Constructive N\'eron Desingularization of algebras with big smooth locus

TL;DR

This paper presents an algorithmic proof of the General Néron Desingularization theorem for morphisms with large smooth loci, extending previous one-dimensional results to higher dimensions.

Contribution

It provides a constructive, algorithmic approach to Néron Desingularization for a broader class of algebraic morphisms.

Findings

Algorithmic proof of the General Néron Desingularization theorem.

Extension of results from one-dimensional to higher-dimensional cases.

Uniform version applicable to morphisms with big smooth locus.

Abstract

An algorithmic proof of the General N\'eron Desingularization theorem and its uniform version is given for morphisms with big smooth locus. This generalizes the results for the one-dimensional case.

Equations

$$ds\text{mbox}{Id}_n=P(y^{\prime })\text{mbox}{Id}_n=\sum _{j=1}^q{G}_j(y^{\prime })H_j(y^{\prime }).$$

$$(\partial f/\partial Y)G_j=(M_j{N}_j\text{mbox}{Id}_r|0).$$

$$G_j(y^{\prime })t_j=M_j(y^{\prime })N_j(y^{\prime })\nu =ds\nu$$

$$s(y-\tilde{y})=d^e\sum _{j=1}^q\omega (G_j(y^{\prime }))t_j.$$

$$h=s(Y-y^{\prime })-d^e\sum _{j=1}^q{G}_j(y^{\prime })T_j,$$

$$s(Y-y^{\prime })\equiv d^e\sum _{j=1}^q{G}_j(y^{\prime })T_j\text{mbox}moduloh$$

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

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

$$g_i=s^p{b}_i+s^p{T}_i+d^{e-1}{Q}_i,i\in [r]$$

$$B\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$(d_1^3,\dots ,d_m^3)A^{\prime n-r}\to \{v^{\prime }\in {\mathrm{Hom}}_A(B,A^{\prime }):v^{\prime }\equiv v\text{mbox}modulo(d_1^3,\dots ,d_m^3)A^{\prime }\}.$$

$${\mathfrak{m}}^{3k}{A}^{\prime n-r}\to \{v^{\prime }\in {\mathrm{Hom}}_A(B,A^{\prime }):v^{\prime }\equiv v\text{mbox}modulo{\mathfrak{m}}^{3k}{A}^{\prime }\}.$$

Full text

*Key words * : Smooth morphisms, regular morphisms

2010 Mathematics Subject Classification: Primary 13B40, Secondary 14B25,13H05,13J15.

Introduction

Motivated to generalize Artin’s Approximation Theorem (cf. [2]) to excellent Henselian rings the third author developed a powerful tool, the General Néron Desingularization (cf. [14]). This result was discussed and used later by many other authors (cf. [1], [20], [19]). The proof of the Desingularization was not constructive. In this paper we want to give an algorithm to compute Néron Desingularization for an important special case. We begin recalling some standard definitions.

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, 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.

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. A localization of a smooth algebra is called essentially smooth. A Henselian Noetherian local ring $A$ is excellent if the completion map $A\rightarrow\hat{A}$ is regular.

Theorem 1**.**

(General Néron Desingularization, Popescu [14], [15], André [1], Swan [20], Spivakovsky [19]) 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 $B\rightarrow C\rightarrow A^{\prime}$.

Constructive General Néron Desingularization for the case when the rings $A$ and $A^{\prime}$ are one-dimensional local rings, is given in [12], [10] and [7], the two dimensional case is partially done in [11]. The purpose of this paper is to find a constructive proof for the case when rings $A$ and $A^{\prime}$ are of dimension $m$ and the smooth locus of $B\rightarrow A^{\prime}$ is big. We proceed using induction on the dimension of rings, with the induction step given in Proposition  3. In the Section 2 we prove a uniform General Néron Desingularization for $m$-dimensional local Cohen-Macaulay rings and some consequences of it. We also give an algorithm to find a uniform General Néron Desingularization using Singular.

1. Constructive Néron Desingularization

Let $u:A\rightarrow A^{\prime}$ be a flat morphism of Noetherian local rings of dimension $m$. Suppose that the maximal ideal ${\mathfrak{m}}$ of $A$ generates the maximal ideal of $A^{\prime}$, $A^{\prime}$ is Henselian and $u$ is a regular morphism.

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 $r\leq n$. $H_{B/A}$ 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 $B=((f):I)\Delta_{f}B$.

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

Theorem 2**.**

Any $A$-morphism $v:B\rightarrow A^{\prime}$ such that $v(H_{B/A}A^{\prime})$ is ${\mathfrak{m}}A^{\prime}$-primary factors through a standard smooth $A$-algebra $B^{\prime}$.

To prove the above theorem we need the following proposition.

Proposition 3**.**

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

1. (1)

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

Proof.

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

Let the map $\bar{B}\rightarrow\bar{D}$ is given by $Y\rightarrow y^{\prime}+d^{2e+1}D$. 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 $y-\tilde{y}=d^{e+1}\nu$ for $\nu\in d^{e}A^{\prime n}$.

We have $M_{j}=\det H_{j}$, where $H_{j}$ is the matrix $(\partial f_{i}/\partial Y_{j^{\prime}})_{i\in[r],{j^{\prime}}\in[n]}$ completed with some $(n-r)$ rows from $0,\ 1$. Since $d\equiv P$ modulo $I$ we get $P(y^{\prime})\equiv d$ modulo $d^{2e+1}$ in $D$ because $I(y^{\prime})\equiv 0$ modulo $d^{2e+1}D$. Thus $P(y^{\prime})=ds$ for 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 so

[TABLE]

But $H_{j}$ is the matrix $(\partial f_{i}/\partial Y_{j^{\prime}})_{i\in[r],{j^{\prime}}\in[n]}$ completed with some $(n-r)$ rows from $0,\ 1$. Especially we obtain

[TABLE]

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

[TABLE]

and so

[TABLE]

Let

[TABLE]

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

[TABLE]

and

[TABLE]

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

[TABLE]

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

[TABLE]

is in the kernel of $\varphi$. Indeed, we have ${s}^{p}f_{i}=d^{e+1}{g}_{i}\ \mbox{modulo}\ {h}$ because of (3) and $P(y^{\prime})=d{s}$. Thus $d^{e+1}\varphi({g})=d^{e+1}{g}(t)\in({h}(y,{t}),f(y))=(0)$ and ${g}({t})\in d^{e}A^{\prime r}$ and so $g(t)\in(0:_{A^{\prime}}d^{e+1})\cap d^{e}A^{\prime}=0$ because $(0:_{A^{\prime}}d^{e})=(0:_{A^{\prime}}d^{e+1})$, the map $u$ being flat. 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 $B\rightarrow B\otimes_{A}D\cong D[Y]/I\rightarrow{E}\xrightarrow{{\psi}}A^{\prime}$.

Now we show that there exist ${s}^{\prime},{s}^{\prime\prime}\in{E}$ such that ${E}_{{s}{s}^{\prime}{s}^{\prime\prime}}$ is standard smooth over $A$ and $\psi$ factors through ${E}_{{s}{s}^{\prime}{s}^{\prime\prime}}$.

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 ${V}=(D[Y,{T}]/({h},{g}))_{{s}{s}^{\prime}}$ is smooth over $D$. We claim 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 $PI\subset({h},{g})D[Y,{T}]_{s}$ and so $P(y^{\prime}+{s}^{-1}d^{e}\sum_{j=1}^{q}G_{j}(y^{\prime})T_{j})I\subset({h},{g})D[Y,{T}]_{s}$. Since $P(y^{\prime}+{s}^{-1}d^{e}\sum_{j=1}^{q}G_{j}(y^{\prime})T_{j})\in P(y^{\prime})+d^{e}({T})D[Y,T]_{s}$ we get

$P(y^{\prime}+{s}^{-1}d^{e}\sum_{j=1}^{q}G_{j}(y^{\prime})T_{j})=d{s}^{\prime\prime}$ for some ${s}^{\prime\prime}\in 1+({T})D[Y,{T}]_{s}$. It follows that ${s}^{\prime\prime}I\subset(({h},{g}):d)D[Y,{T}]_{{s}{s}^{\prime}}$. Thus ${s}^{\prime\prime}IV\subset(0:_{V}d)\cap d^{e}V=0$ because $(0:_{V}d)\cap d^{e}V=0$, and $V$ is flat over $D$ and so over $A$. This shows our claim. It follows that $I\subset({h},{g})D[Y,{T}]_{{s}{s}^{\prime}{s}^{\prime\prime}}$. Thus ${E}_{{s}{s}^{\prime}{s}^{\prime\prime}}\cong{V}_{{s}^{\prime\prime}}$ 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. Thus ${\psi}$ (and so $v$) factors through the standard smooth $A$-algebra $B^{\prime}={E}_{{s}{s}^{\prime}{s}^{\prime\prime}}$.

Proof of Theorem 2

We choose $\gamma_{1},\gamma_{2},\ldots,\gamma_{m}\in v(H_{B/A})A^{\prime}\cap A$ such that $\gamma_{k}$ for $k\in[m]$ is a system of parameters in $A$, and $\gamma_{k}=\sum_{i=1}^{q}v(b_{i})z_{i}^{(k)}$, where $z_{i}^{(k)}\in A^{\prime}$, $b_{i}\in H_{B/A}$. Set $B_{0}=B[Z^{(1)},\ldots,Z^{(m)}]/(f^{(1)},\ldots,f^{(m)})$, where $f^{(k)}=-\gamma_{k}+\sum_{i=1}^{q}b_{i}Z_{i}^{(k)}\in B[Z^{(k)}]$, $Z^{(k)}=(Z_{1}^{(k)},\ldots,Z_{q}^{(k)})$, and let $v_{0}:B_{0}\rightarrow A^{\prime}$ be the map of $B$-algebras given by $Z^{(k)}\rightarrow z^{(k)}$. Changing $B$ by $B_{0}$ we may suppose that $\gamma_{k}\in H_{B/A}$.

As in [10] we need the following lemma.

Lemma 4**.**

1. (1)

(**[13, Lemma 3.4]**) Let $B_{1}$ be the symmetric algebra $S_{B}(I/I^{2})$ of $I/I^{2}$ over222Let $M$ 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 $(B_{1})_{\gamma}$ for any $\gamma\in H_{B/A}$.* 2. (2)

(**[20, 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 $B_{\gamma}$. 3. (3)

(**[16, 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 $I$.

Using $(1)$ of Lemma 4 we reduce our proof to the case when $\Omega_{B_{\gamma_{k}}/A}$ for all $k\in[m]$ are free over $B_{\gamma_{k}}$ respectively.

Let $B_{1}$ be given by Lemma 1. 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 $B_{1},vw$.

Using $(2)$ of Lemma 4 we may reduce to the case when $(I/I^{2})_{\gamma_{k}}$ is free over $B_{\gamma_{k}}$ for all $k\in[m]$.

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

Now using Using $(3)$ of Lemma 4 we will reduce further to the case when a power of $\gamma_{k}$ is in $((f^{(k)}):I)\Delta_{f^{(k)}}$ for some $f^{(k)}=(f_{1}^{(k)},\ldots f_{r_{k}}^{(k)})$, $r_{k}\leq n$ from $I$.

We reduced to the case when $(I/I^{2})_{\gamma_{k}}$ is free over $B_{\gamma_{k}}$. Then it is enough to use Lemma 3.

Replacing $B_{1}$ by $B$ we may assume that a power $d_{k}$ of $\gamma_{k}$ for all $k\in[m]$ has the form $d_{k}\equiv P_{k}=\sum_{i=1}^{q_{k}}M_{i}^{(k)}L_{i}^{(k)}\ \mbox{modulo}\ I$, for some $r_{k}\times r_{k}$ minors $M_{i}^{(k)}$ of $(\partial f^{(k)}/\partial Y)$ and $L_{i}^{(k)}\in((f^{(k)}):I)$.

The Jacobian matrix $(\partial f^{(k)}/\partial Y)$ can be completed with $(n-r_{k})$ rows from $A^{n}$ obtaining a square $n$ matrix $H_{i}^{(k)}$ such that $\det H_{i}^{(k)}=M_{i}^{(k)}$.

This is easy using just the integers $0,1$. Set $d=d_{m}$, $f=f^{(m)}$, $r=r_{m}$, $q=q_{m}$, $M_{i}=M_{i}^{(m)}$, $N_{i}=N_{i}^{(m)}$, $\bar{A}=A/d^{2e+1}$, ${\bar{B}}={\bar{A}}\otimes_{A}B$, ${\bar{A}}^{\prime}=A^{\prime}/(d^{2e+1}A^{\prime})$, ${\bar{v}}={\bar{A}}\otimes_{A}v$. Then we have $d\equiv\sum_{j}M_{j}N_{j}$ modulo $I$. Now we will use the induction on $m$. Case I: $m=0$ If $m=0$ then $A$ and $A^{\prime}$ are Artinian local rings and $u:A\rightarrow A^{\prime}$ is a regular morphism. Then we are done by Corollary 3.3 [13]. Case II: $m>0$ Suppose by the induction hypothesis that we have a standard smooth $\bar{A}$-algebra $\bar{D}\cong(\bar{A}[Z]/(\bar{g}))_{\bar{h}\bar{M}}$, for $Z=(Z_{1},\ldots,Z_{p}),\bar{g}=(\bar{g}_{1},\ldots,\bar{g}_{q})$ with $q\leq p$, $\bar{h}\in\bar{A}[Z]$ and $\bar{M}$ a $q\times q$-minor of $(\frac{\partial\bar{g}}{\partial Z})$, such that the map $\bar{v}:\bar{B}\rightarrow\bar{A^{\prime}}$ factors through $\bar{D}$, let us say $\bar{v}$ is the composite map $\bar{B}\rightarrow\bar{D}\xrightarrow{\bar{\omega}}\bar{A^{\prime}}$.

Now let $g\in A[Z]^{q}$ be a lifting of $\bar{g}$ and $M$ the $q\times q$-minor of $(\frac{\partial g}{\partial Z})$ corresponding $\bar{M}$. Take $h\in A[Z]$ such that $h$ lifts $\bar{h}$. Then $D\cong(A[Z]/(g))_{hM}$ is a standard smooth $A$-algebra and by the Implicit Function Theorem the map ${\bar{\omega}}$ can be lifted to $\omega:D\rightarrow A^{\prime}$ since $A^{\prime}$ is Henselian. It follows that $\operatorname{Im}v\subset\operatorname{Im}\omega+d^{2e+1}A^{\prime}$. Applying Proposition 3 we get a $B$-algebra $C$ smooth over $A$ such that $v$ factors through $C$, $B\rightarrow C\rightarrow A^{\prime}$.

2. A uniform Néron Desingularization

Let $u:A\rightarrow A^{\prime}$ be a regular morphism of Cohen-Macaulay local rings of dimension $m$. Suppose that the maximal ideal ${\mathfrak{m}}$ of $A$ generates the maximal ideal of $A^{\prime}$, $A^{\prime}$ is Henselian and $A$, $A^{\prime}$ have the same completions.

Let $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$, and for $i\in[m]$ let $f^{(i)}=(f_{1}^{(i)},\ldots,f_{r_{i}}^{(i)})$, $r_{i}\leq n$ be a system of polynomials from $I$. Let $M_{i}$ be an $r_{i}\times r_{i}$-minor of the Jacobian matrix $(\partial f^{(i)}/\partial Y)$ and $N_{i}\in((f^{(i)}):I)$, $P_{i}=N_{i}M_{i}$. Let $v:B\rightarrow A^{\prime}/{\mathfrak{m}}^{3k+c}A^{\prime}$ be an $A$-morphism for some $k,c\in\mathbb{N}$. Suppose that $v(N_{1}M_{1},\ldots,N_{m}M_{m})A^{\prime}/{\mathfrak{m}}^{3k+c}A^{\prime}\supset{\mathfrak{m}}^{k}A^{\prime}/{\mathfrak{m}}^{3k+c}A^{\prime}$. Let $y^{\prime}\in A^{n}$ be a lifting of $v(Y)$ to $A$ and let $d_{i}=P_{i}(y^{\prime})$. Then $(d_{1},\ldots,d_{m})A^{\prime}/{\mathfrak{m}}^{3k+c}A^{\prime}\supset{\mathfrak{m}}^{k}A^{\prime}/{\mathfrak{m}}^{3k+c}A^{\prime}$. Note that ${\mathfrak{m}}^{k}\subset(d_{1},\ldots,d_{m})A+{\mathfrak{m}}^{3k+c}\subset(d_{1},\ldots,d_{m})A+{\mathfrak{m}}^{3(3k+c)+c}\subset\ldots$. Thus ${\mathfrak{m}}^{k}\subset(d_{1},\ldots,d_{m})A$ and it follows that $(d_{1},\ldots,d_{m})A^{\prime}\supset{\mathfrak{m}}^{k}A^{\prime}$ . Since $A$ is Cohen-Macaulay we get $d=\{d_{1},\ldots,d_{m}\}$ regular sequence in $A$. Note that $(d_{1},\ldots,d_{m})$ is the ideal corresponding to $v(P_{1},\ldots,P_{m})A^{\prime}$ by the isomorphism $A/{\mathfrak{m}}^{3k+c}\cong A^{\prime}/{\mathfrak{m}}^{3k+c}A^{\prime}$.

Theorem 5**.**

There exists a $B$-algebra $C$ which is standard smooth over $A$ with the following properties.

1. (1)

Every $A$-morphism $v^{\prime}:B\rightarrow A^{\prime}$ with $v^{\prime}\equiv v\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime}$ factors through $C$. 2. (2)

Every $A$-morphism $v^{\prime}:B\rightarrow A^{\prime}$ with $v^{\prime}\equiv v\ \mbox{modulo}\ {\mathfrak{m}}^{3k}A^{\prime}$ factors through $C$. 3. (3)

There exists an $A$-morphism $w:C\rightarrow A^{\prime}$ which makes the following diagram commutative

[TABLE]

Proof.

Let $v^{\prime}:B\rightarrow A^{\prime}$ be an $A$-morphism with $v^{\prime}\equiv v\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime}$. We apply induction on $m$. Case I: $m=1$ If $m=1$ then $A$ and $A^{\prime}$ are Noetherian local rings of dimension $1$ and $u:A\rightarrow A^{\prime}$ is a regular morphism. Then we are done by Theorem 2 [7], with $e=1$. Case II: $m>1$ Now let $\bar{A}=A/(d_{1}^{3},\ldots,d_{m-1}^{3})$, and consider the map $\bar{v^{\prime}}={\bar{A}}{\otimes}_{A}v^{\prime}:\bar{B}={\bar{A}}{\otimes}_{A}B\rightarrow\bar{A}^{\prime}={\bar{A}}{\otimes}_{A}A^{\prime}$. By the induction hypothesis there exists a standard smooth algebra $\bar{D}\cong(\bar{A}[Z]/(\bar{g}))_{\bar{h}\bar{M}}$, for $Z=(Z_{1},\ldots,Z_{p}),\bar{g}=(\bar{g}_{1},\ldots,\bar{g}_{q})$ with $q\leq p$, $\bar{h}\in\bar{A}[Z]$ and $\bar{M}$ a $q\times q$-minor of $(\frac{\partial\bar{g}}{\partial Z})$, such that the map $\bar{v^{\prime}}$ factors through $\bar{D}$, say $\bar{v^{\prime}}$ is the composite map $\bar{B}\rightarrow\bar{D}\xrightarrow{\bar{\omega^{\prime}}}A^{\prime}$.

Now let $g\in A[Z]^{q}$ be a lifting of $\bar{g}$ and $M$ the $q\times q$-minor of $(\frac{\partial g}{\partial Z})$ corresponding $\bar{M}$. Take $h\in A[Z]$ such that $h$ lifts $\bar{h}$. Then $D\cong(A[Z]/(g))_{hM}$ is a standard smooth $A$-algebra and by the Implicit Function Theorem the map ${\bar{\omega^{\prime}}}$ can be lifted to $\omega^{\prime}:D\rightarrow A^{\prime}$ since $A^{\prime}$ is Henselian. It follows that $\operatorname{Im}v^{\prime}\subset\operatorname{Im}\omega^{\prime}+d_{m}^{3}A^{\prime}$. Applying Proposition 3 (with $e=1$) we get a $B$-algebra $C$ standard smooth over $A$ such that $v^{\prime}$ factors through $C$. This proves (1) which obviously implies (2).

Now for (3) take the map ${\hat{w}}:C\cong(D[Y,T]/(I,g,h))_{ss^{\prime}}\rightarrow A^{\prime}/{\mathfrak{m}}^{c}A^{\prime}$ given by $(Y,T)\rightarrow(y^{\prime},0)$. Then the composite map $B\rightarrow C\xrightarrow{\hat{w}}A^{\prime}/{\mathfrak{m}}^{c}A^{\prime}$ is lifted by $v$. Since $C$ is standard smooth, we may lift $\hat{w}$ to an $A$-morphism $w:C\rightarrow A^{\prime}$ by the Implicit Function Theorem. Clearly, $w$ makes the above diagram commutative.

Example 6**.**

(Rond) Let $k$ be a field, $A=k[[x]]$, $x=(x_{1},x_{2},x_{3})$, $B=A[Y]/(f)$, $Y=(Y_{1},\ldots,Y_{4})$, $f=Y_{1}Y_{2}-Y_{3}Y_{4}$. Then $\Delta_{f}=H_{B/A}=(Y)$. Let $p\in{\bf N}$ and set $y^{\prime}_{1}=x_{1}^{p}$, $y^{\prime}_{2}=x_{2}^{p}$, $y^{\prime}_{3}=x_{1}x_{2}-x_{3}^{p}$. Then there exists $y^{\prime}_{4}\in A$ such that $f(y^{\prime})\equiv 0$ modulo $(x)^{p^{2}}$. It follows that $d_{1}=x_{1}^{p}$, $d_{2}=x_{2}^{p}$ and $d_{3}=x_{3}^{p^{2}}$ belongs to $\Delta_{f}(y^{\prime})$ because $x_{1}^{p}x_{2}^{p}-x_{3}^{p^{2}}=y^{\prime}_{3}(x_{1}^{p-1}x_{2}^{p-1}+x_{1}^{p-2}x_{2}^{p-2}x_{3}^{p}+\ldots x_{3}^{p^{2}}$). For $k=2p+p^{2}-2$ we have $(x)^{k}\subset(d_{1},d_{2},d_{3})\subset H_{B/A}(y^{\prime})$. If $f(y^{\prime})\equiv 0$ modulo $(x)^{3k+p+1}$ then by Theorem 5 (3) we could get $y\in A^{4}$ such that $f(y)=0$ and $y\equiv y^{\prime}$ modulo $(x)^{p+1}$. But this is not the case, since $f(y^{\prime})\equiv 0$ modulo $(x)^{p^{2}}$ and we cannot apply the quoted theorem. Thus it is not a surprise that [18, Remark 4.7] says that there exist no $y\in A^{4}$ such that $f(y)=0$ and $y\equiv y^{\prime}$ modulo $(x)^{p+1}$.

Corollary 7**.**

(Elkik) Let $(A,{\mathfrak{m}})$ be a Cohen-Macaulay Henselian local ring of dimension $m$ and $B=A[Y]/I$, $Y=(Y_{1},\ldots,Y_{n})$ an $A$-algbra of finite type. Then for every $k\in\bf N$ there exist two integers $m_{0},p\in\bf N$ such that if $y^{\prime}\in A^{n}$ satisfies $m^{k}\subset H_{B/A}(y^{\prime})$ and $I(y^{\prime})\equiv 0$ modulo ${\mathfrak{m}}^{m}$ for some $m>m_{0}$ then there exists $y\in A^{n}$ such that $I(y)=0$ and $y\equiv y^{\prime}$ modulo ${\mathfrak{m}}^{m-p}$.

Proof.

Suppose that $A^{\prime}=A$. In the notation of Theorem 5 given $k$ set $m_{0}=p=3k$ and suppose that $y^{\prime}\in A^{n}$ satisfies $m^{k}\subset H_{B/A}(y^{\prime})$ and $I(y^{\prime})\equiv 0$ modulo ${\mathfrak{m}}^{m}$ for some $m>m_{0}$. Let $v:B\rightarrow A/{\mathfrak{m}}^{m}$ be given by $Y\rightarrow y^{\prime}$. Set $c=m-p$. By Theorem 5 there exists a smooth $A$-algebra $C$ and a map $w:C\rightarrow A$ which makes the above digram commutative. Let $y$ be the image of $Y$ by the composite map $B\rightarrow C\xrightarrow{w}A$. Then $I(y)=0$ and $y\equiv y^{\prime}$ modulo ${\mathfrak{m}}^{c}={\mathfrak{m}}^{m-p}$.

Corollary 8**.**

With the assumptions and notation of the Theorem 5, let $\rho:B\rightarrow C$ be the structural algebra map. Then $\rho$ induces bijections $\rho^{*}$ given by $\rho^{*}(w)=w\circ\rho$, between

1. (1)

$\{w\in\operatorname{Hom}_{A}(C,A^{\prime}):w\circ\rho\equiv v\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime}\}$* and $\{v^{\prime}\in\operatorname{Hom}_{A}(B,A^{\prime}):v^{\prime}\equiv v\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime}\}$* 2. (2)

$\{w\in\operatorname{Hom}_{A}(C,A^{\prime}):w\circ\rho\equiv v\ \mbox{modulo}\ {\mathfrak{m}}^{3k}A^{\prime}\}$* and $\{v^{\prime}\in\operatorname{Hom}_{A}(B,A^{\prime}):v^{\prime}\equiv v\ \mbox{modulo}\ {\mathfrak{m}}^{3k}A^{\prime}\}$*

Proof.

We will use induction on the dimension of $A$.

1. (1)

Case I: $m=1$ If $m=1$ then $A$ and $A^{\prime}$ are Cohen-Macaulay local rings of dimension $1$. Then we are done by Corollary $8$ [7]. 2. (2)

Case II: $m>1$ By Theorem 5, (1), $\rho^{*}$ is surjective. Let Now let $\bar{A}=A/(d_{1}^{3},\ldots,d_{m-1}^{3})$, and the map $\bar{v^{\prime}}={\bar{A}}{\otimes}_{A}v^{\prime}:\bar{B}={\bar{A}}{\otimes}_{A}B\rightarrow\bar{A}^{\prime}={\bar{A}}{\otimes}_{A}A^{\prime}$. By the induction hypothesis there exists a standard smooth algebra $\bar{D}$ such that the maps $\bar{w}$ and $\bar{w}^{\prime}$ restricted to $\bar{D}$ coincide. This implies that $w\mid_{D}$ and $w^{\prime}\mid_{D}$ lift the same map $\bar{w}\mid_{\bar{D}}$. Thus $w\mid_{D}=w^{\prime}\mid_{D}$ by uniqueness in the Implicit Function Theorem.

By construction $C=E_{ss^{\prime}}$, $E=D[Y,T]/(I,g,h)$ and $H_{m}(y^{\prime})(w(Y)-w^{\prime}(Y))\equiv d_{m}^{2}(w(T)-w^{\prime}(T))\ \mbox{modulo}\ h$. Thus $d_{m}^{2}(w(T)-w^{\prime}(T))=0$ and so $w|_{E}=w^{\prime}|_{E}$ because $d_{m}$ is regular in $A^{\prime}$ since $d_{m}$ is regular in $A$ and $u$ is flat. It follows that $w=w^{\prime}$ 3. (3)

Apply Theorem 5 (2) for the surjectivity. The injectivity follows from above.

Corollary 9**.**

With the assumptions and notation of the above Corollary, the following statements hold:

1. (1)

If there exists an $A$-morphism ${\tilde{v}}:B\rightarrow A^{\prime}$ with ${\tilde{v}}\equiv v\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime}$, then there exists a unique $A$-morphism ${\tilde{w}}:C\rightarrow A^{\prime}$ such that ${\tilde{w}}\circ\rho={\tilde{v}}$. 2. (2)

If there exists an $A$-morphism ${\tilde{v}}:B\rightarrow A^{\prime}$ with ${\tilde{v}}\equiv v\ \mbox{modulo}\ {\mathfrak{m}}^{3k}A^{\prime}$, then there exists a unique $A$-morphism ${\tilde{w}}:C\rightarrow A^{\prime}$ such that ${\tilde{w}}\circ\rho={\tilde{v}}$.

For the proof take ${\tilde{w}}={\rho^{*}}^{-1}(\tilde{v})$, where $\rho^{*}$ is defined in the Corollary 8.

By construction, $C$ has the form $(D[T]/(g))_{Mh}$, where $M=\det(\partial g_{i}/\partial T_{j})_{i,j\in[r]}$ and $h=s^{\prime}\in A[T]$ satisfies ${\tilde{w}}(h)\not\in{\mathfrak{m}}A^{\prime}$.

Lemma 10**.**

There exist canonical bijections

1. (1)

$(d_{1}^{3},\ldots,d_{m}^{3})A^{\prime n-r}\rightarrow\{w^{\prime}\in\operatorname{Hom}_{A}(C,A^{\prime}):w^{\prime}\equiv{\tilde{w}}\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime}\}.$** 2. (2)

${\mathfrak{m}}^{3k}A^{\prime n-r}\rightarrow\{w^{\prime}\in\operatorname{Hom}_{A}(C,A^{\prime}):w^{\prime}\equiv{\tilde{w}}\ \mbox{modulo}\ {\mathfrak{m}}^{3k}A^{\prime}\}.$**

Proof.

Note that $\{w^{\prime}\in\operatorname{Hom}_{A}(C,A^{\prime}):w^{\prime}\equiv{\tilde{w}}\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime}\}$ is in bijection with the set of all $t\in A^{\prime n}$ such that $g(t)=0$ and $t\equiv{\tilde{w}}(T)$ modulo $(d_{1}^{3},\ldots,d_{m}^{3})A^{\prime n}$.

Set $V=(T_{1},\ldots,T_{r})$, $Z=(T_{r+1},\ldots,T_{n})$. Thus $g(U,w^{\prime}(Z))=0$ has a unique solution (namely $U=w^{\prime}(V)$) in ${\tilde{w}}(Z)+(d_{1}^{3},\ldots,d_{m}^{3})A^{\prime n-r}$ by the Implicit Function Theorem. Consequently, $w^{\prime}(V)$ is uniquely defined by $w^{\prime}(Z)$, that is by the restriction $w^{\prime}\mid_{A[Z]}$.

Therefore, $\{w^{\prime}\in\operatorname{Hom}_{A}(C,A^{\prime}):w^{\prime}\equiv{\tilde{w}}\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime n}\}$ is in bijection with $\{w^{\prime\prime}\in\operatorname{Hom}_{A}(A[Z],A^{\prime}):w^{\prime\prime}\equiv{\tilde{w}}|_{A[Z]}\ \mbox{modulo}\ (d_{1}^{3},\ldots,d_{m}^{3})A^{\prime n-r}\}$, the latter set being in bijection with ${\tilde{w}}(Z)+(d_{1}^{3},\ldots,d_{m}^{3})A^{\prime n-r}$, that is with $(d_{1}^{3},\ldots,d_{m}^{3})A^{\prime n-r}$. The proof of (2) goes similarly.

Theorem 11**.**

With the assumptions and notation of Corollary 8 there exist canonical bijections

1. (1)

[TABLE] 2. (2)

[TABLE]

For the proof apply Corollary 8 and the above lemma.

3. Algorithms

In this section we present the algorithms corresponding to the results of Sections 1 and 2. We will use in our algorithm for uniform desingularization the following algorithm for the one dimensional case (cf. [7]):

Next we present the algorithm for uniform desingularization for the higher dimensional case.