# Constructive General Neron Desingularization for one dimensional local rings

Gerhard Pfister  Gerhard Pfister, Department of Mathematics, University of
Kaiserslautern, Erwin-Schrödinger-Str., 67663 Kaiserslautern, Germany
<EMAIL_ADDRESS>Dorin Popescu  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<EMAIL_ADDRESS>
###### Abstract

An algorithmic proof of General Neron Desingularization is given here for one
dimensional local rings and it is implemented in Singular. Also a theorem
recalling Greenberg’ strong approximation theorem is presented for one
dimensional local rings.

###### keywords:

Smooth morphisms, regular morphisms, smoothing ring morphisms.
2010 Mathematics Subject Classification: Primary 13B40, Secondary
14B25,13H05,13J15.

AND

††thanks: This paper was partially written during the visit of the first
author at the Institute of Mathematics Simion Stoilow of the Romanian Academy
supported by a BitDefender Invited Professor Scholarship, 2015. The support
from the Department of Mathematics of the University of Kaiserslautern of the
first author and the support from the project ID-PCE-2011-1023, granted by the
Romanian National Authority for Scientific Research, CNCS - UEFISCDI of the
second author are gratefully acknowledged.

##  1 Introduction

A ring morphism $u:A\rightarrow A^{\prime}$ has regular fibers if for all
prime ideals $P\in\operatorname{Spec}A$ the ring $A^{\prime}/PA^{\prime}$ is a
regular ring, i.e. its localizations are regular local rings. It has
geometrically regular fibers if for all prime ideals
$P\in\operatorname{Spec}A$ and all finite field extensions $K$ of the fraction
field of $A/P$ the ring $K\otimes_{A/P}A^{\prime}/PA^{\prime}$ is regular. If
for all $P\in\operatorname{Spec}A$ the fraction field of $A/P$ has
characteristic $0$ then the regular fibers of $u$ are geometrically regular
fibers. A flat morphism $u$ is regular if its fibers are geometrically
regular. If $u$ is regular of finite type then $u$ is called smooth. A
localization of a smooth algebra is called essentially smooth.

In Artin approximation theory (introduced in [2]) an important result (see
[16]) is the following theorem, generalizing the Neron Desingularization [9],
[2].

######  Theorem 1 (General Neron Desingularization, Popescu [11], [12], [13],
André [1], Swan [19], Spivakovsky [18]).

Let $u:A\rightarrow A^{\prime}$ be a regular morphism of Noetherian rings and
$B$ a finite type $A$-algebra. Then any $A$-morphism $v:B\rightarrow
A^{\prime}$ factors through a smooth $A$-algebra $C$, that is $v$ is a
composite $A$-morphism $B\rightarrow C\rightarrow A^{\prime}$.

The purpose of this paper is to give an algorithmic proof of the above theorem
when $A,A^{\prime}$ are one dimensional local rings, that is Theorem 2. When
$A,A^{\prime}$ are domains such algorithm is given in [10], the case when
$A,A^{\prime}$ are discrete valuation rings proved by Néron [9] is given in a
different way in [15] with applications in arcs frame. The present algorithm
was implemented by the authors in the Computer Algebra system Singular [3] and
will be as soon as possible found in a development version at

    https://github.com/Singular/Sources/blob/spielwiese/Singular/LIB/.

The proof of Theorem 2 splits essentially in three steps. We will give here
the idea in case $A$ and $A^{\prime}$ are domains. In step 1 we reduce the
problem to the case when $H_{B/A}\cap A\not=0$, $H_{B/A}$ being the ideal
defining the nonsmooth locus of $B$ over $A$. Let $0\not=d\in H_{B/A}\cap A$.
This means geometrically that
$\operatorname{Spec}B_{d}\rightarrow\operatorname{Spec}A_{d}$ is smooth. In
the second step we construct a smooth $A$-algebra $D$, $A\subset D\subset
A^{\prime}$ and an $A$-morphism $v^{\prime}:B\rightarrow D/d^{3}D$ such that
$v\equiv v^{\prime}$ modulo $d^{3}A^{\prime}$. If $A^{\prime}$ is the
completion $\hat{A}$ of $A$ we can use $D=A$. The third step resolves the
singularity. If $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$ then we can find
$f=(f_{1},\ldots,f_{r})$, $r\leq n$ a system of polynomials from $I$ (given a
tuple $(b_{1},\ldots b_{s})$ we denote usually by the corresponding unindexed
letter $b$ the vector $(b_{1},\ldots b_{s})$), and an $r\times r$-minor $M$ of
the Jacobian matrix $(\partial f_{i}/\partial Y_{j})$ such that $d\equiv MN$
modulo $I$ for some $N\in((f):I)$, where $(f)$ denotes the ideal generated by
the system $f$. Then $v^{\prime}(MN)=ds$ for some $s\in 1+dD$. Assume that
$M=\det(\partial f_{i}/\partial Y_{j})_{1\leq i,j\leq r}$. Let $H$ be the
matrix obtained by adding to $(\partial f/\partial Y)$ the boarder block
$(0|\mbox{Id}_{n-r})$ and let $G^{\prime}$ be the adjoined matrix of $H$ and
$G=NG^{\prime}$. Consider in $D[Y,T]$, $T=(T_{1},\ldots,T_{n})$, the ideal
$J=((f,s(Y-y^{\prime})-dG(y^{\prime})T):d^{2})$, where $y^{\prime}\in D^{n}$
is lifting $v^{\prime}(Y)$. Then $C$ is a suitable localization of the
$B\otimes_{A}D$-algebra $D[Y,T]/(I,J)$ and $v$ extends to $C$ by
$v(T)=t=(1/d^{2})H(y^{\prime})(v(Y)-y^{\prime})$.

Consider the following example. Let $A={\bf Q}[x]_{(x)}$, $A^{\prime}={\bf
C}[[x]]$, $B=A[Y_{1},Y_{2}]/(Y_{1}^{2}+Y_{2}^{2})$, $a\in{\bf C}$ a
transcendental element over $\bf Q$, ${\bar{u}}\in{\bf C}[[x]]\setminus{\bf
C}[x]_{(x)}$ and $u=a+x^{6}{\bar{u}}$. Let $v$ be given by $v(Y_{1})=xu$,
$v(Y_{2})=xiu$, where $i=\sqrt{-1}$. In step 1 we change $B$ by
$B_{1}=A[Y_{1},Y_{2},Y_{3}]/I$, $I=(Y_{1}^{2}+Y_{2}^{2},x-2Y_{1}Y_{3})$ and
extend $v$ by $v(Y_{3})=1/(2u)$. We have $4Y_{1}^{2}Y_{3}^{2}\in H_{B/A}$
which implies $d=x^{2}\in H_{B/A}\cap A$. We define $D=A[a,a^{-1},i]$ and
$v^{\prime}(Y)=y^{\prime}=(xa,xia,1/(2a))$. This is step two.

To understand step 3 we simplify the example taking
$B=A[Y_{1},Y_{2}]/(Y_{1}Y_{2}-x^{2})$, $u=1+x^{6}{\bar{u}}$ and $v$ given by
$v(Y_{1})=xu$, $v(Y_{2})=x/u$. Then $d=x\in H_{B/A}$, $D=A$,
$y^{\prime}_{1}=x=y^{\prime}_{2}$. We obtain $H=\begin{pmatrix}Y_{2}&Y_{1}\\\
0&1\end{pmatrix}$, $G=G^{\prime}=\begin{pmatrix}1&-Y_{1}\\\
0&Y_{2}\end{pmatrix}$, $N=1$ and
$J=((Y_{1}Y_{2}-x^{2},Y_{1}-x-xT_{1}+x^{2}T_{2},Y_{2}-x-x^{2}T_{2}):x^{2})$.
We have
$J=(xT_{1}T_{2}-x^{2}T_{2}^{2}+T_{1},Y_{1}-x-xT_{1}+x^{2}T_{2},Y_{2}-x-x^{2}T_{2})$
and we obtain that
$C\cong(A[T_{1},T_{2}]/(xT_{1}T_{2}-x^{2}T_{2}^{2}+T_{1}))_{1+xT_{2}}\cong(A[T_{2}])_{1+xT_{2}}$
is a smooth $A$-algebra.

When $A^{\prime}$ is the completion of a Noetherian local ring $A$ of
dimension one we show that we may have a linear Artin function as it happens
in the Greenberg’s case (see [5] and [10, Theorem 18]). More precisely, the
Artin function is given by $c\rightarrow(\rho+1)(e+1)+c$, where $e,\rho$
depend on $A$ and the polynomial system of equations defining $B$ (see Theorem
14).

We thank to both Referees for useful comments and pointing some misprints.

##  2 Theorem 1 in one dimensional local rings

Let $u:A\rightarrow A^{\prime}$ be a flat morphism of Noetherian local rings
of dimension $1$. Suppose that the maximal ideal ${\mathfrak{m}}$ of $A$
generates the maximal ideal of $A^{\prime}$. Moreover suppose that $u$ is a
regular morphism, $k$ is infinite and there exist canonical inclusions
$k=A/{\mathfrak{m}}\subset A$,
$k^{\prime}=A^{\prime}/{\mathfrak{m}}A^{\prime}\subset A^{\prime}$ such that
$u(k)\subset k^{\prime}$.

Let $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$. If $f=(f_{1},\ldots,f_{r})$, $r\leq
n$ is a system of polynomials from $I$ then we can define the ideal
$\Delta_{f}$ generated by all $r\times r$-minors of the Jacobian matrix
$(\partial f_{i}/\partial Y_{j})$. After Elkik [4] let $H_{B/A}$ be the
radical of the ideal $\sum_{f}((f):I)\Delta_{f}B$, where the sum is taken over
all systems of polynomials $f$ from $I$ with $r\leq n$. Then $B_{P}$,
$P\in\operatorname{Spec}B$ is essentially smooth over $A$ if and only if
$P\not\supset H_{B/A}$ by the Jacobian criterion for smoothness. Thus
$H_{B/A}$ measures the non smooth locus of $B$ over $A$. $B$ is standard
smooth over $A$ if there exists $f$ in $I$ as above such that
$B=((f):I)\Delta_{f}B$.

The aim of this paper is to give an easy algorithmic proof of the following
theorem.

######  Theorem 2.

Any $A$-morphism $v:B\rightarrow A^{\prime}$ factors through a standard smooth
$A$-algebra $B^{\prime}$.

We consider in the algorithmic part the following assumption, which we will
keep in the whole paper

$(*)$ $A$ is essentially of finite type over a field $k$, let us say
$A\cong(k[x]/J)_{(x)}$ for some variables $x=(x_{1},\ldots x_{m})$, and the
completion of $A^{\prime}$ is $k^{\prime}[[x]]/(J)$.

When $v$ is defined by polynomials $y$ from $k^{\prime}[x]$ then our problem
is easy. Let $L$ be the field obtained by adjoining to $k$ all coefficients of
$y$. Then $R=(L[x]/(J))_{(x)}$ is a subring of $A^{\prime}$ containing
$\operatorname{Im}v$ which is essentially smooth over $A$. Then we may take
$B^{\prime}$ as a standard smooth $A$-algebra such that $R$ is a localization
of $B^{\prime}$. Thus we will not suppose in this paper that $y$ is polynomial
and therefore $L$ is not necessarily a finite type field extension of $k$.

In the proof we need to know that $v(H_{B/A})$ is not contained in any minimal
prime ideal of $A^{\prime}$. In theory, we may reduce to this case as it
follows. Let ${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$. Since $u$ is
regular, it induces a regular map $u_{{\mathfrak{p}}}:A_{({\mathfrak{p}}\cap
A)}\rightarrow A^{\prime}_{{\mathfrak{p}}}$ of local Artinian rings (in
particular $k({\mathfrak{p}})\otimes_{A_{{\mathfrak{p}}\cap A}}u_{p}$,
$k({\mathfrak{p}})=A_{({\mathfrak{p}}\cap A)}/({\mathfrak{p}}\cap
A)A_{({\mathfrak{p}}\cap A)}$ is a separable field extension). Note that
$A_{({\mathfrak{p}}\cap A)}\supset k({\mathfrak{p}})$ because of $(*)$ and
$A^{\prime}_{{\mathfrak{p}}}$ is a filtered inductive limit of its subrings of
the form $E_{F_{{\mathfrak{p}}}}=A_{({\mathfrak{p}}\cap
A)}\otimes_{k({\mathfrak{p}})}F_{{\mathfrak{p}}}$ for all finite type field
subextension $F_{{\mathfrak{p}}}/k({\mathfrak{p}})$ of
$(A^{\prime}_{{\mathfrak{p}}}/{\mathfrak{p}}A^{\prime}_{{\mathfrak{p}}})/k({\mathfrak{p}})$.
We may change $B$ by a finite type $B$-algebra $\tilde{B}$ of $A^{\prime}$
such that ${\tilde{B}}_{{\mathfrak{p}}\cap{\tilde{B}}}\cong
E_{F_{{\mathfrak{p}}}}$. It follows that
$v(H_{B/A})\not\subset{\mathfrak{p}}A^{\prime}$ for all
${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$. Next we will assume from the
beginning that

$(**)$ $v(H_{B/A})\not\subset{\mathfrak{p}}A^{\prime}$ for all
${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$.

Unfortunately, the computer cannot decide this since we are not able to give
the whole information concerning the coefficients of $y$. But we are able to
decide if $v(H_{B/A})$ is not contained in
${\mathfrak{m}}^{N}A^{\prime}+{\mathfrak{p}}$ for $N>>0$ and one
${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$. In the following we suppose
that $v(H_{B/A})\not\subset{\mathfrak{m}}^{N}A^{\prime}+{\mathfrak{p}}$ for
all ${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$ and a certain $N>>0$.
Choose $\gamma\in H_{B/A}$ such that $v(\gamma)$ is not in
$\cup_{{\mathfrak{p}}\in\operatorname{Min}A^{\prime}}{\mathfrak{p}}+{\mathfrak{m}}^{N}A^{\prime}$.

The idea of the proof of Theorem 2 is to find $f=(f_{1},\ldots,f_{r})$ in $I$
and a $d\in v(((f):I)\Delta_{f})A^{\prime}\cap A$ which is not in
$\cup_{{\mathfrak{p}}\in\operatorname{Min}A^{\prime}}{\mathfrak{p}}$. The
assumption $(**)$ gives just that there exists $f_{{\mathfrak{p}}}$ for any
${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$ such that
$\Delta_{f_{{\mathfrak{p}}}}((f_{{\mathfrak{p}}}):I)\not\subset{\mathfrak{p}}A^{\prime}$
for all ${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$. The main problem is
to reduce to the case when $f_{{\mathfrak{p}}}$ does not depend of
${\mathfrak{p}}$. Actually, this follows if $I_{\gamma}/I^{2}_{\gamma}$ is
free over $B_{\gamma}$. In the next three lemmas containing some results of
Elkik [4] (in the form used in [11], [19], [14]), we see that this is true if
we reduce to the case when $\Omega_{B_{\gamma}}/A$ is free over $B_{\gamma}$.
In general, the last module is projective but not free as the following
example shows.

######  Example 3.

Let $k$ be a subfield of $\bf R$,
$A=(k[x_{1},x_{2},x_{3}]/(x_{1}^{2}-x_{2}x_{3},x_{3}^{2}-x_{1}x_{2}))_{(x_{1},x_{2},x_{3})}$,
and $\alpha=x_{1}Y_{1}^{2}+x_{2}Y_{2}^{2}+x_{3}Y_{3}^{2}-x_{1}-x_{2}-x_{3}\in
A[Y]$, $Y=(Y_{1},Y_{2},Y_{3})$. Set
$f_{1}=x_{2}\alpha=x_{3}^{2}Y_{1}^{2}+x_{2}^{2}Y_{2}^{2}+x_{1}^{2}Y_{3}^{2}-x_{3}^{2}-x_{2}^{2}-x_{1}^{2}$,
$f_{2}=x_{1}\alpha$, $f_{3}=x_{3}\alpha$ and $I=(f)$. Then
$x_{2}I\subset(f_{1})$, $x_{1}I\subset(f_{2})$, $x_{3}I\subset(f_{3})$. Also
note that $x_{2}^{2}Y_{2}\in\Delta_{f_{1}}$,
$x_{1}^{2}Y_{1}\in\Delta_{f_{2}}$, $x_{3}^{2}Y_{3}\in\Delta_{f_{3}}$.

Let $B=A[Y]/I$,
$A^{\prime}=k^{\prime}[[x_{1},x_{2},x_{3}]]/(x_{1}^{2}-x_{2}x_{3},x_{3}^{2}-x_{1}x_{2})$,
where $k\subset k^{\prime}$ is a field extension. Let $u_{1},u_{2}$ be two
algebraically independent elements of $k^{\prime}[[x_{1},x_{2},x_{3}]]$ over
$k[x_{1},x_{2},x_{3}]$. Set $y_{1}=x_{3}u_{1}-1$, $y_{2}=x_{3}u_{2}-1$ and we
may find $y_{3}$ such that
$y_{3}^{2}=1-x_{1}x_{3}u_{1}^{2}+2x_{1}u_{1}-x_{2}x_{3}u_{2}^{2}+2x_{2}u_{2}$.
Clearly, $\alpha(y)=0$, that is $\alpha(y_{i})=0$ for all $1\leq i\leq n$, and
so we get an $A$-morphism $v:B\rightarrow A^{\prime}$ given by $Y\rightarrow
y$.

Note that $H_{B/A}=(x_{1},x_{2},x_{3})$. Take $\gamma=x_{1}+x_{2}+x_{3}$ and
$\beta=f_{1}+f_{2}+f_{3}$. Then
$I_{\gamma}=(\beta)_{\gamma}=(\alpha)_{\gamma}$ and we claim that
$\Omega_{B_{\gamma}/A}$ is projective but not free. Indeed,
$\Omega_{B/A}=BdY_{1}\oplus BdY_{2}\oplus
BdY_{3}/(x_{1}Y_{1}dY_{1}+x_{2}Y_{2}dY_{2}+x_{3}Y_{3}dY_{3})$ and
$\Omega_{B_{\gamma}/A}$ is projective because its Fitting ideal is
$(x_{1}Y_{1},x_{2}Y_{2},x_{3}Y_{3})B_{\gamma}\supset(x_{1}Y_{1}^{2}+x_{2}Y_{2}^{2}+x_{3}Y_{3}^{2})B_{\gamma}=\gamma
B_{\gamma}=B_{\gamma}$ (see e.g. [7, Proposition 1.3.8]).

Now suppose that $\Omega_{B_{\gamma}/A}$ is free over $B_{\gamma}$. Then
$\lambda=x_{1}Y_{1}+x_{2}Y_{2}+x_{3}Y_{3}$ can be included in a basis of
$B_{\gamma}dY_{1}\oplus B_{\gamma}dY_{2}\oplus B_{\gamma}dY_{3}$. More
precisely, there exists a $3\times 3$ invertible matrix $(a_{ij})$,
$a_{ij}=a_{ij}(x,Y)$ over $B_{\gamma}$ with $a_{1j}=x_{j}Y_{j}$ for $j\in[3]$,
let us say $a_{ij}=b_{ij}(x,Y)/c(x)$, $j=2,3$ with $b_{ij}\in k[x,Y]$, $c\in
k[x]$. We may choose some positive real numbers
$x^{\prime}_{1}.x^{\prime}_{2},x^{\prime}_{3}$ such that $x^{\prime
2}_{1}=x^{\prime}_{2}x^{\prime}_{3}$, $x^{\prime
2}_{3}=x^{\prime}_{1}x^{\prime}_{2}$, $c(x^{\prime})\not=0$ and the matrix
$(a_{ij}(x^{\prime},Y))$ invertible in $B^{\prime}={\bf
R}[Y]/(x^{\prime}_{1}Y_{1}^{2}+x^{\prime}_{2}Y_{2}^{2}+x^{\prime}_{3}Y_{3}^{2}-x^{\prime}_{1}-x^{\prime}_{2}-x^{\prime}_{3})$.
It follows that $P^{\prime}=(B^{\prime}dY_{1}\oplus B^{\prime}dY_{2}\oplus
B^{\prime}dY_{3})/<x^{\prime}_{1}Y_{1}dY_{1}+x^{\prime}_{2}Y_{2}dY_{2}+x^{\prime}_{3}Y_{3}dY_{3}>$
is free over $B^{\prime}$. Changing $Y_{i}$ by
$\sqrt{x^{\prime}_{i}/(x^{\prime}_{1}+x^{\prime}_{2}+x^{\prime}_{3})}Y_{i}$ we
see that over $B^{\prime\prime}={\bf R}[Y]/(Y_{1}^{2}+Y_{2}^{2}+Y_{3}^{2}-1)$
the module $P^{\prime\prime}=(B^{\prime\prime}dY_{1}\oplus
B^{\prime\prime}dY_{2}\oplus
B^{\prime\prime}dY_{3})/<Y_{1}dY_{1}+Y_{2}dY_{2}+Y_{3}dY_{3}>$ is free, that
is the tangent bundle over the real sphere is trivial which contradicts for
example [8, page 114].

On the other hand, note that $\Omega_{B_{\gamma}[Y_{4}]/A}=$

$B_{\gamma}[Y_{4}]dY_{1}\oplus B_{\gamma}[Y_{4}]dY_{2}\oplus
B_{\gamma}[Y_{4}]dY_{3}\oplus
B_{\gamma}[Y_{4}]dY_{4}/<x_{1}Y_{1}dY_{1}+x_{2}Y_{2}dY_{2}+x_{3}Y_{3}dY_{3}>$

is free because in $\sum_{i=1}^{4}B_{\gamma}[Y_{4}]dY_{i}$ we have the basis

$\\{x_{1}Y_{1}dY_{1}+x_{2}Y_{2}dY_{2}+x_{3}Y_{3}dY_{3},dY_{1}-Y_{1}dY_{4},dY_{2}-Y_{2}dY_{4},dY_{3}-Y_{3}dY_{4}\\}.$

Let $B_{1}$ be the symmetric algebra $S_{B}(I/I^{2})$ of $I/I^{2}$ over $B$.
Then $(B_{1})_{\gamma}$ is the symmetric algebra
$S_{B_{\gamma}}((I/I^{2})_{\gamma})$ of $(I/I^{2})_{\gamma}$ over
$B_{\gamma}$. But $(I/I^{2})_{\gamma}$ is free generated by $\alpha$ and so
$(B_{1})_{\gamma}\cong B_{\gamma}[Y_{4}]$. Consequently,
$\Omega_{(B_{1})_{\gamma}/A}$ is free from above.

######  Lemma 4.

([11, Lemma 3.4]) Let $B_{1}$ be the symmetric algebra $S_{B}(I/I^{2})$ of
$I/I^{2}$ over111Let $M$ b e a finitely represented $B$-module and
$B^{m}\xrightarrow{(a_{ij})}B^{n}\rightarrow M\rightarrow 0$ a presentation
then $S_{B}(M)=B[T_{1},\ldots,T_{n}]/J$ with
$J=(\\{\sum\limits^{n}_{i=1}a_{ij}T_{i}\\}_{j=1,\ldots,m})$. $B$. Then
$H_{B/A}B_{1}\subset H_{B_{1}/A}$ and $(\Omega_{B_{1}/A})_{\gamma}$ is free
over $(B_{1})_{\gamma}$.

######  Lemma 5.

([19, Proposition 4.6]) Suppose that $(\Omega_{B/A})_{\gamma}$ is free over
$B_{\gamma}$. Let $I^{\prime}=(I,Y^{\prime})\subset A[Y,Y^{\prime}]$,
$Y^{\prime}=(Y^{\prime}_{1},\ldots,Y^{\prime}_{n})$. Then
$(I^{\prime}/I^{\prime 2})_{\gamma}$ is free over $B_{\gamma}$.

######  Lemma 6.

([14, Corollary 5.10]) Suppose that $(I/I^{2})_{\gamma}$ is free over
$B_{\gamma}$. Then a power of $\gamma$ is in $((g):I)\Delta_{g}$ for some
$g=(g_{1},\ldots g_{r})$, $r\leq n$ in $I$.

Step 1. Reduction to the case when $\Omega_{B_{\gamma}/A}$ is free over
$B_{\gamma}$ .

Let $B_{1}$ be given by Lemma 4. The inclusion $B\subset B_{1}$ has a
retraction $w$ which maps $I/I^{2}$ to zero. For the reduction we change $B,v$
by $B_{1},vw$.

Step 2. Reduction to the case when $(I/I^{2})_{\gamma}$ is free over
$B_{\gamma}$.

Since $\Omega_{B_{\gamma}/A}$ is free over $B_{\gamma}$ we see using Lemma 5
that changing $I$ with $(I,Y^{\prime})\subset A[Y,Y^{\prime}]$ we may suppose
that $(I/I^{2})_{\gamma}$ is free over $B_{\gamma}$.

Step 3. Reduction to the case when a power of $\gamma$ is in
$((f):I)\Delta_{f}$ for some $f=(f_{1},\ldots f_{r})$, $r\leq n$ in $I$.

We reduced to the case when $(I/I^{2})_{\gamma}$ is free over $B_{\gamma}$.
Then it is enough to use Lemma 6.

Step 4. The Jacobian matrix $(\partial f/\partial Y)$ can be completed with
$(n-r)$ rows from $k^{n}$ obtaining a square $n$ matrix $H$ with $v(\det
H)\not\in\cup_{{\mathfrak{p}}\in\operatorname{Min}A^{\prime}}{\mathfrak{p}}$.

We may suppose that $r<n$, otherwise there exist nothing to show. Note that
the rows of $(\partial f/\partial Y)$ are mapped by $v$ in $r$ linear
independent vectors from $(A^{\prime}/{\mathfrak{p}})^{n}$ for each
${\mathfrak{p}}\in\operatorname{Min}A^{\prime}$. Fix a ${\mathfrak{p}}$ and
consider the set $\Lambda_{{\mathfrak{p}}}$ of all $(n-r)$ linear independent
vectors from $k^{n}$ which define a basis in
$Q(A^{\prime}/{\mathfrak{p}})^{n}$ together with the rows of $v(\partial
f/\partial Y)$. Clearly $L_{{\mathfrak{p}}}$ is a nonempty open Zariski set of
$k^{n(n-r)}$. Since $k^{n(n-r)}$ is irreducible we get
$\cap_{{\mathfrak{p}}\in\operatorname{Min}A^{\prime}}L_{{\mathfrak{p}}}\not=\emptyset$.
Choosing $(n-r)$ rows from
$\cap_{{\mathfrak{p}}\in\operatorname{Min}A^{\prime}}L_{{\mathfrak{p}}}$ we
may complete $(\partial f/\partial Y)$ to the wanted matrix $H$.

Step 5. Reduction to the case when $((\det H)((f):I))\cap A$ is
${\mathfrak{m}}$ primary.

By Step 3 there exists a polynomial $R^{\prime}\in((f):I)$ such that
$v(R^{\prime})$ is not in $\cup_{p\in\operatorname{Min}A^{\prime}}p$. Set
$P^{\prime}=R^{\prime}\det H$. Then $v(P^{\prime})$ generates in $A^{\prime}$
an ideal of height $1$ which must be ${\mathfrak{m}}A^{\prime}$ primary. Then
$(v(P^{\prime}))\cap A$ is ${\mathfrak{m}}$ primary too and we may choose
$d^{\prime}\in(v(P^{\prime}))\cap A$ such that $d^{\prime}A$ is
${\mathfrak{m}}$ primary, let us say $d^{\prime}=v(P^{\prime})z$ for some
$z\in A^{\prime}$. Set $B_{1}=B[Z]/(f_{r+1})$, where
$f_{r+1}=-d^{\prime}+P^{\prime}Z$ and let $v_{1}:B_{1}\rightarrow A^{\prime}$
be the map of $B$-algebras given by $Z\rightarrow z$. It follows that
$d^{\prime}\in((f,f_{r+1}):(I,f_{r+1}))$ and $d^{\prime}\in\Delta_{f}$,
$d^{\prime}\in\Delta_{f_{r+1}}$. Then $d=d^{\prime 2}\equiv P\ \mbox{modulo}\
(I,f_{r+1})$ for $P=P^{\prime 2}Z^{2}\in H_{B_{1}/A}$. For the reduction
change $B$ by $B_{1}$ and $H$ by $\left(\begin{array}[]{cc}H&0\\\
&P^{\prime}\end{array}\right).$ Note that $d\in((\det H)((f):I))\cap A$. The
determinant of the new $H$ is the determinant of the old $H$ multiplied with
$P^{\prime}$. Thus $P$ is the determinant of the new $H$ multiplied with
$R=R^{\prime}Z^{2}$

######  Remark 7.

In Example 3 the module $\Omega_{B_{\gamma}/A}$ is not free and so we may
apply Step 1. In fact we do not need to apply the Steps 1, 2, 3 because
$(I/I^{2})_{\gamma}$ is already free.

##  3 Proof of the case when $((\det H)((f):I))\cap A$ is ${\mathfrak{m}}$
primary.

Let $(0)=\cap_{{\mathfrak{p}}\in\operatorname{Ass}A}Q_{{\mathfrak{p}}}$ be a
reduced primary decomposition 222The primary decomposition of an ideal in a
polynomial ring and its localizations by a maximal ideal can be computed using
the SINGULAR library primdec.lib. This is a very difficult task in the
computational algebra usually needing a lot of Gröbner basis computations with
respect to the lexicographical ordering. of $(0)$ in $A$, where
$Q_{{\mathfrak{p}}}$ is a primary ideal with
$\sqrt{Q_{{\mathfrak{p}}}}={\mathfrak{p}}$. Let $d$ be defined as in the end
of Section 2. Define $e$ by $(0:_{A}d^{e})=(0:_{A}d^{e+1})$. This equality
happens for example taking $e$ such that ${{\mathfrak{p}}}^{e}\subset
Q_{{\mathfrak{p}}}$ for all ${\mathfrak{p}}\in\operatorname{Ass}A$. Set
$\bar{A}=A/(d^{2e+1})$, $\bar{A}^{\prime}=A^{\prime}/d^{2e+1}A^{\prime}$,
$\bar{u}=\bar{A}\otimes_{A}u$, $\bar{B}=B/d^{2e+1}B$,
$\bar{v}=\bar{A}\otimes_{A}v$. By base change $\bar{u}$ is a regular morphism
of Artinian local rings.

Step 6. There exists a smooth $A$-algebra and an $A$-morphism
$\omega:D\rightarrow A^{\prime}$ such that
$y=v(Y)\in\operatorname{Im}\omega+d^{2e+1}A^{\prime}$.

We extend the proof of [15, Theorem 10] in our case. But now $A^{\prime}$ is
not the completion of $A$, that is the coefficients of $y$ in $x$ are not
necessarily from $k$. Fortunately, as in the proof of [15, Theorem 10] we need
only a finite number of this coefficients, namely those of monomials $x$ which
are not in $d^{2e+1}A^{\prime}$. This is the reason to ask for the existence
of such $D,\omega$.

By [6, 19,7.1.5] for every field extension $L/k$ there exists a flat complete
Noetherian local $\bar{A}$-algebra $\tilde{A}$, unique up to an isomorphism,
such that ${\mathfrak{m}}\tilde{A}$ is the maximal ideal of $\tilde{A}$ and
$\tilde{A}/{\mathfrak{m}}\tilde{A}\cong L$. It follows that $\tilde{A}$ is
Artinian. On the other hand, we may consider the localization $A_{L}$ of
$L\otimes_{k}\bar{A}$ in ${\mathfrak{m}}(L\otimes_{k}\bar{A})$ which is
Artinian and so complete. By uniqueness we see that $A_{L}\cong\tilde{A}$. It
follows that $\bar{A}^{\prime}\cong A_{k^{\prime}}$. Note that $A_{L}$ is
essentially smooth over $A$ by base change and $\bar{A}^{\prime}$ is a
filtered union of sub-$\bar{A}$-algebras $A_{L}$ with $L/k$ finite type field
sub extensions of $k^{\prime}/k$.

Choose $L/k$ a finite type field extension such that $A_{L}$ contains the
residue class ${\bar{y}}\in\bar{A}^{\prime n}$ induced by $y$. In fact
${\bar{y}}$ is a vector of polynomials in the generators of ${\mathfrak{m}}$
with the coefficients $c_{\nu}$ in $k^{\prime}$ and we may take
$L=k((c_{\nu})_{\nu})$. Then $\bar{v}$ factors through $A_{L}$. Assume that
$k[(c_{\nu})_{\nu}]\cong k[(U_{\nu})_{\nu}]/\bar{J}$ for some new variables
$U$ and a prime ideal $\bar{J}\subset k[U]$. We have $H_{L/k}\not=0$ because
$L/k$ is separable. Then we may assume that there exist
$w=(w_{1},\ldots,w_{p})$ in ${\bar{J}}^{p}$ such that $\rho=\det(\partial
w_{i}/\partial U_{\nu})_{i,\nu\in[p]}\not=0$ and a nonzero polynomial
$\tau\in((w):\bar{J})\setminus\bar{J}$ (we set $[p]=\\{1,\ldots,p\\}$).
Actually, we may reduce to the case when $p=1$, but this means a complication
for our algorithm. Thus $L$ is a fraction ring of the smooth $k$-algebra
$(k[U]/(w))_{\rho\tau}$. Note that $w,\rho,\tau$ can be considered in $A[U]$
because $k\subset A$ and $c_{\nu}\in A^{\prime}$ because $k^{\prime}\subset
A^{\prime}$.

Then $\bar{v}$ factors through a smooth $\bar{A}$-algebra
$C\cong(\bar{A}[U]/(w))_{\rho\tau\gamma}$ for some polynomial $\gamma$ which
is not in ${\mathfrak{m}}(\bar{A}[U]/(w))_{\rho\tau}$.

######  Lemma 8.

There exists a smooth $A$-algebra $D$ such that $\bar{v}$ factors through
$\bar{D}=\bar{A}\otimes_{A}D$.

###### Proof..

By our assumptions $u(k)\subset k^{\prime}$. Set
$D=(A[U]/(w))_{\rho\tau\gamma}$ and $\omega:D\rightarrow A^{\prime}$ be the
map given by $U_{\nu}\rightarrow c_{\nu}$. We have
$C\cong{\bar{A}}\otimes_{A}D$. Certainly, $\bar{v}$ factors through
$\bar{\omega}=\bar{A}\otimes_{A}\omega$ but in general $v$ does not factor
through $\omega$.

It is worth recalling the following two remarks from [10].

######  Remark 9.

If $A^{\prime}=\hat{A}$ then $\bar{A}\cong\bar{A}^{\prime}$ and we may take
$D=A$.

######  Remark 10.

If $k\subset A$ but $L\not\subset A^{\prime}$ then
$D=(A[U,Z]/(w-d^{2e+1}Z))_{\rho\tau\gamma}$, $Z=(Z_{1},\ldots,Z_{p})$,
$U=(U_{1},\ldots,U_{q})$ is a smooth $A$-algebra and $\bar{D}\cong C[Z]$. Note
that $\bar{v}$ factors through a map $C\rightarrow\bar{A}^{\prime}$ given let
us say by $U\rightarrow\lambda+d^{2e+1}A^{\prime q}$ for some $\lambda$ in
$A^{\prime q}$. Thus $w(\lambda)=d^{2e+1}z$ for some $z$ in $A^{\prime p}$ and
we get a map $\omega:D\rightarrow A^{\prime}$, $(U,Z)\rightarrow(\lambda,z)$.
As above $\bar{v}$ factors through $\bar{\omega}=\bar{A}\otimes_{A}\omega$ but
in general $v$ does not factor through $\omega$. If also $k\not\subset A$ then
the construction of $D$ goes using a lifting of $w,\tau,\gamma$ from $k[U]$ to
$A[U]$. In both cases we may use $D$ as it follows.

Let $\delta:B\otimes_{A}D\cong D[Y]/ID[Y]\rightarrow A^{\prime}$ be the
$A$-morphism given by $b\otimes\lambda\rightarrow v(b)\omega(\lambda)$.

Step 7. $\delta$ factors through a special finite type $B\otimes_{A}D$-algebra
$E$.

Note that the map $\bar{B}\rightarrow\bar{D}$ is given by $Y\rightarrow
y^{\prime}+d^{2e+1}D$ for some $y^{\prime}\in D^{n}$. Thus
$I(y^{\prime})\equiv 0$ modulo $d^{2e+1}D$. Since $\bar{v}$ factors through
$\bar{\omega}$ we see that $\bar{\omega}(y^{\prime}+d^{2e+1}D)=\bar{y}$. Set
$\tilde{y}=\omega(y^{\prime})$. We get $y-\tilde{y}=v(Y)-\tilde{y}\in
d^{2e+1}A^{\prime n}$, let us say $y-\tilde{y}=d^{e+1}\varepsilon$ for
$\varepsilon\in d^{e}A^{\prime n}$.

Recall that $P=R\det H$ for $R\in((f):I)$ (see Step 5). We have $d\equiv P$
modulo $I$ and so $P(y^{\prime})\equiv d$ modulo $d^{2e+1}$ in $D$ because
$I(y^{\prime})\equiv 0$ modulo $d^{2e+1}D$. Thus $P(y^{\prime})=ds$ for a
certain $s\in D$ with $s\equiv 1$ modulo $d$. Let $G^{\prime}$ be the adjoint
matrix of $H$ and $G=RG^{\prime}$. We have $GH=HG=P\mbox{Id}_{n}$ and so

$ds\mbox{Id}_{n}=P(y^{\prime})\mbox{Id}_{n}=G(y^{\prime})H(y^{\prime}).$

But $H$ is the matrix $(\partial f_{i}/\partial Y_{j})_{i\in[r],j\in[n]}$
completed with some $(n-r)$ rows from $k^{n}$. Especially we obtain

$(\partial f/\partial Y)G=(P\mbox{Id}_{r}|0).$ (1)

Then $t:=H(y^{\prime})\varepsilon\in d^{e}A^{\prime n}$ satisfies

$G(y^{\prime})t=P(y^{\prime})\varepsilon=ds\varepsilon$

and so

$s(y-\tilde{y})=d^{e}\omega(G(y^{\prime}))t.$

Let

$h=s(Y-y^{\prime})-d^{e}G(y^{\prime})T,$ (2)

where $T=(T_{1},\ldots,T_{n})$ are new variables. The kernel of the map
$\varphi:D[Y,T]\rightarrow A^{\prime}$ given by $Y\rightarrow y$,
$T\rightarrow t$ contains $h$. Since

$s(Y-y^{\prime})\equiv d^{e}G(y^{\prime})T\ \mbox{modulo}\ h$

and

$f(Y)-f(y^{\prime})\equiv\sum_{j}\displaystyle\frac{\partial f}{\partial
Y_{j}}(y^{\prime})(Y_{j}-y^{\prime}_{j})$

modulo higher order terms in $Y_{j}-y^{\prime}_{j}$, by Taylor’s formula we
see that for $p=\operatorname{max}_{i}\operatorname{deg}f_{i}$ we have

$s^{p}f(Y)-s^{p}f(y^{\prime})\equiv\sum_{j}s^{p-1}d^{e}\displaystyle\frac{\partial
f}{\partial Y_{j}}(y^{\prime})G_{j}(y^{\prime})T_{j}+d^{2e}Q$ (3)

modulo $h$ where $Q\in T^{2}D[T]^{r}$. We have $f(y^{\prime})=d^{e+1}b$ for
some $b\in d^{e}D^{r}$. Then

$g_{i}=s^{p}b_{i}+s^{p}T_{i}+d^{e-1}Q_{i},\qquad i\in[r]$ (4)

is in the kernel of $\varphi$. Indeed, we have $s^{p}f_{i}=d^{e+1}g_{i}\
\mbox{modulo}\ h$ because of (1) and $P(y^{\prime})=ds$. Thus
$d^{e+1}\varphi(g)=d^{e+1}g(t)\in(h(y,t),f(y))=(0)$. Since $Q\in
T^{2}D[T]^{r}$ and $t\in d^{e}A^{\prime n}$ we get $g(t)\in d^{e}A^{\prime r}$
and so $g(t)\in(0:_{A^{\prime}}d^{e+1})\cap d^{e}A^{\prime}=0$, because $u$ is
flat and $(0:_{A^{\prime}}d^{e})=(0:_{A^{\prime}}d^{e+1})$. Set
$E=D[Y,T]/(I,g,h)$ and let $\psi:E\rightarrow A^{\prime}$ be the map induced
by $\varphi$. Clearly, $v$ factors through $\psi$ because $v$ is the composed
map $B\rightarrow B\otimes_{A}D\cong D[Y]/I\rightarrow
E\xrightarrow{\psi}A^{\prime}$.

Step 8. There exist $s^{\prime},s^{\prime\prime}\in E$ such that
$E_{ss^{\prime}s^{\prime\prime}}$ is smooth over $A$ and $\psi$ factors
through $E_{ss^{\prime}s^{\prime\prime}}$.

Note that the $r\times r$-minor $s^{\prime}$ of $(\partial g/\partial T)$
given by the first $r$-variables $T$ is from $s^{rp}+(T)\subset 1+(d,T)$
because $Q\in(T)^{2}$. Then $V=(D[Y,T]/(h,g))_{ss^{\prime}}$ is smooth over
$D$. We claim that $I\subset(h,g)D[Y,T]_{ss^{\prime}s^{\prime\prime}}$ for
some other $s^{\prime\prime}\in 1+(d,T)D[Y,T]$. Indeed, we have
$PI\subset(h,g)D[Y,T]_{s}$ and so
$P(y^{\prime}+s^{-1}d^{e}G(y^{\prime})T)I\subset(h,g)D[Y,T]_{s}$. Since
$P(y^{\prime}+s^{-1}d^{e}G(y^{\prime})T)\in P(y^{\prime})+d^{e}(T)$ we get
$P(y^{\prime}+s^{-1}d^{e}G(y^{\prime})T)=ds^{\prime\prime}$ for some
$s^{\prime\prime}\in 1+(d,T)D[Y,T]$. It follows that
$s^{\prime\prime}I\subset((h,g):d)D[Y,T]_{ss^{\prime}}$. On the other hand,
$I\equiv I(y^{\prime})\ \mbox{modulo}\ (d^{e},h)D[Y,T]$ and
$I(y^{\prime})\subset d^{2e+1}D$. Thus $s^{\prime\prime}I\subset(0:_{V}d)\cap
d^{e}V=0$ because $(0:_{A}d)\cap d^{e}A=0$ and the maps $A\rightarrow D$,
$D\rightarrow V$ are flat, which shows our claim. It follows that
$I\subset(h,g)D[Y,T]_{ss^{\prime}s^{\prime\prime}}$. Thus
$E_{ss^{\prime}s^{\prime\prime}}\cong V_{s^{\prime\prime}}$ is a $B$-algebra
which is also standard smooth over $D$ and $A$.

As $\omega(s)\equiv 1$ modulo $d$ and
$\psi(s^{\prime}),\psi(s^{\prime\prime})\equiv 1$ modulo $(d,t)$,
$d,t\in{\mathfrak{m}}A^{\prime}$ we see that
$\omega(s),\psi(s^{\prime}),\psi(s^{\prime\prime})$ are invertible because
$A^{\prime}$ is local and $\psi$ (thus $v$) factors through the standard
smooth $A$-algebra $E_{ss^{\prime}s^{\prime\prime}}$.

######  Remark 11.

When $A^{\prime}$ is the completion of $A$ then the algorithmic proof is much
easier (one reason is given by Remark 9) and it is somehow similar to the
proof of Theorem 14. Certainly, in this case the next algorithm could be
substantially easier.

######  Remark 12.

If we want to study the case when $\dim A=2$ then we need first to treat the
case when $\dim A=0$, $\dim A^{\prime}=1$ and after that the case when $A,\
A^{\prime}$ are one dimensional but not local rings. We expect that an
algorithmic proof in the last case is a hard goal. However, if we are lucky to
get such proof it is doubtful that the corresponding algorithm will really
work.

##  4 The algorithm

We obtain the following algorithm:

0: $N\in{\mathbb{Z}}_{>0}$ a
bound$A:=k[x]_{(x)}/J,J=(h_{1},\ldots,h_{g})\subseteq
k[x],x=(x_{1},\ldots,x_{t}),k$ an infinite
field$k^{\prime}:=Q(k[U]/\overline{J}),\overline{J}=(a_{1},\ldots,a_{r})\subseteq
k[U],U=(U_{1},\ldots,U_{w})$ separable over
$k$$B:=A[Y]/I,I=(g_{1},\ldots,g_{l})\subseteq
k[x,Y],Y=(Y_{1},\ldots,Y_{n})$$v:B\rightarrow A^{\prime}\subseteq
K[[x]]/JK[[x]]$ an $A$–morphism, given by
$y^{\prime}_{1},\ldots,y^{\prime}_{n}\in k[U,x]$, approximations
$\operatorname{mod}(x)^{N}$ of $v(Y_{i}),K\supseteq k^{\prime}$ a field.

0: A Neron desingularization of $v:B\rightarrow A^{\prime}$ or the message
”the algorithm fails since the bound $N$ is too small”

1: Compute $P_{1},\ldots,P_{s}$ the minimal associated primes of $A$.

2: Compute $w=(a_{i_{1}},\ldots,a_{i_{p}})$ and a $p$–minor $\rho$ of
$\left(\frac{\partial a_{i_{v}}}{\partial U_{j}}\right)$ such that
$\rho\not\in\overline{J}$.Compute $\tau\in(w):\overline{J}$ such that
$k[U]_{\rho\tau}/(a_{1},\ldots,a_{r})=k[U]_{\rho\tau}/(w)$.$D:=A[U]_{\rho\tau}/(w)$.

3: Compute $H_{B/A}$ and $H_{B/A}\cap A$.

4: If dim$(A/H_{B/A}\cap A)=1$$B:=S_{B}(I/I^{2})$, $v$ trivially extended,
write $B=A[Y]/I$, $n:=\\#Y$$Y:=Y,Z$, $I:=(I,Z)$, $B:=A[Y]/I,\ Z=(Z_{1}\ldots
Z_{n})$ , $v$ trivially extended

5: Compute $f=(f_{1},\ldots,f_{r})$ in $I$ such that
$v(((f):I)\Delta_{f})\not\subseteq P_{i}+(x)^{N}$ for all $i$

6: Complete $\left(\frac{\partial f_{i}}{\partial Y_{j}}\right)$ by random
vectors from $k^{n}$ to obtain a square matrix $H$ with $v(\det(H))\not\in
P_{i}+(x)^{N}$ for all $i$

7: Compute $R\in(f):I$ such that $v(R)\not\in P_{i}+(x)^{N}$ for all $i$.

8: $P:=R\cdot\det(H)$ , compute $(P)\cap A$

9: If dim$(A/(P)\cap A)=1$Compute $d$ in $v(P)\cap A$ an active element.
$f_{r+1}:=PY_{n+1}-d$,$B:=B[Y_{n+1}]/f_{r+1}$ $f:=f,f_{r+1}\ ,\
Y:=Y,Y_{n+1},y^{\prime}:=y^{\prime},\frac{d}{v(P)}\ ,\ n:=n+1$, r:=r+1
$H:=\begin{pmatrix}H&0\\\ \ast&P\end{pmatrix}$ $d:=d^{2}$,$R:=RY^{2}_{n}$,
$P:=R\cdot\det(H)$

10: ElseCompute $d$ in $((P)\cap A)$ an active element, $P:=d$.

11: Compute $e$ such that $(0:d^{e})=(0:d^{e+1})$

12: If $(x)^{N}\not\subseteq(d^{2e+1})$ return ”the algorithm fails since the
bound is too small”

13: Compute $b:=\frac{f(y^{\prime})}{d^{e+1}}$ in $d^{e}D^{r}$

14: Compute $G^{\prime}$ the adjoint matrix of $H\ ,\ G:=RG^{\prime}$

15: Compute $s\in D$ such that
$P(y^{\prime})=ds$$h:=s(Y-y^{\prime})-d^{e}G(y^{\prime})T\ ,\
T=(T_{1},\ldots,T_{n})$

16: $p:=\operatorname{max}\\{\operatorname{deg}(f_{i})\\}_{i=1,\ldots,r}$write
$s^{p}f(Y)-s^{p}f(y^{\prime})=\sum\limits_{j}s^{p-1}d^{e}\frac{\partial
f}{\partial Y_{j}}(y^{\prime})G_{j}(y^{\prime})T_{j}+d^{2e}Q$define
$g_{i}:=s^{p}b_{i}+s^{p}T_{j}+d^{e-1}Q_{i}$ and $E:=D[Y,T]/(I,g,h)$

17: Compute $s^{\prime}$ the $r$–minor of $\left(\frac{\partial g}{\partial
T}\right)$ given by the first $r$ variables of $T$

18: find $s^{\prime\prime}\in 1+(d,T)_{D[Y,T]}$ such that
$I\subset(h,g)D[Y,T]_{ss^{\prime}s^{\prime\prime}}$($s^{\prime\prime}$ is
given by $P(y^{\prime}+s^{-1}d^{e}G(y^{\prime})T)=ds^{\prime\prime})$

19: return $E_{ss^{\prime}s^{\prime\prime}}$

Algorithm 1 Neron Desingularization

######  Example 13.

We assume

$N=12$,

$A={\bf Q}[x_{1},x_{2}]_{(x_{1},x_{2})}/(x_{1}x_{2})$, $J=(x_{1},x_{2})$,

$k^{\prime}={\bf Q}$,

$B=A[Y_{1},Y_{2}]/(x_{2}Y_{1}-x_{1}Y_{2})$, $I=(x_{2}Y_{1},x_{1}Y_{2})$,

$v:B\rightarrow{\bf Q}[[x_{1},x_{2}]]/(x_{1}x_{2})$, $v(Y_{1})=x_{1}u$,
$v(Y_{2})=x_{2}w$,

$u,w\in{\bf Q}[[x_{1},x_{2}]]$ algebraically independent,

$y^{\prime}_{1}=\mbox{jet}(x_{1}u,11)$,
$y^{\prime}_{2}=\mbox{jet}(x_{2}w,11)$.

Now we follow the steps of the algorithm.

1\. $P_{1}=(x_{1})$, $P_{2}=(x_{2})$,

2\. $D=A$,

3\. $H_{B/A}=(x_{1},x_{2})B$, $H_{B/A}\cap A=(x_{1},x_{2})$,

4\. is not true,

5\. $f=x_{2}Y_{1}+x_{1}Y_{2}$,

6\. $H:=\begin{pmatrix}x_{2}&x_{1}\\\ -1&1\end{pmatrix}$

7\. $R=x_{1}+x_{2}$,

8\. $P=(x_{1}+x_{2})^{2}$, $(P)\cap A=(x_{1}+x_{2})^{2}$,

9\. is not true,

10\. $P=d=(x_{1}+x_{2})^{2}$,

11\. $e=1$,

12\. is not true $(x_{1},x_{2})^{12}\subset(x_{1}x_{2},(x_{1},x_{2})^{6})$,

13\. $f(y^{\prime})=0$, $b=0$,

14\. $G^{\prime}:=\begin{pmatrix}1&-x_{1}\\\ 1&x_{2}\end{pmatrix}$,
$G=(x_{1}+x_{2})\begin{pmatrix}1&-x_{1}\\\ 1&x_{2}\end{pmatrix}$,

15\. $s=1$, $h=Y-y^{\prime}-d(x_{1}+x_{2})\begin{pmatrix}1&-x_{1}\\\
1&x_{2}\end{pmatrix}\begin{pmatrix}T_{1}\\\ T_{2}\end{pmatrix}$,

16\. $p=1$,
$x_{2}(Y_{1}-y^{\prime}_{1})+x_{1}(Y_{2}-y^{\prime}_{2})=d^{2}T_{1}\
\mbox{mod}h$,

$g=T_{1}$, $E=A[Y,T]/(I,g,h)=A[Y,T]/(g,h)$,

17\. $s^{\prime}=1$,

18\. $s^{\prime\prime}=1$,

19\. return
$A[Y_{1},Y_{2},T_{2}]/(Y_{1}-y^{\prime}_{1}-x_{1}^{4}T_{2},Y_{2}-y^{\prime}_{2}-x_{2}^{4}T_{2})$.

Thus the General Neron Desingularization $B^{\prime}$ of $B$ could be
$A[T_{2}]$. Let $\tau:B\rightarrow B^{\prime}$ be the map given
$Y_{i}\rightarrow y^{\prime}_{i}+x_{i}^{4}T_{2}$, $i=1,2$. We may find
$\rho_{i}\in(x_{i})Q[[x_{i}]]$ such that
$x_{1}u-y^{\prime}_{1}=x_{1}^{11}\rho_{1}$ and
$x_{2}w-y^{\prime}_{2}=x_{2}^{11}\rho_{2}$. Set
$t_{2}=d^{4}(\rho_{1}+\rho_{2})$ and let $\tau^{\prime}:B^{\prime}\rightarrow
A^{\prime}$ be given by $T_{2}\rightarrow t_{2}$. Clearly,
$v=\tau^{\prime}\tau$, that is $v$ factors through $B^{\prime}$.

##  5 An extension of Greenberg’s theorem on strong approximation

Let $(A,{\mathfrak{m}})$ be a Noetherian local ring of dimension one,
$A^{\prime}={\hat{A}}$ the completion of $A$, $B=A[Y]/I$,
$Y=(Y_{1},\ldots,Y_{n})$ an $A$-algebra of finite type and $c\in\bf N$. If $A$
is Henselian excellent DVR then Greenberg [5] showed that there exists a
linear map $\nu:{\bf N}\rightarrow{\bf N}$ such that for each $A$-morphism
$v:B\rightarrow A/{\mathfrak{m}}^{\nu(c)}$ there exists an $A$-morphism
$v^{\prime}:B\rightarrow A$ such that $v^{\prime}\equiv v$ modulo
${\mathfrak{m}}^{c}$. More general, if $A$ is a DVR then there exists a linear
map $\nu:{\bf N}\rightarrow{\bf N}$ such that for each $A$-morphism
$v:B\rightarrow A/{\mathfrak{m}}^{\nu(c)}$ there exists an $A$-morphism
$v^{\prime}:B\rightarrow A^{\prime}$ such that $v^{\prime}\equiv v$ modulo
${\mathfrak{m}}^{c}A^{\prime}$.

The aim of this section is to give a result of Greenberg’s type for one
dimensional rings when the Jacobian locus is not too small. Let $e$ be as in
the beginning of Section 3 and $\rho\in{\bf N}$. We will show that the map
$\nu$ given by $c\rightarrow(e+1)(\rho+1)+c$ works in the special case below.
Suppose that there exists an $A$-morphism $v:B\rightarrow
A/{\mathfrak{m}}^{(e+1)(\rho+1)+c}$ such that
$(v(((f):I)\Delta_{f}))\supset{\mathfrak{m}}^{\rho}/{\mathfrak{m}}^{(e+1)(\rho+1)+c}$
for some $f=(f_{1},\ldots,f_{r})$, $r\leq n$ in $I$.

######  Theorem 14.

Then there exists an $A$-morphism $v^{\prime}:B\rightarrow{\hat{A}}$ such that
$v^{\prime}\equiv v\ \mbox{modulo}\ {\mathfrak{m}}^{c}$, that is
$v^{\prime}(Y+I)\equiv v(Y+I)\ \mbox{modulo}\ {\mathfrak{m}}^{c}$.

###### Proof..

We note that the proof of Theorem 2 can work somehow in this case. Let
$y^{\prime}\in A^{n}$ be an element inducing $v(Y+I)$. Then
${\mathfrak{m}}^{\rho}\subset(((f):I)\Delta_{f})(y^{\prime}))+{\mathfrak{m}}^{(e+1)(\rho+1)+c}\subset((((f):I)\Delta_{f})(y^{\prime}))+{\mathfrak{m}}^{(e+1)(\rho+2)+2c}\subset\ldots$
by hypothesis. It follows that
${\mathfrak{m}}^{\rho}\subset((((f):I)\Delta_{f})(y^{\prime}))$.

As in Step 4 of Section 1 we may complete the Jacobian matrix $(\partial
f/\partial Y)$ with $n-r$ rows from $k^{n}$ obtaining a square $n$ matrix $H$
with $v(\det H)\not\in\cup_{p\in\operatorname{Min}A^{\prime}}p$

Set $d=(\det H)(y^{\prime})$. Next we follow the proof of Theorem 2 with
$D=A$, $s=1$, $P=L\det H$ and $G$ such that

$GH=HG=P\mbox{Id}_{n}$

and so

$d\mbox{Id}_{n}=P(y^{\prime})\mbox{Id}_{n}=G(y^{\prime})H(y^{\prime}).$

Let

$h=Y-y^{\prime}-d^{e}G(y^{\prime})T,$

where $T=(T_{1},\ldots,T_{n})$ are new variables. We have

$f(Y)-f(y^{\prime})\equiv d^{e}P(y^{\prime})T+d^{2e}Q$

modulo $h$ where $Q\in T^{2}A[T]^{r}$. But
$f(y^{\prime})\in{\mathfrak{m}}^{(e+1)(\rho+1)+c}A^{r}\subset
d^{2}{\mathfrak{m}}^{c+e-1}A^{r}$ and we get $f(y^{\prime})=d^{2}b$ for some
$b\in{\mathfrak{m}}^{c}A^{r}$. Set $g_{i}=b_{i}+T_{i}+d^{e-1}Q_{i}$, $i\in[r]$
and $E=A[Y,T]/(I,h,g)$. We have an $A$-morphism $\beta:E\rightarrow
A/{\mathfrak{m}}^{c}$ given by $(Y,T)\rightarrow(y^{\prime},0)$ because
$I(y^{\prime})\equiv 0\ \mbox{modulo}\ {\mathfrak{m}}^{c}$,
$h(y^{\prime},0)=0$ and $g(0)=b\in{\mathfrak{m}}^{c}A^{r}$.

As in the proof of Theorem 2 we see that
$I\subset(h,g)A[Y,T]_{s^{\prime}s^{\prime\prime}}$ for some
$s^{\prime\prime}\in 1+(d,T)A[Y,T]$. Indeed, we have $PI\subset(h,g)A[Y,T]$
and so $P(y^{\prime}+d^{e}G(y^{\prime})T)I\subset(h,g)A[Y,T]$. Since
$P(y^{\prime}+d^{e}G(y^{\prime})T)\in P(y^{\prime})+d^{e}(T)$ we get
$P(y^{\prime}+d^{e}G(y^{\prime})T)=ds^{\prime\prime}$ for some
$s^{\prime\prime}\in 1+(d,T)A[Y,T]$. It follows that
$s^{\prime\prime}I\subset((h,g):d)A[Y,T]_{s^{\prime}}$. On the other hand,
$I\equiv I(y^{\prime})\ \mbox{modulo}\ (d^{e},h)A[Y,T]$ and
$I(y^{\prime})\subset{\mathfrak{m}}^{e\rho}\subset d^{e}A$. Set
$V=(A[Y,T]/(h,g))_{s^{\prime}}$. Then $s^{\prime\prime}I\subset(0:_{V}d)\cap
d^{e}V=0$ because $(0:_{A}d)\cap d^{e}A=0$ and the map $A\rightarrow V$ is
flat, which shows our claim.

Then $E_{s^{\prime}s^{\prime\prime}}\cong V_{s^{\prime\prime}}$ is a
$B$-algebra which is also standard smooth over $A$ and $\beta$ extends to a
map $\beta^{\prime}:E_{s^{\prime}s^{\prime\prime}}\rightarrow
A/{\mathfrak{m}}^{c}$ as in the proof of Theorem 2. By the Implicit Function
Theorem $\beta^{\prime}$ can be lifted to a map
$w:E_{s^{\prime}s^{\prime\prime}}\rightarrow\hat{A}$ which coincides with
$\beta^{\prime}$ modulo ${\mathfrak{m}}^{c}$. It follows that the composite
map $v^{\prime}$, $B\rightarrow
E_{s^{\prime}s^{\prime\prime}}\xrightarrow{w}\hat{A}$ works.

######  Corollary 15.

In the assumptions of the above theorem, suppose that $(A,{\mathfrak{m}})$ is
excellent Henselian. Then there exists an $A$-morphism
$v^{\prime\prime}:B\rightarrow A$ such that $v^{\prime\prime}\equiv v\
\mbox{modulo}\ {\mathfrak{m}}^{c}$, that is $v^{\prime\prime}(Y+I)\equiv
v(Y+I)\ \mbox{modulo}\ {\mathfrak{m}}^{c}$.

###### Proof..

An excellent Henselian local ring $(A,{\mathfrak{m}})$ has the Artin
approximation property by [12], that is the solutions in $A$ of a system of
polynomial equations $f$ over $A$ are dense in the set of the solutions of $f$
in $\hat{A}$. By Theorem 14 we get an $A$-morphism
$v^{\prime}:B\rightarrow\hat{A}$ such that $v^{\prime}\equiv v\ \mbox{modulo}\
{\mathfrak{m}}^{c}$. Then there exists an $A$-morphism
$v^{\prime\prime}:B\rightarrow A$ such that $v^{\prime\prime}\equiv
v^{\prime}\equiv v\ \mbox{modulo}\ {\mathfrak{m}}^{c}$ by the Artin
approximation property.

######  Remark 16.

If $\dim A>2$ then [17, Remark 4.7] gives an example when there exist no
linear map $\nu$ as above. If $A=k[[x_{1},x_{2}]]$ then Theorem 14 and the
above corollary hold as [17, Theorem 4.3] shows.

######  Theorem 17.

Let $(A,{\mathfrak{m}})$ be a Noetherian local ring of dimension one, $e\in\bf
N$ as in the beginning of Section 2, $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$ an
$A$-algebra of finite type, $\rho\in\bf N$ and $f=(f_{1},\ldots,f_{r})$ a
system of polynomials from $I$. Suppose that $A$ is excellent Henselian and
there exists $y^{\prime}\in A^{n}$ such that $I(y^{\prime})\equiv 0\
\mbox{modulo}\ {\mathfrak{m}}^{\rho}$ and
$((((f):I)\Delta_{f}))(y^{\prime})\supset{\mathfrak{m}}^{\rho}$. Then the
following statements are equivalent:

  1. 1.

there exists $y^{\prime\prime}\in A^{n}$ such that $I(y^{\prime\prime})\equiv
0\ \mbox{modulo}\ {\mathfrak{m}}^{(e+2)(\rho+1)}$ and $y^{\prime\prime}\equiv
y^{\prime}\ \mbox{modulo}\ {\mathfrak{m}}^{\rho}$,

  2. 2.

there exists $y\in A^{n}$ such that $I(y)=0$ and $y\equiv y^{\prime}\
\mbox{modulo}\ {\mathfrak{m}}^{\rho}$.

For the proof apply the above corollary and Theorem 14.

## References

  * [1] M. André, Cinq exposés sur la désingularisation, Handwritten manuscript Ecole Polytechnique Fédérale de Lausanne, (1991).
  * [2] M. Artin, Algebraic approximation of structures over complete local rings, Publ. Math. IHES, 36, (1969), 23-58.
  * [3] W. Decker, G.-M. Greuel, G. Pfister, H. Schönemann:  Singular 4-0-2 — A computer algebra system for polynomial computations.  http://www.singular.uni-kl.de (2015).
  * [4] R. Elkik, Solutions d’équations à coéfficients dans un anneaux hensélien, Ann. Sci. Ecole Normale Sup., 6 (1973), 553-604.
  * [5] M. Greenberg, Rational points in henselian discrete valuation rings, Publ. Math. IHES, 31, (1966), 59-64.
  * [6] A. Grothendieck, J. Dieudonné, Eléments de géometrie algébrique, IV, Part 1, Publ. Math. IHES, 1966.
  * [7] T. De Jong, G. Pfister, Local analytic Geometry, Vieweg, (2000).
  * [8] E. Kunz, Introduction to commutative algebra and algebraic geometry, Birkhäuser, Basel, 1985.
  * [9] A. Néron, Modèles minimaux des varietés abéliennes sur les corps locaux et globaux, Publ. Math. IHES, 21, 1964.
  * [10] A. Popescu, D. Popescu, A method to compute the General Neron Desingularization in the frame of one dimensional local domains, to appear in ”Singularities and Computer Algebra - Festschrift for Gert-Martin Greuel, On the Occasion of his 70th Birthday”, Editors Wolfram Decker, Gerhard Pfister, Mathias Schulze, Springer Monograph., arXiv:AC/1508.05511. Journal of Algebra and Its Applications, 16, No. 2 (2017), arXiv:1504.06938.
  * [11] D. Popescu, General Neron Desingularization, Nagoya Math. J., 100 (1985), 97-126.
  * [12] D. Popescu, General Neron Desingularization and approximation, Nagoya Math. J., 104, (1986), 85-115.
  * [13] D. Popescu, Letter to the Editor. General Neron Desingularization and approximation, Nagoya Math. J., 118 (1990), 45-53.
  * [14] D. Popescu, Artin Approximation, in ”Handbook of Algebra”, vol. 2, Ed. M. Hazewinkel, Elsevier, 2000, 321-355.
  * [15] D. Popescu, Around General Neron Desingularization, Journal of Algebra and Its Applications, 16, No. 2 (2017), arXiv:1504.06938.
  * [16] D. Popescu, Artin approximation property and the General Neron Desingularization, to appear in Revue Roum. Math. Pures et Appl., 2017, arXiv:AC/1511.06967.
  * [17] G. Rond, Lojasiewicz inequality over the ring of power series in two variables, Math. Res. Lett., 20, (2013), 1-13.
  * [18] M. Spivakovsky, A new proof of D. Popescu’s theorem on smoothing of ring homomorphisms, J. Amer. Math. Soc., 294, (1999), 381-444.
  * [19] R. Swan, Neron-Popescu desingularization, in ”Algebra and Geometry”, Ed. M. Kang, International Press, Cambridge, (1998), 135-192.