Properness of Polynomial Maps with Newton Polyhedra

Abstract

We discuss the notion of properness of a polynomial map $\varvec{f}:\mathbb {K}^m\rightarrow \mathbb {K}^n$, $\mathbb {K}=\mathbb {C}$ or $\mathbb {R}$, at a point of the target. We present a method to describe the set of non-proper points of $\varvec{f}$ with respect to Newton polyhedra of $\varvec{f}$. We obtain an explicit precise description of such a set of $\varvec{f}$ when $\varvec{f}$ satisfies certain condition (1.5). A relative version is also given in Sect. 3. Several tricks to describe the set of non-proper points of $\varvec{f}$ without the condition (1.5) is also given in Sect. 5.

We consider a polynomial map $\varvec{f}=(f^1,\dots ,f^n):\mathbb {K}^m\rightarrow \mathbb {K}^n$, defined by

$$\begin{aligned} f^j&=\mathop {\mathrm {\sum }}_{\varvec{\nu }}c^j_{\varvec{\nu }}\varvec{x}^{\varvec{\nu }},\ c^j_{\varvec{\nu }}\in \mathbb {K},\ \varvec{x}^{\varvec{\nu }}=(x_1)^{\nu _1}\cdots (x_m)^{\nu _m},\nonumber \\ \varvec{x}&=(x_1,\dots ,x_m), \varvec{\nu }=(\nu _1,\dots ,\nu _m), \end{aligned}$$

(0.1)

where $\mathbb {K}=\mathbb {C}$ or $\mathbb {R}$. We say that a point $\varvec{y}_0\in \mathbb {K}^n$ is proper for $\varvec{f}$ (or a proper point of $\varvec{f}$) if, for any (algebraic) arc $\varvec{x}(t):\mathbb {K}^*,0\rightarrow \mathbb {K}^m$, $\mathbb {K}^*=\mathbb {K}{\setminus }\{0\}$, the following condition holds:

$$\begin{aligned} \displaystyle {\lim _{t\rightarrow 0}\varvec{f}(\varvec{x}(t))=\varvec{y}_0}\ \Longrightarrow \ \displaystyle {\lim _{t\rightarrow 0}\varvec{x}(t)}\ \text { exists in }\mathbb {K}^m. \end{aligned}$$

Let $S_{\varvec{f}}$ denote the set of points $\varvec{y}_0$ in $\mathbb {K}^n$ which are not proper points of $\varvec{f}$. We say that $\varvec{f}:\mathbb {K}^m\rightarrow \mathbb {K}^n$ is proper if $S_{\varvec{f}}=\emptyset$.

In this paper, we are looking for a method to determine whether a point $\varvec{y}_0$ in $\mathbb {K}^n$ is proper or not. The first statement is Theorem 1.3, which gives a complete description of $S_{\varvec{f}}$ when $\varvec{f}$ satisfies certain non-degeneracy condition with respect to the Newton polyhedron of $\varvec{f}$ (see (1.5) in Theorem 1.3). Our approach is based on simple and careful analysis of $\varvec{f}$ along arcs $\varvec{x}(t)$, which suggests us usefulness of using arcs to describe the set $S_{\varvec{f}}$, even though $\varvec{f}$ is degenerate (Remark 2.6). In Sect. 3, we describe a relative version of our discussion. We present several examples to show how our method works in Sect. 4.

The set $S_{\varvec{f}}$ was introduced by Jelonek [4, 5] and showed that it is empty or a uniruled hypersurface of $\mathbb {K}^n$ when $\mathbb {K}=\mathbb {C}$ and $m=n$. It is thus an interesting problem to seek a method to describe $S_{\varvec{f}}$ in several concrete examples. Chen et al. [2] have investigated the bifurcation locus of a polynomial map $\mathbb {K}^m\rightarrow \mathbb {K}^n$, $m\ge n$, with respect to Newton polyhedron. The bifurcation locus is the minimal locus in the target where the map is not locally trivial, and they show a supset of the bifurcation locus under their non-degeneracy condition. Jelonek and Lasoń [6] called $S_{\varvec{f}}$ as the non-properness set of $\varvec{f}$ and showed that it is covered by parametric curves of degree at most $d-1$ where d is the algebraic degree of $\varvec{f}$ for $\mathbb {K}=\mathbb {C}$. Their words “covered by parametric curves” mean that the set $S_{\varvec{f}}$ has a “$\mathbb {C}$-ruling”. They also discuss real counterpart of their results. Recently, El Hilany [3] has investigated to describe the set $S_{\varvec{f}}$ via the Newton polyhedra of $\varvec{f}$. He calls $S_{\varvec{f}}$ as Jelonek set. He has introduced the notion of T- maps and claimed that $S_{\varvec{f}}$ is described using only the data of $\varvec{f}$ at several faces of its Newton polyhedra. Comparing with these results, our method provides much precise information on the set $S_{\varvec{f}}$ with simple description. For example, Theorem 1.3 shows an explicit decomposition of $S_{\varvec{f}}$ providing an explicit ruling of each component in many cases. In Sect. 3, we present a relative version of our theorem. Namely, we consider the non-properness set $S_{\varvec{f}|_X}$ for $\varvec{f}|_X:X\rightarrow \mathbb {K}^n$ where $\varvec{f}=(f^1,\dots ,f^n):\mathbb {K}^m\rightarrow \mathbb {K}^n$ is a certain polynomial map and $X=(f^{n-k+1},\dots ,f^n)^{-1}(c)$, $c\in \mathbb {K}^k$. In Sect. 5, we present tricks to describe $S_{\varvec{f}}$ for certain degenerate $\varvec{f}$.

We say some words for the definition of $S_{\varvec{f}}$ here. We compactify $\varvec{f}$ as $\bar{\varvec{f}}:X\rightarrow Y$ where X and Y are suitable projective manifolds. We set $X_\infty =X{\setminus }\mathbb {K}^m$, $Y_\infty =Y{\setminus }\mathbb {K}^n$ and we can assume that $X_\infty$ and $Y_\infty$ are simple normal crossing divisors. Then, the condition $\varvec{y}_0\in S_{\varvec{f}}$ is equivalent to one of the following conditions.

• There exists an algebraic arc $\varvec{x}(t):\mathbb {K},0\rightarrow X$, such that

  $$\begin{aligned} \displaystyle {\lim _{t\rightarrow 0}\varvec{x}(t)\in X_\infty },\quad \text { and }\quad \displaystyle {\lim _{t\rightarrow 0}\varvec{f}(\varvec{x}(t))=\varvec{y}_0}. \end{aligned}$$

  (0.2)

• There exists an analytic arc $\varvec{x}(t):\mathbb {K},0\rightarrow X$ defined near 0 with (0.2).

• There exists a sequence $\{\varvec{x}_k\}$ in X, such that $\displaystyle {\lim _{k\rightarrow \infty }\varvec{x}_k}\in X_\infty$ and $\displaystyle {\lim _{k\rightarrow \infty }\varvec{f}(\varvec{x}_k)=\varvec{y}_0}$.

The last condition is equivalent to the condition that $\varvec{y}_0$ is not a proper point of f as a continuous map between metric spaces. We also have

$$\begin{aligned} S_{\varvec{f}}=\bar{\varvec{f}}(X_\infty )\cap \mathbb {K}^n=\bar{\varvec{f}}(X_\infty )\cap (Y{\setminus } Y_\infty ). \end{aligned}$$

Since $\bar{\varvec{f}}$ is proper, the set $\bar{\varvec{f}}(X_\infty )$ is closed in Y and we obtain that $S_{\varvec{f}}$ is closed.

When $\mathbb {K}=\mathbb {C}$ and $m>n$, Noether’s normalization asserts that, for any $\varvec{y}\in \mathbb {C}^n$, there is a linear surjection $p:\varvec{f}^{-1}(\varvec{y})\rightarrow \mathbb {C}^d$, $0\le d\le m$, where $d=\dim _{\mathbb {C}}\varvec{f}^{-1}(\varvec{y})$. If $\varvec{f}^{-1}(\varvec{y})$ is compact, then $p(\varvec{f}^{-1}(\varvec{y}))=\mathbb {C}^d$ is compact and we obtain $d=0$. Since $d\ge m-n$, we conclude that $m\le n$. This implies that $S_{\varvec{f}}$ is the closure of the image of $\varvec{f}$ whenever $m>n$. Therefore, we assume $m\le n$ when $\mathbb {K}=\mathbb {C}$.

When $\mathbb {K}=\mathbb {C}$, Jelonek’s result asserts that $S_{\varvec{f}}$ is Zariski closed. However, if $\mathbb {K}=\mathbb {R}$, $S_{\varvec{f}}$ may not be Zariski closed (for example, $S_{\varvec{f}}=\{(0,y_2)\in \mathbb {R}^{2}: y_2\ge 0\}$ for $\varvec{f}:\mathbb {R}^2\rightarrow \mathbb {R}^2$, $(x_1,x_2)\mapsto (x_1,x_1^2x_2^2)$).

Throughout the paper, we use the following notational convention:

$$\begin{aligned} \mathbb {K}^J\!=\!\{(x_1,\dots ,x_n)\in \mathbb {K}^n:x_i=0, i\not \in J\},\ \mathbb {Z}^J\!=\!\{(\nu _1,\dots ,\nu _n)\in \mathbb {Z}^n:\nu _i\!=\!0, i\not \in J\}. \end{aligned}$$

for a subset J of $\{1,\dots ,n\}$. We set $\mathbb {Z}^J_{\ge 0}=\{(\nu _1,\dots ,\nu _n)\in \mathbb {Z}^J:\nu _i\ge 0, i\in J\}$. We also set $(\mathbb {Z}_{\ge 0})^n=\{(\nu _1,\dots ,\nu _n)\in \mathbb {Z}^n:\nu _i\ge 0, i=1,\dots ,n\}$. We often abbreviate $\mathbb {Z}_{\ge 0}$ as $\mathbb {Z}_\ge$ following custom. We identify $\mathbb {K}^n$ with $\mathbb {K}^J\times \mathbb {K}^{J^c}$ where $J^c=\{1,\dots ,n\}{\setminus } J$ without notice.

1 Newton Polyhedra

Let $\Delta (f^j)$ denote Newton polyhedron of $f^j$, the convex hull of the set $\{\varvec{\nu }:c^j_{\varvec{\nu }}\ne 0\}$, under the notation in (0.1). For $\varvec{p}=(p_1,\dots ,p_m)\in \mathbb {Z}^m$, we define

$$\begin{aligned} d_j(\varvec{p})&=-\min \{\langle \varvec{p},\varvec{\nu }\rangle :\varvec{\nu }\in \Delta (f^j)\}, \end{aligned}$$

(1.1)

$$\begin{aligned} \gamma _j(\varvec{p})&=\{\varvec{\nu }\in \Delta (f^j):\langle \varvec{p},\varvec{\nu }\rangle =-d_j(\varvec{p})\}. \end{aligned}$$

(1.2)

We call $\gamma _j(\varvec{p})$ the face of $\Delta (f^j)$ supported by $\varvec{p}$.

We say which is a face of $\varvec{\Delta }(\varvec{f})=(\Delta (f^1),\dots ,\Delta (f^n))$ if there exist $\varvec{p}\in \mathbb {Z}^m$, so that $\gamma _j$ is a face of $\Delta (f^j)$ supported by $\varvec{p}$. We denote

When we need to mention $\varvec{f}$ explicitly, we denote them by , $\gamma _j(\varvec{f};\varvec{p})$, and so on. We consider Minkowski sum $\Delta (\varvec{f})=\Delta (f^1)+\cdots +\Delta (f^n)$ and its dual fan $\Delta ^*$, which we identify with the set of polyhedral cones. Note that $\gamma (\varvec{p})=\gamma _1(\varvec{p})+\cdots +\gamma _n(\varvec{p})$ is a face of $\Delta (\varvec{f})$. We denote

Lemma 1.1

$\iff$ $\gamma (\varvec{p})=\gamma (\varvec{q})$.

Proof

“$\Longrightarrow$” part is clear, since “ $\iff$ $\gamma _j(\varvec{p})\!=\!\gamma _j(\varvec{q})$ ($j\!=\!1,\dots ,n$)”.

Take $\varvec{\nu }\in \gamma (\varvec{q})$, so that $\varvec{\nu }=\varvec{\nu }_1+\cdots +\varvec{\nu }_n$ for $\varvec{\nu }_j\in \gamma _j(\varvec{q})$. Since $\gamma _j(\varvec{q})\subset \Delta (f^j)$, we have $-d_j(\varvec{p})\le \langle \varvec{p},\varvec{\nu }_j\rangle$. If we assume $\gamma (\varvec{q})\subset \gamma (\varvec{p})$, we then have

$$\begin{aligned} -\mathop {\mathrm {\sum }}_{j=1}^nd_j(\varvec{p})\le \mathop {\mathrm {\sum }}_{j=1}^n\langle \varvec{p},\varvec{\nu }_j\rangle =\langle \varvec{p},\varvec{\nu }\rangle =-\mathop {\mathrm {\sum }}_{j=1}^nd_j(\varvec{p}), \end{aligned}$$

and $\langle \varvec{p},\varvec{\nu }_j\rangle =-d_j(\varvec{p})$, that is, $\varvec{\nu }_j\in \gamma _j(\varvec{p})$. We conclude $\gamma _j(\varvec{q})\subset \gamma _j(\varvec{p})$. By symmetry, we complete the proof of “$\Longleftarrow$”. $\square$

Compositing $\varvec{f}$ with a translation of the target, the set $S_{\varvec{f}}$ is changed by its translation. Without loss of generality, we thus can assume the following condition:

$$\begin{aligned} f^j (j=1,\dots ,n)\text { are non-constant polynomials with }\text { non-zero}\,\text { constant terms.}\nonumber \\ \end{aligned}$$

(1.3)

Throughout the paper, we assume the condition (1.3) unless otherwise stated.

The condition (1.3) implies that $d_j(\varvec{p})\ge 0$ and equality holds if $\varvec{p}\in (\mathbb {Z}_{\ge 0})^n$. For a face of $\varvec{\Delta }(\varvec{f})$, we set . We remark that

(1.4)

Definition 1.2

We say a face of $\varvec{\Delta }(\varvec{f})$ is non-coordinate if there is $\varvec{p}\in \mathbb {Z}^m{\setminus }(\mathbb {Z}_{\ge 0})^m$, so that . Let $\varvec{\Delta }_{\mathrm {nc}}(\varvec{f})$ denote the set of non-coordinate faces of $\varvec{\Delta (f)}$.

For a polynomial map $\varvec{g}=(g^1,\dots ,g^r):\mathbb {K}^m\rightarrow \mathbb {K}^r$, we set

$$\begin{aligned} Z(\varvec{g})&=\{\varvec{x}\in (\mathbb {K}^*)^m:g^j(\varvec{x})=0\ (j=1,\dots ,r)\}, \\ \Sigma (\varvec{g})&=\{\varvec{x}\in (\mathbb {K}^*)^m:\mathop {\mathrm {rank}}{\mathop {\mathrm {Jac}}{(}\varvec{g})(\varvec{x})}<r\}, \end{aligned}$$

where $\mathop {\mathrm {Jac}}{(}\varvec{g})=(\partial _{x_i}g^j)_{i=1,\dots , m; j=1, \dots , r}$. Remark that the codimension of $Z(\varvec{g}){\setminus }\Sigma (\varvec{g})$ is r.

Theorem 1.3

Assume that $\varvec{f}$ is a polynomial map with (1.3) and

(1.5)

for any where . We have

where , , and is the map defined by $\varvec{x}\mapsto (f^j_{\gamma _j}(\varvec{x}))_{j\not \in J}$.

We often say that has the dense nonsingular locus if the condition (1.5) holds.

When $J=\{1,\dots ,n\}$, we have is a constant map, since $\mathbb {K}^\emptyset$ is a one-point set.

Remark 1.4

Chen et al. [2] said that $\varvec{f}$ is non-degenerate if for all . This implies (1.5) for all . However, our condition (1.5) is weaker than their non-degeneracy condition.

Remark 1.5

If , the condition (1.5) implies that is empty. In fact, if we take a nonsingular point , then the condition (1.5) implies that is of codimension n at $\varvec{x}$. This implies that $\varvec{x}$ is isolated in . However, this is impossible, since $f^j_{\gamma _j}$ is weighted homogeneous with respect to the weight $\varvec{p}$.

We also remark that when .

Corollary 1.6

A polynomial map $\varvec{f}$ with (1.3) is proper, if for any none of $\gamma _j$, $j=1,\dots ,n$, contains the origin and has a dense nonsingular locus for any .

Remark 1.7

When $k=\dim \gamma _J=\sum _{j\in J}\dim \gamma _j$, , is a system of polynomials of k Laurent monomials of $\varvec{x}$ and is isomorphic to $X\times (\mathbb {K}^*)^{n-k}$ for some algebraic variety X in $(\mathbb {K}^*)^k$. If , $d_j=0$ and $f^j(x)$ ($j\in J$) is invariant under the natural $\mathbb {K}^*$-action(s). Thus, . When is complete intersection, we have that

where F is a suitable fiber of . We thus have . When $\mathbb {K}=\mathbb {C}$, $S_{\varvec{f}}$ is a hypersurface and should be the union of the closures of with (and $\dim F=0$).

To prove Theorem 1.3, we actually show the following.

Theorem 1.8

If a polynomial map $\varvec{f}:\mathbb {K}^m\rightarrow \mathbb {K}^n$ satisfies (1.3), then

(1.6)

where . Here, ${\overline{Z}}$ denotes the closure of a set Z.

Remark 1.9

Assume that $m=n=2$. Take for a non-degenerate map $\varvec{f}:\mathbb {K}^2\rightarrow \mathbb {K}^2$. We assume that , $\varvec{p}=(p_1,p_2)$ with $p_1<0$ and $p_2>0$.

• If $0\in \gamma _1$ and $0\in \gamma _2$, then $f^1_{\gamma _1}$ and $f^2_{\gamma _2}$ are polynomial of a monomial $u=x_1^{q_1}x_2^{q_2}$. We denote them as $g_1(u)$ and $g_2(u)$. The defining equation of is the resultant of $g_1-y_1$ and $g_2-y_2$ where $(y_1,y_2)$ is a coordinate system of the target.

• If $0\not \in \gamma _1$ and $0\in \gamma _2$, then we can write $f^1_{\gamma _1}=x_1^pg_1(u)$ and $f^2_{\gamma _2}=g_2(u)$ with $u=x_1^{q_1}x_2^{q_2}$ similarly. The defining equation of is the resultant of $g_1$ and $g_2-y_2$.

2 Proof of Theorem 1.8

We are going to evaluate $\varvec{f}(x)=(f^1(x),\dots ,f^n(x))$ along a curve $\varvec{x}(t)$ defined by

$$\begin{aligned} \varvec{x}(t)&=(t^{p_1}v^1(t),\dots , t^{p_m}v^m(t))\quad \text {where}\quad \varvec{p}=(p_1,\dots ,p_m)\in \mathbb {Z}^m, \end{aligned}$$

(2.1)

$$\begin{aligned} \varvec{v}(t)&=(v^1(t),\dots ,v^m(t))=\sum \limits _{i=0}^{\infty } \varvec{v}_i t^i, \ \varvec{v}_i=(v^1_i,\dots ,v^m_i),\ \varvec{v}_0\in (\mathbb {K}^*)^m. \end{aligned}$$

(2.2)

We denote ${\mathcal {A}}(\varvec{p})$ the set of such arcs. It is clear that

$$\begin{aligned} \lim _{t\rightarrow 0}\varvec{x}(t)=\infty \iff \varvec{p}\not \in (\mathbb {Z}_{\ge 0})^m. \end{aligned}$$

(2.3)

We have an obvious decomposition $S_{\varvec{f}}=\bigcup _{\varvec{p}}S_{\varvec{f}}(\varvec{p})$, where

$$\begin{aligned} S_{\varvec{f}}(\varvec{p})=\{\varvec{y}\in \mathbb {K}^n:\exists \varvec{x}(t)\in {\mathcal {A}}(\varvec{p}), \ \lim _{t\rightarrow 0}\varvec{x}(t)=\infty ,\ \lim _{t\rightarrow 0}\varvec{f}(\varvec{x}(t))=\varvec{y}\}. \end{aligned}$$

(2.4)

We have $S_{\varvec{f}}(\varvec{p})=\emptyset$ if $\varvec{p}\in (\mathbb {Z}_{\ge 0})^m$ by (2.3).

Remark 2.1

Observe that the arcs having several components being identically zero are not in ${\mathcal {A}}(\varvec{p})$. However, this does not affect to detect $S_{\varvec{f}}$. Adding the terms $t^l$, $l\gg 1$, to such components does not affect the conditions for $S_{\varvec{f}}$ and we can restrict our attention to ${\mathcal {A}}(\varvec{p})$.

Lemma 2.2

For $\varvec{p}\not \in (\mathbb {Z}_{\ge 0})^m$, we have .

Proof

We express $f^j(\varvec{x}(t))$ as

$$\begin{aligned} f^j(\varvec{x}(t))=t^{-d_j}({\hat{f}}^j_0+{\hat{f}}^j_1t+\cdots +{\hat{f}}^j_{d_j-1}t^{d_j-1}+{\hat{f}}^j_{d_j}t^{d_j}+o(t^{d_j})), \end{aligned}$$

(2.5)

where $d_j=d_j(\varvec{p})\ge 0$. We have ${\hat{f}}^j_0=f^j_{\gamma _j}(\varvec{v}_0)$ where $\gamma _j=\gamma _j(\varvec{p})$. Setting $f^j(x)=\sum _{\varvec{\nu }}c^j_{\varvec{\nu }} \varvec{x}^{\varvec{\nu }}$, more precisely, we have

$$\begin{aligned} f^j(\varvec{x}(t)) =t^{-d_j(\varvec{p})}\mathop {\mathrm {\sum }}_{\varvec{\nu }} c^j_{\varvec{\nu }}\, t^{\langle \varvec{p},\varvec{\nu }\rangle +d_j(\varvec{p})}\varvec{v}(t)^{\varvec{\nu }}. \end{aligned}$$

(2.6)

If $\varvec{y}\in S_{\varvec{f}}(\varvec{p})$, then there exists an arc $\varvec{x}(t)\in {\mathcal {A}}(\varvec{p})$, so that

$$\begin{aligned} \lim _{t\rightarrow 0}\varvec{x}(t)=\infty ,\quad \text {and}\quad \lim _{t\rightarrow 0}\varvec{f}(\varvec{x}(t))=\varvec{y}. \end{aligned}$$

Using the notation in (2.5) and , we have

$$\begin{aligned} {\hat{f}}^j_0={\hat{f}}^j_1=\cdots ={\hat{f}}^j_{d_j-1}=0\quad (j\in J), \end{aligned}$$

and $f^j_{\gamma _j(\varvec{p})}(\varvec{v}_0)=0$ ($j\in J$). This implies that $J^c$ component of $\varvec{y}$ is given by , and we complete the proof. $\square$

Remark 2.3

If $\dim \gamma _j(\varvec{p})=0$ for some $j\in J$, then ${\hat{f}}^j_0\ne 0$, and is empty.

Lemma 2.4

If , then

Proof

Take $\varvec{p}\in \mathbb {Z}^n{\setminus }(\mathbb {Z}_{\ge })^n$ and consider a curve defined by (2.1). We compare (2.5) with (2.6) substituting by (2.2) and taking modulo $t^{i+1}$. Remarking that the terms concerning $\varvec{v}_i$ in ${\hat{f}}^j_i$ depend on the terms in $f^j_{\gamma _j}$ only, we obtain that

$$\begin{aligned} {\hat{f}}^j_l=(df^j_{\gamma _j})_{\varvec{v}_0}(\varvec{v}_l)+r^j_{l}(\varvec{v}_0,\dots ,\varvec{v}_{l-1})\quad (l=1,2,\dots ), \end{aligned}$$

(2.7)

where $r^j_l(\varvec{v}_0,\dots ,\varvec{v}_{l-1})$ is a suitable polynomial of $\varvec{v}_0,\dots ,\varvec{v}_{l-1}$.

Take a point where . Let $(a^j_k)_{j\in J; k\ge 1}$ be any sequence. Suppose we have already taken $\varvec{v}_0$, $\varvec{v}_1$, ..., $\varvec{v}_{l-1}$, so that

$$\begin{aligned} {\hat{f}}^j_k=a^j_k\quad (1\le k<l, j\in J). \end{aligned}$$

By (2.7), there exists $\varvec{v}_l$, so that ${\hat{f}}^j_l=a^j_l$ for $j\in J$, whenever $\mathop {\mathrm {Jac}}{(}(f^j_{\gamma _j})_{j\in J})$ is of full rank at $\varvec{v}_0$. Choose $(a^j_l)_{j\in J;l\ge 0}$, so that $a^j_i=0$ ($0\le i< d_j$). Then, the corresponding curve $\varvec{x}(t)$ has the following property:

$$\begin{aligned} \lim _{t\rightarrow 0}f^j(\varvec{x}(t))= {\left\{ \begin{array}{ll} f^j_{\gamma _j}(\varvec{v}_0)&{}(j\not \in J),\\ a^j_{d_j}&{}(j\in J). \end{array}\right. } \end{aligned}$$

Since one can choose $a^j_{d_j}$ arbitrary, we conclude that . $\square$

In the situation above, we have

Corollary 2.5

for .

Proof

We obtain

since $S_{\varvec{f}}$ is closed. $\square$

Remark 2.6

In the case that $\varvec{f}$ does not satisfy (1.5), we would proceed further analysis using higher order differentials of composite maps. Actually, in the expression (2.5), we have

$$\begin{aligned} {\hat{f}}^j_l=\mathop {\mathrm {\sum }}_{k=0}^l\mathop {\mathrm {\sum }}_{i_1+2i_2+\cdots +ki_k=k} \tfrac{1}{i_1!\cdots i_k!} (d^{i_1+\cdots +i_k}f^j_{l-k})_{\varvec{v}_0}(\overbrace{\varvec{v}_1,\dots ,\varvec{v}_1}^{i_1},\dots ,\overbrace{\varvec{v}_k,\dots ,\varvec{v}_k}^{i_k}), \end{aligned}$$

where

$$\begin{aligned}&f^j(\varvec{x})=f^j_0(\varvec{x})+f^j_1(\varvec{x})+\cdots +f^j_{e_j}(\varvec{x})\text { with }\nonumber \\&f^j_k(t^{p_1}x_1,\dots ,t^{p_m}x_m)=t^{-d_j+k}f^j_k(\varvec{x}). \end{aligned}$$

(2.8)

Here, we use the notation in (2.1), (2.2), and $d^kg$ denotes the symmetric multilinear form defined by kth-order differential of g. The first few of ${\hat{f}}^j_l$ are as follows:

$$\begin{aligned} {\hat{f}}^j_0&=f^j_0(\varvec{v}_0),\nonumber \\ {\hat{f}}^j_1&=f^j_1(\varvec{v}_0)+(df^j_0)_{\varvec{v}_0}(\varvec{v}_1),\nonumber \\ {\hat{f}}^j_2&=f^j_2(\varvec{v}_0)+(df^j_0)_{\varvec{v}_0}(\varvec{v}_2)+(df^j_1)_{\varvec{v}_0}(\varvec{v}_1)+\tfrac{1}{2}(d^2f^j_0)_{\varvec{v}_0}(\varvec{v}_1, \varvec{v}_1),\nonumber \\ {\hat{f}}^j_3&=f^j_3(\varvec{v}_0)+(df^j_0)_{\varvec{v}_0}(\varvec{v}_3)+(df^j_1)_{\varvec{v}_0}(\varvec{v}_2)+(df^j_2)_{\varvec{v}_0}(\varvec{v}_1) \end{aligned}$$

(2.9)

$$\begin{aligned}&+(d^2f^j_0)_{\varvec{v}_0}(\varvec{v}_1,\varvec{v}_2)+\tfrac{1}{2}(d^2f^j_1)_{\varvec{v}_0}(\varvec{v}_1,\varvec{v}_1)+\tfrac{1}{6}(d^3f^j_0)_{\varvec{v}_0}(\varvec{v}_1,\varvec{v}_1,\varvec{v}_1). \end{aligned}$$

(2.10)

The set $S_{\varvec{f}}(\varvec{p})$ is described by eliminating $\varvec{v}_0$, $\varvec{v}_1$, $\varvec{v}_2$, ... from the following system:

$$\begin{aligned} 0={\hat{f}}^j_l\ (l=0,1,2,\dots ,d_j-1), \quad y_j={\hat{f}}^j_{d_j}\quad (j=1,\dots ,n), \end{aligned}$$

where $(y_1,\dots ,y_n)$ denotes the coordinate system of the target.

3 Relative Version

The definition of non-properness set has an obvious generalization for a polynomial map $\varvec{f}:X\rightarrow Y$ between algebraic varieties X and Y defined over $\mathbb {K}$. We say $\varvec{f}$ is not proper at $\varvec{y}_0\in Y$ if there exist an arc $\varvec{x}(t):\mathbb {K}^*,0\rightarrow X$, so that

$$\begin{aligned} \lim _{t\rightarrow 0}\varvec{x}(t)\ \text {does not exist, and } \ \lim _{t\rightarrow 0}\varvec{f}(\varvec{x}(t))=\varvec{y}_0. \end{aligned}$$

We denote by $S_{\varvec{f}}$ the set of non-proper points of $\varvec{f}:X\rightarrow Y$.

Let $\varvec{f}=(f^1,\dots ,f^n):\mathbb {K}^m\rightarrow \mathbb {K}^n$ be a polynomial map with (1.3). Set $\varvec{f}'=(f^1,\dots ,f^{n-k})$ and $\varvec{f}''=(f^{n-k+1},\dots ,f^n)$. Set $\varvec{y}=(\varvec{y}',\varvec{y}'')$ and $\varvec{y}_0=(\varvec{y}'_0,\varvec{y}''_0)$. In this section, we describe a generalization of the discussion above to the map

$$\begin{aligned} \varvec{f}'|_X:X\rightarrow \mathbb {K}^{n-k}\quad \text {where}\ X=(\varvec{f}'')^{-1}(\varvec{y}_0''). \end{aligned}$$

Since $\varvec{f}|_X=(\varvec{f}'|_X,\varvec{y}''_0)$, we identify $\varvec{f}|_X$ with the map $\varvec{f}'|_X$ via the embedding $\mathbb {K}^{n-k}\times \{\varvec{y}''_0\}\subset \mathbb {K}^n$. This means that we identify $\mathbb {K}^{n-k}$ with $\mathbb {K}^{n-k}\times \{\varvec{y}''_0\}$, and we can identify $\varvec{f}^{\{1,\dots ,n-k\}{\setminus } J}$ with $\varvec{f}^{\{1,\dots ,n\}{\setminus } J}$. We call this map by $\varvec{f}^{J^c}$. For a face of $\varvec{\Delta }(\varvec{f})$, we define and . In the same way, we can identify with on the set . So denote its image by as . Set

and .

Under the notation and assumption above, we have the following:

Theorem 3.1

We assume that the nonsingular locus of X is dense in X and X has no component in $\{x_1\cdots x_n=0\}$. Then

(3.1)

If has dense nonsingular loci for , we have equalities in (3.1).

The assumption that X has no component in $\{x_1\cdots x_n=0\}$ comes from Remark 2.1. If there is an arc $\varvec{x}(t)$ in $X\cap \mathbb {K}^I$, $I\subsetneq \{1,\dots ,n\}$, with $\varvec{x}(t)\rightarrow \infty$, and $\varvec{f}(\varvec{x}(t))\rightarrow \varvec{y}_0$ ($t\rightarrow 0$), one can choose $\hat{\varvec{x}}(t)\in {\mathcal {A}}(\varvec{p})$ for some $\varvec{p}$ with $\hat{\varvec{x}}(t)\rightarrow \infty$, and $\varvec{f}(\hat{\varvec{x}}(t))\rightarrow \varvec{y}_0$ $(t\rightarrow 0)$. However, we do not know that $\hat{\varvec{x}}(t)$ can be chosen in X in such a case.

Proof of Theorem 3.1

First, for $\varvec{x}(t)\in {\mathcal {A}}(\varvec{p})$, we can write

$$\begin{aligned} f^j(\varvec{x}(t))&=t^{-d_j}({\hat{f}}^j_0+{\hat{f}}^j_1t+\cdots +{\hat{f}}^j_{d_j-1}t^{d_j-1}\\ {}&\quad +{\hat{f}}^j_{d_j}t^{d_j}+o(t^{d_j})) \quad (j=1,\dots ,n-k),\\ f^j(\varvec{x}(t))-y^j_0&=t^{-d_j}({\hat{f}}^j_0+{\hat{f}}^j_1t+\cdots +{\hat{f}}^j_{d_j-1}t^{d_j-1}\\ {}&\quad +{\hat{f}}^j_{d_j}t^{d_j}+o(t^{d_j})) \quad (j=n-k+1,\dots ,n), \end{aligned}$$

where $d_j=d_j(\varvec{p})$. We remark that $\varvec{y}\in S_{\varvec{f}|_X}$ if and only if there is an arc $\varvec{x}(t):\mathbb {K}^*,0\rightarrow X$, so that $\displaystyle {\lim \nolimits _{t\rightarrow 0}\varvec{x}(t)}=\infty$, and that $\displaystyle {\lim \nolimits _{t\rightarrow 0}\varvec{f}(\varvec{x}(t))}=\varvec{y}$. In a similar way to the proof of Lemma 2.2, we have

This implies that $f^j_{\gamma _j(\varvec{p})}(\varvec{v}_0)=0$ for , and $(f^j-y^j_0)_{\gamma _j(\varvec{p})}(\varvec{v}_0)=0$ for $j=n-k+1,\dots ,n$, which show the second inclusion.

By the discussion similar to the proof of Lemma 2.4, for any $a^j$ ($j\in J$), we can construct a formal power series $\varvec{v}(t)=(v^1(t),\dots ,v^n(t))$, such that

$$\begin{aligned} f^j(t^{p_1}v^1(t),\dots ,t^{p_n}v^n(t))&= a^j+o(t)\quad (j\in J), \end{aligned}$$

(3.2)

$$\begin{aligned} f^j(t^{p_1}v^1(t),\dots ,t^{p_n}v^n(t))&= y^j_0\quad (j=n-k+1,\dots ,n), \end{aligned}$$

(3.3)

where $\varvec{y}_0=(y_0^1,\dots ,y_0^n)$. Remark that we can reduce this system to polynomials by multiplying some power $t^l$. By the approximation theorem of Artin ( [1]), we can take a convergent power series $\varvec{v}(t)=(v^1(t),\dots ,v^n(t))$ which satisfies (3.2) and (3.3). This completes the proof of the first inclusion. $\square$

If X has a component $X_1$ in $\{x_1\cdots x_n=0\}$, we could proceed a similar computation for $\varvec{f}|_{X_1}$ which is a polynomial map with less number of variables and obtain that $S_{\varvec{f}|_{X_1}}\subset S_{\varvec{f}|_X}$.

4 Examples

Example 4.1

Let us start with the simplest example $\mathbb {K}^2\rightarrow \mathbb {K}^2$, $f(x,y)=(x,xy)$. For the assumption (1.3) we consider $\varvec{f}(x,y)=(c_1+x,c_2+xy)$ where $c_1$, $c_2$ are non-zero constants. It suffices to consider only 3 faces below thanks to Remark 2.3. Since

we have $S_{\varvec{f}}=\{(c_1,c_2+xy):(x,y)\in (\mathbb {K}^*)^2\}=\{c_1\}\times \mathbb {K}$.

Example 4.2

Consider the map $\varvec{f}(x,y)=(x^2y^2+xy+y+1,x^2y+y+x+1)$. Since

we have

$$\begin{aligned} S_{\varvec{f}}\!=\!\{x^2y^2+xy+1:xy+1=0\}\times \mathbb {K} \!=\!\{t^2+t+1:t+1=0\}\times \mathbb {K} \!=\!\{1\}\!\times \!\mathbb {K}. \end{aligned}$$

Example 4.3

Consider the map $\varvec{f}(x,y)=(xy+y+1,x^2y^2+y^2+xy+1)$, we have

Therefore, we have

$$\begin{aligned} S_{\varvec{f}}&=\{(X,Y):\exists (x,y)\, \in (\mathbb {K}^*)^2 \ \text {s.t.}\ (X,Y)=(xy+1,x^2y^2+xy+1)\}\\&=\{(X,Y):\exists t \, \in \mathbb {K}^* \ \text {s.t.}\ (X,Y)=(t+1,t^2+t+1)\}\\&=\{Y-X^2+X-1=0\}. \end{aligned}$$

Example 4.4

Consider the map $\varvec{f}:\mathbb {K}^3\rightarrow \mathbb {K}^3$ defined by

$$\begin{aligned} \varvec{f}(x,y,z)\!=\! (1+xy+xz, 1+axz+x(1-xy+xz), 1+bxy+x(1+xy+xz)), \end{aligned}$$

$a\ne 0$, $b\ne 0$. The Newton polyhedra look like

and we obtain the following data.

$\varvec{p}$		$(d_1,d_2,d_3)$
$(1,-1,-1)$	$(1+xy+xz,1+axz,1+bxy)$	(0, 0, 0)	$\emptyset$	$(\mathbb {K}^*)^3$
$(1,-1,-2)$	$(xz,axz,1+bxy+x^2z)$	(1, 1, 0)	$\{1,2\}$	$\emptyset$
$(1,-2,-1)$	$(xy,1-x^2y+axz,bxy)$	(1, 0, 1)	$\{1,3\}$	$\emptyset$
$(0,-1,-1)$	$(x(y+z),x(xz-xy+az),x(xz+xy+by))$	(1, 1, 1)	$\{1,2,3\}$	$\emptyset$
$(-1,1,1)$	$(1+xy+xz,x(1-xy+xz),x(1+xy+xz))$	(0, 1, 1)	$\{2,3\}$

We have

We conclude that .

5 Degenerate Case

We present several tricks to handle the case when (1.5) does not hold for some .

5.1 The First Trick

Let $\varvec{h}:\mathbb {K}^{m+k}\rightarrow \mathbb {K}^{n+k}$ be a polynomial map with (1.5). We assume that $h^{n+i}(x)=\varphi _1(x_1,\dots ,x_m)-x_{m+i}$ for $i=1,\dots ,k$. Let X be a subset of $\mathbb {K}^{m+k}$ defined by

$$\begin{aligned} x_{m+i}=\varphi _i(x_1,\dots ,x_m),\quad i=1,\dots ,k. \end{aligned}$$

The set X is isomorphic to $\mathbb {K}^{m}$ by the map defined by

$$\begin{aligned} (x_1,\dots ,x_m)\mapsto (x_1,\dots ,x_m,\varphi _1(x_1,\dots ,x_m),\dots ,\varphi _k(x_1,\dots ,x_m)). \end{aligned}$$

If $\varvec{f}(x_1,\dots ,x_m)=\varvec{h}(x_1,\dots ,x_m,\varphi _1(x_1,\dots ,x_m),\dots ,\varphi _k(x_1,\dots ,x_m))$, then we have

$$\begin{aligned} S_{\varvec{f}}=S_{\varvec{h}|_X} \end{aligned}$$

via the identification of $\mathbb {K}^n$ with $\mathbb {K}^n\times \{0\}$. When $\varvec{h}$ satisfies the required assumptions, one can use Theorem 3.1 to describe $S_{\varvec{f}}$, even though $\varvec{f}$ does not satisfy (1.5) for some .

Example 5.1

Let $\varvec{h}:\mathbb {K}^3\rightarrow \mathbb {K}^3$ be the map defined by

$$\begin{aligned} \varvec{h}(x,y,z)=(1+x+z^2,1+x^2+z^3,y^2-x^3-z). \end{aligned}$$

Let X be the subset of $\mathbb {K}^3$ defined by $z=y^2-x^3$. Setting $\varvec{f}(x,y)=\varvec{h}(x,y,y^2-x^3)$, we have

$$\begin{aligned} \varvec{f}(x,y)=(1+x+(y^2-x^3)^2,1+x^2+(y^2-x^3)^3). \end{aligned}$$

Applying Theorem 3.1, we conclude that $\varvec{h}$ is proper, and thus so is $\varvec{f}$.

5.2 The Second Trick

We show another trick, which we do not use higher dimension. If , , is not empty for some , we may have a chance to change (resp. ) in (1.6) by a smaller subset of (resp. by a supset of ).

Let $\varvec{f}:\mathbb {K}^m\rightarrow \mathbb {K}^n$ be a polynomial map with (1.3). Set $\varvec{f}^J_l=(f^j_l)_{j\in J}$ for $J\subset \{1,\dots ,n\}$ where $f^j_k$ is defined in (2.8). Remark that $f^j_0(x)=f^j_{\gamma _j}(x)$. We set

where , $\varvec{y}^{J'}=(y^{j'})_{j'\in J'}$,

$$\begin{aligned} J'=J'(\varvec{p})&=\{j\in \{1,\dots ,n\}:d_j(\varvec{p})=1\},\ \text { and }\\ J''=J''(\varvec{p})&=\{j\in \{1,\dots ,n\}:d_j(\varvec{p})\ge 2\}. \end{aligned}$$

Remark that $J=J'\cup J''$ because of (1.4). Under the notations and the assumptions above, we have

Theorem 5.2

If , $\varvec{p}\in \mathbb {Z}^n{\setminus }(\mathbb {Z}_\ge )^n$, is not empty for , then

Moreover, we conclude that $W'(\varvec{p})\subset S_{\varvec{f}}(\varvec{p})$, where

Proof

If $\varvec{y}\in S_{\varvec{f}}(\varvec{p})$, $\varvec{p}\in \mathbb {Z}^n{\setminus }(\mathbb {Z}_\ge )^n$, there exists $\varvec{x}(t)\in {\mathcal {A}}(\varvec{p})$ with $\varvec{f}(\varvec{x}(t))\rightarrow \varvec{y}$ ($t\rightarrow 0$). By (2.9) and (2.10), we have

$$\begin{aligned} 0&=f^j_0(\varvec{v}_0) \quad (j\in J))\\ y^{j'}&=f^{j'}_1(\varvec{v}_0)+(df^{j'})_{\varvec{v}_0}(\varvec{v}_1)\quad (j'\in J')\\ 0&=f^{j''}_1(\varvec{v}_0)+(df^{j''})_{\varvec{v}_0}(\varvec{v}_1)\quad (j''\in J''). \end{aligned}$$

Here, we use the expression in (2.1) and (2.2). This implies that

$$\begin{aligned} \mathop {\mathrm {rank}}{\begin{pmatrix} \partial _{x_i}\varvec{f}^{J'}_0(\varvec{x})\\ \partial _{x_i}\varvec{f}^{J''}_0(\varvec{x}) \end{pmatrix}} = \mathop {\mathrm {rank}}{\begin{pmatrix} \partial _{x_i}\varvec{f}^{J'}_0(\varvec{x})&{}\qquad \varvec{f}^{J'}_1(\varvec{x})-\varvec{y}^{J'} \\ \partial _{x_i}\varvec{f}^{J''}_0(\varvec{x})&{}\qquad \varvec{f}^{J''}_1(\varvec{x}) \end{pmatrix}}, \end{aligned}$$

and we conclude .

Now, we assume $\varvec{y}\in W'(\varvec{p})$. There exists , such that

$$\begin{aligned} \mathop {\mathrm {rank}}{(}\partial _{x_i}\varvec{f}_0^{J''}(\varvec{x}))_{i=1,\dots ,m}=\#J''. \end{aligned}$$

(5.1)

By the discussion in the second paragraph of the proof of Lemma 2.4, we can choose $\varvec{x}(t)$ to attain arbitrary ${\hat{f}}^{j''}_l$ ($j''\in J''$, $l\ge 1$) whenever $\varvec{x}\not \in \Sigma (\varvec{f}^{J''})$. This implies that $W'(\varvec{p})\subset S_{\varvec{f}}(\varvec{p})$. $\square$

We present a trick to describe where $S_{\varvec{f}}(\varvec{p})$ is the set defined by (2.4).

Assume that $\varvec{f}$ does not satisfy (1.5) for some face . We take a primitive $\varvec{p}\in \mathbb {Z}^n{\setminus }(\mathbb {Z}_\ge )^n$, so that . Here, $\varvec{p}$ is primitive means that the greatest common divisor of all components of $\varvec{p}$ is 1. Assume that there exists a rational map $\mathbb {K}^m\times \mathbb {K}^n\rightarrow \mathbb {K}^n$, $(\varvec{x},\varvec{z})\mapsto \Psi (\varvec{x},\varvec{z})$ with the following properties.

• There exist a certain rational map $\varvec{g}:\mathbb {K}^{m}\rightarrow \mathbb {K}^n$, possibly with points of indeterminacy, so that $\varvec{f}(\varvec{x})=\Psi (\varvec{x},\varvec{g}(\varvec{x}))$, and $\varvec{g}$ satisfies (1.5) for the face supported by $\varvec{p}$.

• The limit $\displaystyle {\lim _{t\rightarrow 0}\Psi (\varvec{x}(t),\varvec{z})}$ exists for $\varvec{x}(t)\in {\mathcal {A}}(\varvec{p})$. We assume that this limit depends on $\varvec{v}_0$, and denote the limit by $\Psi _{\varvec{v}_0}(\varvec{z})$, under the notation in (2.1) and(2.2),

• The limit $\displaystyle {\lim _{t\rightarrow 0}\varvec{g}(\varvec{x}(t))}$ for $\varvec{x}(t)\in {\mathcal {A}}(\varvec{p})$ exists.

Theorem 5.3

Under the notations and assumptions above, the set $S_{\varvec{f}}(\varvec{p})$, , is in the image of the following map:

Proof

For $\varvec{p}\in \mathbb {Z}^n{\setminus }(\mathbb {Z}_\ge )^n$

$\square$

Since , this may describe , as we see in the following example.

Example 5.4

Consider the map $\varvec{f}:\mathbb {K}^3\rightarrow \mathbb {K}^3$ defined by

$$\begin{aligned} \varvec{f}(x_1,x_2,x_3)=(1+x_1x_3+x_2x_3,1+x_1(1-x_1x_3+x_2x_3),1+x_2(1-x_1x_3+x_2x_3)). \end{aligned}$$

The map $\varvec{f}$ satisfies the condition (1.5) except the face , as we see in the following data.

$\varvec{p}$		$(d_1,d_2,d_3)$
$(1,1,-1)$	$(1+x_1x_3+x_2x_3,1,1)$	(0, 0, 0)	$\emptyset$
$(1,1,-2)$	$(x_3(x_1+x_2),1+x_1x_3(x_2-x_1),1+x_2x_3(x_2-x_1))$	(1, 0, 0)	$\{1\}$
$(-1,0,1)$	$(1+x_1x_3,x_1(1-x_1x_3),1+x_2-x_1x_2x_3)$	(0, 1, 0)	$\{2\}$
$(0,-1,1)$	$(1+x_2x_3,1+x_1+x_1x_2x_3,x_2(1+x_2x_3))$	(0, 0, 1)	$\{3\}$
$(-1,-1,1)$	$(1+x_1x_3+x_2x_3,x_1(1-x_1x_3+x_2x_3),x_2(1-x_1x_3+x_2x_3))$	(0, 1, 1)	$\{2,3\}$

We easily see that

We also have that

$$\begin{aligned} W(-1,-1,1)&= \Biggl \{ \begin{matrix} (y_1,y_2,y_3) : \exists (x_1,x_2,x_3) \ 1-x_1x_3+x_2x_3=0, y_1=1+x_1x_3+x_2x_3, \\ \mathop {\mathrm {rank}}{ \begin{pmatrix} 1-2x_1x_3+x_2x_3&{}x_1x_3&{}x_1(x_2-x_1)&{}1-y_2\\ x_2x_3&{}1-x_1x_3+2x_2x_3&{}x_2(x_2-x_1)&{}1-y_3 \end{pmatrix}}=1 \end{matrix} \Biggr \}\\&=\{y_1(y_2-y_3)-2y_2+2=0\}. \end{aligned}$$

We will show that this coincides with , considering the rational map

$$\begin{aligned} \Phi :\mathbb {K}^3\times \mathbb {K}^3\rightarrow \mathbb {K}^3, \ (\varvec{x},\varvec{y})\mapsto \varvec{z} =(z_1,z_2,z_3)=\Bigl (y_1,y_2,-\frac{x_2}{x_1}y_2+y_3\Bigr ). \end{aligned}$$

Remark that $\varvec{g}(\varvec{x})=\Phi (\varvec{x},\varvec{f}(\varvec{x}))$ defines the map

$$\begin{aligned} \mathbb {K}^*\!\times \!\mathbb {K}^2\!\rightarrow \!\mathbb {K}^3, \ (x_1,x_2,x_3)\!\mapsto \! (1+x_1x_3+x_2x_3, 1+x_1(1-x_1x_3+x_2x_3), 1\!-\!\tfrac{x_2}{x_1}), \end{aligned}$$

and obtain the following data:

$\varvec{p}$		$(d_1,d_2,d_3)$
$(-1,-1,1)$	$(1+x_1x_2+x_1x_3,x_1(1-x_1x_2+x_2x_3),1-\frac{x_2}{x_1})$	(0, 1, 0)	$\{2\}$

The Newton polyhedra look like

As in the proof of Theorem 1.8, we conclude that

(5.2)

For $(x_1,x_2,x_3)$ in (5.2), we have

$$\begin{aligned} -\frac{x_2}{x_1}=\Bigl (1-\frac{x_2}{x_1}\Bigr )-1=z_3-1=\frac{2}{z_1}-1=\frac{2}{y_1}-1.\ \end{aligned}$$

Setting

$$\begin{aligned} \Psi :\mathbb {K}^3\times \mathbb {K}^3\rightarrow \mathbb {K}^3, \ (\varvec{x},\varvec{y})\mapsto \varvec{z}=\Bigl (y_1,y_2,\frac{x_2}{x_1}y_2+y_3\Bigr ), \end{aligned}$$

we have $\varvec{y}=\Psi (\varvec{x},\Phi (\varvec{x},\varvec{y}))$, and thus, $\varvec{f}(\varvec{x})=\Psi (\varvec{x},\varvec{g}(\varvec{x}))$. The set

and we obtain .

We thus obtain that $S_{\varvec{f}}=\{y_1+y_3=2\}\cup \{y_1y_2-y_1y_3-2y_2+2=0\}$.

Example 5.5

Let us use the first trick to handle Example 5.4. We consider the map $\varvec{h}:\mathbb {K}^4\rightarrow \mathbb {K}^4$ defined by

$$\begin{aligned} \varvec{h}(x_1,x_2,x_3,x_4)=(1+x_1x_3+x_2x_3, 1+x_1x_4, 1+x_2x_4, 1-x_1x_3+x_2x_3-x_4), \end{aligned}$$

since $\varvec{f}(x_1,x_2,x_3)=\varvec{h}(x_1,x_2,x_3,1-x_1x_3+x_2x_3)$. We analyze $\varvec{h}(\varvec{x}(t))$ for $\varvec{x}(t)\in {\mathcal {A}}(-1,-1,1,1)$, because $\varvec{p}=(-1,-1,1,k)$, $k\ge 1$, supports a three-dimensional face of $\varvec{\Delta }(\varvec{g})$ if and only if $k=1$. Setting

$$\begin{aligned} \varvec{x}(t)=( t^{-1}(x^1_0+x^1_1t+\cdots ), t^{-1}(x^2_0+x^1_1t+\cdots ), t(x^3_0+x^3_1t+\cdots ), t(x^4_0+x^4_1t+\cdots )), \end{aligned}$$

we have

$$\begin{aligned} \varvec{h}(\varvec{x}(t))=(1+x^1_0x^3_0+x^2_0x^3_0, 1+x^1_0x^4_0,1+x^2_0x^4_0, 1-x^1_0x^3_0-x^2_0x^3_0-x_0^4)+o(t). \end{aligned}$$

Assuming $\varvec{x}(t)$ is in $X=\{\varvec{x}\in \mathbb {K}^4:1-x_1x_3+x_2x_3=x_4\}$, we obtain that $1-x^1_0x^3_0-x^2_0x^3_0=x_0^4$. Under this condition, we eliminate $x^1_0$, $x^2_0$, $x^3_0$ from the system

$$\begin{aligned} (1+x^1_0x^3_0+x^2_0x^3_0, 1+x^1_0x^4_0,1+x^2_0x^4_0, 1-x^1_0x^3_0-x^2_0x^3_0)=(y_1,y_2,y_3,0), \end{aligned}$$

we conclude that . We thus obtain that $S_{\varvec{f}}=\{y_1+y_3=2\}\cup \{y_1y_2-y_1y_3-2y_2+2=0\}$.