# Néron desingularization of extensions of valuation rings
with an Appendix by Kęstutis Česnavičius

Dorin Popescu  Simion Stoilow Institute of Mathematics of the Romanian
Academy, Research unit 5, University of Bucharest, P.O. Box 1-764, Bucharest
014700, Romania

###### Abstract.

Zariski’s local uniformization, a weak form of resolution of singularities,
implies that every valuation ring containing $\bf Q$ is a filtered direct
limit of smooth $\bf Q$-algebras. Given an extension of valuation rings
$V\subset V^{\prime}$ containing $\bf Q$ we give conditions when $V^{\prime}$
is a filtered direct limit of smooth $V$-algebras. This corrects a paper of us
[23] where we thought that we may reduce to the case when the value groups are
finitely generated. For this correction we use an infinite tower of
ultrapowers construction that rests on results from model theory.

##  1\. A version of local uniformization

Zariski proved in characteristic $0$ in [34], that any integral algebraic
variety $X$ equipped with a dominant morphism
$v:\operatorname{Spec}(V)\rightarrow X$ from a valuation ring $V$ can be
“desingularized along $V$”: there exists a proper birational map
$\tilde{X}\rightarrow X$ for which the lift
$\tilde{v}:\operatorname{Spec}(V)\rightarrow\tilde{X}$ of $v$ supplied by the
valuative criterion of properness factors through the regular locus of
$\tilde{X}$. This implies the following theorem.

######  Theorem 1.

(Zariski) Every valuation ring $V$ containing a field $K$ of characteristic
zero is a filtered direct limit of smooth $K$-subalgebras of $V$(in particular
they are regular rings).

A smooth algebra is here finitely presented. A ring map $A\rightarrow
A^{\prime}$ is _ind-smooth_ if $A^{\prime}$ is a filtered direct limit of
smooth $A$-algebras. Thus the above theorem says in particular that
$K\rightarrow V$ is ind-smooth. A different proof of this theorem is given in
Theorem 36. One goal of this paper is to show:

######  Theorem 2.

Let $V\subset V^{\prime}$ be an extension of valuation rings containing $\bf
Q$. Then $V\subset V^{\prime}$ is ind-smooth if and only if the following
conditions hold:

  1. (1)

for each $q\in\operatorname{Spec}V$ the ideal $qV^{\prime}$ is also prime,

  2. (2)

For any prime ideals $q_{1},q_{2}\in\operatorname{Spec}V$ such that
$q_{1}\subset q_{2}$ and $\operatorname{height}(q_{2}/q_{1})=1$ the extension
$V_{q_{2}}/q_{1}V_{q_{2}}\subset
V^{\prime}_{q_{2}V^{\prime}}/q_{1}V^{\prime}_{q_{2}V^{\prime}}$ of valuation
rings is ind-smooth.

Moreover, if the condition (1) holds then (2) follows if for any prime ideals
$q\in\operatorname{Spec}V$ and $q^{\prime}\in\operatorname{Spec}V^{\prime}$
such that $\operatorname{height}(qV^{\prime}/q^{\prime})=1$ the extension
$V_{q}/(q^{\prime}\cap V)V_{q}\subset
V^{\prime}_{qV^{\prime}}/q^{\prime}V^{\prime}_{qV^{\prime}}$ of one-
dimensional valuation rings has the same value group. If for each
$q\in\operatorname{Spec}V$ the ideal $qV^{\prime}$ is also prime, $\dim V=\dim
V^{\prime}=1$ and the residue field extension of $V\subset V^{\prime}$is an
isomorphism then $V\subset V^{\prime}$ is ind-smooth if and only if the value
group extension of $V\subset V^{\prime}$ is an equality.

The proof follows from Proposition 20, Corollary 22, Lemma 30 and Proposition
39. Let $\Gamma\subset\Gamma^{\prime}$ be the value group extension induced by
$V\subset V^{\prime}$. The above result was stated by mistake in [23] without
condition (2), and assuming that the value group $\Gamma^{\prime}/\Gamma$ is
finitely generated when the fraction field extension of $V\subset V^{\prime}$
is finitely generated. This is not the case as shows [2, Theorem 7.1]. For
solving the general case, that is $\Gamma^{\prime}/\Gamma$ not finitely
generated we had to study cross-sections with the help of Model Theory.

###### 3\. The method of proof.

To achieve the desingularization claimed in Theorem 2, we replace the initial
$V$ by the limit $\tilde{V}$ of a certain countable tower of iterated
ultrapowers of $V$, constructed in such a way that $\tilde{V}$ would, in turn,
be an immediate extension of a filtered increasing union of valuation rings
for which one knows local uniformization. To then conclude, we argue that
large immediate extensions and ultrapowers interact well with
desingularization.

Desingularizing immediate extensions is in fact the first major step. We show
that if $V\subset V^{\prime}$ is a local extension of valuation rings
containing $\bf Q$ that induces isomorphisms on value groups and residue
fields (that is, is immediate) then $V^{\prime}$ is a filtered direct limit of
smooth $V$-algebras.

The techniques we use include extensions of steps of the General Néron
desingularization, notably, Lemma 7 that is also key for reductions to
complete rank $1$ cases. In the purely transcendental case, Kaplansky’s
classification [12] of Ostrowski’s pseudo-convergent sequences plays an
important role.

The utility of desingularizing immediate extensions is evident already in the
case when $V^{\prime}$ is complete of rank $1$ with a finitely generated value
group $\Gamma^{\prime}$. Such a $V^{\prime}$ has a coefficient field $k$, so,
by choosing a presentation $\Gamma^{\prime}\cong{\bf
Z}\mathrm{val}(x_{1})\oplus\cdots\oplus{\bf Z}\mathrm{val}(x_{n})$, one
obtains the immediate extension $V^{\prime}\cap k(x_{1},\dotsc,x_{n})\subset
V^{\prime}$. To show that $V=k\subset V^{\prime}$ is ind-smooth, it remains to
observe that a local uniformization of $V^{\prime}\cap k(x_{1},\dotsc,x_{n})$
may be constructed using Perron’s algorithm in the style of Zariski.

The goal of the tower of ultrapowers argument given in the Appendix is to
overcome the obstacle that in general $\Gamma$ may not be finitely generated
and there may not even be a group section $s:\Gamma\rightarrow K^{*}$ to
$\mathrm{val}:K^{*}\rightarrow\Gamma$ (roughly, such an $s$ suffices).
Nevertheless, $s$ can always be arranged for any finitely generated submonoid
of $\Gamma$, and the idea is to then use the following fact from model theory:
for a system of equations whose finite subsystems have solutions in $V$, the
entire system has a solution in a well-chosen ultrapower of $V$ (see the
Appendix).

This fact, which rests on the Keisler–Kunen theorem about the existence of
good ultrafilters, permits us to obtain $s$ at the expense of passing to an
ultrapower. However, such a passage replaces $\Gamma$ by its corresponding
ultrapower $\Gamma^{*}$ and, in order to extend $s$ to this $\Gamma^{*}$, we
then need another ultrapower and some model-theoretic facts about algebraic
compactness that ensure that $\Gamma\rightarrow\Gamma^{*}$ be a split
injection. Even though the new ultrapower again enlarges the value group, by
repeating the construction countably many times, in the limit we find our
final $s$ and can conclude.

We should mention that the proof of the main part of Theorem 2 uses only Lemma
7, Corollary 19 and Zariski’s Uniformization Theorem. The last sentence from
Theorem 2 needs the method explained above together with some facts from André
homology. Also this method could be used to get an independent proof for a
weak form of Zariski’s Uniformization Theorem, that is Theorem 36 and in some
cases in mixed characteristic (see Proposition 23, Corollary 24 and
Proposition 25).

We owe thanks to Kęstutis Česnavičius especially for the Appendix, but also
for many ideas and his great help on the presentation of the paper. Also we
owe thanks to the referee who hinted several mistakes in a previous version of
the paper.

##  2\. A reduction to the case of complete valuation rings of rank $1$

We begin by reviewing the following class of generators of the singular ideal.

For a finitely presented ring map $A\rightarrow B$, an element $b\in B$ is
_standard over $A$_ if there exists a presentation $B\cong
A[X_{1},\dotsc,X_{m}]/I$ and $f_{1},\dots,f_{r}\in I$ with $r\leq m$ such that
$b=b^{\prime}b^{\prime\prime}$ with $b^{\prime}=\det({(\partial f_{i}/\partial
X_{j}))_{1\leq i,\,j\leq r}}\in A[X_{1},\dotsc,X_{m}]$ and a
$b^{\prime\prime}\in A[X_{1},\dotsc,X_{m}]$ that kills
$I/(f_{1},\dotsc,f_{r})$. Any multiple of an element $b$ standard over $A$ is
standard over $A$. The definition is compatible with base change: more
precisely, for any morphism $A\rightarrow A^{\prime}$, elements of $B$
standard over $A$ map to elements of $B\otimes_{A}A^{\prime}$ standard over
$A^{\prime}$.

######  Lemma 4.

For a finitely presented ring map $A\rightarrow B$, the loci of vanishing of
standard over $A$ elements of $B$ cut out the locus of non-smoothness of
$\operatorname{Spec}(B)\rightarrow\operatorname{Spec}(A)$. The radical of the
ideal generated by the elements of $B$ standard over $A$ is $H_{B/A}$.

###### Proof.

The argument is standard (compare with [8], [33, 4.3]) but we include it due
to the lack of a convenient reference.

If $b\in B$ is standard over $A$, then $B_{b}$ is the localization of the
standard smooth $A$-algebra
$(A[X_{1},\dots,X_{n}]/(f_{1},\dots,f_{r}))_{\det({(\partial f_{i}/\partial
X_{j}))_{1\leq i,\,j\leq r}}}$ (see [32, 00T8]), so is $A$-smooth. Conversely,
if $B_{b}$ is the coordinate ring of a smooth neighborhood of a fixed prime
${\mathfrak{p}}\subset B$, then we may choose a presentation
$B[X_{1},\dotsc,X_{m}]/I$ and $f_{1},\dotsc,f_{r}\in I$ such that, at the
expense of localizing at ${\mathfrak{p}}$ further. The module $(I/I^{2})_{b}$
is a free $B_{b}$-module with a basis given by the classes of
$f_{1},\dotsc,f_{r}$ and
$(I/I^{2})_{b}\rightarrow\bigoplus_{i=1}^{m}B_{b}\cdot dX_{i}$ is a split
injection such that $dX_{r+1},\dots,dX_{m}$ maps to a basis for the quotient.
The first condition and Nakayama’s lemma [32, 00DV] then supply an $i\in I$
with $(1+i/b^{n})I_{b}\subset(f_{1},\dots,f_{r})_{b}$ for some $n>0$. It
follows that $b^{N}(b^{n}+i)$ for some $N>0$ kills $I/(f_{1},\dotsc,f_{n})$
and maps to a power of $b$ in $B$. The second condition implies that
$b^{\prime}=\det{(\partial f_{i}/\partial X_{j}))_{1\leq i,\,j\leq r}}$ is a
unit in $B_{b}$, so that $b^{\prime}$ divides some power of $b$ in $B$. In
conclusion, some power of $b$ is standard over $A$, as desired.

To stress the relevance of the desingularization lemma 7, we recall the
following well-known lemma (see [33, (1.5)] or [32, 07C3]) and definitions,
which will be crucial throughout this paper.

######  Lemma 5.

For a ring $R$ and a set $\mathcal{S}$ of finitely presented $R$-algebras, an
$R$-algebra $R^{\prime}$ is a filtered direct limit of elements of
$\mathcal{S}$ if and only if every $R$-morphism $B\rightarrow R^{\prime}$ with
$B$ a finitely presented $R$-algebra factors as $B\rightarrow S\rightarrow
R^{\prime}$ for some $S\in\mathcal{S}$.

By [26, 1.8] (see also [32, 07GC]), a map of Noetherian rings is ind-smooth if
and only if $A^{\prime}$ is $A$-flat and has geometrically regular $A$-fibers.
In particular, a field extension $K^{\prime}/K$ is ind-smooth if and only if
it is separable.

Concretely, by Lemma 5, a ring map $A\rightarrow A^{\prime}$ is ind-smooth if
and only if every factorization $A\rightarrow B\rightarrow A^{\prime}$ with
$B$ finitely presented over $A$ can be refined to $A\rightarrow B\rightarrow
S\rightarrow A^{\prime}$ with $S$ smooth (or merely ind-smooth) over $A$.
Thus, a finite product or a filtered direct limit of ind-smooth $A$-algebras
is ind-smooth. Evidently, ind-smooth morphisms are stable under base change.
They are also stable under compositions, in fact, we have the following
slightly finer criterion.

######  Lemma 6.

For an ind-smooth map $A\rightarrow A^{\prime}$ and a map
$A^{\prime}\rightarrow A^{\prime\prime}$ such that for every factorization
$A\rightarrow B\rightarrow A^{\prime\prime}$ with $B$ finitely presented over
$A$ the induced factorization $A^{\prime}\rightarrow
A^{\prime}\otimes_{A}B\rightarrow A^{\prime\prime}$ can be refined to
$A^{\prime}\rightarrow A^{\prime}\otimes_{A}B\rightarrow S^{\prime}\rightarrow
A^{\prime\prime}$ for some smooth $A^{\prime}$-algebra $S^{\prime}$, the map
$A\rightarrow A^{\prime\prime}$ is ind-smooth.

###### Proof.

It suffices to argue that the map $A\rightarrow A^{\prime}\rightarrow
S^{\prime}$ is ind-smooth. For this, we express $A^{\prime}$ as a filtered
direct limit of smooth $A$-algebras $S_{i}$, note that $S^{\prime}$ descends
to a smooth $S_{i}$-algebra $S^{\prime}_{i}$ for some $i$, and conclude that
$S^{\prime}$ is then the filtered direct limit of the smooth $A$-algebras
$S^{\prime}_{j}=S_{j}\otimes_{S_{i}}S^{\prime}_{i}$ with $j\geq i$.

The following lemma originates in [24, (7.1)] and its variants have appeared,
for instance, in [33, 18.1], [26, 7.2], [32, 07CT], [15, Proposition 3], and
[28, Proposition 5]. The version below differs in two aspects: we do not
assume Noetherianness and do not require the elements $a$ or $b$ to come from
the base ring $A$. The latter improvement is particularly convenient for our
purposes—we recall that in the General Néron desingularization arranging for
$b$ to come from $A$ is an additional step before one can apply the
desingularization lemma (compare with, for instance, [32, 07F4]).

######  Lemma 7.

For a commutative diagram of ring morphisms

$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{b\,\mapsto\,a}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V}$that
factors as
follows$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V/a^{3}V}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A^{\prime}/a^{3}A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

with $B$ finitely presented over $A$, a $b\in B$ that is standard over $A$,
and a nonzerodivisor $a\in A^{\prime}$ that maps to a nonzerodivisor in $V$
that lies in every maximal ideal of $V$, there is a smooth
$A^{\prime}$-algebra $S$ such that the original diagram factors as follows:

$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V.}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

###### Proof.

The finitely presented $A^{\prime}$-algebra $B\otimes_{A}A^{\prime}$ comes
equipped with a morphism to $V$ and a retraction modulo $b^{3}$ to
$A^{\prime}/a^{3}A^{\prime}$ that sends $b$ to $a$. Moreover, the image of $b$
in $B\otimes_{A}A^{\prime}$ is standard over $A^{\prime}$. Thus, by replacing
$A$ by $A^{\prime}$ and $B$ by $B\otimes_{A}A^{\prime}$, we reduce to the case
$A=A^{\prime}$.

Since the images of $a$ and $b$ in $V$ agree modulo $a^{3}V$, these images are
unit multiples of each other. We write

$B=A[X_{1},\dots,X_{m}]/I\ \ \mbox{and}\ \ f_{1},\dots,f_{r}\in I$

and choose $b^{\prime}=\det({(\partial f_{i}/\partial X_{j})_{1\leq i,\,j\leq
r}})\in A[X_{1},\dotsc,X_{m}]$ and a $b^{\prime\prime}\in
A[X_{1},\dotsc,X_{m}]$ that kills $I/(f_{1},\dotsc,f_{r})$ with
$b=b^{\prime}b^{\prime\prime}$ in $B$. In these coordinates, we fix a map

|
---|---
$\textstyle{A[X_{1},\dotsc,X_{m}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B/b^{3}B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A[X_{1},\dotsc,X_{m}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f\mapsto\tilde{f}}$$\textstyle{A}$that
makes the
diagram$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A/a^{3}A}$

commute, so that $\tilde{f}\in a^{3}A$ for every $f\in I$. In particular, the
assumption $b\mapsto a$ gives $\widetilde{b^{\prime}b^{\prime\prime}}\equiv
a\bmod a^{3}A$. It follows that

$\widetilde{b^{\prime}b^{\prime\prime}}=au\ \ \mbox{for some}\ \ u\in
1+a^{2}A,$

so that, in particular, $u$ maps to a unit in $V$.

We consider the $m\times m$ matrix $\Delta$ given by

$\left({\begin{matrix}\partial f_{1}/\partial X_{1}&\partial f_{1}/\partial
X_{2}&\dotsc&\partial f_{1}/\partial X_{r}&\partial f_{1}/\partial
X_{r+1}&\partial f_{1}/\partial X_{r+2}&\dotsc&\partial f_{1}/\partial
X_{m}\\\ \partial f_{2}/\partial X_{1}&\partial f_{2}/\partial
X_{2}&\dotsc&\partial f_{2}/\partial X_{r}&\partial f_{2}/\partial
X_{r+1}&\partial f_{2}/\partial X_{r+2}&\dotsc&\partial f_{2}/\partial
X_{m}\\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\ \partial
f_{r}/\partial X_{1}&\partial f_{r}/\partial X_{2}&\dotsc&\partial
f_{r}/\partial X_{r}&\partial f_{r}/\partial X_{r+1}&\partial f_{r}/\partial
X_{r+2}&\dotsc&\partial f_{r}/\partial X_{m}\\\ 0&0&\dotsc&0&1&0&\dotsc&0\\\
0&0&\dotsc&0&0&1&\dotsc&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&\dotsc&0&0&0&\dotsc&1\\\ \end{matrix}}\right)$

that satisfies $\det(\Delta)=b^{\prime}$. We let $Ad(\Delta)$ denote the
adjoint matrix, so that

$Ad(\Delta)\cdot\Delta=\Delta\cdot Ad(\Delta)=b^{\prime}\cdot Id_{m\times m}.$

We let $x_{i}$ and $x_{i}^{\prime}$ be the images in $V$ of $X_{i}$ and
$\tilde{X}_{i}$, respectively, so that, by construction,
$x_{i}-x^{\prime}_{i}\in a^{3}V$. Moreover, $a$ is a nonzerodivisor in $V$ and
there we have that

$\left({\begin{smallmatrix}t_{1}\\\ \vdots\\\
t_{m}\end{smallmatrix}}\right)\tilde{\Delta}\ \
\left({\begin{smallmatrix}(x_{1}-x_{1}^{\prime})/a^{2}\\\ \vdots\\\
(x_{m}-x_{m}^{\prime})/a^{2}\end{smallmatrix}}\right)\ \ \mbox{satisfies}\ \
t_{i}\in aV\ \ \mbox{and}\ \
a\widetilde{b^{\prime\prime}Ad(\Delta)}\left({\begin{smallmatrix}t_{1}\\\
\vdots\\\
t_{m}\end{smallmatrix}}\right)=u\left({\begin{smallmatrix}x_{1}-x_{1}^{\prime}\\\
\vdots\\\ x_{m}-x_{m}^{\prime}\end{smallmatrix}}\right).$

We let $T_{1},\ldots,T_{m}$ be new variables and set

$\left({\begin{smallmatrix}h_{1}\\\ \vdots\\\
h_{m}\end{smallmatrix}}\right)=u\left({\begin{smallmatrix}X_{1}-\tilde{X}_{1}\\\
\vdots\\\
X_{m}-\tilde{X}_{m}\end{smallmatrix}}\right)-a\widetilde{b^{\prime\prime}Ad(\Delta)}\left({\begin{smallmatrix}T_{1}\\\
\vdots\\\ T_{m}\end{smallmatrix}}\right),\ \ \mbox{so that}\ \ h_{i}\in
A[X_{1},\dots,X_{m},T_{1},\dots,T_{m}].$

By construction, if we map $T_{i}$ to $t_{i}$ in $V$, then the $h_{i}$ map to
$0$, so we obtain the map

$\varphi:A_{u}[X_{1},\dots,X_{m},T_{1},\dots,T_{m}]/(h_{1},\dots,h_{m})\rightarrow
V\ \ \mbox{given by}\ \ X_{i}\mapsto x_{i},\ \ T_{i}\mapsto t_{i}.$

Since we have inverted $u$, the source of this map may be identified with
$A_{u}[T_{1},\dotsc,T_{m}]$. To proceed further, we will use Taylor’s formula
to express each $f_{i}$ in terms of this identification.

By Taylor’s formula, for any ring $R$, any section
$R[X_{1},\dotsc,X_{m}]\xrightarrow{f\,\mapsto\,\tilde{f}}R$, and any $f\in
R[X_{1},\dotsc,X_{m}]$,

$f-\tilde{f}-\sum_{i=1}^{m}\widetilde{(\partial f/\partial
X_{i})}(X_{i}-\tilde{X_{i}})\in(X_{1}-\tilde{X_{1}},\dots,X_{m}-\widetilde{X_{m}})^{2}\subset
R[X_{1},\dots,X_{m}].$

In particular, by applying this with $R=A[T_{1},\dots,T_{m}]$ and letting $d$
denote the maximal total degree of any monomial that appears in some $f_{i}$,
we obtain

$Q_{i}\in(T_{1},\dotsc,T_{m})^{2}\subset A[T_{1},\dots,T_{m}]$

for which

$\displaystyle u^{d}f_{i}-u^{d}{\tilde{f}}_{i}\equiv
u^{d-1}a\tilde{b^{\prime\prime}}\ \ (\widetilde{(\partial f_{i}/\partial
X_{1})},\dots,\widetilde{(\partial f_{i}/\partial
X_{m})})\widetilde{Ad(\Delta)}\left({\begin{smallmatrix}T_{1}\\\ \vdots\\\
T_{m}\end{smallmatrix}}\right)+$ $\displaystyle
a^{2}Q_{i}\bmod(h_{1},\dots,h_{m})\equiv
u^{d-1}a\tilde{b^{\prime\prime}}\tilde{b^{\prime}}T_{i}+a^{2}Q_{i}\equiv
a^{2}u^{d}T_{i}+a^{2}Q_{i}\bmod(h_{1},\dots,h_{m})$

We have $\tilde{f_{i}}=a^{2}b_{i}$ for some $b_{i}\in aA$, and for $1\leq
i\leq r$ we set

$g_{i}=u^{d}b_{i}+u^{d}T_{i}+Q_{i}\in A[T_{1},\dots,T_{m}],\ \ \mbox{so that}\
\ a^{2}g_{i}\equiv u^{d}f_{i}\bmod(h_{1},\dotsc,h_{m}).$

This achieves the promised expression of $f_{i}$ in terms of the
identification of the source of $\varphi$ with $A[T_{1},\dotsc,T_{m}$] and
simultaneously shows that each $g_{i}$ vanishes in $V$, so that $\varphi$
induces a map

$\varphi\colon
A_{u}[X_{1},\dots,X_{m},T_{1},\ldots,T_{m}]/(I,g_{1},\dots,g_{r},h_{1},\dots,h_{m})\rightarrow
V.$

In $A[X_{1},\dots,X_{m}]$ the element $b^{\prime}b^{\prime\prime}-au$ lies in
the ideal $(X_{1}-\tilde{X}_{1},\dots,X_{m}-\tilde{X}_{m})$, so in the
quotient $A_{u}[X_{1},\dots,X_{m},T_{1},\dots,T_{m}]/(h_{1},\dots,h_{m})$ it
lies in the ideal $a(T_{1},\dots,T_{m})$. It then follows from the definition
of $b^{\prime\prime}$ and the fact that after inverting $u$ and modulo
$(h_{1},\dots,h_{m})$ the ideal $(g_{1},\dots,g_{r})$ contains
$(f_{1},\dots,f_{r})$ that some element from the coset
$a(u+(T_{1},\dots,T_{m}))$ kills the image of $I$ in

$A_{u}[X_{1},\dots,X_{m},T_{1},\dots,T_{m}]/(g_{1},\dots,g_{r},h_{1},\dots,h_{m}).$

Setting $u^{\prime}=\det((\partial g_{i}/\partial T_{j})_{1\leq i,j\leq r})$,
we deduce that the same then holds in the localization

$\displaystyle(A_{u}[X_{1},\dots,X_{m},T_{1},\dots,T_{m}]/$
$\displaystyle(g_{1},\dots,g_{r},h_{1},\dots,h_{m}))_{u^{\prime}}\cong$
$\displaystyle(A_{u}[T_{1},\dots,T_{m}]/(g_{1},\dots,g_{r}))_{u^{\prime}}.$

However, the latter is smooth over $A$, to the effect that $a$ is a
nonzerodivisor in the ring above. It follows that even some element
$u^{\prime\prime}\in u+(T_{1},\dots,T_{m})$ kills the image of $I$ in the ring
above. By construction, both $u^{\prime}$ and $u^{\prime\prime}$ map to units
in $V$ and $\varphi$ factors through the $A$-smooth algebra

$S=(A_{u}[X_{1},\dots,X_{m},T_{1},\dots,T_{m}]/(g_{1},\dots,g_{r},h_{1},\dots,h_{m}))_{u^{\prime}u^{\prime\prime}}.$

In some situations, when applying Lemma 7 we will not initially have a map
$A^{\prime}\rightarrow V$. The following lifting lemma will help to bypass
this obstacle. Its key novel aspect is that the elements $s$, $s^{\prime}$,
and $v$ need not come from the base ring $A$ (compare with [24, (8.1], [33,
(17.1)], or [32, 07CP]).

######  Lemma 8.

For a ring morphism $A\rightarrow V$ with $V$ local, a smooth $A$-algebra $S$,
an element $s\in S$, a nonunit $v\in V$, and a factorization

$A\rightarrow S\xrightarrow{s\,\mapsto\,v}V/v^{n}V\ \ \mbox{for some}\ \ n\geq
2,$

there are a smooth $A$-algebra $S^{\prime}$, an element $s^{\prime}\in
S^{\prime}$, and factorizations

$A\rightarrow S^{\prime}\xrightarrow{s^{\prime}\,\mapsto\,uv}V\ \ \mbox{with}\
\ u\in V^{*}\ \ \mbox{and}\ \ A\rightarrow
S\xrightarrow{s\,\mapsto\,s^{\prime}}S^{\prime}/s^{\prime
n}S^{\prime}\rightarrow V/v^{n}V;$

if $s$ is the image of an element $a\in A$, then one may choose
$s^{\prime}=a$.

###### Proof.

Due to the local structure of smooth and étale morphisms [32, 054L,00UE], by
localizing $S$ around the preimage of the maximal ideal of $V$, we may assume
that $S$ is standard étale over a polynomial $A$-algebra, that is, that

$S\cong(A[X_{1},\dots,X_{d},Y]/(f))_{g\cdot{\partial f}/{\partial Y}}\ \
\mbox{for some}\ \ f,g\in A[X_{1},\dots,X_{d},Y]\ \ \mbox{with $f$ monic in
$Y$.}$

For a suitable $n\in{\bf N}$, some unit multiple of $s\in S$ of the form
$(g\cdot({\partial f}/{\partial Y}))^{N}\cdot s$ lifts to an $\tilde{s}\in
A[X_{1},\dots,X_{d},Y]$. Letting $x_{1},\dots,x_{d},y$ be some lifts to $V$ of
the images of $X_{1},\dots,X_{d},Y$ in $V/v^{n}V$, we find that the
$A$-morphism

$A[X_{1},\dotsc,X_{d},Y]\xrightarrow{X_{i}\,\mapsto\,x_{i},\,Y\,\mapsto\,y}V$

maps $\tilde{s}$ to a unit multiple of $v$ (as may be checked modulo $v^{n}$),
so it maps $f$ to $\tilde{s}^{n}w$ for some $w\in V$. Thus, we obtain the
$A$-morphism

$S^{\prime}=(A[X_{1},\dots,X_{d},Y,W])_{g\cdot{\partial f}/{\partial
Y}\cdot{\partial(f-\tilde{s}^{n}W)}/{\partial
Y}}/(f-\tilde{s}^{n}W)\xrightarrow{W\,\mapsto\,w}V.$

By construction, $S^{\prime}$ is $A$-smooth and, setting
$s^{\prime}=\tilde{s}\cdot(g\cdot{\partial f}/{\partial Y})^{-N}$ in
$S^{\prime}$ we have the identification

$S^{\prime}/s^{\prime
n}S^{\prime}\cong(A[X_{1},\dots,X_{d},Y,W])_{g\cdot{\partial f}/{\partial
Y}\cdot{\partial(f-\tilde{s}^{n}W)}/{\partial Y}}/(f,s^{\prime
n})\cong((S/s^{n}S)[W])_{{\partial(f-\tilde{s}^{n}W)}/{\partial Y}}$

with $s^{\prime}$ corresponding to $s$ and compatibly with the maps to
$V/v^{n}V$. The main part of the claim follows, and for the remaining
assertion about $a$ note that if $s$ is the image of an $a\in A$, then we may
choose $N=0$ and $\tilde{s}=s^{\prime}=a$ above.

For desingularizing valuation rings, the above lemmas will be useful in
several different ways. We illustrate this right away with the following
results that facilitate passage to completions.

######  Proposition 9.

For a ring $A$, a valuation ring $V$ with the fraction field $K$, a ring
morphism $A\rightarrow V$, a finitely presented $A$-algebra $B$, and maps

$A\rightarrow B\rightarrow V\ \ \mbox{such that}\ \ B\rightarrow K\ \
\mbox{factors through some $A$-smooth localization of $B$.}$

If there are a smooth $A$-algebra $S$ that is an integral domain and a
factorization $A\rightarrow B\rightarrow S\rightarrow\tilde{V}$ where
$\tilde{V}$ is the completion of $V$, then there are a smooth $A$-algebra
$S^{\prime}$ and a factorization

$A\rightarrow B\rightarrow S^{\prime}\rightarrow V.$

###### Proof.

Since $V/b^{3}V\cong\tilde{V}/b^{3}\tilde{V}$, Lemma 8 supplies a smooth
$A$-algebra $S^{\prime}_{0}$, an $s^{\prime}\in S$, a factorization
$A\rightarrow S^{\prime}_{0}\rightarrow V$ that sends $s^{\prime}$ to a unit
multiple of $b$ in $V$, and a factorization

$\textstyle{B/b^{3}B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S/b^{3}S\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{b\,\mapsto\,s^{\prime}}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V/b^{3}V.}$$\textstyle{S^{\prime}_{0}/s^{\prime
3}S^{\prime}_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

The local ring of $S_{0}^{\prime}$ at the preimage of the maximal ideal of $V$
is a domain (see [32, 033C], ) and $s^{\prime}$ is nonzero in this local ring,
so it is a nonzerodivisor there. Thus, Lemma 7 applies and supplies an ind-
smooth $A$-algebra $S^{\prime}$ with a factorization $A\rightarrow
B\rightarrow S^{\prime}\rightarrow V$. Since $B$ is finitely presented, a
limit argument then refines this $S^{\prime}$ to a smooth $A$-algebra.

To draw further consequences, we will use the following well-known result of
Nagata (see [20, Theorem 4] or [32, 053E]).

######  Lemma 10.

Any finitely generated, flat (equivalently, torsion free) algebra over a
valuation ring is finitely presented.

######  Corollary 11.

For a local injection $V\rightarrow V^{\prime}$ of valuation rings that
induces a separable extension $K^{\prime}/K$ of fraction fields, if the map
$V\rightarrow\tilde{V^{\prime}}$ is ind-smooth, ${\tilde{V}}^{\prime}$ being
the completion of $V^{\prime}$, then so is $V\rightarrow V^{\prime}$.

###### Proof.

The separability assumption and Lemma 10 imply that Lemma 9 applies to every
finite type $V$-subalgebra $B\subset V^{\prime}$: a limit argument reduces to
showing that the smooth locus of $B_{K}$ is nonempty, which follows from the
separability of $\operatorname{Frac}(B)/K$ thanks to [10, (6.7.4.1) in IV2]
and [10, (17.5.1) in IV4]. It then remains to apply Lemma 5.

The work above allows us to relate certain “formal desingularization”
extensions of valuation rings studied in [25, section 6] to “weak
desingularization” (that is, ind-smooth) extensions as follows.

######  Proposition 12.

Fix a local injection $V\rightarrow V^{\prime}$ of valuation rings with
fraction fields $K\rightarrow K^{\prime}$ such that
$\mathrm{val}(V\setminus\\{0\\})$ is cofinal in
$\mathrm{val}(V^{\prime}\setminus\\{0\\})$ and for each $0\neq v\in V$ the map
$V/vV\rightarrow V^{\prime}/vV^{\prime}$ is ind-smooth, a finitely presented
$A$-algebra $B$, and maps $V\rightarrow B\rightarrow V^{\prime}$ such that the
map $B\rightarrow K^{\prime}$ factors through some $V$-smooth localization of
$B$. There is a smooth $V$-algebra $S$ and a factorization $V\rightarrow
B\rightarrow S\rightarrow V^{\prime}$. If, in addition, $K^{\prime}/K$ is
separable, then $V^{\prime}$ is ind-smooth over $V$.

###### Proof.

In the case when $K^{\prime}/K$ is separable, $B$ could be any finite type
$V$-subalgebra of $V^{\prime}$, which is finitely presented by Lemma 10. So
the last assertion follows from the rest and Lemma 5. For the assertion about
$B$, we use Lemma 4 to choose an element $b\in B$ standard over $V$ that does
not die in $V^{\prime}$. We assume that $b$ is not a unit in $V^{\prime}$ (or
else we may set $S=B_{b}$) and we choose a $0\neq v\in V$ with
$\mathrm{val}(v)>\mathrm{val}(b)$, where $\mathrm{val}(b)$ denotes the
valuation of $b$ considered in $V^{\prime}$. By our assumptions, there are a
smooth $V/v^{3}V$-algebra $\bar{S}$, an $s\in\bar{S}$, and a factorization
$V\rightarrow
B\rightarrow\bar{S}\xrightarrow{s\,\mapsto\,v}V^{\prime}/v^{3}V^{\prime}$ such
that $b\mid s$ in $\bar{S}$. Thus, since $\bar{S}$ lifts to a smooth
$V$-algebra (see [3, (1.3.1)] or [32, 07M8]), Lemma 8 supplies a smooth
$V$-algebra $S^{\prime}$, an $s^{\prime}\in S^{\prime}$, a factorization
$V\rightarrow S^{\prime}\rightarrow V^{\prime}$ that sends $s^{\prime}$ to a
unit multiple of $v$, and a factorization $V\rightarrow
B\rightarrow\bar{S}\xrightarrow{s\,\mapsto\,s^{\prime}}S^{\prime}/s^{\prime
3}S^{\prime}\rightarrow V^{\prime}/v^{3}V^{\prime}.$ Since $b\mid s^{\prime}$
in $S^{\prime}/s^{\prime 3}S^{\prime}$, by replacing $S^{\prime}$ by its
localization by an element of $1+s^{\prime 2}S^{\prime}$ if necessary we may
ensure that $a\mid s^{\prime}$ in $S^{\prime}$ for some lift $a\in S^{\prime}$
of $b\in S^{\prime}/s^{\prime 3}S^{\prime}$. Then we have a factorization

$V\rightarrow
B\xrightarrow{b\,\mapsto\,a}S^{\prime}/a^{3}S^{\prime}\rightarrow
V^{\prime}/a^{3}V^{\prime}.$

As in the proof of Lemma 9, Lemma 7 then supplies a smooth $V$-algebra $S$.

The following localization lemma, a variant of [27, Lemma 2], [33, (12.2)], or
[32, 07F9], will permit us to localize our valuation rings when arguing their
ind-smoothness.

######  Lemma 13.

For ring maps $A\rightarrow B\rightarrow V$ with $B$ of finite type over $A$,
a prime ${\mathfrak{P}}\subset V$ with preimage ${\mathfrak{p}}\subset A$, and
a factorization $A\rightarrow B\rightarrow S^{\prime}\rightarrow
V_{{\mathfrak{P}}}$ for a finitely presented $A_{{\mathfrak{p}}}$-algebra
$S^{\prime}$, there are a finitely presented $A$-algebra $S$, an $s\in S$ with
$S_{s}\otimes_{A}A_{{\mathfrak{p}}}\simeq S^{\prime}[X,X^{-1}]$, and a
factorization

$A\rightarrow B\rightarrow S\rightarrow V\ \ \mbox{such that}\ \ S\rightarrow
V_{{\mathfrak{P}}}\ \ \mbox{factors as}\ \ S\rightarrow
S_{s}\otimes_{A}A_{{\mathfrak{p}}}\rightarrow V_{{\mathfrak{P}}}.$

###### Proof.

Following the argument of [33, (12.2)], we choose a presentation

$S^{\prime}\simeq(B\otimes_{A}A_{{\mathfrak{p}}})[X_{1},\dots,X_{n}]/(f_{1}(X_{1},\dots,X_{n}),\dots,f_{m}(X_{1},\dots,X_{n}))$

(see [32, 00F4]) in which the polynomials $f_{i}$ have coefficients in $B$,
and we set

$S=B[X_{0},X_{1},\dots,X_{n}]/(X_{0}^{N}f_{1}(X_{1}/X_{0},\dots,X_{n}/X_{0}),\dots,X_{0}^{N}f_{m}(X_{1}/X_{0},\dots,X_{n}/X_{0}))$

for a large enough $N>0$ for which each
$X_{0}^{N}f_{i}(X_{1}/X_{0},\dots,X_{n}/X_{0})$ is a (necessarily homogeneous)
polynomial in $X_{0},X_{1},\dotsc,X_{n}$ of positive degree and coefficients
in $B$. We set $s=X_{0}$, so that a desired isomorphism
$S_{s}\otimes_{A}A_{{\mathfrak{p}}}\simeq S^{\prime}[X,X^{-1}]$ is induced by
the change of variables $X_{0}\mapsto X$ and $X_{i}\mapsto XX_{i}$ for $1\leq
i\leq n$. To build the map $S\rightarrow V$, we first choose
$x_{1},\dots,x_{n}\in V$ and $t\in V\setminus{\mathfrak{P}}$ such that $X_{i}$
maps to $x_{i}/t\in V_{{\mathfrak{P}}}$. Continuing to use abusive notation
for homogeneous polynomials, we note that the (“homogeneous” in
$t,x_{1},\dots,x_{n}$) elements $t^{N}f_{i}(x_{1}/t,\dots,x_{n}/t)$ of $V$ die
in $V_{{\mathfrak{P}}}$, so they are killed by some $t^{\prime}\in
V\setminus{\mathfrak{P}}$. Thus, the $B$-morphism

$B[X_{0},X_{1},\dots,X_{n}]\rightarrow V\ \ \mbox{given by}\ \ X_{0}\mapsto
t^{\prime}t,X_{1}\mapsto t^{\prime}x_{1},\dots,X_{n}\mapsto t^{\prime}x_{n}$

factors through $S$. By construction, the resulting morphism $S\rightarrow
V_{{\mathfrak{P}}}$ factors through the localization
$S_{s}\otimes_{A}A_{{\mathfrak{p}}}$ of $S$, as desired.

We are ready for the promised reduction to complete, one-dimensional valuation
rings.

######  Proposition 14.

Consider the following property of a valuation ring $V$ and a subring
$A\subset V$:

$(*)$ every $A\rightarrow B\rightarrow V$ with $B$ a finite type $A$-algebra
such that $B\rightarrow\operatorname{Frac}(V)$ factors through an $A$-smooth
localization of $B$ has a refinement $A\rightarrow B\rightarrow S\rightarrow
V$ with $S$ smooth over $A$.

For a finite dimensional valuation ring $V$ with a subfield $A\subset V$, if
for all consecutive primes
${\mathfrak{q}}^{\prime}\subset{\mathfrak{q}}\subset V$ the complete height
one valuation ring $\widetilde{(V/{\mathfrak{q}}^{\prime})_{{\mathfrak{q}}}}$
satisfies $(*)$, then so does $V$.

###### Proof.

We fix a finite type $A$-algebra $B$ equipped with a factorization
$A\rightarrow B\rightarrow V$ as in $(*)$, which we need to factor further as
$A\rightarrow B\rightarrow S\rightarrow V$ for some smooth $A$-algebra $S$.
When $B\rightarrow V$ itself factors through an $A$-smooth localization of
$B$, there is nothing to show. Otherwise, since $V$ is of finite height, we
may choose the minimal prime ${\mathfrak{q}}\subset V$ whose preimage in $B$
does not lie in the $A$-smooth locus of $\operatorname{Spec}(B)$ and the
largest prime ${\mathfrak{q}}^{\prime}\subsetneq{\mathfrak{q}}\subset V$
properly contained in ${\mathfrak{q}}$ (the assumption in $(*)$ ensures that
${\mathfrak{q}}^{\prime}$ exists). Thanks to Lemma 13, we may replace $V$ by
$V_{{\mathfrak{q}}}$ to reduce to the case when ${\mathfrak{q}}$ is the
maximal ideal (so that
$\widetilde{(V/{\mathfrak{q}}^{\prime})_{{\mathfrak{q}}}}=\widetilde{V/{\mathfrak{q}}^{\prime}}$):
indeed, once we resolve this case, then, by using Lemma 13, we will be able to
refine $B$ to an $A$-algebra that either is smooth or for which
${\mathfrak{q}}$ is strictly larger, and, by iteration, we will then arrive at
a desired $S$.

By Lemma 4, there is an element $b\in B$ standard over $A$ that maps to
${\mathfrak{q}}\setminus{\mathfrak{q}}^{\prime}$. The property from $(*)$ of
$\widetilde{V/{\mathfrak{q}}^{\prime}}$ then supplies a smooth $A$-algebra
$S^{\prime}$, an element $s\in S^{\prime}$ (the image of $b$), and a
factorization

$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{b\,\mapsto\,s}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\widetilde{V/{\mathfrak{q}}^{\prime}}/b^{3}\widetilde{V/{\mathfrak{q}}^{\prime}}\cong
V/b^{3}V.}$$\textstyle{S^{\prime}/s^{3}S^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

Thanks to Lemma 8, we may change $S^{\prime}$ in order to make sure that the
map $S^{\prime}/s^{3}S^{\prime}\rightarrow V/b^{3}V$ lifts to an $A$-morphism
$S^{\prime}\rightarrow V$. This puts us in a situation in which we may apply
Lemma 7 to obtain a smooth $S^{\prime}$-algebra $S$ with a desired
factorization $A\rightarrow B\rightarrow S\rightarrow V$.

##  3\. Ind-smoothness of large immediate extensions of valuation rings

Our next goal is to find a large class of extensions of valuation rings that
are ind-smooth. The argument combines classical results from valuation theory
that go back to Kaplansky, results from [23] (see Lemma 15 and its proof), and
the desingularization lemmas from Section 2.

Consider the case when $V$ is not noetherian and its associated valuation has
rank one. The proofs are similar to the Noetherian case except for the
reduction to the case of immediate extension $V\subset V^{\prime}$. In the
Noetherian case $V$ is dense in $V^{\prime}$, but in general case it need not
be. If $V\supset{\bf Q}$ the problem is solved by Ostrowski’s Defektsatz [22]
but when char $V>0$ immediate algebraic extensions present extra difficulties.

An inclusion $V\subset V^{\prime}$ of valuation rings is an _immediate
extension_ if it is local as a map of local rings and induces isomorphisms
between the value groups and the residue fields of $V$ and $V^{\prime}$. For
such a $V\subset V^{\prime}$, letting $K^{\prime}/K$ be the induced fraction
field extension, we have $V=V^{\prime}\cap K$ (see [5, (4.1) in VI]).
Moreover, for any subextension $K^{\prime}/K^{\prime\prime}/K$ and the
valuation ring $V^{\prime\prime}=V^{\prime}\cap K^{\prime\prime}$, both
$V\subset V^{\prime\prime}$ and $V^{\prime\prime}\subset V^{\prime}$ are then
also immediate extensions (to check the value group requirement one uses that
any $v^{\prime\prime}\in V^{\prime\prime}$ is a unit if and only if so is its
image in $V^{\prime}$).

For example, for any valuation ring $V$, the extension $V\subset\tilde{V}$ is
immediate (see [33]), ${\tilde{V}}^{\prime}$ being the completion of
$V^{\prime}$.

For a valuation ring $V$ with the fraction field $K$, a sequence
$\\{v_{i}\\}_{i<\omega}$ in $K$ indexed by the ordinals $i$ less than a fixed
limit ordinal $\omega$ is _pseudo-convergent_ if

$\mathrm{val}(v_{i}-v_{i^{\prime\prime}})<\mathrm{val}(v_{i^{\prime}}-v_{i^{\prime\prime}})\
\ \mbox{(that is,}\ \
\mathrm{val}(v_{i}-v_{i^{\prime}})<\mathrm{val}(v_{i^{\prime}}-v_{i^{\prime\prime}}))\
\ \mbox{for}\ \ i<i^{\prime}<i^{\prime\prime}<\omega$ (see [12], [33]). A
(possibly nonunique) _pseudo-limit_ of a pseudo-convergent sequence
$\\{v_{i}\\}_{i<\omega}$ is an element $v\in K$ with

$\mathrm{val}(v-v_{i})<\mathrm{val}(v-v_{i^{\prime}})\ \ \mbox{(that is,}\ \
\mathrm{val}(v-v_{i})=\mathrm{val}(v_{i}-v_{i^{\prime}}))\ \ \mbox{for}\ \
i<i^{\prime}<\omega$. A pseudo-convergent sequence $\\{v_{i}\\}_{i<\omega}$ in
$K$ is

  1. (1)

_algebraic_ if some $f\in K[T]$ satisfies
$\mathrm{val}(f(v_{i}))<\mathrm{val}(f(v_{i^{\prime}}))$ for large enough
$i<i^{\prime}<\omega$;

  2. (2)

_transcendental_ if each $f\in K[T]$ satisfies
$\mathrm{val}(f(v_{i}))=\mathrm{val}(f(v_{i^{\prime}}))$ for large enough
$i<i^{\prime}<\omega$.

(Here “large enough” means larger than a fixed ordinal
$\omega^{\prime}<\omega$ that is allowed to depend on $f$.) In both cases,
[12, Theorems 1, 2] describe the valuation of $K^{\prime}$ that extends $V$ of
$K$. For instance, in the transcendental case, by [12, Theorem 2], this
valuation on $K(t)$ is given by setting

$\mathrm{val}(((f(t))/(g(t)))=\mathrm{val}(f(v_{i}))-\mathrm{val}(g(v_{i}))\quad\text{for
large enough}\quad i<\omega.$

These results lead to [12, Theorem 4]: a valuation ring $V$ has no nontrivial
immediate extensions if and only if each pseudo-convergent sequence in its
fraction field $K$ has a pseudo-limit in $K$. If
$\mathrm{val}(v_{i}-v_{i^{\prime}})>\gamma$ for all $\gamma\in\Gamma$, the
value group of $V$, and all $i<i^{\prime}$ large then we call
$\\{v_{i}\\}_{i<\omega}$ fundamental. As in [23, Lemma 3.2] we get the
following lemma.

######  Lemma 15.

For an immediate extension $V\subset V^{\prime}$ of valuation rings and a
transcendental pseudo-convergent sequence $(v_{i})_{i<\alpha}$ in $K$, which
has a pseudo-limit $v^{\prime}$ in $K^{\prime}$ but no pseudo-limit in $K$ the
valuation ring $V^{\prime\prime}=V^{\prime}\cap K(v^{\prime})$ is a filtered
increasing union of smooth $V$-subalgebras.

###### Proof.

For each $i$ set $x_{i}=(v^{\prime}-v_{i})/(v_{i+1}-v_{i}),$ so that $x_{i}$
is an unit in $V^{\prime}$. Let ${\mathfrak{m}}^{\prime}$ be the maximal ideal
of $V^{\prime}$. We show that for every polynomial $0\neq f\in V[t]$ it holds

$f(v^{\prime})\in f(v_{i})\cdot(1+{\mathfrak{m}}^{\prime}\cap
V[x_{i}])\quad\text{for every large enough}\quad i<\omega.$

Since $\\{v_{i}\\}_{i<\omega}$ is transcendental, for each $g(t)\in K[t]$, the
value $\mathrm{val}(g(v_{i}))$ is constant for large $i$. Moreover, for large
$i$ the values $\mathrm{val}(v^{\prime}-v_{i})$ are strictly increasing as $i$
increases. Thus, in the Taylor expansion111The polynomials $D^{(n)}f\in R[t]$
for $f\in R[t]$ make sense for any ring $R$: indeed, one constructs the Taylor
expansion in the universal case $R={\bf
Z}[a_{0},\dots,a_{\operatorname{deg}f}]$ by using the equality
$n!\cdot(D^{(n)}f)=f^{(n)}$.

$\textstyle
f(v^{\prime})=\sum_{n=0}^{\operatorname{deg}f}(D^{(n)}f)(v_{i})\cdot(v^{\prime}-v_{i})^{n}\quad\text{with}\quad
D^{(n)}f\in V[t]$

the values $\mathrm{val}((D^{(n)}f)(v_{i})\cdot(v^{\prime}-v_{i})^{n})$ are
pairwise distinct for every large enough $i$. Consequently, since
$\mathrm{val}(f(v^{\prime}))=\mathrm{val}(f(v_{i}))$ for large $i$, we
conclude that

$\mathrm{val}((D^{(n)}f)(v_{i})\cdot(v^{\prime}-v_{i})^{n})>\mathrm{val}(f(v_{i}))\quad\text{for
every $n>0$ and large enough $i<\omega$.}$

It remains to note that

$(v^{\prime}-v_{i})^{n}=x_{i}^{n}\cdot(v_{i+1}-v_{i})^{n}\ \ \mbox{and}\ \
\mathrm{val}(v^{\prime}-v_{i})=\mathrm{val}(v_{i+1}-v_{i}),$

which is enough for our claim. In particular, we get that $f(v^{\prime})$ is
transcendental over $K$. The element $x_{i}$ is transcendental over $K$, so
$V[x_{i}]\subset V^{\prime}$ is the polynomial algebra. Moreover, for
$i<i^{\prime}<\omega$ we have
$x_{i}=x_{i^{\prime}}\cdot(v_{i^{\prime}+1}-v_{i^{\prime}})/(v_{i+1}-v_{i})+(v_{i^{\prime}}-v_{i})/(v_{i+1}-v_{i})$,
so $V[x_{i}]\subset V[x_{i^{\prime}}]\subset V^{\prime}$. Consequently, we
arrive at a nested sequence

$\\{V[x_{i}]_{{\mathfrak{m}}^{\prime}\cap V[x_{i}]}\\}_{i<\omega}\ \ \mbox{of
ind-smooth $V$-subalgebras of $V^{\prime}$,}$

and, it remains to show that every element of $V^{\prime}$ belongs to some
$V[x_{i}]_{{\mathfrak{m}}^{\prime}\cap V[x_{i}]}$. In fact, it suffices to
show that each $0\neq f\in V[t]$ satisfies

$f(v^{\prime})\in f(v_{i})\cdot(1+{\mathfrak{m}}^{\prime}\cap
V[x_{i}])\quad\text{for every large enough}\quad i<\omega$

which was done above.

######  Lemma 16.

For an immediate extension $V\subset V^{\prime}$ of valuation rings containing
$\bf Q$ with value group $\Gamma\subset{\bf R}$ every algebraic pseudo-
convergent sequence $(v_{i})_{i<\alpha}$ in $K$, which is not a fundamental
sequence but has a pseudo-limit $v^{\prime}$ in $K^{\prime}$ has also a
pseudo-limit in $K$.

###### Proof.

By [12, Theorem 3] there exists an immediate extension of valued fields
$K\subset K(u)$ such that $u$ is algebraic over $K$ and it is a pseudo-limit
of $(v_{i})$ in $K(u)$. As a consequence of Ostrowski’s Defektsatz [22, Sect
9, No 55] (see [23, Corollary 4.2]) we see that $K\subset K(u)$ is dense, that
is, $u$ belongs to the completion of $K$. Thus $(v_{i})$ has a pseudo-limit in
$K$ by [25, Lemma 2.5].

######  Remark 17.

The above lemma is false if the characteristic of the valuation rings is $>0$
(see [25, Example 3.13] inspired by [22, Sect 9, No 57]).

######  Proposition 18.

For an immediate extension $V\subset V^{\prime}$ of valuation rings containing
$\bf Q$ with value group $\Gamma\subset{\bf R}$, $V^{\prime}$ is ind-smooth
over $V$.

###### Proof.

Applying Lemma 15 several times we find a pure transcendental extension
$K^{\prime\prime}\subset K^{\prime}$ of $K$ such that
$V^{\prime\prime}=V^{\prime}\cap K^{\prime\prime}$ is ind-smooth over $V$ and
all transcendental pseudo-convergent sequences in $V^{\prime\prime}$ over $V$,
which are not fundamental sequences, have pseudo-limits in $V^{\prime\prime}$.
By Lemma 16 we see that this holds also for algebraic pseudo-convergent
sequences from $V^{\prime\prime}$, which are not fundamental sequences. It
follows that the extension $V^{\prime\prime}\subset V^{\prime}$ is dense using
[12, Theorem 1]. Now, it is enough to apply Lemma 7 to see that $V^{\prime}$
is ind-smooth over $V^{\prime\prime}$. Indeed, let $B\subset V^{\prime}$ be a
finitely generated $V^{\prime\prime}$-subalgebra and $w$ its inclusion. By
separability, $w(H_{B/V^{\prime\prime}})\not=0$ and choose an element $d\in
V^{\prime\prime}$, which is standard for $B$ over $V^{\prime\prime}$. Then the
composite map $B\rightarrow V^{\prime}\rightarrow
V^{\prime}/d^{3}V^{\prime}\cong V^{\prime\prime}/d^{3}V^{\prime\prime}$
factors obviously to a smooth
$V^{\prime\prime}/d^{3}V^{\prime\prime}$-algebra. By Lemma 7 $w$ factors
through a smooth $V^{\prime\prime}$-algebra.

######  Corollary 19.

For an extension $V\subset V^{\prime}$ of valuation rings containing $\bf Q$
with the same value group $\Gamma\subset{\bf R}$, $V^{\prime}$ is ind-smooth
over $V$.

###### Proof.

By Proposition 9 we may reduce to the case when $V$, $V^{\prime}$ are complete
and so they are Henselian since $\dim V=1$ , that is, they contain their
residue fields $k,k^{\prime}$. Let $K$, $K^{\prime}$ be the fraction fields of
$V$, resp. $V^{\prime}$. Then $V^{\prime}$ is an immediate extension of
$V^{\prime\prime}=V^{\prime}\cap K(k^{\prime})$ and so it is ind-smooth by the
above proposition. Express $k^{\prime}$ as a filtered union of some finitely
generated field extensions $(k_{i})$ of $k$. It is enough to see that
$V_{i}=V^{\prime}\cap K(k_{i})$ is an ind-smooth extension of $V$. But $V_{i}$
is even essentially smooth over $V$ because $k_{i}$ is so over $k$.

##  4\. Extensions of valuation rings

######  Proposition 20.

Let $V\subset V^{\prime}$ be an extension of valuation rings containing $\bf
Q$. Suppose that

  1. (1)

for each $q\in\operatorname{Spec}V$ the ideal $qV^{\prime}$ is also prime,

  2. (2)

For any prime ideals $q_{1},q_{2}\in\operatorname{Spec}V$ such that
$q_{1}\subset q_{2}$ and $\operatorname{height}(q_{2}/q_{1})=1$ the extension
$V_{q_{2}}/q_{1}V_{q_{2}}\subset
V^{\prime}_{q_{2}V^{\prime}}/q_{1}V^{\prime}_{q_{2}V^{\prime}}$ of valuation
rings is ind-smooth.

Then $V^{\prime}$ is ind-smooth over $V$.

###### Proof.

Let $E$ be a $V$-algebra of finite presentation, let us say $E\cong V[Y]/I$,
$Y=(Y_{1}\ldots,Y_{m})$, $I$ being a finitely generated ideal. Let
$w:E\rightarrow V^{\prime}$ be a $V$-morphism. We may suppose that
$\operatorname{Frac}(V)\subset\operatorname{Frac}(V^{\prime})$ is a field
extension of finite type changing $V^{\prime}$ by $V^{\prime}\cap F$ for a
field extension $F/\operatorname{Frac}(V)$ of finite type. By separability (we
may consider $w$ injective) we have $w(H_{E/V})\not=0$, let us assume that
$H_{E/V}V^{\prime}=zV^{\prime}$ for some $z\in{\mathfrak{m}}V^{\prime}$,
$z\not=0$. Also we may assume that for some polynomials
$f=(f_{1},\ldots,f_{r})$ from $I$, we have $z\in NME$ for some $N\in((f):I)$
and a $r\times r$-minor $M$ of the Jacobian matrix $(\partial f_{i}/\partial
Y_{j})$. Let $q_{2}^{\prime}\in\operatorname{Spec}V^{\prime}$, be the minimal
prime ideal of $zV^{\prime}$ and $q_{2}=q_{2}^{\prime}\cap V$.

First assume that $q^{\prime}_{2}\not=q_{2}V^{\prime}$. Thus
$V^{\prime}_{q^{\prime}_{2}}/q_{2}V^{\prime}_{q^{\prime}_{2}}$ is ind-smooth
over the field $k=V_{q_{2}}/q_{2}V_{q_{2}}$ by Zariski’s Uniformization
Theorem (see Theorem 1 or our Theorem 36). Set $E^{\prime}=E[Z]$ and let
$w^{\prime}:E^{\prime}\rightarrow V^{\prime}$ be given by $Z\rightarrow z$.
Thus the composite map
$E^{\prime}\xrightarrow{w^{\prime}}V^{\prime}\rightarrow
V^{\prime}_{q^{\prime}_{2}}/q_{2}V^{\prime}_{q^{\prime}_{2}}$ factors by a
smooth $k$-algebra $G$, let us say it is the composite map
$E^{\prime}\xrightarrow{\alpha}G\xrightarrow{\beta}V^{\prime}_{q^{\prime}_{2}}/q_{2}V^{\prime}_{q^{\prime}_{2}}$.
We may assume that $G=V[U]_{g^{\prime}h}/(g)$, with $U=(U_{1},\ldots,U_{l})$,
$g^{\prime}=\partial g/\partial U_{1}$, $h\in V[U]$ and let $\beta$ be given
by $U\rightarrow u+z^{3}V^{\prime}_{q^{\prime}_{2}}$ for some $u\in V^{\prime
l}$. Choose $P\in V[U]$ inducing $\alpha(Z)$ modulo an invertible element in
$G$. Then $g(u)=P(u)^{3}t$ for some $t\in V^{\prime}_{q^{\prime}_{2}}$. Note
that the composite map $E^{\prime}\rightarrow V^{\prime}\rightarrow
V^{\prime}_{q^{\prime}_{2}}$ factors through the smooth $V_{q_{3}}$-algebra
$D=(V[U,T]/(g-P^{3}T))_{g^{\prime}h}$ modulo $P^{3}$. By Lemma 7 we see that
$E^{\prime}\rightarrow V^{\prime}_{q^{\prime}_{2}}$ factors through a smooth
$D$-algebra $D^{\prime}$. Using Lemma 13 we see that $w^{\prime}$ factors
through a finitely presented $V$-algebra $E^{\prime\prime}$, let us say
through a map $w^{\prime\prime}:E^{\prime\prime}\rightarrow V^{\prime}$ with
$w^{\prime\prime}(H_{E^{\prime\prime}/V})\not\subset q^{\prime}_{2}$.

Next assume that $q_{2}^{\prime}=q_{2}V^{\prime}$ and let
$q_{1}\in\operatorname{Spec}V$ be the biggest prime ideal of $V$ strictly
contained in $q_{2}$. Thus $\operatorname{height}(q_{2}/q_{1})=1$. By
hypothesis (2) the extension $V_{q_{2}}/q_{1}V_{q_{2}}\subset
V^{\prime}{q^{\prime}_{2}}/q_{1}V^{\prime}_{q^{\prime}_{2}}$ is ind-smooth. As
above we find $G$, $D$ and we see that $w$ factors through a finitely
presented $V$-algebra $E^{\prime\prime}$, let us say through a map
$w^{\prime\prime}:E^{\prime\prime}\rightarrow V^{\prime}$ with
$w^{\prime\prime}(H_{E^{\prime\prime}/V})\not\subset q^{\prime}_{2}$.

If the above valuation ring extension above is dense then the result follows
from Proposition 9. Let $\Gamma\subset\Gamma^{\prime}$ be the value group
extension of $V\subset V^{\prime}$. Note that $\Gamma^{\prime}/\Gamma$ has
finite rational rank because the fraction field extension induced by $V\subset
V^{\prime}$ is of finite type. Consequently just for finite prime ideals
$q^{\prime}_{2}$ we are in the case of the hypothesis (2). Thus we arrive in
finite steps in the case when $z$ is a unit, that is we can embed $E$ in a
smooth $V$-algebra. This is enough by Lemma 5.

######  Lemma 21.

Let $V\subset V^{\prime}$ be an extension of valuation rings containing $\bf
Q$, with ${\mathfrak{m}}$,${\mathfrak{m}}^{\prime}={\mathfrak{m}}V^{\prime}$
their maximal ideals and $\dim V=1$. Let
$q^{\prime}\in\operatorname{Spec}V^{\prime}$ be such that
$\operatorname{height}{\mathfrak{m}}^{\prime}/q^{\prime}=1$. Assume that the
extension $V\subset V^{\prime}/q^{\prime}$ has the same value group. Then
$V^{\prime}$ is ind-smooth over $V$.

For the proof apply Corollary 19 to see that $V\subset V^{\prime}/q^{\prime}$
is ind-smooth and follow the above proof using Zariski’s Uniformization
Theorem (see Theorem 1 or our Theorem 36).

######  Corollary 22.

Let $V\subset V^{\prime}$ be an extension of valuation rings containing $\bf
Q$. Suppose that

  1. (1)

for each $q\in\operatorname{Spec}V$ the ideal $qV^{\prime}$ is also prime,

  2. (2)

For any prime ideals $q\in\operatorname{Spec}V$ and
$q^{\prime}\in\operatorname{Spec}V^{\prime}$ such that
$\operatorname{height}(qV^{\prime}/q^{\prime})=1$ the extension
$V_{q}/(q^{\prime}\cap V)V_{q}\subset
V^{\prime}_{qV^{\prime}}/q^{\prime}V^{\prime}_{qV^{\prime}}$ of one
dimensional valuation rings has the same value group

Then $V^{\prime}$ is ind-smooth over $V$.

By the above lemma we see that the condition (2) implies the condition (2)
from Proposition 20 which can be applied.

######  Proposition 23.

Let $V\subset V^{\prime}$ be an extension of valuation rings. Suppose that

  1. (1)

$V$ is a discrete valuation ring an extension of ${\bf Z}_{(p)}$ with $\pi$
its local parameter, and $p$ a prime number.

  2. (2)

$\pi V^{\prime}$ is the maximal ideal of $V^{\prime}$,

  3. (3)

the residue field extension of $V\subset V^{\prime}$ is separable.

Then $V\rightarrow V^{\prime}$ is ind-smooth.

###### Proof.

Like in Proposition 20, we may reduce to the case when the fraction field
extension of $V\subset V^{\prime}$ is of finite type and so to the case when
$\dim V^{\prime}<\infty$. Let $w:E\rightarrow V^{\prime}$ be as in the proof
of Proposition 20 and choose $q^{\prime}_{2}\in\operatorname{Spec}V^{\prime}$
a minimal prime ideal of $w(H_{E/V})V^{\prime}$. If $q^{\prime}_{2}\not=\pi
V^{\prime}$ then using Zariski’s Uniformization Theorem (or Theorem 36) we may
change $E,w$ with some $E^{\prime},w^{\prime}$ such that
$w^{\prime}(H_{E^{\prime}/V})V^{\prime}\not\subset q^{\prime}_{2}$. Step by
step we arrive to the case when either $w(H_{E/V})V^{\prime}=V^{\prime}$, or
$w(H_{E/V})V^{\prime}$ is a $\pi V^{\prime}$-primary ideal. In the first case,
$w$ factors through a localization of $E$ which is smooth. In the second case,
$q^{\prime}_{1}=\cap_{i\in{\bf N}}\pi^{i}V^{\prime}$ is a prime ideal and the
composite map $V\rightarrow V^{\prime}\rightarrow V^{\prime}/q^{\prime}_{1}$
is a regular map of discrete valuation rings and so an ind-smooth map by the
classical Néron Desingularization. The proof ends as in Proposition 20.

######  Corollary 24.

Let $V$ be a discrete valuation ring an extension of ${\bf Z}_{(p)}$ with $p$
a prime number and $V^{\prime}$ an ultrapower of $V$ with respect to a
nonprincipal ultrafilter on $\bf N$. Then $V\subset V^{\prime}$ is ind-smooth.

For the proof note that the maximal ideal of $V$ generates the maximal ideal
of $V^{\prime}$.

######  Proposition 25.

Let $V$ be a discrete valuation ring an extension of ${\bf Z}_{(p)}$ with $p$
a prime number and $V\subset V^{\prime}$ an extension of valuation rings such
that

  1. (1)

$\pi V^{\prime}$ is a ${\mathfrak{m}}^{\prime}$-primary ideal of $V^{\prime}$,
where ${\mathfrak{m}}^{\prime}$ is the maximal ideal of $V^{\prime}$,

  2. (2)

the residue field extension of $V\subset V^{\prime}$ is separable.

Then $V^{\prime}$ is a filtered direct limit of regular local rings
essentially of finite type over $V$.

###### Proof.

As in Proposition 23 we may consider $w:E\rightarrow V^{\prime}$ and we may
reduce to the case when $w^{\prime}(H_{E^{\prime}/V})V^{\prime}$ is
${\mathfrak{m}}^{\prime}$-primary ideal. We may assume that $p^{s}$ is a
standard of $E$ over $V$ for some $s\in{\bf N}$ and as in the proof of [30,
Theorem 3.6] there exists a local essentially smooth $V$-algebra $G$ and $b\in
G$ such that the map $E/p^{3s}E\rightarrow V^{\prime}/p^{3s}V^{\prime}$
factors through $G/(p^{3s},p-b)$. Then a variant of Lemma 7 in the idea of
[30, Proposition 3.4] shows that $w$ factors through a local essentially
smooth $D=G/(p-b)$-algebra $D^{\prime}$. This $D^{\prime}$ is regular local
since $D$ is so.

##  5\. Structure of equicharacteristic valuation rings possessing a cross-
section

Modulo all the reductions and simplifications that go into the overall proof
of Theorem 2, our ultimate source of expressions of valuation rings as
filtered direct limits of smooth rings is Lemma 27 below. This lemma describes
some valuations on an affine space for which local uniformizations can be
constructed by successively blowing up regular centers as in [31, 4.5, 4.19]
following Perron’s algorithm (whose relevance to the resolution of
singularities was explained already in [34]). We present a more direct
argument for this uniformization that is close to [23, Lemma 4.6] and rests on
the following lemma that captures the “combinatorial” part of local
uniformization.

We will need the following lemma (see [7, 2.2], or [23, 4.6.1], or [9,
6.1.30]).

######  Lemma 26.

For a totally ordered abelian group $\Gamma$, the submonoid $\Gamma_{\geq
0}\subset\Gamma$ of nonnegative elements is a filtered increasing union of its
finite free submonoids isomorphic to ${\bf Z}_{\geq 0}^{r}$, where $r\in{\bf
Z}_{\geq 0}$ need not be constant.

We include a mixed characteristic version the following lemma because it
requires virtually no additional effort in comparison to the
equicharacteristic case that we will use below.

######  Lemma 27.

  1. (1)

For a field $\bf F$, a valuation ring ${\bf F}\subset V$ with fraction field
${\bf F}(x_{1},\dotsc,x_{n})$ such that $\mathrm{val}(x_{1})$, $\dots$,
$\mathrm{val}(x_{n})$ are $\bf Z$-linearly independent is a countable direct
union of essentially smooth $\bf F$-algebras.

  2. (2)

For a discrete valuation ring $\Lambda$ with uniformizer $\pi$ and fraction
field $\bf F$, a valuation ring $\Lambda\subset V$ that dominates $\Lambda$
and has fraction field ${\bf F}(x_{1},\dotsc,x_{n})$ such that
$\mathrm{val}(\pi),\mathrm{val}(x_{1}),\dots,\mathrm{val}(x_{n})$ are $\bf
Z$-linearly independent is a countable direct union of regular local
$\Lambda$-algebras of the form

$(\Lambda[Y_{1},\dots,Y_{n+1}]/(\pi-Y_{1}^{b_{1}}\cdots
Y_{n+1}^{b_{n+1}}))_{(Y_{1},\,\dotsc,\,Y_{n+1})}\ \ \mbox{with}\ \
\gcd(b_{1},\dotsc,b_{n+1})=1.$

###### Proof.

To avoid repeating the argument, we will prove both claims simultaneously, so
in (1) we set $\Lambda={\bf F}$ and $\pi=0$ and in both parts we set
$p=$Char$(\Lambda/(\pi))$. By [5, Theorem 1 in VI (10.3)],

$\gamma_{1}=\mathrm{val}(x_{1}),\ \ \dots,\ \ \gamma_{n}=\mathrm{val}(x_{n}),\
\ \gamma_{n+1}=\mathrm{val}(\pi)\ \ (\mbox{resp. ~{}$\gamma_{n+1}=0$ if
$\pi=0$})$

satisfy $\Gamma\cong{\bf Z}\gamma_{1}\oplus\cdots\oplus{\bf Z}\gamma_{n+1}$,
where $\Gamma$ is the value group of $V$. We set $N=n+1$ (resp., $N=n$ if
$\pi=0$) and use Lemma 26 to find a countable sequence
$\Gamma_{0}\subset\Gamma_{1}\subset\dots$ of submonoids of $\Gamma_{\geq 0}$
with $\Gamma_{i}\simeq{\bf Z}^{N}_{\geq 0}$ for each $i$ and $\Gamma_{\geq
0}=\bigcup_{i\geq 0}\Gamma_{i}$. We fix a ${\bf Z}_{\geq 0}$-basis
$\nu_{i1},\dots,\nu_{iN}$ of $\Gamma_{i}$ with
$(\nu_{01},\dots,\nu_{0N})=(\gamma_{1},\dots,\gamma_{N})$, so that the
elements $\nu_{i1},\dots,\nu_{iN}$ are $\bf Z$-linearly independent in
$\Gamma$, and we express them in terms of the fixed $\bf Z$-basis:

$\nu_{ij}=d_{ij1}\gamma_{1}+\dots+d_{ijN}\gamma_{N}\ \ \mbox{for unique}\ \
d_{ij1},\dots,d_{ijN}\in{\bf Z}\ \ \mbox{and every}\ \ j=1,\dots,N.$

We set $x_{n+1}=\pi$ and note that, by construction, for each $i\geq 0$ and
$1\leq j\leq N$, the element

$y_{ij}=x_{1}^{d_{ij1}}\cdots x_{N}^{d_{ijN}}\in{\bf F}(x_{1},\dots,x_{n})\ \
\mbox{has valuation}\ \ \nu_{ij}.$

Since $\Gamma_{i^{\prime}}\subset\Gamma_{i}$ for $i^{\prime}<i$, each
$y_{i^{\prime}j}$ is in a unique way a monomial in the elements
$y_{i1},\dotsc,y_{iN}$:

$\mbox{if we express}\ \
\nu_{i^{\prime}j}=b_{i^{\prime}i1}\nu_{i1}+\dots+b_{i^{\prime}iN}\nu_{iN}\ \
\mbox{with}\ \ b_{i^{\prime}ij}\in{\bf Z}_{\geq 0},\ \ \mbox{then}\ \
y_{i^{\prime}j}=y_{i1}^{b_{i^{\prime}i1}}\cdots y_{iN}^{b_{i^{\prime}iN}}.$

Since the valuations of $y_{i1},\dots,y_{iN}$ are $\bf Z$-linearly
independent, the $\Lambda$-subalgebra $\Lambda[y_{i1},\dots,y_{iN}]$ of ${\bf
F}(x_{1},\dots,x_{n})$ is the regular ring

$\Lambda[Y_{i1},\dots,Y_{iN}]/(\pi-Y_{i1}^{b_{1}}\cdots Y_{iN}^{b_{N}})\ \
\mbox{with}\ \ b_{i}=b_{0Ni}\ \ (\mbox{resp.}\ \ \Lambda[Y_{i1},\dots,Y_{iN}]\
\ \mbox{if}\ \ \pi=0),$

where $\gcd(b_{1},\dots,b_{N})=1$ because $\mathrm{val}(\pi)$ is assumed to be
a primitive element of $\Gamma$ when $\pi\neq 0$. In particular, we obtain a
nested sequence of $\Lambda$-subalgebras

$R_{i}=\Lambda[y_{i1},\dots,y_{iN}]_{(y_{i1},\,\dots,\,y_{iN})}\subset V$

that are regular (resp., essentially smooth if $\pi=0$) and it remains to
argue that every $f\in V$ belongs to some $R_{i}$. For this, we first express
$f$ as a rational function as follows:

$f=(\sum\lambda_{s_{1},\,\dots,\,s_{N}}x_{1}^{s_{1}}\cdots
x_{N}^{s_{N}})/(\sum\lambda^{\prime}_{r_{1},\,\dots,\,r_{N}}x_{1}^{r_{1}}\cdots
x_{N}^{r_{N}})\ \ \mbox{with}\ \
\lambda_{s_{1},\,\dots,\,s_{N}},\lambda^{\prime}_{r_{1},\,\dots,\,r_{N}}\in\Lambda^{\times}\cup\\{0\\}.$

The linear independence of the $\gamma_{i}$ ensures that the valuations of the
monomials that appear in the numerator (resp., denominator) are all distinct.
Thus, by taking out the monomials with minimal valuations, we reduce to
showing that every $x_{1}^{\alpha_{1}}\cdots x_{N}^{\alpha_{N}}$ with
$\alpha_{j}\in{\bf Z}$ and $\alpha_{1}\gamma_{1}+\dots+\alpha_{N}\gamma_{N}>0$
is a product of nonnegative powers of the elements $y_{i1},\dots,y_{iN}$ for
some $i\geq 0$. For this, it suffices to note that
$\alpha_{1}\gamma_{1}+\dots+\alpha_{n}\gamma_{n}$ lies in some $\Gamma_{i}$,
and then to express it as a ${\bf Z}_{\geq 0}$-linear combination of the
$\nu_{i1},\dots,\nu_{iN}$: more precisely, if
$\alpha_{1}\gamma_{1}+\dots+\alpha_{N}\gamma_{N}=c_{1}\nu_{i1}+\dots+c_{N}\nu_{iN}$
with $c_{j}\in{\bf Z}_{\geq 0}$, then

$x_{1}^{\alpha_{1}}\cdots x_{N}^{\alpha_{N}}=y_{i1}^{c_{1}}\cdots
y_{iN}^{c_{N}}.$

For a valuation ring $V$ with the value group $\Gamma$ and the fraction field
$K$, a _cross-section_ of $V$ is a section

$s\colon\Gamma\rightarrow K^{*}\ \ \mbox{in the category of abelian groups of
the valuation map}\ \ \mathrm{val}\colon K^{*}\rightarrow\Gamma.$

(see for details in the Appendix).

######  Proposition 28.

An equicharacteristic valuation ring $V$ that has a cross-section
$s\colon\Gamma\rightarrow K^{*}$ and a subfield $k\subset V$ lifting the
residue field is an immediate extension

$V_{0}=\bigcup_{i}V_{i}\subset V$

of a filtered increasing union of valuation subrings $V_{i}\subset V$
dominated by $V$ such that each $V_{i}$ has a finitely generated value group,
is a countable increasing union of localizations of smooth $k$-subalgebras of
$V$ so $V_{0}$ is ind-smooth over $k$, and has the restriction of $s$ as a
cross-section.

###### Proof.

By Lemma 26, the submonoid $\Gamma_{\geq 0}\subset\Gamma$ of positive elements
is a filtered increasing union $\Gamma_{\geq 0}=\bigcup_{i}\Gamma_{i}$ of
submonoids $\Gamma_{i}\simeq{\bf Z}_{\geq 0}^{d_{i}}$ with $d_{i}\geq 0$.
Thus, the cross-section $s$ gives rise to the filtered increasing system of
subfields $k_{i}=k(s(\gamma)\,|\,\gamma\in\Gamma_{i})$ of the field of
fractions $K$ of $V$. By choosing a ${\bf Z}_{\geq 0}$-basis for $\Gamma_{i}$
and applying [5, Theorem 1, in VI section 10] we see that each $k_{i}$ is a
purely transcendental extension of $k$ and that the value group of the
valuation subring $V_{i}=V\cap k_{i}$ of $V$ is ${\bf Z}^{d_{i}}\simeq{\bf
Z}\Gamma_{i}\subset\Gamma$. By construction, $s$ restricts to a cross-section
of $V_{i}$ and, by Lemma 27, each $V_{i}$ is a filtered increasing union of
localizations of $k$-subalgebras. The construction ensures that $V$ is an
immediate extension of the resulting $V_{0}$.

##  6\.  Counterexamples when the value groups are finitely generated

######  Lemma 29.

Let $V\subset V^{\prime}$ be an extension of valuation rings which is ind-
smooth. Then $\Omega_{V^{\prime}/V}$, that is $H_{0}(V,V^{\prime},V^{\prime})$
in terms of Andre-Quillen homology, is a flat $V^{\prime}$-module and
$H_{1}(V,V^{\prime},V^{\prime})=0$ (the last homology is usually denoted by
$\Gamma_{V^{\prime}/V}$ ).

###### Proof.

Assume that $V^{\prime}$ is the filtered direct limit of some smooth $V$
-algebras $B_{i}$, $i\in I$. Then $\Omega_{B_{i}/V}$ is projective over
$B_{i}$ and $H_{1}(V,B_{i},B_{i})=0$ by e.g [33, Theorem 3.4]. But
$\Omega_{V^{\prime}/V}$ , and $H_{1}(V,V^{\prime},V^{\prime})$ are filtered
direct limits of $V^{\prime}\otimes_{B_{i}}\Omega_{B_{i}/V}$ resp.
$V\otimes_{B_{i}}H_{1}(V,B_{i},B_{i})$ by [33, Lemma 3.2], which is enough.

######  Lemma 30.

Let $V\subset V^{\prime}$ be an extension of valuation rings of dimension one
with the same residue field and let $\Gamma\subsetneq\Gamma^{\prime}$ be their
value group extension. Assume that $\Gamma^{\prime}/\Gamma$ has torsion. Then
the extension $V\subset V^{\prime}$ is not ind-smooth.

###### Proof.

Let $\gamma\in\Gamma^{\prime}\setminus\Gamma$ be such that $n\gamma\in\Gamma$
for some positive integer $n$. Choose an element $x\in V^{\prime}$ such that
$\mathrm{val}(x)=\gamma$. Then $x^{n}=zt$ for some $z\in V$ and an unit $t\in
V^{\prime}$. Thus the system $S$ of polynomials $X^{n}=zT$, $TT^{\prime}=1$
over $V$ has a solution in $V^{\prime}$. If $V^{\prime}$ is ind-smooth over
$V$ then $S$ has a solution in a smooth $V$-algebra and so one
$({\tilde{x}},{\tilde{t}},{\tilde{t}}^{\prime})$ in the completion of $V$. But
then $\gamma=\mathrm{val}(z)/n=\mathrm{val}({\tilde{x}})$ must be in $\Gamma$
which is false.

######  Lemma 31.

Let $V\subset V^{\prime}$ be an extension of valuation rings of dimension one
containing $\bf Q$ having the same residue field $k$. Assume that $V$ contains
$k$ and its value group $\Gamma\subset{\bf R}$ is dense in $\bf R$. Also
assume that the value group $\Gamma^{\prime}\subset{\bf R}$ of $V^{\prime}$ is
finitely generated, $\Gamma\not=\Gamma^{\prime}$ and $\Gamma^{\prime}/\Gamma$
has no torsion. Then the extension $V\subset V^{\prime}$ is not ind-smooth.

###### Proof.

Since $\Gamma$ is free over $\bf Z$ we may take a basis of positive elements
$\gamma_{1},\ldots,\gamma_{m}$ of $\Gamma$ which may be completed with some
positive elements $\gamma_{m+1},\ldots,\gamma_{n}\in\Gamma^{\prime}$ to a
basis of $\Gamma^{\prime}$. Choose $x_{1},\ldots,x_{m}\in V$ and
$x_{m+1},\ldots,x_{n}\in V^{\prime}$ such that
$\mathrm{val}(x_{i})=\gamma_{i}$. Let $V_{0}=V\cap k(x_{1},\ldots,x_{m})$ and
$V^{\prime}_{0}=V^{\prime}\cap k(x_{1},\ldots,x_{n})$.

We will show that $\Omega_{V^{\prime}_{0}/V_{0}}$ has torsion. First assume
that $n=m+1$. We will use the proof of [23, Lemma 7.2]. By Lemma 26
$\Gamma_{+}=\cup_{j\in{\bf N}}\Gamma_{j}$ for some monoids
$\Gamma_{j}\subset\Gamma_{+}$ generated by bases of $\Gamma$, the union being
increasing filtered. We consider as in the quoted lemma two real sequences
$(u_{i})$, $(v_{i})$ which converge in $\bf R$ to $\gamma_{n}$ and such that

1) $u_{j},v_{j}\in\Gamma_{j}$ and
$u_{j+1}-u_{j},v_{j}-v_{j+1}\in\Gamma_{j+1}$,

2) $v_{j}-u_{j}$ is an element of the basis $\nu_{j}$ of $\Gamma$ generating
$\Gamma_{j}$, we may assume $v_{j}-u_{j}=\nu_{j1}.$

3) $u_{j}<\gamma_{n}<v_{j}$ for all $j$.

We may also suppose that $u_{j+1}-u_{j}=v_{j}-v_{j+1}$ if necessary
restricting to a subsequence of $(\Gamma_{j})$. Let $a_{j}$, $b_{j}$ be in $V$
with values $u_{j}$, resp. $v_{j}$, and take $y_{jn}=x_{n}/a_{j}$ and
$z_{jn}=b_{j}/x_{n}$ in $V^{\prime}$. As in the proof of Lemma 4.2 a), we have
$\nu_{ji}=d_{ji1}\gamma_{1}+\ldots d_{jim}\gamma_{m}$ and set
$y_{ji}=x_{1}^{d_{ji1}}\cdots x_{m}^{d_{jim}}\in V$ which has valuation
$\nu_{ji}$, $i\in[m]$. Then $V_{0}$ is a filtered increasing union of
localizations $B_{j}$ of $k[y_{j1},\ldots,y_{jm}]$ and $V^{\prime}_{0}$ is a
filtered increasing union of localizations $C_{j}$ of
$B_{j}[Z_{j},Z^{\prime}_{j}]/(Z_{j}Z^{\prime}_{j}-y_{j1})\cong
B_{j}[z_{j},z^{\prime}_{j}]$, where $z_{j}=x_{n}/a_{j}$ and
$z^{\prime}_{j}=b_{j}/x_{n}$ in $V^{\prime}$. Note that the map
$C_{j}\rightarrow C_{j+1}$ is given by
$Z_{j}\rightarrow(a_{j+1}/a_{j})Z_{j+1}$,
$Z^{\prime}_{j}\rightarrow(b_{j}/b_{j+1})Z^{\prime}_{j+1}$.

We claim that the map
$f_{j}:C_{j+1}\otimes_{C_{j}}\Omega_{C_{j}/B_{j}}\rightarrow\Omega_{C_{j+1}/B_{j+1}}$
given by $dz_{j}\rightarrow(a_{j+1}/a_{j})dz_{j+1}$,
$dz^{\prime}_{j}\rightarrow(b_{j}/b_{j+1})dz^{\prime}_{j+1}$ is injective.
Indeed, an element from $\operatorname{Ker}f_{j}$ induced by $w=\alpha\otimes
dz_{j}+\beta\otimes dz^{\prime}_{j}$, $\alpha,\beta\in C_{j+1}$ must go by
$f_{j}$ in

$\alpha(a_{j+1}/a_{j})dz_{j+1}+\beta(b_{j}/b_{j+1})dz^{\prime}_{j+1}\in<z^{\prime}_{j+1}dz_{j+1}+z_{j+1}dz^{\prime}_{j+1}>$

in $C_{j+1}dz_{j+1}\oplus C_{j+1}dz^{\prime}_{j+1}$. So
$\alpha(a_{j+1}/a_{j})=\mu z^{\prime}_{j+1}$ and $\beta(b_{j}/b_{j+1})=\mu
z_{j+1}$ for some $\mu\in C_{j+1}$. It follows that

$w=(\mu(b_{j+1}/b_{j})(a_{j+1}/a_{j}))\otimes
z^{\prime}_{j}dz_{j}+(\mu(b_{j+1}/b_{j})(a_{j+1}/a_{j}))\otimes
z_{j}dz^{\prime}_{j}$

belongs to $<z^{\prime}_{j}dz_{j}+z_{j}dz^{\prime}_{j}>$, which shows our
claim.

We may assume that $\mathrm{val}(z_{j})\leq\mathrm{val}(z^{\prime}_{j})$. For
some $j^{\prime}\geq j$ we have $z^{\prime}_{j}=t_{j}z_{j}$ in
$C_{j^{\prime}}$ for some $t_{j}$ in $C_{j^{\prime}}$. We have
$dz_{j},dz^{\prime}_{j}$ in $\Omega_{C_{j^{\prime}}/B_{j^{\prime}}}$ and
$z_{j}w^{\prime}_{j}=0$, for $w^{\prime}_{j}=dz^{\prime}_{j}+t_{j}dz_{j}$. So
$C_{j^{\prime}}\otimes_{C_{j}}\Omega_{C_{j}/B_{j}}$ is not torsion free. Here
we should point out that the localizations are given by elements from
$\rho+((y_{j^{\prime}i})_{i},z_{j^{\prime}},z^{\prime}_{j^{\prime}})$,
$\rho\in k$, $\rho\not=0$, which cannot kill $w^{\prime}_{j}$. Since $f_{j}$
are injective we see that $\Omega_{V^{\prime}_{0}/V_{0}}$ has torsion.

Now assume that $n>m+1$ and consider $V^{\prime\prime}_{0}=V^{\prime}\cap
k(x_{1},\ldots,x_{n-1})$. Apply induction on $n-m$, the case $n-m=1$ having
been considered above. By induction hypothesis, we assume that
$\Omega_{V^{\prime\prime}_{0}/V_{0}}$ has torsion. As above
$V^{\prime\prime}_{0}$ is a filtered direct limit of some localizations
${\tilde{B}}_{j}$ of $k[{\tilde{y}}_{j1},\ldots,{\tilde{y}}_{j,n-1}]$ and
$V^{\prime}_{0}$ is the filtered direct limit of some localizations
${\tilde{C}}_{j}$ of
${\tilde{B}}_{j}[{\tilde{z}}_{j},{\tilde{z}}^{\prime}_{j}]$. Set
$I_{j}=(Z_{j}Z^{\prime}_{j}-{\tilde{y}}_{j1})\subset{\tilde{B}}_{j}[Z_{j},Z^{\prime}_{j}]$.
By definition we have the following exact sequence

$0\rightarrow
H_{1}({\tilde{B}}_{j},{\tilde{C}}_{j},{\tilde{C}}_{j})\rightarrow
I_{j}/I_{j}^{2}\xrightarrow{d}{\tilde{C}}_{j}d{\tilde{z}}_{j}\oplus{\tilde{C}}_{j}d{\tilde{z}}^{\prime}_{j}\rightarrow\Omega_{{\tilde{C}}_{j}/{\tilde{B}}_{j}}\rightarrow
0.$

If $h\in I_{j}$ induces an element in $\operatorname{Ker}d$ then we get

$(\partial h/\partial
Z_{j})({\tilde{z}}_{j},{\tilde{z}}^{\prime}_{j})d{\tilde{z}}_{j}\oplus(\partial
h/\partial
Z^{\prime}_{j})({\tilde{z}}_{j},{\tilde{z}}^{\prime}_{j})d{\tilde{z}}^{\prime}_{j}=0.$

But $h={\tilde{h}}(Z_{j}Z^{\prime}_{j}-{\tilde{y}}_{j1})$ for some
${\tilde{h}}\in{\tilde{B}}_{j}[Z_{j},Z^{\prime}j]$ and it follows
${\tilde{h}}({\tilde{z}}_{j},{\tilde{z}}^{\prime}_{j}){\tilde{z}}^{\prime}_{j}=0$,
that is ${\tilde{h}}({\tilde{z}}_{j},{\tilde{z}}^{\prime}_{j})=0$ and so
${\tilde{h}}\in I_{j}$. Hence $d$ is injective and
$H_{1}({\tilde{B}}_{j},{\tilde{C}}_{j},{\tilde{C}}_{j})=0$.

In the Jacobi-Zariski sequence ([33, Theorem 3.3] applied to $V_{0}\rightarrow
V^{\prime\prime}_{0}\rightarrow V^{\prime}_{0}$

$0=H_{1}(V^{\prime\prime}_{0},V^{\prime}_{0},V^{\prime}_{0})\rightarrow
V^{\prime}_{0}\otimes_{V^{\prime\prime}_{0}}\Omega_{V^{\prime\prime}_{0}/V_{0}}\xrightarrow{\lambda}\Omega_{V^{\prime}_{0}/V_{0}}\rightarrow\Omega_{V^{\prime}_{0}/V^{\prime\prime}_{0}}\rightarrow
0$

we see that the map $\lambda$ is injective. It follows that
$\Omega_{V^{\prime}_{0}/V_{0}}$ has torsion which proves our claim.

By Proposition 18 the immediate extension $V_{0}\subset V$ is ind-smooth.
Assume, aiming for contradiction, that $V^{\prime}$ is ind-smooth over $V$.
Then $V^{\prime}$ is ind-smooth over $V_{0}$ and by the above lemma we get
$\Omega_{V^{\prime}/V_{0}}$ flat over $V^{\prime}$. Again by Proposition 18 we
have $V^{\prime}$ ind-smooth over $V^{\prime}_{0}$. As in the above lemma we
obtain that $\Omega_{V^{\prime}/V^{\prime}_{0}}$ is a flat module over
$V^{\prime}$. In the Jacobi-Zariski sequence applied to $V_{0}\rightarrow
V^{\prime}_{0}\rightarrow V^{\prime}$

$H_{1}(V^{\prime}_{0},V^{\prime},V^{\prime})\rightarrow
V^{\prime}\otimes_{V^{\prime}_{0}}\Omega_{V^{\prime}_{0}/V_{0}}\rightarrow\Omega_{V^{\prime}/V_{0}}\rightarrow\Omega_{V^{\prime}/V^{\prime}_{0}}\rightarrow
0$

we have $H_{1}(V^{\prime}_{0},V^{\prime},V^{\prime})=0$ and the last two
modules are flat by the above lemma. We obtain that
$V^{\prime}\otimes_{V^{\prime}_{0}}\Omega_{V^{\prime}_{0}/V_{0}}$ is flat also
over $V^{\prime}$ and so $\Omega_{V^{\prime}_{0}/V_{0}}$ is also flat, which
is not possible because it has torsion. Thus $V^{\prime}$ is not ind-smooth
over $V$.

######  Remark 32.

In the above proof the main point was to show that when
$\Gamma^{\prime}/\Gamma\not=0$ has no torsion then
$\Omega_{V^{\prime}_{0}/V_{0}}$ has torsion.

######  Lemma 33.

Let $V\subset V^{\prime}$ be an extension of valuation rings of dimension one
containing $\bf Q$. Assume that $V$ contains its residue field and its value
group $\Gamma\subset{\bf R}$ is dense in $\bf R$. Also assume that the value
group $\Gamma^{\prime}\subset{\bf R}$ of $V^{\prime}$ is finitely generated
and $\Gamma^{\prime}/\Gamma$, $\Gamma\not=\Gamma^{\prime}$ has no torsion.
Then the extension $V\subset V^{\prime}$ is not ind-smooth.

###### Proof.

In the proof of Lemma 31 take
$W=V^{\prime}\cap\operatorname{Frac}(V)(x_{m+1},\ldots,x_{n})$. Then the
extension $V\subset W$ has the same residue field but the value group
extension is $\Gamma\subset\Gamma^{\prime}$. Then $\Omega_{W/V}$ has torsion
as in the proof of the quoted lemma. In the Jacobi-Zariski sequence applied to
$V,W,V^{\prime}$

$H_{1}(W,V^{\prime},V^{\prime})\rightarrow
V^{\prime}\otimes_{W}\Omega_{W/V}\rightarrow\Omega_{V^{\prime}/V}$

we see that the left module is zero because the valuation extension $W\subset
V^{\prime}$ is ind-smooth (see Lemma 29), having the same value group (see
Corollary 19). It follows that $\Omega_{V^{\prime}/V}$ has torsion. But if
$V\subset V^{\prime}$ is ind-smooth then $\Omega_{V^{\prime}/V}$ is torsion
free. So $V\subset V^{\prime}$ is not ind-smooth.

######  Lemma 34.

Let $V\subset V^{\prime}$ be an extension of valuation rings of dimension one
containing $\bf Q$ with value groups $\Gamma\subset\Gamma^{\prime}$ and having
the same residue field $k$. Assume that $V$ contains $k$ and its value group
$\Gamma\subset{\bf R}$ is dense in $\bf R$. Also assume that the value group
$\Gamma^{\prime}/\Gamma\not=0$ has no torsion and $V^{\prime}$ has a cross-
section $s:\Gamma^{\prime}\rightarrow K^{\prime*}$ such that $s(\Gamma)\subset
K^{*}$, $K,K^{\prime}$ being the fraction fields of $V$, resp. $V^{\prime}$.
Then the extension $V\subset V^{\prime}$ is not ind-smooth.

###### Proof.

We follow the proof of the above lemma. Take $V_{0}=V\cap k(s(\Gamma))$ and
$V^{\prime}_{0}=V^{\prime}\cap k(s(\Gamma^{\prime}))$. For every finitely
generated $\Gamma_{1}\subset\Gamma$ and
$\Gamma_{1}^{\prime}\subset\Gamma^{\prime}$ such that
$\Gamma_{1}\subset\Gamma_{1}^{\prime}\not\subset\Gamma$ we see as in the above
lemma that for $V_{10}=V\cap k(s(\Gamma_{1}))$,
$V^{\prime}_{10}=V^{\prime}\cap k(s(\Gamma^{\prime}_{1}))$ we have a torsion
in $\Omega_{V^{\prime}_{10}/V_{10}}$. Then as in the above lemma we get a
torsion in $\Omega_{V^{\prime}_{0}/V_{0}}$ by [33, Lemma 3.2] and so in
$\Omega_{V^{\prime}/V}$ which implies that $V^{\prime}$ is not ind-smooth over
$V$ by Lemma 30.

######  Lemma 35.

Let $V\subset V^{\prime}$ be an extension of valuation rings of dimension one
containing $\bf Q$ with value groups $\Gamma\subset\Gamma^{\prime}$. Assume
that $V$ contains its residue field and its value group $\Gamma\subset{\bf R}$
is dense in $\bf R$. Also assume that the group $\Gamma^{\prime}/\Gamma\not=0$
has no torsion and $V^{\prime}$ has a cross-section
$s:\Gamma^{\prime}\rightarrow K^{\prime*}$ such that $s(\Gamma)\subset K^{*}$,
$K,K^{\prime}$ being the fraction fields of $V$, resp. $V^{\prime}$. Then the
extension $V\subset V^{\prime}$ is not ind-smooth.

The proof follows as in Lemma 34 using now Lemma 33.

##  7\. The case when the value group is not finitely generated

A weaker form of Theorem 1, but enough for our purpose, is given below with an
independent proof.

######  Theorem 36.

Every valuation ring $V$ containing its residue field $k$ of characteristic
zero is a filtered direct limit of smooth $k$-algebras $(R_{i})_{i}$, that is
the inclusion $k\subset V$ is ind-smooth (in particular, all the $R_{i}$ are
regular rings).

###### Proof.

Let $\Gamma$ be the value group of $V$, $K$ the fraction field of $V$ and
$\tilde{\Gamma}$, $\tilde{V}$, ${\tilde{K}}$,
${\tilde{s}}:{\tilde{\Gamma}}\rightarrow{\tilde{K}}^{*}$ be given by Theorem
A.10. As $k\subset V$ we get ${\tilde{k}}\subset{\tilde{V}}$. By Proposition
28 we see that $\tilde{V}$ is an immediate extension of a valuation ring
${\tilde{V}}_{0}$ which is a filtered increasing union of localizations of
smooth ${\tilde{k}}$-algebras. As ${\tilde{V}}_{0}\subset{\tilde{V}}$ is ind-
smooth by Proposition 18 we see that ${\tilde{k}}\subset{\tilde{V}}$ is ind-
smooth. It follows that $k\subset{\tilde{V}}$ is ind-smooth and so $k\subset
V$ is ind-smooth by [23, Corollary 5.4].

######  Lemma 37.

Let $B$ be an $A$-algebra and $A^{n}$, $B^{n}$ be the product of $n$-copies of
$A$, resp. $B$. Then $\Omega_{B^{n}/A^{n}}\cong\Omega_{B/A}^{n}$ and
$H_{1}(A^{n},B^{n},B^{n})\cong H_{1}(A,B,B)^{n}$ .

###### Proof.

We treat only the case $n=2$. If $B=A[X]/I$, $X=(X_{i})_{i}$ then
$B^{2}=A^{2}[X]/J$, where $J$ is given by polynomials of the form
$h_{f,g}=\sum_{j=(j_{1},\ldots,j_{s})}(a_{j},b_{j})X^{j}$ for some polynomials
$f=\sum a_{j}X_{j}$, $g=\sum b_{j}X^{j}$ from $I$. Then $\Omega_{B^{2}/A^{2}}$
is the cokernel of the map $d:J/J^{2}\rightarrow\oplus_{i}B^{2}dX_{i}$ given
by $h_{f,g}\rightarrow\sum_{i}(\partial f/\partial X_{i},\partial g/\partial
X_{i})dX_{i}=\sum_{i}(\partial f/\partial X_{i}dX_{i},\partial g/\partial
X_{i}dX_{i})$. Also $H_{1}(A^{2},B^{2},B^{2})$ is the kernel of $d$ and note
that $d(h_{(f,g)})=0$ if and only if $d_{0}(f)=0$ and $d_{0}(g)=0$, $d_{0}$
being the map $I/I^{2}\rightarrow\oplus_{i}BdX_{i}$ . Hence
$\Omega_{B^{2}/A^{2}}\cong\Omega_{B/A}^{2}$ and $H_{1}(A^{2},B^{2},B^{2})\cong
H_{1}(A,B,B)^{2}$.

######  Lemma 38.

Let $B$ be an $A$-algebra, $\mathcal{U}$ and ultrafilter on a set $U$ and
$\tilde{B}$, resp. $\tilde{A}$ the ultrapowers of $B$ (see the Appendix for
the details), resp. $A$ with respect to $\mathcal{U}$. Then
$\Omega_{{\tilde{B}}/{\tilde{A}}}$ (resp.
$H_{1}({\tilde{A}},{\tilde{B}},{\tilde{B}})$) is the corresponding ultrapower
of $\Omega_{B/A}$ (resp. $H_{1}(A,B,B)$) with respect to $\mathcal{U}$. In
particular, $\Omega_{B/A}$ has torsion if and only if
$\Omega_{{\tilde{B}}/{\tilde{A}}}$ has torsion and
$H_{1}({\tilde{A}},{\tilde{B}},{\tilde{B}})=0$ if and only if
$H_{1}(A,B,B)=0$.

######  Proposition 39.

Let $V\subset V^{\prime}$ be an extension of valuation rings of dimension one
containing $\bf Q$. Assume that the value group $\Gamma\subset{\bf R}$ of $V$
is dense in $\bf R$ and the factor of the value groups
$\Gamma^{\prime}/\Gamma\not=0$ has no torsion. Then the extension $V\subset
V^{\prime}$ is not ind-smooth.

###### Proof.

Using Lemma 7 as in Lemma 11 we may suppose that $V,V^{\prime}$ are complete.
So $V$ contains its residue field. By Variant A.11 we find an extension of
valuation rings ${\tilde{V}}\subset{\tilde{V}}^{\prime}$ such that there
exists a cross-section
${\tilde{s}}:{\tilde{\Gamma}}^{\prime}\rightarrow({\tilde{K}}^{\prime})^{*}$
such that ${\tilde{s}}({\tilde{\Gamma}})\subset{\tilde{K}}$. We remind that we
wrote $\Gamma^{\prime}$ as a filtered increasing union of finitely generated
subgroups $\Gamma^{\prime}_{i})_{i\in I}$ and
$\Gamma_{i}=\Gamma^{\prime}_{i}\cap\Gamma$. Set $V_{0i}=V\cap
s^{\prime}(\Gamma_{i})=V^{\prime}\cap s^{\prime}(\Gamma_{i})$,
$V^{\prime}_{0i}=V^{\prime}\cap s^{\prime}(\Gamma^{\prime}_{i})$. By Remark 32
the modules $\Omega_{V^{\prime}_{0i}/V_{0i}}$, $i\in I$ have torsion. Note
that the filtered increasing union $V_{01}$ of $V_{0i}$, $i\in I$ is a
valuation ring with value group $\Gamma$, and similarly consider
$V^{\prime}_{01}$ which has the value group $\Gamma^{\prime}$. Clearly,
$\Omega_{V^{\prime}_{01}/V_{01}}$ has torsion since it is the limit of
$V^{\prime}_{01}\otimes_{V^{\prime}_{01}}\Omega_{V^{\prime}_{0i}/V_{0i}}$. Set
${\tilde{V}}_{0}={\tilde{V}}\cap s^{\prime}({\tilde{\Gamma}})$,
${\tilde{V}}^{\prime}_{0}={\tilde{V}}^{\prime}\cap
s^{\prime}({\tilde{\Gamma}}^{\prime})$. By iteration we define the extensions
$V_{0n}\subset V_{n}$ and $V^{\prime}_{0n}\subset V^{\prime}_{n}$ with the
same value group $\Gamma_{n}$, $\Gamma^{\prime}_{n}$ obtained taking
$n$-ultrapowers of $\Gamma$, resp. $\Gamma^{\prime}$ and we see that
$\Omega_{V^{\prime}_{0n}/V_{0n}}$ has torsion. Then
$\Omega_{{\tilde{V}}^{\prime}_{0}/{\tilde{V}}_{0}}$ has torsion since it is
the limit of
${\tilde{V}^{\prime}}\otimes_{V^{\prime}_{0n}}\Omega_{V^{\prime}_{0n}/V_{0n}}$.

Assume that $V\subset V^{\prime}$ is ind-smooth. In the Jacobi-Zariski
sequence applied to ${\tilde{V}}_{0},{\tilde{V}},{\tilde{V}}^{\prime}$

$H_{1}({\tilde{V}},{\tilde{V}}^{\prime},{\tilde{V}}^{\prime})\rightarrow{\tilde{V}}^{\prime}\otimes_{\tilde{V}}\Omega_{{\tilde{V}}/{\tilde{V}}_{0}}\rightarrow\Omega_{{\tilde{V}}^{\prime}/{\tilde{V}}_{0}}\rightarrow\Omega_{{\tilde{V}}^{\prime}/{\tilde{V}}}\rightarrow
0$

we see that the left module is zero and the last module has no torsion by
Lemmas 29, 38 because $\Omega_{V^{\prime}/V}$ has no torsion and
$H_{1}(V,V^{\prime})=0$, $V\subset V^{\prime}$ being ind-smooth.

Also $\Omega_{{\tilde{V}}/{\tilde{V}}_{0}}$ is flat (so it has no torsion)
since the extension ${\tilde{V}}_{0}\subset{\tilde{V}}$ is ind-smooth having
the same value group (see Proposition 20). It follows that
$\Omega_{{\tilde{V}}^{\prime}/{\tilde{V}}_{0}}$ has also no torsion.

Now, in the Jacobi-Zariski sequence applied to
${\tilde{V}}_{0},{\tilde{V}}^{\prime}_{0},{\tilde{V}}^{\prime}$

$H_{1}({\tilde{V}},{\tilde{V}}^{\prime}_{0},{\tilde{V}}^{\prime}_{0})\rightarrow{\tilde{V}}^{\prime}\otimes_{{\tilde{V}}^{\prime}_{0}}\Omega_{{\tilde{V}}^{\prime}_{0}/{\tilde{V}}_{0}}\rightarrow\Omega_{{\tilde{V}}^{\prime}/{\tilde{V}}_{0}}$

we see that the left module is zero by Lemmas 30, 38 because
$V^{\prime}_{0}\subset V^{\prime}$ is ind-smooth by Proposition 20. As above
$\Omega_{{\tilde{V}}^{\prime}_{0}/{\tilde{V}}_{0}}$ has torsion and so
$\Omega_{{\tilde{V}}^{\prime}/{\tilde{V}}_{0}}$ has torsion too, which is
false.

Acknowledgements: This work has been partially elaborated in the frame of the
International Research Network ECO-Math.

## Appendix. Cross-sections via infinite towers of ultrapowers
by Kęstutis Česnavičius222CNRS, UMR 8628, Laboratoire de Mathématiques
d’Orsay, Univ. Paris-Sud, Université Paris-Saclay, 91405 Orsay, France. Email:
<EMAIL_ADDRESS>
I thank Dorin Popescu for helpful discussions and for hosting me during a
visit to Romania in March 2019. I thank Matthias Aschenbrenner for helpful
comments. This project has received funding from l’Agence Universitaire de la
Francophonie and the European Research Council (ERC) under the European
Union’s Horizon 2020 research and innovation programme (grant agreement No.
851146).

The goal of this Appendix is to show that by replacing a valuation ring by the
limit of an infinite tower of its suitable ultrapowers one may arrange the
valuation map $\mathrm{val}\colon V\setminus\\{0\\}\rightarrow\Gamma$ to admit
a multiplicative section (see Theorem A.10). For this, we use techniques from
model theory, specifically, the Keisler–Kunen theorem about the existence of
good ultrafilters:333After this appendix was written, we learned of a much
simpler way to deduce sharper versions of Theorem A.10 and Variant A.11 from
the results reviewed in §A.8, see [1, 3.3.39, 3.3.40]. We left this appendix
in place in case the method used here would prove useful for other purposes.
the idea is that constructing a section amounts to solving a system of
equations for which any finite subsystem has a solution, and such systems
always have solutions in well-chosen ultrapowers. For instance, if the system
is countable, then solutions exists in any nonprincipal ultrafilter on $\bf
N$, and in general the main subtlety is in constructing the ultrafilter
(within ZFC).

###### A.1. Cross-sections of valuation rings.

For a valuation ring $V$ with the value group $\Gamma$ and the fraction field
$K$, a _cross-section_ of $V$ is a section

$s\colon\Gamma\rightarrow K^{*}\ \ \mbox{in the category of abelian groups to
the valuation map}\ \ \mathrm{val}\colon K^{*}\rightarrow\Gamma.$

Concretely, $s$ is a group homomorphism such that
$\mathrm{val}(s(\gamma))=\gamma$ for $\gamma\in\Gamma$. For a submonoid
$M\subset\Gamma$, a _partial cross-section_ defined on $M$ is a monoid
morphism $s\colon M\rightarrow K^{*}$ (concretely, $s(0)=1$ and
$s(m+m^{\prime})=s(m)s(m^{\prime})$) with $\mathrm{val}(s(m))=m$ for $m\in M$.
Evidently, partial cross-sections exist for $M\simeq{\bf Z}_{\geq 0}^{r}$ and
correspond to choices of tuples of elements of $K^{*}$ whose valuations form
the standard basis of ${\bf Z}_{\geq 0}^{r}$.

######  Example A.2.

Cross-sections exist when $\Gamma$ is free as a $\bf Z$-module, for instance,
when it is finitely generated. As we now explain, they also exist when $V$ is
strictly Henselian of residue characteristic $p$ and there is a free subgroup
$\Gamma_{0}\subset\Gamma$ such that $\Gamma/\Gamma_{0}$ is torsion with
$(\Gamma/\Gamma_{0})[p^{\infty}]=0$. Indeed, we first define $s$ on
$\Gamma_{0}$ and then, by Zorn’s lemma, reduce to the situation in which $s$
is already defined on some subgroup $\Gamma^{\prime}\supset\Gamma_{0}$ and
needs to be extended to a $\Gamma^{\prime\prime}\supsetneq\Gamma^{\prime}$
with $\Gamma^{\prime\prime}/\Gamma^{\prime}$ cyclic of order $n$ prime to $p$.
For the latter, we first choose an $x\in V$ such that $\mathrm{val}(x)$ lies
in $\Gamma^{\prime\prime}$ and generates the quotient
$\Gamma^{\prime\prime}/\Gamma^{\prime}$, which gives the following equation in
$V$:

$x^{n}=u\cdot s(n\cdot\mathrm{val}(x))\quad\text{for some}\quad u\in V^{*}.$

Since $p\nmid n$, Hensel’s lemma [10, IV, 18.5.17] (which is the Implicit
Function Theorem in this context) now implies that the equation $X^{n}=u$ has
a solution in $V$, so we may adjust $x$ to assume that $u=1$. Granted this,
$s$ then extends to $\Gamma^{\prime\prime}$ by setting $s(\mathrm{val}(x))=x$:
indeed, any relation $N\cdot\mathrm{val}(x)=\gamma$ with $N\in{\bf Z}$ and
$\gamma\in\Gamma^{\prime}$ must be a multiple of such a relation with $N=n$,
so $s(N\cdot\mathrm{val}(x))=s(\gamma)$.

###### A.3. Ultrafilters and ultraproducts.

We recall that an _ultrafilter_ on a nonempty set $U$ is a set $\mathscr{U}$
of subsets of $U$ that is closed under finite intersections, closed under
taking supersets, does not contain the empty set, and for every
$U^{\prime}\subset U$ contains either $U^{\prime}$ or $U\setminus U^{\prime}$.
Such a $\mathscr{U}$ is _principal_ if it consists of all the subsets
containing some fixed $u\in U$, and is _nonprincipal_ otherwise. An
ultrafilter $\mathscr{U}$ is _countably incomplete_ if some countable
collection of elements of $\mathscr{U}$ has an empty intersection. Such a
$\mathscr{U}$ is also nonprincipal and it exists whenever $U$ is infinite (see
[4, §A.3, 8.4]). We view any ultrafilter $\mathscr{U}$ as a partially ordered
set, where $U^{\prime}\leq U^{\prime\prime}$ if $U^{\prime}\supseteq
U^{\prime\prime}$.

For any category $\mathcal{C}$ that has small products and filtered direct
limits, the _ultraproduct_ of a set $\\{C_{u}\\}_{u\in U}$ of objects of
$\mathcal{C}$ with respect to an ultrafilter $\mathscr{U}$ on $U$, which we
denote abusively by $\prod_{\mathscr{U}}C_{u}$, is

$\textstyle\varinjlim_{U^{\prime}\in{\mathscr{U}}}(\prod_{u\in
U^{\prime}}C_{u})\quad\text{where transition maps are projections onto partial
products}$

(the limit is filtered because $\mathscr{U}$ is closed under finite
intersections). In the case when all the $C_{u}$ are the same object
$C\in{\mathcal{C}}$, we call $\prod_{\mathscr{U}}C$ an _ultrapower_ of $C$.

###### A.4. Ultraproducts of valuation rings.

We will work with ultraproducts of rings or modules. For instance, an
ultraproduct of fields is again a field: every nonzero element is invertible
(thanks to the axiom that $U^{\prime}\in{\mathscr{U}}$ or $U\setminus
U^{\prime}\in{\mathscr{U}}$). Likewise, an ultraproduct
$\prod_{\mathscr{U}}V_{u}$ of valuation rings $\\{V_{u}\\}_{u\in U}$ with
fraction fields $\\{K_{u}\\}_{u\in U}$ is a valuation ring with fraction field
$\prod_{\mathscr{U}}K_{u}$: for any nonzero element $v$ of the latter, either
$v$ or $v^{-1}$ lies in $\prod_{\mathscr{U}}V_{u}$. We see similarly that

  1. (1)

the maximal ideal of $\prod_{\mathscr{U}}V_{u}$ is the ultraproduct
$\prod_{\mathscr{U}}{\mathfrak{m}}_{u}$ of the maximal ideals;

  2. (2)

the residue field of $\prod_{\mathscr{U}}V_{u}$ is the ultraproduct
$\prod_{\mathscr{U}}k_{u}$ of the residue fields;

  3. (3)

the value group of $\prod_{\mathscr{U}}V_{u}$ is the ultraproduct
$\prod_{\mathscr{U}}\Gamma_{u}$ of the value groups;

  4. (4)

the monoid of nonnegative elements $(\prod_{\mathscr{U}}\Gamma_{u})_{\geq 0}$
is identified with $\prod_{\mathscr{U}}(\Gamma_{u})_{\geq 0}$.

The existence of “well-chosen” ultrapowers mentioned above rests on the
Keisler–Kunen theorem from model theory that we recall in the following lemma.
Keisler proved it in [13] assuming the Generalized Continuum Hypothesis and
Kunen gave an unconditional proof in [16, Theorem 3.2].

######  Lemma A.5 ([6], Theorem 6.1.4).

For an infinite set $U$, there is a countably incomplete ultrafilter
$\mathscr{U}$ on $U$ such that for any inclusion-reversing function

$f\colon\\{\mbox{finite subsets of }U\\}\rightarrow{\mathscr{U}},\ \
\mbox{there is a function}\ \ f_{0}\colon\\{\mbox{finite subsets of
}U\\}\rightarrow{\mathscr{U}}$

that is also inclusion-reversing and satisfies

$\ \ f_{0}(U^{\prime})\subset f(U^{\prime})\ \ \mbox{and}\ \
f_{0}(U^{\prime}\cup U^{\prime\prime})=f_{0}(U^{\prime})\cap
f_{0}(U^{\prime\prime})\ \ \mbox{for all finite subsets}\ \
U^{\prime},U^{\prime\prime}\subset U.$

Of course, the requirement that $f_{0}$ be inclusion-reversing is superfluous:
it is a special case of the requirement that $f_{0}$ transform finite unions
into intersections.

We now verify that the ultrapowers that result from the ultrafilters supplied
by Lemma A.5 have the promised property of solvability of systems of
equations.

######  Proposition A.6.

For an infinite cardinal $\kappa$, every ultrafilter $\mathscr{U}$ supplied by
Lemma A.5 for a set $U$ of cardinality $\kappa$ is such that: for any ring $R$
(resp., and any left $R$-module $M$), any polynomial (resp., linear) system of
equations

$\textstyle\\{g_{i}(\\{X_{\sigma}\\}_{\sigma})=0\\}_{i\in
I}\quad\text{{(}resp.,}\quad\\{\sum_{\sigma}r_{i,\,\sigma}X_{\sigma}=m_{i}\\}_{i\in
I})\ \ \mbox{with}\ \ \\#I\leq\kappa$

in variables $\\{X_{\sigma}\\}_{\sigma}$ and coefficients in
$\prod_{\mathscr{U}}R$ (resp., $r_{i,\,\sigma}\in\prod_{\mathscr{U}}R$ and
$m_{i}\in\prod_{\mathscr{U}}M$) has a solution in $\prod_{\mathscr{U}}R$
(resp., $\prod_{\mathscr{U}}M$) as soon as so do all its finite subsystems.

###### Proof.

The assertion is a concrete case of the model-theoretic [6, Theorem 6.1.8],
and the latter is sharper in multiple aspects. For convenience, we recall the
argument.

For brevity, we denote the system in question by $\\{g_{i}=0\\}_{i\in I}$ and
we lift it to a system $\\{\widetilde{g}_{i}=0\\}_{i\in I}$ with coefficients
in $\prod_{u\in U}R$ (resp., and in $\prod_{u\in U}M$) and the same variables
$\\{X_{\sigma}\\}_{\sigma}$ by lifting the nonzero coefficients along the
surjection

$\textstyle\prod_{u\in
U}R\twoheadrightarrow\prod_{\mathscr{U}}R\quad\text{{(}resp.,~{}and
along}\quad\prod_{u\in U}M\twoheadrightarrow\prod_{\mathscr{U}}M\quad\text{for
the $m_{i}$}).$

Since $\mathscr{U}$ is countably incomplete, we may fix a decreasing sequence

$\textstyle U\supset U_{0}\supset U_{1}\supset U_{2}\supset\cdots\quad\text{of
sets in $\mathscr{U}$ with}\quad\bigcap_{n\geq 0}U_{n}=\emptyset.$

We then define a function

$f\colon\\{\mbox{finite subsets of $I$}\\}\rightarrow{\mathscr{U}}$

by letting $\mathrm{pr}_{u}$ denote the projection onto the $u$-th factor of
$\prod_{u\in U}$ and setting

$f(I^{\prime}):=U_{\\#I^{\prime}}\cap\\{u\in U\,|\,\mbox{the system
}\\{\mathrm{pr}_{u}(\widetilde{g}_{i})=0\\}_{i\in I^{\prime}}\mbox{ is
solvable in $R$ (resp.,~{}$M$)}\\}.$

The well-definedness of $f$ follows from the solvability of the subsystem
$\\{g_{i}=0\\}_{i\in I^{\prime}}$ in $\prod_{\mathscr{U}}R$ (resp.,
$\prod_{\mathscr{U}}M$) and from the stability of $\mathscr{U}$ under
supersets. By construction, $f(I^{\prime})\supset f(I^{\prime\prime})$
whenever $I^{\prime}\subset I^{\prime\prime}$, so, since $\\#I\leq\\#U$, Lemma
A.5 supplies a function

$f_{0}\colon\\{\mbox{finite subsets of $I$}\\}\rightarrow{\mathscr{U}}\ \
\mbox{such that}\ \ f_{0}(I^{\prime})\subset f(I^{\prime}),\
f_{0}(I^{\prime}\cup I^{\prime\prime})=f_{0}(I^{\prime})\cap
f_{0}(I^{\prime\prime})$

for all finite subsets $I^{\prime},I^{\prime\prime}\subset I$ (technically, to
apply Lemma A.5 we first embed $I$ into $U$ as a subset and then extend $f$ to
finite subsets $U^{\prime}\subset U$ by the rule $U^{\prime}\mapsto
f(U^{\prime}\cap I)$).

For each $u\in U$, we set

$I_{u}:=\\{i\in I\,|\,u\in f_{0}(\\{i\\})\\}.$

Whenever, $i_{1},\dots,i_{n}\in I_{u}$ are pairwise distinct, we have

$u\in f_{0}(\\{i_{1}\\})\cap\dots\cap
f_{0}(\\{i_{n}\\})=f_{0}(\\{i_{1},\dots,i_{n}\\})\subset
f(\\{i_{1},\dots,i_{n}\\})\subset U_{n},$

so, since the $U_{n}$ have empty intersection, each $I_{u}$ is finite. Then
the preceding display applied to an enumeration of $I_{u}$ shows that $u\in
f(I_{u})$, to the effect that the system
$\\{\mathrm{pr}_{u}(\widetilde{g}_{i})=0\\}_{i\in I_{u}}\ \ \mbox{has a
solution}\ \ \\{x_{\sigma,\,u}\\}_{\sigma}\ \ \mbox{in $R$ (resp.,~{}$M$)}.$

We claim that $\\{x_{\sigma}:=(x_{\sigma,\,u})_{u\in U}\\}_{\sigma}$ gives a
solution in $\prod_{\mathscr{U}}R$ (resp., $\prod_{\mathscr{U}}M$) to the
system $\\{g_{i}=0\\}_{i\in I}$. Indeed, for every $i\in I$ we have
$f_{0}(\\{i\\})\in{\mathscr{U}}$ and for every $u\in f_{0}(\\{i\\})$ we have
$i\in I_{u}$, so $\widetilde{g}_{i}(\\{x_{\sigma}\\}_{\sigma})=0$ in the
projection on $\prod_{u\in f_{0}(\\{i\\})}$.

The argument is not specific to rings or modules, and it also shows the
following.

######  Variant A.7.

For an infinite cardinal $\kappa$, every ultrafilter $\mathscr{U}$ supplied by
Lemma A.5 for a set $U$ of cardinality $\kappa$ is such that: for any monoid
$G$, any system

$\\{g_{i}(\\{X_{\sigma}\\}_{\sigma})=g_{i}^{\prime}(\\{X_{\sigma}\\}_{\sigma})\\}_{i\in
I}\quad\text{with}\quad\\#I\leq\kappa$

of monomial equations in variables $\\{X_{\sigma}\\}_{\sigma}$ and
coefficients in $\prod_{\mathscr{U}}G$ has a solution in
$\prod_{\mathscr{U}}G$ as soon as so does each of its finite subsystems.

As we now review, Proposition A.6 supplies algebraically compact ultrapowers.

###### A.8. Algebraic compactness.

We fix a ring $R$ and recall that a map $M\rightarrow M^{\prime}$ of left
$R$-modules is _pure_ if the map $M^{\prime\prime}\otimes_{R}M\rightarrow
M^{\prime\prime}\otimes_{R}M^{\prime}$ is injective for every right $R$-module
$M^{\prime\prime}$. An $R$-module $M$ is _algebraically compact_ (or _pure-
injective_) if every pure map $M\rightarrow M^{\prime}$ of $R$-modules is a
split injection. For example, if $M$ is an algebraically compact abelian group
(so $R={\bf Z}$), then every short exact sequence

$0\rightarrow M\rightarrow M^{\prime}\rightarrow M^{\prime}/M\rightarrow
0\quad\text{of abelian groups
with}\quad(M^{\prime}/M)_{\mathrm{tors}}=0\quad\text{splits.}$

A concrete criterion for algebraic compactness is given by [11, 7.1 (with
6.5)]: a left $R$-module $M$ is algebraically compact if every system of
equations

$\textstyle\\{\sum_{\sigma}r_{i,\,\sigma}X_{\sigma}=m_{i}\\}_{i\in
I}\quad\text{with}\quad r_{i,\,\sigma}\in R\quad\text{and}\quad m_{i}\in M$

has a solution in $M$ as soon as so do all its finite subsystems. Moreover, by
[11, 7.28, 7.29], it suffices to consider systems with cardinality
$\\#I\leq\operatorname{max}(\\#R,\\#{\bf Z})$. In particular, thanks to
Proposition A.6, there is an ultrafilter $\mathscr{U}$ such that for any
$R$-module $M$, the $R$-module $\prod_{\mathscr{U}}M$ is algebraically
compact.

With model-theoretic input in place, we turn to the tower of ultrapowers
argument in Theorem A.10. The final input is the following lemma proved in [7,
2.2], [23, 4.6.1], or [9, 6.1.30] that captures the “combinatorial” part of
local uniformization.

######  Lemma A.9.

For a totally ordered abelian group $\Gamma$, the submonoid $\Gamma_{\geq
0}\subset\Gamma$ of nonnegative elements is a filtered increasing union of its
finite free submonoids isomorphic to ${\bf Z}_{\geq 0}^{r}$ (where $r\in{\bf
Z}_{\geq 0}$ need not be constant).

######  Theorem A.10.

For a valuation ring $V$ with value group $\Gamma$, there is a countable
sequence of ultrafilters ${\mathscr{U}}_{1},{\mathscr{U}}_{2},\dots$ on some
respective sets $U_{1},U_{2},\dots$ for which the valuation rings
$\\{V_{n}\\}_{n\geq 0}$ defined inductively by $V_{0}:=V$ and
$V_{n+1}:=\prod_{{\mathscr{U}}_{n+1}}V_{n}$ are such that the valuation ring

$\textstyle\widetilde{V}:=\varinjlim_{n\geq 0}V_{n}\quad\text{has a cross-
section}\quad\widetilde{s}:\widetilde{\Gamma}\rightarrow\widetilde{K}^{*},$

where $\widetilde{K}$ and $\widetilde{\Gamma}$ are the fraction field and the
value group of $\widetilde{V}$.

###### Proof.

We let $K_{n}$ and $\Gamma_{n}$ denote the fraction field and the value group
of $V_{n}$, so that $\Gamma_{n+1}\cong\prod_{{\mathscr{U}}_{n+1}}\Gamma_{n}$
and $K_{n+1}=\prod_{{\mathscr{U}}_{n+1}}K_{n}$ (see §A.4) with

$\textstyle\widetilde{\Gamma}\cong\varinjlim_{n\geq 0}\Gamma_{n}\ \
\mbox{and}\ \ \widetilde{K}\cong\varinjlim_{n\geq 0}K_{n}.$

The idea is to build ultrafilters ${\mathscr{U}}_{n}$ one by one using Lemma
A.5 in such a way that a desired cross-section

$\textstyle\widetilde{s}\colon\widetilde{\Gamma}\rightarrow\widetilde{K}^{*}\
\ \mbox{would be the limit of compatible partial cross-sections}\ \
s_{n}\colon\Gamma_{n}\rightarrow K_{n+1}^{*}.$

For this, as an initial step, we replace $V$ by a suitable ultrapower to
ensure that the abelian group $\Gamma$ is algebraically compact (see §A.8).
Granted this, it suffices to carry out the inductive step: setting
$\Gamma_{-1}:=0$ for convenience and assuming that we have already constructed
$s_{n-1}$ and $V_{n}$ for some $n\geq 0$ in such a way that the abelian groups
$\Gamma_{n-1}$ and $\Gamma_{n}$ are algebraically compact, it suffices to
construct $V_{n+1}$ with $\Gamma_{n+1}$ algebraically compact in such a way
that $s_{n-1}$ extends to an $s_{n}$.

The role of algebraic compactness is to split the map
$\Gamma_{n-1}\hookrightarrow\Gamma_{n}\cong\prod_{{\mathscr{U}}_{n}}\Gamma_{n-1}$
whose cokernel is torsion free:

$\Gamma_{n}\cong\Gamma_{n-1}\oplus G\ \ \mbox{for some subgroup}\ \
G\subset\Gamma_{n}.$

Thanks to this splitting, we only need to build an ultrafilter
${\mathscr{U}}_{n+1}$ and a partial cross-section
$s_{G}:G\rightarrow(\prod_{{\mathscr{U}}_{n+1}}K_{n})^{*}$ such that
$\prod_{{\mathscr{U}}_{n+1}}\Gamma_{n}$ is algebraically compact. In fact, we
let ${\mathscr{U}}_{n+1}$ be any ultrafilter as in Lemma A.5 applied to the
cardinal $\operatorname{max}(\\#\Gamma_{n},\\#{\bf Z})$. Then
$\prod_{{\mathscr{U}}_{n+1}}\Gamma_{n}$ is necessarily algebraically compact
by the criterion reviewed in §A.8 and Proposition A.6.

The subgroup $G$ inherits a total order from $\Gamma_{n}$, and any partial
cross-section

$\textstyle s_{G_{\geq 0}}:G_{\geq
0}\rightarrow(\prod_{{\mathscr{U}}_{n+1}}V_{n})\setminus\\{0\\}\quad\text{will
give rise to a desired}\quad s_{G}.$

For each $g\in G_{>0}$, we fix a $v_{g}\in V_{n}$ with
$\mathrm{val}(v_{g})=g$. Then $s_{G_{\geq 0}}$ amounts to a solution in
$\prod_{{\mathscr{U}}_{n+1}}V_{n}$ of the following system of equations in
variables $\\{X_{g},U_{g},U^{\prime}_{g}\\}_{g\in G_{>0}}$:

$\textstyle\\{X_{g+g^{\prime}}=X_{g}X_{g^{\prime}},\ \ X_{g}U_{g}=v_{g},\ \
U_{g}U^{\prime}_{g}=1\\}_{g,\,g^{\prime}\in G_{>0}}.$

Likewise, for any submonoid $G^{\prime}\subset G_{\geq 0}$, the restriction of
$s_{G_{\geq 0}}|_{G^{\prime}}$, that is, a partial cross-section defined on
$G^{\prime}$, amounts to a solution in $\prod_{{\mathscr{U}}_{n+1}}V_{n}$ of
the subsystem consisting of those equations that only involve the variables
$\\{X_{g},U_{g},U^{\prime}_{g}\\}_{g\in G^{\prime}}$. However, a partial
cross-section $G^{\prime}\rightarrow\prod_{{\mathscr{U}}_{n+1}}V_{n}$ (and
even $G^{\prime}\rightarrow V_{n}$) certainly exists if $G^{\prime}\simeq{\bf
Z}_{\geq 0}^{d}$, and, by Lemma A.9, the monoid $G_{\geq 0}$ is a filtered
increasing union of such $G^{\prime}$. This implies that every finite
subsystem of the above system has a solution in
$\prod_{{\mathscr{U}}_{n+1}}V_{n}$ (and even in $V_{n}$). Then, by Proposition
A.6, the entire system has a solution in $\prod_{{\mathscr{U}}_{n+1}}V_{n}$,
which completes the inductive step.

######  Variant A.11.

For every faithfully flat map $V\subset V^{\prime}$ of valuation rings with
value groups $\Gamma\subset\Gamma^{\prime}$ such that $\Gamma^{\prime}/\Gamma$
is torsion free, there is a countable sequence of ultrafilters
${\mathscr{U}}_{1},{\mathscr{U}}_{2},\dots$ on some respective sets
$U_{1},U_{2},\dots$ for which the valuation rings $\\{V_{n}\\}_{n\geq 0}$ and
$\\{V^{\prime}_{n}\\}_{n\geq 0}$ defined inductively by $V_{0}:=V$ and
$V^{\prime}_{0}:=V^{\prime}$ with $V_{n+1}:=\prod_{{\mathscr{U}}_{n+1}}V_{n}$
and $V^{\prime}_{n+1}:=\prod_{{\mathscr{U}}_{n+1}}V^{\prime}_{n}$ are such
that the valuation ring

$\textstyle\widetilde{V}^{\prime}=\varinjlim_{n\geq
0}V^{\prime}_{n}\quad\text{with}\quad\widetilde{K}^{\prime}:=\operatorname{Frac}(\widetilde{V}^{\prime})\quad\text{has
a cross-
section}\quad\widetilde{s}:\widetilde{\Gamma}^{\prime}\rightarrow\widetilde{K}^{\prime*}$

whose restriction to the value group $\widetilde{\Gamma}$ of
$\widetilde{V}:=\varinjlim_{n\geq 0}V_{n}$ lands in
$\widetilde{K}:=\operatorname{Frac}(\widetilde{V})$.

###### Proof.

By replacing $V$ and $V^{\prime}$ by suitable large ultrapowers if needed and
using §A.8 with Proposition A.6, we may ensure that $\Gamma$ is algebraically
compact. Then, thanks to the torsion-freeness assumption on
$\Gamma^{\prime}/\Gamma$, the inclusion $\Gamma\rightarrow\Gamma^{\prime}$
splits. A choice of a splitting induces a compatible splitting on any
ultrapower, so the proof of Theorem A.10 continues to give the claimed
variant.

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