Normalization: the general setup

We now discuss normalization in any number of variables, in the axiomatic setting suggested by this corollary. From now on, we set $X = (X_1, \dots, X_n)$ and $X’:= (X_1, \dots, X_{n-1})$.

One of the key ingredients to do this is that normalization of one series implies normalization of finitely many series:

Definition
Let $F_1, \dots, F_k \in \Ps{R}{X}$. We call $\{F_1, \dots, F_k\}$ normal if $F_i = X^{\alpha_i} U_i$ for each $i$, where the $U_i \in \Ps{R}{X}$ are units and the $\alpha_i \in \NN^n$ are such that the set of monomials $\set{X^{\alpha_1}, \dots, X^{\alpha_n}}$ is linearly ordered by divisibility.

Exercise 15
Let $F_1, \dots, F_k \in \Ps{R}{X}$.

1. Prove that $F_1 \cdots F_k$ is normal if and only if each $F_i$ is normal.
2. If each $F_i$ and each $F_i-F_j$ is normal, show that $\{F_1, \dots, F_k\}$ is normal. [Hint: use Lemma 4.7 in Bierstone and Milman’s paper.]

Let now $F \in \Ps{R}{X}$ and write $$F = \sum_{k=0}^\infty F_k\left(X’\right) X_n^k,$$ with each $F_k \in \Ps{R}{X’}$. The first problem not encountered in the two-variable case is that we cannot simply factor out a monomial in $X’$ to obtain a series that has finite order in $X_n$. Luckily,

a simple linear substitution (see below) can be used to reduce to finite order in $X_n$.

So assume now that $F$ has finite order in $X_n$. As in the two-variable case, if $0 \lt d:= \ord_{X_n}(F) \lt \infty$, there is a unit $U \in \Ps{R}{X}$ such that $$F = \sum_{k-0}^{d-1} F_k\left(X’\right) X^k + X^d U,$$ with $F_k(0) = 0$ for $k=0, \dots, d-1$. The next obstacle to normalization of $F$ as in the two-variable case is that the series $F_k$ may not be normal and, even if they are, there still may not be a single monomial in $X’$ dividing all of them. However, Exercise 15(2) shows how to deal with this problem:

We will proceed by induction on $n$, normalizing the set $\{F_0, \dots, F_{d-1}\}$ before continuing as in the two-variable case.

Then there is the (minor) issue of which variables to use for blow-up substitutions at which time. The previously described step will use blow-up substitutions involving pairs of variables of $X’$, while the other steps will involve pairing $X_n$ with one of the variables of $X’$.

Which of the $X’$ variables to pair $X_n$ with will be determined based on which of these variables is “closest to disappearing from one of the $F_i$s” after the blow-up substitution,

similarly to the way the $X$ variable was treated in the two-variable case.

Therefore, we define the following substitutions:

Linear substitutions
For $i\gt 1$ and $c = (c_{1}, \dots, c_{i-1}) \in \RR^{i-1}$, we let $l_{i,c}:\Ps{R}{X} \into \Ps{R}{X}$ be the ring homomorphism given by
$$
l_{i,c}(X_{j}):= \begin{cases}
X_{j} + c_{j} X_{i} &\text{if } 1 \leq j \lt i \\
X_{j} &\text{otherwise}.
\end{cases}
$$

Power substitutions
For $1 \leq i \leq n$ and an integer $q \gt 0$ we let $p^{+}_{i,q}, p^{-}_{i,q}:\Ps{R}{X} \into \Ps{R}{X}$ be the ring homomorphisms defined by
$$
p^{+}_{i,q}(X_{j}) := \begin{cases}
X_{i}^{q} &\text{if } j = i, \\
X_{j} &\text{otherwise,}
\end{cases}
$$
and
$$
p^{-}_{i,q}(X_{j}) := \begin{cases}
-X_{i}^{q} &\text{if } j = i, \\
X_{j} &\text{otherwise}.
\end{cases}
$$
The reason for considering these two power substitutions is that, as maps from $\RR^n$ into $\RR^n$, while each of them is a homeomorphism, only their restrictions to the half-space $\{x_i \gt 0\}$ are diffeomorphisms. While the former is enough to, say, count numbers of connected components, the latter is needed to keep track of various smoothness properties in the normalization process.

Translation substitutions
For $1 \lt i \leq n$ and $\alpha \in \Ps{R}{X_1, \dots, X_{i-1}}$ such that $\alpha(0) = 0$, we let $t_{\alpha}:\Ps{R}{X} \into \Ps{R}{X}$ be the ring homomorphism given by
$$
t_{\alpha}(X_{j}):= \begin{cases}
X_{i} + \alpha(X_{1}, \dots, X_{i-1}) &\text{if } j=i, \\
X_{j} &\text{otherwise}.
\end{cases}
$$

Blow-up substitutions
For $1 \leq i \lt j \leq n$ and $\lambda \in \RR$, we let $\bl_\lambda^{i,j}: \Ps{R}{X} \into \Ps{R}{X}$ be the ring homomorphism given by
$$
\bl_\lambda^{i,j}(X_k):= \begin{cases}
X_i (\lambda + X_j) &\text{if } k=j \\
X_k &\text{otherwise},
\end{cases}
$$
and we let $\bl_\infty^{i,j}:\Ps{R}{X} \into \Ps{R}{X}$ be the ring homomorphism defined by
$$
\bl_\infty^{i,j}(X_k):= \begin{cases}
X_i X_j &\text{if } k=i \\
X_k &\text{otherwise}.
\end{cases}
$$

We address our first point above with the following:

Lemma
Let $F \in \Ps{R}{X}$ be nonzero. Then there exists $c = (c_1, \dots, c_{n-1}) \in \RR^{n-1}$ such that $l_{n,c} F$ has finite order in $X_n$.

Proof. We write $F = \sum_{k \ge d} F^{(k)}$, where $F^{(k)}$ denotes the homogenous part of $F$ of degree $k$ and $d:= \ord(F)$. Then $$l_{n,c} F = p(c) X_n^d + X_n^{d+1} G,$$ where $p \in \RR[Y_1, \dots, Y_{n-1}]$ is not zero and only depends on $F$, but not on $c$, and where $G \in \Ps{R}{X}$. Therefore, any $c \in \RR^{n-1}$ that is not a zero of $p$ will do. $\qed$

Since for any finite number of nonzero polynomials in $\RR[Y_1, \dots, Y_{n-1}]$, there exists $c \in \RR^{n-1}$ at which none of these finitely many polynomials is zero, we obtain:

Corollary
Let $F_1, \dots, F_k \in \Ps{R}{X}$ be nonzero. Then there exists $c \in \RR^{n-1}$ such that each $l_{n,c} F_i$ has finite order in $X_n$. $\qed$