# A method to compute the General Neron Desingularization in the frame of one
dimensional local domains

Adrian Popescu, Dorin Popescu  Adrian Popescu, Department of Mathematics,
University of Kaiserslautern, Erwin-Schrödinger-Str., 67663 Kaiserslautern,
Germany<EMAIL_ADDRESS>Dorin Popescu, Simion Stoilow Institute
of Mathematics , 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 domains and it is implemented in Singular. Also a theorem
recalling Greenberg’ strong approximation theorem is presented for one
dimensional Cohen-Macaulay local rings.

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

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.
Both authors thank CIRM, Luminy who provided excellent conditions and
stimulative atmosphere in the main stage of our work.

## 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.

In Artin approximation theory [2] an important result is the following theorem
generalizing the Neron Desingularization [7], [2].

######  Theorem 1 (General Neron Desingularization, Popescu [8], [9], [10],
Andre [1], Swan [13], Spivakovski [12]).

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 domains and
$A\supset\mathbb{Q}$. This proof is somehow presented by the second author in
a lecture given in the frame of a semester of Artin Approximation and
Singularity Theory organized in 2015 by CIRM in Luminy (see
http://hlombardi.free.fr/Popescu-Luminy2015.pdf). The 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 as the library
`GND.lib` at

    https://github.com/Singular/Source.

We may take the same General Neron Desingularization for
$v,v^{\prime}:B\rightarrow A^{\prime}$ if they are closed enough as Examples
4, 10 show. The last section computes the General Neron Desingularization in
several examples. We should point that the General Neron Desingularization is
not unique and it is better to speak above about a General Neron
Desingularization.

When $A^{\prime}$ is the completion of a Cohen-Macaulay 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]). More precisely, the Artin function is given
by $c\rightarrow 2e+c$, where $e$ depends from the polynomial system of
equations defining $B$ (see Theorem 20).

##  1\. The theorem

Let $u:A\rightarrow A^{\prime}$ be a flat morphism of Noetherian local domains
of dimension $1$. Suppose that $A\supset\mathbb{Q}$ and the maximal ideal $m$
of $A$ generates the maximal ideal of $A^{\prime}$. Then $u$ is regular
morphism. Moreover, we suppose that there exist canonical inclusions
$k=A/m\rightarrow A$, $k^{\prime}=A^{\prime}/mA^{\prime}\rightarrow
A^{\prime}$ 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
$\left(\displaystyle\frac{\partial f_{i}}{\partial Y_{j}}\right)$. After Elkik
[4] let $H_{B/A}$ be the radical of the ideal
$\sum_{f}\big{(}(f):I\big{)}\Delta_{f}B$, 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 $A$.

######  Definition 2.

$B$ is standard smooth over $A$ if there exists $f$ in $I$ as above such that
$1\in\big{(}(f):I\big{)}\Delta_{f}B$.

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

######  Theorem 3.

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

If $A$ is essentially of finite type over $\mathbb{Q}$, then the ideal
$H_{B/A}$ can be computed in Singular by following its definition but it is
easier to describe only the ideal $\sum_{f}\big{(}(f):I\big{)}\Delta_{f}B$
defined above. This is the case considered in our algorithmic part, let us say
$A\cong k[x]/F$ for some variables $x=(x_{1},\ldots x_{t})$, and the
completion of $A^{\prime}$ is $K\llbracket x\rrbracket/(F)$ for some field
extension $k\subset K$. When $v$ is defined by polynomials $y$ from $K[x]$
then our problem is easy. Let $L$ be the field obtained by adjoining to $k$
all coefficients of $y$. Then $R=L[x]/(F)$ 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 $B^{\prime}$. Consequently we suppose usually that $y$ is not
polynomial defined and moreover $L$ is not a finite type field extension of
$k$.

##  2\. Reduction to the case when $H_{B/A}\cap A\not=0$.

We may suppose that $v(H_{B/A})\not=0$. Indeed, if $v(H_{B/A})=0$ then $v$
induces an $A$-morphism $v^{\prime}:B^{\prime}=B/H_{B/A}\rightarrow
A^{\prime}$ and we may change $(B,v)$ by $(B^{\prime},v^{\prime})$. Applying
this trick several times we reduce to the case $v(H_{B/A})\not=0$. However the
fraction field of $\operatorname{Im}v$ is essentially smooth over $A$ by
separability, that is $H_{\operatorname{Im}v/A}A^{\prime}\not=0$ and in the
worst case our trick will change $B$ by $\operatorname{Im}v$ after several
steps.

Choose $P^{\prime}\in\big{(}\Delta_{f}((f):I)\big{)}\setminus I$ for some
system of polynomials $f=(f_{1},\ldots,f_{r})$ from $I$ and
$d^{\prime}\in\big{(}v(P^{\prime})A^{\prime}\big{)}\cap A$,
$d^{\prime}\not=0$. Moreover we may choose $P^{\prime}$ to be from
$M\big{(}(f):I\big{)}$ where $M$ is a $r\times r$-minor of
$\left(\displaystyle\frac{\partial f}{\partial Y}\right)$ . Then
$d^{\prime}=v(P^{\prime})z\in\big{(}v(H_{B/A})\big{)}\cap A$ 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
$d^{\prime}\in\big{(}(f,f_{r+1}):(I,f_{r+1})\big{)}$ 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 the Jacobian matrix
$J=(\partial f/\partial Y)$ will be now the new $J$ given by
$\left(\begin{array}[]{cc}J&0\\\ &P^{\prime}\end{array}\right).$ Note that
$d\in H_{B/A}\cap A$.

######  Example 4.

Let $a_{1},a_{2}\in\mathbb{C}$ be two elements algebraically independent over
$\mathbb{Q}$ and $\rho$ a root of the polynomial $T^{2}+T+1$ in $\mathbb{C}$.
Then
$k^{\prime}=\displaystyle\frac{\mathbb{Q}(a_{1},a_{2})[a_{3}]}{(a_{3}^{2}+a_{3}+1)}\cong\mathbb{Q}(\rho,a_{1},a_{2})$.
Let
$A=\left(\displaystyle\frac{\mathbb{Q}[x_{1},x_{2}]}{(x_{1}^{3}-x_{2}^{2})}\right)_{(x_{1},x_{2})}$
and $B=\displaystyle\frac{A[Y_{1},Y_{2},Y_{3}]}{(Y_{1}^{3}-Y_{2}^{3})}$,
$A^{\prime}=\displaystyle\frac{k^{\prime}\llbracket
x_{1},x_{2}\rrbracket}{(x_{1}^{3}-x_{2}^{2})}$ and the map $v$ defined as

$\textstyle{v:}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A^{\prime}}$$\textstyle{Y_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}x_{2}}$$\textstyle{Y_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}a_{3}x_{2}}$$\textstyle{Y_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}\displaystyle\sum_{i=0}^{30}\displaystyle\frac{x_{1}^{i}}{i!}+a_{2}x_{2}\displaystyle\sum_{i=31}^{50}\displaystyle\frac{x_{1}^{i}}{i!}}$

This is an easy example. Indeed, let
$v^{\prime\prime}:B^{\prime\prime}=A[a_{3},a_{1}x_{2},v(Y_{3})]\rightarrow
A^{\prime}$ be the inclusion. We have $\operatorname{Im}v\subset
B^{\prime\prime}\cong\displaystyle\frac{A[T,Y_{1},Y_{3}]}{(T^{2}+T+1)}$ and
$B^{\prime\prime}_{2a_{3}+1}\cong\left(\displaystyle\frac{A[T,Y_{1},Y_{3}]}{(T^{2}+T+1)}\right)_{2T+1}$
is a smooth $A$-algebra, which could be taken as a General Neron
Desingularization of $B$. Applying our algorithm we will get more complicated
General Neron Desingularizations but useful for an illustration of our
construction.

Then $\operatorname{Im}v$, the new $B$ will be
$\displaystyle\frac{B}{\operatorname{Ker}v}$, where the kernel is generated by
the following polynomial:

    ker[1]=Y1^2+Y1*Y2+Y2^2

Next we choose $f=Y_{1}^{2}+Y_{1}Y_{2}+Y_{2}^{2}$ and we have $M=2Y_{2}+Y_{1}$
and $1\in\big{(}(f):I\big{)}$ and hence $P^{\prime}=Y_{1}+2Y_{2}$. Therefore
$v(P^{\prime})=(2a_{1}a_{3}+a_{1})\cdot x_{2}$ and $d^{\prime}=x_{2}$,
$z=\displaystyle\frac{1}{2a_{1}a_{3}+a_{1}}$. Therefore $d=d^{\prime
2}=x_{2}^{2}$.

To be able to construct
$\mathbb{Q}\left[\displaystyle\frac{1}{2a_{1}a_{3}+a_{1}}\right][x]$ in
Singular we will add a new variable $a$ and we will factorize with the
corresponding polynomial $2a_{1}a_{3}\cdot a+a_{1}\cdot a-1$. We will see this
$a$ as a new parameter from $k^{\prime}\subset A^{\prime}$. Then we change $B$
by $B_{1}=\displaystyle\frac{B[Y_{4}]}{(-d^{\prime}+P^{\prime}Y_{4})}$ and
extend $v$ to a map $v_{1}:B_{1}\rightarrow A^{\prime}$ given by
$Y_{4}\rightarrow a$. Changing $B$ by $B_{1}$ we may assume that $d\in
H_{B/A}\cap A$.

######  Example 5.

Note that we could use $B$ instead $\operatorname{Im}v$. In this case we
choose $f=Y_{1}^{3}-Y_{2}^{3}$ and take $M=3Y_{2}^{2}$ and
$1\in\big{(}(f):I\big{)}$. Therefore we obtain $P^{\prime}=3Y_{2}^{2}$,
$d^{\prime}=x_{2}^{2}$, $d=x_{2}^{4}$ and the next computations are harder as
we will see in the Examples 17 and 25.

######  Remark 6.

We would like to work above with
$A^{\prime\prime}=\displaystyle\frac{\mathbb{C}\llbracket
x_{1},x_{2}\rrbracket}{(x_{1}^{3}-x_{2}^{2})}$ instead $A^{\prime}$, $v$ being
given by $v(Y_{2})=a_{1}\rho x_{2}$. But this is hard since we cannot work in
Singular with an infinite set of parameters. We have two choices. If the
definition of $v$ involves only a finite set of parameters then we proceed as
Example 4 using some $A^{\prime}\supset\operatorname{Im}v.$ Otherwise, we will
see later that in the computation of the General Neron Desingularization we
may use only a finite number of the coefficients of the formal power series
defining $v(Y)$ and so this computation works in Singular.

######  Remark 7.

As we may see our algorithm could go also when $A^{\prime}$ is not a domain,
but there exist $P\in M((f):I)$ as above and a regular element $d\in m$ with
$d\equiv P\ \mbox{modulo}\ I$. If $A$ is Cohen-Macaulay we may reduce to the
case when there exists a regular element $d\in H_{B/A}\cap A$. However, it is
hard usually to reduce to the case when $d\equiv P\ \mbox{modulo}\ I$ for some
$P\in M((f):I)$. Sometimes this is possible as shows the following example.

######  Example 8.

Let $a_{1},a_{2}\in\mathbb{C}$ be two elements algebraically independent over
$\mathbb{Q}$. Consider
$A=\left(\displaystyle\frac{\mathbb{Q}[x_{1},x_{2},x_{3}]}{(x_{2}^{3}-x_{3}^{2},x_{1}^{3}-x_{3}^{2})}\right)_{(x_{1},x_{2},x_{3})}$
and $B=\displaystyle\frac{A[Y_{1},Y_{2},Y_{3}]}{(Y_{1}^{3}-Y_{2}^{3})}$,
$K^{\prime}=\displaystyle\frac{\mathbb{Q}(a_{1},a_{2})[a_{3}]}{(a_{3}^{2}-a_{1}a_{2})}$,
$A^{\prime}=\displaystyle\frac{K^{\prime}\llbracket
x_{1},x_{2},x_{3}\rrbracket}{(x_{2}^{3}-x_{3}^{2},x_{1}^{3}-x_{3}^{2})}$ and
the map $v$ defined as

$\textstyle{v:}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A^{\prime}}$$\textstyle{Y_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{3}x_{1}}$$\textstyle{Y_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{3}x_{2}}$$\textstyle{Y_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}\displaystyle\sum_{i=0}^{30}\displaystyle\frac{x_{3}^{i}}{i!}+a_{2}\displaystyle\sum_{i=31}^{50}\displaystyle\frac{x_{3}^{i}}{i!}}$

Then $\operatorname{Im}\ v$, the new $B$, will be
$\displaystyle\frac{B}{\operatorname{Ker}\ v}$, where the kernel is generated
by six polynomials:

    ker[1]=x2*Y1-x1*Y2
    ker[2]=Y1^3-Y2^3
    ker[3]=x1*Y1^2-x2*Y2^2
    ker[4]=x1^2*Y1-x2^2*Y2
    ker[5]=x1*x2^2*Y2-x3^2*Y1
    ker[6]=x1^2*x2*Y2^2-x3^2*Y1^2

Next we choose $f=x_{2}Y_{1}-x_{1}Y_{2}$ and we have $M=-x_{1}$. We may take
$N=-x_{3}^{2}\in\big{(}(f):I\big{)}$ and $P^{\prime}=x_{1}x_{3}^{2}$. Note
that $x_{1}-x_{2}$ is a zero divisor in $A$ but $d^{\prime}=P^{\prime}$ is
regular in $A$. In this example we may take $d=d^{\prime}=P^{\prime}=P$.

######  Remark 9.

Changing $B$ by $\operatorname{Im}v$ can be a hard goal if let us say
$A^{\prime}$ is a factor of the power series ring over $\mathbb{C}$ in some
variables $x$ and $v(Y)$ is defined by formal power series whose coefficients
form an infinite field extension $F$ of $\mathbb{Q}$. If $v(Y)$ are
polynomials in $x$ as in Examples 4, 8 then it is trivial to find the General
Neron Desingularization of $B$ as we explained already in the last sentences
of Section 1. For instance in Example 8, $B^{\prime}$ could be a localization
of $K^{\prime}\otimes_{\mathbb{Q}}A$. Thus Examples 4, 8 have no real
importance, they being useful only for an illustration of our algorithm. This
is the reason that in the next examples the field $L$ obtained by adjoining to
$k$ of all coefficients of $y$ will be an infinite type field extension of $k$
and $v(Y)$ are not all polynomials in $x$.

However, this will complicate the algorithm because we are not able to tell to
the computer who is $v(Y)$ and so how to get $d^{\prime}$. We may choose an
element $a\in m$ and find a minimal $c\in\mathbb{N}$ such that
$a^{c}\in(v(M))+(a^{2c})$ (this is possible because $\dim A=1$). Set
$d^{\prime}=a^{c}$. It follows that $d^{\prime}\in(v(M))+(d^{\prime
2})\subset(v(M))+(d^{\prime 4})\subset\ldots$ and so $d^{\prime}\in(v(M))$,
that is $d^{\prime}=v(M)z$ for some $z\in A^{\prime}$. Certainly we cannot
find precisely $z$ but later it is enough to know just a kind of truncation of
it modulo $d^{\prime 6}$.

######  Example 10.

Let $a_{i}\in\mathbb{C}$, $i\in\mathbb{N}$ be elements algebraically
independent over $\mathbb{Q}$ and $\rho$ a root of the polynomial $T^{2}+T+1$
in $\mathbb{C}$. Let
$A=\left(\displaystyle\frac{\mathbb{Q}[x_{1},x_{2}]}{(x_{1}^{3}-x_{2}^{2})}\right)_{(x_{1},x_{2})}$
and
$B=\displaystyle\frac{A[Y_{1},Y_{2},Y_{3}]}{(Y_{1}^{2}+Y_{1}Y_{2}+Y_{2}^{2})}$,
$A^{\prime}=\displaystyle\frac{\mathbb{C}\llbracket
x_{1},x_{2}\rrbracket}{(x_{1}^{3}-x_{2}^{2})}$ and the map $v$ defined as

$\textstyle{v:}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A^{\prime}}$$\textstyle{Y_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}\left(x_{2}+\displaystyle\sum_{i\geq
7}a_{i}x_{2}^{i}\right)}$$\textstyle{Y_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}a_{3}\left(x_{2}+\displaystyle\sum_{i\geq
7}a_{i}x_{2}^{i}\right)}$$\textstyle{Y_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}\displaystyle\sum_{i=0}^{9}\displaystyle\frac{x_{1}^{i}}{i!}+x_{2}\sum_{i=10}^{\infty}a_{i-8}\displaystyle\frac{x_{1}^{i}}{i!}}$

As in Example 4 we may take $d^{\prime}=x_{2}$, $d=d^{\prime 2}$ and $a$. Our
algorithm goes exactly as in Examples 4, 16, 24 providing the same General
Neron Desingularization. This time we cannot find an easy General Neron
Desingularization as in the first part of Example 4.

######  Example 11.

Let
$A=\displaystyle\frac{\mathbb{Q}[x_{1},x_{2}]_{(x_{1},x_{2})}}{(x_{1}^{2}-x_{2}^{3})}$,
$A^{\prime}=\displaystyle\frac{\mathbb{C}\llbracket
x_{1},x_{2}\rrbracket}{(x_{1}^{2}-x_{2}^{3})}$. Then the inclusion $A\subset
A^{\prime}$ is regular. Let
$\theta_{i}=\displaystyle\sum_{j=0}^{\infty}\alpha_{ij}x_{2}^{j}+x_{1}\displaystyle\sum_{j=0}^{\infty}\beta_{ij}x_{2}^{j}\in\mathbb{C}\llbracket
x_{1},x_{2}\rrbracket$ for $i=3,4$ with $\alpha_{i0}=1$ and
$y_{1}=\displaystyle\frac{\theta_{3}^{3}}{\theta_{4}^{2}}$,
$y_{2}=\displaystyle\frac{\theta_{4}^{2}}{\theta_{3}}$,
$y_{3}=x_{2}\theta_{3}$, $y_{4}=x_{2}\theta_{4}$. Let
$f_{1}=Y_{3}^{2}-x_{2}^{2}Y_{1}Y_{2}$, $f_{2}=Y_{4}^{2}-x_{2}Y_{2}Y_{3}$ be
polynomials in $A[Y]$, $Y=(Y_{1},\ldots,Y_{4})$ and set $B=A[Y]/(f)$,
$f=(f_{1},f_{2})$.

If $R$ is a domain and $u\in R$ is such that $Y^{2}-u\in R[Y]$ has no
solutions in $Q(R)$ then it is easy to see that $R[Y]/(Y^{2}-u)$ is a domain
too. In our case we get that $R=A[Y_{1},Y_{2},Y_{3}]/(f_{1})$ and
$B=R[Y_{4}]/(f_{2})$ are domains too. Then the map $v:B\rightarrow A^{\prime}$
given by $Y\rightarrow y=(y_{1},\ldots,y_{4})$ is injective if we suppose that
$\theta_{3},\theta_{4}$ are algebraically independent over $A$. This follows
since $B$ is a domain and $\dim
B=\operatorname{tr\,deg}_{Q(A)}Q(B)=\operatorname{tr\,deg}_{Q(A)}Q(\operatorname{Im}v)=2=\dim\operatorname{Im}v$.
Moreover we will assume that the fields
$L_{i}=\mathbb{Q}\big{(}(\alpha_{ij},\beta_{ij})_{j}\big{)}$, $i=3,4$ have
infinite transcendental degree over $\mathbb{Q}$. The Jacobian matrix
$\left(\displaystyle\frac{\partial f}{\partial Y}\right)$ have a $2\times
2-$minor $M=\det\left(\displaystyle\frac{\partial f_{i}}{\partial
Y_{j}}\right)_{\begin{subarray}{c}1\leq i\leq 2\\\ 3\leq j\leq
4\end{subarray}}=4Y_{3}Y_{4}\not\in(f)$. Note that $v(M)=x_{2}^{2}y_{5}$,
where $y_{5}=1/(4\theta_{3}\theta_{4})$. Then we may take
$B_{1}=B[Y_{5}]/(f_{3})$, $f_{3}=-x_{2}^{2}+MY_{5}$ and $v_{1}$ given by
$Y_{5}\rightarrow y_{5}$. Clearly, $P=M^{2}Y_{5}^{2}\in H_{B_{1}/A}$ and
$0\neq d=x_{2}^{4}=v_{1}(P)\in A$.

##  3\. Proof of the case when $H_{B/A}\cap A\not=0$.

Thus we may suppose that there exists $f=(f_{1},\ldots,f_{r})$, $r\leq n$ a
system of polynomials from $I$, a $r\times r$-minor $M$ of the Jacobian matrix
$(\partial f_{i}/\partial Y_{j})$ and $N\in((f):I)$ such that $0\not=d\equiv
MN\ \mbox{modulo}\ I$. Set $\bar{A}=A/(d^{3})$,
$\bar{A}^{\prime}=A^{\prime}/d^{3}A^{\prime}$, $\bar{u}=\bar{A}\otimes_{A}u$,
$\bar{B}=B/d^{3}B$, $\bar{v}=\bar{A}\otimes_{A}v$. Clearly, $\bar{u}$ is a
regular morphism of Artinian local rings.

######  Remark 12.

The whole proof could work with $\bar{A}=A/d^{2}u$ for any $u\in m$. We prefer
to take $u=d$ as is done in [9] and [11] but we could choose $u\not=d$, $u\in
m\setminus m^{2}$ for easy computations.

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 $m\tilde{A}$ is the maximal ideal of $\tilde{A}$ and
$\tilde{A}/m\tilde{A}\cong L$. 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 $m(L\otimes_{k}\bar{A})$ which is Artinian and so complete. By uniqueness
we see that $A_{L}\cong\tilde{A}$. Set $k^{\prime}=A^{\prime}/mA^{\prime}$. 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 $k^{\prime}/k$.

Let $v$ be given by $Y\rightarrow y\in A^{\prime n}$. 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 $m$ 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 $\omega=(\omega_{1},\ldots,\omega_{p})$ in
$\bar{J}$ such that $\rho=\det(\partial\omega_{i}/\partial
U_{\nu})_{i,\nu\in[p]}\not=0$ and a nonzero polynomial
$\tau\in((\omega):\bar{J})\setminus\bar{J}$. Thus $L$ is a fraction ring of
the smooth $k$-algebra $(k[U]/(\omega))_{\rho\tau}$. Note that
$\omega,\rho,\tau$ can be considered in $A$ 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
$C\cong(\bar{A}[U]/(\omega))_{\rho\tau\gamma}$ for some polynomial $\gamma$
which is not in $m(\bar{A}[U]/(\omega))_{\rho\tau}$.

######  Lemma 13.

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

######  Remark 14.

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

######  Remark 15.

Suppose that $k\subset A$ but $L\not\subset A^{\prime}$ and so
$k^{\prime}\not\subset A^{\prime}$. Then
$D=(A[U,Z]/(\omega-d^{3}Z))_{\rho\tau\gamma}$, $Z=(Z_{\nu})$ is a smooth
$A$-algebra and $\bar{D}\cong C[Z]$. Since $\bar{v}$ factors through a map
$C\rightarrow\bar{A}^{\prime}$ given let us say by
$U\rightarrow\lambda+d^{3}A^{\prime}$ for some $\lambda$ in $A^{\prime}$ we
see that $\omega(\lambda)\equiv 0$ modulo $d^{3}$, that is
$\omega(\lambda)=d^{3}z$ for some $z$ in $A^{\prime}$. Let $w:D\rightarrow
A^{\prime}$ be the $A$-morphism given by $(U,Z)\rightarrow(\lambda,z)$.
Certainly, $\bar{v}$ factors through $\bar{w}=\bar{A}\otimes_{A}w$ but in
general $v$ does not factor through $w$. If also $k\not\subset A$ then the
construction of $D$ goes as above but using a lifting of $\omega,\tau,\gamma$
from $k[U]$ to $A[U]$. In both cases we may use $D$ as it follows.

######  Example 16.

We reconsider Example 4. We already know that $d=x_{2}^{2}$. The algorithm
gives us the following output:

    This is C:
    //   characteristic : 0
    //   number of vars : 5
    //        block   1 : ordering dp
    //                  : names    a1 a3 a x1 x2
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=3*a1*a+2*a3+1
    _[2]=a3^2+a3+1
    _[3]=x1^3-x2^2
    _[4]=x2^6
    This is D:
    //   characteristic : 0
    //   number of vars : 5
    //        block   1 : ordering dp
    //                  : names    a1 a3 a x1 x2
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=3*a1*a+2*a3+1
    _[2]=a3^2+a3+1
    _[3]=x1^3-x2^2

Indeed,

$C=\displaystyle\frac{\bar{A}[a_{1},a_{3},a]}{\left(3a_{1}a+2a_{3}+1,a_{3}^{2}+a_{3}+1,x_{2}^{6}\right)}$

and

$D=\displaystyle\frac{A[a_{1},a_{3},a]}{\left(3a_{1}a+2a_{3}+1,a_{3}^{2}+a_{3}+1\right)}.$

Note that the first polynomial from $C$ comes from the standard basis
computation of the ideal
$\left(2a_{1}a_{3}a+a_{1}a-1,a_{3}^{2}+a_{3}+1\right)$.

######  Example 17.

Now we reconsider Example 5. We know that $d=x_{2}^{4}$. The algorithm gives
us the following output:

    This is C:
    //   characteristic : 0
    //   number of vars : 5
    //        block   1 : ordering dp
    //                  : names    a1 a3 a x1 x2
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=a3^2+a3+1
    _[2]=x1^3-x2^2
    _[3]=a1^2*a-a3
    _[4]=x2^12

    This is D:
    //   characteristic : 0
    //   number of vars : 5
    //        block   1 : ordering dp
    //                  : names    a1 a3 a x1 x2
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=a3^2+a3+1
    _[2]=x1^3-x2^2
    _[3]=a1^2*a-a3

Indeed,

$C=\displaystyle\frac{\bar{A}[a_{1},a_{3},a]}{\left(a_{3}^{2}+a_{3}+1,a_{1}^{2}a-a_{3},x_{2}^{12}\right)}$

and

$D=\displaystyle\frac{A[a_{1},a_{3},a]}{\left(a_{3}^{2}+a_{3}+1,a_{1}^{2}a-a_{3}\right)}.$

######  Example 18.

In the case of Example 8 we obtain the following output:

    This is C:
    //   characteristic : 0
    //   number of vars : 5
    //        block   1 : ordering dp
    //                  : names    a1 a3 x1 x2 x3
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=x2^3-x3^2
    _[2]=x1^3-x3^2
    _[3]=x3^8
    This is D:
    //   characteristic : 0
    //   number of vars : 5
    //        block   1 : ordering dp
    //                  : names    a1 a3 x1 x2 x3
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=x2^3-x3^2
    _[2]=x1^3-x3^2

Indeed this is the case since we have $d=x_{1}x_{3}^{2}$ and hence

$C=\displaystyle\frac{\bar{A}[a_{1},a_{3}]}{\left(x_{3}^{8}\right)}$

and

$D=A[a_{1},a_{3}].$

######  Example 19.

In Example 11 we consider $a_{1},a_{2}$ algebraically independent over
$\mathbb{Q}$ and set $\theta_{3}^{\prime}=1+a_{1}x_{2}$ and
$\theta_{4}^{\prime}=1+a_{2}x_{2}^{2}$. Suppose that
$\theta_{i}^{\prime}\equiv\theta_{i}\ \mbox{modulo}\ x_{2}^{12}$. We have
$y_{3}=x_{2}\theta_{3}$, $y_{4}=x_{2}\theta_{4}$,
$y_{1}=\displaystyle\frac{\theta_{3}^{3}}{\theta_{4}^{2}}$,
$y_{2}=\displaystyle\frac{\theta_{4}^{2}}{\theta_{3}}$,
$y_{5}=\displaystyle\frac{1}{(4\theta_{3}\theta_{4})}$. Choose
$y^{\prime}_{i}$, $i\in[5]$ polynomials with degrees $\leq 11$ in $x_{2}$ and
linear in $x_{1}$ such that $y^{\prime}_{i}\equiv
y_{i}\operatorname{mod}(x_{1}^{2},x_{2}^{12})$. We get $y^{\prime}_{1}\equiv
y_{1}=\theta_{3}^{3}/\theta_{4}^{2}\equiv{\theta^{\prime}}_{3}^{3}/{\theta^{\prime}}_{4}^{2}\operatorname{mod}(x_{1}^{2},x_{2}^{12})$
and similarly for $y^{\prime}_{i}$, $i>1$. Here we use the fact that
$\theta_{4}^{-2}=\displaystyle\sum_{j=1}^{e}(1-\theta_{4}^{2})^{j}$ for some
$e>>0$ because $1-\theta_{4}^{2}$ is nilpotent in the ring
$\bar{A}[a_{1},a_{2},a_{3},a_{4}]$. Thus the coefficients of $y^{\prime}_{i}$,
$i\in[5]$ belong to the field $L$ obtained by adjoining to $\mathbb{Q}$ the
coefficients of $\theta^{\prime}_{3},\theta^{\prime}_{4}$. Note that in this
case $L=Q\left(\mathbb{Q}[a_{1},\ldots,a_{4}]\right)$. Then we obtain the
following output:

    This is C:
    //   characteristic : 0
    //   number of vars : 4
    //        block   1 : ordering dp
    //                  : names    a1 a2 x1 x2
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=x2^3-x1^2
    _[2]=x1^8
    This is D:
    //   characteristic : 0
    //   number of vars : 4
    //        block   1 : ordering dp
    //                  : names    a1 a2 x1 x2
    //        block   2 : ordering C
    // quotient ring from ideal
    _[1]=x2^3-x1^2

Thus
$C=\displaystyle\frac{\bar{A}[a_{1},\ldots,a_{4}]}{(x_{2}^{12})}\cong{\bar{A}}[a_{1},\ldots,a_{4}]$
which is smooth over $\bar{A}$. Then $D$ is equal with
$A[a_{1},\ldots,a_{4}]$.

Back to our proof note that the composite map $\bar{B}\rightarrow
C\rightarrow\bar{D}$ is given by $Y\rightarrow y^{\prime}+d^{3}D$ for some
$y^{\prime}\in D^{n}$. Thus $I(y^{\prime})\equiv 0$ modulo $d^{3}D$. Since
$\bar{v}$ factors through $\bar{w}$ we see that
$\bar{w}(y^{\prime}+d^{3}D)=\bar{y}$. Set $\tilde{y}=w(y^{\prime})$. We get
$y-\tilde{y}\in d^{3}A^{\prime n}$, let us say $y-\tilde{y}=d^{2}\varepsilon$
for $\varepsilon\in dA^{\prime n}$.

We have $d\equiv P$ modulo $I$ and so $P(y^{\prime})\equiv d$ modulo $d^{3}$
in $D$ because $I(y^{\prime})\equiv 0$ modulo $d^{3}D$. Thus
$P(y^{\prime})=ds$ for a certain $s\in D$ with $s\equiv 1$ modulo $d$. Assume
that $P=NM$ for some $N\in((f):I)$. Recall from beginning of Section 2 that
the new $M$ is now the old one multiplied with $P^{\prime}$ and the new $N$ is
the old one multiplied with $Z^{2}$. Let $H$ be the $n\times n$-matrix
obtained adding down to $(\partial f/\partial Y)$ as a border the block
$(0|\mbox{Id}_{n-r})$. Let $G^{\prime}$ be the adjoint matrix of $H$ and
$G=NG^{\prime}$. We have

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

and so

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

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

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

and so

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

Let

(1) $h=s(Y-y^{\prime})-dG(y^{\prime})T,$

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

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

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

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

modulo $h$ where $Q\in T^{2}D[T]^{r}$. This is because $(\partial f/\partial
Y)G=(P\mbox{Id}_{r}|0)$. We have $f(y^{\prime})=d^{2}b$ for some $b\in
dD^{r}$. Then

(3) $g_{i}=s^{p}b_{i}+s^{p}T_{i}+Q_{i},\qquad i\in[r]$

is in the kernel of $\varphi$ because
$d^{2}\varphi(g)=d^{2}g(t)\in(h(y,t),f(y))=(0)$. 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 $B\rightarrow
B\otimes_{A}D\cong D[Y]/I\rightarrow E\xrightarrow{\psi}A^{\prime}$.

Note that the $r\times r$-minor $s^{\prime}$ of $(\partial g/\partial T)$
given by the first $r$-variables $T$ is from $s^{rp}+(T)\subset 1+(d,T)D[Y,T]$
because $Q\in(T)^{2}$. Then $U=(D[Y,T]/(h,g))_{ss^{\prime}}$ 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
$P(y^{\prime}+s^{-1}dG(y^{\prime})T)I\subset(h,g)D[Y,T]_{s}$. Since
$P(y^{\prime}+s^{-1}dG(y^{\prime})T)\in P(y^{\prime})+d(T)$ we get
$P(y^{\prime}+s^{-1}dG(y^{\prime})T)=ds^{\prime\prime}$ for some
$s^{\prime\prime}\in 1+(d,T)D[Y,T]$. It follows that
$s^{\prime\prime}I\subset(h,g)D[Y,T]_{ss^{\prime}}$ because $d$ is regular in
$U$, the map $D\rightarrow U$ being flat, and so
$I\subset(h,g)D[Y,T]_{ss^{\prime}s^{\prime\prime}}$. Thus
$E_{ss^{\prime}s^{\prime\prime}}\cong U_{s^{\prime\prime}}$ is a $B$-algebra
which is also standard smooth over $D$ and $A$.

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

##  4\. A theorem of Greenberg’s type

Let $(A,m)$ be a Cohen-Macaulay local ring (for example a reduced ring) of
dimension one, $A^{\prime}={\hat{A}}$ the completion of $A$, $B=A[Y]/I$,
$Y=(Y_{1},\ldots,Y_{n})$ a finite type $A$-algebra and $c,e\in\mathbb{N}$.
Suppose that there exist $f=(f_{1},\ldots,f_{r})$ in $I$, a $r\times r$-minor
$M$ of the Jacobian matrix $(\partial f/\partial Y)$, $N\in((f):I)$ and an
$A$-morphism $v:B\rightarrow A/m^{2e+c}$ such that $(v(MN))\supset
m^{e}/m^{2e+c}$.

######  Theorem 20.

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

###### Proof.

We note that the proof of Theorem 3 can work somehow in this case. Let
$y^{\prime}\in A^{n}$ be an element inducing $v(Y+I)$. Then
$m^{e}\subset((MN)(y^{\prime}))+m^{2e+c}\subset((MN)(y^{\prime}))+m^{3e+2c}\subset\ldots$
by hypothesis. It follows that $m^{e}\subset((MN)(y^{\prime}))$. Since $A$ is
Cohen-Macaulay we see that $m^{e}$ contains a regular element of $A$ and so
$(MN)(y^{\prime})$ must be regular too.

Set $d=(MN)(y^{\prime})$. Next we follow the proof of Theorem 3 with $D=A$,
$s=1$, $P=MN$ and $H$, $G$ such that .

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

Let

$h=Y-y^{\prime}-dG(y^{\prime})T,$

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

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

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

As in the proof of Theorem 3 we have $E_{s^{\prime}s^{\prime\prime}}\cong
U_{s^{\prime\prime}}$, where $U=(A[Y,T]/(g,h))_{s^{\prime}}$. This isomorphism
follows because $d$ is regular in $A$ and so in $U$. Consequently,
$E_{s^{\prime}s^{\prime\prime}}$ is smooth over $A$. Note that $\beta$ extends
to a map $\beta^{\prime}:E_{s^{\prime}s^{\prime\prime}}\rightarrow A/m^{c}$.
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 $m^{c}$. It follows that the composite map
$v^{\prime}$, $B\rightarrow
E_{s^{\prime}s^{\prime\prime}}\xrightarrow{w}\hat{A}$ works.

######  Corollary 21.

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

###### Proof.

An excellent Henselian local ring $(A,m)$ has the property of Artin
approximation by [9], 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 20 we get an $A$-morphism
$v^{\prime}:B\rightarrow\hat{A}$ such that $v^{\prime}\equiv v\ \mbox{modulo}\
m^{c}$. Then there exists an $A$-morphism $v^{\prime\prime}:B\rightarrow A$
such that $v^{\prime\prime}\equiv v^{\prime}\equiv v\ \mbox{modulo}\ m^{c}$ by
the property of Artin approximation.

######  Theorem 22.

Let $(A,m)$ be a Cohen-Macaulay local ring of dimension one, $B=A[Y]/I$,
$Y=(Y_{1},\ldots,Y_{n})$ a finite type $A$-algebra, $e\in\mathbb{N}$ and
$f=(f_{1},\ldots,f_{r})$ a system of polynomials from $I$. Suppose that $A$ is
excellent Henselian and there exist a $r\times r$-minor $M$ of the Jacobian
matrix $(\partial f/\partial Y)$, $N\in((f):I)$ and $y^{\prime}\in A^{n}$ such
that $I(y^{\prime})\equiv 0\ \mbox{modulo}\ m^{e}$ and
$((NM)(y^{\prime}))\supset m^{e}$. Then the following statements are
equivalent:

  1. (1)

there exists $y^{\prime\prime}\in A^{n}$ such that $I(y^{\prime\prime})\equiv
0\ \mbox{modulo}\ m^{3e}$ and $y^{\prime\prime}\equiv y^{\prime}\
\mbox{modulo}\ m^{e}$,

  2. (2)

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

For the proof apply the above corollary and Theorem 20.

##  5\. Computation of the General Neron Desingularization in Examples 4, 5,
8, 11

######  Example 23.

We would like to compute Example 4 in Singular using `GND.lib`. We quickly
recall the example.

Let $a_{1},a_{2}\in\mathbb{C}$ be two elements algebraically independent over
$\mathbb{Q}$ and $\rho$ a root of the polynomial $T^{2}+T+1$ in $\mathbb{C}$.
Then
$k^{\prime}=\displaystyle\frac{\mathbb{Q}(a_{1},a_{2})[a_{3}]}{(a_{3}^{2}+a_{3}+1)}\cong\mathbb{Q}(\rho,a_{1},a_{2})$.
Let
$A=\left(\displaystyle\frac{\mathbb{Q}[x_{1},x_{2}]}{(x_{1}^{3}-x_{2}^{2})}\right)_{(x_{1},x_{2})}$
and $B=\displaystyle\frac{A[Y_{1},Y_{2},Y_{3}]}{(Y_{1}^{3}-Y_{2}^{3})}$,
$A^{\prime}=\displaystyle\frac{k^{\prime}\llbracket
x_{1},x_{2}\rrbracket}{(x_{1}^{3}-x_{2}^{2})}$ and the map $v$ defined as

$\textstyle{v:}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A^{\prime}}$$\textstyle{Y_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}x_{2}}$$\textstyle{Y_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}a_{3}x_{2}}$$\textstyle{Y_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a_{1}\displaystyle\sum_{i=0}^{30}\displaystyle\frac{x_{1}^{i}}{i!}+a_{2}x_{2}\displaystyle\sum_{i=31}^{50}\displaystyle\frac{x_{1}^{i}}{i!}}$

For this we do the following:

    LIB "GND.lib";                              //load the library
    ring All = 0,(a1,a2,a3,x1,x2,Y1,Y2,Y3),dp;  //define the ring
    int nra = 3;                                //number of a’s
    int nrx = 2;                                //number of x’s
    int nry = 3;                                //number of Y’s
    ideal xid = x1^3-x2^2;                      //define the ideal from A
    ideal yid = Y1^3-Y2^3;                      //define the ideal from B
    ideal aid = a3^2+a3+1;                      //define the ideal from k’
    poly y;
    int i;
    for(i=0;i<=30;i++)
    {
      y = y + a1*x1^i/factorial(i);
    }
    for(i=31;i<=50;i++)
    {
      y = y + a2*x2*x1^i/factorial(i);
    }
    ideal f = a1*x2,a1*a3*x2,y;                //define the map v
    desingularization(All, nra,nrx,nry,xid,yid,aid,f,"debug");

######  Example 24.

We continue on the idea of Examples 4, 16. The bordered matrix $H$ defined
above is equal to

$H=\left(\begin{array}[]{cccc}2Y_{1}+Y_{2}&Y_{1}+2Y_{2}&0&0\\\ 0&0&1&0\\\
1&0&0&0\\\ Y_{4}&2Y_{4}&0&Y_{1}+2Y_{2}\\\ \end{array}\right)$

and hence $G=N\cdot G^{\prime}$ is equal to

$G=Y_{4}^{2}\cdot\left(\begin{array}[]{cccc}0&0&Y_{1}^{2}+4Y_{1}Y_{2}+4Y_{2}^{2}&0\\\
Y_{1}+2Y_{2}&0&-2Y_{1}^{2}-5Y_{1}Y_{2}-2Y_{2}^{2}&0\\\
0&Y_{1}^{2}+4Y_{1}Y_{2}+4Y_{2}^{2}&0&0\\\
-2Y_{4}&0&3Y_{1}Y_{4}&Y_{1}+2Y_{2}\\\ \end{array}\right)$

and $s=1$. Using the definition of $h$ in Equation 1, we get that

$\begin{array}[]{ccl}h_{1}&=&Y_{1}-\left(x_{2}^{4}\right)\cdot
T_{3}-(a_{1}x_{2}),\\\
h_{2}&=&Y_{2}-\displaystyle\frac{x_{2}^{3}}{2a_{1}a_{3}+a_{1}}\cdot
T_{1}+\displaystyle\frac{a_{3}x_{2}^{4}+2x_{2}^{4}}{2a_{3}+1}\cdot
T_{3}-(a_{1}a_{3}x_{2}),\\\ h_{3}&=&Y_{3}-\left(x_{2}^{4}\right)\cdot
T_{2}-\left(\displaystyle\frac{1}{6!}a_{1}x_{1}^{6}+\displaystyle\frac{1}{5!}a_{1}x_{1}^{5}+\displaystyle\frac{1}{4!}a_{1}x_{1}^{4}+\displaystyle\frac{1}{3!}a_{1}x_{1}^{3}+\displaystyle\frac{1}{2}a_{1}x_{1}^{2}+\right.\\\
&&\left.a_{1}x_{1}+a_{1}\right),\\\
h_{4}&=&Y_{4}+\displaystyle\frac{2x_{2}^{2}}{\left(2a_{1}a_{3}+a_{1}\right)^{3}}\cdot
T_{1}-\displaystyle\frac{3x_{2}^{3}}{a_{1}^{2}\left(2a_{3}+1\right)^{3}}\cdot
T_{3}-\displaystyle\frac{x_{2}^{3}}{2a_{1}a_{3}+a_{1}}\cdot
T_{4}+\displaystyle\frac{1}{2a_{1}a_{3}+a_{1}}.\end{array}$

From Equation 2 we get that

$\begin{array}[]{ccl}Q_{1}&=&\displaystyle\frac{x_{2}^{2}}{\left(2a_{1}a_{3}+a_{1}\right)^{2}}\cdot
T_{1}^{2}-\displaystyle\frac{3x_{2}^{3}}{a_{1}\left(2a_{3}+1\right)^{2}}\cdot
T_{1}T_{3}+\displaystyle\frac{3a_{3}^{2}x_{2}^{4}+3a_{3}x_{2}^{4}+3x_{2}^{4}}{\left(2a_{3}+1\right)^{2}}\cdot
T_{3}^{2},\\\
Q_{2}&=&-\displaystyle\frac{4x_{2}}{\left(2a_{1}a_{3}+a_{1}\right)^{4}}\cdot
T_{1}^{2}+\displaystyle\frac{12x_{2}^{2}}{a_{1}^{3}\left(2a_{3}+1\right)^{4}}\cdot
T_{1}T_{3}-\displaystyle\frac{9x_{2}^{3}}{a_{1}^{2}\left(2a_{3}+1\right)^{4}}\cdot
T_{3}^{2}+\\\
&&\displaystyle\frac{2x_{2}^{2}}{\left(2a_{1}a_{3}+a_{1}\right)^{2}}\cdot
T_{1}T_{4}-\displaystyle\frac{3x_{2}^{3}}{a_{1}\left(2a_{3}+1\right)^{2}}\cdot
T_{3}T_{4}\end{array}$

and therefore following the definition of $g$ in Equation 3 we have

$\begin{array}[]{ccl}g_{1}&=&Q_{1}+T_{1}+(a_{1}^{2}a_{3}^{2}+a_{1}^{2}a_{3}+a_{1}^{2}),\\\
g_{2}&=&Q_{2}+T_{2}.\end{array}$

We print now the algorithm’s debug output using the line codes from Example
23.

    This is the bordered matrix H:
    2*Y1+Y2,Y1+2*Y2,0,0,
    0,      0,      1,0,
    1,      0,      0,0,
    Z,      2*Z,    0,Y1+2*Y2
    This is G:
    0,                0,     G[1,3],   0,
    Y1*Y4^2+2*Y2*Y4^2,0,     G[2,3],   0,
    0,                G[3,2],0,        0,
    -2*Y4^3,          0,     3*Y1*Y4^3,Y1*Y4^2+2*Y2*Y4^2

    G[1,3]=Y1^2*Y4^2+4*Y1*Y2*Y4^2+4*Y2^2*Y4^2
    G[2,3]=-2*Y1^2*Y4^2-5*Y1*Y2*Y4^2-2*Y2^2*Y4^2
    G[3,2]=Y1^2*Y4^2+4*Y1*Y2*Y4^2+4*Y2^2*Y4^2

    s = 1
    h =
    _[1]=Y1+(-x2^4)*T3+(-a1*x2)
    _[2]=Y2+(-x2^3)/(2*a1*a3+a1)*T1+(a3*x2^4+2*x2^4)/(2*a3+1)*T3+
    Ψ(-a1*a3*x2)
    _[3]=Y3+(-x2^4)*T2+(-a1*x1^6-6*a1*x1^5-30*a1*x1^4-120*a1*x1^3
    Ψ-360*a1*x1^2-720*a1*x1-720*a1)/720
    _[4]=Y4+(2*x2^2)/(8*a1^3*a3^3+12*a1^3*a3^2+6*a1^3*a3+a1^3)*T1+
    Ψ(-3*x2^3)/(8*a1^2*a3^3+12*a1^2*a3^2+6*a1^2*a3+a1^2)*T3+
    Ψ(-x2^3)/(2*a1*a3+a1)*T4-1/(2*a1*a3+a1)

    m = 2
    QT =
    _[1]=(x2^2)/(4*a1^2*a3^2+4*a1^2*a3+a1^2)*T1^2+
    Ψ(-3*x2^3)/(4*a1*a3^2+4*a1*a3+a1)*T1*T3+
    Ψ(3*a3^2*x2^4+3*a3*x2^4+3*x2^4)/(4*a3^2+4*a3+1)*T3^2

    _[2]=(-4*x2)/(16*a1^4*a3^4+32*a1^4*a3^3+24*a1^4*a3^2+8*a1^4*a3+a1^4)
    Ψ*T1^2+(12*x2^2)/(16*a1^3*a3^4+32*a1^3*a3^3+24*a1^3*a3^2+8*a1^3*a3
    Ψ+a1^3)*T1*T3+(-9*x2^3)/(16*a1^2*a3^4+32*a1^2*a3^3+24*a1^2*a3^2
    Ψ+8*a1^2*a3+a1^2)*T3^2+(2*x2^2)/(4*a1^2*a3^2+4*a1^2*a3+a1^2)*T1*T4
    Ψ+(-3*x2^3)/(4*a1*a3^2+4*a1*a3+a1)*T3*T4

    f =
    f[1]=Y1^2+Y1*Y2+Y2^2
    f[2]=Y1*Y4+2*Y2*Y4+(-x2^2)

    g =
    _[1]=(x2^2)/(4*a1^2*a3^2+4*a1^2*a3+a1^2)*T1^2+(-3*x2^3)/(4*a1*a3^2+
    Ψ4*a1*a3+a1)*T1*T3+(3*a3^2*x2^4+3*a3*x2^4+3*x2^4)/(4*a3^2+4*a3+1)*T3^2
    Ψ+T1+(a1^2*a3^2+a1^2*a3+a1^2)
    _[2]=(-4*x2)/(16*a1^4*a3^4+32*a1^4*a3^3+24*a1^4*a3^2+8*a1^4*a3+a1^4)
    Ψ*T1^2+(12*x2^2)/(16*a1^3*a3^4+32*a1^3*a3^3+24*a1^3*a3^2+8*a1^3*
    Ψa3+a1^3)*T1*T3+(-9*x2^3)/(16*a1^2*a3^4+32*a1^2*a3^3+24*a1^2*a3^2
    Ψ+8*a1^2*a3+a1^2)*T3^2+(2*x2^2)/(4*a1^2*a3^2+4*a1^2*a3+a1^2)*T1*T4
    Ψ+(-3*x2^3)/(4*a1*a3^2+4*a1*a3+a1)*T3*T4+T2

Thus the General Neron Desingularization is a localization of
$D[Y,T]/(h,g)\cong D[T]/(g)$.

######  Example 25.

In the case of Example 5 and 17 we obtain that the bordered matrix

$H=\left(\begin{array}[]{cccc}3Y_{1}^{2}&-3Y_{2}^{2}&0&0\\\ 0&0&1&0\\\
1&0&0&0\\\ 0&-6Y_{2}Y_{4}&0&-3Y_{2}^{2}\\\ \end{array}\right)$

and hence $G=N\cdot G^{\prime}$ is equal to

$G=Y_{4}^{2}\cdot\left(\begin{array}[]{cccc}0&0&9Y_{2}^{4}&0\\\
-3Y_{2}^{2}&0&-3Y_{1}^{2}Y_{2}^{2}&0\\\ 0&9Y_{2}^{4}&0&0\\\
6Y_{2}Y_{4}&0&-18Y_{1}^{2}Y_{2}Y_{4}&-3Y_{2}^{2}\\\ \end{array}\right)$

and $s=1$. Using the definition of $h$ in Equation 1, we get that

$\begin{array}[]{ccl}h_{1}&=&Y_{1}-\left(x_{2}^{8}\right)\cdot
T_{3}+\left(-a_{1}x_{2}\right),\\\
h_{2}&=&Y_{2}+\displaystyle\frac{x_{2}^{6}}{3a_{1}^{2}a_{3}^{2}}\cdot
T_{1}-\displaystyle\frac{x_{2}^{8}}{a_{3}^{2}}\cdot
T_{3}-\left(a_{1}a_{3}x_{2}\right),\\\
h_{3}&=&Y_{3}-\left(x_{2}^{8}\right)\cdot
T_{2}-\left(\displaystyle\frac{1}{12!}a_{1}x_{1}^{12}+\displaystyle\frac{1}{11!}a_{1}x_{1}^{11}+\displaystyle\frac{1}{10!}a_{1}x_{1}^{10}+\displaystyle\frac{1}{9!}a_{1}x_{1}^{9}+\displaystyle\frac{1}{8!}a_{1}x_{1}^{8}\right.\\\
&&\left.+\displaystyle\frac{1}{7!}a_{1}x_{1}^{7}+\displaystyle\frac{1}{6!}a_{1}x_{1}^{6}+\displaystyle\frac{1}{5!}a_{1}x_{1}^{5}+\displaystyle\frac{1}{4!}a_{1}x_{1}^{4}+\displaystyle\frac{1}{3!}a_{1}x_{1}^{3}+\displaystyle\frac{1}{2}a_{1}x_{1}^{2}+a_{1}x_{1}+a_{1}\right),\\\
h_{4}&=&Y_{4}+\displaystyle\frac{2x_{2}^{5}}{9a_{1}^{5}a_{3}^{5}}\cdot
T_{1}-\displaystyle\frac{2x_{2}^{7}}{3a_{1}^{3}a_{3}^{5}}\cdot
T_{3}+\displaystyle\frac{x_{2}^{6}}{3a_{1}^{2}a_{3}^{2}}\cdot
T_{4}+\displaystyle\frac{1}{3a_{1}^{2}a_{3}^{2}}.\end{array}$

From Equation 2 we get that

$\begin{array}[]{ccl}Q_{1}&=&\displaystyle\frac{x_{2}^{10}}{27a_{1}^{6}a_{3}^{6}}\cdot
T_{1}^{3}-\displaystyle\frac{x_{2}^{12}}{3a_{1}^{4}a_{3}^{6}}\cdot
T_{1}^{2}T_{3}+\displaystyle\frac{x_{2}^{14}}{a_{1}^{2}a_{3}^{6}}\cdot
T_{1}T_{3}^{2}+\displaystyle\frac{a_{3}^{6}x_{2}^{16}-x_{2}^{16}}{a_{3}^{6}}\cdot
T_{3}^{3}-\\\ &&\displaystyle\frac{x_{2}^{5}}{3a_{1}^{3}a_{3}^{3}}\cdot
T_{1}^{2}+\displaystyle\frac{2x_{2}^{7}}{a_{1}a_{3}^{3}}\cdot
T_{1}T_{3}+\displaystyle\frac{3a_{1}a_{3}^{3}x_{2}^{9}-3a_{1}x_{2}^{9}}{a_{3}^{3}}\cdot
T_{3}^{2},\\\
Q_{2}&=&\displaystyle\frac{2ax_{2}^{9}}{27a_{1}^{9}a_{3}^{9}}\cdot
T_{1}^{3}-\displaystyle\frac{2x_{2}^{11}}{3a_{1}75a_{3}^{9}}\cdot
T_{1}^{2}T_{3}+\displaystyle\frac{2x_{2}^{13}}{a_{1}^{5}a_{3}^{9}}\cdot
T_{1}T_{3}^{2}-\displaystyle\frac{2x_{2}^{15}}{a_{1}^{3}a_{3}^{9}}\cdot
T_{3}^{3}+\\\ &&\displaystyle\frac{x_{2}^{10}}{9a_{1}^{6}a_{3}^{6}}\cdot
T_{1}^{2}T_{4}-\displaystyle\frac{2x_{2}^{12}}{3a_{1}^{4}a_{3}^{6}}\cdot
T_{1}T_{3}T_{4}+\displaystyle\frac{x_{2}^{14}}{a_{1}^{2}a_{3}^{6}}\cdot
T_{3}^{2}T_{4}-\displaystyle\frac{x_{2}^{4}}{3a_{1}^{6}a_{3}^{6}}\cdot
T_{1}^{2}+\\\ &&\displaystyle\frac{2x_{2}^{6}}{a_{1}^{4}a_{3}^{6}}\cdot
T_{1}T_{3}-\displaystyle\frac{3x_{2}^{8}}{a_{1}^{2}a_{3}^{6}}\cdot
T_{3}^{2}-\displaystyle\frac{2x_{2}^{5}}{3a_{1}^{3}a_{3}^{3}}\cdot
T_{1}T_{4}+\displaystyle\frac{2x_{2}^{7}}{a_{1}a_{3}^{3}}\cdot
T_{3}T_{4}\end{array}$

and therefore following the definition of $g$ in Equation 3 we have

$\begin{array}[]{ccr}g_{1}&=&Q_{1}+T_{1},\\\ g_{2}&=&Q_{2}+T_{2}\end{array}$

To obtain this with Singular, we use the same code lines as in Example 23, but
we change the last one with

    desingularization(All, nra,nrx,nry,xid,yid,aid,f,"injective","debug");

Doing this, the algorithm will not compute the kernel because of the
`injective` argument.

######  Example 26.

We do now the same computations for Examples 8, 18. The bordered matrix $H$
defined above is equal to

$H=\left(\begin{array}[]{cccc}x_{2}&-x_{1}&0&0\\\ 0&0&1&0\\\ 1&0&0&0\\\
0&0&0&-x_{1}x_{3}^{2}\\\ \end{array}\right)$

and hence $G=N\cdot G^{\prime}$ is equal to

$G=Y_{4}^{2}\cdot\left(\begin{array}[]{cccc}0&0&x_{1}^{2}x_{3}^{4}&0\\\
-x_{1}x_{3}^{4}&0&x_{1}x_{2}x_{3}^{4}&0\\\ 0&x_{1}^{2}x_{3}^{4}&0&0\\\
0&0&0&-x_{1}x_{3}^{2}\\\ \end{array}\right)$

and $s=1$. Using the definition of $h$ in Equation 1, we get that

$\begin{array}[]{ccl}h_{1}&=&Y_{1}+\left(x_{1}^{3}x_{3}^{6}\right)\cdot
T_{3}-(a_{3}x_{1}),\\\ h_{2}&=&Y_{2}-\left(x_{1}^{2}x_{3}^{6}\right)\cdot
T_{1}+\left(x_{1}^{2}x_{2}x_{3}^{6}\right)\cdot T_{3}-(a_{3}x_{2}),\\\
h_{3}&=&Y_{3}+\left(x_{1}^{3}x_{3}^{6}\right)\cdot
T_{2}-\left(\displaystyle\frac{1}{7!}a_{1}x_{3}^{7}+\displaystyle\frac{1}{6!}a_{1}x_{3}^{6}+\displaystyle\frac{1}{5!}a_{1}x_{3}^{5}\right.\\\
&&\left.+\displaystyle\frac{1}{4!}a_{1}x_{3}^{4}+\displaystyle\frac{1}{3!}a_{1}x_{3}^{3}+\displaystyle\frac{1}{2!}a_{1}x_{3}^{2}+a_{1}x_{3}+a_{1}\right),\\\
h_{4}&=&Y_{4}+\left(x_{1}^{2}x_{3}^{4}\right)\cdot T_{4}+1.\end{array}$

From Equation 2 we get that

$\begin{array}[]{ccl}Q_{1}&=&0,\\\ Q_{2}&=&0\end{array}$

and therefore following the definition of $g$ in Equation 3 we have

$\begin{array}[]{ccr}g_{1}&=&T_{1}\\\ g_{2}&=&T_{2}.\end{array}$

To compute this with the library we do the following:

    ring All = 0,(a1,a2,a3,x1,x2,x3,Y1,Y2,Y3),dp;
    int nra = 3;
    int nrx = 3;
    int nry = 3;
    ideal xid = x2^3-x3^2,x1^3-x3^2;
    ideal yid = Y1^3-Y2^3;
    ideal aid = a3^2-a1*a2;
    poly y;
    int i;
    for(i=0;i<=30;i++)
    {
      y = y + a1*x3^i/factorial(i);
    }
    for(i=31;i<=50;i++)
    {
      y = y + a2*x3^i/factorial(i);
    }
    ideal f = a3*x1,a3*x2,y;
    desingularization(All, nra,nrx,nry,xid,yid,aid,f,"debug");

The algorithm’s output is as expected:

    This is the nice bordered matrix H:
    (x2),(-x1),0,0,
    0,   0,    1,0,
    1,   0,    0,0,
    0,   0,    0,(-x1*x3^2)
    This is G:
    0,              0,               (x1^2*x3^4)*Y4^2, 0,
    (-x1*x3^4)*Y4^2,0,               (x1*x2*x3^4)*Y4^2,0,
    0,              (x1^2*x3^4)*Y4^2,0,                0,
    0,              0,               0,                (-x1*x3^2)*Y4^2

    s = 1

    h =
    h[1]=Y1+(x1^3*x3^6)*T3+(-a3*x1)
    h[2]=Y2+(-x1^2*x3^6)*T1+(x1^2*x2*x3^6)*T3+(-a3*x2)
    h[3]=Y3+(x1^3*x3^6)*T2+(-a1*x3^7-7*a1*x3^6-42*a1*x3^5-210*a1*x3^4
         -840*a1*x3^3-2520*a1*x3^2-5040*a1*x3-5040*a1)/5040
    h[4]=Y4+(-x1^2*x3^4)*T4+1

    m = 1

    QT =
    QT[1]=0
    QT[2]=0
    f =
    f[1]=(x2)*Y1+(-x1)*Y2
    f[2]=(x1*x3^2)*Y4+(-x1*x3^2)

    g
    _[1]=T1
    _[2]=T2

Thus the General Neron Desingularization is a localization of
$D[Y,T_{3},T_{4}]/(h)\cong D[T_{3},T_{4}]$.

######  Example 27.

We do now the same computations for Example 19. In this example, the
computations are much more complicated. The output is unfortunately too big
but we will try to describe the result.

The bordered matrix $H$ defined above is equal to

$H=\left(\begin{array}[]{ccccc}0&x_{2}\cdot Y_{3}&x_{2}\cdot Y_{2}&-2\cdot
Y_{4}&0\\\ x_{1}^{2}\cdot Y_{2}&x_{1}^{2}\cdot Y_{1}&-2\cdot Y_{3}&0&0\\\
0&1&0&0&0\\\ 1&0&0&0&0\\\ 0&0&4Y_{4}Y_{5}&4Y_{3}Y_{5}&4Y_{3}Y_{4}\\\
\end{array}\right)$

and hence $G=N\cdot G^{\prime}$ is equal to

$G=Y_{5}^{2}\cdot\left(\begin{array}[]{ccccc}0&0&0&16Y_{3}^{2}Y_{4}^{2}&0\\\
0&0&16Y_{3}^{2}Y_{4}^{2}&0&0\\\ 0&-8Y_{3}Y_{4}^{2}&8x_{1}^{2}\cdot
Y_{1}Y_{3}Y_{4}^{2}&8x_{1}^{2}\cdot Y_{2}Y_{3}Y_{4}^{2}&0\\\
-8Y_{3}^{2}Y_{4}&-4x_{2}\cdot Y_{2}Y_{3}Y_{4}&G[4,3]&4x_{1}^{2}x_{2}\cdot
Y_{2}^{2}Y_{3}Y_{4}&0\\\ 8Y_{3}^{2}Y_{5}&x_{2}\cdot
Y_{2}Y_{3}Y_{5}+2Y_{4}^{2}Y_{5}&G[5,3]&G[5,4]&4Y_{3}Y_{4}\\\
\end{array}\right),$

where

$G[4,3]=4x_{1}^{2}x_{2}\cdot Y_{1}Y_{2}Y_{3}Y_{4}+2x_{2}\cdot Y_{3}^{3}Y_{4},$
$G[5,3]=-4x_{1}^{2}x_{2}\cdot Y_{1}Y_{2}Y_{3}Y_{5}-2x_{2}\cdot
Y_{3}^{3}Y_{5}-2x_{1}^{2}\cdot Y_{1}Y_{4}^{2}Y_{5}\textnormal{ and }$
$G[5,4]=-4x_{1}^{2}x_{2}\cdot Y_{2}^{2}Y_{3}Y_{5}-2x_{1}^{2}\cdot
Y_{2}Y_{4}^{2}Y_{5}$

and

$\begin{array}[]{cl}s=&a_{1}^{8}a_{2}^{2}x_{2}^{12}-2a_{1}^{5}a_{2}^{4}x_{2}^{13}+a_{1}^{2}a_{2}^{6}x_{2}^{14}-2a_{1}^{6}a_{2}^{3}x_{2}^{12}+2a_{1}^{3}a_{2}^{5}x_{2}^{13}-a_{1}^{4}a_{2}^{4}x_{2}^{12}+2a_{1}a_{2}^{6}x_{2}^{13}+\\\
&2a_{1}^{8}a_{2}x_{2}^{10}-4a_{1}^{5}a_{2}^{3}x_{2}^{11}+4a_{1}^{2}a_{2}^{5}x_{2}^{12}-4a_{1}^{6}a_{2}^{2}x_{2}^{10}+4a_{1}^{3}a_{2}^{4}x_{2}^{11}+a_{2}^{6}x_{2}^{12}+2a_{1}a_{2}^{5}x_{2}^{11}+\\\
&a_{1}^{8}x_{2}^{8}-2a_{1}^{5}a_{2}^{2}x_{2}^{9}+3a_{1}^{2}a_{2}^{4}x_{2}^{10}-2a_{1}^{6}a_{2}x_{2}^{8}+2a_{1}^{3}a_{2}^{3}x_{2}^{9}+a_{1}^{4}a_{2}^{2}x_{2}^{8}-2a_{1}^{4}a_{2}x_{2}^{6}+\\\
&2a_{1}a_{2}^{3}x_{2}^{7}+2a_{1}^{2}a_{2}^{2}x_{2}^{6}+2a_{2}^{3}x_{2}^{6}-2a_{1}^{4}x_{2}^{4}+2a_{1}a_{2}^{2}x_{2}^{5}+2a_{1}^{2}a_{2}x_{2}^{4}+1\end{array}$

Using the definition of $h$ in Equation 1, we get that

$\left(\begin{array}[]{c}h_{1}\\\ h_{2}\\\ h_{3}\\\ h_{4}\\\
h_{5}\end{array}\right)=s\cdot\left(\begin{array}[]{c}Y_{1}-y_{1}^{\prime}\\\
Y_{2}-y_{2}^{\prime}\\\ Y_{3}-y_{3}^{\prime}\\\ Y_{4}-y_{4}^{\prime}\\\
Y_{5}-y_{5}^{\prime}\end{array}\right)-x_{2}^{4}G(y^{\prime})\cdot\left(\begin{array}[]{c}T_{1}\\\
T_{2}\\\ T_{3}\\\ T_{4}\\\ T_{5}\end{array}\right),$

where

$\begin{array}[]{ccl}y_{1}^{\prime}&=&-18a_{1}^{2}a_{2}^{5}x_{2}^{12}+5a_{1}^{3}a_{2}^{4}x_{2}^{11}+7a_{2}^{6}x_{2}^{12}-18a_{1}a_{2}^{5}x_{2}^{11}+15a_{1}^{2}a_{2}^{4}x_{2}^{10}-4a_{1}^{3}a_{2}^{3}x_{2}^{9}-\\\
&&6a_{2}^{5}x_{2}^{10}+15a_{1}a_{2}^{4}x_{2}^{9}-12a_{1}^{2}a_{2}^{3}x_{2}^{8}+3a_{1}^{3}a_{2}^{2}x_{2}^{7}+5a_{2}^{4}x_{2}^{8}-12a_{1}a_{2}^{3}x_{2}^{7}+9a_{1}^{2}a_{2}^{2}x_{2}^{6}-\\\
&&2a_{1}^{3}a_{2}x_{2}^{5}-4a_{2}^{3}x_{2}^{6}+9a_{1}a_{2}^{2}x_{2}^{5}-6a_{1}^{2}a_{2}x_{2}^{4}+a_{1}^{3}x_{2}^{3}+3a_{2}^{2}x_{2}^{4}-6a_{1}a_{2}x_{2}^{3}+3a_{1}^{2}x_{2}^{2}-\\\
&&2a_{2}x_{2}^{2}+3a_{1}x_{2}+1\\\
y_{2}^{\prime}&=&a_{1}^{12}x_{2}^{12}+2a_{1}^{10}a_{2}x_{2}^{12}-a_{1}^{11}x_{2}^{11}+a_{1}^{8}a_{2}^{2}x_{2}^{12}-2a_{1}^{9}a_{2}x_{2}^{11}+a_{1}^{10}x_{2}^{10}-a_{1}^{7}a_{2}^{2}x_{2}^{11}+\\\
&&2a_{1}^{8}a_{2}x_{2}^{10}-a_{1}^{9}x_{2}^{9}+a_{1}^{6}a_{2}^{2}x_{2}^{10}-2a_{1}^{7}a_{2}x_{2}^{9}+a_{1}^{8}x_{2}^{8}-a_{1}^{5}a_{2}^{2}x_{2}^{9}+2a_{1}^{6}a_{2}x_{2}^{8}-a_{1}^{7}x_{2}^{7}+\\\
&&a_{1}^{4}a_{2}^{2}x_{2}^{8}-2a_{1}^{5}a_{2}x_{2}^{7}+a_{1}^{6}x_{2}^{6}-a_{1}^{3}a_{2}^{2}x_{2}^{7}+2a_{1}^{4}a_{2}x_{2}^{6}-a_{1}^{5}x_{2}^{5}+a_{1}^{2}a_{2}^{2}x_{2}^{6}-2a_{1}^{3}a_{2}x_{2}^{5}+\\\
&&a_{1}^{4}x_{2}^{4}-a_{1}a_{2}^{2}x_{2}^{5}+2a_{1}^{2}a_{2}x_{2}^{4}-a_{1}^{3}x_{2}^{3}+a_{2}^{2}x_{2}^{4}-2a_{1}a_{2}x_{2}^{3}+a_{1}^{2}x_{2}^{2}+2a_{2}x_{2}^{2}-a_{1}x_{2}+\\\
&&1\\\ y_{3}^{\prime}&=&a_{1}x_{2}^{2}+x_{2}\\\
y_{4}^{\prime}&=&a_{2}x_{2}^{3}+x_{2}\\\
y_{5}^{\prime}&=&\displaystyle\frac{a_{2}^{2}}{4}x_{2}^{4}-\displaystyle\frac{a_{1}^{3}+a_{1}a_{2}}{4}x_{2}^{3}+\displaystyle\frac{a_{1}^{2}-a_{2}}{4}x_{2}^{2}-\displaystyle\frac{a_{1}}{4}x_{2}+\displaystyle\frac{1}{4}.\end{array}$

However, the output is too big to be printed. Following the idea in the above
examples, we compute $Q$ and $g$. This is even bigger than $h$ so we print the
numerators and denominators of the coefficients just till degree 10 in the
$x_{i}$’s. However in some cases even the terms till degree 10 will be too
many to write down and hence we will print just the first terms and “$\ldots$”
.

As a small remark, $Q_{3}$ contains also terms in degree 3 in the $T_{i}$ but
the numerator of the coefficients have power greater than 10 and therefore
they do not appear in our shortcutting.

$\begin{array}[]{crc}Q_{1}=&\displaystyle\frac{\scriptstyle
3a_{1}^{2}x_{1}^{2}x_{2}^{8}-2a_{1}x_{1}^{2}x_{2}^{7}+4a_{2}x_{1}^{2}x_{2}^{8}+x_{1}^{2}x_{2}^{6}}{\scriptstyle
4a_{1}a_{2}^{2}x_{2}^{5}+8a_{1}a_{2}x_{2}^{3}+4a_{1}x_{2}+4a_{2}^{2}x_{2}^{4}+8a_{2}x_{2}^{2}+4}\cdot
T_{1}T_{4}-\displaystyle\frac{\scriptstyle x_{2}^{6}}{\scriptstyle
4a_{2}^{2}x_{2}^{4}+8a_{2}x_{2}^{2}+4}\cdot&T_{1}^{2}\\\
&+\displaystyle\frac{\scriptstyle-
a_{1}^{4}x_{2}^{10}+a_{1}^{3}x_{2}^{9}-2a_{1}^{2}a_{2}x_{2}^{10}-a_{1}^{2}x_{2}^{8}+2a_{1}a_{2}x_{2}^{9}+a_{1}x_{2}^{7}-a_{2}^{2}x_{2}^{10}-2a_{2}x_{2}^{8}-x_{2}^{6}}{\scriptstyle
4a_{1}a_{2}^{2}x_{2}^{5}+8a_{1}a_{2}x_{2}^{3}+4a_{1}x_{2}+4a_{2}^{2}x_{2}^{4}+8a_{2}x_{2}^{2}+4}\cdot&T_{1}T_{2}\\\
&+\displaystyle\frac{\scriptstyle-5a_{1}^{4}x_{2}^{10}+4a_{1}^{3}x_{2}^{9}-12a_{1}^{2}a_{2}x_{2}^{10}-3a_{1}^{2}x_{2}^{8}+8a_{1}a_{2}x_{2}^{9}+2a_{1}x_{2}^{7}-6a_{2}^{2}x_{2}^{10}-4a_{2}x_{2}^{8}-x_{2}^{6}}{\scriptstyle
16a_{1}^{2}a_{2}^{2}x_{2}^{6}+32a_{1}^{2}a_{2}x_{2}^{4}+16a_{1}^{2}x_{2}^{2}+32a_{1}a_{2}^{2}x_{2}^{5}+64a_{1}a_{2}x_{2}^{3}+32a_{1}x_{2}+16a_{2}^{2}x_{2}^{4}+32a_{2}x_{2}^{2}+16}\cdot&T_{2}^{2}\\\
&+\displaystyle\frac{\scriptstyle
a_{1}^{2}x_{1}^{2}x_{2}^{8}+2a_{1}^{2}x_{2}^{10}+2a_{1}x_{1}^{2}x_{2}^{7}+4a_{1}x_{2}^{9}+x_{1}^{2}x_{2}^{6}+2x_{2}^{8}}{\scriptstyle
4a_{1}a_{2}^{2}x_{2}^{5}+8a_{1}a_{2}x_{2}^{3}+4a_{1}x_{2}+4a_{2}^{2}x_{2}^{4}+8a_{2}x_{2}^{2}+4}\cdot&T_{1}T_{3}\\\
&+\displaystyle\frac{\scriptstyle
a_{1}x_{1}^{2}x_{2}^{7}-2a_{1}x_{2}^{9}+2a_{2}x_{1}^{2}x_{2}^{8}-4a_{2}x_{2}^{10}+x_{1}^{2}x_{2}^{6}-2x_{2}^{8}}{\scriptstyle
8a_{1}^{2}a_{2}^{2}x_{2}^{6}+16a_{1}^{2}a_{2}x_{2}^{4}+8a_{1}^{2}x_{2}^{2}+16a_{1}a_{2}^{2}x_{2}^{5}+32a_{1}a_{2}x_{2}^{3}+16a_{1}x_{2}+8a_{2}^{2}x_{2}^{4}+16a_{2}x_{2}^{2}+8}\cdot&T_{2}T_{3}\\\
&+\displaystyle\frac{\scriptstyle-
x_{1}^{4}x_{2}^{6}+4x_{1}^{2}x_{2}^{8}-4x_{2}^{10}}{\scriptstyle
16a_{1}^{2}a_{2}^{2}x_{2}^{6}+32a_{1}^{2}a_{2}x_{2}^{4}+16a_{1}^{2}x_{2}^{2}+32a_{1}a_{2}^{2}x_{2}^{5}+64a_{1}a_{2}x_{2}^{3}+32a_{1}x_{2}+16a_{2}^{2}x_{2}^{4}+32a_{2}x_{2}^{2}+16}\cdot&T_{3}^{2}\\\
&+\displaystyle\frac{\scriptstyle
6a_{1}^{2}x_{1}^{2}x_{2}^{8}-3a_{1}x_{1}^{2}x_{2}^{7}+6a_{2}x_{1}^{2}x_{2}^{8}+x_{1}^{2}x_{2}^{6}}{\scriptstyle
8a_{1}^{2}a_{2}^{2}x_{2}^{6}+16a_{1}^{2}a_{2}x_{2}^{4}+8a_{1}^{2}x_{2}^{2}+16a_{1}a_{2}^{2}x_{2}^{5}+32a_{1}a_{2}x_{2}^{3}+16a_{1}x_{2}+8a_{2}^{2}x_{2}^{4}+16a_{2}x_{2}^{2}+8}\cdot&T_{2}T_{4}\\\
&+\displaystyle\frac{\scriptstyle-
x_{1}^{4}x_{2}^{6}+2x_{1}^{2}x_{2}^{8}}{\scriptstyle
8a_{1}^{2}a_{2}^{2}x_{2}^{6}+16a_{1}^{2}a_{2}x_{2}^{4}+8a_{1}^{2}x_{2}^{2}+16a_{1}a_{2}^{2}x_{2}^{5}+32a_{1}a_{2}x_{2}^{3}+16a_{1}x_{2}+8a_{2}^{2}x_{2}^{4}+16a_{2}x_{2}^{2}+8}\cdot&T_{3}T_{4}\\\
&+\displaystyle\frac{\scriptstyle-x_{1}^{4}x_{2}^{6}}{\scriptstyle
16a_{1}^{2}a_{2}^{2}x_{2}^{6}+32a_{1}^{2}a_{2}x_{2}^{4}+16a_{1}^{2}x_{2}^{2}+32a_{1}a_{2}^{2}x_{2}^{5}+64a_{1}a_{2}x_{2}^{3}+32a_{1}x_{2}+16a_{2}^{2}x_{2}^{4}+32a_{2}x_{2}^{2}+16}\cdot&T_{4}^{2}\\\
Q_{2}=&\displaystyle\frac{\scriptstyle-x_{2}^{6}}{\scriptstyle
4a_{1}^{2}x_{2}^{2}+8a_{1}x_{2}+4}\cdot
T_{2}^{2}+\displaystyle\frac{\scriptstyle
3a_{1}^{2}x_{1}^{2}x_{2}^{8}+3a_{1}x_{1}^{2}x_{2}^{7}-2a_{2}x_{1}^{2}x_{2}^{8}+x_{1}^{2}x_{2}^{6}}{\scriptstyle
2a_{1}^{2}x\textasciiacute_{2}^{2}+4a_{1}x_{2}+2}\cdot
T_{2}T_{3}+\displaystyle\frac{\scriptstyle-x_{1}^{4}x_{2}^{6}}{\scriptstyle
4a_{1}^{2}x_{2}^{2}+8a_{1}x_{2}+4}\cdot&T_{3}^{2}\\\
&+\displaystyle\frac{\scriptstyle
a_{1}^{2}x_{1}^{2}x_{2}^{8}-a_{1}x_{1}^{2}x_{2}^{7}+2a_{2}x_{1}^{2}x_{2}^{8}+x_{1}^{2}x_{2}^{6}}{\scriptstyle
2a_{1}^{2}x_{2}^{2}+4a_{1}x_{2}+2}\cdot
T_{2}T_{4}+\displaystyle\frac{\scriptstyle-
x_{1}^{4}x_{2}^{6}+2x_{1}^{2}x_{2}^{8}}{\scriptstyle
2a_{1}^{2}x_{2}^{2}+4a_{1}x_{2}+2}\cdot
T_{3}T_{4}+\displaystyle\frac{\scriptstyle-x_{1}^{4}x_{2}^{6}}{\scriptstyle
4a_{1}^{2}x_{2}^{2}+8a_{1}x_{2}+4}\cdot&T_{4}^{2}\\\
Q_{3}=&\displaystyle\frac{\scriptstyle
2a_{1}^{3}x_{2}^{9}-4a_{1}^{2}a_{2}x_{2}^{10}-2a_{1}^{2}x_{2}^{8}+2a_{1}a_{2}x_{2}^{9}+2a_{1}x_{2}^{7}-2a_{2}x_{2}^{8}-2x_{2}^{6}}{\scriptstyle
4a_{1}a_{2}^{3}x_{2}^{7}+12a_{1}a_{2}^{2}x_{2}^{5}+12a_{1}a_{2}x_{2}^{3}+4a_{1}x_{2}+4a_{2}^{3}x_{2}^{6}+12a_{2}^{2}x_{2}^{4}+12a_{2}x_{2}^{2}+4}\cdot&T_{1}T_{2}\\\
&+\displaystyle\frac{\scriptstyle
7a_{1}^{3}x_{2}^{9}-28a_{1}^{2}a_{2}x_{2}^{10}-7a_{1}^{2}x_{2}^{8}+21a_{1}a_{2}x_{2}^{9}+7a_{1}x_{2}^{7}-21a_{2}^{2}x_{2}^{10}-21a_{2}x_{2}^{8}-7x_{2}^{6}}{\scriptstyle\ldots+48a_{1}^{2}x_{2}^{2}+48a_{1}a_{2}^{3}x_{2}^{7}+144a_{1}a_{2}^{2}x_{2}^{5}+144a_{1}a_{2}x_{2}^{3}+48a_{1}x_{2}+16a_{2}^{3}x_{2}^{6}+48a_{2}^{2}x_{2}^{4}+48a_{2}x_{2}^{2}+16}\cdot&T_{2}^{2}\\\
&+\displaystyle\frac{\scriptstyle
2a_{1}^{2}x_{1}^{2}x_{2}^{8}+2a_{1}^{2}x_{2}^{10}+4a_{1}x_{1}^{2}x_{2}^{7}+4a_{1}x_{2}^{9}-2a_{2}x_{1}^{2}x_{2}^{8}-2a_{2}x_{2}^{10}+2x_{1}^{2}x_{2}^{6}+2x_{2}^{8}}{\scriptstyle
4a_{1}a_{2}^{3}x_{2}^{7}+12a_{1}a_{2}^{2}x_{2}^{5}+12a_{1}a_{2}x_{2}^{3}+4a_{1}x_{2}+4a_{2}^{3}x_{2}^{6}+12a_{2}^{2}x_{2}^{4}+12a_{2}x_{2}^{2}+4}\cdot&T_{1}T_{3}\\\
&+\displaystyle\frac{\scriptstyle
7a_{1}^{2}x_{1}^{2}x_{2}^{8}+4a_{1}^{2}x_{2}^{10}+14a_{1}x_{1}^{2}x_{2}^{7}+8a_{1}x_{2}^{9}+7a_{2}x_{1}^{2}x_{2}^{8}+4a_{2}x_{2}^{10}+7x_{1}^{2}x_{2}^{6}+4x_{2}^{8}}{\scriptstyle\ldots+24a_{1}^{2}x_{2}^{2}+24a_{1}a_{2}^{3}x_{2}^{7}+72a_{1}a_{2}^{2}x_{2}^{5}+72a_{1}a_{2}x_{2}^{3}+24a_{1}x_{2}+8a_{2}^{3}x_{2}^{6}+24a_{2}^{2}x_{2}^{4}+24a_{2}x_{2}^{2}+8}\cdot&T_{2}T_{3}\\\
&+\displaystyle\frac{\scriptstyle-7x_{1}^{4}x_{2}^{6}-8x_{1}^{2}x_{2}^{8}-4x_{2}^{10}}{\scriptstyle\ldots+48a_{1}^{2}x_{2}^{2}+48a_{1}a_{2}^{3}x_{2}^{7}+144a_{1}a_{2}^{2}x_{2}^{5}+144a_{1}a_{2}x_{2}^{3}+48a_{1}x_{2}+16a_{2}^{3}x_{2}^{6}+48a_{2}^{2}x_{2}^{4}+48a_{2}x_{2}^{2}+16}\cdot&T_{3}^{2}\\\
&+\displaystyle\frac{\scriptstyle
6a_{1}^{2}x_{1}^{2}x_{2}^{8}-4a_{1}x_{1}^{2}x_{2}^{7}+6a_{2}x_{1}^{2}x_{2}^{8}+2x_{1}^{2}x_{2}^{6}}{\scriptstyle
4a_{1}a_{2}^{3}x_{2}^{7}+12a_{1}a_{2}^{2}x_{2}^{5}+12a_{1}a_{2}x_{2}^{3}+4a_{1}x_{2}+4a_{2}^{3}x_{2}^{6}+12a_{2}^{2}x_{2}^{4}+12a_{2}x_{2}^{2}+4}\cdot&T_{1}T_{4}\\\
&+\displaystyle\frac{\scriptstyle
21a_{1}^{2}x_{1}^{2}x_{2}^{8}-14a_{1}x_{1}^{2}x_{2}^{7}+35a_{2}x_{1}^{2}x_{2}^{8}+7x_{1}^{2}x_{2}^{6}}{\scriptstyle\ldots+24a_{1}^{2}x_{2}^{2}+24a_{1}a_{2}^{3}x_{2}^{7}+72a_{1}a_{2}^{2}x_{2}^{5}+72a_{1}a_{2}x_{2}^{3}+24a_{1}x_{2}+8a_{2}^{3}x_{2}^{6}+24a_{2}^{2}x_{2}^{4}+24a_{2}x_{2}^{2}+8}\cdot&T_{2}T_{4}\\\
&+\displaystyle\frac{\scriptstyle-7x_{1}^{4}x_{2}^{6}-4x_{1}^{2}x_{2}^{8}}{\scriptstyle\ldots+24a_{1}^{2}x_{2}^{2}+24a_{1}a_{2}^{3}x_{2}^{7}+72a_{1}a_{2}^{2}x_{2}^{5}+72a_{1}a_{2}x_{2}^{3}+24a_{1}x_{2}+8a_{2}^{3}x_{2}^{6}+24a_{2}^{2}x_{2}^{4}+24a_{2}x_{2}^{2}+8}\cdot&T_{3}T_{4}\\\
\end{array}$

$\begin{array}[]{crc}\phantom{Q_{3}=}&+\displaystyle\frac{\scriptstyle-7x_{1}^{4}x_{2}^{6}}{\scriptstyle\ldots+48a_{1}^{2}x_{2}^{2}+48a_{1}a_{2}^{3}x_{2}^{7}+144a_{1}a_{2}^{2}x_{2}^{5}+144a_{1}a_{2}x_{2}^{3}+48a_{1}x_{2}+16a_{2}^{3}x_{2}^{6}+48a_{2}^{2}x_{2}^{4}+48a_{2}x_{2}^{2}+16}\cdot&T_{4}^{2}\\\
&+\displaystyle\frac{\scriptstyle
a_{1}^{4}x_{2}^{10}-a_{1}^{2}a_{2}x_{2}^{10}-a_{2}^{2}x_{2}^{10}+a_{2}x_{2}^{8}-x_{2}^{6}}{\scriptstyle
4a_{2}^{3}x_{2}^{6}+12a_{2}^{2}x_{2}^{4}+12a_{2}x_{2}^{2}+4}\cdot
T_{1}^{2}+\displaystyle\frac{\scriptstyle-x_{2}^{6}}{\scriptstyle
2a_{2}^{2}x_{2}^{4}+4a_{2}x_{2}^{2}+2}\cdot&T_{1}T_{5}\\\
&+\displaystyle\frac{\scriptstyle-3a_{2}^{2}x_{2}^{10}-6a_{2}x_{2}^{8}-3x_{2}^{6}}{\scriptstyle
4a_{1}^{2}a_{2}^{2}x_{2}^{6}+8a_{1}^{2}a_{2}x_{2}^{4}+4a_{1}^{2}x_{2}^{2}+8a_{1}a_{2}^{2}x_{2}^{5}+16a_{1}a_{2}x_{2}^{3}+8a_{1}x_{2}+4a_{2}^{2}x_{2}^{4}+8a_{2}x_{2}^{2}+4}\cdot&T_{2}T_{5}\\\
&+\displaystyle\frac{\scriptstyle
9a_{1}^{2}x_{1}^{2}x_{2}^{8}+6a_{1}^{2}x_{2}^{10}+9a_{1}x_{1}^{2}x_{2}^{7}+6a_{1}x_{2}^{9}+3x_{1}^{2}x_{2}^{6}+2x_{2}^{8}}{\scriptstyle
4a_{1}^{2}a_{2}^{2}x_{2}^{6}+8a_{1}^{2}a_{2}x_{2}^{4}+4a_{1}^{2}x_{2}^{2}+8a_{1}a_{2}^{2}x_{2}^{5}+16a_{1}a_{2}x_{2}^{3}+8a_{1}x_{2}+4a_{2}^{2}x_{2}^{4}+8a_{2}x_{2}^{2}+4}\cdot&T_{3}T_{5}\\\
&+\displaystyle\frac{\scriptstyle
3a_{1}^{2}x_{1}^{2}x_{2}^{8}-3a_{1}x_{1}^{2}x_{2}^{7}+12a_{2}x_{1}^{2}x_{2}^{8}+3x_{1}^{2}x_{2}^{6}}{\scriptstyle
4a_{1}^{2}a_{2}^{2}x_{2}^{6}+8a_{1}^{2}a_{2}x_{2}^{4}+4a_{1}^{2}x_{2}^{2}+8a_{1}a_{2}^{2}x_{2}^{5}+16a_{1}a_{2}x_{2}^{3}+8a_{1}x_{2}+4a_{2}^{2}x_{2}^{4}+8a_{2}x_{2}^{2}+4}\cdot&T_{4}T_{5}\par\par\par\par\\\
\end{array}$

Having $Q_{i}$ we obtain $g_{i}$:

$\begin{array}[]{rcrc}g_{1}=Q_{1}+&{\big{(}\scriptstyle\ldots-6a_{1}^{4}x_{2}^{4}+30a_{1}^{3}a_{2}^{3}x_{2}^{9}+45a_{1}^{2}a_{2}^{4}x_{2}^{10}+6a_{1}^{2}a_{2}^{2}x_{2}^{6}+6a_{1}^{2}a_{2}x_{2}^{4}+6a_{1}a_{2}^{3}x_{2}^{7}+6a_{1}a_{2}^{2}x_{2}^{5}+6a_{2}^{3}x_{2}^{6}+1\big{)}}&\cdot
T_{1}\\\
g_{2}=Q_{2}+&{\big{(}\scriptstyle\ldots-6a_{1}^{4}x_{2}^{4}+30a_{1}^{3}a_{2}^{3}x_{2}^{9}+45a_{1}^{2}a_{2}^{4}x_{2}^{10}+6a_{1}^{2}a_{2}^{2}x_{2}^{6}+6a_{1}^{2}a_{2}x_{2}^{4}+6a_{1}a_{2}^{3}x_{2}^{7}+6a_{1}a_{2}^{2}x_{2}^{5}+6a_{2}^{3}x_{2}^{6}+1\big{)}}&\cdot
T_{2}\\\
+&{\big{(}\scriptstyle\ldots+6a_{1}^{6}x_{2}^{4}-30a_{1}^{5}a_{2}^{3}x_{2}^{9}-45a_{1}^{4}a_{2}^{4}x_{2}^{10}-6a_{1}^{4}a_{2}^{2}x_{2}^{6}-6a_{1}^{4}a_{2}x_{2}^{4}-6a_{1}^{3}a_{2}^{3}x_{2}^{7}-6a_{1}^{3}a_{2}^{2}x_{2}^{5}-6a_{1}^{2}a_{2}^{3}x_{2}^{6}-a_{1}^{2}\big{)}}&\\\
g_{3}=Q_{3}+&{\big{(}\scriptstyle\ldots-6a_{1}^{4}x_{2}^{4}+30a_{1}^{3}a_{2}^{3}x_{2}^{9}+45a_{1}^{2}a_{2}^{4}x_{2}^{10}+6a_{1}^{2}a_{2}^{2}x_{2}^{6}+6a_{1}^{2}a_{2}x_{2}^{4}+6a_{1}a_{2}^{3}x_{2}^{7}+6a_{1}a_{2}^{2}x_{2}^{5}+6a_{2}^{3}x_{2}^{6}+1\big{)}}&\cdot
T_{3}\\\
+&{\big{(}\scriptstyle\ldots+24a_{1}^{2}a_{2}^{5}x_{2}^{10}+18a_{1}^{2}a_{2}^{4}x_{2}^{8}+a_{1}^{2}a_{2}^{2}x_{2}^{4}+a_{1}^{2}a_{2}x_{2}^{2}+12a_{1}a_{2}^{5}x_{2}^{9}+a_{1}a_{2}^{3}x_{2}^{5}+a_{1}a_{2}^{2}x_{2}^{3}+6a_{2}^{6}x_{2}^{10}+a_{2}^{3}x_{2}^{4}\big{)}}.&\\\
\end{array}$

The General Neron Desingularization is a localization of $D[Y,T]/(h,g)$. For
this example we will need a function

    invp(poly p, int bound,string param,string variab)

which computes computes the inverse of `p" till order ``bound" in
$\mathbb{Q}($``param"$)[$``variab"$]$. The input for this example is the
following: \begin{verbatim} ring All = 0,(a1,a2,x1,x2,Y1,Y2,Y3,Y4),dp; int nra
= 2; int nrx = 2; int nry = 4; ideal xid = x1^2-x2^3; ideal yid =
Y3^2-x1^2*Y1*Y2,Y4^2-x2*Y2*Y3; ideal aid = 0; poly y1,y2,y3,y4; y3 = 1+a1*x2;
y4 = 1+a2*x2^2; string as,xs; if(nra != 0) { as = string(var(1)); for( int
i=2;i<=nra;i++) { as = as+`,”+string(var(i));

if(nrx!=0) xs = string(var(nra+1)); for(int i=nra+2;i¡=nra+nrx;i++) xs =
xs+”,”+string(var(i)); y1 = y3^3*invp(y4^2,12,as,xs); y2 =
y4^2*invp(y3,12,as,xs); y3 = x2*y3; y4 = x2*y4; ideal f = y1,y2,y3,y4;
desingularization(All, nra,nrx,nry,xid,yid,aid,f,”injective”,”debug”);

######  Remark 28.

Our algorithm works mainly for local domains of dimension one. If $A^{\prime}$
is not a domain but a Cohen-Macaulay ring of dimension one then we can build
an algorithm in the idea of the proof of Theorem 20. In this case it is
necessary to change $B$ by an Elkik’s trick [4] (see [8, Lemma 3.4], [13,
Proposition 4.6], [11, Corollary 5.10]). The algorithm and as well Theorem 20
might be also build when $A^{\prime}$ is not Cohen-Macaulay substituting in
the proofs $d$ by a certain power $d^{r}$ such that
$(0:_{A}d^{r})=(0:_{A}d^{r+1})$. Such algorithm could be too complicated to
work really. On the other hand, if we restrict our present algorithm to the
case when $A^{\prime}$ is the completion of $A$ then we might get a faster
algorithm using the idea of the proof of Theorem 20. This algorithm could be
useful in the arc frame.

## References

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