A first measure describing how far from normal a series $F \in \Ps{R}{X,Y}$ in two indeterminates $X$ and $Y$ is is given by $$\ord_Y(F):= \ord(F(0,Y)).$$

**Exercise**

Show that $\ord_Y(F) = 0$ if and only if $F$ is a unit, while, if $\ord_Y(F) = d>0$, there exists a unit $U \in \Ps{R}{X,Y}$ such that $$F = \sum_{k=0}^{d-1} \frac1{k!}\frac{\partial^k F}{\partial Y^k}(X,0)Y^k + U \cdot Y^d.$$

**Example 1**

We saw that the series $$F = X + Y + Y^2$$ is not normal; it satisfies $$\ord_Y(F) = 1.$$ However, not only is $\bl_{-1} F$ not normal, we actually have $\ord_Y(\bl_{-1}(F)) = \infty$! On the other hand, we have $$\bl_{-1}(F) = X \cdot G,$$ with $G = X + (1-2X)Y + XY^2$; in particular, $$\ord_Y(G) = 1,$$ which is better–but still of the same order in $Y$ as $F$.

As the example suggests, we will always factor out the highest order monomial possible before computing $\ord_Y$. This is justified by the following observation:

**Lemma 1**

*Let $F,G \in \Ps{R}{X,Y}$ be normal and $\lambda \in \RR \cup \{\infty\}$. Then $F \cdot G$ and $\bl_\lambda(F)$ are normal.*

**Example 1 (continued)**

Since $G$ is not normal, we can try to use another round of blow-up substitutions to further lower $\ord_Y$. However, we get $\bl_{-1}(G) = X \cdot H$ with $\ord_Y(H) = 1$ again!

So Example 1 suggests that we need to find another way to deal with $F$: note that $F(0,0) = 0$, while $(\partial F/\partial Y)(0,0) = 1$; so there exists $\alpha \in \Ps{R}{X}$ such that $\alpha(0) = 0$ and $$F(X,\alpha(X)) = 0.$$ Now consider the substitution $t_\alpha:\Ps{R}{X,Y} \into \Ps{R}{X,Y}$ defined by $$t_\alpha(X) := X$$ and $$t_\alpha(Y):= Y + \alpha(X):$$ we get $$t_\alpha(F) = X + \alpha(X) + Y + (\alpha(X) + Y)^2.$$ Since $F(X,\alpha(X)) = 0$, this simplifies to $$t_\alpha(G) = Y(1+ 2\alpha(X) +Y),$$ which is normal.

**Definition**

A **translation substitution** is a substitution $t_\alpha:\Ps{R}{X,Y} \into \Ps{R}{X,Y}$ as above, with $\alpha \in \Ps{R}{X}$ such that $\alpha(0)=0$.

Note that the translation substitution is not the magic bullet either; it just happened to finish the job in Example 1.

In general, we use it the following way: let $F \in \Ps{R}{X,Y}$ be of order $d:= \ord_Y(F)> 0$ in $Y$. By the Taylor expansion in $Y$ around $(X,0)$, we have $$F = \sum_{k=0}^{\infty} \frac1{k!}\frac{\partial^k F}{\partial Y^k}(X,0)Y^k.$$ Note that $\ord_Y(F) = d$ means, in particular, that $$\frac{\partial^{d-1}F}{\partial Y^{d-1}}(0,0) = 0, \quad\text{but } \frac{\partial^{d}F}{\partial Y^{d}}(0,0) \ne 0.$$ So there exists $\alpha \in \Ps{R}{X}$ such that $\alpha(0)=0$ and $$\tag{$\ast$} \frac{\partial^{d-1}F}{\partial Y^{d-1}}(X,\alpha(X)) = 0.$$ Then, by Taylor expansion in $Y$ around each point $(X,\alpha(X))$ we have, because of $(\ast)$, that $$\tag{$\ast\ast$} t_\alpha(F) = F(X,\alpha(X)+Y) = \sum_{k=0}^{d-2} \frac1{k!} \frac{\partial^k F}{\partial Y^k}(X,\alpha(X))Y^k + VY^d,$$ where $V \in \Ps{R}{X,Y}$ is again a unit, because $\left(\partial^d F/\partial Y^d\right)(0,\alpha(0)) = \left(\partial^d F/\partial Y^d\right)(0,0) \ne 0$.

This translation substitution, for the particular $\alpha$ chosen for $F$ above, is called a **Tschirnhausen transformation**. What are its effect on $F$? Examining $(\ast\ast)$, we see that:

**Lemma 2**

*We have $\ord_Y(t_\alpha F) \le \ord_Y(F)$ and, considered as a series in $Y$ with coefficients in $\Ps{R}{X}$, the coefficient of $t_\alpha F$ corresponding to $Y^{d-1}$ is $0$.* $\qed$

Using translation substitutions to study $\Pc{R}{X,Y}$-sets is OK:

**Exercise**

Let $r>0$ and $\alpha \in \Pc{R}{X}_r$. Show that the map $t_\alpha: [-r,r] \times \RR \into [-r,r] \times \RR$ defined by $t_\alpha(x,y):= (x,y+\alpha(x))$ is a homeomorphism.

**Definition**

Let $F \in \Ps{R}{X,Y}$. We call $F$ **Tschirnhausen prepared** if, written as a series in $\Ps{R}{X}[\![Y]\!]$, there is a $d \in \NN$ such that $$F = \sum_{k=0}^{d-2} F_k(X) Y^k + Y^d U,$$ where $U \in \Ps{R}{X,Y}$ is a unit and $F_k(0) = 0$ for $k=0, \dots, d-2$.

**Remark**

In the situation of the previous definition, we have $\ord_Y(F) = d$.

What happens to $\bl_\lambda F$ if $F$ is Tschirnhausen prepared?

**Example 2**

Consider $F = X^4 + X^3 Y + Y^3$; this $F$ is Tschirnhausen prepared with $\ord_Y(F) = 3$. Then $\bl_\infty F$ is normal and, for $0 \ne \lambda \in \RR$, $\bl_\lambda F$ is normal. However, $$\bl_0 F = X^3 G \quad\text{with } G:= \left(X + XY + Y^3\right).$$ But while $G$ is still Tschirnhausen prepared, its order in $Y$ is still 3. However, the powers of $X$ in the various terms have gone down: replacing $Y$ by $XY$ raises the power of $X$ in front of $Y^0$ by $0$, the power of $X$ in front of $Y^1$ by $1$ and the power of $X$ in front of $Y^3$ by $3$; factoring out $X^3$ therefore *lowers* the power of $X$ in front of $Y^0$ by $3 = 3-0$, the power of $X$ in front of $Y^1$ by $2 = 3-1$ and the power of $X$ in front of $Y^3$ by $0 = 3-3$.

This happens in general (with $d$ in place of $3$), as long as the power of $X$ in front of $Y^k$ is larger than $d-k$, for $k=0, \dots, d-2$.

How can we guarantee that the power of $X$ in front of $Y^k$ is smaller than $d-k$ if and only if it is $0$, for $k=0, \dots, d-2$? Before starting to use blow-up substitutions, we use one of the following:

**Definition**

A **power substitution** is a substitution $p_r:\Ps{R}{X,Y} \into \Ps{R}{X,Y}$ defined by $p_r(X):= X^r$ and $p_r(Y):= Y$, where $r \in \NN$.

**Exercise**

Show that we can use power substitutions to count the number of connected components of $\Pc{R}{X,Y}$-sets.

Thus, before starting to use blow-up substitutions, we perform the power substitution $p_{d!}$, where $d = \ord_Y(F)$ and $F$ is Tschirnhausen prepared. The resulting algorithm and measure of “distance from normality” are given in the next post.