideal $\mathfrak{p}$ of $R$, the completion morphism $R_{\mathfrak{p}}\to\widehat{R}_{\mathfrak{p}}$ is a regular homomorphism.

Definition 8.2**.**

Let $R$ be a local ring. Then $R$ is quasi-excellent if $R$ is a G-ring.* *

More generally, if $A$ is a ring, then $A$ is quasi-excellent if $A$ is a local G-ring whose regular locus is open for all $A$-algebra of finite type.

Proposition 8.3**.**

[38]** A local noetherian ring $R$ is quasi-excellent if the completion morphism $R\to\widehat{R}$ is regular.

Remark 8.4*.*

Let $R$ be a local ring. If $R$ is a G-ring, then its regular locus is open. Since the class of $G$-rings is stable under passing to algebras of finite type, for every $R$-algebra $A$ of finite type, the set $\mathrm{Reg}(A)$ is open.

Definition 8.5**.**

We call the implicit prime ideal $H$ of $R$ the ideal $H=\bigcap\limits_{\beta\in\nu\left(R\setminus\left\{0\right\}\right)}P_{\beta}\widehat{R}$. The ideal $H$ is composed of the elements of $\widehat{R}$ whose value is greater than every element of $\Gamma_{1}$.

Furthermore, the valuation $\nu$ extends uniquely to a valuation $\widetilde{\nu}$ centered in $\frac{\widehat{R}}{H}$ ([47]).

Proposition 8.6**.**

Let $R$ be a quasi-excellent local ring. Then $\widehat{R}_{H}$ is regular.

Proof.

The ring $R$ is a G-ring. Then for every prime ideal $\mathfrak{p}$ of $R$, we have the injective map $\kappa\left(\mathfrak{p}\right)\hookrightarrow\kappa\left(\mathfrak{p}\right)\otimes_{R}\widehat{R}$ such that the fiber $\kappa\left(\mathfrak{p}\right)\otimes_{R}\widehat{R}$ is geometrically regular over $\kappa\left(\mathfrak{p}\right)$, where $\kappa\left(\mathfrak{p}\right):=\frac{R_{\mathfrak{p}}}{\mathfrak{p}R_{\mathfrak{p}}}$. Since $R$ is a domain, $\left(0\right)$ is a prime ideal of $R$.

We write $K:=\mathrm{Frac}\left(R\right)$, then we have the injective map $K\hookrightarrow K\otimes_{R}\widehat{R}$ such that the fiber $K\otimes_{R}\widehat{R}$ is geometrically regular over $\mathrm{K}$. In other words the morphism $K\hookrightarrow K\otimes_{R}\widehat{R}$ is regular.

But $R\setminus\left\{0\right\}$ and $\widehat{R}\setminus H$ are two multiplicative subsets of $\widehat{R}$ such that $R\setminus\left\{0\right\}\subseteq\widehat{R}\setminus H$, since $R\cap H=\left\{0\right\}$. Then, $\widehat{R}_{H}$ is a localisation of $\widehat{R}_{R\setminus\left\{0\right\}}$. If we show that $\widehat{R}_{R\setminus\left\{0\right\}}$ is regular, then $\widehat{R}_{H}$ will be also regular as a localization of a regular ring. By the universal property of tensor product, the ring $\widehat{R}_{R\setminus\left\{0\right\}}$ is isomorphic to $K\otimes_{R}\widehat{R}$, which is regular by hypothesis. This completes the proof. ∎

8.2. Numerical characters associated to a singular local noetherian ring.

Let $\left(S,\mathfrak{q},L\right)$ be a local noetherian ring and $\mu$ a valuation centered in $S$. We write $\mu=\mu_{2}\circ\mu_{1}$ with $\mu_{1}$ of rank $1$. The valuation $\mu_{2}$ is trivial if and only if $\mu$ is also of rank $1$. We denote by $G$ the value group of $\mu$ and by $G_{1}$ the value group of $\mu_{1}$. In fact $G_{1}$ is the smallest isolated subgroup non-trivial of $G$. We set $I:=\left\{x\in S\text{ such that }\mu(x)\notin G_{1}\right\}$, and then $\mu_{1}$ induces a valuation of rank $1$ over $\frac{S}{I}$. Let $\overline{J}$ be the implicit prime ideal of $\frac{\hat{S}}{I\hat{S}}$ for the valuation $\mu_{1}$ and $J$ its preimage in $\hat{S}$.

Definition 8.7**.**

We set

[TABLE]

We assume that $I\subseteq\mathfrak{q}^{2}$. Let $v=\left(v_{1},\dots,v_{n}\right)$ be a minimal set of generators of $\mathfrak{q}$. We have $\mu\left(v_{j}\right)\in G_{1}$ for every index $j$.

Definition 8.8**.**

We have $\sum\limits_{j=1}^{n}\mathbb{Q}\mu\left(v_{j}\right)\subseteq G_{1}\otimes\mathbb{Q}$ and we set

[TABLE]

Remark 8.9*.*

We have $r\left(S,v,\mu\right)\leq e\left(S,\mu\right)$.

Now we consider $M\subset\left\{1,\dots,n\right\}$ and

[TABLE]

a framed blow-up along $\left(v_{M}\right)$. We set $C^{\prime}=\left\{1,\dots,n_{1}\right\}\setminus D_{1}$, where $D_{1}$ is as in 3.3.

If the elements of $v_{M}$ are $L$-linearly independent in $\frac{\mathfrak{q}\hat{S}}{J+\mathfrak{q}^{2}\hat{S}}$, then there exists a partition of $A$ that we denote by $A^{\prime}\sqcup A"$. This partition is such that $v_{M}\cup v_{A^{\prime}}$ are $L$-linearly independent modulo $J+\mathfrak{q}^{2}\hat{S}$ and $v_{A"}$ is in the space generated by $v_{J}\cup v_{A^{\prime}}$ over $L$ modulo $J+\mathfrak{q}^{2}\hat{S}$. As we know that $v^{\prime}_{A\cup B\cup\left\{j\right\}}=v_{D_{1}}^{(1)}$, we can identify $A^{\prime}\cup B\cup\left\{j\right\}$ with a subset of $D_{1}$.

Now we set $I_{1}:=\left\{x\in S_{1}\text{ such that }\mu(x)\notin G_{1}\right\}$ and we consider $\overline{J}_{1}$ the implicit prime ideal of $\frac{\hat{S}_{1}}{I_{1}\hat{S}_{1}}$ with respect to $\mu_{1}$ and $J_{1}$ its preimage in $\hat{S}_{1}$. We call $\mathfrak{q}_{1}$ the maximal ideal of $S_{1}$ and $L_{1}$ its residue field.

Remark 8.10*.*

We have $e\left(S,\mu\right)=n$ if and only if the elements of $v$ are $L$-linarly independent in $\frac{\mathfrak{q}\hat{S}}{J+\mathfrak{q}^{2}\hat{S}}$.

Theorem 8.11**.**

If $e\left(S,\mu\right)=n$, then:

[TABLE]

This inequality is strict once the elements of $v_{A^{\prime}\cup B\cup\left\{j\right\}\cup C^{\prime}}^{(1)}$ are $L_{1}$-linearly dependent in $\frac{\mathfrak{q_{1}}\hat{S}_{1}}{J_{1}+\mathfrak{q}_{1}^{2}\hat{S}_{1}}$.

Proof.

By definition, $v^{(1)}$ generates the maximal ideal $\mathfrak{q}_{1}$ of $S_{1}$, and so induces a set of generators of $\mathfrak{q}_{1}\frac{\widehat{S_{1}}}{J_{1}}$. Since $n_{1}\leq n$, by definition of a framed blow-up, we know that $\sharp C^{\prime}\leq\sharp C$.

Furthermore, we have $e\left(S,\mu\right)=\sharp M+\sharp A^{\prime}$. We also know that $v_{D_{1}\setminus\left(A^{\prime}\cup B\cup\left\{j\right\}\right)}^{(1)}$ is in the $L$-vector space of $v_{A^{\prime}\cup B\cup\left\{j\right\}\cup C^{\prime}}^{(1)}$ modulo $J_{1}+\mathfrak{q}_{1}^{2}\hat{S}_{1}$.

So:

[TABLE]

If in addition the elements of $v_{A^{\prime}\cup B\cup\left\{j\right\}\cup C^{\prime}}^{(1)}$ are $L_{1}$-linearly dependents in $\frac{\mathfrak{q_{1}}\hat{S}_{1}}{J_{1}+\mathfrak{q}_{1}^{2}\hat{S}_{1}}$, then we have $e\left(S_{1},\mu\right)<\sharp A^{\prime}+\sharp B+\sharp\left\{j\right\}+\sharp C^{\prime}$ and so $e\left(S_{1},\mu\right)<e\left(S,\mu\right)$. ∎

Theorem 8.12**.**

We have $r\left(S_{1},v^{(1)},\mu\right)\geq r\left(S,v,\mu\right)$.

Proof.

This is induced by the two last points of Proposition 4.5. ∎

Corollary 8.13**.**

Once $e\left(S,\mu\right)=n$, we have

[TABLE]

The inequality is strict if $e\left(S_{1},\mu\right)<n$.

Remark 8.14*.*

We are doing an induction on the dimension $n$. We saw that this dimension decreases by the sequence of blow-ups.

If it decreases strictly, then it will happen a finite number of time and the proof is finished.

Then, after now, we assume this dimension to be constant by blow-up. In other words for all framed sequence $S\to S_{1}$, we assume that $e\left(S,\mu\right)=e\left(S_{1},\mu\right)=n$.

Similarly, we may assume that $r\left(S,v,\mu\right)=r\left(S_{1},v^{\left(1\right)},\mu\right)$.

9. Implicit ideal.

Let $\left(R,\mathfrak{m},k\right)$ be a local quasi excellent ring equicharacteristic and let $\nu$ be a valuation of rank $1$ of its field of fractions, centered in $R$ and of value group $\Gamma_{1}$. We denote by $H$ the implicit prime ideal of $R$ for the valuation $\nu$.

By the Cohen structure Theorem, there exists an epimorphism $\Phi$ from a complete regular local ring $A\simeq k\left[\left[u_{1},\dots,u_{n}\right]\right]$ of field of fractions $K$ into $\frac{\widehat{R}}{H}$. Its kernel $I$ is a prime ideal of $A$.

We consider $\mu$ a monomial valuation with respect to a regular system of parameters of $A_{I}$. It is a valuation on $A$ centered in $I$ such that $k_{\mu}=\kappa\left(I\right)$ where $\kappa\left(I\right)$ is the residue field of $I$. Then we set $\widehat{\nu}:=\widetilde{\nu}\circ\mu$, hence we define a valuation on $A$. Let $\Gamma$ be the value group of $\widehat{\nu}$.

Then, $\Gamma_{1}$ is the smallest non-trivial isolated subgroup of $\Gamma$ and we have:

[TABLE]

Definition 9.1**.**

Let $\pi\colon\left(A,u\right)\to\left(A^{\prime},u^{\prime}\right)$ be a framed blow-up and $\sigma\colon A^{\prime}\to\widehat{A^{\prime}}$ be the formal completion of $A^{\prime}$. The composition $\sigma\circ\pi$ is called formal framed blow-up.

A composition of such blow-ups is called a formal framed sequence.

Let $\left(A,u\right)\to\left(A_{1},u^{(1)}\right)\to\cdots\to\left(A_{l},u^{(l)}\right)$ a formal sequence, that we denote by $(\ast)$.

Definition 9.2**.**

The formal sequence $\left(A,u\right)\to\left(A_{1},u^{(1)}\right)\to\cdots\to\left(A_{l},u^{(l)}\right)$ is said *defined on $\Gamma_{1}$ *if for every integers $i\in\left\{0,\dots,l-1\right\}$ and $q\in J_{i}$, we have $\nu\left(u_{q}^{(i)}\right)\in\Gamma_{1}$.

Now we consider $A_{i}\simeq k_{i}\left[\left[u_{1}^{(i)},\dots,u_{n}^{(i)}\right]\right]$ and we denote by $I_{i}^{\mathrm{strict}}$ the strict transform of $I$ in $A_{i}$.

Definition 9.3**.**

We call formal transformed of $I$ in $A_{i}$, and we denote it by $I_{i}$, the preimage in $A_{i}$ of the implicit ideal of $\frac{A_{i}}{I_{i}^{\mathrm{strict}}}$.

Let $v_{i}$ be the greatest integer of $\left\{r,\dots,n\right\}$ such that

[TABLE]

and we set

[TABLE]

Definition 9.4**.**

Let $P$ be a prime ideal of $A$. We call $\ell$-th symbolic power of $P$ the ideal $P^{(\ell)}:=\left(P^{\ell}A_{P}\right)\cap A$.

Equivalently, we have $P^{(\ell)}=\left\{x\in A\text{ such that }\exists y\in A\setminus P\text{ such that }xy\in P^{\ell}\right\}$.

It is the set composed by the elements that vanish with order at least $\ell$ in the generic point of $\mathrm{V}\left(P\right)$.

Let $G$ be a complete ring of dimension strictly less than $n$ and let $\theta$ be a valuation centered in $G$, of value group $\widetilde{\Gamma}$.

We consider $\widetilde{\Gamma}_{1}$ the first non trivial isolated subgroup of $\widetilde{\Gamma}$ and $\mathfrak{g}:=\left\{g\in G\text{ such that }\theta\left(g\right)\notin\widetilde{\Gamma}_{1}\right\}$.

The next result will help us to prove the simultaneous local uniformization by induction.

Proposition 9.5**.**

Assume that:

1. (1)

In the formal sequence $\left(A,u\right)\to\left(A_{1},u^{(1)}\right)\to\cdots\to\left(A_{l},u^{(l)}\right)$, there exists a formal framed subsequence

[TABLE]

such that $v_{i}<n-1$. 2. (2)

For every ring $G$ as above, every element in $G\setminus\mathfrak{g}^{(2)}$ is monomializable by a formal framed sequence defined on $\widetilde{\Gamma}_{1}$.

Then for every element $f$ of $A\setminus I^{(2)}$, there exists a formal sequence

[TABLE]

defined over $\Gamma_{1}$ such that $f$ can be written as a monomial in $u_{1}^{(l)},\dots,u_{n}^{(l)}$ multiplied by an element of $A_{l}^{\times}$.

Proof.

We assume that there exists a formal framed sequence

[TABLE]

such that $v_{i}<n-1$. It means that $v_{i}+1<n$. By definition of $v_{i}$, we know that $\mathfrak{g}_{i}:=I_{i}\cap k_{i}\left[\left[u_{1}^{(i)},\dots,u_{v_{i}+1}^{(i)}\right]\right]\neq\left(0\right)$. So we consider an element $g$ in $\mathfrak{g}_{i}\setminus\mathfrak{g}_{i}^{(2)}\subseteq C_{i}\setminus\mathfrak{g}_{i}^{(2)}$, where $C_{i}:=k_{i}\left[\left[u_{1}^{(i)},\dots,u_{v_{i}+1}^{(i)}\right]\right]$. Since $v_{i}+1<n$, the ring $C_{i}$ is of dimension strictly less than $n$. So we can use the second hypothesis on the element $g$ in the ring $C_{i}$.

Hence there exists a formal sequence defined over $\Gamma_{1}$

[TABLE]

where $v^{\prime}\leq v_{i}+1$, and such that $g$ can be written as a monomial in $u^{\prime}_{1},\dots,u^{\prime}_{v^{\prime}}$ multiplied by an element of $S^{\prime\times}$.

Since $g\in\mathfrak{g}_{i}$, there exists a regular parameter of $S^{\prime}$, say $u^{\prime}_{v^{\prime}}$, such that $\nu\left(u^{\prime}_{v^{\prime}}\right)\notin\Gamma_{1}$. Indeed, $g\in\mathfrak{g}_{i}=I_{i}\cap C_{i}$, so $g\in I_{i}$ hence it belongs to $I$. Equivalently, it satisfies $\widehat{\nu}(g)\notin\Gamma_{1}$. Since $g$ can be written as a monomial in the generators of the maximal ideal of $S^{\prime}$, one of these generators which appears in the factorization of $g$ must be in $I$. Hence $e\left(S^{\prime},\widehat{\nu}_{|_{S^{\prime}}}\right)<v_{i}+1$.

Replacing every ring $O$ which appears in

[TABLE]

by $O\left[\left[u_{v_{i}+2}^{(i)},\dots,u_{n}^{(i)}\right]\right]$, we obtain a formal sequence

[TABLE]

independent of $u_{v_{i}+2}^{(i)},\dots,u_{n}^{(i)}$, with $A_{l}=S^{\prime}\left[\left[u_{v_{i}+2}^{(i)},\dots,u_{n}^{(i)}\right]\right]$. But we know that $e\left(S^{\prime},\widehat{\nu}_{|_{S^{\prime}}}\right)<v_{i}+1$, and so $e\left(A_{l},\widehat{\nu}\right)<n$.

Let $f$ be an element of $A\setminus I^{(2)}$. Its image under $\pi^{\prime}\circ\pi$ is an element of $A_{l}$, whose dimension is strictly less than $n$. Since all the $A_{i}$ are quasi-excellent, we have $f\notin A_{i}\setminus I_{i}^{\left(2\right)}$ and we can use again the second hypothesis. Hence we constructed a formal sequence $\pi^{\prime}\circ\pi$ such that $f$ can be written as a monomial in the generators of the maximal ideal of $A_{l}$ multiplied by a unit of $A_{l}$. This completes the proof. ∎

Now, we assume that for every formal sequence $\left(A,u\right)\to\left(A_{1},u^{(1)}\right)\to\cdots\to\left(A_{l},u^{(l)}\right)$ and for every integer $i$, we have $v_{i}\in\left\{n-1,n\right\}$.

So for every integer $i$, we have $I_{i}\cap k_{i}\left[\left[u_{1}^{(i)},\dots,u_{n-1}^{(i)}\right]\right]=(0)$.

We consider a complete local ring $G$ of dimension strictly less than $n$ and a valuation $\theta$ of rank $1$ centered in $G$.

Lemma 9.6**.**

Assume that for every ring $G$ as above, there exists a formal framed sequence that monomializes every element of $G$.

Then $I$ is of height at most $1$.

Proof.

If $I=(0)$, the proof is finished. So we assume $I\neq(0)$ and we consider $f\in I\setminus\left\{0\right\}$. We write

[TABLE]

with $a_{j}\in k\left[\left[u_{1},\dots,u_{n-1}\right]\right]$. We consider an integer $N$ big enough such that every $a_{j}$ with $j>N$ is in the ideal generated by $\left(a_{0},\dots,a_{N}\right)$. Now let us consider

[TABLE]

We set $\overline{u}:=\left(u_{1},\dots,u_{n-1}\right)$ and $B:=k\left[\left[\overline{u}\right]\right]$. Since $B$ is a complete local ring of dimension strictly less than $n$, by hypothesis we can construct a formal sequence $\left(B,\overline{u}\right)\to\left(B^{\prime},\overline{u}^{\prime}\right)$ such that for every $j\in\left\{0,\dots,N\right\}$, the element $a_{j}$ is a monomial in $\overline{u}^{\prime}$. By Propositions 4.4 and 4.7, we can construct a local framed sequence $\left(B^{\prime},\overline{u}^{\prime}\right)\to\left(B",\overline{u}"\right)$ such that $a_{\delta}\mid a_{j}$ for every $j\in\left\{0,\dots,N\right\}$ in $B"$, since $a_{\delta}$ has minimal value. So we have a sequence

[TABLE]

We compose with the formal completion and obtain

[TABLE]

in which we still have $a_{\delta}\mid a_{j}$ for every $j\in\left\{0,\dots,N\right\}$.

We replace again all the rings $O$ of the sequence $\left(B,\overline{u}\right)\to\left(\widehat{B"},\overline{u}"\right)$ by $O\left[\left[u_{n}\right]\right]$, and obtain a sequence $\left(A,u\right)\to\left(A^{\prime},u^{\prime}\right)$ independent of $u_{n}$ and in which we still have $a_{\delta}\mid a_{j}$ for every $j\in\left\{0,\dots,N\right\}$.

We recall that for every index $i$, we have

[TABLE]

If we denote by $I^{\prime}$ the formal transform of $I$ in $A^{\prime}$, we obtain $I^{\prime}\cap\widehat{B"}=(0)$. We know that $\frac{f}{a_{\delta}}\in I^{\prime}$, and by Weierstrass preparation Theorem, $\frac{f}{a_{\delta}}=xy$ where $x$ is a unit of $A^{\prime}$, and $y$ is a monic polynomial in $u_{n}$ of degree $\delta$. Then the morphism $\widehat{B"}\to\frac{A^{\prime}}{I^{\prime}}$ is injective and finite.

Hence $\dim\left(\frac{A^{\prime}}{I^{\prime}}\right)=\dim\left(\widehat{B"}\right)=n-1$. Since $\dim\left(A^{\prime}\right)=n$, we have $\mathrm{ht}\left(I\right)\leq\mathrm{ht}\left(I^{\prime}\right)=\dim\left(A^{\prime}\right)-\dim\left(\frac{A^{\prime}}{I^{\prime}}\right)=n-(n-1)=1$. This completes the proof. ∎

Corollary 9.7** (of Lemma 9.6.).**

We keep the same hypothesis as in Lemma 9.6. Let $I=\left(h\right)$.

There exists a formal framed sequence $\left(A,u\right)\to\left(A^{\prime},u^{\prime}\right)$ such that in $A^{\prime}$, the strict transform of $h$ is a monic polynomial of degree $\delta$.

From now on, we assume that $h$ is a monic polynomial of degree $\delta$.

Proposition 9.8**.**

We keep the same hypothesis as in Lemma 9.6. Let $I=\left(h\right)$. The polynomial $h$ is a key polynomial.

Proof.

By definition, $I=\left\{f\in A\text{ such that }\widehat{\nu}\left(f\right)\notin\Gamma_{1}\right\}$, so $\widehat{\nu}\left(h\right)\notin\Gamma_{1}$ . Furthermore, for every non-zero integer $b$, we have $\widehat{\nu}\left(\partial_{b}h\right)\in\Gamma_{1}$ since $h$ is a generator of $I$, hence has the smallest degree among all the elements of $I$ and so $\partial_{b}h\notin I$.

Then $\epsilon\left(h\right)\notin\Gamma_{1}$.

Let $P$ be a polynomial such that $\deg\left(P\right)<\deg\left(h\right)$. To show that $h$ is a key polynomial, it remains to prove that $\epsilon\left(P\right)<\epsilon\left(h\right)$.

By the minimality of $\deg\left(h\right)$, we still have $P\notin I$ and so $\widehat{\nu}\left(P\right)\in\Gamma_{1}$. So for every non-zero integer $b$, we also have $\widehat{\nu}\left(\partial_{b}P\right)\in\Gamma_{1}$. Then $\epsilon\left(P\right)\in\Gamma_{1}$.

Assume, aiming for contradiction, that $\epsilon\left(P\right)\geq\epsilon\left(h\right)$.

Then $-\epsilon\left(P\right)\leq\epsilon\left(h\right)\leq\epsilon\left(P\right)$ and since $\Gamma_{1}$ is an isolated subgroup, $\Gamma_{1}$ is a segment and so $\epsilon\left(h\right)\in\Gamma_{1}$. Contradiction.

Hence, $\epsilon\left(P\right)<\epsilon\left(h\right)$ and $h$ is a key polynomial. ∎

Now we are going to monomialize the key polynomial $h$.

As in the previous part, we construct a sequence $(Q_{i})_{i\geq 1}$ of key polynomials such that for each $i$ the polynomial $Q_{i+1}$ is either an optimal or a limit immediate successor of $Q_{i}$ that begins with $x$ and ends with $h$. So since $\epsilon\left(h\right)$ is maximal in $\epsilon\left(\Lambda\right)$, we stop. Then we have a finite sequence $(Q_{i})_{i\geq 1}$ of key polynomials such that for each $i$ the polynomial $Q_{i+1}$ is either an optimal or a limit immediate successor of $Q_{i}$ that begins with $x$ and ends with $h$.

In the case $I=\left(0\right)$, we construct again a sequence $(Q_{i})_{i\geq 1}$ of key polynomials such that for each $i$ the polynomial $Q_{i+1}$ is either an optimal or a limit immediate successor of $Q_{i}$ such that $\epsilon\left(\mathcal{Q}\right)$ is cofinal in $\epsilon\left(\Lambda\right)$.

Since we don’t assume $k=k_{\nu}$ in this part, we need a generalization of the monomialization Theorems of the Part $3$, paragraph $7$.

10. Monomialization of key polynomials.

Here we consider the ring $A\simeq k\left[\left[u_{1},\dots,u_{n}\right]\right]$ and a valuation $\nu$ centered in $A$ of value group $\Gamma$. For more clarity, we recall some previous notation.

Let $r$ be the dimension of $\sum\limits_{i=1}^{n}\mathbb{Q}\nu(u_{i})$ in $\Gamma\otimes_{\mathbb{Z}}\mathbb{Q}$. Renumbering if necessary, we may assume that $\nu\left(u_{1}\right),\dots,\nu\left(u_{r}\right)$ are rationaly independent and we consider $\Delta$ the subgroup of $\Gamma$ generated by $\nu(u_{1}),\ldots,\nu(u_{r})$.

We set $E:=\left\{1,\dots,r,n\right\}$ and

[TABLE]

So $\overline{\alpha^{(0)}}\nu(u_{n})=\sum\limits_{j=1}^{r}\alpha_{j}^{(0)}\nu(u_{j})$ with

[TABLE]

and

[TABLE]

We set

[TABLE]

and

[TABLE]

with $t=n-r-1$.

We write $x_{i}=\mathrm{in}_{\nu}u_{i}$, and so $x_{1},\ldots,x_{r}$ are algebraically independent over $k$ in $G_{\nu}$. Let $\lambda_{0}$ be the minimal polynomial of $x_{n}$ over $k[x_{1},\ldots,x_{r}]$, of degree $\alpha$. If $x_{n}$ is transcental, we set $\lambda_{0}:=0$.

We consider

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Let $d_{0}:=\frac{\alpha}{\overline{\alpha^{(0)}}}\in\mathbb{N}$.

If $\lambda_{0}\neq 0$, we have

[TABLE]

where $c_{q}\in k$ , $c_{d}=1$ and $\sum\limits_{q=0}^{d_{0}}c_{q}Z^{q}$ is the minimal polynomial of $z$ over $G_{\nu}$.

We are going to show that there exists a formal framed sequence that monomializes all the $Q_{i}$. We have $Q_{1}=u_{n}$ so we have to begin by monomializing $Q_{2}$.

First, let us consider

[TABLE]

where $b_{q}\in R$ such that $b_{q}\equiv c_{q}$ modulo $\mathfrak{m}$ and $a_{q}\in A^{\times}$.

Then we will show that we can reduce the problem to this special case.

Let

[TABLE]

and

[TABLE]

We have

[TABLE]

and

[TABLE]

So $\frac{w^{\delta}}{w^{\gamma}}=\frac{w_{n}^{\overline{\alpha^{(0)}}}}{\prod\limits_{j=1}^{r}w_{j}^{\alpha_{j}^{(0)}}}=\overline{z}$.

Let us compute the value of $w^{\delta}$.

[TABLE]

Theorem 10.1**.**

There exists a local framed sequence

[TABLE]

with respect to $\nu$, independent of $v$, that has the following properties:

For every integer $i\in\{1,\dots,l\}$, we write $u^{(i)}=\left(u_{1}^{(i)},\dots,u_{n_{i}}^{(i)}\right)$ and denote by $k_{i}$ the residue field of $A_{i}$.

1. (1)

The blow-ups $\pi_{0},\dots,\pi_{l-2}$ are monomial. 2. (2)

We have $\overline{z}\in A_{l}^{\times}$. 3. (3)

We have

[TABLE] 4. (4)

We set

[TABLE]

For every integer $j\in\left\{1,\dots,r,n\right\}$, $w_{j}$ is a monomial in $w_{1}^{(l)},\dots,w_{r}^{(l)}$ multiplied by an element of $A_{l}^{\times}$. And for every integer $j\in\left\{1,\dots,r\right\}$, $w_{j}^{(l)}=w^{\eta}$ where $\eta\in\mathbb{Z}^{r+1}$. 5. (5)

If $\lambda_{0}\neq 0$, then $Q=w_{n}^{(l)}\times\overline{y}^{d_{0}}$.

Proof.

We apply Proposition 4.4 to $(w^{\delta},w^{\gamma})$ and obtain a local framed sequence for $\nu$, independent of $v$, such that $w^{\gamma}\mid w^{\delta}$ in $A_{l}$.

By Proposition 4.7 and the fact that $w^{\delta}$ and $w^{\gamma}$ have same value, we have $w^{\delta}\mid w^{\gamma}$ in $R_{l}$. In fact $\overline{z},\overline{z}^{-1}\in A_{l}^{\times}.$ So we have the point $(2)$.

We choose the sequence to be minimal, it means that the sequence composed by $\pi_{0},\ldots,\pi_{l-2}$ does not satisfy the conclusion of Proposition 4.4 for $(w^{\delta},w^{\gamma})$. We are now going to show that this sequence satisfies the conclusion of Theorem 10.1. Let $i\in\left\{0,\dots,l\right\}$. We write $w^{(i)}=\left(w_{1}^{(i)},\dots,w_{r_{i}}^{(i)},w_{n_{i}}^{(i)}\right)$, with $r_{i}=n_{i}-t-1>0$. For every integers $i\in\left\{1,\dots,l\right\}$ and $j\in\left\{1,\dots,n_{i}\right\}$, we write $\beta_{j}^{(i)}=\nu\left(u_{j}^{(i)}\right)$. For all $i<l$, $\pi_{i}$ is a blow-up along an ideal of the form $\left(u_{J_{i}}^{(i)}\right)$. Renumbering if necessary, we may assume that $1\in J_{i}$ and that $A_{i+1}$ is a localization of $A_{i}\left[\frac{u_{J_{i}}^{(i)}}{u_{1}^{(i)}}\right]$. Hence, $\beta_{1}^{(i)}=\min\limits_{j\in J_{i}}\left\{\beta_{j}^{(i)}\right\}$.

Lemma 10.2**.**

Let $i\in\left\{0,\dots,l-1\right\}$. We assume that the sequence $\pi_{0},\dots,\pi_{i-1}$ of 10.1 is monomial.

We write $w^{\gamma}=\left(w^{(i)}\right)^{\gamma^{(i)}}$ and $w^{\delta}=\left(w^{(i)}\right)^{\delta^{(i)}}$. Then:

1. (1)

$r_{i}=r$, 2. (2)

[TABLE] 3. (3)

$\mathrm{gcd}\left(\gamma_{1}^{(i)}-\delta_{1}^{(i)},\dots,\gamma_{r}^{(i)}-\delta_{r}^{(i)},\gamma_{n}^{(i)}-\delta_{n}^{(i)}\right)=1,$ 4. (4)

Every $\mathbb{Z}$-linear dependence relation between $\beta_{1}^{(i)},\dots,\beta_{r}^{(i)},\beta_{n}^{(i)}$ is an integer multiple of (10.2).

Proof.

1. (1)

It is enough to do an induction on $i$ and use Remark 3.6. 2. (2)

We have $\nu\left(w^{\gamma}\right)=\nu\left(w^{\delta}\right)$, in other words $\nu\left(\left(w^{(i)}\right)^{\gamma^{(i)}}\right)=\nu\left(\left(w^{(i)}\right)^{\delta^{(i)}}\right)$. Since $w^{(i)}=\left(w_{1}^{(i)},\dots,w_{r_{i}}^{(i)},w_{n_{i}}^{(i)}\right)$, we have:

[TABLE]

So we have

[TABLE]

By definition of $w^{(i)}$, for every integer $j\in\left\{1,\dots,r_{i},n_{i}\right\}$, we have $w_{j}^{(i)}=u_{j}^{(i)}$. So $\nu\left(w_{j}^{(i)}\right)=\beta_{j}^{(i)}$. Then:

[TABLE]

Hence $\sum\limits_{j\in\left\{1,\dots,r_{i},n_{i}\right\}}\left(\gamma_{j}^{(i)}-\delta_{j}^{(i)}\right)\beta_{j}^{(i)}=0$.

But $r_{i}=n_{i}-t-1=r$, so $n_{i}=r+t+1=n$, and:

[TABLE] 3. (3)

Same proof as in Theorem 6.4. 4. (4)

Same proof as in Theorem 6.4.

∎

Lemma 10.3**.**

The sequence $\left(A,u\right)\overset{\pi_{0}}{\to}\left(A_{1},u^{(1)}\right)\overset{\pi_{1}}{\to}\cdots\overset{\pi_{l-1}}{\to}\left(A_{l},u^{(l)}\right)$ of Theorem 10.1 is not monomial.

Proof.

Same proof as Lemma 6.7. ∎

Lemma 10.4**.**

Let $i\in\{0,\dots,l-1\}$ and we assume that $\pi_{0},\dots,\pi_{i-1}$ are all monomial. Then following properties are equivalent:

1. (1)

The blow-up $\pi_{i}$ is not monomial. 2. (2)

There exists a unique index $q\in J_{i}\setminus\left\{1\right\}$ such that $\beta_{q}^{(i)}=\beta_{1}^{(i)}$. 3. (3)

We have $i=l-1$.

Proof.

Same proof as Lemma 6.8. ∎

Using induction on $i$ and Lemma 10.4, we conclude that $\pi_{0},\dots,\pi_{l-2}$ are monomial. This proves the first point of the Theorem.

It remains to prove the last three points.

By Lemma 10.4 we know that there exists a unique element $q\in J_{l-1}\setminus\left\{j_{l-1}\right\}$ such that $\beta_{q}^{(l-1)}=\beta_{1}^{(l-1)}$, hence we are in the case $\sharp B_{l-1}+1=\sharp J_{l-1}-1$. We now have to see if $t_{k_{l-1}}=0$ or $1$.

We recall that $w_{1}^{(l-1)}=w^{\epsilon}$ and $w_{q}^{(l-1)}=w^{\mu}$ where $\epsilon$ and $\mu$ are two columns of a unimodular matrix such that $\mu-\epsilon=\pm(\gamma-\delta)$. So $x_{1}^{(l-1)}=x^{\epsilon}$ and $x_{q}^{(l-1)}=x^{\mu}$, then

[TABLE]

In other words

[TABLE]

So we can assume $\frac{x_{q}^{(l-1)}}{x_{1}^{(l-1)}}=z$ .

The case $t_{k_{l-1}}=1$ corresponds to the fact that $z$ is transcendantal over $k$, in other words $\lambda_{0}=0$. The case $t_{k_{l-1}}=0$ corresponds to the fact that $z$ is algebraic over $k$, in other words $\lambda_{0}\neq 0$. The third point of the Theorem is then a consequence of 3.9.

Since $\beta_{1}^{(l-1)},\dots,\beta_{r}^{(l-1)}$ are linearly independent, we have $q=n$. By 3.9, if $\lambda_{0}\neq 0$, we have

[TABLE]

Remark 10.5*.*

We have $\overline{\lambda_{0}}\left(\overline{z}\right)=\sum\limits_{i=0}^{d}c_{i}b_{i}\overline{z}^{i}$ where $c_{i}$ are units. Then we choose to set $c_{i}=a_{i}$ for every index $i$.

But since $\overline{z}=\frac{w_{n}^{\overline{\alpha^{(0)}}}}{\overline{y}}$, we have

[TABLE]

and the point $(5)$ is proven.

So it remains to prove the point $\left(4\right)$.

We apply Proposition 4.5 to $i=0$ and $i^{\prime}=l$. By the monomiality of $\pi_{0},\dots,\pi_{l-2}$, we know that $D_{i}=\{1,\dots,n\}$ for every $i\in\{1,\dots,l-1\}$.

We know that $D_{l}=\{1,\dots,n\}$ if $\lambda\neq 0$ and $D_{l}=\{1,\dots,n-1\}$ otherwise. Here we set again $u_{T}=v$.

By Proposition 4.5, for every $j\in\left\{1,\dots,r,n\right\}$, $w_{j}=u_{j}$ is a monomial in $w_{1}^{(l)},\dots,w_{r}^{(l)}$ (or equivalently in $u_{1}^{(l)},\dots,u_{r}^{(l)}$) multiplied by an element of $A_{l}^{\times}$.

Same thing for the fact that for every integer $j\in\{1,\dots,r\}$, we have $w_{j}^{(l)}=w^{\eta}$. This completes the proof. ∎

Remark 10.6*.*

In the case $Q_{2}=Q$, the we constructed a local framed sequence such that the total transform of $Q_{2}$ is a monomial. We will bring us to this case.

Definition 10.7**.**

[24] A local framed sequence that satisfies Theorem 10.1 is called a* $n$-generalized Puiseux package*.

Let $j\in\left\{r+1,\dots,n\right\}$. A *$j$-generalized Puiseux package *is a $n$-generalized Puiseux package replacing $n$ by $j$ in Theorem 10.1.

Remark 10.8*.*

We consider $\left(A,u\right)\to\dots\to\left(A_{i},u^{(i)}\right)\to\dots$ a $j$-generalized Puiseux package, with $j\in\left\{r+1,\dots,n\right\}$. We replace each ring of this sequence by its formal completion, hence we obtain o formal framed sequence that we call a formal $j$-Puiseux package. So Theorem 10.1 induces a formal $n$-Puiseux package that satisfies the same conclusion as in Theorem 10.1.

Since we want to to an induction, now we will assume until the end of Theorem 10.14, that we know how to monomialize every complete local equicharacteristic quasi excellent ring $G$ of dimension strictly less than $n$ equipped with a valuation of rank $1$ centered in $G$ by a formal framed sequence. This hypothesis is called $H_{n}$.

Lemma 10.9**.**

Let $P=\sum\limits_{j\in S_{u_{n}}\left(P\right)}c_{j}u_{n}^{j}$ the $u_{n}$-expansion of an optimal immediat successor key element of $u_{n}$.

There exists a formal framed sequence $\left(A,u\right)\to\left(A_{l},u^{(l)}\right)$ that transforms each coefficient $c_{j}$ in a monomial in $\left(u_{1}^{(l)},\dots,u_{r}^{(l)}\right)$, multiplied by a unit of $A_{l}$.

Hence, after this sequence, $P$ can be written like $\sum\limits_{i=0}^{d_{0}}a_{i}b_{i}\overline{y}^{d_{0}-i}\left(w_{n}^{\overline{\alpha^{(0)}}}\right)^{i}$.

Proof.

We will prove a more general result in 10.12. ∎

Theorem 10.10**.**

If $u_{n}\ll_{\lim}P$, then $P$ is monomializable.

Proof.

Same proof as Theorem 6.14. ∎

Lemma 10.11**.**

There exists a formal framed sequence

[TABLE]

such that in $A_{l}$, the strict transform of the polynomial $Q_{2}$ is a monomial.

Proof.

If $u_{n}<Q_{2}$, we use Lemma 10.9 and Theorem 10.1 to conclude. Otherwise, $u_{n}<_{\lim}Q_{2}$ and so we use Theorem 6.14. ∎

We constructed a formal framed sequence that monomializes $Q_{2}$. But we want one that monomializes all the key polynomials of $\mathcal{Q}$.

Now we are going to show that if we constructed a formal framed sequence $\left(A,u\right)\to\left(A_{l},u^{\left(l\right)}\right)$ that monomializes $Q_{i}$, then we can associate another $\left(A_{l},u^{\left(l\right)}\right)\to\left(A_{s},u^{\left(s\right)}\right)$ such that in $A_{s}$, the strict transform of $Q_{i+1}$ is also a monomial.

Let $\Delta_{l}$ be the group $\nu\left(k_{l}\left(u_{1}^{\left(l\right)},\dots,u_{n-1}^{\left(l\right)}\right)\setminus\left\{0\right\}\right)$ and

[TABLE]

We set $X_{j}=\mathrm{in}_{\nu}\left(u_{j}^{(l)}\right)$, $W_{j}=w_{j}^{(l)}$ and $\lambda_{l}$ the minimal polynomial of $X_{n}$ over $\mathrm{gr}_{\nu}k_{l}\left(u_{1}^{\left(l\right)},\dots,u_{n-1}^{\left(l\right)}\right)$ of degree $\alpha_{l}$.

We know that $Q_{i}=\overline{\omega}w_{n}^{(l)}$ with $\overline{\omega}$ a monomial in $W_{1},\dots,W_{r}$ multiplied by a unit. We set $\omega:=\mathrm{in}_{\nu}\left(\overline{\omega}\right)$.

If $Q_{i}<_{\lim}Q_{i+1}$, we use Theorem 10.10 and the proof is finished. So we assume that $Q_{i+1}$ is an optimal immediate successor of $Q_{i}$.

We write $Q_{i+1}=\sum\limits_{j\in S_{Q_{i}}(Q_{i+1})}a_{j}Q_{i}^{j}=\sum\limits_{j=0}^{s}a_{j}Q_{i}^{j}$ the $Q_{i}$-expansion of $Q_{i+1}$ in $k_{l}\left(u_{1}^{\left(l\right)},\dots,u_{n-1}^{\left(l\right)}\right)\left(u_{n}^{\left(l\right)}\right)$ .

We have $Q_{i+1}=Q_{i}^{s}+a_{s-1}Q_{i}^{s-1}+\cdots+a_{0}$ and since $Q_{i}=\overline{\omega}w_{n}^{(l)}$, we have

[TABLE]

We know that for every index $j$ such that $a_{j}\neq 0$, we have

[TABLE]

So all non-zero terms of the $Q_{i}$-expansion of $Q_{i+1}$ have same value. Then, by hypothesis $H_{n}$, all these terms are divisible by the same power of $\overline{\omega}$ after an appropriate sequence of blow-ups $(\ast_{i})$ independent of $u_{n}^{(l)}$.

We denote by $\widetilde{Q}_{i+1}$ the strict transform of $Q_{i+1}$ by the composition of $\left(\ast_{i}\right)$ with the sequence $\left(\ast^{\prime}_{i}\right)$ that monomializes $Q_{i}$. We denote this composition by $\left(c_{i}\right)$.

We know that $\widetilde{Q}_{i}$, the strict transform of $Q_{i}$ by $\left(c_{i}\right)$, is a regular parameter of the maximal ideal of $A_{l}$. Indeed, by Proposition 4.5, we know that each $u_{j}$ of $A$ can be written as a monomial on $w_{1}^{(l)},\dots,w_{r}^{(l)}$. In fact, the reduced exceptional divisor of this sequence is exactly $\mathrm{V}\left(\overline{\omega}\right)_{\mathrm{red}}$. Hence, as we know that $Q_{i}=w_{n}^{(l)}\overline{\omega}$, we do have that the strict transform of $Q_{i}$ is $\widetilde{Q}_{i}=w_{n}^{(l)}=u_{n}^{(l)}$. So it is a key polynomial in the extension $k_{l}\left(u_{1}^{(l)},\dots,u_{n-1}^{(l)}\right)\left(u_{n}^{(l)}\right)$.

Let us show that $\widetilde{Q}_{i+1}=\frac{Q_{i+1}}{\overline{\omega}^{s}}$.

We have $a_{s}=1$ and $Q_{i}^{s}=\overline{\omega}^{s}\left(u_{n}^{(l)}\right)^{s}$ and also $u_{n}^{(l)}\nmid\overline{\omega}$, so $\overline{\omega}^{s}$ divides the term $a_{s}Q_{i}^{s}$ and so all the non-zero terms of the $Q_{i}$-expansion of $Q_{i+1}$. Furthermore, it is the biggest power of $\overline{\omega}$ that divides each term, hence $\frac{Q_{i+1}}{\overline{\omega}^{s}}\left(u_{n}^{(l)}\right)^{s}+\frac{a_{s-1}}{\overline{\omega}}\left(u_{n}^{(l)}\right)^{s-1}+\cdots+\frac{a_{0}}{\overline{\omega}^{s}}$ is $\widetilde{Q}_{i+1}$ the strict transform of $Q_{i+1}$ by the sequence of blow-ups, that satisfies $\widetilde{Q}_{i}\ll\widetilde{Q}_{i+1}$ by hypothesis.

Let $G$ be a complete local equicharaceristic ring of dimension strictly less than $n$ equipped with a valuation centered in $G$.

Lemma 10.12**.**

We assume that for every ring $G$ as above, every element of $G$ is monomializable.

Assume that$Q_{i}<Q_{i+1}$ in $\mathcal{Q}$.

Then there exists a local framed sequence $\left(A_{l},u^{(l)}\right)\to\left(A_{e},u^{(e)}\right)$ such that in $A_{e}$, the strict transform of $Q_{i+1}$ is of the form $\sum\limits_{q=0}^{s}\tau_{q}\eta_{q}X_{n}^{q}$

where $\tau_{q}\in R_{e}^{\times}$ and $\eta_{q}$ are monomials in $u_{1}^{\left(e\right)},\dots,u_{r}^{\left(e\right)}$.

Proof.

By hypothesis, after a sequence of blow-ups independent of $u_{n}^{(l)}$, we can monomialize the $a_{j}$ and assume that they are monomials in $\left(u_{1}^{(l)},\dots,u_{n-1}^{(l)}\right)$ multiplied by units of $A_{l}$.

For every $g\in\left\{r+1,\dots,n-1\right\}$, we do a generalized $g$-Puiseux package as in Theorem 10.1, hence we have a sequence

[TABLE]

such that each $u_{g}^{(l)}$ is a monomial in$\left(u_{1}^{(t)},\dots,u_{r}^{(t)}\right)$.

In fact we can assume that the $a_{j}$ are monomials in $\left(u_{1}^{(l)},\dots,u_{r}^{(l)}\right)$ multiplied by units of $A_{l}$.

Since the strict transform

[TABLE]

is an immediate successor key element of $\widetilde{Q}_{i}$, this completes the proof. ∎

Remark 10.13*.*

Lemma 10.9 is a particular case of Lemma 10.12.

Theorem 10.14**.**

We still assume $H_{n}$.

We recall that $\mathrm{car}\left(k_{\nu}\right)=0$. If $Q_{i}$ is monomializable, then there exists a formal framed sequence

[TABLE]

that monomializes $Q_{i+1}$.

Proof.

There are two cases.

The first one: $Q_{i}<Q_{i+1}$.

Then the strict transform $\widetilde{Q}_{i+1}$ of $Q_{i+1}$ by the sequence $\left(A,u\right)\to\left(A_{l},u^{(l)}\right)$ that monomializes $Q_{i}$ is an immediate successor key element of $\widetilde{Q}_{i}=u_{n_{l}}^{(l)}$, and by Lemma 10.12 we just saw that we can bring us to the hypothesis of Theorem 10.1. So we use Theorem 10.1 replacing $Q_{1}$ by $\widetilde{Q}_{i}$ and $Q_{2}$ by $\widetilde{Q}_{i+1}$.

The last one: $Q_{i}<_{\lim}Q_{i+1}$.

We apply Theorem 10.10 replacing $u_{n}$ by $\widetilde{Q}_{i}$ and $P$ by $\widetilde{Q}_{i+1}$. ∎

As in the previous part, we consider, for every integer $j$, the countable sets

[TABLE]

and

[TABLE]

assuming that for every $i\in\left\{1,\dots,n\right\}$, $u_{i}^{(0)}=u_{i}$.

The set $\widetilde{\mathscr{S}}_{j}$ is countable for every $j$, so we can number its elements, and set $\widetilde{\mathscr{S}}_{j}:=\left\{s_{m}^{(j)}\right\}_{m\in\mathbb{N}}$. Now we consider the finite set

[TABLE]

Hence $\bigcup\limits_{j\in\mathbb{N}}\left(\mathscr{S}_{j}\times\mathscr{S}_{j}\right)=\bigcup\limits_{j\in\mathbb{N}}\widetilde{\mathscr{S}}_{j}=\bigcup\limits_{j\in\mathbb{N}}\mathscr{S}^{\prime}_{j}$ is a countable union of finite sets.

Since we consider all the elements according uniquely to the variable $u_{n}$, and more generally according to $u_{n}^{(i)}$, and since we do an induction on the dimension, we have to know how to monomialize the elements of $B_{i}:=k\left[u_{1}^{(i)},\dots,u_{n-1}^{(i)}\right]$.

Theorem 10.15**.**

Let $A\simeq k\left[\left[u_{1},\dots,u_{n}\right]\right]$ equipped with a valuation $\nu$ centered in $A$.

We recall that $\mathrm{car}\left(k_{\nu}\right)=0$. There exists a formal sequence

[TABLE]

that monomializes all the key polynomials of $\mathcal{Q}$ and all the elements of the $B_{i}$ for all $i$. Furthermore, the sequence has the property:

[TABLE]

In other words for every index $l$, there exists an index $p_{l}$ such that in $A_{p_{l}}$, $Q_{l}$ is a monomial in $u^{\left(p_{l}\right)}$ multiplied by a unit of $A_{p_{l}}$.

Proof.

To show that we can choose the sequence (10.4) such that

[TABLE]

and that all the elements of the $B_{i}$ are monomialized, we do the same thing than in Theorem 6.21.

Then we do an induction on the dimension $n$ and on the index $i$ and we iterate the above process. ∎

Corollary 10.16**.**

Let $A\simeq k\left[\left[u_{1},\dots,u_{n}\right]\right]$ equipped with a valuation $\widehat{\nu}$ centered in $A$, of value group $\Gamma$. We assume

[TABLE]

where $\Gamma_{1}$ is the smallest isolated subgroup of $\Gamma$. We recall that $\mathrm{car}\left(k_{\nu}\right)=0$.

There exists a formal framed sequence

[TABLE]

such that in $A_{l},$ the polynomial $h$ can be written as a monomial multiplied by a unit.

Proof.

The sequence $\mathcal{Q}$ has been constructed to contain $h$, so we just have to use Theorem 10.15. ∎

11. Reduction.

Let $\left(R,\mathfrak{m},k\right)$ be a local quasi excellent equicharacteristic ring and let $\nu$ be a valuation of its field of fractions, of rank $1$, centered in $R$ and of value group $\Gamma_{1}$. We denote by $\overline{H}$ the implicit ideal of $R$.

We are going to see that in this case, we just have to regularise $\frac{\widehat{R}}{\overline{H}}$.

We consider $\mathcal{F}:=\left\{f_{1},\dots,f_{s}\right\}\subseteq\mathfrak{m}$, and assume that $f_{1}$ has minimal value.

Remark 11.1*.*

We consider $R\to\widehat{R}\to R_{1}\to\widehat{R}_{1}$ a formal framed blow-up and we denote by $H^{\prime}$ the strict transformed of $\overline{H}$ in $R_{1}$.

Then we define $\overline{H_{1}}$ as the preimage in $\widehat{R}_{1}$ of the implicit ideal of $\frac{\widehat{R}_{1}}{H^{\prime}\widehat{R}_{1}}$.

We iterate this contruction for every formal framed sequence.

Theorem 11.2**.**

We recall that $\mathrm{car}\left(k_{\nu}\right)=0$. There exists a formal framed sequence

[TABLE]

such that:

1. (1)

The ring $\frac{\widehat{R_{i}}}{\overline{H_{i}}}$ is regular, 2. (2)

For every index $j$, we have that f_{j}\mathrm{\mod}\left(\overline{H_{i}}\right) is a monomial in $u^{(i)}$ multiplied by a unit of $\frac{\widehat{R_{i}}}{\overline{H_{i}}}$, 3. (3)

For every index $j$, we have $f_{1}\mathrm{mod}\left(\overline{H_{i}}\right)\mid f_{j}\mathrm{mod}\left(\overline{H_{i}}\right)$ in $\frac{\widehat{R_{i}}}{\overline{H_{i}}}$.

Proof.

Set $n:=e\left(R,\nu\right)$ and $u:=\left(y,x\right)$ with

[TABLE]

and

[TABLE]

such that the images of the $x_{j}$ in $\frac{\widehat{R}}{\overline{H}}$ induce a minimal set of generators of $\frac{\mathfrak{m}}{\overline{H}}$ and such that $y$ generates $\overline{H}$.

We do an induction on $\left(n_{i},n_{i}-r_{i},v_{i}\right)$.

We saw the existence of the surjection $\Phi$ from $A\simeq k\left[\left[x_{1},\dots,x_{n}\right]\right]$ to $\frac{\widehat{R}}{\overline{H}}$, of kernel $I=\left\{f\in A\text{ such that }\widehat{\nu}\left(f\right)\notin\Gamma_{1}\right\}\in\mathrm{Spec}\left(A\right)$ where $\widehat{\nu}$ is defined as in section 9. We denote by $L$ the field of fractions of $A$.

If $v_{0}<n-1$, then we do the same thing as in Proposition 9.5 and we strictly decrease $e\left(A,\widehat{\nu}\right)$.

The we can assume $v_{0}\in\left\{n-1,n\right\}$.

Assume $v_{0}=n-1$.

Then we know that $I=\left(h\right)$ and that there exists a formal framed sequence $\left(A,x\right)\to\left(A_{\ell},x^{\left(\ell\right)}\right)$ that monomializes $h$ by Corollary 10.16. So one of the generators that appears in its decomposition must be in $I_{\ell}$. Hence there exists $x_{p}^{\left(\ell\right)}$ such that $\widehat{\nu}\left(x_{p}^{\left(\ell\right)}\right)\notin\Gamma_{1}$. So by Theorems 9.5 and 8.11, there exists a local framed sequence that decreases strictly $e\left(A,\widehat{\nu}\right)$, so this case can happen a finite number of time, and we bring us at the case $I=\left(0\right)$. It means the case where $\frac{\widehat{A}}{I}$ is regular.

Case $I=\left(0\right)$. For every $f_{j}$, we have $\widehat{\nu}\left(f_{j}\right)\in\Gamma_{1}$ . So the element $f_{j}$ is a non-zero formal series and by Weierstrass preparation Theorem, we know that we can see it like a polynomial in $x_{l}$ with coefficients in $k\left[\left[x_{1},\dots,x_{n-1}\right]\right]$. We construct a sequence of key polynomials in the extension $k\left(\left(x_{1},\dots,x_{n-1}\right)\right)\left(x_{n}\right)$ as in previous section. In other words this sequence is a sequence of optimal (possibly limit) immediate successors which is cofinal in $\epsilon\left(\Lambda\right)$, where $\Lambda$ is the set of key polynomials. So the element $f_{j}$ is non-degenerate with respect of one of these polynomials that all are monomializable by the above part. Hence there exists a local framed sequence $\left(A,x\right)\to\left(A_{i},x^{(i)}\right)$ such that in $A_{i}$, the strict transform of $f_{j}$ is a monomial in $x^{(i)}$ multiplied by a unit of $A_{i}$.

If there exists a formal framed sequence such that $v_{i}<n-1$, then by Proposition 9.5, we can conclude by induction.

Iterating the case $I=\left(0\right)$, we assure the existence of a local framed sequence such that all the strict transforms of the $f_{j}$ are monomials multiplied by units. Doing another blow-up if necessary, we assume that there exists of a local framed sequence $\left(A,x\right)\to\left(A^{\prime},x^{\prime}\right)$ such that all the strict transforms of the $f_{j}$ are monomials only in $x^{\prime}_{1},\dots,x^{\prime}_{r}$.

By Proposition 4.4, we can assume that for every $j$ and every $p$, we have either $f_{j}\mid f_{p}$ or $f_{p}\mid f_{j}$.

So we have a local framed sequence

[TABLE]

that monomializes the $f_{j}$ and such that for all $j$ and $q$, we have $f_{j}\mid f_{q}$ or the converse.

By the minimality of $\nu\left(f_{1}\right)$, in $A_{i}$, we have $f_{1}\mid f_{j}$ for every $j$.

We have also two maps

[TABLE]

and we know that $\frac{A}{I}\simeq\frac{\widehat{R}}{\overline{H}}$ since $I=\mathrm{Ker}\left(\Phi\right)$. Hence, looking at the strict transform of $\frac{A}{I}$ at each step of the sequence $\left\{\rho_{j}\right\}_{0\leq j\leq i}$, we obtain a local framed sequence

[TABLE]

So we have the diagram:

[TABLE]

Similarly, either $\frac{A}{I}$ is regular, or the sequence $\left\{\rho_{j}\right\}$ can be chosen such that $e\left(R,\mu\right)$ strictly decreases.

So after a finite sequence of blow-ups, we bring us to the case where $\frac{\widehat{R_{i}}}{\overline{H_{i}}}$ is regular. Hence we can assume $\frac{\widehat{R_{i}}}{\overline{H_{i}}}$ regular and consider $f_{1},\dots,f_{s}$ elements of $R\setminus\left\{0\right\}$ such that $\nu\left(f_{1}\right)=\min\limits_{1\leq j\leq s}\left\{\nu\left(f_{j}\right)\right\}$. We know that the $f_{j}$ are all monomials in the $u^{(i)}$ and that $f_{1}\mathrm{mod}\left(\overline{H_{i}}\right)\mid f_{j}\mathrm{mod}\left(\overline{H_{i}}\right)$. This completes the proof. ∎

Theorem 11.3**.**

Let $R$ be a local quasi excellent domain and $H$ be his implicit prime ideal. We assume that $\frac{\widehat{R}}{H}$ is regular.

We recall that $\mathrm{car}\left(k_{\nu}\right)=0$. There exists a sequence of blow-ups defined over $R$ that resolves the singularities of R.

Proof.

The ring $\widehat{R}_{H}$ is regular by Proposition 8.6. So we know that there exist elements $\left(\widetilde{y}_{1},\dots,\widetilde{y}_{g}\right)$ of $H\widehat{R}_{H}$ that form a regular system of parameters of $\widehat{R}_{H}$.

By definition of $H\widehat{R}_{H}$, it means that there exist $y_{1},\dots,y_{g}$ elements of $H$ and $b_{1},\dots,b_{g}$ elements of $\widehat{R}\setminus H$ such that for every index $i$, we have $\widetilde{y}_{i}=\frac{y_{i}}{b_{i}}$.

The $b_{i}$ are elements of $\widehat{R}_{H}^{\times}$, so

[TABLE]

Then we have some elements $\left(y_{1},\dots,y_{g}\right)$of $H$ that form a regular system of parameters of $\widehat{R}_{H}$.

Now we consider $\left(x_{1},\dots,x_{t}\right)$ some elements of $\widehat{R}\setminus H$ whose images $\left(\overline{x}_{1},\dots,\overline{x}_{t}\right)$ modulo $H$ form a regular system of parameters of $\frac{\widehat{R}}{H}$.

If $\left(y_{1},\dots,y_{g}\right)$ generate $H$, then $\widehat{R}$ is regular. Indeed, in this case, $\left(y_{1},\dots,y_{g},x_{1},\dots,x_{t}\right)$ generate $\widehat{\mathfrak{m}}=\mathfrak{m}\otimes_{R}\widehat{R}$, which is the maximal ideal of $\widehat{R}$.

So

[TABLE]

We know that

[TABLE]

and

[TABLE]

Then

[TABLE]

Then $\dim\left(\widehat{R}\right)=g+t$ and $\left(y_{1},\dots,y_{g},x_{1},\dots,x_{t}\right)$ is a minimal set of generators of $\widehat{\mathfrak{m}}$, and so $\widehat{R}$ is regular.

Now we assume that $\left(y_{1},\dots,y_{g}\right)$ do not generate $H$ in $\widehat{R}$. So let us set $\left(y_{1},\dots,y_{g},y_{g+1},\dots,y_{g+s}\right)$ some elements that generate $H$ in $\widehat{R}$.

We consider $V:=\frac{H\widehat{R}_{H}}{H^{2}\widehat{R}_{H}}$ that is a vector space of dimension $g=\mathrm{ht}\left(H\right)$ over the residue field of $H$ since $\widehat{R}_{H}$ is regular.

We know that $y_{1},\dots,y_{g+s}$ generate $V$ and that

[TABLE]

so there exist elements $a_{1},\dots,a_{g+s}$ of $\widehat{R}$ such that

[TABLE]

In other words there exist $a_{1},\dots,a_{g+s}$ in $\widehat{R}$ and $\left(b_{i,j}\right)_{1\leq i,j\leq g+s}$ in $\widehat{R}_{H}$ such that

[TABLE]

We may assume

[TABLE]

and also that for every $i$, the element $a_{i}$ is not in $H$ or is zero.

Since the $a_{i}$ are in $\widehat{R}$, we look at them modulo $H$. By Theorem 6.21, we know that the classes $\overline{a_{i}}$ of $a_{i}$ modulo $H$ are monomialisable in $\frac{\widehat{R}}{H}$ and that for every $i$, we have $\overline{a_{1}}\mid\overline{a_{i}}.$

Hence after a sequence of blow-ups, we have that $\overline{a_{1}}$ is a monomial $w=\prod\limits_{i=1}^{t}x_{i}^{c_{i}}$ in $x$ multiplied by a unit.

If we can show that $a_{1}$ divides all the $b_{i,j}$, then we could generate $H$ in $\widehat{R}$ by $\left(y_{2},\dots,y_{g+s}\right)$.

Iterating, we could generate $H$ in $\widehat{R}$ by $g$ elements, and it would be over.

So let us show that we can do a sequence of blow-ups such that at the end $a_{1}$ divides all the $b_{i,j}$.

For every index $i\in\left\{1,\dots,g+s\right\}$, there exists $n_{i}\in\mathbb{N}_{>1}$ such that $y_{i}\in\widehat{\mathfrak{m}}^{n_{i}-1}\setminus\widehat{\mathfrak{m}}^{n_{i}}$. We set $N:=\max\limits_{i\in\left\{1,\dots,g+s\right\}}\left\{n_{i}\right\}$, and then for every $i\in\left\{1,\dots,g+s\right\}$, $y_{i}\notin\widehat{\mathfrak{m}}^{N}$.

We have a map $R\to\widehat{R}$ and we know that for every integer $c$, we have $\widehat{\mathfrak{m}}^{c}\cap R=\mathfrak{m}^{c}$. Hence we have an isomorphism $\frac{R}{\mathfrak{m}^{c}}\to\frac{\widehat{R}}{\widehat{\mathfrak{m}}^{c}}$.

So for all $i\in\left\{1,\dots,g+s\right\}$, there exists $z_{i}\in R$ whose class modulo $\mathfrak{m}^{N+2}$ is sent on $y_{i}$ by this map. Hence $z_{i}\mathrm{\text{ }mod}\left(\mathfrak{m}^{N+2}\right)=y_{i}$. Increasing $N$ if necessary, we may assume $\nu\left(\widehat{\mathfrak{m}}^{N}\right)>\nu\left(a_{1}\right)$.

More precisely $y_{i}=z_{i}+h_{i}+\zeta_{i}$ where $h_{i}\in\left(z_{1},\dots,z_{g+s}\right)^{2}$ and $\zeta_{i}\in\left(x_{1},\dots,x_{t}\right)^{N}$.

After a sequence of blow-ups independent of $\left(z_{1},\dots,z_{g+s}\right)$, we may assume that $w$, and so $a_{1}$, divides all the $\zeta_{i}$.

We do $c_{1}$ blow-ups of $\left(z_{1},\dots,z_{g+s},x_{1}\right)$. Each $z_{i}$ is transformed in a $z_{i}^{\prime}$ which is of the form $\frac{z_{i}}{x_{1}^{c_{1}}}$.

We do $c_{2}$ blow-ups of $\left(z_{1}^{\prime},\dots,z_{g+s}^{\prime},x_{2}\right)$. Each $z_{i}^{\prime}$ is transformed in a $z_{i}"$ which is of the form $\frac{z_{i}^{\prime}}{x_{2}^{c_{2}}}=\frac{z_{i}}{x_{1}^{c_{1}}x_{2}^{c_{2}}}$.

We iterate until doing $c_{t}$ blow-ups of

[TABLE]

So we transformed $z_{i}$ in $z_{i}^{(t)}$ which is of the form $\frac{z_{i}}{a_{1}}$.

Then $a_{1}$ divides all the $z_{i}^{(t)}$, and so all the $h_{i}^{(t)}$ and the $y_{i}^{(t)}$. The $b_{i,j}$ are elements of $\widehat{R}_{H}$, so after this sequence of blow-ups, since the strict transform of $H$ is generated by the $y_{i}^{(t)}$, we have that $a_{1}$ divides all the $b_{i,j}$, and the proof is finished. ∎

12. Conclusion.

We know are going to give the principal results of this part. First we recall a fundamental result of Novacoski and Spivakovsky ([42]).

Theorem 12.1**.**

Let $S$ be a noetherian local ring. If the local uniformization Theorem is true for every valuation of rank $1$ centered in $S$, then it is true for any valuation centered in $S$.

So we just have to consider valuations of rank $1$.

Theorem 12.2**.**

Let $S$ be a noetherian equicharacteristic quasi excellent singular local ring of characteristic zero. We consider $\mu$ a valuation of rank $1$ centered in $S$.

There exists a formal framed sequence

[TABLE]

such that for $j$ big enough, $S_{j}$ is regular and for every element $s$ of $S$, there exists $i$ such that in $S_{i}$, $s$ is a monomial.

Proof.

We consider $\widehat{S}$ the formal completion of $S$ and $H$ its implicit prime ideal. By Cohen structure Theorem, there exists an epimorphism $\Phi$ from a complete regular local ring $R$ in $\widehat{S}$. We consider $\overline{H}$ the preimage of $H$ in $R$. We extend now $\mu$ to a valuation $\nu$ centered in $R$ by composition with a valuation centered in $\overline{H}$.

By Proposition 8.6 we know that $\widehat{S}_{H}$ is regular, and by Theorem 11.3 it is enough to show that $\frac{\widehat{S}}{H}$ is also regular.

We know that $\frac{\widehat{S}}{H}\simeq\frac{R}{\overline{H}}$, so we just have to regularize $\frac{R}{\overline{H}}$. We conclude with Theorem 10.15. ∎

Now we prove the principal result of this part: the simultaneous embedded local uniformization for local noetherian quasi excellent equicharacteristic rings.

Theorem 12.3**.**

Let $R$ be a local noetherian quasi excellent complete regular ring and $\nu$ be a valuation centered in $R$.

Assume that $\nu$ is of rank $1$ or $2$ but composed of a valuation $\left(f\right)$-adic where $f$ is an irreducible element of $R$. We assume $\mathrm{car}\left(k_{\nu}\right)=0$.

There exists a formal framed sequence

[TABLE]

such that for every element $g$ of $R$, there exists $i$ such that in $R_{i}$, $g$ is a monomial.

Proof.

We consider the ring $A=\frac{R}{\left(f\right)}$. The valuation $\nu$ is of rank $2$ composed of valuation $\left(f\right)$-adic, so $\nu$ can be written $\mu\circ\theta$ where $\theta$ is the valuation $\left(f\right)$-adic.

So we have a valuation $\mu$ centered in $A$ of rank $1$. By Theorem 12.2, we can regularize $A$, and so there exists a local framed sequence $\left(R,u\right)\to\dots\to\left(R_{i},u^{(i)}\right)$ such that in $R_{i}$, $f$ is a monomial. In $R_{i}$, we also have that every element $g$ of $R$ can be written $g=\left(u_{n}^{\left(i\right)}\right)^{a}h$ where $u_{n}^{\left(i\right)}$ is the strict transform of $f$ and $h$ is not divisible by $u_{n}^{\left(i\right)}$. We apply another time Theorem 12.2 to construct a local framed sequence which monomialize $h$. This completes the proof. ∎

Corollary 12.4**.**

We keep the same notations and hypothesis as in the previous Theorem.

Then $\lim\limits_{\rightarrow}R_{i}$ is a valuation ring.

Remark 12.5*.*

The restriction on the rank of the valuation was setted to give an autosufficient proof. Otherwise, there exists a countable sequence of polynomials $\chi_{i}$ such that every $\nu$-ideal $P_{\beta}$ is generated by a subset of the $\chi_{i}$. Assume the embedded local uniformization Theorem.

Then there exists a local (respectively formal) framed sequence $\left(R,u\right)\to\dots\to\left(R_{i},u^{\left(i\right)}\right)\to\dots$ that has following properties:

1. (1)

For $i$ big enough, $R_{i}$ is regular. 2. (2)

For every finite set $\left\{f_{1},\dots,f_{s}\right\}\subseteq\mathfrak{m}$ there exists $i$ such that in $R_{i}$, every $f_{j}$ is a monomial and $f_{1}\mid f_{j}$.

Then for every element $g$ in $R$, there exists $i$ such that in $R_{i}$, $g$ is a monomial.

Index

• $\epsilon_{\nu}\left(P\right)$ §1
• Framed blow-up Definition 3.3
• G-ring Definition 8.1
• Generalized Puiseux package Definition 10.7
• Immediate successor key polynomial Definition 2.1
• Immediate successors key elements Definition 3.13
• Independent framed blow-up Definition 3.7
• Key element Definition 3.11
• Key polynomial Definition 1.7
• Limit immediat successor key polynomial Definition 2.10
• Limit immediate successor key elements Definition 3.14
• Limit key polynomial Definition 5.2
• Monomial framed blow-up Definition 3.5
• Monomializable Definition 6.3
• Non-degenerate elements Definition 3.1
• Optimal immediate successor key polynomial Definition 2.8
• Puiseux package Definition 6.11
• Quasi excellent ring Definition 8.2
• Truncated valuation Definition 1.9
• Valuation Definition 1.1