Assumptions 3.1. Throughout this section, we assume that for any $k_1,\ldots , k_m\in \{1,\ldots , k_0\}$ , the germ of the variety $V(f^{k_1},\ldots ,f^{k_m})$ at $\mathbf {0}$ is the germ of a nondegenerate complete intersection variety.

Proof of Theorem 3.5

We argue by contradiction. By [Reference Milnor6, Cor. 2.8], for $\delta>0$ small enough, the fibers $f^{-1}(\eta )\cap \mathring {B}_\varepsilon $ are nonsingular for any $\eta $ , $0<|\eta |\leq \delta $ . It follows that if the assertion in Theorem 3.5 is not true, then, by the Curve Selection Lemma (see [Reference Hamm3], [Reference Milnor6]), there exist a real analytic curve $\mathbf {z}(s)=(z_1(s),\ldots ,z_n(s))$ in $\mathbb {C}^n$ , $0\leq s\leq 1$ , and a family of complex numbers $\lambda (s)$ , $0<s\leq 1$ , satisfying the following three conditions:

1. (i) $\frac {\partial f}{\partial z_i}(\mathbf {z}(s)) = \lambda (s) \bar z_i(s)$ for $1\leq i\leq n$ and $s\not =0$ .

2. (ii) $f(\mathbf {z}(0))=0$ , but $f(\mathbf {z}(s))$ is not constantly zero.

3. (iii) There exists $\varepsilon '>0$ such that $\varepsilon '\leq \Vert \mathbf {z}(s)\Vert \leq \varepsilon $ .

Note that, by (i) and (ii), $\lambda (s)\not \equiv 0$ and we can express it in a Laurent series

$$ \begin{align*} \lambda(s)=\lambda_0 s^c+\cdots, \end{align*} $$

where $\lambda _0\in \mathbb {C}^*$ . Throughout, the dots “ $\cdots $ ” stand for the higher-order terms. Let $I:=\{i\, ;\, z_i(s)\not \equiv 0\}$ . By (ii), $I\in \mathcal {I}(f)$ , and hence $I\in \mathcal {I}(f^{1})\cap \cdots \cap \mathcal {I}(f^{k_0})$ . For each $i\in I$ , consider the Taylor expansion

$$ \begin{align*} z_i(s)=a_i s^{w_i}+\cdots, \end{align*} $$

where $a_i\in \mathbb {C}^*$ and $w_i\in \mathbb {N}$ .

Hereafter, to simplify the notation, we shall assume that $I=\{1,\ldots ,n\}$ , so that the function $f^{k,I}$ is simply written as $f^k$ , the intersection $I\cap I(\mathbf {w})$ is written as $I(\mathbf {w})$ (where, as in Lemma 3.3, $I(\mathbf {w})$ is the set of all indexes $i\in \{1,\ldots ,n\}$ for which $w_i=0$ ), and so on. The argument for a general I is completely similar.

Before proving Claim 3.6, let us first complete the proof of Theorem 3.5. For that purpose, we look at the set consisting of all integers k for which $f^{k}_{\mathbf {w}}(\mathbf {a})=0$ , which is not empty by Claim 3.6. For simplicity again, we shall assume

$$ \begin{align*} \begin{aligned} & f^{k}_{\mathbf{w}}(\mathbf{a})=0 &&\mbox{ for } && 1\leq k\leq k_0'\leq k_0;\\ & f^{k}_{\mathbf{w}}(\mathbf{a})\not=0 &&\mbox{ for } && k^{\prime}_0+1\leq k\leq k_0. \end{aligned} \end{align*} $$

Write $f=f^1\cdots f^{k_0'}\cdot h$ , where $h:=f^{k_0'+1}\cdots f^{k_0}$ if $k_0'\leq k_0-1$ and $h:=1$ if $k_0'=k_0$ . Then, for all $1\leq i\leq n$ , we have

(3.1) $$ \begin{align} \frac{\partial f}{\partial z_i}(\mathbf{z}(s)) = \sum_{k=1}^{k_0'} \bigg( \frac{\partial f^{k}}{\partial z_i}(\mathbf{z}(s))\cdot h(\mathbf{z}(s))\cdot\prod_{\genfrac{}{}{0pt}{}{1\leq \ell\leq k_0'}{\ell\not=k}} f^{\ell}(\mathbf{z}(s))\bigg) +\frac{\partial h}{\partial z_i}(\mathbf{z}(s))\cdot\prod_{1\leq k\leq k_0'} f^{k}(\mathbf{z}(s)). \end{align} $$

For each $1\leq k\leq k_0'$ , if $o_k\equiv \mbox {ord}\, f^{k}(\mathbf {z}(s))$ denotes the order (in s) of $f^{k}(\mathbf {z}(s))$ and if $e_k:=d(\mathbf {w};f^{k})-o_k+\sum _{\ell =1}^{k_0'} o_{\ell }$ , then

(3.2) $$ \begin{align} \mbox{ord}\bigg( \frac{\partial f^{k}}{\partial z_i}(\mathbf{z}(s))\cdot h(\mathbf{z}(s))\cdot\prod_{\genfrac{}{}{0pt}{}{1\leq \ell\leq k_0'}{\ell\not=k}} f^{\ell}(\mathbf{z}(s))\bigg)\geq d(\mathbf{w};h)-w_i+e_k, \end{align} $$

and the equality holds if and only if $\frac {\partial f^{k}_{\mathbf {w}}}{\partial z_i}(\mathbf {a})\not =0$ . Since $o_k>d(\mathbf {w};f^{k})$ for $1\leq k\leq k^{\prime }_0$ , we also have

(3.3) $$ \begin{align} \mbox{ord}\bigg(\frac{\partial h}{\partial z_i}(\mathbf{z}(s))\cdot\prod_{\ell=1}^{k_0'} f^{\ell}(\mathbf{z}(s))\bigg) \geq d(\mathbf{w};h)-w_i+\sum_{\ell=1}^{k_0'} o_{\ell}> d(\mathbf{w};h)-w_i+e_k \end{align} $$

for all $1\leq k\leq k_0'$ . Still for simplicity, let us assume that

$$ \begin{align*} e_{\mbox{min}}:=e_1=\cdots=e_{k_0"}<e_{k_0"+1}\leq \cdots\leq e_{k_0'}. \end{align*} $$

The relations (3.1)–(3.3) show that there exist $\mu _1,\ldots ,\mu _{k^{\prime \prime }_0}\in \mathbb {C}^*$ such that for any $1\leq i\leq n$ ,

$$ \begin{align*} \frac{\partial f}{\partial z_{i}}(\mathbf{z}(s))=\sum_{k=1}^{k^{\prime\prime}_0}\frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_{i}} (\mathbf{a})\cdot \mu_k\cdot s^{d(\mathbf{w};h)-w_{i}+e_{\text{min}}}+\cdots, \end{align*} $$

and hence, by multiplying both sides of the relation (i) by $s^{w_i}$ ,

(3.4) $$ \begin{align} \sum_{k=1}^{k^{\prime\prime}_0}\frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_{i}} (\mathbf{a})\cdot \mu_k\cdot s^{d(\mathbf{w};h)+e_{\text{min}}}+\cdots = \lambda_0 \bar a_{i}s^{c+2w_{i}}+\cdots. \end{align} $$

Note that the coefficient $\lambda _0 \bar a_i$ of $s^{c+2w_i}$ on the right-hand side of (3.4) being nonzero, we must have $d(\mathbf {w};h)+e_{\text {min}}\leq c+2w_i$ for any $1\leq i\leq n$ , and since $I(\mathbf {w})\not =\emptyset $ (by (iii)), in fact, we have $d(\mathbf {w};h)+e_{\text {min}}\leq c$ . It follows that for any $i\notin I(\mathbf {w})$ , the sum

$$ \begin{align*} S_i:=\sum_{k=1}^{k^{\prime\prime}_0} \mu_k \frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_{i}} (\mathbf{a}) \end{align*} $$

vanishes. (Indeed, if there exists $i_0\notin I(\mathbf {w})$ such that $S_{i_0}\not =0$ , then $ c+2w_{i_0}=d(\mathbf {w};h)+e_{\text {min}}\leq c, $ which is a contradiction.) Now, if we also have $S_i=0$ for all $i\in I(\mathbf {w})$ , then the condition (2) of Lemma 3.3 is satisfied. (Note that the complex number denoted by $\lambda $ in Lemma 3.3 may vanish.) However, the relation (iii) implies

(3.5) $$ \begin{align} \mathbf{a}\in\bigg\{\mathbf{z}\in \mathbb{C}^{*n};\sum_{i\in I(\mathbf{w})}|z_i|^2\leq \varepsilon^2\bigg\}, \end{align} $$

which contradicts the conclusion of this lemma. If there exists $i_0\in I(\mathbf {w})$ such that $S_{i_0}\not =0$ , then it follows that $S_i\not =0$ for any $i\in I(\mathbf {w})$ , so that for all such i’s,

$$ \begin{align*} S_i\equiv\sum_{k=1}^{k^{\prime\prime}_0} \mu_k \frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_i} (\mathbf{a}) = \lambda_0 \bar a_i. \end{align*} $$

Thus, the condition (2) of Lemma 3.3 is satisfied in this case too, and again the relation (iii) (which implies (3.5)) leads to a contradiction with the conclusion of the lemma. So, up to Claim 3.6, the theorem is proved.

To complete the proof of the theorem, it remains to prove Claim 3.6.

Proof of Claim 3.6

Again, to simplify, we assume $I=\{1,\ldots ,n\}$ , so that $f^{k,I}=f^k$ , $I\cap I(\mathbf {w})=I(\mathbf {w})$ , and so on. We argue by contradiction. Suppose $f^{k}_{\mathbf {w}}(\mathbf {a})\not =0$ for all $1\leq k\leq k_0$ . Then $d(\mathbf {w};f^{k})=o_k$ for all $1\leq k\leq k_0$ , where $o_k$ is the order of $f^{k}(\mathbf {z}(s))$ . Furthermore, note that, by (ii), there exists $1\leq k_1\leq k_0$ such that $f^{k_1}(\mathbf {z}(0))=0$ . If $I(\mathbf {w})=\{1,\ldots ,n\}$ , then $d(\mathbf {w};f^{k_1})=0$ and

$$ \begin{align*} f^{k_1}(\mathbf{z}(s))=f^{k_1}_{\mathbf{w}}(\mathbf{a})\, s^0+\cdots, \end{align*} $$

and therefore $0=f^{k_1}(\mathbf {z}(0))=f^{k_1}_{\mathbf {w}}(\mathbf {a})$ , which is a contradiction. So, from now on, suppose that $I(\mathbf {w})$ is a proper subset of $\{1,\ldots ,n\}$ and $d(\mathbf {w};f^{k_1})\not =0$ . Put $e:=\sum _{k=1}^{k_0} o_{k}$ . Then, as above, there exist nonzero complex numbers $\mu _1,\ldots ,\mu _{k_0}$ (actually, here, for each k, $\mu _k=\prod _{\ell \not = k}f^{\ell }_{\mathbf {w}}(\mathbf {a})$ ) such that for any $1\leq i\leq n$ ,

$$ \begin{align*} \frac{\partial f}{\partial z_{i}}(\mathbf{z}(s))=\sum_{k=1}^{k_0}\frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_{i}} (\mathbf{a})\cdot \mu_k\cdot s^{-w_{i}+e}+\cdots, \end{align*} $$

and hence, by multiplying both sides of the relation (i) by $s^{w_i}$ ,

$$ \begin{align*} \sum_{k=1}^{k_0}\frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_{i}} (\mathbf{a})\cdot \mu_k\cdot s^{e}+\cdots = \lambda_0 \bar a_{i}s^{c+2w_{i}}+\cdots. \end{align*} $$

Again, since $\lambda _0 \bar a_i\not =0$ and $I(\mathbf {w})\not =\emptyset $ , we have $e\leq c$ and the sum $\sum _{k=1}^{k_0}\mu _k\frac {\partial {f^{k}_{\mathbf {w}}}}{\partial z_{i}} (\mathbf {a})$ vanishes for all $i\in I(\mathbf {w})^c:=\{1,\ldots ,n\}\setminus I(\mathbf {w})$ . As $f^{k}_{\mathbf {w}}$ is weighted homogeneous, this, together with the Euler identity, implies that

$$ \begin{align*} 0 & = \sum_{i\in I(\mathbf{w})^c}a_i w_i\bigg(\underbrace{\sum_{k=1}^{k_0}\frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_{i}} (\mathbf{a})\cdot \mu_k}_{=0}\bigg) = \sum_{k=1}^{k_0}\bigg(\prod_{\genfrac{}{}{0pt}{}{1\leq \ell\leq k_0}{\ell\not=k}}f^{\ell}_{\mathbf{w}}(\mathbf{a})\cdot \sum_{i\in I(\mathbf{w})^c}a_i w_i \frac{\partial {f^{k}_{\mathbf{w}}}}{\partial z_{i}} (\mathbf{a})\bigg)\\ & = \sum_{k=1}^{k_0}\bigg(\prod_{\genfrac{}{}{0pt}{}{1\leq \ell\leq k_0}{\ell\not=k}}f^{\ell}_{\mathbf{w}}(\mathbf{a})\bigg)\cdot d(\mathbf{w};f^{k})\cdot f^{k}_{\mathbf{w}}(\mathbf{a}) = \bigg(\prod_{\ell=1}^{k_0}f^{\ell}_{\mathbf{w}}(\mathbf{a})\bigg)\cdot\sum_{k=1}^{k_0} d(\mathbf{w};f^{k})\not=0, \end{align*} $$

which is a contradiction too.

This completes the proof of Theorem 3.5 (up to Lemma 3.3).

Proof of Lemma 3.3

First, observe that if the assertion fails for some $k_1,\dots ,k_m$ , I, and $\Delta (\mathbf {w};f^{k_1}),\ldots ,\Delta (\mathbf {w};f^{k_m})$ such that $I\cap I(\mathbf {w})=\emptyset $ , then for any $\varepsilon>0$ , the set

$$ \begin{align*} \bigg\{\mathbf{z}\in \mathbb{C}^{*I}\, ;\, \sum_{i\in I\cap I(\mathbf{w})=\emptyset}|z_i|^2\leq \varepsilon^2\bigg\} \end{align*} $$

is nothing but $\mathbb {C}^{*I}$ and there exists a point $\mathbf {a}$ in it that satisfies the conditions (1) and (2) of the lemma; in particular, $\mathbf {a}\in V^{*I}(f^{k_1}_{\mathbf {w}},\ldots ,f^{k_m}_{\mathbf {w}})$ and the vectors $\mathbf {z}^{k_{1}}(\mathbf {a}),\ldots ,\mathbf {z}^{k_{m}}(\mathbf {a})\in \mathbb {C}^I$ whose ith coordinates ( $i\in I$ ) are

$$ \begin{align*} \frac{\partial f_{\mathbf{w}}^{k_{1},I}}{\partial z_{i}}(\mathbf{a}),\ldots,\frac{\partial f_{\mathbf{w}}^{k_{m},I}}{\partial z_{i}}(\mathbf{a}), \end{align*} $$

respectively, are linearly dependent, that is,

$$ \begin{align*} df^{k_1,I}_{\mathbf{w}}(\mathbf{a})\wedge\cdots\wedge df^{k_m,I}_{\mathbf{w}}(\mathbf{a})=0. \end{align*} $$

However, since $I\in \mathcal {I}(f^{k_1})\cap \cdots \cap \mathcal {I}(f^{k_m})$ and $I\cap I(\mathbf {w})=\emptyset $ , this contradicts Assumptions 3.1, which imply that

$$ \begin{align*} df^{k_1,I}_{\mathbf{w}}(\mathbf{p})\wedge\cdots\wedge df^{k_m,I}_{\mathbf{w}}(\mathbf{p})\not=0 \end{align*} $$

for any $\mathbf {p}\in V^{*I}(f^{k_1}_{\mathbf {w}},\ldots ,f^{k_m}_{\mathbf {w}})$ (see Remark 2.3).

Now, assume that the assertion in Lemma 3.3 fails for some $k_1,\dots ,k_m$ , I and $\Delta (\mathbf {w};f^{k_1}),\ldots , \Delta (\mathbf {w};f^{k_m})$ such that $I\cap I(\mathbf {w})\not =\emptyset $ . Again, without loss of generality, and in order to simplify the notation, we assume that $I=\{1,\ldots ,n\}$ , so that $f^{k,I}_{\mathbf {w}}=f^{k}_{\mathbf {w}}$ , $I\cap I(\mathbf {w})=I(\mathbf {w})$ , $\mathbb {C}^{*I}=\mathbb {C}^{*n}$ , and so on. Then there is a sequence $\{\mathbf {p}_q\}_{q\in \mathbb {N}}$ of points in $\mathbb {C}^{*n}$ and a sequence $\{\lambda _q\}_{q\in \mathbb {N}}$ of complex numbers such that:

1. (1) $f_{\mathbf {w}}^{k_1}(\mathbf {p}_q)=\cdots =f_{\mathbf {w}}^{k_m}(\mathbf {p}_q)=0$ for all $q\in \mathbb {N}$ .

2. (2) There exists a sequence $\{(\mu _{k_1,q},\ldots ,\mu _{k_m,q})\}_{q\in \mathbb {N}}$ of points in $\mathbb {C}^m\setminus \{\mathbf {0}\}$ such that for all $q\in \mathbb {N}$ and all $1\leq i\leq n$ ,

   $$ \begin{align*} \sum_{j=1}^m \mu_{k_j,q}\frac{\partial f_{\mathbf{w}}^{k_j}}{\partial z_i}(\mathbf{p}_q)=\left\{ \begin{aligned} & \lambda_q \, \bar p_{q,i}, &&\mbox{if} && i\in I(\mathbf{w}),\\ & 0, &&\mbox{if} && i\notin I(\mathbf{w}), \end{aligned} \right. \end{align*} $$

   where, for each $1\leq i\leq n$ , $\bar p_{q,i}$ denotes the conjugate of the ith coordinate $p_{q,i}$ of $\mathbf {p}_q$ .

3. (3) $\sum _{i\in I(\mathbf {w})} |p_{q,i}|^2\to 0$ as $q\to \infty $ .

For any $\zeta \in \mathbb {C}$ and any $\mathbf {z}\in \mathbb {C}^n$ , let $\zeta *\mathbf {z}=((\zeta *\mathbf {z})_1,\ldots ,(\zeta *\mathbf {z})_n)$ be the point of $\mathbb {C}^n$ defined by

$$ \begin{align*} (\zeta*\mathbf{z})_i:= \zeta^{w_i}z_i= \left\{ \begin{aligned} &z_i, &&\mbox{for} && i\in I(\mathbf{w}),\\ & \zeta^{w_i}z_i, &&\mbox{for} && i\notin I(\mathbf{w}).\\ \end{aligned} \right. \end{align*} $$

Then pick a sequence $\{\zeta _q\}_{q\in \mathbb {N}}$ of points in $\mathbb {C}^{*}$ that converges to zero sufficiently fast so that the sequence $\{\zeta _q*\mathbf {p}_q\}_{q\in \mathbb {N}}$ converges to the origin of $\mathbb {C}^{n}$ . Clearly, $\{\zeta _q*\mathbf {p}_q\}_{q\in \mathbb {N}}$ also satisfies the above properties (1)–(3). Indeed, for any $1\leq j\leq m$ , we have

$$ \begin{align*} f_{\mathbf{w}}^{k_j}(\zeta_q*\mathbf{p}_q)=\zeta_q^{d(\mathbf{w};f^{k_j})}f_{\mathbf{w}}^{k_j}(\mathbf{p}_q)=0, \end{align*} $$

so $\{\zeta _q*\mathbf {p}_q\}_{q\in \mathbb {N}}$ satisfies (1). For each $1\leq i\leq n$ , we also have

$$ \begin{align*} \frac{\partial f_{\mathbf{w}}^{k_j}}{\partial z_i}(\zeta_q*\mathbf{p}_q)= \zeta_q^{d(\mathbf{w};f^{k_j})-w_i} \, \frac{\partial f_{\mathbf{w}}^{k_j}}{\partial z_i}(\mathbf{p}_q), \end{align*} $$

and since $\zeta _q^{w_i}=1$ for all $i\in I(\mathbf {w})$ and $\zeta _q^{w_i}$ (which is nonzero) is independent of the index j ( $1\leq j\leq m$ ) for all $i\notin I(\mathbf {w})$ , it follows that

$$ \begin{align*} \sum_{j=1}^m \frac{\mu_{k_j,q}}{\zeta_q^{d(\mathbf{w};f^{k_j})}} \frac{\partial f_{\mathbf{w}}^{k_j}}{\partial z_i}(\zeta_q*\mathbf{p}_q)=\left\{ \begin{aligned} & \lambda_q \, \bar p_{q,i}, &&\mbox{for} && i\in I(\mathbf{w}),\\ & 0, &&\mbox{for} && i\notin I(\mathbf{w}), \end{aligned} \right. \end{align*} $$

so that the sequence $\{\zeta _q*\mathbf {p}_q\}_{q\in \mathbb {N}}$ satisfies (2) with the complex numbers $\mu _{k_j,q} / \zeta _q^{d(\mathbf {w};f^{k_j})}$ ( $1\leq j\leq m$ ). Finally,

$$ \begin{align*} \sum_{i\in I(\mathbf{w})} |(\zeta_q*\mathbf{p}_q)_i|^2 = \sum_{i\in I(\mathbf{w})} |p_{q,i}|^2\to 0, \end{align*} $$

as $q\to \infty $ , so $\{\zeta _q*\mathbf {p}_q\}_{q\in \mathbb {N}}$ also satisfies (3). Altogether, $\{\zeta _q*\mathbf {p}_q\}_{q\in \mathbb {N}}$ satisfies the properties (1)–(3). Therefore, we can apply the Curve Selection Lemma to this situation in order to find a real analytic curve $\mathbf {a}(s)=(a_1(s),\ldots ,a_n(s))$ in $\mathbb {C}^n$ , $0\leq s\leq 1$ , and a family of complex numbers $\lambda (s)$ , $0<s\leq 1$ , such that:

1. (1^′) $f_{\mathbf {w}}^{k_1}(\mathbf {a}(s))=\cdots =f_{\mathbf {w}}^{k_m}(\mathbf {a}(s))=0$ for all $s\not =0$ .

2. (2^′) There exists a real analytic curve $(\mu _{k_1}(s),\ldots ,\mu _{k_m}(s))$ in $\mathbb {C}^m\setminus \{\mathbf {0}\}$ , $0< s\leq 1$ , such that for all $s\not =0$ and all $1\leq i\leq n$ ,

   $$ \begin{align*} \sum_{j=1}^m \mu_{k_j}(s)\, \frac{\partial f_{\mathbf{w}}^{k_j}}{\partial z_i}(\mathbf{a}(s))=\left\{ \begin{aligned} & \lambda(s)\, \bar a_{i}(s), &&\mbox{if} &&i\in I(\mathbf{w}),\\ & 0, &&\mbox{if} &&i\notin I(\mathbf{w}). \end{aligned} \right. \end{align*} $$

3. (3^′) $\mathbf {a}(0)=\mathbf {0}$ and $\mathbf {a}(s)\in \mathbb {C}^{*n}$ for $s\not =0$ .

For each $1\leq i\leq n$ , consider the Taylor expansion

$$ \begin{align*} a_i(s)=b_i s^{v_i}+\cdots, \end{align*} $$

where $b_i\in \mathbb {C}^*$ and $v_i\in \mathbb {N}^*$ . Since the $v_i$ ’s are all positive, for each $1\leq j\leq m$ , the face $\Delta \big (\mathbf {v};f_{\mathbf {w}}^{k_j}\big )$ is a compact face of $\Delta (\mathbf {w};f^{k_j})$ , and hence $\Delta \big (\mathbf {v};f_{\mathbf {w}}^{k_j}\big )$ is a face of $\Gamma (f^{k_j})$ , where $\mathbf {v}$ is the point of $\mathbb {N}^{*n}$ whose ith coordinate is $v_i$ . Furthermore, note that for each j, we have $d\big (\mathbf {v};f_{\mathbf {w}}^{k_j}\big )>0$ , and since

$$ \begin{align*} 0=f_{\mathbf{w}}^{k_j}(\mathbf{a}(s))=\big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}(\mathbf{b})\, s^{d\big(\mathbf{v};f_{\mathbf{w}}^{k_j}\big)}+\cdots \end{align*} $$

for all $s\not =0$ , we also have $\big (f_{\mathbf {w}}^{k_j}\big )_{\mathbf {v}}(\mathbf {b})=0$ , where $\mathbf {b}$ is the point of $\mathbb {C}^{*n}$ whose ith coordinate is $b_i$ . (As usual, $\big (f_{\mathbf {w}}^{k_j}\big )_{\mathbf {v}}$ is the face function of $f_{\mathbf {w}}^{k_j}$ with respect to $\mathbf {v}$ .)

Write $\mu _{k_j}(s)=\mu _{k_j} s^{g_j}+\cdots $ , where $\mu _{k_j}\not =0$ . If $\mu _{k_j}(s)\equiv 0$ , then $g_j=\infty $ . Let

$$ \begin{align*} \delta:=\mbox{min}\{d\big(\mathbf{v};f_{\mathbf{w}}^{k_1}\big)+g_1,\ldots,d\big(\mathbf{v};f_{\mathbf{w}}^{k_m}\big)+g_m\}, \end{align*} $$

and put

$$ \begin{align*} \tilde\mu_{k_j} = \left\{ \begin{aligned} & \mu_{k_j}, &&\mbox{if} && d\big(\mathbf{v};f_{\mathbf{w}}^{k_j}\big)+g_j=\delta,\\ & 0, &&\mbox{if} && d\big(\mathbf{v};f_{\mathbf{w}}^{k_j}\big)+g_j>\delta. \end{aligned} \right. \end{align*} $$

Proof. First, observe that for all $1\leq j\leq m$ and all $1\leq i\leq n$ ,

$$ \begin{align*} \frac{\partial f_{\mathbf{w}}^{k_j}}{\partial z_{i}}(\mathbf{a}(s)) = \frac{\partial \big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}}{\partial z_{i}}(\mathbf{b})\, s^{d\big(\mathbf{v};f_{\mathbf{w}}^{k_j}\big)-v_i} + \cdots. \end{align*} $$

Thus, if the assertion in Claim 3.7 fails, then the sum

$$ \begin{align*} \sum_{j=1}^m \tilde\mu_{k_j} \frac{\partial \big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}}{\partial z_{i}}(\mathbf{b}) \end{align*} $$

vanishes for all $i\in I(\mathbf {w})$ , and so, by ( $2'$ ), it vanishes for all $1\leq i\leq n$ . In other words, if $k_{j_1},\ldots ,k_{j_p}$ are the elements of the set $\{k_1,\ldots ,k_m\}$ for which $d\big (\mathbf {v};f_{\mathbf {w}}^{k_{j_\ell }}\big )+g_{j_\ell }=\delta $ , $1\leq \ell \leq p$ , then the vectors

$$ \begin{align*} \bigg(\frac{\partial \big(f_{\mathbf{w}}^{k_{j_1}}\big)_{\mathbf{v}}}{\partial z_{1}}(\mathbf{b}),\ldots,\frac{\partial \big(f_{\mathbf{w}}^{k_{j_1}}\big)_{\mathbf{v}}}{\partial z_{n}}(\mathbf{b})\bigg) ,\ldots, \bigg(\frac{\partial \big(f_{\mathbf{w}}^{k_{j_p}}\big)_{\mathbf{v}}}{\partial z_{1}}(\mathbf{b}),\ldots,\frac{\partial \big(f_{\mathbf{w}}^{k_{j_p}}\big)_{\mathbf{v}}}{\partial z_{n}}(\mathbf{b})\bigg) \end{align*} $$

of $\mathbb {C}^n$ are linearly dependent, that is,

$$ \begin{align*} d\big(f_{\mathbf{w}}^{k_{j_1}}\big)_{\mathbf{v}}(\mathbf{b})\wedge\cdots\wedge d\big(f_{\mathbf{w}}^{k_{j_p}}\big)_{\mathbf{v}}(\mathbf{b})=0. \end{align*} $$

As $\big (f_{\mathbf {w}}^{k_{j_\ell }}\big )_{\mathbf {v}} = f_{\mathbf {v}+\nu \mathbf {w}}^{k_{j_\ell }}$ for any sufficiently large integer $\nu \in \mathbb {N}$ (so that $\big (f_{\mathbf {w}}^{k_{j_\ell }}\big )_{\mathbf {v}}$ is the face function of $f^{k_{j_\ell }}$ with respect to the weight vector $\mathbf {v}+\nu \mathbf {w}$ ) and $\big (f_{\mathbf {w}}^{k_{j_\ell }}\big )_{\mathbf {v}}(\mathbf {b})=0$ for $1\leq \ell \leq p$ , this contradicts the nondegeneracy of $V(f^{k_{j_1}},\ldots ,f^{k_{j_p}})$ (see Assumptions 3.1).

Combined with ( $2'$ ) again, Claim 3.7 implies that $\lambda (s)$ is not constantly zero. Write it as a Laurent series $\lambda (s)=\lambda _0 s^c+\cdots $ , where $\lambda _0\not =0$ . Then, still from ( $2'$ ), we deduce that for all $1\leq i\leq n$ ,

(3.6) $$ \begin{align} \sum_{j=1}^m \tilde\mu_{k_j} \frac{\partial \big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}}{\partial z_i}(\mathbf{b})\, s^{\delta} + \cdots = \left\{ \begin{aligned} & \lambda_0 \bar b_i\, s^{c+2v_i}+\cdots, &&\mbox{if} &&i\in I(\mathbf{w}),\\ & 0, &&\mbox{if} &&i\notin I(\mathbf{w}). \end{aligned} \right. \end{align} $$

Put $S_i:=\sum _{j=1}^m \tilde \mu _{k_j} \frac {\partial \big (f_{\mathbf {w}}^{k_j}\big )_{\mathbf {v}}}{\partial z_i}(\mathbf {b})$ , and define

(3.7) $$ \begin{align} v_0:=\mbox{min}\{v_i\, ;\, i\in I(\mathbf{w})\} \quad\mbox{and}\quad I_0 :=\{i\in I(\mathbf{w}) \, ;\, v_i=v_0\}. \end{align} $$

Since the coefficient $\lambda _0\bar b_i$ on the right-hand side of (3.6) is nonzero and the set of indexes $i\in I(\mathbf {w})$ such that $S_i\not =0$ is not empty (see Claim 3.7), we have $\delta =c+2v_0$ and $S_i\not =0$ for any $i\in I_0$ . In fact, for any $1\leq i\leq n$ , the following equality holds:

(3.8) $$ \begin{align} S_i\equiv\sum_{j=1}^m \tilde\mu_{k_j} \frac{\partial \big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}}{\partial z_i}(\mathbf{b})= \left\{ \begin{aligned} & \lambda_0 \bar b_i, && \mbox{if} && i\in I_0,\\ & 0, && \mbox{if} && i\notin I_0. \end{aligned} \right. \end{align} $$

Since $I_0\not =\emptyset $ and $\big (f_{\mathbf {w}}^{k_j}\big )_{\mathbf {v}}(\mathbf {b})=0$ ( $1\leq j\leq m$ ), combined with the Euler identity, the relation (3.8) implies

$$ \begin{align*} 0 & = \sum_{j=1}^{m}\tilde\mu_{k_j}\cdot d\big(\mathbf{v};f_{\mathbf{w}}^{k_j}\big) \cdot \big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}(\mathbf{b}) = \sum_{j=1}^{m}\tilde\mu_{k_j}\bigg(\sum_{i=1}^n v_i b_i \frac{\partial \big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}}{\partial z_i}(\mathbf{b})\bigg) \\ & = \sum_{i\in I_0}v_i b_i\bigg( \sum_{j=1}^{m}\tilde\mu_{k_j} \frac{\partial \big(f_{\mathbf{w}}^{k_j}\big)_{\mathbf{v}}}{\partial z_i}(\mathbf{b}) \bigg) =\lambda_0\cdot\sum_{i\in I_0} v_i |b_i|^2\not=0, \end{align*} $$

which is a contradiction. This completes the proof of Lemma 3.3.

Assumptions 4.1. Throughout this section, we suppose that for any sufficiently small t (say, $|t|\leq \tau _0$ for some $\tau _0>0$ ), the following two conditions hold true:

1. (1) For any $1\leq k\leq k_0$ , the Newton boundary $\Gamma (f^k_t)$ is independent of t. (We may still have $\Gamma (f_t^{k})\not =\Gamma (f_t^{k'})$ for $k\not =k'$ .).

2. (2) For any $k_1,\ldots ,k_m\in \{k_1,\ldots ,k_0\}$ , the germ at $\mathbf {0}$ of the variety $V(f^{k_1}_t,\ldots ,f^{k_m}_t)$ is the germ of a nondegenerate complete intersection variety.

Assumptions 5.1. Throughout this section, we suppose that the following two conditions hold true:

1. (1) For any $1\leq k\leq k_0$ , the Newton boundaries $\Gamma (f^k)$ and $\Gamma (g^k)$ coincide.

2. (2) For any $k_1,\ldots ,k_m\in \{k_1,\ldots ,k_0\}$ , the germs at $\mathbf {0}$ of the varieties $V(f^{k_1},\ldots ,f^{k_m})$ and $V(g^{k_1},\ldots ,g^{k_m})$ are the germs of nondegenerate complete intersection varieties.

Proof. For any $1\leq k\leq k_0$ and any $t\in D_1:=\{t\in \mathbb {C}\, ;\, \vert t\vert \leq 1\}$ , we consider the polynomial functions

$$ \begin{align*} f_t^k(\mathbf{z}):=(1-t)f^k+t F^k \quad\mbox{and}\quad g_t^k(\mathbf{z}):=(1-t)g^k+t G^k, \end{align*} $$

where

$$ \begin{align*} F^k(\mathbf{z}):=\sum_{\alpha\in\Gamma(f^k)}c_\alpha\, \mathbf{z}^\alpha \quad\mbox{and}\quad G^k(\mathbf{z}):=\sum_{\alpha\in\Gamma(g^k)}c^{\prime}_\alpha\, \mathbf{z}^\alpha \end{align*} $$

are the Newton principal parts of

$$ \begin{align*} f^k(\mathbf{z}):=\sum_{\alpha\in\mathbb{N}^n}c_\alpha\, \mathbf{z}^\alpha \quad\mbox{and}\quad g^k(\mathbf{z}):=\sum_{\alpha\in\mathbb{N}^n}c^{\prime}_\alpha\, \mathbf{z}^\alpha, \end{align*} $$

respectively.

Proof. First, observe that for any $1\leq k\leq k_0$ and any positive weight vector $\mathbf {w}$ , we have

$$ \begin{align*} f_{t,\mathbf{w}}^k(\mathbf{z}):=((1-t)f^k+t F^k)_{\mathbf{w}}=(1-t)f^k_{\mathbf{w}}+t F^k_{\mathbf{w}}=f^k_{\mathbf{w}}. \end{align*} $$

From this observation and Assumptions 5.1, we deduce that $V(f_t^{k_1},\ldots ,f_t^{k_m})$ is a nondegenerate complete intersection variety for any $t\in D_1$ , and since $\Gamma (f^k_t)=\Gamma (f^k)$ , it follows from Theorem 4.8 that the functions $f_t(\mathbf {z}):=f_t^1(\mathbf {z})\cdots f_t^{k_0}(\mathbf {z})$ and $f_0(\mathbf {z}):=f_0^1(\mathbf {z})\cdots f_0^{k_0}(\mathbf {z})=f^1(\mathbf {z})\cdots f^{k_0}(\mathbf {z})$ have isomorphic Milnor fibrations for any $t\in D_1$ . In particular, taking $t=1$ gives that $F^1\cdots F^{k_0}$ and $f^1\cdots f^{k_0}$ have isomorphic Milnor fibrations as announced.

The proof of this claim is given below. Of course, Theorem 5.2 follows from Claims 5.3 and 5.4.

Proof of Claim 5.4

For each $1\leq k\leq k_0$ , let $\nu _{k,1},\ldots ,\nu _{k,n_k}$ be the integral points of $\Gamma (f^k)=\Gamma (F^k)$ , and for any $\mathbf {c}_k=(c_{k,1},\ldots ,c_{k,n_k})\in \mathbb {C}^{n_k}$ , put

$$ \begin{align*} h^k_{\mathbf{c}_k}(\mathbf{z}):=\sum_{j=1}^{n_k} c_{k,j}\, \mathbf{z}^{\nu_{k,j}}. \end{align*} $$

Now, consider the set U of points $(\mathbf {c}_1,\ldots ,\mathbf {c}_{k_0})$ in $\mathbb {C}^{n_1}\times \cdots \times \mathbb {C}^{n_{k_0}}$ such that:

1. (1) $\Gamma (h^k_{\mathbf {c}_k})=\Gamma (f^k)$ for any $1\leq k\leq k_0$ .

2. (2) For any $1\leq k_1,\ldots ,k_m\leq k_0$ , the variety $V(h^{k_1}_{\mathbf {c}_{k_1}},\ldots ,h^{k_m}_{\mathbf {c}_{k_m}})$ is a nondegenerate complete intersection variety.

The special case $k_0=1$ in Claim 5.5 is treated in the Appendix of [Reference Oka7]. Before proving this claim in the general case, we complete the proof of Claim 5.4.

For each $1\leq k\leq k_0$ , let

$$ \begin{align*} \mathbf{c}_{k}(F^k):=(c_{k,1}(F^k),\ldots,c_{k,n_k}(F^k)) \quad\mbox{and}\quad \mathbf{c}_{k}(G^k):=(c_{k,1}(G^k),\ldots,c_{k,n_k}(G^k)) \end{align*} $$

be the points defined by

$$ \begin{align*} F^k(\mathbf{z}):=h^k_{\mathbf{c}_{k}(F^k)}(\mathbf{z}) :=\sum_{j=1}^{n_k} c_{k,j}(F^k)\, \mathbf{z}^{\nu_{k,j}} \quad\mbox{and}\quad G^k(\mathbf{z}):=h^k_{\mathbf{c}_{k}(G^k)}(\mathbf{z}) :=\sum_{j=1}^{n_k} c_{k,j}(G^k)\, \mathbf{z}^{\nu_{k,j}}. \end{align*} $$

By Claim 5.5, we can choose a finite sequence of (say, $p_0$ ) $k_0$ -tuples

$$ \begin{align*} (\mathbf{c}_1(1),\ldots,\mathbf{c}_{k_0}(1)), \ldots, (\mathbf{c}_1(p_0),\ldots,\mathbf{c}_{k_0}(p_0)) \end{align*} $$

in U, starting at $(\mathbf {c}_1(F^1),\ldots ,\mathbf {c}_{k_0}(F^{k_0}))$ and ending at $(\mathbf {c}_1(G^1),\ldots ,\mathbf {c}_{k_0}(G^{k_0}))$ , such that for each $1\leq p\leq p_0-1$ , the straight-line segment

$$ \begin{align*} \ell_{p}(t):=(1-t)\, (\mathbf{c}_1(p),\ldots,\mathbf{c}_{k_0}(p))+t\, (\mathbf{c}_1(p+1),\ldots,\mathbf{c}_{k_0}(p+1)) \end{align*} $$

( $1\leq t\leq 1$ ) is contained in U. For each $1\leq p\leq p_0-1$ , we consider the family $\big \{h_{\ell _{p}(t)}\big \}_{0\leq t\leq 1}$ of polynomial functions defined by

$$ \begin{align*} h_{\ell_{p}(t)}(\mathbf{z}):=h^1_{\ell^1_{p}(t)}(\mathbf{z})\cdots h^{k_0}_{\ell^{k_0}_{p}(t)}(\mathbf{z}), \end{align*} $$

where $\ell ^{k}_{p}(t):=(1-t)\mathbf {c}_{k}(p)+t\mathbf {c}_{k}(p+1)$ is the kth coordinate of $\ell _{p}(t)$ . By Theorem 4.8, the Milnor fibrations of $h_{\ell _{p}(0)}$ and $h_{\ell _{p}(1)}$ at $\mathbf {0}$ are isomorphic. Claim 5.4 then follows from the equalities

$$ \begin{align*} h_{\ell_{1}(0)}=F^1\cdots F^{k_0} \quad\mbox{and}\quad h_{\ell_{p_0-1}(1)}=G^1\cdots G^{k_0}. \end{align*} $$

This completes the proof of Claim 5.4 (up to Claim 5.5).

Proof of Claim 5.5

For any $1\leq k\leq k_0$ and any positive weight vector $\mathbf {w}$ defining a (compact) face $\Delta (\mathbf {w};f^k)$ of $\Gamma (f^k)$ with maximal dimension, let us denote by $\theta _{k,1},\ldots ,\theta _{k,q_k}$ the integral points of $\Delta (\mathbf {w};f^k)$ . Then, for any $\mathbf {a}_k=(a_{k,1},\ldots ,a_{k,q_k})\in \mathbb {C}^{q_k}$ , put

$$ \begin{align*} \phi^k_{\mathbf{a}_k}(\mathbf{z}):=\sum_{j=1}^{q_k} a_{k,j}\, \mathbf{z}^{\theta_{k,j}}. \end{align*} $$

Note that $\phi ^k_{\mathbf {a}_k}$ is weighted homogeneous with respect to $\mathbf {w}$ . Now, consider the set $U_{\mathbf {w}}$ consisting of the points $(\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0})$ in $\mathbb {C}^{q_1}\times \cdots \times \mathbb {C}^{q_{k_0}}$ satisfying the following two properties:

1. (1) $\Gamma (\phi ^k_{\mathbf {a}_k})=\Delta (\mathbf {w};f^k)$ for any $1\leq k\leq k_0$ .

2. (2) For any $1\leq k_1,\ldots ,k_m\leq k_0$ , the variety $V(\phi ^{k_1}_{\mathbf {a}_{k_1}},\ldots ,\phi ^{k_m}_{\mathbf {a}_{k_m}})$ is a nondegenerate complete intersection variety.

To prove Claim 5.5, it suffices to show that $U_{\mathbf {w}}$ is a Zariski open set. To do that, we first observe that since $\phi ^{k_j}_{\mathbf {a}_{k_j}}$ ( $1\leq j\leq m$ ) is weighted homogeneous, there exist $\lambda _1,\ldots ,\lambda _n\in \mathbb {N}^*$ such that the polynomial

$$ \begin{align*} \Phi^{k_j}_{\mathbf{a}_{k_j}}(z_1,\ldots,z_n):=\phi^{k_j}_{\mathbf{a}_{k_j}}(z_1^{\lambda_1},\ldots,z_n^{\lambda_n}) \end{align*} $$

is homogeneous. Then, since $V(\phi ^{k_1}_{\mathbf {a}_{k_1}},\ldots ,\phi ^{k_m}_{\mathbf {a}_{k_m}})$ is nondegenerate if and only if $V(\Phi ^{k_1}_{\mathbf {a}_{k_1}},\ldots ,\Phi ^{k_m}_{\mathbf {a}_{k_m}})$ is nondegenerate, we may assume that $\phi ^{k_j}_{\mathbf {a}_{k_j}}$ is homogeneous for any $1\leq j\leq m$ . Now, observe that for any positive weight vector $\mathbf {w}'$ , the set $\Delta (\mathbf {w}';f^k_{\mathbf {w}})$ is a (compact) face of $\Delta (\mathbf {w};f^k)$ , and then consider the set $V_{\mathbf {w}}(\mathbf {w}')$ made up of all the points $(\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0},\mathbf {z})$ in $\mathbb {P}^{q_1-1}\times \cdots \times \mathbb {P}^{q_{k_0}-1}\times \mathbb {P}^{n-1}$ for which there exists a subset $K\subseteq \{1,\ldots , k_0\}$ such that

$$ \begin{align*} \forall k\in K,\ \phi^{k}_{\mathbf{a}_{k},\mathbf{w}'}(\mathbf{z})=0 \quad\mbox{and}\quad \bigwedge_{k\in K} d\phi^{k}_{\mathbf{a}_{k},\mathbf{w}'}(\mathbf{z})=0, \end{align*} $$

where we still denote by $\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0}$ and $\mathbf {z}$ the classes of $\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0}$ and $\mathbf {z}$ in the projective spaces $\mathbb {P}^{q_1-1},\ldots ,\mathbb {P}^{q_{k_0}-1}$ and $\mathbb {P}^{n-1}$ , respectively. (Once more, let us recall that $\phi ^{k}_{\mathbf {a}_{k},\mathbf {w}'}\equiv (\phi ^{k}_{\mathbf {a}_{k}})_{\mathbf {w}'}$ denotes the face function of $\phi ^{k}_{\mathbf {a}_{k}}$ with respect to the weight vector $\mathbf {w}'$ .) Let $\bar V_{\mathbf {w}}(\mathbf {w}')$ be the closure of $V_{\mathbf {w}}^*(\mathbf {w}'):=V_{\mathbf {w}}(\mathbf {w}')\cap \{z_1\cdots z_n\not =0\}$ in $\mathbb {P}^{q_1-1}\times \cdots \times \mathbb {P}^{q_{k_0}-1}\times \mathbb {P}^{n-1}$ . Then $\bar V_{\mathbf {w}}(\mathbf {w}')$ is an algebraic set of dimension $\dim V^*(\mathbf {w}')$ (see [Reference Whitney12, Lem. 3.9]). Let

$$ \begin{align*} \pi\colon (\mathbb{P}^{q_1-1}\times \cdots\times \mathbb{P}^{q_{k_0}-1})\times \mathbb{P}^{n-1}\to \mathbb{P}^{q_1-1}\times \cdots\times \mathbb{P}^{q_{k_0}-1} \end{align*} $$

be the standard projection, and let

$$ \begin{align*} W_{\mathbf{w}}^*:=\pi(V_{\mathbf{w}}^*) \quad\mbox{and}\quad \bar W_{\mathbf{w}}:=\pi(\bar V_{\mathbf{w}}), \end{align*} $$

where

$$ \begin{align*} V_{\mathbf{w}}^*:=\bigcup_{\mathbf{w}'\in\mathbb{N}^{*n}} V_{\mathbf{w}}^*(\mathbf{w}') \quad\mbox{and}\quad \bar V_{\mathbf{w}}:=\bigcup_{\mathbf{w}'\in\mathbb{N}^{*n}} \bar V_{\mathbf{w}}(\mathbf{w}'). \end{align*} $$

Clearly, $U_{\mathbf {w}}$ is the complement of $(p_1\times \cdots \times p_{k_0})^{-1}(W_{\mathbf {w}}^*)\cup \{\mathbf {0}\}$ , where $p_k\colon \mathbb {C}^{q_k}\setminus \{\mathbf {0}\}\to \mathbb {P}^{q_k-1}$ is the standard canonical map. By the proper mapping theorem (see [Reference Remmert11, Satz 23]), $\bar W_{\mathbf {w}}$ is an algebraic set containing $W_{\mathbf {w}}^*$ . In fact, we are going to prove that $W_{\mathbf {w}}^*=\bar W_{\mathbf {w}}$ , which implies that $U_{\mathbf {w}}$ is a Zariski open set. To show the equality $W_{\mathbf {w}}^*=\bar W_{\mathbf {w}}$ , we argue by contradiction. Suppose that $W_{\mathbf {w}}^*\subsetneq \bar W_{\mathbf {w}}$ . Then there exists $(\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0},\mathbf {z})\in \bar V_{\mathbf {w}}$ such that $(\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0})\in \bar W_{\mathbf {w}}\setminus W_{\mathbf {w}}^*$ . By the Curve Selection Lemma, there exist a real analytic curve

$$ \begin{align*} \rho(s)=(\mathbf{a}_1(s),\ldots,\mathbf{a}_{k_0}(s),\mathbf{z}(s)), \end{align*} $$

$0\leq s\leq 1$ , and a positive weight vector $\mathbf {w}'\in \mathbb {N}^{*n}$ such that $\rho (s)\in V^*_{\mathbf {w}}(\mathbf {w}')$ for $s>0$ and $\rho (0)=(\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0},\mathbf {z})$ . For each $1\leq k\leq k_0$ , write

$$ \begin{align*} \mathbf{a}_k(s)=\mathbf{a}_k+\mathbf{a}_{k,1}s+\cdots \quad\mbox{and}\quad \mathbf{z}(s)=(b_1s^{w^{\prime\prime}_1}+\cdots,\ldots,b_ns^{w^{\prime\prime}_n}+\cdots). \end{align*} $$

By the assumption, $b_i\in \mathbb {C}^*$ , $w^{\prime \prime }_i\in \mathbb {N}$ ( $1\leq i\leq n$ ) and $\mbox {max}\{w^{\prime \prime }_i\, ;\, 1\leq i\leq n\}>0$ . Moreover, for any $s\not =0$ , there exists $K(s)\subseteq \{1,\ldots ,k_0\}$ such that

$$ \begin{align*} \forall k\in K(s),\ \phi^{k}_{\mathbf{a}_{k}(s),\mathbf{w}'}(\mathbf{z}(s))=0 \quad\mbox{and}\quad \bigwedge_{k\in K(s)} d\phi^{k}_{\mathbf{a}_{k}(s),\mathbf{w}'}(\mathbf{z}(s))=0. \end{align*} $$

By looking at the leading terms (with respect to s) in the above expressions, it follows that there exists a subset $K(0)\subseteq \{1,\ldots ,k_0\}$ such that

(5.1) $$ \begin{align} \forall k\in K(0),\ \big(\phi^{k}_{\mathbf{a}_{k},\mathbf{w}'}\big)_{\Delta}(\mathbf{b})=0 \quad\mbox{and}\quad \bigwedge_{k\in K(0)} d\big(\phi^{k}_{\mathbf{a}_{k},\mathbf{w}'}\big)_{\Delta}(\mathbf{b})=0, \end{align} $$

where $\mathbf {b}:=(b_1,\ldots ,b_n)$ , $\Delta $ is the (compact) face of $\Delta (\mathbf {w}';f^k_{\mathbf {w}})$ on which the linear form

$$ \begin{align*} \alpha\in\Delta(\mathbf{w}';f^k_{\mathbf{w}})\mapsto \sum_{i=1}^n \alpha_i w_i"\in\mathbb{R} \end{align*} $$

takes its minimal value, and $\big (\phi ^{k}_{\mathbf {a}_{k},\mathbf {w}'}\big )_{\Delta }$ is the corresponding face function. However, since $b_i\in \mathbb {C}^*$ for all $1\leq i\leq n$ , the relations (5.1) imply $(\mathbf {a}_1,\ldots ,\mathbf {a}_{k_0})\in W^*_{\mathbf {w}}$ , which is a contradiction.