#  Simple General Neron Desingularization in local $\mathbb{Q}$-algebras.

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.

We give here a easier proof of the so-called General Neron Desingularization
in the frame of local algebras.

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

## Introduction

A ring morphism $u:A\rightarrow A^{\prime}$ of Noetherian rings has regular
fibers if for all prime ideals $p\in\operatorname{Spec}A$ the ring
$A^{\prime}/pA^{\prime}$ is a regular ring. 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. A flat morphism of
Noetherian rings $u$ is regular if its fibers are geometrically regular. If
$u$ is regular of finite type then $u$ is called smooth.

The following theorem extends Néron’s desingularization (see [11], [7]) and it
was used to solve different problems concerning projective modules over
regular rings, or in Artin Approximation Theory (see [2], [16], [18], [19],
[20], [24], [22], [4]).

######  Theorem 1.

(General Néron Desingularization, Popescu [16], [17], [18], André [1], Swan
[24]) Let $u:A\rightarrow A^{\prime}$ be a regular morphism of Noetherian
rings and $B$ an $A$-algebra of finite type. Then any $A$-morphism
$v:B\rightarrow A^{\prime}$ factors through a smooth $A$-algebra $C$, that is
$v$ is a composite $A$-morphism $B\rightarrow C\rightarrow A^{\prime}$.

Other authors gave similar proofs (see [23], [21], [9]) but most of them were
not constructive. In [14], [12], [13], [5], [6], [7], [8], there exist some
constructive proofs with algorithms in some special cases. We mention that all
these proofs are simpler when we deal with $\mathbb{Q}$-algebras because in
this case all residue field extensions are separable and it is easier to
handle the regularity of the morphisms.

Let $A$ be a Noetherian ring, $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$. 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 [3] 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 $r\leq n$.
$H_{B/A}$ defines the non smooth locus of $B$ over $A$. $B$ is standard smooth
over $A$ if there exists $f$ in $I$ as above such that $B=((f):I)\Delta_{f}B$.

In [8], it is given the following constructive theorem.

######  Theorem 2.

Let $A$ and $A^{\prime}$ be Noetherian local rings, $u:A\rightarrow
A^{\prime}$ be a regular morphism and $B$ an $A$-algebra of finite type.
Suppose that $A^{\prime}$ is Henselian and the maximal ideal ${\mathfrak{m}}$
of $A$ generates the maximal ideal of $A^{\prime}$. Any $A$-morphism
$v:B\rightarrow A^{\prime}$ such that $v(H_{B/A}A^{\prime})$ is
${\mathfrak{m}}A^{\prime}$-primary ideal factors through a standard smooth
$A$-algebra $B^{\prime}$.

The proof was based on the following proposition (see [8, Proposition 3],
where the conditions ”$A^{\prime}$ is Henselian, and $A,A^{\prime}$ have
dimensions $m$ are superfluous” and we remove them here).

######  Proposition 3.

Let $A$ and $A^{\prime}$ be Noetherian local rings and $u:A\rightarrow
A^{\prime}$ be a regular morphism. Let $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$,
$f=(f_{1},\ldots,f_{r})$, $r\leq n$ be a system of polynomials from $I$ as
above, $(M_{j})_{j\in[l]}$ some $r\times r$-minors 111 We use the notation
$[l]=\\{1,\ldots,l\\}$. of the Jacobian matrix $(\partial f_{i}/\partial
Y_{j^{\prime}})$, $(N_{j})_{j\in[l]}\in((f):I)$ and set
$P:=\sum_{j=1}^{l}N_{j}M_{j}$. Let $v:B\rightarrow A^{\prime}$ be an
$A$-morphism. Suppose that

  1. (1)

there exists an element $d\in A$ such that $d\equiv P$ modulo $I$ and

  2. (2)

there exist a smooth $A$-algebra $D$ and an $A$-morphism $\omega:D\rightarrow
A^{\prime}$ such that
$\operatorname{Im}v\subset\operatorname{Im}\omega+d^{2e+1}A^{\prime}$ and for
${\bar{A}}=A/(d^{2e+1})$ (defining e by $(0:_{A}d^{e})=(0:_{A}d^{e+1})$) the
map
$\bar{v}={\bar{A}}{\otimes}_{A}v:\bar{B}=B/d^{2e+1}B\rightarrow\bar{A}^{\prime}=A^{\prime}/d^{2e+1}A^{\prime}$
factors through $\bar{D}=D/d^{2e+1}D$.

Then there exists a $B$-algebra $B^{\prime}$ which is standard smooth over $A$
such that $v$ factors through $B^{\prime}$.

It is the purpose of our paper to provide a easier proof (see Theorem 9) of
Theorem 1 in the local $\mathbb{Q}$-algebras following the ideas from Theorem
2 and Proposition 3. Here $\mathbb{Q}$ denotes the field of rational numbers.
The proof is not constructive (see Remark 7), but Proposition 10 shows a case
when this is so. We mention that the proof of Theorem 9 is easier, because
mainly we had to apply only Proposition 5, which concentrates in short the
ideas of the previous proofs from [15], [18] in the frame of
$\mathbb{Q}$-algebras.

##  1\. A simple proof of the General Neron Desingularization

Let $q\in\operatorname{Spec}A^{\prime}$ be a minimal prime associated ideal of
$h_{B}=\sqrt{v(H_{B/A})A^{\prime}}$. After [24] we say that $A\rightarrow
B\rightarrow A^{\prime}\supset q$ is resolvable if there exists an $A$-algebra
of finite type $C$ such that $v$ factors through $C$, let us say $v=\beta t$,
$t:B\rightarrow C$, $\beta:C\rightarrow A^{\prime}$ and $h_{B}\subset
h_{C}=\sqrt{\beta(H_{C/A})A^{\prime}}\not\subset q$.

The following lemma is in fact [18, Lemma 6.6] which unifies [24, Lemmas 12.2,
12.3]. We give here its proof in sketch for the sake of completeness.

######  Lemma 4.

Suppose that $\operatorname{height}q=0$ and $A_{p}\rightarrow
B_{p}=A_{p}\otimes_{A}B\otimes A^{\prime}_{q}\supset qA^{\prime}_{q}$,
$p=u^{-1}(q)$ is resolvable. Then $A\rightarrow B\rightarrow A^{\prime}\supset
q$ is resolvable too.

###### Proof.

By hypothesis there exists an $A_{p}$-algebra $D$ such that
$v_{p}=A_{p}\otimes_{A}v$ factors through $D$, let us say $v_{p}$ is the
composite map $B_{p}\rightarrow D\xrightarrow{\beta_{p}}A^{\prime}_{q}$ and
$\beta_{p}(H_{D/A_{p}})=A^{\prime}_{q}$. Moreover we may take $D$ to be smooth
over $A_{p}$ using for instance the easy lemma [15, Lemma 2.4]

Let $D=B_{p}[Z]/(g)$, $Z=(Z_{1},\ldots,Z_{s})$, $g=(g_{1},\ldots,g_{e})\in
B^{e}$ and $\beta_{p}$ be given by $Z\rightarrow z/z_{0}$ for some $z\in
A^{\prime s}$, $z_{0}\in A^{\prime}\setminus q$ Take the homogeneous
polynomials $G(Z,Z_{0})=Z_{0}^{c}g(Z/Z_{0})$ for some $c>>1$. We have
$G(z,z_{0})=0$ in $A^{\prime}_{q}$ and so $aG(z,z_{0})=0$ in $A^{\prime}$ for
some $a\in A^{\prime}\setminus q$. Changing $z,z_{0}$ by $az,az_{0}$ we may
assume that $G(z,z_{0})=0$ in $A^{\prime}$. Let $C=B[Z,Z_{0}]/(G)$ and
$\beta:C\rightarrow A^{\prime}$ be the extension of $v$ by
$(Z,Z_{0})\rightarrow(z,z_{0})$. We get a $B_{p}$-isomorphism
$\nu:D[Z_{0},Z_{0}^{-1}]\rightarrow(B_{p}\otimes_{B}C)_{Z_{0}}$ by
$(Z,Z_{0})\rightarrow(Z/Z_{0},Z_{0})$ and the map $C_{p,Z_{0}}\rightarrow
A^{\prime}_{q}$ induced by $\beta$, factors through $\nu$. We may change $D$
by the smooth $A_{p}$-algebra $D[Z_{0},Z_{0}^{-1}]$, resp. $g$ by $G$, $C$ by
$B[Z]/(g)$. We have $h_{C}=\sqrt{\beta(H_{C/A})A^{\prime}}\not\subset q$, let
us say $\beta(t)\not\in q$ for some $t\in H_{C/A}$.

If $h_{B}\subset h_{C}$ there exists nothing to show. Otherwise, $h_{B}^{k}=0$
in $A^{\prime}_{q}$, that is $wh_{B}^{k}=0$ in $A^{\prime}$ for some $w\in
A^{\prime}\setminus q$. Let $b_{1},\ldots,b_{s}$ be a system of generators of
$h_{B}$ and set $f_{j}=g_{j}-\sum_{i}U_{i}V_{ij}$, for some new variables
$U=(U_{1},\ldots,U_{s})$, $V=(V_{ij})_{i\in[s],j\in[e]}$. Then

$E=B[Z,U,V,W]/(f,WU_{1},\ldots,WU_{s})$

and $\delta:E\rightarrow A^{\prime}$ extending $v$ by $U\rightarrow b$,
$V\rightarrow 0$, $W\rightarrow w$ resolves $A\rightarrow B\rightarrow
A^{\prime}\supset q$. Indeed, $E_{U_{i}}$ is a polynomial algebra over
$B[U_{i},U_{i}^{-1}]$, so smooth over $B$. Thus $E_{b_{i}U_{i}}$ is smooth
over $A$ and $b_{i}U_{i}\in H_{E/A}$, that is $h_{B}\subset
h_{E}=\sqrt{\delta(H_{E/A})A^{\prime}}$. Note that $E_{W}\cong
B[Z,V,W,W^{-1}]/(g)\cong C[V,W,W^{-1}]$ and so $E_{tW}$ is smooth over $A$.
Hence $\beta(t)w\in h_{E}\setminus q$.

The proof of Proposition 3 works for the extension given below (its idea comes
from [15, Proposition 7.1] and [18, Lemma 7.2]).

######  Proposition 5.

Let $A$ and $A^{\prime}$ be Noetherian rings and $u:A\rightarrow A^{\prime}$
be a flat morphism. Suppose that $A^{\prime}$ is local. Let $B=A[Y]/I$,
$Y=(Y_{1},\ldots,Y_{n})$, $f=(f_{1},\ldots,f_{r})$, $r\leq n$ be a system of
polynomials from $I$ as above, $(M_{j})_{j\in[l]}$ some $r\times r$-minors of
the Jacobian matrix $(\partial f_{i}/\partial Y_{j^{\prime}})$,
$(N_{j})_{j\in[l]}\in((f):I)$ and set $P:=\sum_{j=1}^{l}N_{j}M_{j}$. Let
$v:B\rightarrow A^{\prime}$ be an $A$-morphism. Suppose that

  1. (1)

there exists an element $d\in A$ such that $d\equiv P$ modulo $I$ and

  2. (2)

there exist a flat $A$-algebra $D$ of finite type and an $A$-morphism
$\omega:D\rightarrow A^{\prime}$ such that
$\operatorname{Im}v\subset\operatorname{Im}\omega+d^{2e+1}A^{\prime}$ and for
${\bar{A}}=A/(d^{2e+1})$ (defining e by $(0:_{A}d^{e})=(0:_{A}d^{e+1})$) the
map
$\bar{v}={\bar{A}}{\otimes}_{A}v:\bar{B}=B/d^{2e+1}B\rightarrow\bar{A}^{\prime}=A^{\prime}/d^{2e+1}A^{\prime}$
factors through $\bar{D}=D/d^{2e+1}D$.

Then there exists a smooth $D$-algebra $B^{\prime}$ such that $v$ factors
through $B^{\prime}$, let us say $v$ is the composite map $B\rightarrow
B^{\prime}\xrightarrow{w}A^{\prime}$ and
$h_{D}=\sqrt{\omega(H_{D/A})A^{\prime}}\subset
h_{B^{\prime}}=\sqrt{w(H_{B^{\prime}/A})A^{\prime}}$.

###### Proof.

(Sketch after [8, Proposition 3]) Let $\delta:B\otimes_{A}D\cong
D[Y]/ID[Y]\rightarrow A^{\prime}$ be the $A$-morphism given by
$b\otimes\lambda\rightarrow v(b)\omega(\lambda)$. We show that $\delta$
factors through a special $B\otimes_{A}D$-algebra $E$ of finite type.

By hypothesis $\bar{v}$ factors through a map $\bar{B}\rightarrow\bar{D}$
given, let us say, by $Y\rightarrow y^{\prime}+d^{2e+1}D$, $y^{\prime}\in
D^{n}$. Thus $I(y^{\prime})\equiv 0$ modulo $d^{2e+1}D$ and
${\tilde{y}}:=\omega(y^{\prime})\equiv v(Y)=:y$ modulo $d^{2e+1}A^{\prime n}$,
let us say $y-\tilde{y}=d^{e+1}\nu$ for some $\nu\in d^{e}A^{\prime n}$.

For $j\in[q]$ we may complete the matrix $(\partial f_{i}/\partial
Y_{j^{\prime}})_{i\in[r],{j^{\prime}}\in[n]}$ with some $(n-r)$ rows from $0,\
1$ to get a square matrix $H_{j}$ with $M_{j}=\det H_{j}$. Since $d\equiv P$
modulo $I$ it follows that and so $P(y^{\prime})=ds$ for some $s\in D$ with
$s\equiv 1$ modulo $d$. Let $G^{\prime}_{j}$ be the adjoint matrix of $H_{j}$
and $G_{j}=N_{j}G^{\prime}_{j}$. We have
$G_{j}H_{j}=H_{j}G_{j}=M_{j}N_{j}\mbox{Id}_{n}$ and

$ds\mbox{Id}_{n}=P(y^{\prime})\mbox{Id}_{n}=\sum_{j=1}^{q}G_{j}(y^{\prime})H_{j}(y^{\prime}).$

We obtain

(1) $(\partial f/\partial Y){G_{j}}=(M_{j}N_{j}\mbox{Id}_{r}|0).$

${t_{j}}:=H_{j}(y^{\prime})\nu\in d^{e}A^{\prime n}$ satisfies

${G}_{j}(y^{\prime}){t_{j}}=M_{j}(y^{\prime})N_{j}(y^{\prime})\nu=d{s}\nu$

and

${s}(y-\tilde{y})=d^{e}\sum_{j=1}^{q}\omega({G}_{j}(y^{\prime})){t_{j}}.$

Let

(2) ${h}={s}(Y-y^{\prime})-d^{e}\sum_{j=1}^{q}{G}_{j}(y^{\prime}){T_{j}},$

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

${s}(Y-y^{\prime})\equiv d^{e}\sum_{j=1}^{q}{G}_{j}(y^{\prime})T_{j}\
\mbox{modulo}\ {h}$

and

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

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

(3)
${s}^{p}f(Y)-{s}^{p}f(y^{\prime})\equiv\sum_{j^{\prime}}{s}^{p-1}d^{e}(\partial
f/\partial
Y_{j^{\prime}})(y^{\prime})\sum_{j=1}^{q}{G}_{jj^{\prime}}(y^{\prime}){T}_{jj^{\prime}}+d^{2e}{Q}$

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

(4) ${g}_{i}={s}^{p}{b}_{i}+{s}^{p}{T}_{i}+d^{e-1}{Q}_{i},\qquad i\in[r]$

is in the kernel of $\varphi$ as in the proof of [8, Proposition 3].

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 ${s}^{rp}+({T})\subset 1+(d,{T})$ because
$Q\in({T})^{2}$. Then $V:=(D[Y,{T}]/({h},{g}))_{{s}{s}^{\prime}}$ is smooth
over $D$.

As in the proof of [8, Proposition 3] we see that
$I\subset({h},{g})D[Y,{T}]_{{s}{s}^{\prime}{s}^{\prime\prime}}$ for some other
${s}^{\prime\prime}\in 1+(d,{T})D[Y,{T}]$. Indeed, we have
$(({h},{g})D[Y,{T}]_{s}):I$ contains $P$ and so
$P(y^{\prime}+{s}^{-1}d^{e}\sum_{j=1}^{q}G_{j}(y^{\prime})T_{j})$ from
$P(y^{\prime})+d^{e}(T)D[Y,T]_{s}$. Thus $(({h},{g})D[Y,{T}]_{s}):I$ contains
an element of type $ds^{\prime\prime}$ for some $s^{\prime\prime}\in
s+d^{e-1}(T)D[Y,T]_{s}$ and we get ${s}^{\prime\prime}IV\subset(0:_{V}d)\cap
d^{e}V=0$ because $V$ is flat over $D$ and so over $A$ by hypothesis. Then
${E}_{{s}{s}^{\prime}{s}^{\prime\prime}}\cong{V}_{{s}^{\prime\prime}}$ is a
$B$-algebra which is also smooth over $D$.

As $\omega({s}),{\psi}({s}^{\prime}),{\psi}({s}^{\prime\prime})\equiv 1$
modulo $(d)$ we see that
$\omega({s}),{\psi}({s}^{\prime}),{\psi}({s}^{\prime\prime})$ are invertible
because $A^{\prime}$ is local. Thus ${\psi}$ (and so $v$) factors through the
$A$-algebra $B^{\prime}={E}_{{s}{s}^{\prime}{s}^{\prime\prime}}$. Clearly,
$h_{D}\subset h_{B^{\prime}}$.

Next theorem is an extension of Theorem 2.

######  Theorem 6.

Let $u:A\rightarrow A^{\prime}$ be a flat morphism of Noetherian rings and
$q\in\operatorname{Spec}A^{\prime}$. Suppose that $A^{\prime}$ is local,
$p=u^{-1}(q)\in\operatorname{Spec}A$ generates the maximal ideal
$qA^{\prime}_{q}$ of $A^{\prime}_{q}$ and $u$ induces a regular morphism
$A_{p}\rightarrow A^{\prime}_{q}$. Any $A$-morphism $v:B\rightarrow
A^{\prime}$ such that $v(H_{B/A})A^{\prime}_{q}$ is $qA^{\prime}_{q}$-primary
ideal factors through an $A$-algebra of finite type $B^{\prime}$, let us say
$v$ is the composite map $B\rightarrow B^{\prime}\xrightarrow{w}A^{\prime}$,
with $h_{B}\subset
h_{B^{\prime}}=\sqrt{w(H_{B^{\prime}/A})A^{\prime}}\not\subset q$, that is
$A\rightarrow B\rightarrow A^{\prime}\supset q$ is resolvable.

###### Proof.

Let $m=\operatorname{height}\ p=\operatorname{height}\ q$ and choose
$\gamma_{1},\ldots,\gamma_{m}\in u^{-1}(v(H_{B/A}A^{\prime}_{q})$, which
induces a system of parameters in $A^{\prime}_{q}$. The proof goes with
$A^{\prime}_{q}$ instead $A^{\prime}$ as in [8, Theorem 2] up to the ”Cases”,
where we proceed as follows.

Case I If $m=0$ then $A_{p},\ A^{\prime}_{q}$ are Artinian local rings and by
[15, Corollary 3.3] we get that $A_{p}\rightarrow A_{p}\otimes_{A}B\rightarrow
A^{\prime}_{q}\supset qA^{\prime}_{q}$ is resolvable. Then by Lemma 4 it
follows that $A\rightarrow B\rightarrow A^{\prime}\supset q$ is resolvable.

Case II If $m>0$ we assume by induction hypothesis as in [8, Theorem 2]
$\gamma=\gamma_{m}$, $d=d_{m}$ and the existence of an
$\bar{A}=A/(d^{2e+1})$-algebra of finite type
${\bar{D}}=D/(d^{2e+1})D\cong({\bar{A}}[Z]/({\bar{g}}))_{{\bar{h}}{\bar{M}}}$,
$Z=(Z_{1},\ldots,Z_{r})$, ${\bar{g}}=({\bar{g}}_{1},\ldots,{\bar{g}}_{s})$
such that $\bar{v}=\bar{A}\otimes_{A}v$ factors through $\bar{D}$, let us say
$\bar{v}$ is the composite map
$\bar{B}\rightarrow\bar{D}\xrightarrow{\bar{\omega}}\bar{A}^{\prime}$ and
$h_{B}{\bar{A}}^{\prime}\subset
h_{{\bar{D}}}=\sqrt{{\bar{\omega}}(H_{\bar{D}/\bar{A}}){\bar{A}}^{\prime}}\not\subset
q{\bar{A}}^{\prime}$.

Now, let $g,M,h$ be some liftings of $\bar{g},\bar{M},\bar{h}$ in $A[Z]$,
$f_{i}=g_{i}-d^{2e+1}T_{i}$, $T=(T_{1},\ldots,T_{s})$ and set $D=A[Z,T]/(f)$.
Let $t\in A^{\prime s}$ be such that $g(z)=d^{2e+1}t$. The map
$\omega:D\rightarrow A^{\prime}$ given by $(Z,T)\rightarrow(z,t)$ lifts
$\bar{\omega}$. It follows that $\operatorname{Im}v\subset\operatorname{Im}\
\omega+d^{2e+1}A^{\prime}$. Note that $d\in
h_{D}=\sqrt{\omega(H_{D/A})A^{\prime}}$. It follows that $h_{B}\subset
h_{D}\not\subset q$ because ${\bar{A}}\otimes_{A}D\cong{\bar{D}}[T]$ is smooth
over $\bar{D}$ and $h_{B}{\bar{A}}^{\prime}\subset h_{{\bar{D}}}\not\subset
q{\bar{A}}^{\prime}$.

Applying Proposition 5 we see that $v$ factors through an $A$-algebra of
finite type $B^{\prime}$, let us say $v$ is the composite map $B\rightarrow
B^{\prime}\xrightarrow{w}A^{\prime}$ and $h_{D}\subset
h_{B^{\prime}}=\sqrt{w(H_{B^{\prime}/A})A^{\prime}}$. It follows that
$A\rightarrow B\rightarrow A^{\prime}\supset q$ is resolvable.

######  Remark 7.

The above proof of the theorem seems to be not constructive as it is Theorem 2
(more details in Proposition 10). If $A^{\prime}={\bf Q}[[x_{1}.x_{2}]]$ and
$\operatorname{height}q=1$ then to resolve $A\rightarrow B\rightarrow
A^{\prime}\supset q$ we might need to define precisely $v$ modulo a power of
$q$, that is to give formal power series in $x$ with infinite number of
coefficients. Indeed, our procedure is based in resolving
$A/(d^{2e+1})\rightarrow B/d^{2e+1}B\rightarrow
A^{\prime}/d^{2e+1}A^{\prime}\supset q/d^{2e+1}A^{\prime}$ for some $d\in A$
with $dA^{\prime}\subset\sqrt{v(H_{B/A})A}^{\prime}$. As in the proof of
Proposition 5 we had to find $y^{\prime}$ in $D$ with
$\omega(y^{\prime})\equiv y$ modulo $d^{2e+1}A^{\prime}$, or modulo an ideal
generated by several powers of such $d$, or almost modulo a high enough power
of a minimal prime ideal $q$ associated to $v(H_{B/A})A^{\prime}$. When
$A^{\prime}/q$ is not Artinian then it is not enough to use a high jet of $y$.
When $q=(x_{1},x_{2})$ then to define $v$ modulo a power of $q$ means only to
give a finite number of such coefficients (this follows from the proof of
Theorem 2).

######  Lemma 8.

Let $u:A\rightarrow A^{\prime}$ be a regular morphism of Noetherian rings, $B$
an $A$-algebra of finite type, $v:B\rightarrow A^{\prime}$ an $A$-algebra
morphism and $q\in\operatorname{Spec}A^{\prime}$ a minimal prime associated
ideal of $h_{B}$. Suppose that $A^{\prime}$ is local,
$\operatorname{height}q\geq 1$ and char $A_{p}/pA_{p}=0$, $p=u^{-1}(q)$. Then
$A\rightarrow B\rightarrow A^{\prime}\supset q$ is resolvable, that is $v$
factors through an $A$-algebra of finite type $C$, let us say $v$ is the
composite map $B\rightarrow C\xrightarrow{w}A^{\prime}$, such that
$h_{B}\subset h_{C}=\sqrt{w(H_{C/A})A^{\prime}}\not\subset q$.

###### Proof.

Apply Theorem 6 if $t:=\operatorname{height}q-\operatorname{height}p=0$.
Otherwise, $A^{\prime}_{q}/pA^{\prime}_{q}$ is a regular local ring of
dimension $t>0$ by the regularity of $u$. Suppose that
$z=(z_{1},\ldots,z_{t})\in A^{\prime t}$ induces a regular system of
parameters in $A^{\prime}_{q}/pA^{\prime}_{q}$. Let $A[Z]$ be the polynomial
$A$-algebra in the variables $Z=(Z_{1},\ldots,Z_{t})$. By the local criterion
of flatness [10], the map $u_{1}:A_{1}=A_{p}[Z]_{(p,Z)}\rightarrow
A^{\prime}_{q}$ given by $Z\rightarrow z$ is flat, even regular since
$A^{\prime}_{q}/pA^{\prime}_{q}$ is regular and char $A_{p}/pA_{p}=0$. Let
$v_{1}:B_{1}=A_{1}\otimes_{A}B\rightarrow A^{\prime}_{q}$ be the extension of
$v$ given by $Z\rightarrow z$. Then $v_{1}(H_{B_{1}/A_{1}})A^{\prime}_{q}$ is
a $qA^{\prime}_{q}$-primary ideal and applying Theorem 6 we see that
$A[Z]\rightarrow B[Z]\rightarrow A^{\prime}\supset q$ is resolvable and so
$A\rightarrow B\rightarrow A^{\prime}\supset q$ is resolvable too.

######  Theorem 9.

Let $u:A\rightarrow A^{\prime}$ be a regular morphism of Noetherian
$\mathbb{Q}$-algebras, $B$ an $A$-algebra of finite type and $v:B\rightarrow
A^{\prime}$ an $A$-algebra morphism. Suppose that $A^{\prime}$ is a local
ring. Then $v$ factors through a standard smooth $A$-algebra.

###### Proof.

As in [15, Corollary 3.3] we may suppose that
$\operatorname{height}v(H_{B/A})A^{\prime}\geq 1$. This could be done in a
constructive way (see for instance the beginning of Section 2 of [14] in the
case when $A^{\prime}$ is a domain). Let $q\in\operatorname{Spec}A^{\prime}$
be a minimal prime associated ideal of $v(H_{B/A})A^{\prime}$. Applying Lemma
8 we see that $v$ factors through an $A$-algebra of finite type $B_{1}$, let
us say $v$ is the composite map $B\rightarrow
B_{1}\xrightarrow{w_{1}}A^{\prime}$, such that
$v(H_{B/A})A^{\prime}\subset\sqrt{w_{1}(H_{B_{1}/A})A^{\prime}}\not\subset q$.
Applying again Lemma 8 for $B_{1}$ and a certain minimal prime associated
ideal $q_{1}$ of $w_{1}(H_{B_{1}/A})A^{\prime}$ we see that $w_{1}$ factors
through an $A$-algebra of finite type $B_{2}$, let us say $w_{1}$ is the
composite map $B_{1}\rightarrow B_{2}\xrightarrow{w_{2}}A^{\prime}$, such that
$w_{1}(H_{B_{1}/A})A^{\prime}\subset\sqrt{w_{2}(H_{B_{2}/A})A^{\prime}}\not\subset
q_{1}$. By Noetherianity this procedure should end since the strictly
increasing chain
$v(H_{B/A})A^{\prime}\subset\sqrt{w_{1}(H_{B_{1}/A})A^{\prime}}\subset\sqrt{w_{2}(H_{B_{2}/A})A^{\prime}}\subset\ldots$
should stop. This could happens only when
$w_{i}(H_{B_{i}/A})A^{\prime}=A^{\prime}$ for some $i$. Using [15, Lemma 2.4]
we may replace $B_{i}$ by an $A$-algebra $C$ of finite type with $H_{C/A}=C$,
that is $C$ is a smooth algebra over $A$. We may choose $C$ to be standard
smooth over $A$ using [15, Lemma 3.4] (see also [8, Lemma 4]).

Next, let $u:A\rightarrow A^{\prime}$ be a regular morphism of Noetherian
local rings of dimension two, $B$ an $A$-algebra of finite type and
$v:B\rightarrow A^{\prime}$ an A-morphism. If
$h_{B}=\sqrt{v(H_{B/A})A^{\prime}}$ is the maximal ideal of $A^{\prime}$ then
we have a constructive General Neron Desingularization using Theorem 2 (see
also [13]). Always we may reduce to the case $\operatorname{height}h_{B}\geq
1$ as we noticed in the proof of Theorem 9. The difficulties to construct a
General Neron Desingularization could appear only when we show that
$A\rightarrow B\rightarrow A^{\prime}\supset q$ is resolvable for a minimal
prime associated ideal $q$ of $h_{B}$ with $\operatorname{height}q=1$.

######  Proposition 10.

If there exists a minimal associated ideal $q$ of $h_{B}$ such that
$u^{-1}(q)$ has height $1$ then $A\rightarrow B\rightarrow A^{\prime}\supset
q$ is resolvable in a constructive way. Consequently, in this case Theorem 9
holds in a constructive way.

###### Proof.

Let $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$ and $q$ be a minimal associated ideal
of $h_{B}$ such that $u^{-1}(q)$ has height $1$. Choose $\gamma\in
u^{-1}(v(H_{B/A})A^{\prime})$ such that $\operatorname{height}(\gamma
A^{\prime})=1$. As in the proof of [8, Theorem 2] we may change $B$ such that
a power $d$ of $\gamma$ is in $((f):I)\Delta_{f}$ for some polynomials
$f=(f_{1},\ldots,f_{r})$, $r\leq n$ from $I$. Choosing $e$ as in Proposition 5
it is enough to see as in Case 2 from Theorem 6 that
${\bar{A}}=A/(d^{2e+1})\rightarrow\bar{B}={\bar{A}}\otimes_{A}B\rightarrow{\bar{A}}^{\prime}=A^{\prime}/d^{2e+1}A^{\prime}\supset
q{\bar{A}}^{\prime}$ is resolvable. But this is done in [12] because we are
now in dimension one.

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