Nested Artin Strong Approximation Property

TL;DR

This paper explores a new perspective on the Artin Approximation property, establishing a nested strong approximation result for algebraic power series rings over a field, expanding understanding of approximation in algebraic geometry.

Contribution

It introduces a nested Artin Strong Approximation property for algebraic power series rings, providing a novel approach to approximation constraints in algebraic geometry.

Findings

Established a nested Artin Strong Approximation property.

Extended approximation results to algebraic power series rings.

Provided new tools for algebraic geometry and power series analysis.

Abstract

We study the Artin Approximation property with constraints in a different frame. As a consequence we give a nested Artin Strong Approximation property for algebraic power series rings over a field.

Equations

$$y_j(x)\in K[[x_1,\dots ,x_{s_j}]]\forall j\in [p]$$

$$y_j^{\prime }(x)\in k\{x_1,\dots ,x_{s_j}\}\forall j\in [p].$$

$$\hat{u}=\sum _{j=1}^r{w}_j{\hat{v}}_j$$

$$T\tilde{v}=\tilde{u},$$

$$f(Y)=\sum _{j\in {\mathbb{N}}^n,|j|<\alpha }{w}_j{Y}^j,$$

$$\sum _{j\in {\mathbb{N}}^n,|j|<\alpha }{w}_j{\hat{y}}^j=0$$

$${\mathbf{Q}}^{\prime }(x)\hat{\mathbf{Q}}(x{)}^{\alpha }(\sum _{j\in {\mathbb{N}}^n,|j|<\alpha }{w}_j{\hat{y}}^j)=0.$$

$$\{q\in {\mathbb{N}}^n:y_{iq}\ne 0\}=\{q\in {\mathbb{N}}^n:{\hat{y}}_{iq}\ne 0\}$$

$$K[x]\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$G^r({\Pi }_{i=1}^p{G}_i(a_i^{\prime }{)}^r{f}^{\prime }(b^{\prime },y_1^{\prime }(a_1^{\prime }),\dots ,y_p^{\prime }(a_p^{\prime }))=\tilde{f}(b^{\prime },{\bar{y}}_1^{\prime }(a_1^{\prime }),\dots ,{\bar{y}}_p^{\prime }(a_p^{\prime }),G_1(a_1^{\prime }),\dots ,G_p(a_p^{\prime })),$$

$$V\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$(\sum _{k=1}^m{\bar{G}}_k{T}^k{)}^r\sum _{l_i=1}^{d_i}{\bar{F}}_{i,l_i}(\sum _{s_i=1}^d{a}_{i,s_i}^{\prime }{T}^{s_i}{)}^{l_i}=FP$$

$$(\sum _{k\le m,k^{\prime }\in {\mathbb{N}}^n,|k^{\prime }|<\alpha }{\bar{G}}_{k,k^{\prime }}{x}^{k^{\prime }}{T}^k{)}^r(\sum _{l_i<d_i,l_i^{\prime }\in {\mathbb{N}}^n,|l_i^{\prime }|<\alpha }{\bar{F}}_{i,l_i,l_i^{\prime }}{x}_{J_i}^{l_i^{\prime }}(\sum _{s_i<d,s_i^{\prime }\in {\mathbb{N}}^n,|s_i^{\prime }|<\alpha }{\bar{a}}_{i,s_i,s_i^{\prime }}^{\prime }{x}^{s_i^{\prime }}{T}^{s_i}{)}^{l_i})=$$

$$(\sum _{l<d,l^{\prime }\in {\mathbb{N}}^n,|l^{\prime }|<\alpha }{\bar{F}}_{l,l^{\prime }}{x}^{l^{\prime }}{T}^l)(\sum _{c\le m,c^{\prime }\in {\mathbb{N}}^n,|c^{\prime }|<\alpha }{\bar{P}}_{c,c^{\prime }}{x}^{c^{\prime }}{T}^c)$$

$$(\sum _{k_i\le m,k_i^{\prime }\in {\mathbb{N}}^n,|k_i^{\prime }|<\alpha }{\bar{G}}_{i,k_i,k_i^{\prime }}{x}^{k_i^{\prime }}(\sum _{s_i<d,s_i^{\prime }\in {\mathbb{N}}^n,|s_i^{\prime }|<\alpha }{\bar{a}}_{i,s_i,s_i^{\prime }}^{\prime }{x}^{s_i^{\prime }}{T}^{s_i}{)}^{k_i})(\sum _{q\le m,q^{\prime }\in {\mathbb{N}}^n,|q^{\prime }|<\alpha }{\bar{Q}}_{q,q^{\prime }}{x}^{q^{\prime }}{T}^q)=$$

$$(\sum _{k\le m,k^{\prime }\in {\mathbb{N}}^n,|k^{\prime }|<\alpha }{\bar{G}}_{k,k^{\prime }}{x}^{k^{\prime }}{T}^k{)}^r+(\sum _{l<d,l^{\prime }\in {\mathbb{N}}^n,|l^{\prime }|<\alpha }{\bar{F}}_{l,l^{\prime }}{x}^{l^{\prime }}{T}^l)(\sum _{t\le m,t^{\prime }\in {\mathbb{N}}^n,|t^{\prime }|<\alpha }{\bar{L}}_{t,t^{\prime }}{x}^{t^{\prime }}{T}^t)$$

$$(\sum _{k=1}^m{\bar{G}}_k{T}^k{)}^r\sum _{\tilde{l}=1}^{\tilde{d}}{\overline{\stackrel{~}{F}}}_{\tilde{l}}(\sum _{\tilde{s}=1}^d{b}_{\tilde{s}}^{\prime }{T}^{\tilde{s}}{)}^{\tilde{l}}=F\tilde{P}$$

$$(\sum _{k\le m,k^{\prime }\in {\mathbb{N}}^n,|k^{\prime }|<\alpha }{\bar{G}}_{k,k^{\prime }}{x}^{k^{\prime }}{T}^k{)}^r(\sum _{\tilde{l}=1}^{\tilde{d}}(\sum _{{\tilde{l}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{l}}^{\prime }|<\alpha }{\overline{\stackrel{~}{F}}}_{\tilde{l}{\tilde{l}}^{\prime }}{x}^{{\tilde{l}}^{\prime }}(\sum _{\tilde{s}<d,{\tilde{s}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{s}}^{\prime }|<\alpha }{\bar{b}}_{\tilde{s},\widetilde{s^{\prime }}}^{\prime }{x}^{\widetilde{s^{\prime }}}{T}^{\tilde{s}}{)}^{\tilde{l}}))=$$

$$(\sum _{l<d,l^{\prime }\in {\mathbb{N}}^n,|l^{\prime }|<\alpha }{\bar{F}}_{l,l^{\prime }}{x}^{l^{\prime }}{T}^l)(\sum _{\tilde{c}\le m,{\widetilde{c^{\prime }}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{c}}^{\prime }|<\alpha }{\overline{\stackrel{~}{P}}}_{\tilde{c},{\tilde{c}}^{\prime }}{x}^{{\tilde{c}}^{\prime }}{T}^{\tilde{c}}).$$

$$\sum _{\tilde{k}\le \tilde{m},{\tilde{k}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{k}}^{\prime }|<\alpha }{\overline{\stackrel{~}{G}}}_{\tilde{k},{\tilde{k}}^{\prime }}{x}^{{\tilde{k}}^{\prime }}(\sum _{\tilde{s}<d,{\tilde{s}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{s}}^{\prime }|<\alpha }{\bar{b}}_{\tilde{s},{\tilde{s}}^{\prime }}^{\prime }{x}^{{\tilde{s}}^{\prime }}{T}^{\tilde{s}}{)}^{\tilde{k}}.(\sum _{\tilde{q}\le m,{\tilde{q}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{q}}^{\prime }|<\alpha }{\overline{\stackrel{~}{Q}}}_{\tilde{q},{\tilde{q}}^{\prime }}{x}^{{\tilde{q}}^{\prime }}{T}^{\tilde{q}})$$

$$(\sum _{k\le m,{\tilde{k}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{k}}^{\prime }|<\alpha }{\bar{G}}_{k,k^{\prime }}{x}^{k^{\prime }}{T}^k{)}^r+(\sum _{l<d,l^{\prime }\in {\mathbb{N}}^n,|l^{\prime }|<\alpha }{\bar{F}}_{l,l^{\prime }}{x}^{l^{\prime }}{T}^l)(\sum _{\tilde{t}\le m,{\tilde{t}}^{\prime }\in {\mathbb{N}}^n,|{\tilde{t}}^{\prime }|<\alpha }{\overline{\stackrel{~}{L}}}_{\tilde{t},{\tilde{t}}^{\prime }}{x}^{{\tilde{t}}^{\prime }}{T}^{\tilde{t}})$$

$$\tilde{f}(b^{\prime },{\bar{y}}_1^{\prime }(a_1^{\prime }),\dots ,{\bar{y}}_p^{\prime }(a_p^{\prime }),G_1(a_1^{\prime }),\dots ,G_p(a_p^{\prime }))=$$

$$(\sum _{l<d,l^{\prime }\in {\mathbb{N}}^n,|l^{\prime }|<\alpha }{\bar{F}}_{l,l^{\prime }}{x}^{l^{\prime }}{T}^l)(\sum _{c\le m,c^{\prime }\in {\mathbb{N}}^n,|c^{\prime }|<\alpha }{\bar{P}}_{c,c^{\prime }}{x}^{c^{\prime }}{T}^c)$$

$${\Gamma }_{ij}((Y_{i,k_i,q_i}^{\prime }{)}_{k_i\le d_i,|q_i|<\omega },(Z_{ik}{)}_{|k|<\omega })=0,|j|<c_i,$$

$${\Gamma }_{i\gamma }((Y_{i,k_i,q_i}^{\prime }{)}_{k_i\le d_i,|q_i|<\omega },(Z_{ik}{)}_{|k|<\omega })Z_{i\gamma }^{\prime \prime }=1.$$

$${\eta }_i:(K⟨x_1,\dots ,x_{s_i}⟩{)}^{\ast }\to (K⟨x_1,\dots ,x_{s_i}⟩{)}^{\ast }/\bigcap _{l\in \mathbb{N}}(x_1,\dots ,x_{s_i}{)}^l,$$

$$(K⟨x_1,\dots ,x_{s_i}⟩{)}^{\ast }\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$(K⟨x_1,\dots ,x_{s_i}⟩{)}^{\ast }/\bigcap _{l\in \mathbb{N}}(x_1,\dots ,x_{s_i}{)}^l\cong K^{\ast }[[x_1,\dots ,x_{s_i}]]$$

$$(K⟨x⟩{)}^{\ast }/\bigcap _{l\in \mathbb{N}}(x{)}^l\cong K^{\ast }[[x]].$$

Full text

Key words: Henselian rings, Etale neighborhood, Algebraic power series rings, Nested Artin approximation property, Artin approximation with constraints.

2010 Mathematics Subject Classification: Primary 13B40, Secondary : 13J05, 14B12.

Introduction

Let $K$ be a field and $R=K\langle x\rangle$, $x=(x_{1},\ldots,x_{n})$ be the ring of algebraic power series in $x$ over $K$, that is the algebraic closure of the polynomial ring $K[x]$ in the formal power series ring ${\hat{R}}=K[[x]]$. Let $f=(f_{1},\ldots,f_{q})$ be a system of polynomials in $Y=(Y_{1},\ldots,Y_{p})$ over $R$ and $\hat{y}$ be a solution of $f$ in the completion ${\hat{R}}$ of $R$.

Theorem 1** (M. Artin [2]).**

For any $c\in{\bf N}$ there exists a solution $y^{(c)}$ in $R$ such that $y^{(c)}\equiv{\hat{y}}$ mod $(x)^{c}$.

Also M. Artin proved before (see [1]) a similar statement for the ring $R$ of complex convergent power series and later (see [3, p.7]) asked, whether, given $c\in\mathbb{N}$ and a formal solution $y(x)=(y_{1}(x),\ldots,y_{p}(x))\in\mathbb{C}[[x]]^{p}$ satisfying

[TABLE]

for some integers $s_{j}\in[n]$, there exists a convergent solution $y^{\prime}(x)$ of $f$ such that $y^{\prime}(x)\equiv y(x)$ mod $(x)^{c}$ and

[TABLE]

Shortly after, A. Gabrielov [8] (see also [20, Example 5.3.1]) gave an example showing that the answer to the previous question is negative in general. On the other hand, since Theorem 1 remains valid if we replace convergent power series by algebraic power series, the question of M. Artin is also relevant in this context and it appears that in this case the question has a positive answer as it is shown in [17, Theorem 3.7] (see also [18, Corollary 3.7], [20, Theorem 5.2.1]).

Question 2**.**

(Artin Approximation with constraints [20, Problem 1, page 65]) Let $R$ be an excellent local subring of $K[[x]]$, $x=(x_{1},\ldots,x_{n})$ such that the completion of $R$ is $K[[x]]$ and $f\in R[Y]^{q}$, $Y=(Y_{1},\ldots,Y_{p})$. Assume that there exists a formal solution $\hat{y}\in K[[x]]^{p}$ of $f=0$ such that $\hat{y}_{i}\in K[[\{x_{j}:j\in J_{i}\}]]$ for some subset $J_{i}\subset[n]$, $i\in[p]$. Is it possible to approximate $\hat{y}$ by a solution $y\in R^{p}$ of $f=0$ such that $y_{i}\in R\cap K[[\{x_{j}:j\in J_{i}\}]]$, $i\in[p]$?

Similarly, we considered below the following question.

Question 3**.**

Let $K\subset K^{\prime}$ be a field extension, $R=K[x]_{(x)}$, $R^{\prime}=K^{\prime}[x]_{(x)}$ (resp. $R=K\langle x\rangle$, $R^{\prime}=K^{\prime}\langle x\rangle$) and $f\in R[Y]^{p}$, $Y=(Y_{1},\ldots,Y_{m})$. Assume that there exists a solution $\hat{y}\in R^{\prime p}$ of $f=0$ such that $\hat{y}_{i}\in K^{\prime}[\{x_{j}:j\in J_{i}\}]_{(x_{J_{i}})}$ (resp. $K^{\prime}\langle\{x_{j}:j\in J_{i}\}\rangle$) for some subset $J_{i}\subset[n]$, $i\in[m]$. Is it possible to find a solution $y\in R^{p}$ of $f=0$ such that $y_{i}\in K[\{x_{j}:j\in J_{i}\}]_{(x_{J_{i}})}$ (resp, $K\langle\{x_{j}:j\in J_{i}\}\rangle$), and $\operatorname{ord}y_{i}=\operatorname{ord}\hat{y}_{i}$, $i\in[p]$?

We show (see Proposition 7 and Theorem 9) that Question 3 has a positive answer when the field extension $K\subset K^{\prime}$ is algebraically pure. A ring morphism $u:A\rightarrow B$ is called algebraically pure (see [16]), if every finite system of polynomial equations over $A$ has a solution in $A$ if it has a solution in $B$. A finite type ring morphism is algebraically pure if and only if it has a retraction and a filtered inductive limit of algebraically pure morphisms is an algebraically pure morphism by [16] (see also [14, Theorem 1.10]). It is easy to see that a field extension of an algebraically closed field is algebraically pure and the ultrapower of fields define algebraically pure field extension (see the proof of Theorem 11).

The above questions are related with the so called Artin approximation property. There exists also a strong approximation property (see [10], [2], [15], [11], [16], [17], [18], [20], [19]).

Question 4**.**

(Strong Artin Approximation with constraints [20, Problem 2, page 65]) Let us consider $f(y)\in K[[x]][y]^{q}$ and $J_{i}\subset[n]$, $i\in[p]$. Does there exist a function $\nu:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $c\in\mathbb{N}$ and all $\hat{y}_{i}(x)\in K[[x_{J_{i}}]]$, $j\in J_{i}$, $i\in[p]$, such that $f(\hat{y}(x))\in(x)^{\nu(c)}$, $\hat{y}(x)=(\hat{y}_{1}(x),\ldots,\hat{y}_{p}(x))$, there exist $y_{i}(x)\in K[[x_{J_{i}}]]$, $i\in[m]$ such that $f(y(x))=0$, $y(x)=(y_{1}(x),\ldots,y_{p}(x))$ and $\hat{y}_{i}(x)\equiv y_{i}(x)$ mod $(x)^{c}$, $i\in[p]$?

This question has a positive answer when $K=\mathbb{C}$ but a negative one when $K=\mathbb{Q}$ (see [4], [20, Proposition 3.3.4] and [20, Example 5.4.5])). Similarly, we considered below the following nested question.

Question 5**.**

Let us consider a field $K$, $f=(f_{1},\ldots,f_{q})\in K\langle x\rangle[Y]^{q}$, $Y=(Y_{1},\ldots,Y_{p})$ and $0\leq s_{1}\leq\ldots\leq s_{p}\leq n$ be some non-negative integers. Does there exist a function $\nu:\mathbb{N}^{p}\rightarrow\mathbb{N}$ such that for all $c\in\mathbb{N}^{p}$ and all $\hat{y}_{i}(x)\in K[x_{1},\ldots,x_{s_{i}}]$, such that $\operatorname{ord}\hat{y}_{i}(x)=c_{i}$, $i\in[p]$ and $f(\hat{y}(x))\in(x)^{\nu(c)}$, $\hat{y}(x)=(\hat{y}_{1}(x)\ldots,\hat{y}_{p}(x))$, there exist $y_{i}(x)\in K\langle x_{1},\ldots,x_{s_{i}}\rangle$, $i\in[p]$ such that $f(y(x))=0$, $y(x)=(y_{1}(x),\ldots,y_{p}(x))$ and $\operatorname{ord}y_{i}(x)=c_{i}$, $i\in[p]$?

Theorem 11 shows a positive answer to this question. The proof uses the ultrapower methods (see [4], [16], [17], [18]) and so it is not constructive. We should mention that there exists a stronger result (see Remark 13).

Artin approximation property with constraints is necessary in CR Geometry (see [5] and [12]), the nested case appears in the construction of the analytic deformations of a complex analytic germ when it has an isolated singularity (see [9], [11], [19]). It is also used to prove that analytic set germs are homeomorphic to set germs (see [13]), or to polynomial germs (see [6]) having no assumption on the singular locus.

We owe thanks to G. Rond who hinted us the Remark 13 and to a referee who showed us several misprints and had useful comments.

1. Properties on polynomial rings similar to the Artin approximation with constraints.

Question 2 together with [7, Theorem 3.4] give the idea of the following proposition.

Proposition 6**.**

Let $M\subset K[x]^{p}$ be a finitely generated $K[x]$-submodule and $K\rightarrow K^{\prime}$ a field extension. Let $J_{i}$, $i\in[p]$ be subsets of $[n]$, $x_{J_{i}}=(x_{k})_{k\in J_{i}}$ and $A_{i}=K[x_{J_{i}}]$, $A^{\prime}_{i}=K^{\prime}[x_{J_{i}}]$, $i\in[p]$. Set $\mathcal{N}=M\cap(A_{1}\times\cdots\times A_{p})$ and $\mathcal{N^{\prime}}=(K^{\prime}[x]M)\cap(A^{\prime}_{1}\times\cdots\times A^{\prime}_{p}).$ If $\hat{u}=\sum_{j=1}^{r}w_{j}\hat{v}_{j}\in\mathcal{N^{\prime}}$ for some $w_{j}\in M$ and $\hat{v}_{j}\in K^{\prime}[x]$ then there exist $v_{j}\in K[x]$ such that $u=\sum_{j=1}^{r}w_{j}v_{j}\in\mathcal{N}$ and $\operatorname{min}_{i\in[p]}\operatorname{ord}u_{i}=\operatorname{min}_{i\in[p]}\operatorname{ord}\hat{u}_{i}$.

Proof.

Let $w_{1},\ldots,w_{r}$ be the generators of $M$ and ${\hat{u}}\in\mathcal{N^{\prime}}$. We have

[TABLE]

has this form for some polynomials $\hat{v}_{j}\in K^{\prime}[x]$. Let $\hat{u}_{i}=\sum_{l\in\mathbb{N}^{n},|l|\leq\alpha}\hat{u}_{il}x^{l}$, $w_{ij}=\sum_{t\in\mathbb{N}^{n},|t|\leq\beta}w_{i,j,t}x^{t}$, $i\in[p]$ and $\hat{v}_{j}=\sum_{q\in\mathbb{N}^{n},|q|\leq\gamma}\hat{v}_{j,q}x^{q}$, where $\alpha=\beta+\gamma$. The components of equation (1) give a system of $(p(\sum_{1\leq s\leq\beta}{s+n-1\choose s}))$-linear equations as

[TABLE]

where $T$ is a $(p(\sum_{1\leq s\leq\beta}{s+n-1\choose s}))\times(r(\sum_{1\leq s^{\prime}\leq\gamma}{s^{\prime}+n-1\choose s^{\prime}}))$ matrix of entries of coefficients of $w_{j}$ from $K$, $\tilde{v}$ is a vector of entries from the coefficients of $\hat{v}_{j}$ from $K^{\prime}$ and $\tilde{u}$ is a vector of entries from the coefficients of $\hat{u}_{i}$ from $K^{\prime}$. Now this is a system $L$ of linear equations with coefficients in $K$ and has a solution $(\hat{v}_{jq}),(\hat{u}_{il})$ in $K^{\prime}$. We may consider in $L$ only variables corresponding to those $(\hat{v}_{jq}),(\hat{u}_{il})$ which are not zero. Since $K\rightarrow K^{\prime}$ is a flat morphism there exists a solution of $L$ in $K$ say $(v_{jq}),(u_{il})$. Thus we have $u=\sum_{l\in\mathbb{N}^{n},|l|\leq\alpha}{u}_{l}x^{l}\in K[x]^{p}$, and ${v}_{j}=\sum_{q\in\mathbb{N}^{n},|q|\leq\gamma}{v}_{jq}x^{q}\in K[x]$ such that $u=\sum_{j=1}^{r}w_{j}v_{j}$.

Assume that $\operatorname{min}_{i\in[p]}\operatorname{ord}\hat{u}_{i}=\operatorname{ord}\hat{u}_{i_{0}}=c$ for some $i_{0}\in[p]$. Then there exists $\nu_{1},\ldots,\nu_{n}\in\mathbb{N}$ such that $\sum_{i}\nu_{i}=c$ and $\hat{u}_{i_{0}\nu}\neq 0$. If we have $u_{i_{0}\nu}=0$ for all solutions of $L$ from $K$ then we get a contradiction because $(\hat{u}_{il})$ is generated by the solutions of $L$ from $K$. Hence $u_{i_{0}\nu}\not=0$. Since the coefficients of $u_{i}$ are zero when the corresponding coefficients of $\hat{u}_{i}$ are zero we see that $u\in\mathcal{N}$ and it has no coefficients corresponding to ${\hat{u}}_{il}$ when $|l|<c$. Hence we have $\operatorname{min}_{i}\operatorname{ord}u_{i}=\operatorname{ord}u_{i_{0}}=c$.

Proposition 7**.**

Let $K\rightarrow K^{\prime}$ be an algebraically pure morphism of fields and $x=(x_{1},\ldots,x_{n})$. Let $J_{i}$, $i\in[p]$ be subsets of $[n]$, $x_{J_{i}}=(x_{k})_{k\in J_{i}}$ and $A_{i}=K[x_{J_{i}}]_{(x_{J_{i}})}$, $A^{\prime}_{i}=K^{\prime}[x_{J_{i}}]_{(x_{J_{i}})}$, $i\in[p]$. Set $\mathcal{N}=A_{1}\times\cdots\times A_{p}$ and $\mathcal{N^{\prime}}=A^{\prime}_{1}\times\cdots\times A^{\prime}_{p}$. Let $f$ be a system of polynomials from $K[x]_{(x)}[Y]$, $Y=(Y_{1},\ldots,Y_{p})$, and ${\hat{y}}\in\mathcal{N^{\prime}}$, such that $f({\hat{y}})=0$. Then there exists $y\in\mathcal{N}$ such that $f(y)=0$ and $\operatorname{ord}y_{i}=\operatorname{ord}\hat{y}_{i}$ for $i\in[p]$.

Proof.

Let $f\in K[x]_{(x)}[Y]$ be of the form

[TABLE]

where $w_{j}$ are the coefficients of $f$ in $Y$ and belong to $K[x]_{(x)}$. Since ${\hat{y}}\in\mathcal{N^{\prime}}$ is a solution of $f$, we get

[TABLE]

Let $\hat{y}=\frac{\hat{P}(x)}{\hat{Q}(x)}\in\mathcal{N^{\prime}}$, where $\hat{P}(x)=(\hat{P}_{1}(x),\ldots,\hat{P}_{p}(x))$, $\hat{Q}(x)=(\hat{Q}_{1}(x),\ldots,\hat{Q}_{p}(x))$, $\hat{P}_{i}(x),\hat{Q}_{i}(x)\in K^{\prime}[x_{J_{i}}]$, and $\hat{Q}_{i}(x)\notin(x_{J_{i}})$ for all $i\in[p]$. Let $w_{j}=(w_{1j},\ldots,w_{pj})$, $w_{ij}=\frac{P^{\prime}_{ij}(x)}{Q^{\prime}_{ij}(x)}\in K[x]_{(x)}$, where $j\in\mathbb{N}^{n},|j|<\alpha$, $Q^{\prime}_{ij}(x)\in K[x]$ and $Q^{\prime}_{ij}(x)\notin(x)$. Let $\mathbf{Q}^{\prime}(x)$ be the least common multiple of $((Q^{\prime}_{ij})(x))$ for all $j\in\mathbb{N}^{n},|j|<\alpha$ and $\mathbf{\hat{Q}}(x)=\Pi_{i\in[p]}(\hat{Q}_{i}(x))$. If we multiply the equation (2) by $\mathbf{Q}^{\prime}(x)\mathbf{\hat{Q}}(x)^{\alpha}$ we get

[TABLE]

From equation (3) we see that $f$ will be transformed in a different system of equations in $\hat{P}_{i},\hat{Q}_{i}$, that is we change $f$ by another system in more $\hat{y}$ (in fact $2p$ will be the new $p$) but this time they are from $K^{\prime}[x_{J_{1}}]\times\ldots\times K^{\prime}[x_{J_{p}}]$ for some bigger $p$ and we look for a solution from $K[x_{J_{1}}]\times\ldots\times K[x_{J_{p}}]$. This new system will give a system of equations $F$ in the nonzero coefficients $(\hat{y}_{iq})$ of $\hat{y}_{i}$ in $x$. Moreover we add for each $i,q$ with $\hat{y}_{iq}\not=0$ a new equation $G_{iq}=Y_{iq}Y_{iq}^{{}^{\prime}-1}-1$ which has a solution in $K^{\prime}$ given by $\hat{y}_{iq},\hat{y}_{iq}^{-1}$. Thus $F,G=(G_{iq})$ must have a solution $(y_{iq}),(y^{\prime}_{iq})$ also in $K$. Then

[TABLE]

for any $i\in[p]$. It follows that the new $y$ given by $y_{i}=\sum_{q}y_{iq}x^{q}$ satisfies $y\in K[x_{J_{1}}]\times\ldots\times K[x_{J_{p}}]\subset\mathcal{N}$, $f(y)=0$ and $\operatorname{ord}y_{i}=\operatorname{ord}\ \hat{y}_{i}$ for all $i\in[p]$.

Corollary 8**.**

Let $K$ be an algebraically closed field, $K\subset K^{\prime}$ a field extension and $x=(x_{1},\ldots,x_{n})$. Let $J_{i}$, $i\in[p]$ be subsets of $[n]$, $x_{J_{i}}=(x_{k})_{k\in J_{i}}$ and $A_{i}=K[x_{J_{i}}]_{(x_{J_{i}})}$, $A^{\prime}_{i}=K^{\prime}[x_{J_{i}}]_{(x_{J_{i}})}$, $i\in[p]$. Set $\mathcal{N}=A_{1}\times\cdots\times A_{p}$ and $\mathcal{N^{\prime}}=A^{\prime}_{1}\times\cdots\times A^{\prime}_{p}$. Let $f$ be a system of polynomials from $K[x]_{(x)}[Y]$, $Y=(Y_{1},\ldots,Y_{p})$, and ${\hat{y}}\in\mathcal{N^{\prime}}$, such that $f({\hat{y}})=0$. Then there exist $y\in\mathcal{N}$ such that $f(y)=0$ and $\operatorname{ord}y_{i}=\operatorname{ord}\hat{y}_{i}$ for $i\in[p]$.

For the proof note that a field extension of an algebraically closed field is an algebraically pure field extension (see [16, Corollary 1.8]).

2. Properties on algebraic power series similar to the Artin approximation with constraints.

Theorem 9**.**

Let $K\rightarrow K^{\prime}$ be an algebraically pure morphism of fields and $x=(x_{1},\ldots,x_{n})$. Let $J_{i}$, $i\in[p]$ be subsets of $[n]$, $x_{J_{i}}=(x_{k})_{k\in J_{i}}$ and $A_{i}=K\langle x_{J_{i}}\rangle$, resp. $A^{\prime}_{i}=K^{\prime}\langle x_{J_{i}}\rangle$, $i\in[p]$ be the algebraic power series in $x_{J_{I}}$ over $K$ resp. $K^{\prime}$. Set $\mathcal{N}=A_{1}\times\cdots\times A_{p}$ and $\mathcal{N^{\prime}}=A^{\prime}_{1}\times\cdots\times A^{\prime}_{p}$. Let $f$ be a system of polynomials from $K\langle x\rangle[Y]$, $Y=(Y_{1},\ldots,Y_{p})$, and ${\hat{y}}\in\mathcal{N^{\prime}}$, such that $f({\hat{y}})=0$. Then there exist $y\in\mathcal{N}$ such that $f(y)=0$ and $\operatorname{ord}y_{i}=\operatorname{ord}\hat{y}_{i}$ for $i\in[p]$.

Proof.

As $\hat{y}_{i}\in A^{\prime}_{i}$ and $A^{\prime}_{i}$ is a filtered inductive limit of etale neighborhoods of $K^{\prime}[x_{J_{i}}]_{(x_{J_{i}})}$ we may find $y^{\prime}_{i}$ in an etale $K^{\prime}[x_{J_{i}}]$-algebra $U_{i}$ such that $\hat{y}_{i}$ is the image of $y^{\prime}_{i}$ by the limit map $\rho_{i}:U_{i}\rightarrow K^{\prime}\langle x_{J_{i}}\rangle$. Let us say $y^{\prime}_{i}\in U_{i}=(K^{\prime}[x_{J_{i}}][T_{i}]/(F_{i}))_{G_{i}}$, where $F_{i}$ is monic in $T_{i}$ and $G_{i}$ is a multiple of $\partial F_{i}/\partial T_{i}$. The images of $\rho_{i}$ are contained in an etale neighborhood of $K^{\prime}[x]_{(x)}$ and so they factor through an etale $K^{\prime}[x]$-algebra, let us say through $U=(K^{\prime}[x][T]/(F))_{G}$, where $F$ is monic in $T$ and $G$ is a multiple of $\partial F/\partial T$. Suppose that the $K^{\prime}[x_{J_{i}}]$-morphism $\varphi_{i}:U_{i}\rightarrow U$ is given by $T_{i}\rightarrow a^{\prime}_{i}\in U$. It follows that $F_{i}(a^{\prime}_{i})=0$ and $G_{i}(a^{\prime}_{i})$ is invertible in $U$, that is

i) $G^{r}F_{i}(a^{\prime}_{i})\in(F)$,

ii) $G_{i}(a^{\prime}_{i})|G^{r}$ modulo $(F)$

for some $r\in\mathbb{N}$.

Note that the coefficients of $f$ belong to an etale neighborhood of $K[x]_{(x)}$ and so we may consider the coefficients of $f$ as images of some elements of an etale $K[x]$-algebra $V$ by the limit map $\theta:V\rightarrow K\langle x\rangle$. Let us say $V=(K[x,\tilde{T}]/(\tilde{F}))_{\tilde{G}}$, where $\tilde{F}$ is monic in $\tilde{T}$ and $\tilde{G}$ is a multiple of $\partial\tilde{F}/\partial\tilde{T}$. More precisely, assume that $f=\sum_{j<\beta}w_{j}Y^{j}$, $w_{j}\in K\langle x\rangle$ and choose $w^{\prime}_{j}\in V$ with $\theta(w^{\prime}_{j})=w_{j}$. Set $f^{\prime}=\sum_{j<\beta}w^{\prime}_{j}Y^{j}\in V[Y]$. We may enlarge $U$ such that the composite map $V\rightarrow K\langle x\rangle\rightarrow K^{\prime}\langle x\rangle$ factors through $U$, let us say the $K[x]$-morphism $\psi:V\rightarrow U$ is given by $\tilde{T}\rightarrow b^{\prime}\in U$.

Let $\rho:U\rightarrow K^{\prime}\langle x\rangle$ be the corresponding limit map. We have the following commutative diagram.

[TABLE]

We have

i’) $G^{r}\tilde{F}(b^{\prime})\in(F)$,

ii’) $\tilde{G}(b^{\prime})|G^{r}$ modulo $(F)$

possibly by increasing $r$. By multiplying $f^{\prime}$ by a certain power of $\tilde{G}$ we may consider that the coefficients of $f^{\prime}$ are in $K[x,\tilde{T}]$. Changing again $U$ if necessary, we have

$f^{\prime}(\psi(\tilde{T}),\varphi_{1}(y^{\prime}_{1}),\ldots,\varphi_{p}(y^{\prime}_{p}))=0$ in $U$. Replacing $y^{\prime}_{i}$ by $\bar{y}^{\prime}_{i}/G_{i}^{r_{i}}$ modulo $F_{i}$ for some $r_{i}\in{\mathbb{N}}$, $\bar{y}^{\prime}_{i}\in K^{\prime}[x_{J_{i}},T_{i}]$ we may change $f^{\prime}$ into a system of polynomials $\tilde{f}$ in $Y,Y^{\prime}=(Y^{\prime}_{1},\ldots,Y^{\prime}_{p})$ such that

[TABLE]

increasing again $r$ if necessary. It follows that

iii) $\tilde{f}(b^{\prime},\bar{y}^{\prime}_{1}(a^{\prime}_{1}),\ldots,\bar{y}^{\prime}_{p}(a^{\prime}_{p}),G_{1}(a^{\prime}_{1}),\ldots,G_{p}(a^{\prime}_{p}))\in(F).$

Note that $\tilde{f}$ depends on $\bar{y}^{\prime}_{i},a^{\prime}_{i},G_{i}$. We have the following commutative diagram.

[TABLE]

Next we note that i),ii), i’),ii’),iii) means that the coefficients from $K^{\prime}$ of $y^{\prime}_{i}$, $a^{\prime}_{i}$, $b^{\prime}$, $G_{i}$, $G$, $F_{i}$, $F$ are solutions in $K^{\prime}$ of a certain system $H$ of polynomial equations over $K$.

Let $F_{i}=\sum_{l_{i}=1}^{d_{i}}\bar{F}_{i,l_{i}}T_{i}^{l_{i}}$ where $\bar{F}_{i,l_{i}}\in K^{\prime}[x_{J_{i}}]$ with $\bar{F}_{i,d_{i}}=1$, $F=\sum_{l=1}^{d}\bar{F}_{l}T^{l}$ where $\bar{F}_{l}\in K^{\prime}[x]$ with $\bar{F}_{d}=1$, and $G_{i}=\sum_{k_{i}=1}^{m}\bar{G}_{i,k_{i}}T_{i}^{k_{i}}$ where $\bar{G}_{i,k_{i}}\in K^{\prime}[x_{J_{i}}]$, $G=\sum_{k=1}^{m}\bar{G}_{k}T^{k}$ where $\bar{G}_{k}\in K^{\prime}[x]$ for some $m>>0$.

i) Since we have $G^{r}F_{i}(a^{\prime}_{i})\in(F)$, where $a^{\prime}_{i}=\sum_{s_{i}=1}^{d-1}a^{\prime}_{i,s_{i}}T^{s_{i}}$, $a^{\prime}_{i,s_{i}}\in K^{\prime}[x]$ it follows that

[TABLE]

for some $P=\sum_{c=1}^{m}{\bar{P}}_{c}T^{c}\in K^{\prime}[x][T]$ increasing $m$ if necessary. This equation has the form:

[TABLE]

[TABLE]

for some $\alpha>>0$. Here $\bar{G}_{k}=\sum_{k^{\prime}\in\mathbb{N}^{n},|k^{\prime}|<\alpha}\bar{G}_{kk^{\prime}}x^{k^{\prime}}$, $\bar{G}_{kk^{\prime}}\in K^{\prime}$ and similarly for others. This gives a system of polynomial equations $H_{i}$ over $K$ which has a solution in $K^{\prime}$, namely $({\bar{G}}_{k,k^{\prime}}),({\bar{F}}_{i,l_{i},l^{\prime}_{i}}),(\bar{a}^{\prime}_{i,s_{i},s^{\prime}_{i}}),({\bar{F}}_{l,l^{\prime}}),({\bar{P}}_{c,c^{\prime}})$.

ii) Since $G_{i}(a^{\prime}_{i})|G^{r}$ modulo $(F)$, there exist $Q,L\in K[x][T]$ such that $G_{i}(a^{\prime}_{i}).Q=G^{r}+F.L$. We may assume that $Q=\sum_{q=1}^{m}{\bar{Q}}_{q}T^{q}$ and $L=\sum_{t=1}^{m}{\bar{L}}_{t}T^{t}$ increasing $m$ if necessary. Then we get:

[TABLE]

[TABLE]

increasing $m,\alpha$ if necessary. This gives a system of polynomial equations $H_{ii}$ over $K$ which has a solution in $K^{\prime}$, namely $({\bar{G}}_{i,k_{i},k^{\prime}_{i}}),(\bar{a}^{\prime}_{i,s_{i},s^{\prime}_{i}}),({\bar{Q}}_{q,q^{\prime}}),({\bar{F}}_{l,l^{\prime}}),({\bar{L}}_{t,t^{\prime}}),({\bar{G}}_{k,k^{\prime}})$.

Let $\tilde{F}=\sum_{\tilde{l}=1}^{\tilde{d}}\bar{\tilde{F}}_{\tilde{l}}T^{\tilde{l}}$ where $\bar{\tilde{F}}_{l}\in K[x]$ with $\bar{\tilde{F}}_{\tilde{d}}=1$, $\tilde{G}=\sum_{\tilde{k}=1}^{m}\bar{\tilde{G}}_{\tilde{k}}T^{\tilde{k}}$ where $\bar{\tilde{G}}_{\tilde{k}}\in K^{\prime}[x]$.

i’) Since we have $G^{r}\tilde{F}(b^{\prime})\in(F)$, where $b^{\prime}=\sum_{\tilde{s}=1}^{d-1}b^{\prime}_{\tilde{s}}T^{\tilde{s}}$, we have

[TABLE]

for some $\tilde{P}=\sum_{\tilde{c}=1}^{m}\bar{\tilde{P}}_{\tilde{c}}T^{\tilde{c}}\in K^{\prime}[x][T]$ increasing $m$ if necessary. This equation has the form:

[TABLE]

[TABLE]

Here $\bar{\tilde{P}}_{\tilde{c}}=\sum_{\tilde{c^{\prime}}^{\prime}\in\mathbb{N}^{n},|\tilde{c}^{\prime}|<\alpha}\bar{\tilde{P}}_{\tilde{c}\tilde{c}^{\prime}}x^{\tilde{c}^{\prime}}$, $\bar{\tilde{P}}_{\tilde{c}\tilde{c}^{\prime}}\in K^{\prime}[x]$ and similarly for others. It gives a system of polynomial equations $H_{i^{\prime}}$ over $K$ which has a solution in $K^{\prime}$, namely $(\bar{G}_{k,k^{\prime}}),(\bar{b}^{\prime}_{\tilde{s},\tilde{s^{\prime}}}),(\bar{F}_{l,l^{\prime}}),(\bar{\tilde{P}}_{\tilde{c},\tilde{c}^{\prime}})$.

ii’) Since $\tilde{G}(b^{\prime})|G^{r}$ modulo $(F)$, there exist $\tilde{Q},\tilde{L}\in K^{\prime}[x][T]$ such that $\tilde{G}(b^{\prime}).\tilde{Q}=G^{r}+F.\tilde{L}$. We may choose $\tilde{Q}=\sum_{\tilde{q}=1}^{m}\bar{\tilde{Q}}_{\tilde{q}}T^{\tilde{q}}$ and $\tilde{L}=\sum_{\tilde{t}=1}^{m}\bar{\tilde{L}}_{\tilde{t}}T^{\tilde{t}}$ increasing $m$ if necessary. Now this equation has the following form:

[TABLE]

=

[TABLE]

increasing $\alpha$ if necessary. This gives a system of polynomial equations $H_{ii^{\prime}}$ over $K$ which has a solution in $K^{\prime}$, namely $(\bar{b}^{\prime}_{\tilde{s},\tilde{s}^{\prime}}),(\bar{F}_{l,l^{\prime}}),(\bar{\tilde{L}}_{\tilde{t},\tilde{t}^{\prime}}),(\bar{G}_{k,k^{\prime}}),(\bar{\tilde{Q}}_{\tilde{q},\tilde{q}^{\prime}})$ .

iii) Since $\tilde{f}(b^{\prime},\bar{y}^{\prime}_{1}(a^{\prime}_{1}),\ldots,\bar{y}^{\prime}_{p}(a^{\prime}_{p}),G_{1}(a^{\prime}_{1}),\ldots,G_{p}(a^{\prime}_{p}))\in(F)$ in $K^{\prime}[x,T]$, there exists $P=\sum_{c\leq m,c^{\prime}\in\mathbb{N}^{n},|c^{\prime}|<\alpha}{\bar{P}}_{c,c^{\prime}}x^{c^{\prime}}T^{c}\in K^{\prime}[x,T]$ (increasing $m,\alpha$ if necessary) such that

[TABLE]

[TABLE]

where $y^{\prime}_{i}=\sum_{j_{i},q_{i}\in\mathbb{N}^{n},|q_{i}|<\alpha,k_{i}<d_{i}}{\bar{y}}^{\prime}_{i,k_{i},q_{i}}x^{q_{i}}T_{i}^{k_{i}}$, ${\bar{y}}^{\prime}_{i,k_{i},q_{i}}\in K^{\prime}$. This gives a system of polynomial equations $H_{iii}$ over $K$ which has a solution in $K^{\prime}$, namely $(\bar{G}_{i,k_{i},k^{\prime}_{i}}),(\bar{F}_{l,l^{\prime}}),$ $(\bar{b}^{\prime}_{\tilde{s},\tilde{s^{\prime}}}),(\bar{a}^{\prime}_{i,s_{i},s^{\prime}_{i}})$, $(\bar{\tilde{P}}_{\tilde{c},\tilde{c}^{\prime}})$, $(\bar{y}^{\prime}_{i,k_{i},q_{i}})$.

A solution of $H=H_{i}\cup H_{ii}\cup H_{i^{\prime}}\cup H_{ii^{\prime}}\cup H_{iii}$ in $K$ will define some elements $\tilde{a}_{i}$, $\tilde{b}$, $\tilde{y}_{i}$ over $K$, some polynomials $\tilde{F}_{i}$, $\tilde{G}_{i}$, $\tilde{F}^{\prime}$, $\tilde{G}^{\prime}$ over $K$, some etale algebras $\tilde{U}_{i}=(K[x_{J_{i}}][T_{i}]/(\tilde{F}_{i}))_{\tilde{G}_{i}}$, $\tilde{U}=(K[x][T]/(\tilde{F}^{\prime}))_{\tilde{G}^{\prime}}$ and some maps $\tilde{\varphi}_{i}:\tilde{U}_{i}\rightarrow\tilde{U}$, $\tilde{\rho}_{i}:\tilde{U}_{i}\rightarrow K\langle x_{J_{i}}\rangle$, $\tilde{\rho}:\tilde{U}\rightarrow K\langle x\rangle$, and similar $\tilde{\psi}$, $\tilde{\varphi}$ which make commutative two diagrams similar as above but written for the case $K=K^{\prime}$ if we can show that $\tilde{\rho}_{i}$ and $\tilde{\theta}$ are given by $\tilde{\rho}\tilde{\varphi}_{i}$, resp. $\tilde{\rho}\tilde{\psi}$.

Suppose that $\rho_{i}$, $\rho$, $\theta$ are given by $\rho_{i}(T_{i})=\hat{z}_{i}=\sum_{k}\hat{z}_{ik}x^{k}$, $\rho(T)=\bar{z}=\sum_{k}\bar{z}_{k}x^{k}$, $\theta(\tilde{T})=z^{\prime}=\sum_{k}z^{\prime}_{k}x^{k}$. Then we have $a^{\prime}_{i}(\bar{z})=\hat{z}_{i}$, $b^{\prime}(\bar{z})=z^{\prime}$ and so $(\bar{G}_{i,k_{i},k^{\prime}_{i}}),(\bar{F}_{l,l^{\prime}}),$ $(\bar{b}^{\prime}_{\tilde{s},\tilde{s^{\prime}}}),(\bar{a}^{\prime}_{i,s_{i},s^{\prime}_{i}})$, $(\bar{\tilde{P}}_{\tilde{c},\tilde{c}^{\prime}})$, $(\bar{y}^{\prime}_{i,k_{i},q_{i}})$, $(\bar{z}_{k})$, $(\hat{z}_{k})$, $(\bar{b}^{\prime}_{s,s^{\prime}})$, $(z^{\prime}_{k})$ is a solution of a certain system of polynomial equations $(\Lambda_{k})_{k\in\mathbb{N}^{n}}$. A solution of $H\cup(\Lambda_{k})_{k\in\mathbb{N}^{n}}$ will define the maps $\tilde{\rho}$, $\tilde{\rho}_{i}$, $\tilde{\theta}$, $\tilde{\psi}$, $\tilde{\varphi}$ which make indeed commutative the two diagrams above written for $K^{\prime}=K$.

But $(\Lambda_{k})_{k\in\mathbb{N}^{n}}$ is an infinite set of equations and we have to see that just a finite set of them are actually enough. Set $\operatorname{ord}\hat{z}_{i}=\nu_{i}$, $\operatorname{ord}\bar{z}=\bar{\nu}$, $\operatorname{ord}z^{\prime}=\nu^{\prime}$ and $\mu=\operatorname{max}\{(\nu_{i}),\bar{\nu},\nu^{\prime}\}$. Using the unicity of the Implicit Function Theorem ($V,U,U_{i}$ are etale) we see that $(\hat{z}_{ik})$, $\bar{z}_{k}$, $(z^{\prime}_{k})$ are uniquelly defined by $(\hat{z}_{ik})_{|k|\leq\mu}$, $(\bar{z}_{k})_{|k|\leq\mu}$ $(z^{\prime}_{k})_{|k|\leq\mu}$. Taken $\Lambda=(\Lambda_{k})_{|k|\leq\mu}$ we see that the two above diagrams are commutative, that is $a^{\prime}_{i}(\bar{z})=\hat{z}_{i}$, $b^{\prime}(\bar{z})=z^{\prime}$ if and only if the solution of $H$ in $K$ is also a solution of $\Lambda$, providing that the order of the corresponding $\tilde{\rho}(T)$, $\tilde{\rho}(T_{i})$, $\tilde{\theta}(\tilde{T})$ is $\leq\mu$. Assume that $\hat{z}_{i,\hat{\tau}_{i}}\not=0$, $\bar{z}_{\bar{\tau}}\not=0$, $z^{\prime}_{\tau^{\prime}}\not=0$ for some $\hat{\tau}_{i}$, $\bar{\tau}$, $\tau^{\prime}$ with $|\hat{\tau}_{i}|=\nu_{i}$, $|\bar{\tau}|=\bar{\nu}$, $|\tau^{\prime}|=\nu^{\prime}$. Then $\hat{z}_{i,\hat{\tau}_{i}}$, $\bar{z}_{\bar{\tau}}$, $z^{\prime}_{\tau^{\prime}}$ and their inverses satisfy also the system $\Delta$ of equations $\hat{Z}_{i,\hat{\tau}_{i}}\hat{Z}_{-1}=1$, $\bar{Z}_{\bar{\tau}}\bar{Z}_{-1}=1$, $Z^{\prime}_{\tau^{\prime}}Z^{\prime}_{-1}=1$. Asking for a solution of $H\cup\Lambda\cup\Delta$ in $K$ means to get the maps $\tilde{\rho}$, $\tilde{\rho}_{i}$, $\tilde{\theta}$, $\tilde{\psi}$, $\tilde{\varphi}$ satisfying the two above commutative diagrams written for $K^{\prime}=K$.

Then $y_{i}=\tilde{\rho}_{i}(\tilde{y}_{i})$ form a solution of $f$ in $K\langle x\rangle$ which is in $\mathcal{N}$. Unfortunately, $\operatorname{ord}y_{i}\not=\operatorname{ord}\hat{y}_{i}$ in general and to get equality we must choose the solution of $H\cup\Lambda\cup\Delta$ more carefully, that is satisfying also some other system of equations.

Note that there exists a system of polynomials $(\Gamma_{ij})_{j\in\mathbb{N}^{n}}$ in some variables $Y^{\prime},Z$ such that $\hat{y}_{i}=\rho_{i}(y^{\prime}_{i})=y^{\prime}_{i}(T_{i}=\hat{z})=\sum_{j\in\mathbb{N}^{n}}\Gamma_{ij}((\bar{y}^{\prime}_{i,k_{i},q_{i}}),(\hat{z}_{k}))x^{j}$. Then

$\Gamma_{ij}((\bar{y}^{\prime}_{i,k_{i},q_{i}}),(\hat{z}_{ik}))=0$ for all $j$ with $|j|<c_{i}$ and $\Gamma_{i\gamma}((\bar{y}^{\prime}_{i,k_{i},q_{i}}),(\hat{z}_{ik}))\not=0$ for some $\gamma$ with $|\gamma|=c_{i}$. Note that only a finite number $(\bar{y}^{\prime}_{i,k_{i},q_{i}}),(\hat{z}_{ik})$, let us say for $q_{i},k\in\mathbb{N}^{n}$,$|q_{i}|,|k|<\omega$, will enter in $\Gamma_{ij}$ when $|j|\leq c_{i}$. We may suppose that $\omega\geq\mu$. Then $(\bar{y}^{\prime}_{i,k_{i},q_{i}})_{k_{i}\leq d_{i},|q_{i}|<\omega}$, $(\hat{z}_{ik})_{|k|<\omega}$ and some $\hat{z}_{i\gamma}^{\prime\prime}$ is a solution of the system $\Gamma_{i}$ given by the polynomial equations

[TABLE]

[TABLE]

Thus $(\bar{G}_{i,k_{i},k^{\prime}_{i}})$, $(\bar{F}_{l,l^{\prime}}),$ $(\bar{G}_{kk^{\prime}})$, $(\bar{F}_{i,l_{i},l^{\prime}_{i}}),$ $(\bar{G}_{kk^{\prime}}$ $(\bar{b}^{\prime}_{\tilde{s},\tilde{s^{\prime}}}),(\bar{a}^{\prime}_{i,s_{i},s^{\prime}_{i}})$, $(\bar{\tilde{P}}_{\tilde{c},\tilde{c}^{\prime}})$, $(\bar{\tilde{L}}_{\tilde{t}\tilde{t}^{\prime}})$, $(\bar{\tilde{Q}}_{\tilde{q}\tilde{q}^{\prime}})$ $(\bar{y}^{\prime}_{i,k_{i},q_{i}})_{|q_{i}|\leq\omega}$, $(\hat{z}_{ik})_{|k|<\omega}$, $(\hat{z}^{\prime\prime}_{i\gamma})$, $(\bar{z}_{k})$, $(z^{\prime}_{k})$, $\hat{z}_{-1}$,$\bar{z}_{-1}$, $z^{\prime}_{-1}$ is a solution of $H\cup\Lambda\cup\Delta\cup\Gamma$, $\Gamma=(\Gamma_{i})_{i\in[p]}$ in $K^{\prime}$. Choosing a solution of $H\cup\Lambda\cup\Delta\cup\Gamma$ in $K$ we see that $y_{i}=\tilde{\rho}_{i}(y^{\prime}_{i})$ satisfies also $\operatorname{ord}y_{i}=\operatorname{ord}\hat{y}_{i}$, $i\in[p]$.

Corollary 10**.**

Let $K$ be an algebraically closed field, $K\subset K^{\prime}$ a field extension and $x=(x_{1},\ldots,x_{n})$. Let $J_{i}$, $i\in[p]$ be subsets of $[n]$, $x_{J_{i}}=(x_{k})_{k\in J_{i}}$ and $A_{i}=K\langle x_{J_{i}}\rangle$, resp. $A^{\prime}_{i}=K^{\prime}\langle x_{J_{i}}\rangle$, $i\in[p]$ be the algebraic power series in $x_{J_{i}}$ over $K$ resp. $K^{\prime}$. Set $\mathcal{N}=A_{1}\times\cdots\times A_{p}$ and $\mathcal{N^{\prime}}=A^{\prime}_{1}\times\cdots\times A^{\prime}_{p}$. Let $f$ be a system of polynomials from $K\langle x\rangle[Y]$, $Y=(Y_{1},\ldots,Y_{p})$, and ${\hat{y}}\in\mathcal{N^{\prime}}$, such that $f({\hat{y}})=0$. Then there exist $y\in\mathcal{N}$ such that $f(y)=0$ and $\operatorname{ord}y_{i}=\operatorname{ord}\hat{y}_{i}$ for $i\in[p]$.

For the proof note that a field extension of an algebraically closed field is an algebraically pure field extension (see [16, Corollary 1.8]).

3. Ultrapower and Nested Strong Artin Approximation

A filter on $\mathbb{N}$ is a non-empty family $D$ of subsets of $\mathbb{N}$ satisfying

1. (1)

$\emptyset\notin D$, 2. (2)

if $s,t\in D$ then $s\cap t\in D$, 3. (3)

if $s\in D$ and $s\subset t\subset\mathbb{N}$ then $t\in D$.

An ultrafilter on $\mathbb{N}$ is a maximal filter in the set of filters on $\mathbb{N}$ with respect to inclusion. A filter $D$ is an ultrafilter if and only if $\mathbb{N}\setminus s\in D$ for each subset $s\subset\mathbb{N}$ such that $s\notin D$. It follows that $s\cup t=\mathbb{N}$ implies $s\in D$ or $t\in D$ because $D$ is an ultrafilter. An ultrafilter is called nonprincipal if there exist no $r$ such that $D=\{s|r\in s\subset\mathbb{N}\}$. More precisely, an ultrafilter is nonprincipal if and only if it contains the filter of all cofinite sets of $\mathbb{N}$.

Let $(A_{i})_{i\in\mathbb{N}}$ be a family of rings and $\mathcal{I}=\{(x_{i})_{i\in\mathbb{N}}\in\prod_{i\in\mathbb{N}}A_{i}|\{i|x_{i}=0\}\in D\}$ an ideal in $\prod_{i\in\mathbb{N}}A_{i}$. We call the factor ring $\prod_{i\in\mathbb{N}}A_{i}/\mathcal{I}$ the ultraproduct of the family $(A_{i})_{i\in\mathbb{N}}$ with respect to the ultrafilter $D$. If $A_{i}=A$ for all $i\in\mathbb{N}$, then denote $A^{*}=\prod_{i\in\mathbb{N}}A_{i}/\mathcal{I}$ and call $A^{*}$ the ultrapower of $A$ with repect to the ultrafilter $D$. An element $a\in A^{*}$ has the form $[(a_{i})_{i\in\mathbb{N}}]$, $a_{i}\in A$, where “[ ]” means the class modulo $\mathcal{I}$. We recall below some properties of the ultrapower given in [4], [16], [17], [18].

If $A$ is a local ring with ${\mathfrak{m}}$ its maximal ideal, then ${\mathfrak{m}}^{*}=\{[(x_{i})_{i\in\mathbb{N}}]|\{i_{i}\in\mathbb{N}|x_{i}\in{\mathfrak{m}}\}\in D\}$ is a maximal ideal in $A^{*}$ and the unique one. If $K$ is the residue field of $A$ then the residue field of $A^{*}$ is the ultrapower $K^{*}$ of $K$ with respect to the ultrafilter $D$.

Theorem 11**.**

Let $K$ be a field, $A=K\langle x\rangle$, $x=(x_{1},\ldots,x_{n})$, $f=(f_{1},\ldots,f_{r})\in K\langle x\rangle[Y]^{r}$, $Y=(Y_{1},\ldots,Y_{p})$ and $0\leq s_{1}\leq\ldots\leq s_{p}\leq n$ be some non-negative integers. Then there exists a map $\nu:\mathbb{N}^{p}\rightarrow\mathbb{N}$ such that if $y^{\prime}=(y^{\prime}_{1},\ldots,y^{\prime}_{p})$, $y^{\prime}_{i}\in K[x_{1},\ldots,x_{s_{i}}]$, $i\in[p]$ satisfies $f(y^{\prime})\equiv 0$ modulo $(x)^{\nu(c)}$ for some $c=(c_{1},\ldots,c_{p})\in\mathbb{N}^{p}$ and $\operatorname{ord}y^{\prime}_{i}=c_{i}$, $i\in[p]$ then there exists $y_{i}\in K\langle x_{1},\ldots,x_{s_{i}}\rangle$ for all $i\in[p]$ such that $y=(y_{1},\ldots,y_{p})$ is a solution of $f$ in $A$ and $\operatorname{ord}y_{i}=c_{i}$ for all $i\in[p]$.

Proof.

We will show that given $c=(c_{1},\ldots,c_{p})\in\mathbb{N}^{p}$ there exists an integer $k_{c}\in\mathbb{N}$ such that if $y^{\prime}=(y^{\prime}_{1},\ldots,y^{\prime}_{p})$, $y^{\prime}_{i}\in K[x_{1},\ldots,x_{s_{i}}]$ satisfies $f(y^{\prime})\equiv 0$ modulo $(x)^{k_{c}}$ and $\operatorname{ord}y^{\prime}_{i}=c_{i}$, $i\in[p]$ then there exists $y_{i}\in K\langle x_{1},\ldots,x_{s_{i}}\rangle$ for all $i\in[p]$ such that $y=(y_{1},\ldots,y_{p})$ is a solution of $f$ in $A$ and $\operatorname{ord}y_{i}=c_{i}$ for all $i\in[p]$.

Assume that this is false. Thus there exists $c$ such that for all $k\in\mathbb{N}$ there exists $y^{\prime(k)}=(y^{\prime(k)}_{1}\ldots,y^{\prime(k)}_{p})$, $y^{\prime(k)}_{i}\in K[x_{1},\ldots,x_{s_{i}}]$ with $f(y^{\prime(k)})\equiv 0$ modulo $(x)^{k}$ and $\operatorname{ord}y^{\prime(k)}_{i}=c_{i}$, $i\in[p]$, but there exists no $y=(y_{1},\ldots,y_{p})$, $y_{i}\in K\langle x_{1},\ldots,x_{s_{i}}\rangle$ for all $i\in[p]$ which is a solution of $f$ in $A$ and $\operatorname{ord}y_{i}=c_{i}$ for all $i\in[p]$.

Let $D$ be a nonprincipal ultrafilter on $\mathbb{N}$ and $(K\langle x\rangle)^{*}$ be the ultrapower of $K\langle x\rangle$ with respect to $D$. Take ${y^{\prime}}_{i}^{*}=[({y^{\prime}}_{i}^{(k)})_{k\in\mathbb{N}}]\in(K[x_{1},\ldots,x_{s_{i}}])^{*}\subset(K\langle x_{1},\ldots,x_{s_{i}}\rangle)^{*}\subset(K\langle x\rangle)^{*}$. Consider the canonical surjection

[TABLE]

and set $\hat{{y^{\prime}}_{i}^{*}}=\eta_{i}({y^{\prime}}_{i}^{*})$. We have the following commutative diagram:

[TABLE]

By [16, Proposition 2.3] (see also [17], [18, Theorem 2.5], [20, Lemma 3.3.2]) we know that

[TABLE]

and similarly

[TABLE]

Thus we may consider $\hat{{y^{\prime}}^{*}_{i}}\in K^{*}[[x_{1},\ldots,x_{s_{i}}]]\subset K^{*}[[x]]$ and note that

$\hat{{y^{\prime}}^{*}}=(\hat{{y^{\prime}}_{1}^{*}},\ldots,\hat{{y^{\prime}}_{p}^{*}})$ is a solution of $f$ in $K^{*}[[x]]$. Indeed, we have $f(y^{\prime(k)})\equiv 0$ modulo $(x)^{k}$ for all $k$. Fix $t\in\mathbb{N}$. we have $f(y^{\prime(k)})\in(x)^{t}$ for all $k\geq t$ and so $\{k\in\mathbb{N}:f(y^{\prime(k)})\in(x)^{t}\}\in D$ because $D$ contains cofinite sets, being a nonprincipal ultrafilter. It follows that $f({y^{\prime}}^{*})\in(x)^{t}(K\langle x_{1},\ldots,x_{n}\rangle)^{*}$ for all $t$. So we have $f({y^{\prime}}^{*})\in\bigcap_{t\in\mathbb{N}}(x)^{t}(K\langle x_{1},\ldots,x_{n}\rangle)^{*}$ and hence $\hat{{y^{\prime}}^{*}}=(\hat{{y^{\prime}}_{1}^{*}},\ldots,\hat{{y^{\prime}}_{p}^{*}})$ gives a solution of $f$ in $K^{*}[[x]]$.

We claim that $\operatorname{ord}\hat{{y^{\prime}}^{*}}_{i}=c_{i}$. Indeed, set $s_{ij}=\{k\in\mathbb{N}:{y^{\prime}}^{(k)}_{ij}\not=0\}$. By assumption $\operatorname{ord}{y^{\prime}}^{(k)}_{i}=c_{i}$, $i\in[p]$ and so $\cup_{j\in\mathbb{N}^{n},|j|=c_{i}}s_{ij}=\mathbb{N}$ which implies $s_{ij_{i}}\in D$ for some $j_{i}$ because $D$ is an ultrafilter. It follows that ${y^{\prime}}^{*}_{ij_{i}}\not=0$ and so $\hat{{y^{\prime}}^{*}}_{ij_{i}}\not=0$, which shows our claim.

By Nested Artin Approximation Property [17, Theorem 3.7] (see also [18, Corollary 3.7], [20, Theorem 5.2.1]) this implies that for all $u\in\mathbb{N}$ there exists a solution $\tilde{y}=(\tilde{y}_{1},\ldots,\tilde{y}_{p})\in K^{*}\langle x\rangle$ of $f$ with $\tilde{y}_{i}\in K^{*}\langle x_{1},\ldots,x_{s_{i}}\rangle$ and $\tilde{y}_{i}\equiv\hat{{y^{\prime}}_{i}^{*}}$ mod $x^{u}$ for all $i$. Take $u>c_{i}$ for all $i$. This implies that ord $\tilde{y}_{i}$= ord $\hat{{y^{\prime}}_{i}^{*}}$, $i\in[p]$. Since $K\rightarrow K^{*}$ is algebraically pure we get by Theorem 9 a $y^{\prime\prime}$ in $K\langle x\rangle$ such that $y^{\prime\prime}_{i}\in K\langle x_{1},\ldots,x_{s_{i}}\rangle$ with ord $y^{\prime\prime}_{i}$= ord $\tilde{y}_{i}$ and $f(y^{\prime\prime})=0$ , a contradiction.

Remark 12**.**

The above theorem does not hold if $A$ is the complex convergent power series ring in $x$. Indeed, Gabrielov’s example (see [8], or [20, Example 5.3.1]) gives a nested formal solution of a certain polynomial equation $f$, which has no nested convergent solutions. Clearly, the formal solution defines some polynomials $y^{(k)}\in\mathbb{C}[x]$ such that $f(y^{(k)}\equiv 0$ mod $(x)^{k}$ for all $k\in\mathbb{N}$. Thus Question 5 has a negative answer for $A$.**

Remark 13**.**

(Rond) In fact it is true a stronger result, namely: There exist a map $\nu^{\prime}:\mathbb{N}^{p}\rightarrow\mathbb{N}$ such that if $y^{\prime}=(y^{\prime}_{1},\ldots,y^{\prime}_{p})$, $y^{\prime}_{i}\in K[x_{1},\ldots,x_{s_{i}}]$, $i\in[p]$ satisfies $f(y^{\prime})\equiv 0$ modulo $(x)^{\nu^{\prime}(c)}$ for some $c\in\mathbb{N}$ then there exists $y_{i}\in K\langle x_{1},\ldots,x_{s_{i}}\rangle$ for all $i\in[p]$ such that $y=(y_{1},\ldots,y_{p})$ is a solution of $f$ in $A$ and $y_{i}\equiv y^{\prime}_{i}$ modulo $(x)^{c}$ for all $i\in[p]$.

This is stated in [20, Corollary 5.2.4]. Its proof is done in a particular case [4, Theorem 4.2] (see also [4, Remark, page 199]) but the general case follows in the same way using the Nested Approximation property [17, Theorem 3.7].**

Acknowledgements The first author gratefully acknowledges the support from the ASSMS GC. University Lahore, for arranging her visit to Bucharest, Romania and she is also grateful to the Simion Stoilow Institute of the Mathematics of the Romanian Academy for inviting her.