#  Nested Artin approximation

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.

A short proof of the linear nested Artin approximation property of the
algebraic power series rings is given here.

Key words  : Henselian rings, Algebraic power series rings, Nested Artin
approximation property
2010 Mathematics Subject Classification: Primary 13B40, Secondary 14B25,13J15,
14B12.

The support from the the project ID-PCE-2011-1023, granted by the Romanian
National Authority for Scientific Research, CNCS - UEFISCDI is gratefully
acknowledged.

## Introduction

The solution of an old problem (see for instance [3]), the so-called nested
Artin approximation property, is given in the following theorem.

######  Theorem 1 ([4], [5, Theorem 3.6]).

Let $k$ be a field, $A=k\langle x\rangle$, $x=(x_{1},\ldots,x_{n})$ the
algebraic power series over $k$, $f=(f_{1},\ldots,f_{s})\in A[Y]^{s}$,
$Y=(Y_{1},\ldots,Y_{m})$ and $0<r_{1}\leq\ldots\leq r_{m}\leq n$, $c$ be some
non-negative integers. Suppose that $f$ has a solution
${\hat{y}}=({\hat{y}}_{1},\ldots,{\hat{y}}_{m})$ in $k[[x]]$ such that
${\hat{y}}_{i}\in k[[x_{1},\ldots,x_{r_{i}}]]$ for all $1\leq i\leq m$ (that
is a so-called nested formal solution). Then there exists a solution
$y=(y_{1},\ldots,y_{m})$ of $f$ in $A$ such that $y_{i}\in k\langle
x_{1},\ldots,x_{r_{i}}\rangle$ for all $1\leq i\leq m$ and $y\equiv{\hat{y}}\
\ \mbox{mod}\ \ (x)^{c}k[[x]]$.

The proof relies on an idea of Kurke from 1972 and the Artin approximation
property of rings of type $k[[u]]\langle x\rangle$ (see [4], [9], [5]). When
$f$ is linear, interesting relations with other problems and a description of
many results on this topic are nicely explained in [2]. Also a proof of the
above theorem in the linear case can be found in [2] using just the classical
Artin approximation property of $A$ (see [1]). Unfortunately, we had some
difficulties in reading [2], but finally we noticed a shorter proof using
mainly the same ideas. This proof is the content of the present note.

We owe thanks to G. Rond who noticed a gap in a previous version of this note.

##  1\. Linear nested Artin approximation property

We start recalling [8, Lemma 9.2] with a simplified proof.

######  Lemma 2.

Let $(A,{\mathfrak{m}})$ be a complete normal local domain,
$u=(u_{1},\dots,u_{n})$, $v=(v_{1},\ldots,v_{m})$, $B=A[[u]]\langle v\rangle$
be the algebraic closure of $A[[u]][v]$ in $A[[u,v]]$ and $f\in B$. Then there
exists $g$ in the algebraic closure $A\langle v,Z\rangle$ of $A[v,Z]$,
$Z=(Z_{1},\ldots,Z_{s})$ in $A[[v,Z]]$ for some $s\in\bf N$ and ${\hat{z}}\in
A[[u]]^{s}$ such that $f=g(\hat{z})$.

###### Proof.

Changing $f$ by $f-f(u=0)$ we may assume that $f\in(u)$. Note that $B$ is the
Henselization of $C=A[[u]][v]_{({\mathfrak{m}},u,v)}$ by [7] and so there
exists some etale neighborhood of $C$ containing $f$. Using for example [9,
Theorem 2.5] there exists a monic polynomial $F$ in $X$ over $A[[u]][v]$ such
that $F(f)=0$ and $F^{\prime}(f)\not\in({\mathfrak{m}},u,v)$, let us say
$F=\sum_{i,j}F_{ij}v^{i}X^{j}$ for some $F_{ij}\in A[[u]]$.

Set ${\hat{z}}_{ij}=F_{ij}-F_{ij}(u=0)\in(u)A[[u]]$,
${\hat{z}}=({\hat{z}}_{ij})$ and $G=\sum_{ij}(F_{ij}(u=0)+Z_{ij})v^{i}X^{j}$
for some new variables $Z=(Z_{ij})$. We have $G({\hat{z}})=F$. Set
$G^{\prime}=\partial G/\partial X$. As

$G(Z=0)=G({\hat{z}}(u=0))\equiv F(f)=0\ \mbox{ modulo}\ (u),$
$G^{\prime}(Z=0)=G^{\prime}({\hat{z}}(u=0))\equiv F^{\prime}(f)\not\equiv 0\
\mbox{modulo}\ ({\mathfrak{m}},u,v)$

we get $G(X=0)\equiv 0$, $G^{\prime}(X=0)\not\equiv 0$ modulo
$({\mathfrak{m}},v,Z)$. By the Implicit Function Theorem there exists
$g\in({\mathfrak{m}},v,Z)A\langle v,Z\rangle$ such that $G(g)=0$. It follows
that $G(g({\hat{z}}))=0$. But $F=G({\hat{z}})=0$ has just a solution $X=f$ in
$({\mathfrak{m}},u,v)B$ by the Implicit Function Theorem and so
$f=g({\hat{z}})$.

######  Lemma 3.

Let $(A,{\mathfrak{m}})$ be a Noetherian local ring, $f\in A[U]$,
$U=(U_{1},\ldots,U_{s})$ a linear system of polynomial equations, $c\in\bf N$
and ${\hat{u}}$ a solution of $f$ in the completion $\hat{A}$ of $A$. Then
there exists a solution $u\in A^{s}$ of $f$ such that $u\equiv{\hat{u}}\
\mbox{modulo}\ {\mathfrak{m}}^{c}\hat{A}$.

###### Proof.

Let $B=A[U]/(f)$ and $h:B\rightarrow{\hat{A}}$ be the map given by
$U\rightarrow{\hat{u}}$. By [4, Lemma 4.2] (or [6, Proposition 36]) $h$
factors through a polynomial algebra $A[Z]$, $Z=(Z_{1},\ldots,Z_{s})$, let us
say $h$ is the composite map $B\xrightarrow{t}A[Z]\xrightarrow{g}{\hat{A}}$.
Choose $z\in A^{s}$ such that $z\equiv g(Z)\ \mbox{modulo}\
{\mathfrak{m}}^{c}\hat{A}$. Then $u=t(cls\ U)(z)$ is a solution of $f$ in $A$
such that $u\equiv{\hat{u}}\ \mbox{modulo}\ {\mathfrak{m}}^{c}\hat{A}$.

######  Proposition 4.

Let $k\langle x,y\rangle$, $x=(x_{1},\ldots,x_{n})$, $y=(y_{1},\ldots,y_{m})$
be the ring of algebraic power series in $x,y$ over a field $k$ and $M\subset
k\langle x,y\rangle^{p}$ a finitely generated $k\langle x,y\rangle$-submodule.
Then

$k[[x]](M\cap k\langle x\rangle^{p})=(k[[x,y]]M)\cap k[[x]]^{p},$

equivalently $M\cap k\langle x\rangle^{p}$ is dense in $(k[[x,y]]M)\cap
k[[x]]^{p}$, that is for all ${\hat{v}}\in(k[[x,y]]M)\\\ \cap k[[x]]^{p}$ and
$c\in\bf N$ there exists $v_{c}\in(M\cap k\langle x\rangle)^{p}$ such that
$v_{c}\equiv{\hat{v}}\ \mbox{modulo}\ (x)^{c}k[[x]]^{p}$. Moreover, if
$c\in\bf N$ and ${\hat{v}}=\sum_{i=1}^{t}{\hat{u}}_{i}a_{i}$ for some
$a_{i}\in M$, ${\hat{u}}_{i}\in k[[x,y]]$ then there exist $u_{ic}\in k\langle
x,y\rangle$ such that $u_{ic}\equiv{\hat{u}}_{i}\ \mbox{modulo}\
(x)^{c}k[[x,y]]$, $v_{c}=\sum_{i=1}^{t}u_{ic}a_{i}\in(M\cap k\langle
x\rangle^{p})$ and $\hat{v}$ is the limit of $(v_{c})_{c}$ in the $(x)$-adic
topology.

###### Proof.

Let ${\hat{v}}\in(k[[x,y]]M)\cap k[[x]]^{p}$, let us say
${\hat{v}}=\sum_{i=1}^{t}{\hat{u}}_{i}a_{i}$ for some $a_{i}\in M$ and
${\hat{u}}_{i}\in k[[x,y]]^{p}$. By flatness of $k[[x]]\langle y\rangle\subset
k[[x,y]]$ we see that there exist ${\tilde{u}}_{i}\in k[[x]]\langle y\rangle$
such that ${\hat{v}}=\sum_{i=1}^{t}{\tilde{u}}_{i}a_{i}$. Moreover by Lemma 3
we may choose ${\tilde{u}}_{i}$ such that ${\tilde{u}}_{i}\equiv{\hat{u}}_{i}\
\mbox{modulo}\ (x)^{c}k[[x,y]]$. Then using Lemma 2 there exist $g_{i}\in
k\langle y,Z\rangle$, $i\in[t]$ for some variables $Z=(Z_{1},\ldots,Z_{s})$
and ${\hat{z}}\in k[[x]]^{s}$ such that ${\tilde{u}}_{i}=g_{i}({\hat{z}})$.
Note that $a_{i}=\sum_{r\in{\bf N}^{m}}a_{ir}y^{r}$, $g_{i}=\sum_{r\in{\bf
N}^{m}}g_{ir}y^{r}$ with $a_{ir}\in k\langle x\rangle^{p}$, $g_{ir}\in
k\langle Z\rangle$.

Clearly, ${\hat{z}},{\hat{v}}$ is a solution in $k[[x]]$ of the system of
polynomial equations $V=\sum_{i=1}^{t}a_{i}g_{i}(Z)$, $V=(V_{1},\ldots,V_{p})$
if and only if it is a solution of the infinite system of polynomial equations

($*$) $V=\sum_{i=1}^{t}a_{i0}g_{i0}(Z)$,
$\sum_{i=1}^{t}\sum_{r+r^{\prime}=e}a_{ir}g_{ir^{\prime}}(Z)=0$, $e\in{\bf
N}^{m}$, $e\not=0.$

Since $k\langle x,Z,V\rangle$ is Noetherian we see that it is enough to
consider in ($*$) only a finite set of equations, let us say with
$e\leq\omega$ for some $\omega$ high enough. Applying the Artin approximation
property of $k\langle x\rangle$ (see [1]) we can find for any $c\in\bf N$ a
solution $v_{c}\in k\langle x\rangle^{p}$, $z_{c}\in k\langle x\rangle^{s}$ of
($*$) such that $v_{c}\equiv{\hat{v}}\ \mbox{modulo}\ (x)^{c}k[[x]]^{p}$,
$z_{c}\equiv{\hat{z}}\ \mbox{modulo}\ (x)^{c}k[[x]]^{s}$. Then
$v_{c}=\sum_{i=1}^{t}a_{i}g_{i}(z_{c})\in M\cap k\langle x\rangle^{p}$, and
$u_{ic}=g_{i}(z_{ic})\in k\langle x,y\rangle$ satisfies
$u_{ic}\equiv{\tilde{u}}_{i}\equiv{\hat{u}}_{i}\ \mbox{modulo}\
(x)^{c}k[[x,y]]$. Clearly $\hat{v}$ is the limit of $(v_{c})_{c}$ in the
$(x)$-adic topology and belongs to $k[[x]](M\cap k\langle x\rangle^{p})$.

######  Remark 5.

When $p=1$ then the above $M$ is an ideal and we get the so-called (see [2])
strong elimination property of the algebraic power series.

The following proposition is partially contained in [2, Lemma 4.2].

######  Proposition 6.

Let $M\subset k\langle x\rangle^{p}$ be a finitely generated $k\langle
x\rangle$-submodule and $1\leq r_{1}<\ldots<r_{e}\leq n$, $p_{1},\ldots,p_{e}$
be some positive integers such that $p=p_{1}+\ldots+p_{e}$. Then

$T=M\cap(k\langle x_{1},\ldots,x_{r_{1}}\rangle^{p_{1}}\times\ldots\times
k\langle x_{1},\ldots,x_{r_{e}}\rangle^{p_{e}})$

is dense in

${\hat{T}}=(k[[x]]M)\cap(k[[x_{1},\ldots,x_{r_{1}}]]^{p_{1}}\times\ldots\times
k[[x_{1},\ldots,x_{r_{e}}]]^{p_{e}}).$

Moreover, if $c\in\bf N$ and
${\hat{v}}=\sum_{i=1}^{t}{\hat{u}}_{i}a_{i}\in{\hat{T}}$ for some $a_{i}\in
M$, ${\hat{u}}_{i}\in k[[x]]$ then there exist $u_{ic}\in k\langle x\rangle$
such that $u_{ic}\equiv{\hat{u}}_{i}\ \mbox{modulo}\ (x)^{c}k[[x]]$,
$v_{c}=\sum_{i=1}^{t}u_{ic}a_{i}\in T$ and $\hat{v}$ is the limit of
$(v_{c})_{c}$ in the $(x)$-adic topology.

###### Proof.

Apply induction on $e$, the case $e=1$ being done in Proposition 4. Assume
that $e>1$. We may reduce to the case when $r_{e}=n$ replacing $M$ by $M\cap
k\langle x_{1},\ldots,x_{r_{e}}\rangle^{p}$ if $r_{e}<n$. Let

$q:\Pi_{i=1}^{p_{e}}k[[x_{1},\ldots,x_{r_{i}}]]^{p_{i}}\rightarrow\Pi_{i=1}^{p_{e-1}}k[[x_{1},\ldots,x_{r_{i}}]]^{p_{i}},$
$q^{\prime}:\Pi_{i=1}^{p_{e}}k[[x_{1},\ldots,x_{r_{i}}]]^{p_{i}}\rightarrow
k[[x_{1},\ldots,x_{r_{e}}]]^{p_{e}}$

be the canonical projections,
${\hat{v}}=({\hat{v}_{1}},\ldots,{\hat{v}}_{p})\in{\hat{T}},$ and
$M_{1}=q(M)$. Assume that ${\hat{v}}=\sum_{i=1}^{t}{\hat{u}}_{i}a_{i}$ for
some ${\hat{u}}_{i}\in k[[x]]$, $a_{i}\in M$. By induction hypothesis applied
to $M_{1}$, $q(\hat{v})$ given $c\in\bf N$ there exists $u_{ic}\in k\langle
x\rangle$ with $u_{ic}\equiv{\hat{u}}_{i}\ \mbox{modulo}\ (x)^{c}k[[x]]$ such
that $v^{\prime}_{c}=\sum_{i=1}^{t}u_{ic}q(a_{i})\in q(T)$ and $q({\hat{v}})$
is the limit of $(v^{\prime}_{c})_{c}$ in the $(x)$-adic topology.

Now, let $v^{\prime\prime}_{c}=\sum_{i=1}^{t}u_{ic}q^{\prime}(a_{i})\in
k\langle x_{1},\ldots,x_{n}\rangle^{p_{e}}$. We have
$v^{\prime\prime}_{c}\equiv q^{\prime}({\hat{v}})\\\ \mbox{modulo}\
(x)^{c}k[[x]]^{p_{e}}$. Then
$v_{c}=(v^{\prime}_{c},v^{\prime\prime}_{c})=\sum_{i=1}^{t}u_{ic}a_{i}\in T$,
$v_{c}\equiv{\hat{v}}\ \mbox{modulo}\ (x)^{c}k[[x]]^{p}$ and $\hat{v}$ is the
limit of $(v_{c})_{c}$ in the $(x)$-adic topology.

######  Corollary 7.

Theorem 1 holds when $f$ is linear.

###### Proof.

If $f$ is homogeneous then it is enough to apply Proposition 6 for the module
$M$ of the solutions of $f$ in $A=k\langle x\rangle$. Suppose that $f$ is not
homogeneous, let us say $f$ has the form $g+a_{0}$ for some system of linear
homogeneous polynomials $g\in A[Y]^{s}$ and $a_{0}\in A^{s}$. The proof in
this case follows [2, page 7] and we give it here only for the sake of
completeness. Change $f$ by the homogeneous system of linear polynomials
${\bar{f}}=g+a_{0}Y_{0}$ from $A[Y_{0},Y]^{s}$. A nested formal solution
$\hat{y}$ of $\bar{f}$ in $k[[x]]$ with ${\hat{y}}_{i}\in
K[[x_{1},\ldots,x_{r_{i}}]]$, $1\leq i\leq m$ induces a nested formal solution
$({\hat{y}}_{0},{\hat{y}})$, ${\hat{y}}_{0}=1$ of $\bar{f}$ with
$r_{0}=r_{1}$. As above, for all $c\in\bf N$ we get a nested algebraic
solution $(y_{0},y)$ of $\bar{f}$ with $y_{i}\in k\langle
x_{1},\ldots,x_{r_{i}}\rangle$ and $y_{i}\equiv{\hat{y}}_{i}\ \mbox{modulo}\
(x)^{c}k[[x]]$ for all $0\leq i\leq m$. It follows that $y_{0}$ is invertible
and clearly $y_{0}^{-1}y$ is the wanted nested algebraic solution of $f$.

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