$\newcommand{\kk}{\mathbb{K}} \newcommand{\rr}{\mathbb{R}} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\supp}{supp}$
Let $f$ be a power series in $n$ variables $x_1, \ldots, x_n$ over a field $\kk$. If $f$ is regular in $x_n$, i.e. $f(0, \ldots, 0, x_n) \not\equiv 0$, then there is a unique Weierstrass polynomial (with respect to $x_n$), say $w$, and an invertible power series $u$ associated to $f$ such that $f = uw$. In this post we will prove a simple observation relating $\supp(f)$ with $\supp(u)$ and $\supp(w)$.
For convenience we will use $y$ in place of $x_n$ and write $x := (x_1, \ldots, x_{n-1})$. Let $d := \ord_y(f(0, \ldots, 0, y))$. Then $w = y^d + \sum_{i=1}^d w_i(x)y^{d-i}$, such that none of the $w_i$ is invertible. Let $C$ be a cone in $\rr^n$ with vertex at $(0, \ldots, 0, d)$ such that $\supp(f) \subseteq C + (0,\ldots, 0, d)$.
Theorem. $\supp(u) \subseteq C$ and $\supp (w) \subseteq C + (0, \ldots, 0, d)$.
Proof. It follows from the definition of $d$ that $f = u_0y^d + \sum_{i=1}^d w’_{0,i}(x)y^{d – i}$ for an invertible $u_0 \in \kk[[x_1, \ldots, x_{n-1}, y]]$ and non-invertible $w’_{0,i} \in \kk[[x_1, \ldots, x_{n-1}]]$. Since $\supp(u_0) \subseteq C$ and $C$ is a cone, it follows that $\supp(u_0^{-1}) \subseteq C$ as well. Moreover, $\supp(w’_{0,i}y^{-i}) \subseteq C$, so that $\supp(u_0^{-1}w’_{0,i}y^{-i}) \subseteq C$, and consequently, $\supp(u_0^{-1}w’_{0,i}y^{d-i}) \subseteq C+(0, \ldots, 0, d)$. Write $w_{0,i} := u_0^{-1}w’_{0,i}$ and $f_0 := y^d + \sum_{i=1}^d w_{0,i}y^{d-i}$, so that $f = u_0f_0$. Note that
- $\supp(u_0) \subseteq C$, $\supp(f_0) \subseteq C + (0,\ldots, 0, d)$
Write $w_{0,i} = \sum_{j \geq 0} w_{0,i,j}(x)y^j$. Now, gather all monomials in $f_0$ divisible by $y^d$ to form $f_0 = u_1y^d + \sum_i w’_{1,i}(x)y^{d-i}$. Note that
$
\begin{align*}
u_1 &= (1 + \sum_{i=1}^d w_{0,i,i}(x)) + y\sum_{i=1}^d w_{0,i,i+1}(x) + y^2\sum_{i=1}^d w_{0,i,i+2}(x) + \cdots \\
w’_{1,1} &= w_{0,1,0} + w_{0,2,1} + \cdots + w_{0,d,d-1} \\
w’_{1,2} &= w_{0,2,0} + w_{0,3,1} + \cdots + w_{0,d,d-2} \\
\vdots \\
w’_{1,d} &= w_{0,d,0}
\end{align*}
$
Now $f_0 = u_1f_1$, where $f_1 := u_1^{1}f_0 = y^d + \sum_{i=1}^d w_{1,i}(x)y^{d-i}$, where $w_{1,i} := u_1^{-1} w’_{1,i}$. Note that
- $\supp(u_1) \subseteq C$, $\supp(f_1) \subseteq C + (0,\ldots, 0, d)$
Write $u_{1,j}(x) := \sum_{i=1}^d w_{0,i,i+j}(x)$, so that $u_1 = 1 + \sum_j u_{1,j}(x)y^j$. Note that
- $\ord(u_{1,j}) \geq 1$ for each $j$.
It follows that
$
\begin{align*}
w_{1,i}
&= w’_{1,i}(x)(1 – \sum_j u_{1,j}(x)y^j + ( \sum_j u_{1,j}(x)y^j)^2 + \cdots )\\
&= \sum_{j\geq 0} w_{1,i,j}(x)y^j
\end{align*}
$
where
- $\ord(w_{1,i,0}) \geq 1$,
- $\ord(w_{1,i,j}) \geq 2$ for all $j \geq 1$.
Now continue as above to form $u_2, f_2$ such that $f_1 = u_2f_2$ and in addition,
- $\supp(u_2) \subseteq C$, $\supp(f_2) \subseteq C + (0,\ldots, 0, d)$
- $u_2 = 1 + \sum_j u_{2,j}(x)y^j$, where
- each monomial in the support of $u_{2,j}$ is also in the support of $w_{1,i,i+j}$ for some $i$, and consequently,
- $\ord(u_{2,j}) \geq 2$ for each $j$;
- $f_2 = y^d + \sum_{i=1}^d w_{2,i}(x)y^{d-i}$, where
- $w_{2,i} = (u_2)^{-1}$ times a power series consisting of a subset of monomial terms of $w_{1,1}, \ldots, w_{1,d}$,
- $w_{2,i} = \sum_{j \geq 0} w_{2,i,j}(x)y^j$, where
- $\ord(w_{2,i,0}) \geq 1$,
- $\ord(w_{2,i,j}) \geq 3$ for all $j \geq 1$.
It is then straightforward to check that this process can be continued inductively, and $u := \prod_{i=0}^\infty u_i$ and $w := \lim_{i\to \infty} f_i$ are well defined, and in addition,
- $\supp(u) \subseteq C$, $\supp(w) \subseteq C + (0,\ldots, 0, d)$
- $u$ is invertible,
- $w$ is a Weierstrass polynomial in $y$ of degree $d$.
This completes the proof.