Remark on a theorem of H. Hauser on textile maps

TL;DR

This paper challenges a recent theorem on Artin approximation by providing a counterexample, then offers a corrected statement and proof to clarify the theorem's validity.

Contribution

It identifies an error in a recent survey's theorem on textile maps and supplies a corrected version with a valid proof.

Findings

Counterexample disproves the recent theorem

Corrected theorem statement provided

Validated proof of the corrected theorem

Abstract

We give a counter example to the new theorem that appeared in the survey \cite{H} on Artin approximation. We then provide a correct statement and a proof of it.

Equations

$$G(y(x))\equiv 0\text{ mod }(x{)}^l,$$

$$G\left(\sum _{k\ge 0}{y}_k{x}^k\right)=\sum _{l\ge 1}\left(\left(y_0-{\alpha }_{l-1}\right)y_l-1\right)x^l.$$

$$y_N(x)={\alpha }_N+\sum _{k=1}^N\frac{1}{{\alpha }_N-{\alpha }_{k-1}}{x}^k.$$

$$(y_0-{\alpha }_{l-1})y_l=1\forall l\ge 1.$$

$$0=(y_0-{\alpha }_{l_0+1-1})y_{l_0+1}=1$$

$$G(y(x))=\left(\sum _{\beta \in {\mathbb{N}}^n}{G}_{1,\beta }{x}^{\beta },\dots ,\sum _{\beta \in {\mathbb{N}}^n}{G}_{q,\beta }{x}^{\beta }\right)$$

$${\pi }_k:\mathbb{K}[[x]{]}^m⟶{\mathbb{K}[x{]}_{<k}}^m$$

$${\pi }_{k,l}:{\mathbb{K}[x{]}_{<k}}^m⟶{\mathbb{K}[x{]}_{<l}}^m$$

$$X_N:={\pi }_{D_N}\left(\left\{y(x)\in \mathbb{K}[[x]{]}^m\mid G(y(x))\in (x{)}^N\right\}\right)$$

$$\left\{y\in {\mathbb{A}}_{\mathbb{K}}^{m\left(\genfrac{}{}{0ex}{}{n+D_N-1}{D_N-1}\right)}\mid G_{j,\beta }(y)=0,j=1,\dots ,q,|\beta |<N\right\}.$$

$$C_N^k={\pi }_{D_N,k}(X_N).$$

$$C_{N+1}^k\subset C_N^k.$$

$$F_{N_0}^k=F_N^k={\cup }_i{W}_i^N.$$

$$F\backslash V_N\subset C_N^k\forall N\ge N_0.$$

$$\bigcup _{N\ge N_0}{V}_N⊊F.$$

$$A_k:=\bigcap _N{C}_N^k\ne \varnothing .$$

$${\pi }_{l,k}(A_l)=A_k\forall l\ge k.$$

$${\pi }_{l+1,l}(y^{(l+1)})=y^{(l)}.$$

Full text

The aforementioned theorem is the following one:

Strong Approximation Theorem for Textile Maps. [H] Assume that $\mathbb{K}$ is an algebraically closed field, and let $x$ denote again a vector of variables. Let $G:\mathbb{K}[[x]]^{m}\longrightarrow\mathbb{K}[[x]]^{q}$ be a textile map. There exists an $l\in\mathbb{N}$ depending on $G$ such that, if $G=0$ admits an approximate solution $y(x)\in\mathbb{K}[[x]]^{m}$ up to degree $l$,

[TABLE]

then there exists an exact solution $y(x)\in\mathbb{K}[[x]]^{m}$: $G(y(x))=0.$

Let us recall that the map $G$ is textile if $G(y(x))$ is a vector of power series whose coefficients are polynomials in the coefficients of $y(x)$.

This statement is incorrect as shown by the following example:

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For every integer $N\geq 1$ let us set

[TABLE]

Then $G(y_{N}(x))\equiv 0$ modulo $(z)^{N+1}$. But there is no $y(x)=\sum_{k\geq 0}y_{k}x^{k}$ with $G(y(x))=0$. Indeed if such $y(x)$ would exist then we would have

[TABLE]

But $y_{0}\in\overline{\mathbb{Q}}$ so $y_{0}=\alpha_{l_{0}}$ for some $l_{0}\geq 0$. Thus (0.1) for $l=l_{0}+1$ would give

[TABLE]

which is impossible.

In fact with the additional assumption that $\mathbb{K}$ is uncountable the theorem is true:

For a textile map $G:\mathbb{K}[[x]]^{m}\longrightarrow\mathbb{K}[[x]]^{q}$ two situations may occur: either $G=0$ has an exact solution, either it does not. So the theorem is equivalent to say if $G=0$ has no exact solution then there exists a $l_{0}$ such that $G$ has no approximate solution up to degree $l_{0}$. So the theorem is equivalent to say that if for every $l$ there exists $y(x)\in\mathbb{K}[[x]]^{m}$ such that $G(y(x))\in(x)^{l}$, then there exists an exact solution $y(x)\in\mathbb{K}[[x]]^{m}$: $G(y(x))=0$. This is what we are going to prove.

Let us fix a textile map $G$ which has approximate solutions up to every degree and let $x=(x_{1},\ldots,x_{n})$.

The map $G$ being textile, for any $y(x)=(y_{1}(x),\ldots,y_{m}(x))$, with $y_{i}(x)=\sum_{\alpha\in\mathbb{N}^{n}}y_{i,\alpha}x^{\alpha}$, we have that

[TABLE]

where the $G_{j,\beta}$ are polynomials in the $y_{i,\alpha}$. For any $N\in\mathbb{N}$ let $D_{N}\geq N$ be an integer such that the polynomials $G_{j,\beta}$, for $|\beta|<N$, depend only on the $y_{i,\alpha}$ for $|\alpha|<D_{N}$.

For every integer $k\geq l$ we consider the truncation maps:

[TABLE]

[TABLE]

where $\mathbb{K}[x]_{<k}$ denote the set of polynomials in $x$ of degree $<k$. We identify ${\mathbb{K}[x]_{<k}}^{m}$ to the affine space $\mathbb{A}_{\mathbb{K}}^{m\binom{n+k-1}{k-1}}$ by identifying a vector of polynomials with the vectors of the coefficients of these polynomials. With such an identification the maps $\pi_{k,l}$ correspond to projection maps.

For every positive integer $N$ the set

[TABLE]

is exactly the affine variety

[TABLE]

For every positive integers $N\geq k$ we define

[TABLE]

These sets are constructible subsets of $\mathbb{A}_{\mathbb{K}}^{m\binom{n+k-1}{k-1}}$ since $\mathbb{K}$ is algebraically closed (by Chevalley’s Theorem). Let us recall that a constructible set is a finite union of sets of the form $W\backslash V$ where $W$ and $V$ are Zariski closed subsets of $\mathbb{A}_{\mathbb{K}}^{m\binom{n+k-1}{k-1}}$.

Let us fix $k\geq 1$. Since $(x)^{N+1}\subset(x)^{N}$ for every positive integer $N$ and $\pi_{D_{N}}=\pi_{D_{N+1},D_{N}}\circ\pi_{D_{N+1}}$ we have that

[TABLE]

Thus the sequence $(C_{N}^{k})_{N}$ is a decreasing sequence of constructible subsets of $\mathbb{K}^{m\binom{n+k-1}{k-1}}$. Let $F_{N}^{k}$ denote the Zariski closure of $C_{N}^{k}$. Then the sequence $(F_{N}^{k})_{N}$ is a decreasing sequence of Zariski closed subsets of $\mathbb{A}_{\mathbb{K}}^{n\binom{n+k-1}{k-1}}$. By Noetherianity this sequence stabilizes, i.e. $F_{N}^{k}=F_{N_{0}}^{k}$ for every $N\geq N_{0}$ and some positive integer $N_{0}\geq k$. By assumption $C_{N_{0}}^{k}\neq\emptyset$ so $F_{N_{0}}^{k}\neq\emptyset$. Let $F$ be an irreducible component of $F_{N_{0}}^{k}$.

Since $C_{N}^{k}$ is constructible, $C_{N}^{k}=\cup_{i}W^{N}_{i}\backslash V^{N}_{i}$ for a finite number of Zariski closed sets $W^{N}_{i}$ and $V^{N}_{i}$ with $W^{N}_{i}\backslash V^{N}_{i}\neq\emptyset$. So for $N\geq N_{0}$ we have that

[TABLE]

But $F$ being irreducible, for every $N\geq N_{0}$ one of the $W_{i}^{N}$ has to be equal to $F$. Thus for every $N\geq N_{0}$ there exists a closed proper subset $V_{N}\subset F$ such that

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Since $\mathbb{K}$ is uncountable

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This is a well known fact (see for instance Exercice 5.10, [L] p. 76). This implies that

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By definition $y^{(k)}\in A_{k}$ if and only if for every $l\geq k$ there exists $y^{(l)}\in{\mathbb{K}[x]_{<l}}^{m}$ such that $\pi_{l,k}(y^{(l)})=y^{(k)}$. In particular we have that

[TABLE]

Thus for a given $y^{(k)}\in A_{k}$, there exists an element $y^{(k+1)}\in A_{k+1}$ such that $\pi_{k+1,k}(y^{(k+1)})=y^{(k)}$. By induction there exists a sequence of $y^{(l)}\in A_{l}$, for every $l\geq k$, such that

[TABLE]

At the limit we obtain $y\in\mathbb{K}[[x]]^{m}$ such that $G(y)\in(x)^{l}$ for every $l\geq k$, i.e. $G(y)=0$. ∎