Let $R$ be as in Proposition \ref{oneequicompletion}, and let
$\iota\colon R\rightarrow\hat R^\nu$ be the completion morphism. Let $\nu_1$ be
a valuation with which $\nu$ is composed, and denote by $\hat \nu_1$ its extension
to $\hat R^\nu$ according to the first Corollary. Denote by $\overline \nu$ the
residual valuation on $R/(m_{\nu_1}\cap R)$. After subsection \ref{compargr}, we
know that the $\overline \nu$-adic topology on $R/(m_{\nu_1}\cap R)$ is the
quotient of the $\nu$-adic topology of $R$. Even if $R$ is complete for the $\nu$-adic topology,
the quotient $R/(m_{\nu_1}\cap R)$ is not complete for the $\overline\nu$-adic
topology in general (see example 2) above). However, we have\footnote{In the classical theory of
the completion of valued fields, it is the $\nu$-adic completions of the valuation rings which
play a role, more or less explicitly. They are not n\oe therian in general, and a result such as this
Proposition is not true; I believe this complicates the study of {\it compl\'etion par \'etages}; see
[Ri].}  %---------------------------------------------------------------------
\begin{proposition}\label{residcomp} If the quotient ring $\hat R^\nu /(m_{\hat \nu_1}\cap \hat
R^\nu)$ is complete for the\break $m_{\hat
\nu}\cap\hat R^\nu/(m_{\hat \nu_1}\cap \hat R^\nu)$-adic topology, the morphism
$$R/(m_{\nu_1}\cap R)\rightarrow \hat R^\nu/(m_{\hat \nu_1}\cap \hat R^\nu)$$ induced by $\iota$ is
the completion morphism of the first ring with respect to the $\overline \nu$-adic
topology, quotient of the $\nu$-adic topology of $R$.
\end{proposition}
%----------------------------------------------------------------------
\begin{proof} Since $R$ is n\oe therian, the value semigroup $\Gamma$ of $\nu$ on $R$ is
well ordered (Proposition \ref{wellord}), so that the the $\nu$-adic topology may be
defined by a decreasing chain of ideals. Moreover,  if $\delta_1$ denotes the smallest
non-zero element of $\Gamma_1=\\ \nu_1 (R\setminus \{ 0\})$, which is also well ordered,
the ideal $m_{\nu_1}\cap R$ is equal to $\{x\in R\mid\nu_1 (x)\geq \delta_1\}$, so that its
closure in $\hat R^\nu$ is $m_{\hat\nu_1}\cap \hat R^\nu$. Now we can apply
Theorem 8.1, page 56 of [Ma], which states the equality we seek.
\end{proof}
%---------------------------------------------------------------------
For the next two propositions, given $R\subset R_\nu$, remember that there is a unique
sequence of consecutive valuation rings
$$R_{\nu}\subset R_{\nu_1}\subset\cdots\subset R_{\nu_{h-1}}$$ ending at the ring of the
valuation of height one
$\nu_{h-1}$ with which $\nu$ is composed, and by the results of subsection \ref{speciaval} we can
produce a sequence of consecutive specializations of the valuation $\nu$ having a center in $R$, of the
form
$$R\subset R_{\nu_{-s}}\subset R_{\nu_{-s+1}} \subset\cdots\subset R_{\nu_{-1}}\subset R_\nu$$ and
such that
$R_{\nu_{-s}}$ dominates $R$.  \par\noindent 
\begin{proposition}\label{hensel} Let $R$ be a n\oe therian equicharacteristic analytically
irreducible local ring, and $\nu$ a valuation of its field of fractions, non negative on $R$. If, for
all pairs of consecutive valuation rings $R_\mu\subset R_{\mu_1}$ appearing in the sequence of
consecutive valuation rings which connects $R_\nu$ to $R_{\nu_{h-1}}$, the quotient $
R/(m_{\mu_1}\cap R )$ is complete for the $m_{\mu}\cap  R /(m_{ \mu_1}\cap R) $-adic topology and
$R$ is complete for  the ${\mathbf p}$-adic topology, where ${\mathbf p}$ is the center of the valuation
of height one with which $\nu$ is composed, each pair $( R  ,m_{\nu_i}\cap R )$ is
henselian. If the assumption of completeness can be extended to a sequence of specializations of $\nu$ ending at a
valuation ring dominating $R$, the local ring $( R , m)$ is henselian.
\end{proposition}
%--------------------------------------------------------------------
 \begin{proof} Notice first that by definition of
restricted power series, for $m_{ \nu_{j}}\cap R\subseteq m_{ \nu_{j-1}}\cap
R$, the natural map of polynomial rings extends to a map of rings of restricted power
series $$\bigl( R/(m_{\nu_j} \cap R)\bigr)\{ X \}\rightarrow \bigl(
R/(m_{\nu_{j-1}} \cap R)\bigr)\{ X \}.$$ \par\noindent We prove that $(R,\ m_{ \nu_i}
\cap R )$ satisfies Hensel's Lemma in the sense of Bourbaki ({\it cf } [B3], Chap. 3, \S 4,
No.3), which we now recall.  \par\medskip\noindent {\it Let $F\in R \{ X\}$ be a
restricted power series, and assume that the image of $F$ modulo $m_{\nu_i} \cap R $, denoted by $\overline F$, admits a
factorization
$\overline F=\overline P.\overline Q$, where $\overline P\in \bigl(R/(m_{\nu_i}
\cap R)\bigr)[X]$ is a unitary polynomial, and $\overline Q$ a restricted power
series $\overline Q\in \bigl(R/(m_{\nu_{j-1}} \cap R)\bigr)\{ X \}$, with
$\overline P,\ \overline Q$ strongly relatively prime. Then they lift in a unique way to a
unitary polynomial $P\in R [X]$ and a restricted power series $Q\in R\{ X\}$,
strongly relatively prime, such that $F=P.Q$. If $F$ is a polynomial, so is
$Q$.} \par\medskip\noindent To prove this, note that the ideals $m_{\nu_k} \cap R$ are closed for the $\nu_k$-adic topologies,
which all coincide. By Theorem 1 of the quoted paragraph of [B3], the result is true for $\nu_{h-1}$, which is a valuation of height
one, since the elements of $m_{\nu_{h-1}} \cap R$ are topologically nilpotent
as we saw in Remark 1) following Proposition
\ref{oneequicompletion}.\par\noindent To prove the general case, it suffices to note that the
statement is amenable to induction on the height and that if $\hbox{\rm h}(\nu_{j-1})>1$, there
is a unique (up to isomorphism) valuation $\nu_j$ of $R$ which is the immediate successor of
$\nu_{j-1}$. The valuation induced by
$\nu_{j-1}$ on the quotient ring $R/(m_{\nu_j}\cap R ) $ is of height
at most one, according to Section \ref{resigroup}. Assuming that we have a decomposition in
$\bigl(R/(m_{{\nu}_{j-1}}\cap R )\bigr) \{ X\}$ as in Hensel's lemma, by
using the case of height one, we can lift it to a decomposition in $\bigl(R/(m_{
{\nu}_j}\cap R )\bigr) \{ X\}$, and continue this until we have a decomposition in
$\bigl( R/{\mathbf p}\bigr)\{ X\}$, where ${\mathbf p}$ is the center of the
valuation $\nu_{h-1}$ of height one with which $ \nu$ is composed, and we are reduced
to the case of height one. The last sentence is proved by applying this result to the valuation
ring $R_{\nu_{-s}}$ which dominates $(R,m)$.
\end{proof}
\begin{proposition}\label{fieldrep} Let $R$ be a n\oe therian equicharacteristic analytically
irreducible local ring, and $\nu$ a valuation of its field of fractions, non negative on $R$. If for
all pairs of consecutive valuation rings $R_\mu\subset R_{\mu_1}$ appearing in a sequence of nested
valuation rings which contains $R_\nu$ and connects a valuation ring dominating $R$ to the ring of  a
valuation of height one the quotient
$ R /(m_{\mu_1}\cap  R  )$ is complete for the $m_{\mu}\cap R/(m_{\mu_1}\cap R)$-adic topology and
$R$ is complete for the ${\mathbf p}$-adic topology, where ${\mathbf p}$ is the center of the valuation of height
one with which $\nu$ is composed, the local ring $R$ admits a field of representatives.
\end{proposition}
%----------------------------------------------------------
\begin{proof}  We have a sequence of
quotient rings $$R/(m_{\nu_{h-1}}\cap R )
\to\cdots\to R/(m_{\nu_{-s+1}}\cap  R )\to
R/(m_{\nu_{-s}}\cap R)=k_R ,$$ where some of the maps may be
isomorphisms. I need the following:
%----------------------------------------------------
\begin{lemma}\label{onerep} Let $S$ be a n\oe therian
equicharacteristic local ring and $I$ an ideal of $S$. Assume that the quotient $S/I$ admits a
field of representatives and that $S$ is complete for the $I$-adic topology. Then
$S$ admits a field of representatives.
\end{lemma} 
%-----------------------------------------------
\begin{proof} This follows from Cohen's 
Theorem as found in
[Z-S], Vol. 2, Chap. VIII, \S 12 after making the following remark.
\par\noindent Let $I$ be
an ideal of $S$ such that $S/I$ admits a field of representatives. Let $\phi\colon S\to S/I$
be the natural map. Set $B=\phi^{-1}(k)$, where $k\subset S/I$ is a field of
representatives.  We have
$I=m\cap B$, where
$m$ is the maximal ideal of $S$. \par\noindent It follows from this that $B$ is closed in $S$ for
the $I$-adic topology, so that it is complete for its $m_B=m\cap B$-adic topology, and by 
Cohen's Theorem it has a field of representatives, which is also a field of representatives for
$S$.\end{proof} \par\medskip\noindent
I apply the lemma to the maps $$R/(m_{
\nu_j}\cap R ) \to R/(m_{\nu_{j-1}}\cap R ). $$ Each of them
satisfies the hypotheses of the Lemma since by subsection \ref{resigroup} the valuation
induced by $\nu_{j-1}$ on $R/(m_{\nu_j}\cap R)$ is of height
at most one, and so the residual topology coincides with
the $(m_{\nu_{j-1}}\cap R )/(m_{\nu_j}\cap R )$-adic
topology, for which this ring is complete by our assumption. So we can step by step lift $k$ to a
field of representatives of $R/(m_{\nu_{h-1}}\cap R ) $. Since the valuation
$\nu_{h-1}$ is of height one, and we assume that the ring $R$ is complete for the ${\mathbf p}$-adic
topology, where ${\mathbf p}=m_{ \nu_{h-1}}\cap R$, we can apply the lemma one last
time to lift the field of representatives to $R$.
\end{proof}
%------------------------------------------------------------
\par\medskip\noindent
{\textbf {Question.}} Can one extend this result to the case where $R$ is not
equicharacteristic, to prove under similar assumptions the existence of an injection $W(k)\to R$ for a
n\oe therian analytically irreducible local ring $R$ of unequal characteristic with a perfect
residue field $k$ and a valuation $\nu$, where $W(k)$ is the ring of Witt vectors relative to the residue field
$k$ of $S$ ([B3], Chap. IX, \S\S 1,2, [Ei], Chap.7)? \par\medskip\noindent
\begin{remark} In order to be able to apply Propositions \ref{residcomp}, \ref{hensel} and \ref{fieldrep} 
the easiest way is to deal with a local ring which is complete with respect to the ${\mathbf p}$-adic topologies 
for primes ideals ${\mathbf p}$ which are the centers of the valuations with which $\nu$ is composed. Then $R$ and its quotients by the
centers of valuations with which $\nu$ is composed will be complete and separated with respect to the $\nu$-adic topology and its
images in these quotients. This will be the case if $R$ is $m$-adically complete, by [Z-S], {\it loc.cit}, and then of course the two
propositions above are not needed in view of Hensel's lemma and Cohen's theorem; they will be used elsewhere. \end{remark}



 Suppose that there is a regular local ring
$R_0\subset R$ such that $R$ is a finite $R_0$-algebra. Then the completion $\hat R^{m_0}_0$ is isomorphic to $k[[x_1,\ldots
,x_d]]\subset\hat R^m$. Consider the restriction $\nu^{(0)}$ of $\nu$ to $R_0$,  which is centered at $m_0$.  If we can prove
the proposition for $\nu^{(0)}$, obtaining an ideal $H^{(0)}$ of $\hat R^{m_0}$ and an extension $\hat\nu^{(0)}$ to $\hat
R^{m_0}/H^{(0)}$, the completeness of $\hat R^{m_0}$ will ensure that we can extend the valuation $\hat\nu^{(0)}$ to a
valuation $\hat \nu$ of
$\hat R^m/H^{(0)}\hat R^m$. It is then not difficult to check the assertion on the graded rings.\par\noindent
I may therefore assume that $R$ is regular, and consider the valuations with which $\nu$ is composed. The case of height
one was settled above. Let us consider two successive valuations $\mu$ and $\mu_1$ with which $\nu$ is composed, with
centers ${\mathbf p}$ and ${\mathbf p}_1$ respectively, and denote by $\Psi$ the kernel  of the corresponding map
$\lambda\colon\Phi_{i-1}\to
\Phi_i$ of value groups. Assume that we have built an ideal
$H_1\subset
\hat R^m$ and a pseudo-valuation $\hat\mu_1$ on
$\hat R^m/H_1$ which  extends $\mu_1$, with the properties that $H_1$ is a prime ideal, $H_1\cap R=(0)$ and $$\hbox{\rm
gr}_{\hat \mu_1}\hat R^m/H_1=\hbox{\rm gr}_{\mu_1}R\otimes _{R/{\mathbf p}_1} \hat R^m/{\mathbf p}_1\hat R^m.$$
 Reasoning by induction on the 
dimension, and using the results of section
\ref{proof} for rings of smaller dimension than
$R$, we may assume embedded local uniformization for the image
$\overline\nu$ of $\nu$ in the ring $R/{\mathbf p}_1$  by a toric map of $R$. Thus, after replacing $R$ by a toric modification
$R'$ of $R$, which is still a regular local ring, we may assume that both
$R$ and $R/{\mathbf p}_1$ are regular. For the next argument we write $R$ instead of $R'$, for simplicity. The  valuation
$\overline\mu$, which is of height at most one, may be extended to the quotient of $\hat R^m/{\mathbf p}_1\hat R^m$ by
an ideal $\overline H=\bigcap_{\psi\in \Psi_+}\P_\psi \hat R^m/{\mathbf p}_1\hat R^m$. We may for the same reason assume
that $\overline H$ is generated by a regular sequence $(\overline b_1,\ldots ,\overline b_k)$ in
$\hat R^m/{\mathbf p}_1^m\hat R^m$.  Representatives $( b_1,\ldots , b_k)$ in $\hat R^m$ generate an ideal
$\tilde H\subset \hat R^m$. The ideal ${\mathbf p}_1 \hat R^m$ is generated by $(x_1,\ldots ,x_c)$, with
$x_i\in R$. Assume first that $c=1$. Then we have $\tilde H\cap R=(0)$. Indeed, let $a\in \tilde H\cap R$, say
$a=\sum_{j=1}^k\lambda_jb_j$. The image of $a$ in $R/{\mathbf p}_1$ is in $\overline H\cap (R/{\mathbf p}_1)=(0)$
so that $a\in \tilde H\cap {\mathbf p}_1\hat R^m$. The two ideals are transversal (their intersection is equal to their
product) so that
$a=\lambda x_1a'$ with $a'\in \tilde H\cap R$; by iteration this shows that $a\in \bigcap_{i=1}^\infty x_1^i\hat
R^m=(0)$. Using this argument inductively to lift the generators $(\overline b_1,\ldots ,\overline b_k)$ of $\overline H$ to the
quotients $\hat R^m/(x_1,\ldots ,x_j)\hat R^m$ we finally get that the ideal $\tilde H$ satisfies $\tilde H\cap R=(0)$, and
$\tilde H\subset {\mathbf p}\hat R^m$. Now we can extend $\mu$ to $\hat R^m/H=\hat R^m/H_1/\tilde H(\hat R^m/H_1)$ as
follows: given an element $y\in
(\hat R^m/H)\setminus\{ 0\}$, we can lift it to $\tilde y\in
\hat R^m/H_1$ which  has a pseudo-valuation $\hat\mu_1 (\tilde y)\in \Phi_i$. Since $\tilde y\notin \bigcap_{\psi\in
\Psi_+}\P_\psi \hat R^m/H_1$, it follows from ([Z-S], Vol.II, Appendix 3, Lemma 4) that there is a $\phi\in \Phi_{i-1+}$ such
that $\tilde y\notin \P_\phi\hat R^m/H_1$, and $\lambda (\phi )=\hat \mu_1 (\tilde y )$. By the argument invoked at the
beginning, there is a largest $\eta\in
\Phi_{i-1}$ such that $\tilde y\in \P_\eta$, and $\lambda (\eta )=\hat \mu_1(\tilde y)$. Setting $\hat \mu (y)=\eta$ defines a
pseudo-valuation $\hat\mu$ on $\hat R^m/H$ with value group $\Phi_{i-1}$. We have $H\subset {\mathbf p}$ so the same argument as
above  shows that
$$\hbox{\rm gr}_{\hat \mu}\hat R^m/H=\hbox{\rm gr}_{\mu}R\otimes _{R/{\mathbf p}} (\hat R^m/{\mathbf p}\hat
R^m)^{(\overline \nu)}.$$ The map $R\to \hat R^m/H$ induces the map $$\hbox{\rm gr}_{\mu}R\to\hbox{\rm
gr}_{\mu}R\otimes _{R/{\mathbf p}} 1\subset
\hbox{\rm gr}_{\mu}R\otimes _{R/{\mathbf p}} (\hat R^m/{\mathbf p}\hat R^m)^{(\overline \nu)},$$ which shows that
$H(\hat R^m/H_1)\bigcap R=(0)$. The point is that the ideals of the form
${\mathbf p}^N\P_{\phi_0}$ for some $\phi_0$ with
$\lambda(\phi_0)=\phi_1$ are cofinal in  the ideals
$(\P_\phi)_{\lambda (\phi )=\phi_1}$, as are the ideals $(\P_\psi\P_{\phi_0})_{\psi\in \Psi_+}$.\par\noindent Following this inductive
procedure, we continue until the valuation
$\mu$ is our original valuation
$\nu$ with center $m$. But now the graded ring $\hbox{\rm gr}_{\hat \nu}\hat R^m/H$ is equal to $\hbox{\rm gr}_\nu R$,
so that it  is an integral domain, and $\hat \nu$ is a valuation and not only a pseudo-valuation, so that the images in $\hat
R^m/H$ of all the $\hat
\mu$ are also valuations, those with which $\hat\nu$ is composed.\par\noindent
Now we remember that this was done in a toric modification (and localization at the point picked by $\nu$) $R'$ of $R$.
Denote by $H'\subset \hat {R'}^{m'}$ the ideal built in the preceding argument. Denote by $\tau\colon\hat R^m \to \hat
{R'}^{m'}$ the  natural map of complete local rings, and set $H=\tau^{-1}(H')=H'\cap \hat R^m$. The valuation $\nu$ extends to
a valuation $\hat\nu$ of $\hat R^m/H$, with group $\Phi$ and this proves the Proposition.
\end{proof} 
%--------------------------------------------------------------------------------------------------
\begin{remark} 1) It would be preferable to show that one can reduce to the case  where ${\mathbf p}_1$ is generated by a
regular sequence by using the graded algebra of the valuation,  and thus remove the assumption that $R$ is
finite over a regular local ring.\par\noindent

Let us set $K={\mathbf q}(\infty)$ and note that we have by construction, since $\P_\phi\hat R_1\supset\P^+_{\phi_1}\hat R_1$ for all $\phi$ such
that $\lambda (\phi)=\phi_1$
$$\frac{\bigcap_{\lambda (\phi)=\phi_1}\P_\phi\hat R_1}{\P^+_{\phi_1}\hat R_1}\subset (K\cap \P_{\phi_1}\hat R_1)\hat R_1/\P^+_{\phi_1}\hat
R_1\subset
\frac{(K+\P_{\phi_1}^+)\cap \P_{\phi_1}\hat R_1}{\P^+_{\phi_1}\hat R_1}.$$
from this we deduce
$$\bigcap_{\lambda (\phi)=\phi_1}\P_\phi\hat R_1\subset (K+\P_{\phi_1}^+)\cap \P_{\phi_1}\hat R_1+\P_{\phi_1}^+\hat R_1$$
By the computation we made above, the image in $\hat R_1/K$ of $\P_{\phi_1}\hat R_1$ is 
$\bigcap_{\lambda(\phi)=\phi_1}\P_\phi\hat R_1+\P_{\phi_1}^+\hat R_1$, so that
if we pass to the quotient $\hat R_0=\hat R_1/K$  we get 
$$\bigcap_{\lambda (\phi)=\phi_1}\P_\phi\hat R_0/\P_{\phi_1}^+\hat R_0=(0).$$

\par\noindent From this follows, by Lemma \ref{grim}
$$\frac{\bigcap_{\lambda(\phi)=\phi_1}\P_\phi\hat R_1/H}{\P^+_{\phi_1}\hat R_1/H}=(0).$$
Remark also that we have $\frac{\hat R_1/H}{{\mathbf p}_1\hat R_1/H}=\frac{\hat R_1/{\mathbf p}_1\hat R_1}{\overline H}=\hat{\overline
R}_1/\overline H$.
\par\medskip\noindent
Let us now consider the ring $\hat{\overline R}_0=\hat{\overline R}_1/\overline H$ and the $\hat{\overline R}_0$-modules
$N_\phi:=\frac{\P_\phi\hat{\overline R}_0}{\P^+_{\phi_1}\hat{\overline R}_0}\subset M:=\frac{\P_{\phi_1}\hat{\overline
R}_0}{\P^+_{\phi_1}\hat{\overline R}_0}$; we have just seen that $\bigcap_{\lambda(\phi)=\phi_1}N_\phi=(0)$. It follows from lemma \ref{inter} that
we have $(N_\phi :M)=\P_{\psi(\phi)}\hat{\overline R}_0\subset {\mathbf p}\hat{\overline R}_0$.\par\noindent
Since the ring $R$ is assumed to be excellent, the ideal ${\mathbf p}\hat R^m$ is the intersection of prime ideals of the same height, and therefore the same
is true for
${\mathbf p}\hat{\overline R}_0$ since $H\subset {\mathbf p}\hat R_1$. (see [EGA], \S 7, scholie 7.8.3, (x), p. 216). Let us consider the multiplicative
subset $T$ of  $\hat{\overline R}_0$ which is the complement of the union of the prime ideal of ${\mathbf p}\hat{\overline R}_0$. The rings
$(\hat{\overline R}_0)_T/{\mathbf p}^n(\hat{\overline R}_0)_T$ are artinian. Moreover, since $(N_\phi :M)\subset
{\mathbf p}\hat{\overline R}_0$, we have that $\bigcap_{\lambda(\phi)=\phi_1}(N_\phi\otimes_{\hat{\overline R}_0} (\hat{\overline
R}_0)_T)=(0)$.\par\noindent Now we can apply the linear compactness of complete filtered modules (see [B3}, Chap III, \S 2, No. 7), or generalized Chevalley
theorem. It gives us that for every integer $n$ there exists $\phi\in \lambda^{-1}(\phi_1)$ such that $N_\phi \otimes_{\hat{\overline R}_0} 
(\hat{\overline R}_0)_T\subset {\mathbf p}^n M\otimes_{\hat{\overline R}_0}  (\hat{\overline R}_0)_T$. Finally we can deduce from this the 
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{Hgodown} Keeping the notations just introduced, let $R\to R'$ be a rational toric extension subordinate to $\nu$ associated to a finite
family
$(\xi_j)_{j\in F}$ of elements such that $\nu (\xi_j)=0$ for $j\in F$. The equality
$$\P_\phi (R')\hat R'^{m'}\cap \hat R^m=\P_\phi (R)\hat R^m$$ holds for every $\phi\in \Gamma$ and in particular, setting
$H'=\bigcap_{\phi\in
\Phi_+}\P_\phi (R')
\hat R'^{m'}$, we have via the natural inclusion $\hat R^m\subset \hat R'^{m'}$,
$$H'\cap
\hat R^m=H.$$\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} The inclusion $\P_\phi (R)\hat R^m\subset \P_\phi (R')\hat R'^{m'}\cap \hat R^m$ is clear. Let us prove the inclusions
$\P_\phi (R')\hat R'^{m'}\cap \hat R^m\subset\P_\phi(R)\hat R^m$. In view of our assumption, denoting by ${\mathbf p}'$ the center of $\nu$ in $R'$,
we have a natural birational inclusion $R/{\mathbf p}\subset R'/{\mathbf p}'$. Moreover, the map of $R/{\mathbf p}$-modules $$\P_\phi
(R)/\P^+_\phi(R)\to\P_\phi (R')/\P^+_\phi(R')$$ is injective. Since the $m'$-adic topology of $R'/{\mathbf p}'$
induces the $m$-adic topology on
$R/{\mathbf p}$, the map of completions $$\P_\phi
(R)\hat R^m/\P^+_\phi(R)\hat R^m\to\P_\phi (R')\hat R^{m'}/\P^+_\phi(R')\hat R^{m'}$$ is also injective. If an element $x$ of $\P_\phi (R')\hat
R'^{m'}\cap\hat R^m$ belonged to
$\P_{\phi_1}(R)\setminus \P^+_{\phi_1}(R)$ for
$\phi_1<\phi$, it would not belong to $\P^+_{\phi_1}(R')\hat R'^{m'}$, which contains $\P_\phi (R')\hat R'^{m'}$, and this is a contradiction.
\end{proof}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{definition}
Assertion TLU(d) (Toric local uniformization in dimension $\leq d$) is the following: For every local n\oe therian excellent
equicharacteristic integral domain
$R$ of dimension
$\leq d$, given a rational valuation $\nu$ on $R$, there is a rational toric extension $R\subset R'\subset R_\nu$ such that $R'$ is a regular local ring
, $\nu$
extends to a valuation $\hat\nu$ of a regular quotient $\hat {R'}^{m'}/H'$ of the $m'$-adic completion of $R'$ in such a way that the
induced map of graded rings
$$\hbox{\rm gr}_\nu R'\to \hbox{\rm gr}_{\hat\nu} \hat {R'}^{m'}/H'$$ is an isomorphism; in particular, $H'\cap R'=(0)$
\par\noindent 


2) One can slightly generalize the concept of a rational valuation as follows \par\noindent
\begin{definition}
 The valuation $\nu$ is said to be generically rational if
 the valuation induced by
$\nu$ on the local ring $R_{\mathbf p}$ is rational, where ${\mathbf p}$ denotes the center of $\nu$ in $R$.\end{definition}
The following result helps to explain the presence of binomial relations in certain non-rational cases.
%----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proposition}\label{nonratbin} Let $R\subset R_{\nu_1}$ be a generically rational valuation of $R$, and let $(\overline\eta_j)_{j\in J}$ be a set of
generators of the
$R/{\mathbf p}_1$-algebra $\hbox{\rm gr}_{\nu_1}R$. Then the map of $R/{\mathbf p}_1$-algebras sending $W_j$ to $\overline\eta_j$ induces a
presentation
$$\hbox{\rm gr}_{\nu_1}R=R/{\mathbf p}_1[(W_j)_{j\in J}]/\bigl((\lambda_m W^m-\lambda_nW^n)_{(m,n)\in E}\bigr)\ ,\ \ \ \ \ \
\lambda_m,\lambda_n\in R/{\mathbf p}_1\setminus\{0\}.$$
%----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\end{proposition}
%----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof}
By our hypothesis, the valuation induced by $\nu_1$ on $R_{{\mathbf p}_1}$ is rational. On the other hand we
have $\P_{\phi_1}(R_{{\mathbf p}_1})=\P_{\phi_1}(R)R_{{\mathbf p}_1}$ and, setting $\kappa({\mathbf p}_1)=R_{{\mathbf p}_1}/{\mathbf
p}_1R_{{\mathbf p}_1}$,
$$\hbox{\rm gr}_{\nu_1}R_{{\mathbf p}_1}=\hbox{\rm gr}_{\nu_1}R\otimes_{R/{\mathbf p}_1}\kappa({\mathbf p}_1).$$The
elements $\overline\eta_j\otimes 1$ generate the $\kappa ({\mathbf p}_1)$-algebra $\hbox{\rm gr}_{\nu_1}R_{{\mathbf p}_1}$, so by Proposition
\ref {binomialideal} it is of the form $\kappa ({\mathbf p}_1)[(W_j)_{j\in
J}]/(W^m-\lambda_{mn}W^n)$, where
 $\lambda_{mn}\in \kappa({\mathbf p}_1)^*$.  If we remember that each
$\P_{\phi_1}/\P^+_{\phi_1}$ is a torsion-free $R/{\mathbf p}_1$-module the result follows from the flatness of $\kappa({\mathbf p}_1)$
over $R/{\mathbf p}_1$, writing
$\lambda_{mn}=\lambda^{-1}_m\lambda_n$ with
$\lambda_m,\ \lambda_n\in R/{\mathbf p}_1\setminus \{0\}$.\end{proof}


\subsection{More speculation}
Keeping the notations of the previous subsection and of subsection \ref{abyssalph}, remark that if the images of the $(\xi_i)_{i\in F}$ generate the
maximal ideal of $\hat R^{(\nu)}$, the (\xi_i)_{i\in F}$ must generate the maximal ideal of $R$. Let us denote by
$\tilde\nu$ the valuation of
$R$ which is monomial in the variables
$(\xi_i)_{i\in F}$ and satisfies $\tilde\nu (\xi_i)=\nu(\xi_i)$ for $i\in F$. The valuation $\tilde\nu$ extends as a monomial valuation to $\hat
R^{(\nu)}$ and a toric map which uniformizes as above the valuation $\tilde\nu$ via a uniformisation of its extension to $\hat R^{(\nu)}$ also uniformizes
$\nu$. Now the set of rational valuations which are uniformized by a given process is open in the patch topology of the Zariski-Riemann manifold. 
By the compactness, it would follow that every rational valuation is resolved by a toric map 
$\tilde\nu$

First, we have in any basic extension $\hat{\overline\nu} (\overline\lambda)<\infty$
unless $\lambda \in {\mathbf p_{h-1}}'(\hat R^m/H)'$, and in that case, after some calculation, we see that $x\in
\hat\P^+_{\phi_{h-1}}$. Let us show that any element of $\hat\P_{\phi_{h-1}}/\hat\P^+_{\phi_{h-1}}$ is in one of the
$\hat\P_\phi/\hat P_{\phi_{h-1}}$. Let us write $x=\sum_{i=1}^rc_ie_i$ a representative of such an element. Since all the $e_i$ have the
same $\nu_{h-1}$-valuation, by considering 






$\overline R\to
\overline R^{(\overline\nu)}=\overline R^m/\overline H$ 

\end{proof}\par\noindent

%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
In the proof, I consider the valuations $(\nu_i)_{0\leq i\leq h-1}$ with which the rational valuation $\nu$ is composed, with $\nu_0=\nu$ and
$\nu_{h-1}$ of height one. The idea is to build a sequence of ideals $H_i$ and extensions of the $\nu_i$ by loose valuations of the $\hat R^m/H_i$, by
descending induction on $i$. 
%---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proposition}\label{pseudoval} Let $(R,\nu)$ be a valued n\oe therian local integral domain and $H$ an ideal of $R$. Let us assume that $H$ is
closed in
$R$ for the $\nu$-adic topology, that is
$$\bigcap_{\phi\in \Phi_+}(H+\P_\phi)=H$$ and that in addition, if we denote by $$\Phi=\Phi_0\to \Phi_1\to\cdots \to \Phi_i\to\Phi_{i+1}\to \cdots\to
\Phi_{h-1}\to (0)$$ the sequence of the groups of the valuations with which $\nu$ is composed, for each $\lambda_i\colon \Phi_i\to \Phi_{i+1}$
we have for each $\phi_{i+1}\in \Phi_{i+1+}$
$$\bigcap_{\phi_i\in\lambda_i^{-1}(\phi_{i+1})}(H+\P_{\phi_i})=H+\P^+_{\phi_{i+1}}.$$ Then the valuation $\nu$ gives rise
to a loose valuation in $R/H$ whose group is contained in $\Phi$.\end{proposition}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} Let us first assume that $\nu$ is of height one. Given a nonzero element $\overline x\in R/H$, I claim that the valuations of all the
elements in $R$ whose image in $R/H$ is $\overline x$ are bounded. Fix such an element $x_0$; if the valuations are not bounded, we have $x_0\in
\bigcap_{\phi\in \Phi_+}(H+\P_\phi)$, which is equal to $H$ by assumption, a contradiction. Since $\Phi$ is of height one, there is a
largest element among these valuations, which we choose as $\overline\nu (\overline x)$. \par\noindent
If $\overline z=\overline x +\overline y$, we may choose representatives $x$ and $y$ with the largest valuations, but this does not guarantee that
$x+y$ gives the largest valuation in its class mod $H$, so that indeed\par\noindent
$\overline\nu (\overline z)\geq \nu (x+y)\geq \hbox{\rm min}(\overline \nu (\overline x),\overline \nu (\overline y))$.\par\noindent
I leave it as an exercise to check that if $\overline\nu (x)\neq \overline\nu (y)$ we have equalities. Remark also that the ideal of
elements of $R/H$ having valuation $\geq \phi$ is $\P_\phi R/H$.\par\noindent Similarly, choosing representatives of maximal valuation for
$\overline x,\overline y$ gives\par\noindent
$\overline\nu (\overline {xy})\geq \overline \nu (\overline x)+\overline \nu (\overline y)$.\par\noindent
Let us now proceed by induction and assume that we have defined a loose valuation $\overline\nu_{i+1}$ on $R/H$. Given an element $\overline
x\in R/H$, let $\overline\nu_{i+1}(\overline x)$ be its valuation. I claim that the valuations $\nu_i (x)$ of the representatives of $\overline x$ in
$R$ such that $\nu_{i+1}(x)=\overline\nu_{i+1}(\overline x)$ are bounded. Denote this last valuation by $\phi_{i+1}$. Let $x\in R$ be a
representative of $\overline x$; if the valuations were not bounded in $\lambda_i^{-1}(\phi_{i+1})$, we would have the inclusion 
$$x\in\bigcap_{\phi_i\in\lambda_i^{-1}(\phi_{i+1})}(H+\P_{\phi_i})=H+\P^+_{\phi_{i+1}},$$
where the last equality follows from our assumptions. But this means that $\overline x$ has a representative $y$ of valuation
$\nu_{i+1}(y)>\phi_{i+1}$, which contradicts our definitions. Since our set of valuations is bounded in $\lambda_i^{-1}(\phi_{i+1})$, by the
reference given above, the set of valuations $\nu_i(y)\in\lambda_i^{-1}(\phi_{i+1})$ of representatives $y$ of $\overline x$ is finite, and we
choose its largest element as
$\overline \nu_i (\overline x)$. The proof that it is a loose valuation is almost the same as above.\end{proof}
%--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{Remark} 1) The result still holds, with the same proof, if $\nu$ is a loose valuation with values in a group of height one, and more
generally if $\nu$ is a loose valuation ``composed'' in the same sense as valuations are, through a sequence of monotone maps
$\lambda_i\colon\Phi_i\to\Phi_{i+1}$, with loose valuations.
\par\noindent 2) It would be useful to know when $\overline \nu$ is a valuation apart from the cases where $H$ is the center of a
valuation with which $\nu$ is composed and $\overline \nu$ the corresponding residual valuation.\end{Remark}
%--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

\par\noindent Let us consider the case $i=h-1$; denote temporarily by $\nu$ the valuation of height one
$\nu_{h-1}$, and let $H$ stand for $H_{h-1}=\bigcap_{\phi\in \Phi_+}\P_\phi \hat R^m$. The valuation
$\nu$ extends to a loose valuation $\hat\nu$ of $\hat R^m/H$ as follows:
For an element $x\in (\hat R^m/H)\setminus \{ 0\}$ there is a representative $\tilde x\in \hat R^m$ and
there is a $\phi\in \Phi_+$ such that $\tilde x\notin \P_{\phi}\hat R^m$; since $\Phi$ is archimedian and $R$  is n\oe 
therian, the elements of $\nu (R\setminus \{ 0\})$ which
are smaller than $\phi$ (see [Z-S], {\it loc. cit}) are finite in number, so that there is an element $\eta\in
\nu(R\setminus \{ 0\})$ such that $\tilde x\in\P_\eta\hat R^m\setminus\P^+_\eta\hat R^m$.\par\noindent Let us check that this $\eta$ is independant of
the choice of the representative of $x$; if $\tilde x-\tilde y\in H$, assuming that $\eta (\tilde x)< \eta (\tilde y)$, we must have $\eta (\tilde x-\tilde
y)=\eta(\tilde x)$ and this contradicts the fact that $\tilde x-\tilde y\in H$.\par\noindent We may now set
$\hat\nu (x)=\eta$; one verifies as in the proof of proposition \ref{pseudoval} that this defines a loose valuation on $\hat R^m/H$. Finally, we have
$H\cap R=(0)$ since $\P_\phi\hat R^m\cap R=\P_\phi$ and $\bigcap_{\phi\in \Gamma}\P_\phi=(0)$.
 \par Now let us remark that if
${\mathbf p}_\nu$ is the center of $\nu$, we have the inclusions
${\mathbf p}_\nu\P_\phi\subset\P_\phi^+$ and the equalities $\P_\phi \hat R^m=\P_\phi
\otimes_R\hat R^m$ by the flatness of $\hat R^m$ over $R$. We therefore have the following
equalities after remarking that $H\subset \P_\phi\hat R^m$ for all $\phi\in \Phi_+$:  $$(\hbox{\rm gr}_{\hat \nu}\hat
R^m/H)_\phi=\frac{\P_\phi \hat R^m/H}{\P^+_\phi\hat R^m/H}=\frac{\P_\phi\hat R^m}{\P^+_\phi \hat R^m+{\mathbf
p}_\nu\P_\phi \hat R^m}=\frac{\P_\phi}{\P^+_\phi}\otimes_R(\hat R^m/{\mathbf p}_\nu\hat R^m).$$ 
Note that the graded
algebra $\hbox{\rm gr}_{\hat \nu}\hat R^m/H$ is not necessarily an integral domain, except in the case where $\nu$ has center $m$ studied in [S2]. In
this case, we have, setting $\hat R_{h-1}=\hat R^m/H$, the equalities $\P_\phi/\P_\phi^+=\P_\phi \hat R_1/\P_\phi^+\hat R_1$, so that for any
element $x_1\in \P_\phi\hat R_1$, there is an element $x\in \P_\phi$ such that $x_1-x\in \P_\phi^+\hat R_1$. This suffices, since $\nu$ is a
valuation, to show that $\hat\nu (x_1y_1)=\nu (xy)=\nu (x)+\nu (y)=\hat\nu (x_1)+\hat\nu (y_1)$, so that $\hat \nu$ is a valuation and $H_1$ is
prime.
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{remark} In general, the ideal $H$ is equal to its radical, as one easily verifies, so that $\hat\nu_1$ is a loose valuation on a reduced but not
necessarily integral n\oe therian local ring.\end{remark}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\par\noindent Let us now consider the case where
$\nu$ is of height $>1$. We proceed by induction on the height of $\nu$ and for a while I stop assuming that $\nu$ is rational; the valuation $\nu$ now
stands for one of the valuations with which my original rational valuation is composed, and I assume by induction that I have extended the valuation
$\nu_1$ of height
$\hbox{\rm h}(\nu )-1$ with which $\nu$ is composed, of center ${\mathbf p}_{\nu_1}$. It means that we have an ideal $H_1\subset \hat R^m$ with
$H_1\cap R=(0)$ and
$H_1\subset {\mathbf p}_{\nu_1}\hat R^m$, and such that $\nu_1$ extends to a loose valuation $\hat \nu_1$ of $ \hat R^m/H_1$.\par\noindent  If the
centers of
$\nu$ and $\nu_1$ coincide, there are finitely many ideals $\P_\phi$ between $\P_{\phi_1}$ and $\P^+_{\phi_1}$, so that, given $x\in \hat R_1$ with
$\hat\nu_1 (x)=\phi_1$, we have a largest
$\phi$ such that $x\in \P_\phi\hat R_1$, and this defines a loose valuation on $\hat R_1$ extending $\nu$.\par
I therefore assume, until the end of this proof, that the centers of $\nu$ and $\nu_1$ are distinct, and try to construct an ideal $\tilde H$ of $\hat
R_1$ such that $\nu$ extends to a loose valuation of $\hat R_1/\tilde H$. I denote by $\Psi$ the group of height one which is the kernel of the map
from the group $\Phi$ of $\nu$ to the group $\Phi_1$ of $\nu_1$. The main difficulty is that the ideals
$\bigcap_{\lambda (\phi)=\phi_1}\P_\phi\hat R_1$ and $\P^+_{\phi_1}\hat R_1$ will in general be different. We must arrange that they become equal
after passing to $\hat R_1/\tilde H$. 
\par\noindent  Let now
$S$ be an $R$-algebra and $I$ an ideal of $S$. Given an ideal $\P$ of $R$, I denote as usual by $\P S$ the ideal generated by the image of $\P$ in $S$.
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{grim} a) For each $\phi\in\lambda^{-1}(\phi_1)$ we have an exact sequence of $S/\P_{\phi_1}^+S$-modules
$$0\to\frac{I\cap\P_\phi S+\P^+_{\phi_1}S}{\P^+_{\phi_1}S}\to\frac{\P_{\phi}S}{\P^+_{\phi_1}S} \to\frac{\P_\phi S}{\P_\phi S\cap
(I+\P^+_{\phi_1}S)}\to 0,$$ and the equalities
$$\frac{\P_{\phi}S/I}{\P^+_{\phi_1}S/I}=\frac{\P_\phi S}{\P_\phi S\cap (I+\P^+_{\phi_1}S)}=\frac{\P_\phi S}{I\cap\P_\phi S+\P^+_{\phi_1}S}.$$ 
These $S/\P_{\phi_1}^+$-modules are actually $S/{\mathbf p}_{\nu_1}S$-modules since ${\mathbf p}_{\nu_1}\P_{\phi}S\subset
\P^+_{\phi_1}S$.\par\noindent
b) Similarly, for $\phi\in \Phi_+\cup\{0\}$, we have the equalities
$$\frac{\P_{\phi}S/I}{\P^+_{\phi}S/I}=\frac{\P_\phi S}{\P_\phi S\cap (I+\P^+_{\phi}S)}=\frac{\P_\phi S}{I\cap\P_\phi S+\P^+_{\phi}S}.$$ 
\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} This is just a repeated application of the Dedekind-N\oe ther modular law ([Z-S], vol. II, Chap. 3, p.137).\end{proof}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{quotgrad} Let ${\mathbf p}$ be the center of the valuation $\nu$ in $R$, and $I$ be an ideal of the $m$-adic completion $\hat R^m$
of $R$.
\par\noindent a) The graded algebra 
$$\bigoplus_{\phi\in \Phi_+\cup\{0\}}\frac{\P_\phi .(\hat R^m/I)}{\P_{\phi}^+.(\hat R^m/I)}$$ is a quotient of the graded algebra  $$\bigoplus_{\phi\in
\Phi_+\cup\{0\}}\frac{\P_\phi\hat R^m}{\P_{\phi}^+\hat R^m}=\hbox{\rm gr}_\nu R\otimes_{R/{\mathbf p}}\hat R^m/{\mathbf p}\hat R^m$$ by the
ideal $$\bigoplus_{\phi\in
\Phi_+\cup\{0\}}\frac{I\cap\P_\phi\hat R^m+\P^+_\phi\hat R^m}{\P^+_\phi\hat R^m}.$$\par\noindent b) If for all $\phi\in\Phi_+$ we have
$(I+\P_\phi\hat R^m)\cap R=\P_\phi$, the intersection of this ideal with the subalgebra
$\hbox{\rm gr}_\nu R\otimes_{R/{\mathbf p}}R/{\mathbf p}$ is equal to $(0)$.\end{lemma}
\begin{proof} The proof of a) is the same as for lemma \ref{grim}, and the proof of b) is a direct computation. One may also refer to [B3], Chap.
III, \S 2, No. 8, Theorem 1.\end{proof}
 Note that $\hat R_1/{\mathbf p}_{\nu_1}\hat R_1=\hat R^m/{\mathbf p}_{\nu_1}\hat R^m$ is equal to the $m$-adic completion
of
$\overline{R_1}=R/{\mathbf p}_{\nu_1}$. In this last ring, we have the valuation $\overline \nu$ of height one with group $\Psi=\hbox{\rm Ker}\lambda$,
where
$\lambda\colon\Phi\to \Phi_1$ is the map corresponding to the composition of valuations as in section \ref{valco}. \par\noindent By the height one case,
there is an ideal $\overline H=\bigcap_{\psi\in
\Psi_+}\overline\P_\psi (\hat R/{\mathbf p}_{\nu_1}\hat R)$, with $\overline\P_\psi=\P_\psi/{\mathbf p}_{\nu_1}$ and such that $\overline
\nu$ extends to a loose valuation of
$(\hat R/{\mathbf p}_{\nu_1}\hat R)/\overline H$.\par\noindent
Let us choose a system of generators $(\overline b_1,\ldots ,\overline b_s)$ of $\overline H$ and lift them to elements $( b_1,\ldots
, b_s)$ of $\hat R_1$. \par\noindent  Let us denote by $\tilde H$ the ideal of $\hat R_1$ which they generate. \par\noindent For simplicity let us
write ${\mathbf p}_1$ for ${\mathbf p}_{\nu_1}$ \par Remark that since
$\overline H\subset  {\mathbf p}\hat{\overline R}_1$, where $\hat{\overline R}_1=\hat R_1/{\mathbf p}_1\hat R_1$, we have the inclusion
$$\tilde H\subset {\mathbf p}\hat R_1.$$\par\medskip\noindent
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 For a given $\phi_1\in \Gamma_1\subset \Phi_{1+}\cap\{ 0\}$, and $\phi\in \Gamma\cap\lambda^{-1}(\phi_1)$, let us define the ideal 
$$\overline T_\phi =\{z\in \overline{R_1}/z\frac{\P_{\phi_1}}{\P_{\phi_1}^+}\subset\frac
{\P_{\phi}}{\P_{\phi_1}^+}\}=(\frac{\P_{\phi}}{\P_{\phi_1}^+}:\frac{\P_{\phi_1}}{\P_{\phi_1}^+}).$$
The last notation is that of [Z-S] and [B3], Chap. 1, \S 2, No. 10, for the transporter of an $\overline{R_1}$-module, say, into one of its submodules.
 Let us denote by $\tilde\phi_1$ the smallest
element of
$\Gamma\cap\lambda^{-1}(\phi_1)$, which exists since $\Gamma$ is well ordered and is equal to $\nu
(\P_{\phi_1})$, and, for $\phi\in\lambda^{-1}(\phi_1)$, by
$\psi(\phi)$ the smallest element of $\Gamma\cap \Psi$ which is at least equal to $\phi-\tilde\phi_1$, so that we have
$\P_{\psi(\phi)}=\P_{\phi-\tilde\phi_1}$.
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{TT} a) The ideal $T_\phi=(\P_\phi:\P_{\phi_1})$ is equal to $\P_{\psi (\phi)}$. \par\noindent b) The ideal $\overline T_\phi$
is the image of $T_\phi$ in $\overline R_1$; setting
$\overline
\P_\psi=\P_\psi\overline{R_1}$, the ideals
$\overline T_\phi$ coincide in
$\overline{R_1}$ with some of the ideals
$\overline
\P_\psi$; they define the same topology as the ideals $(\overline\P_\psi)_{\psi\in \Gamma\cap
\Psi}$.\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} We have $z\P_{\phi_1}\subset \P_\phi$ if and only if $\nu(z)\geq \phi-\tilde\phi_1$, so that $ T_\phi=
\P_{\phi-\tilde\phi_1}=\P_{\psi (\phi)}$. Since
$\P^+_{\phi_1}\subset \P_\phi$, the image of $T_\phi$ in $R_1/\P_{\phi^+_1}$ is the transporter of the $R_1/\P_{\phi^+_1}$-module
$\frac{\P_{\phi_1}}{\P_{\phi_1}^+}$ into $\frac{\P_{\phi}}{\P_{\phi_1}^+}$. Since they are both in fact $R_1/{\mathbf p}_1R_1$-modules, the
transporter acts through its image $\overline T_\phi$ in $R_1/{\mathbf p}_1R_1$. The result on topologies follows from Proposition
\ref{almostone} which implies that for each $\psi\in \Gamma\cap\Psi_+$ there are finitely many $\phi$'s in $\lambda^{-1}(\phi_1)$ such that
$\phi-\tilde\phi_1<\psi$.\end{proof}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
I can now apply Corollary 8, Vol. 2, p. 267, of [Z-S], which tells us that the formation of the transporter of a module into a submodule commutes with
$m$-adic completion in this case, since $\P_{\phi_1}/\P_{\phi_1}^+$ is an $\overline{R_1}$-module of finite type. It gives the following 
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{transport} a) We have the equality $$(\P_\phi \hat R^m:\P_{\phi_1}\hat R^m)=\P_{\psi(\phi)}\hat
R^m.$$\par\noindent b) Given
$\phi_1\in
\lambda (\Gamma)$, for
$\phi\in
\Gamma\cap\lambda^{-1}(\phi_1)$, we have 
$$\{z\in \frac{\hat R^m}{\P^+_{\phi_1}\hat R^m}/z\frac{\P_{\phi_1}\hat R^m}{\P_{\phi_1}^+\hat R^m}\subset \frac{\P_\phi\hat
R^m}{\P_{\phi_1}^+\hat R^m}\}= P_{\psi(\phi)}\hat R^m/\P^+_{\phi_1}\hat R^m.$$ 
\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} The first statement follows from the reference just given, and the second from the fact that when we pass from $\hat R^m$ to $\hat
R^m/\P_{\phi_1}\hat R^m$ the image of the transporter is equal to the transporter of the images since $\P_{\phi_1}\hat R^m\subset \P_\phi\hat
R^m$.
\end{proof}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Let us remark here, as in Lemma \ref{TT}, that the elements of  $\hat R^m/\P^+_{\phi_1}\hat R^m$ act on $\frac{\P_{\phi_1}\hat
R^m}{\P_{\phi_1}^+\hat R^m}$ via their images in $\hat R^m/{\mathbf p}_1\hat R^m$. Therefore in statement b) above we may replace
$\P_{\psi(\phi)}\hat R^m$ by $\tilde\P_{\psi (\phi)}+\P^+_{\phi_1}\hat R^m$, where $\tilde\P_{\psi (\phi)}$ is any other ideal of $\hat R^m$ having
the same image in $\hat R^m/{\mathbf p}_1\hat R^m$. From now on
$\tilde\P_{\psi (\phi)}$ denotes such an ideal, which is {\it not} a valuation ideal in general.\par\medskip By [Z-S], Vol. II, Appendix 3, Proposition 1, p.
342 and Proposition 2, p. 345, a primary decomposition for
$\phi\in
\Gamma$ is of the form
$$\P_\phi={\mathbf q}\cap{\mathbf q}_{h_1}\cap\cdots \cap{\mathbf q}_{h_q}$$  for $\P_\phi$ in $R$, where the $\sqrt {\mathbf q}_{h_i}={\mathbf
p}_{h_i}$ are among the centers in $R$ of the valuations with which $\nu$ is composed and ${\mathbf p}$ is the center of $\nu$.\par\noindent  I assume
them ordered as 
$${\mathbf p}\supsetneq{\mathbf p}_{h_1}\supsetneq\cdots \supsetneq {\mathbf p}_{h_q}.$$
By {\it loc.cit}, the ideal ${\mathbf q}_{h_1}\cap\cdots \cap{\mathbf q}_{h_q}$ is a valuation ideal, equal to $(\P_\phi :{\mathbf q})$ and which has to be
equal to
$\P_{\phi_1}$, where
$\phi_1=\lambda (\phi)$, since it is contained in the center ${\mathbf p}_1$ of $\nu_1$, which we assumed to be distinct from ${\mathbf p}$, and its
$\nu_1$-valuation is
$\phi_1$. Although the ideal ${\mathbf q}$ is not uniquely defined, we may for each $\phi\in \lambda^{-1}(\phi_1)$ choose one such ideal and write
$$\P_\phi={\mathbf q}(\phi)\cap\P_{\phi_1},\ \ \forall \phi\in\lambda^{-1}(\phi_1).$$
Since $\hat R^m$ is a flat $R$-algebra, we have $$\P_\phi\hat R^m= {\mathbf q}(\phi)\hat R^m\cap\P_{\phi_1}\hat R^m.$$ Note that we have
also
$$\P^+_{\phi_1}\hat R^m\subseteq {\mathbf q}(\phi)\hat R^m.$$ Using Lemma \ref{transport}, we obtain for each $\phi\in
\lambda^{-1}(\phi_1)$ the inclusion
$${\mathbf q}(\phi)\hat R^m\subseteq \tilde\P_{\psi (\phi)}\hat R^m+\P_{\phi_1}^+\hat R^m,$$ since ${\mathbf q}(\phi)$ is contained in the
transporter of $\P_{\phi_1}$ into $\P_\phi$. This gives
$$\leqno{(1)}\ \ \ \ \ \ \ \ \ \ \ \ \ \P_\phi\hat R^m\subseteq (\tilde\P_{\psi (\phi)}\hat R^m+\P^+_{\phi_1}\hat R^m)\cap \P_{\phi_1}\hat R^m.$$
The next step is to use Chevalley's theorem to show that the $\tilde\P_\psi$ approximate $\tilde H$.
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{cheva} Let $\overline H\subset \hat R^m/{\mathbf p}_1\hat R^m$ be the ideal $\bigcap_{\psi\in \Psi} \overline \P_\psi \hat
R^m/{\mathbf p}_1\hat R^m$ associated to the valuation $\overline\nu$ of $R/{\mathbf p}_1R$ in the height one case. There is a function $\psi\mapsto
t(\psi)\in
\N$ tending to infinity with $\psi$ such that for all $\psi\in \Psi_+$ the inclusions
$$ \overline \P_\psi \hat R^m/{\mathbf p}_1\hat R^m\subseteq \overline H+ m^{t(\psi)}\hat R^m/{\mathbf p}_1\hat R^m$$ hold.\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} By the height one case the valuation $\overline \nu$ of $R/{\mathbf p}_1$ extends to a valuation of height one $\hat{\overline \nu}$ of of
$(\hat R/{\mathbf p}_1)/\overline H$; we have $\bigcap_{\psi\in \Psi_+}\P_\psi(\hat R/{\mathbf p}_1)/\overline H=(0)$, and the valuation ideals form
a simple decreasing sequence of ideals. By Chevalley's theorem ([Z-S], Chap. VIII, \S 5, Th. 13)  there exists a map $t\colon \Psi_+\to \N$ such that
$t(\psi)$ tends to infinity with
$\psi$ and such that $\P_\psi(\hat R^m/{\mathbf p}_1\hat R^m)/\overline H\subset m^{t(\psi)}(\hat R^m/{\mathbf p}_1\hat R^m)/\overline H$.
\end{proof}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
As a consequence of this lemma, we can lift to $\hat R_1$ generators of the ideals $\overline\P_\psi\hat R^m/{\mathbf p}_1\hat R^m$ in such a way that
the ideals $\tilde\P_\psi\hat R_1$ which these lifted elements generate in $\hat R_1$ satisfy
$$\leqno{(2)}\ \ \ \ \ \ \ \ \ \ \ \ \  \tilde H\subset \tilde\P_\psi\hat R_1\subseteq \tilde H+m^{t(\psi)}\hat R_1.$$
The second inclusion follows from the lemma, and we may then add $\tilde H$ to each $\tilde\P_\psi$ in order to satisfy the first.\par\noindent
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Now if we go back to our quotient $\hat R_1=\hat R^m/H_1$ of $\hat R^m$, to which the valuation $\nu_1$ extends as a loose valuation, it follows
from the inclusions $(1)$ and $(2)$ above that we have, for $\phi\in \lambda^{-1}(\phi_1)$, the inclusions \par
$$\leqno{(3)}\ \ \ \ \P_\phi\hat R_1\subseteq (\tilde\P_{\psi (\phi)}\hat R_1+\P^+_{\phi_1}\hat R_1)\cap \P_{\phi_1}\hat
R_1\subseteq (\tilde H+\P^+_{\phi_1}\hat R_1+m^{t(\psi(\phi))}\hat R_1)\cap\P_{\phi_1}\hat
R_1.$$
 This implies in turn, taking intersections and using the modular equality , that 
$$\bigcap_{\lambda(\phi)=\phi_1}\P_\phi \hat R_1\subseteq \tilde H\cap\P_{\phi_1}\hat R_1+\P^+_{\phi_1}\hat R_1.$$ 
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{inter} We have the inclusions
$$\tilde H\P_{\phi_1}\hat R_1+\P^+_{\phi_1}\hat R_1\subseteq \bigcap_{\lambda(\phi)=\phi_1}\P_\phi \hat R_1\subseteq \tilde H\cap
\P_{\phi_1}\hat R_1+\P^+_{\phi_1}\hat R_1.$$\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof}  Assume by induction that $H_1\subset {\mathbf
p}_1\hat R^m$. Then the first
inclusion follows from Lemma \ref{transport} since the image of $\tilde H$ in $\hat
R^m/{\mathbf p}_1\hat R^m$ is $\overline H$, and we have just proved the second one.
\end{proof}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Let us now check the 
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma} The loose valuation $\hat\nu_1$ descends to a loose valuation of $\hat R_1/\tilde H$.\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} Let us first prove that $\tilde H$ is closed in $\hat R_1$ for the
topology of the
$\P_{\phi_1}\hat R_1$. This topology is the same as that defined by the loose valuation of height one with which $\hat\nu_1$ is composed. But the
loose valuation ideals for that loose valuation form a simple decreasing sequence with intersection $(0)$. By Chevalley's Theorem each is contained in
an increasing power of the maximal ideal of $\hat R_1$; since $\hat R_1$ is n\oe therian $\tilde H$ is closed for the $m$-adic topology, which shows that it
is closed for the $\hat \nu_1$-adic topology. If now we go to $\lambda_i\colon \Phi_i\to \Phi_{i+1}$ using the notations of Proposition \ref{pseudoval},
we can apply Chevalley's theorem to $\hat R_1/\P_{\phi_{i+1}^+}\hat R_1$ and obtain, for $\phi_i\in\lambda_i^{-1}(\phi_{i+1})$,
$$\P_{\phi_i}\hat R_1\subseteq \P^+_{\phi_{i+1}}\hat R_1+m^{s(\phi_i)}\hat R_1,$$ from which follows, since $s(\phi_i)$ tends to infinity as $\phi_i$ grows in
$\lambda_i^{-1}(\phi_{i+1})$.
$$\bigcap_{\phi_i\in\lambda_i^{-1}(\phi_{i+1})}(\tilde H+\P_{\phi_i}\hat R_1)\subseteq\bigcap_{\phi_i\in\lambda_i^{-1}(\phi_{i+1})}(\tilde
H+\P^+_{\phi_{i+1}}\hat R_1+m^{s(\phi_i)}\hat R_1)=\tilde H+\P^+_{\phi_{i+1}}\hat R_1.$$ The result now follows from Proposition
\ref{pseudoval}.\end{proof}
 In particular, if we set
$S=\hat R_1/\tilde H$, we have, since
$\bigcap_{\lambda(\phi)=\phi_1}\tilde\P_{\psi (\phi)}S=(0)$,
$$\bigcap_{\lambda(\phi)=\phi_1}\P_\phi S=\P^+_{\phi_1}S.$$ 
This implies that the valuation $\nu$ extends as a loose valuation $\hat \nu$ on $S$. Let us denote by
$H$ the inverse image of $\tilde H$ in $\hat R^m$. We have by construction that $$H_1\subset H\subset {\mathbf p}\hat R^m$$ and the 
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{goodint} {\rm (Assuming that TLU($\hbox{\rm dim}R-1$) holds)} For each $\phi_1\in \Phi_1$, we have $(\tilde H+\P_{\phi_1}\hat
R_1)\cap R=\P_{\phi_1}$.\end{lemma}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proof} Fix $x\in R$, and let us first assume that the ideal $\overline H$ is generated by a regular sequence in the ring $\hat R^m/{\mathbf p}_1\hat
R^m$ and that there exists a monomial $\overline\xi\in R/{\mathbf p}_1$ such that the $(R/{\mathbf p}_1)_{(\overline\xi)}$-module
$\P_\phi/\P^+_\phi\otimes_{R/{\mathbf p}_1}(R/{\mathbf p}_1)_{(\overline\xi)}$ is free for all $\phi\leq\nu(x)$. Then, if $x\in \tilde H$, for some
$n$ we have, after choosing a representative
$\xi\in R$ of $\overline\xi$, 
$\hbox{\rm in}_{\nu_1}(\xi ^n x)\in \hbox{\rm gr}_{\nu_1}R\otimes_{R/{\mathbf p}_1}\overline H$. \par
To see this, fix a regular sequence of generators of $\overline H$ and lift them as generators $(b_1,\ldots, b_s)$ of $\tilde H$ in $\hat R_1$. Write
$x=\sum_1^s\lambda_ib_i$. By induction on the height of the valuation, we may assume that if $\nu_1$ has an immediate succcessor $\nu_2$, the
minimum of the $\nu_2(\lambda_ib_i)$ is equal to $\nu_2(x)$. If
$\hbox{\rm min}_{1\leq i\leq s}\nu_1(\lambda_i)=\nu_1(x)$, the initial form is
$\sum_{i\in M}\overline b_i\hbox{\rm in}_{\nu_1}\lambda_i$, where $M$ is the set of indices where the minimum of valuations is attained, and we are
done. Otherwise, the minimum of the valuations is $<\nu_1(x)$ and we have a relation $\sum_{i\in M}\overline b_i\hbox{\rm in}_{\nu_1}\lambda_i=0$.
Since we assume that the
$\overline\lambda_i$ form a regular sequence and $\P_\phi/\P^+_\phi\otimes_{R/{\mathbf p}_1}(R/{\mathbf p}_1)_{(\overline\xi)}$ is free, after
multiplication by a suitable power $\overline\xi^{n_1}$ of $\overline\xi$ we have expressions $\overline\xi^{n_1}\hbox{\rm in}_{\nu_1}\lambda_i=
\sum_{k=1}^s\overline b_k\hbox{\rm in}_{\nu_1}\mu_{ik}$ with
$\hbox{\rm in}_{\nu_1}\mu_{ik}\in
\P_\phi/\P^+_\phi\otimes_{R/{\mathbf p}_1}(R/{\mathbf p}_1)$ and $\mu_{ki}=-\mu_{ik}$. We may now write
$$\xi^{n_1}x=\sum_{i\in M}(\xi^{n_1}\lambda_i-\sum_kb_k\mu_{ki})b_i+\sum_{i\notin M}\xi^{n_1}\lambda_ib_i,$$ and the minimum of the $\nu_1$
valuations of the coefficients has increased. Since $\nu_2$ is fixed, in a finite number of such steps, we reach the situation where the minimum of the
valuation of the coefficients is $\nu_1(x)$, and we are done.
Now if we assume $TLU(d)$ for $d<\hbox{\rm dim} R$, since $R$ is excellent by assumption, so is $R/{\mathbf p}_1$ and we can reach the situation taken
as a hypothesis above by a toric modification of
$R$ as in Lemma
\ref{torictensor} which makes
$R/{\mathbf p}_1$ regular and $\overline H$ generated by part of a system of coordinates of $\hat R^m/{\mathbf p}_1\hat R^m$, and therefore by a
regular sequence.\par  Without affecting this, we may by a toric modification $R\to R'$ of the same type flatten the image of
$\P_{\phi_1}(R)/\P^+_{\phi_1}(R)$ modulo torsion by principalizing a suitable Fitting ideal (see [G-R]) and this, by Lemma \ref{torictensor}, gives us the
freeness of
$\P_{\phi_1}(R')/\P^+_{\phi_1}(R')$ after localization by some
$(\xi)$. In a toric modification such as $R/{\mathbf p}_1\to R'/{\mathbf p}'_1$, if we denote by $\overline H'$ the ideal
$\bigcap_{\psi\in\Psi_+}\P_\psi(R')\hat {R'}^{m'}/{\mathbf p}'_1\hat {R'}^{m'}$ of $\hat {R'}^{m'}/{\mathbf p}'_1\hat {R'}^{m'}$, we have $\overline
H'\cap (\hat R^m/{\mathbf p}_1\hat R^m)=\overline H$. From this follows the inclusion $\tilde H\subseteq\tilde H'\cap\hat R_1$.\par Now we observe
that for the $\phi_1$ which we are considering, we have\break
$(\tilde H'+\P_{\phi_1}(R')\hat R'_1)\cap R'=\P_{\phi_1}(R')$; if this were not the case, there would be an element $x'$ in the left hand side of valuation
$<\phi_1$. Then its initial form would be that of an element of $\tilde H'$ so that, by what we have just proved, we would have, remembering that
$\nu_1(\xi)=0$, the inclusion
$\hbox{\rm in}_{\nu_1}(\xi^nx')\in \hbox{\rm gr}_{\nu_1}R'\otimes_{R'/{\mathbf p}'_1}\overline H'$. But since $\xi^nx'\in R'$ and $\overline H'\cap
(R'/{\mathbf p}'_1)=(0)$, this is impossible, which proves the result.
Now since $\P_{\phi_1}(R')\cap R=\P_{\phi_1}$ and $\tilde H\subseteq\tilde H'\cap \hat R_1$, we have the inclusion\break $(\tilde H+\P_{\phi_1}\hat
R_1)\cap R\subseteq \P_{\phi_1}$, and the result since the reverse inclusion is clear.\end{proof}
\begin{lemma}\label{zeroint} We have $H\cap R=(0)$\end{lemma}
 \begin{proof} It suffices to prove that $\tilde H\cap R=(0)$ since we know by induction that $H_1\cap R=(0)$. By Lemma \ref{goodint}, we have $\tilde
H\cap R\subset \bigcap_{\phi_1\in \Phi_{1+}}\P_{\phi_1}=(0)$.\end{proof}
Finally, we have proved that $H\cap R=(0)$ and $\nu$ extends
to a loose valuation of $\hat R^m/H$ composed with $\hat\nu_1$. This also validates our induction hypothesis.\par Moreover, by lemmas
\ref{grim} and \ref{goodint}, the graded algebra
$\hbox{\rm gr}_{\hat
\nu_1}\hat R^m/H$ is a quotient of
$$\hbox{\rm gr}_{\nu_1} R\otimes_{R/{\mathbf p}_1}\hat R^m/{\mathbf p_1}\hat R^m$$ by an homogeneous ideal whose intersection
with
$\hbox{\rm gr}_{\nu_1} R\otimes_{R/{\mathbf p}_1}R/{\mathbf p}_1$ is zero.\par\medskip Going back to our rational valuation
$\nu$ composed with
$\nu_1,\nu_2,\ldots ,\nu_{h-1}$, we see that we can  build a sequence of ideals
$H_{h-1}\subset\cdots \subset H_0=H$ of $\hat R^m$ such that $H_i\cap R=(0)$ and, if we set $\nu_0=\nu$, the valuation $\nu_i$ extends as a loose
valuation of
$\hat R^m/H_i$ for $0\leq i\leq h-1$. However, since $\nu$ is rational, the last graded ring is, in view of Lemma \ref{quotgrad}, a quotient of  $\hbox{\rm
gr}_\nu R\otimes_{R/m}k=
\hbox{\rm gr}_\nu R$ which contains $\hbox{\rm gr}_\nu R$ since $H\cap R=(0)$ and the loose valuation extends $\nu$, and so has to be equal to
$\hbox{\rm gr}_\nu R$. Since that graded ring is an integral domain, this implies that $H_0$ is a prime ideal of $\hat R^m$ and that
$\hat\nu$ is a valuation of $\hat R^m/H_0$, and therefore all the extensions $\hat\nu_i$ to $\hat R^m/H_0$ have to be
valuations too, those with which $\hat \nu$ is composed.\par\noindent
This end the proof of Proposition \ref{madicomp}.



By the regularity of $\hat{\overline R}^m$, we may choose a representative $\hat{\overline R}^m\subset \hat
R^m$ of the quotient $\hat{\overline R}^m=\hat R^m/{\mathbf p}_{h-1}\hat R^m$. Define the ideal
$$H=\bigcap_{\phi_{h-1}\in \Phi_{h-1+}}\bigl(\overline H\hat R^m+\P_{\phi_{h-1}}\hat R^m\bigr)\subset \hat
R^m.$$ 
The claim now is that the valuation $\nu$ extends to $\hat R^m/H$ in the way required by the Proposition. The proof is in several steps:
\par\noindent
First, there is a candidate extension of $\nu_{h-1}$ to $\hat R^m/H$, obtained by defining the candidate valuation ideals
$$\hat\P_{\phi_{h-1}}=\bigl(\overline H\hat R^m+\P_{\phi_{h-1}}\hat R^m\bigr)/H=P_{\phi_{h-1}}\hat R^m/H.$$



%----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{proposition}\label{madicomp} {\rm *}Let $R$ be a local equicharacteristic excellent integral domain with maximal ideal $m$ and an
algebraically closed residue field, and let
$\nu$ a rational valuation on $R$ with value group $\Phi$.\par\noindent
 a) There exists a prime ideal $H$ of the $m$-adic completion $\hat R^m$ of $R$ such that $\nu$ extends to a valuation $\hat\nu$ of 
$\hat R^m/H$ in such a way that:\par\noindent b) the associated graded map
$$\hbox{\rm gr}_{\nu} R\to \hbox{\rm gr}_{\hat {\nu}}\hat R^m/H$$ is a scalewise birational extension of $\Phi$-graded $
R/m$-algebras, uniquely determined by $(R,\nu)$.\par\noindent If the valuation $\nu$ is of height one, the map $\hbox{\rm gr}_{\nu} R\to \hbox{\rm
gr}_{\hat {\nu}}\hat R^m/H$ is an isomorphism.{\rm *}
\end{proposition}
%-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
\begin{Remark} 1) The proof depends on toric local uniformization, an argument for which is only sketched at the end of the paper. \par\noindent 
2) The fact that $\hat\nu$ extends $\nu$ implies that
$H\cap R=(0)$.\par\noindent
3) The assumption that the residue field is algebraically closed is due to the use of local uniformization in lower dimensions
in the proof. It should eventually disappear when the method is generalized to excellent equicharacteristic local rings. 
\par\noindent 4) Although the ideal $H$ is unique in the case considered by Zariski and
Spivakovsky, it is not uniquely determined in general; it is only the graded map associated to the map
$R\to
\hat R^m/H$ which is uniquely determined. Once $H$ is fixed, the extension of $\nu$ is unique. A quotient $\hat R^m/H$ as in the Proposition will
nevertheless be denoted by $\hat R^{(\nu)}$.
\par 4) The fact that the extension of the graded rings is scalewise birational implies that any toric modification we make in $\hat R^{(\nu)}$ with
respect to representatives $(\eta_j)_{j\in J}$ of the generators of the graded algebra $\hbox{\rm gr}_{\hat\nu}\hat R^{(\nu)}$ can be viewed as
coming from a toric modification in $R$ with respect to the chosen system of representatives
$(\xi_i)_{i\in I}$.\end{Remark}

\begin{proof}(Sketch) The method is to work by induction on the dimension of $R$. Let us therefore assume first that $R$ is a
one-dimensional n\oe therian local domain with a rational valuation $\nu$, which is necessarily of height one. Let us set
$H=\bigcap_{\phi\in \Phi_+}\P_\phi (R)\hat R^m$. Then we have $H\cap R=\{0\}$ and if we set $\hat \P_\phi=\P_\phi\hat R^m/H$, we
see that $\hat \P_\phi\cap R=\P_\phi$ since by the faithful flatness of the extension $R\to\hat R^m$ we have $ \P_\phi\hat R^m\cap R=\P_\phi$, and that
the natural inclusion
$$\hbox{\rm gr}_\nu R\subset \bigoplus_{\phi\in
\Phi_+}\frac{\hat\P_\phi}{\hat\P^+_\phi} $$ is an equality. This shows that $\hat R^m/H$ is an integral domain, since its associated graded ring for the
filtration by the $\hat\P_\phi$ is an integral graded algebra, and that the
$\hat\P_\phi$ are the valuation ideals of a valuation
$\hat\nu$ on $\hat R^m/H$ which extends $\nu$ through the inclusion $R\subset \hat R^m/H$. Since $H\cap R=(0)$, the ideal $H$ is a
minimal prime ideal of $\hat R^m$ uniquely determined by the valuation
$\nu$ and the quotient $\hat R^m/H$ is a one-dimensional local domain; we may call it the formal branch of the curve $\hbox{\rm Spec}R$ corresponding to
$\nu$.  The proof in the dimension one case is partly inspired by a proof in [S2], Theorem 4.3. \par Assume now that the result is true for excellent local
domains of dimension $<d=\hbox{\rm dim}R$. Let us choose a set of elements $(\xi_i)_{i\in I}$ of $R$ whose images in $\hbox{\rm gr}_\nu R$ generate it
as a $k$-algebra. It may happen that the $(\xi_i)_{i\in I}$ do not generate the maximal ideal of $R$. We choose a minimal set of elements
$y_1,\ldots ,y_a$ such that the  $(\xi_i)_{i\in I}, y_1,\ldots, y_a$ generate the maximal ideal. Our toric modifications will always be in terms of the
$(\xi_i)_{i\in I}$. Let
$\nu_{h-1}$ denote the valuation of height one with which $\nu$ is composed, and ${\mathbf p}_{h-1}$
denote its center in $R$. The valuation $\nu$ projects as a rational valuation
$\overline\nu$ of height $\hbox{\rm h}(\nu)-1$ and value group $\Psi_1$ of the ring
$\overline R=R/{\mathbf p}_{h-1}$, and we assume the result to be true for $\overline \nu$.\par  The proof now
uses TLU(d-1) and TP(d-1). By using TLU(d-1) we may make a toric birational
extension $\overline R\to
\overline R'$ of $\overline R$ in the coordinates which are the images of the $\{\xi_i)\vert\nu(\xi_i)\in \Psi_1\}$ such that the
conclusion of TLU is satisfied for $\overline R'$. Such a toric modification lifts in a natural way to a birational
toric modification
$R\to R'$. By applying TP to the relevant Fitting ideals as $\overline R$-modules of any finite number of the
quotients
$\P_{\phi_{h-1}}/\P^+_{\phi_{h-1}}$ we may assume that in addition their transforms modulo torsion, in the
sense of Lemma \ref {torictensor}, are free $\overline R'$-submodules of
$\P_{\phi_{h-1}}(R')/\P^+_{\phi_{h-1}}(R')$ with a cokernel annihilated by some monomials in the images in
$\overline R'$ of the $\{\xi_i)\vert\nu(\xi_i)\in \Psi_1\}$.\par
In order to extend the valuation $\nu$ on $R$ to a quotient $\hat R^m/H$ in the way required by the proposition it
suffices to do the same for $R'$; the ideal $H\in \hat R^m$ will then be the kernel of the map $\hat R^m\to \hat {R'}^{m'}/H'$. From now on, I assume
therefore that
$\overline R$ is regular, that
$\overline H\subset \hat{\overline R}^m$ is generated by part of a system of coordinates, and that
$\P_{\phi_{h-1}}/\P^+_{\phi_{h-1}}$ contains a free submodule such that the quotient is annihilated by
monomials in the $\{\xi_i)\vert\nu(\xi_i)\in \Psi_1\}$. Moreover, the ideal $\overline H$ is generated by elements $\overline b_1,\ldots ,\overline b_s$ of
the form $\overline y_k-a_k(\overline \eta)$ where the $\overline \eta$ are representatives in $\hat{\overline R}^m$ of generators of the graded
algebra
$\hbox{\rm gr}_{\hat{\overline\nu}}\hat{\overline R}^{(\overline\nu)}$.
\par\noindent Denote by $\hat{\mathbf p}$ the kernel of the natural map $\hat R^m\to \hat {\overline R}^{(\overline \nu)}$.
Let $\overline b_1,\ldots, \overline b_t$ be the generators of $\overline H$ which are not contained in ${\mathbf p}_{h-1}\hat R^m_{\hat {\mathbf p}}\cap
\hat R^m/{\mathbf p}_{h-1}\hat R^m$. We can lift them to elements $b_1,\ldots ,b_t$ of $\hat R^m$, of the form $y_k-a_k(\eta)$. Now set
$$H=(b_1,\ldots ,b_t)\hat R^m+\bigcap_{\phi_{h-1}\in \Phi_{h-1 +}}\bigl(\P_{\phi_{h-1}}\hat R^m_{\hat {\mathbf p}}\cap
\hat R^m\bigr).$$
The idea now is to extend the valuation $\nu_{h-1}$ to a valuation $\hat \nu_{h-1}$ of the ring $(\hat R^m/H)_{\hat {\mathbf p}/H}$ which induces a
valuation with center $\hat {\mathbf p}/H$ in the ring $\hat R^m/H$. The valuation $\hat \nu$ which we seek will be the composition of $\hat{\overline
\nu}$ and 
$\hat\nu_{h-1}$.\par\noindent
For each $\phi_{h-1}\in \Phi_{h-1+}$ let us set $$\hat\P_{\phi_{h-1}}=(\P_{\phi_{h-1}}\hat R^m_{\hat {\mathbf p}}\cap
\hat R^m+(b_1,\ldots ,b_t)\hat R^m)/H.$$
%-------------------------------------------------------------------------------------------------------------------------------------
Let us first prove that the graded algebra $\bigoplus_{\phi_{h-1}\in \Phi_{h-1}}\hat \P_{\phi_{h-1}}/\hat \P_{\phi_{h-1}+}$ is a $\hat{\overline
R}^{(\overline \nu)}$-algebra, which is a birational extension of $\hbox{\rm gr}_{\nu_{h-1}}R\otimes_{\overline R}\hat{\overline
R}^{(\overline \nu)}$ and is an integral domain. We see that $\hat {\mathbf p}=(b_1,\ldots ,b_t)\hat R^m+{\mathbf p}_{h-1}\hat R^m_{\hat {\mathbf p}}\cap
\hat R^m$, and it follows from this that in $\hat R^m/H$ we have $\hat {\mathbf p}\hat \P_{\phi_{h-1}}\subset\hat \P_{\phi_{h-1}+}$, so that our algebra
is indeed a $\hat{\overline R}^{(\overline \nu)}$-algebra. By construction, it is contained in $\hbox{\rm gr}_{\nu_{h-1}}R\otimes_{\overline R}
\kappa (\hat {\mathbf p}_{h-1})$, where $\kappa (\hat {\mathbf p}_{h-1})$ is the field of fractions of $\hat{\overline R}^{(\overline \nu)}$, so that it is
a birational extension of $\hbox{\rm gr}_{\nu_{h-1}}R\otimes_{\overline R}\hat{\overline R}^{(\overline \nu)}$. \par
%-------------------------------------------------------------------------------------------------------------------------------------
Now we have a candidate graded algebra of $\hat R^m/H$ with respect to the $\hat\nu_{h-1}$ valuation. We are going to prove that it is an integral domain
by building a filtration on it such that the associated graded ring is a domain. This filtration will turn out to be the filtration associated to a valuation $\hat
\nu$ extending $\nu$ to $\hat R^m/H$, and its associated graded ring will be a scalewise birational extension of $\hbox{\rm gr}_\nu R$.\par\noindent
First, one must make a distinction concerning birational extensions of rings such as $\hat R^m/H$; this distinction is
essentially due to the fact that if $R\to R'\subset R_\nu$ is a birational extension, the extension $R/{\mathbf p}_{h-1}\to
R'/{\mathbf p}'_{h-1}$ is not in general birational:\par\noindent Given a birational toric extension
$\hat{\overline R}^{(\overline\nu)}\to (\hat{\overline R}^{(\overline\nu)})'$ of $\hat{\overline R}^{(\overline\nu)}$, we can lift it to a
birational toric extension of $\hat R^m/H$ with respect to representatives $(\eta_j)_{j\in J}$ in $\hat R^m/H$ of the $(\overline \eta_j)_{j\in J}$.
Let us call such an extension
$\hat R^m/H\to (\hat R^m/H)'$ a basic birational extension of
$\hat R^m/H$. There is a natural map $(\hat R^m/H)'\to  (\hat{\overline R}^{(\overline\nu)})'$. We use the fact that
after a toric modification $\overline R\to \overline R'$, for all further toric extensions $\overline R'\to \overline R''$ the ideal $\overline
H''$ is the strict transform of $\overline H'$ under the map $\hat{\overline R'}\to\hat{\overline R''}$. This is part of the embedded resolution
''package'' in lower dimension, see Proposition \ref{transform}.\par\noindent There is another type of birational extension:\par\noindent Given
$\phi_{h-1}$, let us choose elements $( e_1,\ldots,  e_r)$ of $\P_{\phi_{h-1}}(R)$ such that $\nu (e_1)<\nu (e_2)<\ldots <\nu (e_r)$ and their images modulo
$\P^+_{\phi_{h-1}}$ form a system of generators
$(\overline e_1,\ldots, \overline e_r)$ of the $R/{\mathbf p}_{h-1}$-module
$\P_{\phi_{h-1}}/\P^+_{\phi_{h-1}}$ .\par\noindent
Let $\kappa\colon R\to R'$ be the morphism of local rings obtained by blowing up the ideal $\P_{\phi_{h-1}}(R)$ in $\hbox{\rm Spec}R$ and localizing
at the point picked by the valuation $\nu$. We have a map (not necessarily injective)
$$\hat R^m\to \hat {R'}^{m'}.$$ \par\noindent The idea now
is that to an element $x\in\hat\P_{\phi_{h-1}}$ may be attributed a $\hat\nu$-valuation as follows:\par\noindent
There is an element $s\notin \hat{\mathbf p}$ such that we may write $sx=\sum_{i=1}^s\overline c_ie_i$ with $\overline c_i\in \hat R^m/H$; in the ring
$\hat {R'}^{m'}$, we may write
$e_i=d_ie_1\
\ (2\leq i\leq s)$, and therefore
$sx=(c_1+\sum_{i=2}^sc_id_i)e_1$.\par By our induction hypothesis, the valuation $\overline \nu'$ on $R'/{\mathbf p}'_{h-1}$ induced by the valuation
$\nu$ on $R'$ extends to a valuation
$\hat{\overline
\nu'}$ with value group $\Psi_1$ on $\widehat {R'/{\mathbf p}'_{h-1}}^{(\overline \nu')}$, which is a quotient of $\hat {R'}^{m'}$.\par
 Now we may view $c=c_1+\sum_{i=2}^s\lambda_ic_i$ as an element of $\hat
{R'}^{m'}$; its image in $\widehat {R'/{\mathbf p}'_{h-1}}^{(\overline \nu')}$ is not zero by the construction of the ideals $\overline H'$ and we can define the
$\hat\nu$-valuation of
$sx$ as the sum of
$\nu (e_1)$ and the
$\hat{\overline\nu '}$-valuation of the image of
$c$ in $\widehat {R'/{\mathbf p}'_{h-1}}^{(\overline \nu')}$. This is well defined since $H'\hat{ R'}^{m'}\cap \hat R^m$ and $H$ are
equal modulo ${\mathbf p}_{h-1}\hat R^m$. Then define the $\hat\nu$ valuation of $x$ as $\hat\nu (sx)-\hat\nu (s)$. If we replace $R'$ by an $R''$ such
that
${\mathbf p}_{h-1}R''=e_1R''$, the
$\hat{\overline\nu'}$ value of $c$ and its $\hat{\overline\nu"}$ value are equal. \par\noindent
For $\phi\in \Phi_+\cup\{0\}$ having image $\phi_{h-1}$ in $\Phi_{h-1}$, define the ideal $\hat \P_\phi\subset
\hat\P_{\phi_{h-1}}\subset\hat R^m/H$ to be the set of those elements $x\in\hat\P_{\phi_{h-1}}$ such that in the local ring of some (not
necessarily basic) birational extension $(\hat R^m/H)'$ of $\hat R^m/H$ we have $sx=c e_1$, with $s\notin \hat{\mathbf p}'/H'$ and
$\phi\leq\hat{\overline\nu} (\overline c)+\nu (e_1)-\hat{\overline\nu} (\overline s)$, where $\overline c$ is the image of $c$ in $(\hat{\overline
R}^{(\overline\nu)})'$. By what we have just seen, the $\hat\P_\phi/\hat\P_{\phi_{h-1}}$ form an exhaustive filtration of the quotient
$\hat\P_{\phi_{h-1}}/\hat\P^+_{\phi_{h-1}}$. The quotient $\hat\P_\phi/\hat\P^+_\phi$ is either $(0)$ or $k$. Given $x\in\hat\P_{\phi_{h-1}}$
and $y\in\hat\P_{\psi_{h-1}}$, let $f_1,g_1$ be generators of minimal valuation of $\P_{\psi_{h-1}}$ and $\P_{\phi_{h-1}+\psi_{h-1}}$
respectively;  after blowing up the product of ideals
$\P_{\phi_{h-1}}\P_{\psi_{h-1}}\P_{\phi_{h-1}+\psi_{h-1}}$ and localizing at the point picked by $\nu$, we may write $e_1f_1=qg_1$, together with
$x=ce_1, y=df_1$. Using the definitions above, we see that $\hat\nu(xy)=\hat{\overline\nu}(\overline {cdq})+\nu (g_1)$, which means $\hat
\nu(xy)=\hat\nu (x)+\hat\nu (y)$. This shows that the graded algebra $\bigoplus_{\phi\in \Phi_+\cup\{0\}}\hat\P_\phi/\hat\P^+_\phi$ is a graded
domain; it is therefore the graded algebra $\hbox{\rm gr}_{\hat\nu} \hat R^m/H$ with respect to the valuation $\hat\nu$. 
In particular, the ideal $\hbox{\rm gr}_{\nu_{h-1}}R\otimes_{\overline R}\overline H$ of $\hbox{\rm gr}_{\nu_{h-1}}R\otimes_{\overline R}\hat
{\overline R}^m$ is prime.\par\medskip\noindent
%-----------------------------------------
Now we have to show that the valuation $\hat\nu$ induces $\nu$ on $R$. Let us first prove that for all $\phi_{h-1}\in \Phi_{h-1}$  we have that the inverse
image in
$R$ of
$\hat\P_{\phi_{h-1}}2$ is equal to $\P_{\phi_{h-1}}$.\par\noindent  The inverse image clearly contains $\P_{\phi_{h-1}}$. If we did not have the
opposite inclusion, there would be an element
$x\in R$ such that $\nu_{h-1}(x)=\mu_{h-1}<\phi_{h-1}$ and an element $s\in \hat R^m\setminus \hat {\mathbf p}$ such that $sx\in (b_1,\ldots ,b_t)\hat
R^m+\P^+_{\phi_{h-1}}\hat R^m$. The image of $sx$ in
$$\P_{\mu_{h-1}}\hat R^m/\P^+_{\mu_{h-1}}\hat R^m=\P_{\mu_{h-1}}/\P^+_{\mu_{h-1}}  \otimes_{\overline R}\hat R^m/{\mathbf
p}_{h-1}\hat R^m$$ has to come from an element
$y$ of $(b_1,\ldots ,b_t)\hat R^m$. Let us write $y=\sum_{i=1}^t b_ic_i$ with $c_i\in \hat R^m$. If the least
$\P_{\phi}\hat R^m$ to which all $c_i$ belong is $\mu_{h-1}$, the image of $y$ in $\P_{\mu_{h-1}}/\P^+_{\mu_{h-1}}\otimes_{\overline R}\hat
R^m/{\mathbf p}_{h-1}\hat R^m$ is in $\P_{\mu_{h-1}}/\P^+_{\mu_{h-1}}\otimes_{\overline R}\overline H$. Now $\overline s\notin
\overline H$ since $s\notin \hat{\mathbf p}$ and so we must have that the initial form of $x$ is in
$\P_{\mu_{h-1}}/\P^+_{\mu_{h-1}}\otimes_{\overline R}\overline H$. Since
$\overline H\cap (R/{\mathbf p}_{h-1})=(0)$, and the image of $y$ is also that of $x\in R$, this is impossible, and so an element such as $x$ cannot
exist.\par\noindent If the smallest $\phi$ such that  $c_i\in \P_\phi$ for all $i$ is $\kappa< \mu_{h-1}$, there must be a relation $$\sum_{j\in J\subset
\{1,\ldots ,t\}} \overline b_j\overline c_j=0,$$ where
$\overline c_j$ is the image of $c_j$ modulo $\P^+_\kappa\hat R^m$. After possibly making a new toric modification we may assume, since
$\nu_{h-1}$ is of height one, that all $\P_\kappa/\P^+_\kappa$ for $\kappa\leq \phi_{h-1}$ contain a free submodule $M_\kappa$ such
that the quotient by this submodule is annihilated by monomials in the $\{\xi_i\vert\nu(\xi_i)\in \Psi_1\}$. Multiplying $x$ by a suitable
monomial $\xi^e$, and since $\xi^ex\in R$, we may assume that the $\overline c_j$ are in $M_\kappa$, in view of the fact that the $\hat
R^m/{\mathbf p}_{h-1}\hat R^m$-module $\P_\kappa\hat R^m/\P^+_\kappa\hat R^m$ is torsion-free. Since $\overline H$ is generated
by a regular sequence, the relation between the $\overline c_j$ implies that for all $j\in J$ we can write $\overline
c_j=\sum_k\overline e_{jk}\overline b_k$ with $\overline e_{jk}=-\overline e_{kj}$. Taking representatives $ e_{jk}\in \hat
R^m$ such that $ e_{jk}=-e_{kj}$, we can finally write
$$\xi^ex=\sum_{i\notin J}\xi^ec_ib_i+\sum_{j\in J}(\xi^ec_j-\sum_ke_{jk}b_k)b_j,$$
and the minimum of the $\nu_{h-1}$-valuations has increased; in a finite number of such steps, we reach the situation where the
minimum valuation of the coefficients is $\mu_{h-1}$ and a contradiction as above.\par\noindent
This also shows that $H\cap R=\bigcap_{\phi_{h-1}\in \Phi_{h-1+}}\P_{\phi_{h-1}}=(0)$. Finally, the valuation $\hat\nu$ on $\hat R^m/H$ is indeed the
composition of the extension $\hat\nu_{h-1}$ of $\nu_{h-1}$ to $\hat R^m/H$ and $\hat{\overline\nu}$ on $\hat{\overline R}^{(\overline \nu)}$.\par

Since we have by the induction hypothesis that the graded algebra
$\hbox{\rm gr}_{\hat{\overline \nu'}}(\hat{\overline R'}^{(\overline\nu)})'$ is a scalewise birational extension of $\hbox{\rm
gr}_{\overline \nu}(\overline R')$, the group of values of $\hat{\overline\nu '}$ is $\Psi_1$, and therefore the group of values of
$\hat\nu$ is $\Phi$. The inclusion
$\hbox{\rm gr}_\nu R\subset \bigoplus_{\phi\in
\Phi_+\cup\{0\}}\hat\P_\phi/\hat P^+_\phi$ is scalewise graded. From this and Lemma \ref{scalbir} follows that the inclusion
$$\hbox{\rm gr}_\nu R\subset \hbox{\rm gr}_{\hat\nu} \hat R^m/H$$ is scalewise birational. \par

From the fact that the graded algebra $\hbox{\rm gr}_{\hat\nu_{h-1}}\hat R^m/H$ is independent from the choice of represntatives and that inside
each of its homogeneous components the definition of $\hat\P_\phi/\hat\P_{\phi_{h-1}}$ is itself by construction independent of the choice of the
representatives $\xi_i$, we see that the inclusion
$\hbox{\rm gr}_\nu R\subset \bigoplus_{\phi\in
\Phi_+\cup\{0\}}\hat\P_\phi/\hat P_{\phi_{h-1}}$ is independent of the choice of the representatives\par
If the valuation $\nu$ is of height one, we set $H=\bigcap_{\phi\in \Phi_+}\P_\phi\hat R^m$ and $\hat\P_\phi=\P_\phi\hat R^m/H$. Since for all $\phi$
the ideal $\hat\P_\phi$ contains $H$ and $m\P_\phi\subset\P^+_\phi$, one sees that $\hat\P_\phi/\hat\P^+_\phi=\P_\phi/\P^+_\phi$, and the fact that in
this case the graded algebra does not change.
\end{proof}
