In this post, we introduce a type of substitutions to be used in our normalization algorithm: let $X$ and $Y$ be two single indeterminates.

**Definition**

For $\lambda \in \RR$, we let $\bl_\lambda:\Ps{R}{X,Y} \into \Ps{R}{X,Y}$ be the **blow-up substitution** defined by $$\bl_\lambda(X):= X \quad\text{and}\quad \bl_\lambda(Y):= X(\lambda+Y).$$ We also let $\bl_\infty:\Ps{R}{X,Y} \into \Ps{R}{X,Y}$ be the blow-up substitution defined by $$\bl_\infty(X):= XY \quad\text{and}\quad \bl_\infty(Y):= Y.$$ Note that each blow-up sustitution corresponds to composing on the right with some polynomial; therefore, the restriction of each $\bl_\lambda$ to $\Pc{R}{X,Y}$ maps convergent power series to convergent power series.

**Why can’t we just use the blow-up substitution $\bl_0$, as in Example 3?**

The answer is that, for general normal series, we only get finitely many connected components, by this corollary, *inside* a box $B(r,s)$ for some $r,s>0$.

This means that, if $F \in \Pc{R}{X,Y}$ is such that $\bl_0(F)$ is normal, there are $r,s>0$ such that the set $$\{\bl_0(F) = 0\} \cap B(r,s)$$ has finitely many connected components. Also denoting by $\bl_0:\RR^2 \into \RR^2$ the map $(x,y) \mapsto (x,xy)$, it follows that the set $$\{F = 0\} \cap \bl_0(B(r,s))$$ has finitely many connected components. However, the set $\bl_0(B(r,s))$ is not a neighbourhood of the origin!

**Exercise**

For $\lambda \in \RR \cup \{\infty\}$, let $r_\lambda, s_\lambda>0$.

- Describe the sets $\bl_\lambda(B(r_\lambda,s_\lambda))$, for $\lambda \in \RR \cup \{\infty\}$.
- Prove that there exist $k \in \NN$ and $\lambda_1, \dots, \lambda_k \in \RR \cup \{\infty\}$ such that the set $$\bl_{\lambda_1}\left(B\left(r_{\lambda_1},s_{\lambda_1}\right)\right) \cup \cdots \cup \bl_{\lambda_k}\left(B\left(r_{\lambda_k},s_{\lambda_k}\right)\right)$$ contains a neighbourhood of the origin.

**Lemma**

*Let $F \in \Pc{R}{X,Y}$ be such that $\,\bl_\lambda(F)$ is normal, for $\lambda \in \RR \cup \{\infty\}$. Then there exists a set $B \subseteq \RR^2$ such that $(0,0) \in \ir(B)$ and the set $\{F=0\} \cap B$ has finitely many connected components.*

**Proof.**

By this corollary, for each $\lambda$, there exist $r_\lambda,s_\lambda>0$ such that the set $\{\bl_\lambda(F) = 0\} \cap B\left(r_{\lambda}, s_{\lambda}\right)$ has finitely many connected components. By the exercise above, there exist $k \in \NN$ and $\lambda_1, \dots, \lambda_k \in \RR \cup \{\infty\}$ such that the set $$B:= \bl_{\lambda_1}\left(B\left(r_{\lambda_1},s_{\lambda_1}\right)\right) \cup \cdots \cup \bl_{\lambda_k}\left(B\left(r_{\lambda_k},s_{\lambda_k}\right)\right)$$ contains a neighbourhood of the origin. Since $$\{F = 0\} \cap B = \bigcup_{i=1}^k \bl_{\lambda_i} \left(\{\bl_{\lambda_i}(F)=0\} \cap B\left(r_{\lambda_k},s_{\lambda_k}\right)\right),$$ the lemma follows. $\qed$

**Remark**

The crucial assumption of the previous lemma is that $\bl_\lambda(F)$ is normal for *all* $\lambda$.

The hypothesis of the lemma is not usually satisfied: for $F(X,Y) = X+Y +Y^2$ and $\lambda = -1$, the series $\bl_{-1}F$ is not normal.

Therefore, we need to refine our strategy: instead of directly normalizing using blow-up substitutions, we introduce a measure of complexity associated to series in $\Ps{R}{X}$ that

- takes values in a well-ordered set;
- decreases every time a $\bl_\lambda$ is applied;
- when equal to 0, implies that the series is normal.

Such a measure does not actually exist (to my knowledge); instead, we need to introduce two additional types of substitutions to handle the cases when blow-up substitutions don’t work.