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

- Prove that $F_1 \cdots F_k$ is normal if and only if each $F_i$ is normal.
- 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 $X_n$.* $\qed$

Here is an attempt of the proof for exercise 15:

1. If each $F_i$ is normal, it is clear that $\prod_{i} F_i$ is normal.

If one of the $F_i$’s is not normal, we can show that the product will not be normal.

For $k=1$, it is trivial. Assume that the property holds for $n\leq k-1$. If $F_k$ is not normal but $\prod_{i} F_i$ is normal then there exists $\alpha \in \mathbb{N}^n$, a unit $U$ such that $\prod_{i=1}^{k} F_i=X^{\alpha}U$.

By the inductive hypothesis, $\prod_{i=1}^{k-1} F_i$ is normal so there exists $\alpha’ \in \mathbb{N}^n$, a unit $U’$ such that $\prod_{i=1}^{k-1} F_i=X^{\alpha’}U’$.

Since $X^{\alpha’}$ divides $\prod_{i=1}^{k} F_i=X^{\alpha}U$, we must have $\alpha’\leq\alpha$ i.e. $\alpha-\alpha’ \in \mathbb{N}^n$ and since $U’$ is a unit, we obtain $F_k= X^{\alpha-\alpha’}U U’^{-1}$ which contradicts the non-normality of $F_k$.

2. Assume that the property holds for $n\leq k-1$ (the case k=1 is trivial), then in particular, $F_k, F_{k-1}$ and $F_k – F_{k-1}$ are normal. Then there exist $\alpha, \alpha_k, \alpha_{k-1} \in \mathbb{N}^n$ and units $U, U_k, U_{k-1}$ such that $F_k-F_{k-1}=X^{\alpha}U$, i.e., $X^{\alpha_k}U_k-X^{\alpha_{k-1}}U_{k-1} =X^{\alpha}U$.

Using the indicated lemma, either $\alpha_{k-1}\leq \alpha_k$ and we are done, or $\alpha_{k}\leq \alpha_{k-1}$ and we can apply the inductive hypothesis to $\{F_1, \dots, F_{k-2}, F_k \}$.

Good!