We are now ready to give a measure of “distance from normality” for series in $\Ps{R}{X,Y}$.
Definition
We call $F \in \Ps{R}{X,Y}$ blow-up prepared if $F$ is Tschirnhausen prepared of order $d$ in $Y$ and, for $k=0, \dots, d-2$, the series $\left(\partial^k F/\partial Y^k\right)(X,0)$ is divisible by $X^{d-k}$.
Let now $F = \sum_{k=0}^\infty F_k(X) Y^k \in \Ps{R}{X,Y}$. Then each $F_k$ is of the form $X^{r_k} G_k$, where $G_k \in \Ps{R}{X}$ is a unit and $r_k \in \NN$. We define
- $\tp(F):= 0$ if $F$ is Tschirnhausen prepared, and $\tp(F):= 1$ otherwise;
- $\bp(F):= 0$ if $F$ is blow-up prepared, and $\bp(F):= 1$ otherwise;
- $\rd(F):= \infty$ if $F$ is not blow-up prepared, and if $F$ is blow-up prepared and $d:= \ord_Y(F) < \infty$, we set $$\rd(F):= \min\left(\set{\frac{r_k}{d-k}:\ k=0, \dots, d-2} \cup \{0\}\right).$$
Finally, we set $$\height_2(F):= \left(\ord_Y(F), \tp(F), \bp(F), \rd(F)\right),$$ considered as an element of $\NN^4$ with its lexicographic ordering.
Summarizing our discussion so far, we obtain:
Proposition
Let $F \in \Ps{R}{X,Y}$ be nonzero and not normal. Then $\height_2(F) \gt 0$ and one of the following statements holds:
- there exist $r \in \NN$ and $G \in \Ps{R}{X,Y}$ such that $\,\height_2(G) \lt \height_2(F)$ and $F = X^r G$;
- there exists $\alpha \in \Ps{R}{X}$ such that $\alpha(0) = 0$ and $\,\height_2(t_\alpha F) \lt \height_2(F)$;
- there exists $p \in \NN$ such that $\,\height_2(p_r F) \lt \height_2(F)$;
- $\bl_\infty F$ is normal and, for $\lambda \in \RR$, there exist $r \in \NN$ and $G \in \Ps{R}{X,Y}$ such that $\,\height_2(G) \lt \height_2(F)$ and $\,\bl_\lambda F = X^r G$.
Case 1: $d = \infty$. Then Statement 1 holds with $r:= \min\{r_k:\ k \in \NN\} > 0$.
Case 2: $d \lt \infty$, but $F$ is not Tschirnhausen prepared. Then Statement 2 holds by Lemma 2.
Case 3: $d \lt \infty$ and $F$ is Tschirnhausen prepared, but $F$ is not blow-up prepared. Set $p:= d!$; then $\height_2(p_r F) \le \height_2(F)$ and $p_rF$ is blow-up prepared, so Statement 3 holds.
Case 4: $d \lt \infty$ and $F$ is blow-up prepared. If $d=0$, we are done, so we assume $d\gt 0$. Since $F$ is not normal, we have $F_k \ne 0$ for some $k \in \{0, \dots, d-2\}$ (in particular, $d\gt 1$), and we set $$K:= \set{k \in \{0, \dots, d-2\}:\ F_k \ne 0}.$$ Then $$F = \sum_{k \in K} X^{r_k} U_k(X) Y^k + Y^d U,$$ where $U \in \Ps{R}{X,Y}$ is a unit. Hence $r_k \ge d-k$ for $k \in K$, so that $$\bl_\infty F = \sum_{k \in K} X^{r_k}Y^{r_k} U_k(XY) Y^k + Y^d \bl_\infty U$$ $$= Y^d \left(\bl_\infty U + \sum_{k \in K} X^{r_k} Y^{r_k+k-d} U_k(XY) \right).$$ Since $\bl_\infty U$ is a unit and $r_k\gt 0$ for $k \in K$, it follows that $\bl_\infty F$ is normal, as required.
On the other hand, for $\lambda \in \RR$, we get $$\bl_\lambda F = \sum_{k \in K} X^{r_k + k} U_k(X) (\lambda + Y)^k + X^d(\lambda +Y)^d \bl_\lambda U.$$ Since $r_k+k \ge d$ for $k \in K$, this means that $\bl_\lambda F = X^d G$ with $$G:= \sum_{k \in K} X^{r_k – (d-k)} U_k(X) (\lambda + Y)^k + (\lambda +Y)^d \bl_\lambda U.$$ Thus, if $\lambda = 0$ and $\rd(F) \gt 1$, then $\bl_0 F$ is still blow-up prepared with $\rd(\bl_0 F) = \rd(F)-1$; while, if $\rd(F)=1$, then the exponent $r_k-(d-k)$ is equal to $0$ for some $k \in K$, which means that $\ord_Y(G) \lt d$, as required.
So it remains to consider the case $\lambda \in \RR$ not zero: since $U = u_0 + Y V$, where $0 \ne u_0 \in \RR$ and $V \in \Ps{R}{X,Y}$, we have $$\bl_\lambda U = u_0 + X(\lambda+Y) \bl_\lambda V.$$ Therefore, writing $G = \sum_{k=0}^\infty G_k(X) Y^k$, we have for $k=d-1$ that $$G_{d-1}(X) = \begin{pmatrix} d \\ d-1 \end{pmatrix} \lambda u_0 + X H \quad\text{with}\quad H \in \Ps{R}{X},$$ that is, we have $\ord_Y(G) \le d-1$, as required. $\qed$
Remark
Let $\D_1 \subseteq \Ps{R}{X}$ and $\D_2 \subseteq \Ps{R}{X,Y}$ be subrings such that:
- $\RR[X] \subseteq \D_1$, $\RR[X,Y] \subseteq \D_2$ and $\D_1 \subseteq \D_2$;
- if $m,n \in \{1,2\}$, $F \in \D_m$ and $G_1, \dots, G_m \in \D_n$ are such that $G_i(0) = 0$ for each $i$, then $F(G_1, \dots, G_m) \in \D_n$;
- both $\D_1$ and $\D_2$ are closed under differentiation;
- if $n \in \{1,2\}$, $Z \in \{X,Y\}$ and $F \in \D_n$ are such that $F = Z G$ for some $G \in \Ps{R}{X,Y}$, then $G \in \D_n$;
- if $F \in \D_2$ is such that $F(0,0) = 0$ and $(\partial F/\partial Y)(0,0) \ne 0$, there exists $\alpha \in \D_1$ such that $\alpha(0) = 0$ and $F(X,\alpha(X)) = 0$.
Then the proposition holds with $\D_1$ in place of $\Ps{R}{X}$ and $\D_2$ in place of $\Ps{R}{X,Y}$.
Corollary
The proposition holds with $\Pc{R}{X}$ in place of $\Ps{R}{X}$ and $\Pc{R}{X,Y}$ in place of $\Ps{R}{X,Y}$. $\qed$