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COUNTING ISOLATED POINTS OUTSIDE THE IMAGE OF A POLYNOMIAL MAP

BOULOS EL HILANY

ABSTRACT. We consider a generic family of polynomial maps f := (f1, f2) : C2 → C2 with

given supports of polynomials, and degree deg f := max(deg f1,deg f2). We show that the (non-)

properness of maps f in this family depends uniquely on the pair of supports and that the set of

isolated points in C2 \ f(C2) has a size of at most 6 deg f . This improves an existing upper bound

(deg f − 1)2 proven by Jelonek. Moreover, for each n ∈ N, we construct a dominant map f above,

with deg f = 2n+ 2, and having 2n isolated points in C2 \ f(C2).Our proofs are constructive and can be adapted to a method for computing isolated missing

points of f . As a byproduct, we describe those points in terms of singularities of the bifurcation set

of f .

1. INTRODUCTION

A polynomial map f := (f1, . . . , fn) : Cn → Cn is said to be dominant if it sends Cn onto adense subset of Cn. Little is known about the set Cn \ f(Cn) of missing points for a dominantmap f . The scarcity of tools for describing the topology of a map leaves several good questionsunanswered. Notably, those concerning the geometrical aspects of the set of missing points. Forexample, how many equi-dimensional components can this set have? What kind of singularitiescan appear?

The set of missing points measures how far a map is from being surjective. Characterizingsuch topological properties is a classical problem with numerous applications, especially sincepolynomial maps appear in algebraic statistics [3], computer vision [14], robotics [20], opti-mization [21] and other areas.

A polynomial map usually constitutes a model for above applications in which the input datais a point p in Cn, and the solution to the problem is the output f−1(p). Consequently, one avoidsthe vicinity of the set of missing points in order to guarantee a manageable output. This bringsforward the following problem: What would be an effective method to compute the set of missingpoints and to find interesting examples?

This work is a further step (c.f. [5]) towards understanding those maps whose topology pos-sesses an extremal structure. The particular motivation here lies in the two-dimensional setupwhere the set of missing points consists of zero- and one-dimensional components: We are in-terested in counting the isolated points in C2 \ f(C2). The proofs in this paper are constructive,can be generalized and aimed towards the computational aspects of the set of missing points.

1.1. Main results. The preimages under f around generic points in Cn form a locally analyticalcovering of constant degree µf . In other words, µf is the number of preimages of a map overa generic point in Cn. Such is not the case around the set of missing points of f . These arecontained in the set Jf of points in Cn at which f is non-proper (see Definition 2.2). Jelonek

Part of this work was supported by the Austrian Science Fund (FWF) P33003.

MSC: Primary 14E05, Secondary: 12D10, 52B20.

Key words: Polynomial maps on the plane, the Jelonek set, Newton polytopes.

1

2

showed [15] that Jf (also referred to as the Jelonek set) is an algebraic hypersurface ruled byregular rational curves, for which he provided upper bounds on the degree.

Above findings, together with other results [17], play an important role in describing the setof missing points. In fact, Jelonek later showed in [16] that for planar maps f := (f1, f2) :

C2 → C2, the set I∅f of isolated points in C2 \ f(C2) is either empty, or has a size of at most

deg f1 deg f2 − µf − 1. Moreover, if deg f := max(deg f1,deg f2), then |I∅f | ≤ (deg f − 1)2. Our

main result is the improvement (from quadratic to linear) of these upper bounds for a large classof maps.

The support of a polynomial is the set of exponent vectors, in the lattice of integers, of mono-mial terms appearing in the polynomial with non-zero coefficients. The supports of polynomialsin a planar map f form a pair of finite subsets in N2, also called the support of f . We use thepolyhedral properties of the latter to study the geometry of the Jelonek set and of C2 \ f(C2). Inorder to relate these two notions, a genericity assumption on f must be assumed.

Consider the three curves in C2, {f1 = 0}, {f2 = 0} and the set C(f) of critical points of f .We say that f is generically non-proper (see Definition 2.1) if the number of isolated points in theintersection of any of these two curves is maximal with respect to the support of f . This meansthat no other map with the same support has more of such isolated points.

Theorem 1.1. Let f : C2 → C2 be a generically non-proper map. Then, the number of isolatedpoints in C2 \ f(C2) cannot exceed any of the two values 6 deg f and

(1.1)deg f1 deg f2µf (µf − 1)

+ 2(deg f1 + deg f2).

We will emphasize the importance of this Theorem from two standpoints: Our upper boundshold true for very large classes of polynomial maps (Corollary 1.3), and are close to beingoptimal (Theorem 1.4).

We use C[A] to denote the space of all polynomial maps C2 → C2 having a support containedin a pair A of finite subsets of N2. This means that coordinates of any point C[A] give a list ofcoefficients that identifies a planar map (see Example 4.4).

Theorem 1.2. Let A be a pair of finite subsets of N2. Then, all maps in C[A] that are either proper,or generically non-proper, form a dense subset.

Since proper maps have no missing points, we obtain the following immediate consequence.

Corollary 1.3. Let A be as in Theorem 1.2. Then, there exists a dense subset of maps in C[A] forwhich the bounds in Theorem 1.1 hold true.

Our next result how far bounds presented in Theorem 1.1 are from being sharp.

Theorem 1.4. There exists a pair A of finite subsets in N2, such that all generically non-properpolynomial maps in C[A] have degree 2n + 2, and have 2n missing isolated points.

With generic enough coefficients, a well-chosen collection of supports can be sufficient to con-struct a map with an interesting topology [5]. Theorem 1.4 is another instance of this approach.

All the main results in this paper are obtained by carefully analyzing the pair of supports of apolynomial map. As a byproduct, we discover a universality result (See Lemmas 2.6 and 2.7).

Theorem 1.5. Let k be a positive integer. Then, there exists a polynomial map f : C2 → C2 such

that µf = k + 1 and I∅f 6= ∅ if and only if k ≥ 1.

3

Another byproduct of our study is a set of examples suggesting that the missing points of amap can be described uniquely from the geometry of the Jelonek set and the discriminant of amap (that is, the set f(C(f))). The below map C2

u,v → C2s,t, is taken from the proof of Lemma 2.7

for k = 2, has (0, 1) as the only missing point, and its Jelonek set has a singularity at (0, 1) formedby a horizontal line and a node of the curve {1− 2t+ t2 − s2 + s3 = 0}:

(1.2) (u, v) 7→ (1− u2v2 + u2v3, 1 + uv − u3v3 + 2u3v4).

Missing isolated points can be degenerate intersections among components of the Jelonek setand the discriminant: The map

(1.3) (u, v) 7→ ((1− uv)2, 1 + v + uv(1− uv)2)

has (0, 1) as the only missing point. Its Jelonek set is {1 + 2s− 2t+ t2 − s2 − 2st+ s3 = 0}, havinga cusp at (0, 1), that belongs to the discriminant {s = 0}.

Such degenerate intersections/singularities can be milder, and still correspond to missingisolated points: The map

(1.4) (u, v) 7→ (uv, v2 − uv3 + 2v − uv2 + uv)

from Section 2 has {(1, 1), (2, 2)} as the set of missing isolated points. Its Jelonek set {s− t = 0} ∪{s−1 = 0} has a singularity (1, 1), and intersects tangentially the subset {−4−4t+8s−5s2+4st =0} of the discriminant at (2, 2).

Some of those phenomena will be explained in Sections 5 and 7. Others will be thoroughlyexamined and generalized in a future work.

1.2. On attaining the upper bounds. One deduces from the definitions that the set I∅f of iso-

lated missing points is contained in the Jelonek set Jf of f : C2 → C2. The straightforward

approach, considered by Jelonek in [16] to approximate I∅f is to study Jf .

Although we follow a similar strategy, our methods are more refined than those in [16]; wetake into account the combinatorics of the support A of f .

A more general form of our approach has proven to be powerful in describing mixed discrim-inants [13, 6, 2], resultants [24] and the bifurcation set BF of a polynomial map F : Cn →Cp [22, 26, 7]. This is the complement to the maximal open set S ⊂ Cp, such that the restrictionof F to the preimage F−1(S) is a locally trivial fibration.

In our setting, the Jelonek set forms half of the bifurcation set Bf , for it describes the atypicalbehavior of f at infinity. This interpretation allows one to identify a generic p ∈ Jf with a pair

of subsets in the boundary of A, called faces[4, 7] (see also Section 3.1). Points in I∅f are not

generic in Jf , hence computing them requires a thorough examination of f at the faces of A.

Our analysis is based on the following observation in Section 5.2: The size of I∅f is propor-

tional to the integer length (Section 5.1.1) of some faces of A. This in turn is proportional todeg f (Lemma 5.2), and to the topological degree µf (Lemma 4.5). However, the latter quantity

is disproportional to the size of I∅f .

This delicate relation among deg f , µf and |I∅f | is summarized in Theorem 1.1. Only a small

class of supports A have long faces and support maps with small topological degrees. Thesewere used to prove Theorem 1.4.

After concluding this section with future directions, we organize the rest of the paper asfollows.

In Section 2 we prove Theorems 1.4 and 1.5. We also divide Theorem 1.1 into three Proposi-tions. These will be proven in Sections 5 – 7.

4

In Section 3, we introduce the main notations and preliminary results (Proposition 3.6). Wedefine relevant faces in Section 3 (Definition 4.2). We show that these play an important role incomputing the Jelonek set and in detecting isolated missing points for generically non-propermaps. We prove Theorem 1.2 at the end of that Section.

1.3. Future directions. Inequality (1.1) can be sharpened by refining the analysis made inSections 5 - 7. Such improvement is not included here since it makes the proofs too cumbersomeand would not add significant content to the goals of this work.

Regarding the validity of upper bounds in Theorem 1.1 for maps that are not generically non-proper, the problem remains open. However, the analysis made here is beneficial to identify

potential candidates for maps f : C2 → C2 whose size of I∅f is proportional to deg2 f . This

direction is the subject of our future work.Methods presented here can be generalized for higher dimensions and for rational maps.

This is because one can still describe the Jelonek set in these settings using faces of tuples ofpolytopes [4, 7].

Another equally important future aim is to tackle this problem for the case where the map andthe spaces are real. This is more pertinent to applications mentioned in the beginning. Relevantto this direction are ample works of Fernando, Gamboa and Ueno [8, 9, 10, 11, 12] on describingthe possible images that Rn can take under a polynomial map. In fact, Fernando showed thatthe complement of any finite set S in Rn is the image of a polynomial map Rm → Rn for somem ≥ n [10]. The polynomial map constructed for proving this highly non-trivial result restrictsto a map similar to the one for Theorem 1.4. Results in the same vain exist for when the aboveset S is a convex polyhedron [11], or the complement of one [12].

Concerning the computation of the Jelonek set for real maps, Jelonek described its geome-try [19], and gave a method for computing it for some classes of maps R2 → R2 [18]. Stasica-Valette provided a different technique for computing the Jelonek set of the latter planar mapsthat have finite fibers [25].

Acknowledgements. The author is grateful to Zbigniew Jelonek for introducing him to theproblem and for his helpful remarks. The author would also like to thank Elias Tsigaridas forfruitful discussions. The author thanks the anonymous referee for their valuable suggestions ona earlier versions of the manuscript, for pointing out mistakes therein and mentioning works ofFernando, Gamboa and Ueno on the subject. The author is grateful to the Mathematical Instituteof the Polish Academy of Sciences in Warsaw for their financial support and hospitality.

2. PROOF OF THEOREMS 1.1, 1.4 AND 1.5

Consider any pair of subsets A := (A1, A2) in C2. Bernstein [1] has shown that there exists apositive integer V (A) (Section 3.1.2) such that any pair of polynomials supported on A cannothave more than V (A) isolated solutions in (C∗)2. Moreover, equality holds if those polynomialsare generic.

Consider a dominant polynomial map f := (f1, f2) : C2 → C2 having A as support (Sec-tion 3.2). Let C(f) denote the set of points in C2 at which the Jacobian matrix of f is singular,and let Σ denote the support of the determinant | Jac f | of this Jacobian.

5

Definition 2.1. We say that f is generically non-proper if it is dominant, non-proper, satisfiesf(0, 0) ∈ (C∗)2 and systems

(2.1)f1 = f2 = 0,f1 = | Jac f | = 0,f2 = | Jac f | = 0,

have respectively V (A), V (A1,Σ) and V (A2,Σ) isolated solutions in (C∗)2. 7

2.1. Proof of Theorem 1.1. We start with a definition [15, 17].

Definition 2.2. Given two affine varieties, X and Y , and a map F : X → Y , we say that F isnon-proper at a point y ∈ Y , if there is no neighborhood U ⊂ Y of y, such that the preimageF−1(U ) is compact, where U is the Euclidean closure of U . In other words, F is non-proper at yif there is a sequence of points {xk} in X such that ‖xk‖ → +∞ and f(xk) → y. The Jelonek setof F , JF , consists of all points y ∈ Y at which F is non-proper. 7

Let Kf ⊂ Jf denote the Jelonek set of the restricted map

f|C(f) : C(f) → f(C(f)).

For example, Kf coincides with {(0, 1)} in the map (1.3) and with {(2, 2)} for the map in (1.4).

We prove the following result in Section 7. Recall that I∅f denotes the set of isolated points in

C2 \ f(C2).

Proposition 2.3. Let f : C2 → C2 be a generically non-proper map. Then, we have

|I∅f ∩ Kf | ≤ deg f1 + deg f2.

Let C+f be cross in C2 centered at f(0, 0). The next result will be proven in Section 6.

Proposition 2.4. Let f : C2 → C2 be a generically non-proper map. Then, we have

|I∅f ∩ C+

f \ Kf | ≤ deg f1 + deg f2.

Write C0f for the complement of C+

f in C2. The following result will be proven in Section 5.

Proposition 2.5. Let f : C2 → C2 be a generically non-proper map. Then, we have

(2.2) |I∅f ∩ C◦

f \ Kf | ≤3 deg f1 · deg f24µf (µf − 1)

,

(2.3) |I∅f ∩ C◦

f \ Kf | ≤ 2max(deg f1,deg f2).

The bound in Theorem 1.1 is obtained by observing that I∅f can be written as the disjoint

union of the three sets appearing in the above three propositions.

2.2. Proof of Theorem 1.4. Let P,Q be two univariate polynomials such that: degP = degQ =n, gcd(P,Q) = 1 and P (0) ·Q(0) 6= 0. Define f to be the map

(u, v) 7→ (uv, v2 · P (uv) + v ·Q(uv) + uv).

We will show that all maps defined this way form the family proving Theorem 1.4.Let y ∈ (C∗)2 and let (u, v) ∈ C2 be a solution to f − y = 0, that is,

f1(u, v) − y1 = 0

f2(u, v) − y2 = 0.

6

Then, we have uv = y1 6= 0 and

(2.4) v2P (y1) + vQ(y1) + y1 − y2 = 0.

If P (y1) = 0 (resp. Q(y1) = 0), then Q(y1) 6= 0 (resp. P (y1) 6= 0). Therefore, PQ = 0 ⇒ y1 6= y2since otherwise v = 0, a contradiction to uv = y1 6= 0. This shows that for any y ∈ (C∗)2, suchthat PQ = 0, we have f−1(y) = ∅ ⇔ y1 = y2. Otherwise, if PQ 6= 0, then (2.4) has a non-zerosolution v and u = v/y1 6= 0. This shows that y ∈ (C∗)2, y1 6= y2 ⇒ f−1(y) 6= ∅.

We conclude from this analysis that

(C∗)2 \ f(C2) =⋃

P (a)·Q(a)=0{(a, a)}.

The size of this set is equal to degP + degQ = 2n. Furthermore, one can easily check thatf−1(y) 6= ∅ for any y ∈ C2 \ (C∗)2.

Up to now, we have constructed a map having degree 2n + 2 and #I∅f = 2n. Obviously, all

polynomials P,Q above form a dense family of pairs of univariate polynomials in Cn[uv]2. This

shows that the set of polynomial maps f constructed this way forms a dense subset in C[A].Finally, since f is non-proper, Theorem 1.2 yields the proof.

2.3. Proof of Theorem 1.5. We split the proof into two statements. Recall that µf is the number

of preimages of a map over a generic point in C2

Lemma 2.6. Let f : C2 → C2 be a dominant map such that µf = 1. Then, we have I∅f = ∅.

Proof. For any y ∈ Jf , the set f−1(y) is either empty or has positive dimension. Recall that

Jf is a curve in C2 [15]. Then, it is enough to show that the preimage under f has a positivedimension over finitely-many points.

If the preimage of some point y has positive dimension, then it forms a union of distinctirreducible components. Therefore, the set f−1(y) shares a component C with the set C(f) ofcritical points of f .

This set C(f) has finitely-many components C arising this way. Each of them is mapped to apoint under f . �

Lemma 2.7. Let k be a positive integer. Then, there exists a polynomial map f : C2 → C2 such

that µf = k + 1 and I∅f 6= ∅.

Proof. Let P be a univariate complex polynomial of degree k such that P (0) 6= 0. We show that

(u, v) 7→(P (uv) + ukvk+1, 1 + uvP (uv) + 2uk+1vk+2

).

satisfies the claims of the Lemma.To compute µf , we fix an arbitrary y ∈ C2 and solve f − y = 0 for u, v. Make the formal

substitution s = uv, t = v, and eliminate t, we obtain

(2.5) s(P (s)− 2y1) + y2 − 1 = 0.

There are k+1 distinct values s ∈ C∗ satisfying (2.5) for a generic choice of P and y. This makesfor k + 1 distinct couples (uv, v) ∈ (C∗)2. Therefore, we have µf = k + 1.

One can check that if P (0) 6= 0, then f − y = 0 has no solutions only when y = (0, 1). �

7

3. FACES OF FINITE SETS AND RESTRICTED SYSTEMS

In Section 3.1, we consider pairs of finite sets in a two-dimensional lattice. We introducethe notions of a face, independence, Minkowski sum and mixed volumes. In Section 3.2, werecall Bernstein’s results in [1], relating those combinatorial notions to the number of solutionsto polynomial equations.

In Section 3.3, we introduce a monomial change of variables that depends on the faces of pairsof supports. We then illustrate its usefulness in keeping track of solutions outside the complextorus to a polynomial system of equations (Proposition 3.6). Such a description is important incomputing isolated missing points of dominant polynomial maps, and will be used in the stepsthat follows in proving the main upper bounds.

3.1. Pairs of finite sets, faces and mixed volume. A polytope ∆ in R2 is a bounded intersectionof closed half-planes of the form {α0 + α1X1 + α2X2 ≥ 0}, for some α0, α1, α2 ∈ R. These arecalled the supporting half-spaces of ∆, and their boundary intersects ∂∆ at a connected set of ∆called face.

Any face F of ∆ minimizes a function α∗ : ∆ → R, (X1,X2) 7→ α0 + α1X1 + α2X2. In thiscase, we say that α = (α1, α2) supports F .

We formulate this terminology in terms of finite sets in Z2.

3.1.1. Faces of finite sets. A face φ of a finite subset Σ ⊂ Z2 is the intersection of Σ with a face ofits convex hull. The supporting vector of φ ⊂ Σ is the supporting vector of the convex hull of φ.

The following notation is taken from [7, Section 2.2]. Recall that the Minkowski sum A + Bof two subsets A,B ⊂ Rn is the set {a + b | a ∈ A, b ∈ B}. In Figure 1, the Minkowski sum ofthe two subsets on the left is the set on the right.

A pair of finite sets in Z2 refers to a pair A := (A1, A2), in which each member Ai (i = 1, 2) isa finite set of Z2. A face Γ of a pair A, denoted by Γ ≺ A, is a pair Γ := (Γ1,Γ2) such that Γi is aface of Ai, i = 1, 2 and Γ1 + Γ2 is a face of A1 +A2. The pair of sets on the left of Figure 1 havethe pair (γ, δ) as a face.

The dimension of a face, denoted dimΓ, is the dimension of the convex hull of Γ1 + Γ2. InFigure 1, both (γ, δ) and (a,�) have dimension one, (⋆,�) has dimension zero.

Since a supporting vector α ∈ Q2 of Γ1 + Γ2 is also one for the face Γi ⊂ Ai (i = 1, 2) (thefunction α∗ is linear), we say that α supports Γ ≺ A . In Figure 1, (−2, 1) supports (γ, δ) and(3,−1) supports (a,�).

We use the convention that a face of a pair is the pair itself if and only if the vector (0, 0)supports it. We refere to such face as trivial.

3.1.2. Mixed volume of finite sets. Given a convex set ∆ ⊂ R2, let Vol(∆) denote its fixed,translation-invariant Lebesgue measure endowed on R2. Recall that Minkowski’s mixed volumeis the unique real-valued multi-linear (with respect to the Minkowski sum) function of twoconvex sets ∆1,∆2 ⊂ R2, whose value at two copies of ∆ equals 2Vol(∆). It is known thatV (∆1,∆2) can be expressed as

Vol(∆1 +∆2)−Vol(∆1)−Vol(∆2).

One can check that if ∆1 +∆2 is a line segment, or one of ∆i is a point, then V (∆1,∆2) = 0.The other direction holds true as well, and is a particular case of Minkowski’s Theorem for thehigher-dimensional mixed volume. We say that (∆1,∆2) is independent if V (∆1,∆2) 6= 0, or,equivalently, if dim(

∑i∈I ∆i) ≥ |I| for all I ⊂ 1, 2. A dependent couple above is one that is not

independent.

8

(−2, 1)

(3,−1)

⋆

N

a

b

γ

�

�

c

d

δ

γ + δ

FIGURE 1. (L): Pair (A1, A2) of finite subsets. (R): Their Minkowski sum.

A couple A of finite sets in Z2 is said to be independent if the convex hulls of its members forman independent pair. Similarly, the notation V (A) will refer to the mixed volume of the pairconsisting of the convex hulls corresponding to A’s members.

3.2. The number of roots to a system of equations. A bivariate complex polynomial P ∈C[x1, x2] is written as a finite linear combination

∑

w∈N2

cwxw,

of monomials xw := xw1

1 xw2

2 , where w := (w1, w2) and cw ∈ C. The support suppP of P is the

set {w ∈ Z2 | cw 6= 0}.A pair ϕ := (ϕ1, ϕ2) of polynomials in C[x1, x2] identify a polynomial map ϕ : C2 → C2. The

support of ϕ is the pair (suppϕ1, suppϕ2). The polynomial system

(3.1)ϕ1(x1, x2) = 0,ϕ2(x1, x2) = 0,

is denoted by ϕ = 0 and its set of solutions (x1, x2) in C2 is denoted by V(ϕ). We use V◦(ϕ)to denote the subset V(ϕ) ∩ (C∗)2. We will abuse notations by writing #V(ϕ) and #V◦(ϕ) inreference to the number of isolated solutions to (3.1) counted with multiplicities in C2 and (C∗)2

respectively.A consequence of Bernstein’s Theorem A is

#V◦(ϕ) ≤ V (suppϕ).

In particular, the set of isolated solutions is empty if the support is dependent.Another result of Bernstein will be used throughout this paper goes as follows.Given a face Γ ≺ suppϕ, the restriction of the polynomial ϕi (i = 1, 2) to monomial terms

cwxw satisfying w ∈ Γi, is denoted by ϕi,Γ. We write ϕΓ to denote the pair (ϕ1,Γ, ϕ2,Γ).

Theorem 3.1 (Theorem B of [1] for n = 2). Let ϕ be a pair of bivariate polynomials. Then, allsolutions to ϕ = 0 are isolated and

#V◦(ϕ) = V (suppϕ)

if and only if for any non-trivial face Γ ≺ suppϕ, we have #V◦(ϕΓ) = ∅.

9

3.3. Toric change of variables. Following [1], a unimodular toric change of variables on x =(x1, x2) is a map (C∗)2 → (C∗)2, x 7→ z = (z1, z2), such that

x1 = zu11

1 zu21

2 , x2 = zu12

1 zu2

2 ,

where U =( u11 u1,2u2,1 u2,2

)∈ SL(2,Z). This transformation induces an isomorphism between polyno-

mials, which by abuse of notation, we also denote by U ; it is

U : C[x1, x2] → C[z±11 , z±1

2 ],

taking the monomial xa to xUa, where a := (a1, a2).Hence, for any polynomial P ∈ C[x1, x2], we have

supp(UP ) = U(suppP ),

and the zero locus V◦(P ) is isomorphic to V◦(UP ).

Remark 3.2. For Uϕ := (Uϕ1, Uϕ2), we have

#V◦(ϕ) = #V◦(Uϕ).

Example 3.3. Let ϕ be the pair of polynomials(1 + 2uv2 + 3u2v4 + 4uv3 + 5u2v5, −1− 2u− 3u2v2

)

supported on the pair of sets in Figure 1 on the left. If U =(

3 −1−2 1

), then Uϕ is written as

(1 + 2s + 3s2 + 4s2t+ 5s3t, −1− 2s−1t−2 − 3t−1

)

7

A transformed pair Uϕ may have monomials with negative exponents. We can transform itto a pair of polynomials by multiplying by a suitable monomial. We denote by Uϕ the pair(Uxr1ϕ1, Ux

r2ϕ2), where the coordinates of r1, r2 ∈ N2 are the minimal ones that allow to cleardenominators. In Example 3.3, we have r1 = (0, 0) and r2 = (1, 2).

3.3.1. Base-change using faces. We will introduce an above toric change of coordinates U ∈SL(2,Z) that will help us later deduce a useful description of points in the preimage, under amap, that escape to infinity. It turns out that the best choice for U is one whose entries dependon some faces of supports.

Let A = (A1, A2) be a pair of finite sets in N2 and let Γ = (Γ1,Γ2) be a face of A (seeSection 3.1.1) such that dimΓ = 1. Our goal is to construct a pair of vectors e := (e1, e2) asfollows.

Let a ∈ N2 be the closest point in Γ1 + Γ2 to (0, 0). Set e1 to be the primitive integer vectorspanning Γ1 + Γ2 away from a (i.e. the first coordinate of e1 is positive). In the Example ofFigure 1, we have γ + δ is the set of black dots to the right, a = (1, 0) and e1 is the red vector(1, 2).

We choose e2 such that

(i) e is the basis of the lattice Z2.(ii) The Minkowski sum A1+A2+{−a} is contained in the cone {b1e1+ b2e2 | b1, b2 ∈ R≥0}.

Or in other words, the basis e spans positively the lattice of points A1 +A2 − a.

This represents a linear transformation U : Z2 → Z2 taking e to the canonical basis of Z2. In theExample of Figure 1, e2 is the red vector (−1,−1).

We consider the matrix T , where

E =(e11 e12e21 e22

)=

(v1 v2w1 w2

)∈ SL(2,Z).

10

Then, T corresponds to the following change of variables z = xE(1

1). Thus the transformation U

that we are looking for, is the inverse of this map, that is

x1 7→ zw2/D1 z

−v2/D2 , x2 7→ z

−w1/D1 z

v1/D2 ,

where D = det(E) = ±1 is the determinant of T .From the properties (i) and (ii) of T ∈ SL(2,Z), we deduce that E, and thus also U , depends

on Γ and A. We denote the subset of SL(2,Z) of all such U by TrsfΓ(Z2).

We also have the following immediate consequence.

Lemma 3.4. Let α ∈ Z2 be the primitive integer vector supporting the Γ ≺ A. Then, for anyU ∈ TrsfΓ(Z

2), UΓ is a face of UA. Moreover, the vector U · αTr = (0, 1) and it supports UΓ.

Remark 3.5. For any ϕ ∈ C[x1, x2]2, where A = supp(ϕ), the following hold:

• For any U ∈ TrsfΓ(Z2), the system U

⋆fΓ = 0 is univariate and U

⋆f = 0 is bivariate.

• U⋆fΓ = 0 has a solution ρ ∈ C∗ iff U

⋆f = 0 has a solution (ρ, 0) ∈ C∗ × {0} iff fΓ = 0 has

a solution in (C∗)2.

Proposition 3.6. Let ϕ be a pair of bivariate polynomials. Let S denote the set of non-trivial faces

Γ ≺ suppϕ for which there exists a matrix U ∈ TrsfΓ(Z2), such that U

⋆ϕ = 0 has mΓ > 0 solutions

in C∗ × {0}, counted with multiplicities. Then, we have

(3.2) #V◦(ϕ) = V (suppϕ)−∑

Γ∈S

mΓ.

Proof. Let A denote suppϕ. We proceed using similar arguments as in the proof of [1, TheoremB]. Consider the parametrized polynomial system

(3.3) ϕ+ tψ = 0,

where ψ := (ψ1, ψ2) ∈ C[A] and t ∈]0, 1[. Using [1, Theorem A], one can choose the pair ψ sothat (3.3) has V (A) parametrized isolated solutions x(t) ∈ (C∗)2. The result follows by provingthe claim: There exists a bijection x(t) 7→ ρ between the set of solutions to (3.3) in (C∗)2, escaping(C∗)2 as t → 0, and the multiset defined as the union of all solutions in C∗ × 0, counted with

multiplicities, to systems U⋆ϕ = 0 above.

A solution x(t) to (3.3) can be presented as a function in t, where its j-th coordinate (j = 1, 2)is written as the Puiseux series

ajtαj + higher order terms in t,

where aj ∈ C∗ and αj ∈ Q.If we plug x(t) into ϕi + tψi and set the coefficient of the smallest power p of t equal to zero,

we obtain ϕi,Γ(a) = 0 for some Γ ≺ A. Indeed, the value p is the minimum min(〈α, q〉 | q ∈ Ai),which is reached only for points q in the face of Ai supported by the vector α := (α1, α2).Therefore, a := (a1, a2) ∈ (C∗)2 is a solution to

ϕΓ = 0.

Assuming in what follows that #V◦ < V (A) (the result follows automatically from Theorem 3.1if we do not make this assumption).

Let us choose one x(t) above that escapes (C∗)2 for small enough t.On the one hand, the vector β := U · αtr supporting the face UΓ of UA is equal to (0, 1) (see

Lemma 3.4). On the other hand, the point z(t), defined as

(3.4) zU (t) = x(t)

11

is a solution to

(3.5) U⋆(ϕ+ tψ) = 0.

Then, for j = 1, 2, we have

zj(t) = bjtβj + higher order terms in t,

for some b := (b1, b2) ∈ (C∗)2. Therefore, the limit z(0) belongs to C∗ × {0}, and is a solution to

U⋆ϕ = 0.This description implies the following: Each x(t) escaping (C∗)2 determines a unique proper

face Γ ≺ A, together with a matrix U ∈ TrsfΓ(Z2) such that U

⋆ϕ = 0 has a solution ρ ∈ C∗×{0}

satisfying ρ ∈ C∗ × {0}, and ρ = limt→0 z(t), where z(t), x(t) satisfy (3.4).The multiplicity mρ of ρ is no less than the number Nρ of distinct solutions z(t) to (3.5),

converging to ρ.Finally, our choice of ψ in the beginning implies that for any t, the system

U⋆(ϕ+ tψ)Γ = 0

has no solutions in (C∗)2 for any proper Γ. Then, (3.5) has no solutions in C∗ × {0} for anymatrix U ∈ TrsfΓ(Z

2). Therefore, we have mρ = Nρ. �

4. RELEVANT FACES AND PROOF OF THEOREM 1.2

One does not require the data of all monomial terms appearing in a polynomial map in orderto compute its isolated missing points. In this section, we point out those faces that are pertinentto such computation. We will utilize all notations and Proposition 3.6 in the previous section toprove some important technical results. These will be used to prove Theorem 1.2 and the resultsin Sections 5 – 7.

Consider a dominant polynomial map f := (f1, f2) : C2 → C2 and let A = (A1, A2) denotethe pair supp f . For any y = (y1, y2) ∈ C2, we denote by f − y the pair (f1 − y1, f2 − y2), and byf − y = 0 the system

f1(x1, x2)− y1 = 0,f2(x1, x2)− y2 = 0.

We are interested in describing the set C2 \ f(C2). This problem is invariant under translationf+c for any c ∈ C2. Therefore, we will assume in what follows that f(0, 0) ∈ (C∗)2. This impliesthat each member of A contains (0, 0) (see e.g. Example 3.3 and Figure 1).

Remark 4.1. Note that f maps C2 onto a line if Ai = {(0, 0)} for some i ∈ {1, 2}, and onto a curveif dimA1 + A2 = 1. Therefore, it is necessary for A to be independent (see Section 3.1.2) in orderfor f to be dominant.

4.1. Face-classification and generic non-properness. We will distinguish several types of facesof A (see the diagram in Figure 2).

Definition 4.2. Let A be an independent pair of finite sets in N2. A face Γ = (Γ1,Γ2) ≺ A is

• semi-origin if at least one of its members contains the origin, that is (0, 0) ∈ Γ1, or(0, 0) ∈ Γ2.

- Γ is origin if both its members contain (0, 0).- Γ is half-origin otherwise.

12

Face Γ ≺ A

Not semi-origin Irrelevant

Semi-origin

Origin

Relevant

Coordinate Irrelevant

Half-originRelevant

Irrelevant

FIGURE 2. Diagram classifying types of faces

• coordinate if all of its supporting vectors α = (α1, α2) (see Section 3.1) have one coordi-nate that is positive and another that is zero.

• relevant if it is semi-origin, not coordinate, and no member of Γ is a point different from(0, 0).

• irrelevant if it is not relevant.

7

Example 4.3. The pair of sets on the left of Figure 1 give the following classification for its faces:

• Origin faces: (⋆,�), (a,�), (⋆, d)• Half-origin faces: (b,�), (N,�), (N, c), (γ, δ), (⋆, ◦)• Coordinate faces: (⋆, d)• Relevant faces: (a,�), (b,�), (⋆,�), (γ, δ)• The face (N, c) is semi-origin, not coordinate, but not relevant. This is since dimN = 0,

but N 6= {(0, 0)}. The same goes for (⋆, ◦)

7

Recall the notation in Section 3.

Lemma 4.4. If #V(f) < V (A), then there exists a face Γ ≺ A that is not coordinate, such that thesystem fΓ = 0 has a solution in (C∗)2.

Proof. Split the isolated points in #V(f) into two subsets: Those in (C∗)2 and those outside it.The former has size equal to

(4.1) V (A) −∑

mΓ,

for all Γ as in Proposition 3.6 and the latter has size equal to

(4.2)∑

Γ≺AΓ coordinate

mΓ.

We sum up (4.1) and (4.2) to deduce that mΓ > 0 for some Γ ≺ A that is not coordinate. Thisyields the proof. �

Recall Definition 2.1 for generically non-proper maps.

Lemma 4.5. Let f be generically non-proper. Then, we have

µf = V (A).

Proof. We have #f−1(0) = V (A) from Definition 2.1, µf ≤ V (A) from Theorem 3.1 and

#f−1(0) ≤ µf from f being dominant. �

13

Lemma 4.6. Let f be generically non-proper. If for some face Γ ≺ A, there exists y ∈ C2 such thatthe system (f − y)Γ = 0 has a solution in (C∗)2, then Γ is a semi-origin face.

Proof. Let y be any point in C2, and let Γ be a face of A that is not semi-origin. Then, we have(f − y)Γ = fΓ. Therefore, if (f − y)Γ = 0 has a solution in (C∗)2, then so does fΓ = 0.

We deduce from Theorem 3.1 that #V◦(f) < V (A). This contradicts Lemma 4.5. �

Recall Definition 2.2 for the Jelonek set Jf of f . The following result will be used in Section 6.

Lemma 4.7. Let f be generically non-proper. Then, we have y ∈ Jf if and only if for some relevant

Γ ≺ A, the system (f − y)Γ = 0 has a solution in (C∗)2.

Proof. For the first direction, assume that y ∈ Jf . Then, #V(f − y) < µf . Lemma 4.4 implies

that there exists a non-coordinate face Γ ≺ A such that (f − y)Γ = 0 has a solution in (C∗)2. Wecan deduce that Γ is relevant from Lemma 4.6.

We prove the second direction. Let y be a point outside Jf . Then, Lemma 4.5 shows that

(4.3) V (A) = #(V(f − y) \ V◦(f − y)

)+#V◦(f − y).

Assume that y 6= f(0, 0). Then, using the notations of Proposition 3.6, the first summandin (4.3) is accounted for by

∑mΓ, where Γ runs over all (at most two) coordinate faces of

A. Equation (3.2) shows that for any other faces Γ, we have mΓ = 0. Therefore, the system(f − y)Γ = 0 has no solutions if Γ is relevant.

Finally, one can check that y = f(0, 0) ∧ y /∈ Jf ⇒ A has no relevant faces. �

Recall the set Kf defined in Section 2.1. The following result will be used in Sections 5 – 7.

Lemma 4.8. Let f be generically non-proper. Then, we have y ∈ Kf if and only if there exists a

relevant face Γ ≺ A and a matrix U ∈ TrsfΓ(Z2) such that U

⋆(f−y) = 0 has a solution in C∗×{0}

of multiplicity ≥ 2.

Proof. Let y ∈ Kf . Then, there exists a sequence {xk}k≥0 ⊂ C(f) ⊂ C2 that converges to infinity,and f(xk) converges to y [15]. In particular, this is a sequence of double roots to f − f(xk) = 0,converging to infinity.

Note that any toric transformation U preserves the number of solutions in (C∗)2 counted withmultiplicities (Remark 3.2). Therefore, using the curve selection Lemma, we follow closely theproof of Proposition 3.6 to deduce the following claim:

There exists a sequence {xk}k≥0 ⊂ C2 of double roots to

U⋆(f − f(xk)) = 0,

converging to a double root ρ ∈ C∗ × {0} to U⋆(f − y) = 0. This proves the first direction.

We omit the proof of the other direction since it is similar to the first one. �

4.2. Proof of Theorem 1.2. Recall that each member of the pair A contains (0, 0). For i = 1, 2,

let C[Ai] ∼= C|Ai| denote the space of all polynomials P : C2 → C such that suppP ⊂ Ai (apolynomial becomes identified with its coefficients). The space C[A1] ⊕ C[A2] is denoted byC[A].

Example 4.9. Let A be as in Figure 3 defined by the pair A1 ={(

00

),(11

),(12

)},

A2 ={(

00

),(10

),(11

)}. Then, any complex map

(4.4) (u, v) 7→ (a0 + a1uv + a2uv2, b0 + b1u+ b2uv),

is identified with (a0, a1, a2, b0, b1, b2) ∈ C[A] ∼= C6. 7

14

FIGURE 3. Support of the map in Example 4.9

Proposition 4.10. Let A be an independent pair of finite subsets in N2. If A has no relevantfaces, then proper maps form a dense subset in C[A], and no map in C[A] is generically non-proper.Otherwise, all maps in C[A] are non-proper, and generically non-proper maps form a dense subsetof C[A].

Proof. For any face Γ ≺ A, let ResΓ(A) denote the multivariate resultant of Γ in C[A]. This is,the subset of all maps f ∈ C[A] such that fΓ = 0 has a solution in (C∗)2. Multivariate resultantswere introduced in [13], and it was shown in [24, Theorem 1.1] that ResΓ(A) is a Zariski closedsubset of C[A].

For any f ∈ C[A] and any y ∈ Jf , we have #V(f − y) < V (A). Then, Lemma 4.4 shows that

there exists a non-coordinate face Γ ≺ A such that (f − y)Γ = 0 has a solution in (C∗)2.If A has no relevant faces, then either Jf = ∅, or (f − y)Γ = fΓ = 0. In the firt case, f is

proper (in particular, not generically non-proper). In the second case, we have f ∈ ResΓ(A) andLemma 4.6 shows that f is not generically non-proper.

Assume that A has a relevant face Γ. Then, for any f ∈ C[A], the set f−1(f(0, 0)) containsone of the two coordinate axes of C2. We deduce that any f is non-proper.

To show that generically non-proper maps form a dense subset of C[A], we proceed as follows.Let | Jac f | denote the determinant of the Jacobian matrix of f and let Σ denote its support.

Clearly, we have B := (A1,Σ) is independent of f as long as f is generically chosen.The coefficients of | Jac f | are polynomials in the coefficients of f . Moreover, recall that for

any face Λ ≺ B, the set ResΛ(B) is a Zariski closed subset of C[B]. Therefore, the set

SΛ := {(f1, f2) ∈ C[A] | (f1, | Jac f |) ∈ ResΛ(B)}

is a Zariski closed subset of C[A]. Note that the same holds true if we replace B by (A2,Σ).Finally, we conclude from the definition of generic non-properness (Definition 2.1) and Propo-

sition 3.6 that it enough for f ∈ C[A] to be outside all subsets ResΓ(A) and SΛ for it to begenerically non-proper. �

Consider the pair A in Figure 3 corresponding to the map f in Example 4.9. For any choiceof coefficients, f is non-proper since {u = 0} ⊂ f−1(a0, b0). Indeed, A has a relevant face; it issupported by (2,−1).

If (γ, δ) ≺ A is supported by the vector (−1, 0), then, f(γ,δ)(u, v) = 0 has solutions in (C∗)2 if

and only if a1b2− b1a2 = 0. Therefore, the map is generically non-proper inside the dense subset{a1b2 − b1a2 6= 0} of C6.

5. LONG FACES AND PROOF OF PROPOSITION 2.5

In this section, we prove Proposition 2.5. To do this, we need a few extra technical results.These are gathered in Section 5.1.

5.1. Lengths of faces. In this part, we will introduce two invariants for faces of finite pairs inZ2: integer length and dimensional length.

15

5.1.1. Integer length. For any set σ ⊂ R2, such that dimconv(σ) = 1, we use ℓ(σ) to denote thenumber | conv(σ) ∩ Z2| − 1, where conv(·) denotes the convex hull of any set in Rn.

Lemma 5.1. Let S1, S2 ⊂ R2 be two bounded segments having rational slopes. Then, we have

ℓ(S1) · ℓ(S2) ≤ V (S1, S2).

Proof. From dimS1 = dimS2 = 1, we have Vol(S1) = Vol(S2) = 0, and thus V (S1, S2) =Vol(S1 + S2). Let Lσ(S2) be the union of lines in R2, parallel to S1, and intersecting S2 atS2 ∩ N2 and Lθ(S1) be the analogous union of lines, but with S1 and S2 switched. Then, theset Lθ(S1) ∪ Lσ(S2) subdivides S1 + S2 into ℓ(S1) · ℓ(S2) parallelograms Pij , i = 1, . . . , ℓ(S1),j = 1, . . . , ℓ(S2), from which we obtain

Vol(S1 + S2) =∑

i,j

Vol(Pij).

From the construction of Pij , for any i, j we get VolPij = |det(α, β)|, where α and β are thedirectional vectors for S1 and S2 respectively. The result follows from |det(α, β)| ∈ N∗. �

Let f : C2 → C2 be a dominant polynomial map, and let A denote the support of f (SeeSection 3.2). Recall that, for each face Γ ≺ A, dimΓ = 1, we have TrsfΓ(Z

2) refers to the spaceof transformations in SL(2,Z), constructed using Γ as in Section 3.3.1.

Lemma 5.2. Let Γ ≺ A be a relevant face (see Definition 4.2), and let U be a matrix in TrsfΓ(Z2).

Assume that for some i ∈ {1, 2}, there exists a subset σ ⊂ UAi such that U(fi)σ is a polynomial Pin z1, up to monomial multiplication, with P (0) = 0. Then, we have

(5.1) degP = ℓ(ξ) ≤ deg fi/2,

where ξ ⊂ Ai and σ = Uξ.

Proof. The assumptions of the Lemma imply that both conv(σ) and conv(ξ) have dimension one,and that ℓ(σ) = ℓ(ξ).

To show that the equality part of (5.1), it suffices to notice that degP = ℓ(σ).Finally, Γ being relevant implies that conv(ξ) is not horizontal, nor vertical. One can check

(Figure 4) that

| conv(ξ) ∩Ai| ≤ deg fi/2.

This finishes the inequality part of (5.1). �

(deg fi, 0)

conv ξ

U

conv σ

FIGURE 4. An instance of Lemma 5.2.

16

5.1.2. Dimensional length. We also have the following refinement for types of faces of A. Wesay that a face Γ ≺ A is long if dimΓ1 = dimΓ2 = 1 and short otherwise.

Recall that a supporting vector α = (α1, α2) ∈ R2 of a relevant face of A satisfies α1α2 < 0. Ifα1 > 0, we say that it is a left relevant face and it is a right one otherwise.

In Figure 1 to the left, faces (a,�) and (b,�) of the pair of sets appearing are left and areshort. The face (γ, δ), is right and long.

Lemma 5.3. Assume that the map f is generically non-proper. Then, its support A has at most oneright/left long relevant face.

Proof. Assume without loss of generality that Γ is right, and that the first member Γ1 ⊂ A1 ofΓ contains (0, 0) (see Definition 4.2). If α = (α1, α2) ∈ Q2 denotes the supporting vector of Γ1,then (α1, α2 + q) supports the vertex (0, 0) ∈ A1 for any q > 0. This implies that any other rightbad face of A has to be a short one. �

5.2. Proof of Proposition 2.5. Assume in what follows that f is generically non-proper. We

will use I∅ := I∅f to denote the set of isolated points in C2 \ f(C2), and recall the sets C◦ := C◦

f

and K := Kf defined in Section 2.1 for dominant maps f : C2 → C2.We start by proving the following Lemma.

Lemma 5.4. For any q ∈ I∅ ∩ C◦ \ K, one of the following claims holds true.

• There exists a long origin relevant face Γ ≺ A and a matrix U ∈ TrsfΓ(Z2) such that

U⋆(f − q)Γ = 0

has µf distinct solutions in C∗.• There exist two long relevant faces Γ(1),Γ(2) ≺ A, together with two matrices U1 ∈TMΓ(1)(Z

2) and U2 ∈ TMΓ(2)(Z2), such that

U i(f − q)Γ(i) = 0

has Ni solutions (i = 1, 2) in C∗, and N1 +N2 = µf .

Proof. The system f − q = 0 has no solutions in C2. Then, Theorem 3.1 and Remark 3.5 show

that U⋆(f − y) = 0 has a solution in C∗ × {0}.

The proof will follow from Proposition 3.6 and Lemma 5.3 once we show the following claim:

All such faces Γ can only be long and relevant. Moreover, solutions to U⋆(f − y) = 0 in C∗ ×{0}

are simple.Since f − q = 0 has no solutions in C2 (in particular, no solutions in C2 \ (C∗)2), Γ cannot be

a coordinate face. Then, Lemma 4.6 yields that Γ is pertinent.To deduce that Γ is long, notice that Γ short ⇒ fi(0, 0) = qi (for some i ∈ {1, 2}) ⇒ q /∈ C◦.

Finally, we have q /∈ K ⇒ all solutions to U⋆(f − y) = 0 are simple (see Lemma 4.8). This

proves the two statements of the claim. �

Assume that I∅ ∩ C◦ \ K 6= ∅, and consider the set

{q1, . . . , qN} := I∅ ∩ C◦ \ K.

Lemma 5.4 shows that A has a long relevant face Γ. We decompose {1, . . . , N} into a disjointunion I1 ⊔ · · · ⊔ Iλ of non-empty subsets, satisfying k ∈ Ii ⇔ there are exactly mi ≥ 0 distincta ∈ C∗ such that for some matrix U ∈ TrsfΓ(Z

2), we have

(5.2) U⋆(f − qk)Γ(a) = 0.

Note that this decomposition is not unique.

17

Example 5.5. Consider the map f constructed in Section 4.2. We have C◦ = (C∗)2, |I∅| = 2nand µf = 2. Choose Γ to be the pair as in Figure 5.

{(00

),(11

)}, {∪n

i=1(i− 1, i+ 1)} .

For any q ∈ I∅, we have

U⋆(f − q)Γ = (t− q, b0 + b1t+ · · ·+ bnt

n),

where the choice of (b0, b1, . . . , bn) is so that b0 + b1t + · · · + bntn has exactly n simple roots in

C∗. Then, we have λ = 1, with m1 = 1. 7

Γ1

Γ2

FIGURE 5. The couple of supports in Example 5.5.

Let g1, g2 ∈ C[z1] denote the polynomials U⋆f1,Γ, U

⋆f2,Γ respectively and letG(r, s) = 0 denote

the polynomial system

(5.3)g1(r)− g1(s) = 0,g2(r)− g2(s) = 0.

Then, Equation (5.2) shows that for any Ii with mi ≥ 1, points a1, . . . , ami∈ C∗ above satisfy

(5.4) {(ak, al) ∈ V◦(G) | k, l = 1, . . . ,mi}.

The following observation is obvious.

Claim 5.6. The set of isolated points in V◦(G) satisfies r 6= s.

This shows that (5.4) identifies a distinguished subset of m2i −mi isolated points in V◦(G).

We proceed as follows: Each set Ii distinguishes a unique subset

(5.5) {qk}k∈Ii

in I∅. Since the solutions to U⋆(f − q)Γ = 0 vary with q in (5.5), each qk distinguishes a unique

set {a1, . . . , ami} ⊂ C∗. This in turn distinguishes a subset ofmi(mi−1) isolated points in V◦(G).

Therefore, each Ii gives rise to mi(mi − 1)|Ii| distinct points in V◦(G). That is, we have

(5.6)∑

i

mi(mi − 1)|Ii| ≤ #V◦(G).

In order to provide an upper bound on N above, we consider two cases according to Lemma 5.4.In what follows, we use µ as shorthand to µf .

Assume that Γ is the only long relevant face. Then, Lemma 5.4 shows that λ = 1, withm1 = µ, and Equation (5.6) becomes

(5.7) µ(µ− 1)N ≤ #V◦(G).

18

Proof of Proposition 2.5 (2.2). Bézout’s theorem shows that

(5.8) #V◦(G) ≤ deg g1 · deg g2,

and Lemma 5.2 shows that for j = 1, 2 we have

(5.9) deg gj = ℓ(Γj) ≤ deg fj/2.

Therefore, we have N ≤ deg f1 · deg f2/(4µ(µ − 1)). �

Proof of Proposition 2.5 (2.3). Remark 4.1 shows that for some i ∈ {1, 2}, say i = 1, we haveAi 6= Γi.

For any w ∈ A1 \ Γ1, we have conv(σ) ⊂ conv(A1) and conv(Γ2) ⊂ conv(A2). Then, theinequality

(5.10) V (σ,Γ2) ≤ V (A) = µ

is deduced from Lemma 4.5 and from the following fact.

Fact 5.7 (monotonicity, see [23], Chapter 5.25). If L1, L2,∆1,∆2 ⊂ R2 are convex bodies suchthat L1 ⊂ ∆1 and L2 ⊂ ∆2, then V (L1, L2) ≤ V (∆1,∆2).

Finally, Lemma 5.1 shows that

(5.11) ℓ(Γ2) ≤ ℓ(σ) · ℓ(Γ2) ≤ V (σ,Γ1).

Therefore, we conclude

N(5.7)≤

V◦(G)

µ(µ− 1)

(5.8), (5.9)

≤ℓ(Γ1) · ℓ(Γ1)

µ(µ− 1)

(5.10), (5.11)

≤ ℓ(Γ1)Lemma 5.2

≤d12.

�

Assume that there is another long bad face Λ ≺ A. Lemma 5.4 shows that for each Ii, thereexists V ∈ TrsfΛ(Z

2), such that we have the following setup. For each q ∈ {qk}k∈Ii , there existsa set Aq of mi values a ∈ C∗, and a set Bq of ni distinct values b ∈ C∗ such that mi + ni = µ and

(5.12)U

⋆(f − q)Γ(a) = 0,

V (f − q)Λ(b) = 0.

Example 5.8 (5.5 continued). The face Λ here is({(

00

),(11

)}),({(

00

),(11

)}), and V (f − q)

is the pair (t − q1, t − q2). From Example 5.5, we have λ = 1, and m1 = 1. Then, for eachq ∈ ∪t∈V(P ){(t, t)} (see Section 4.2), we have n1 = 1, Aq = Bq = {t}. 7

For j = 1, 2, we define gj and hj to be U⋆fj,Γ and V gj,Γ respectively. Then, the sets

(5.13) Aq ×Aq, Bq × Bq, Aq × Bq,

form subsets of solutions to the systems G = 0, H = 0 and K = 0 respectively, defined by

(5.14)gj(s) − gj(t) = 0, j = 1, 2hj(u) − hj(v) = 0, j = 1, 2gj(s) − hj(u) = 0, j = 1, 2.

The following Claim is similar to Claim 5.6.

Claim 5.9. The set of non-isolated points in V◦(G),V◦(H), or V◦(K) satisfy s = t.

19

This shows that for each above q, there exists a unique subset (5.13) identifying a subset of

m2i −mi + n2i − ni +mini

isolated points in V◦(G) ∪ V◦(H) ∪ V◦(K). Then, the above discussion shows that

(5.15)∑

i

(m2i −mi + n2i − ni +mini) · |Ii| ≤ #V◦(G) + #V◦(H) + #V◦(K).

We proceed by computing the upper bounds in Proposition 2.5.Using Bézout’s theorem, on the systems (5.14) and the inequality of Lemma 5.2, we bound

the value in (5.15) by

(5.16) ℓ(Γ1) · ℓ(Γ2) + ℓ(Λ1) · ℓ(Λ2) + max(ℓ(Γ1), ℓ(Λ1)) ·max(ℓ(Γ2), ℓ(Λ2)).

Moreover, the equality part in Lemma 5.2 produces the upper bound

(5.17) 3 deg f1 deg f2/4

for (5.15). Next, we use N =∑Ii, mi = µ − ni in the l.h.s of (5.15), expand it, then divide by

µ2 to obtain

(5.18) N +∑

i(n2iµ2

−niµ

−1

µ

)≤

k

µ2,

where k is the minimum of (5.16) and (5.17).Identities ni/µ ≤ 1 and N =

∑Ii imply

∑i(n2iµ2

−niµ

−1

µ

)≤ −

N

µ.

Therefore, (5.18) yields

(5.19) N ≤k

µ(µ− 1).

Proof of Proposition 2.5 (2.2). It follows from this inequality for k = (5.17) and µ ≥ 2. �

Proof of Proposition 2.5 (2.3). Without loss of generality, we suppose that dimA2 = 2 (this ispossible since (A1, A2) has two relevant long faces). Moreover, we may also suppose that ℓ(Γ1) ≥ℓ(Λ1).

Then, there exists a subset σ ∈ A2 such that dim(σ + γ1) = 2. We get

ℓ(σ) · ℓ(Γ1)Lemma 5.1

≤ V (σ,Γ1)Fact 5.7

≤ V (A)Lemma 4.5

= µ.

Replacing k by (5.16), above inequalities together with (5.19) yield

N ≤1

(µ− 1)ℓ(Γ1) · ℓ(σ)

(ℓ(Γ1) · ℓ(Γ2) + ℓ(Γ1) · ℓ(Λ2) + ℓ(Γ1) ·max(ℓ(Γ2), ℓ(Λ2))

).

Lemma 5.2≤ 3 deg f/2(µ − 1).

�

20

5.3. Geometry of the Jenonek set and missing points of maps. Upper bounds for #V◦(G) +#V◦(H) + #V◦(K) appearing above estimate the number of some nodes of the Jelonek set. Aninstance of this is when there exists only one long face Γ ≺ A that happens to be origin. Then,

{y ∈ C2 | U

⋆(f − y)Γ(a) = 0 for some a ∈ C∗

}

describes a part of a rational curve

C :={(g1, g2)(s) ∈ C2 | s ∈ C∗

}.

Therefore, the set V◦(G) describes all self-intersections (nodes) that C can have in (C∗)2. Namely,if a point y ∈ C also belongs to I∅ ∩ C◦ \ K, then it is a node of multiplicity µ. This explains thephenomenon in the map (4.4) of Section 1.

As for the case where there are two long relevant faces Γ,Λ ≺ A, the set V◦(G) ∪ V◦(H) ∪V◦(K) describes both the intersection locus C∩C ′ (with C ′ similarly defined as C, using H) andthe nodes of the curves. Here, each of C and C ′ can be a rational curve, or a union of vertical(horizontal) lines in C2. In the map (1.4), the set I∅

f ∩C◦ \K results from intersections of vertical

lines with the line {y1 = y2}.

6. PROOF OF PROPOSITION 2.4

We consider the generically non-proper map f from the previous section. Recall from Lemma 4.8that K coincides with the set of all y ∈ C2 at which there exists a face Γ ≺ A and a matrix

U ∈ TrsfΓ(Z2) such that U

⋆(f − y) = 0 has a double solution in C∗ × {0}.

In this section, we prove the inequality

(6.1) |I∅ ∩ C+ \ K| ≤ deg f1 + deg f2,

where C+ := C+f is the cross centered at f(0, 0) (also, it is the set C2 \ C◦).

In the notations of Proposition 3.6, for any y ∈ I∅\K, there are µ points in C∗×{0} distributed

among simple solutions to systems U⋆(f − y) = 0.

We deduce from y ∈ C2 \ f(C2) and from Lemma 4.7 that all above matrices U ∈ TrsfΓ(Z2)

correspond to only relevant faces Γ ≺ A.

Claim 6.1. For each y ∈ I∅ \ K, at least one of the above relevant faces of A is long.

Proof. Lemma 5.4 takes care of the case where y ∈ C◦.Assume that y ∈ C+, and let Γ be a short pertinent face of A. One can check that the set of all

y ∈ C2 such that U⋆(f − y)Γ = 0 has a solution in C∗, is a vertical/horizontal line in C+.

If all faces Γ are short, then there exists µ solutions in C∗ to systems above determining thesame line L ⊂ C+. Therefore, L ⊂ C2 \ f(C2), and y is a missing point that is not isolated. �

Thanks to the above Claim, it is enough to give an upper bound on the total number of distinct

y ∈ C+ such that U⋆(f − y)Γ = 0 has a solution in C∗ for some long relevant face Γ ≺ A.

We assume that y belongs to the horizontal line H of C+. Then, the second coordinate ofy equals to f2(0, 0). To determine the first coordinate, we solve the (now, triangular) square

system U⋆(f − y)Γ = 0, with y2 = f2(0, 0).

Lemma 5.2 applied to U⋆(f2 − f2(0))Γ shows that we obtain at most deg f2/2 distinct points

in H for each system U⋆(f − y)Γ = 0. Lemma 5.3 shows that there are at most two such systems

with Γ being long relevant. This amounts to the upper bound deg f2 if y ∈ H. By symmetry onthe vertical line of C+, we obtain the upper bound deg f1 + deg f2.

21

7. CRITICAL POINTS AT INFINITY AND PROOF OF PROPOSITION 2.3

Keeping with the same notations as in the previous section, let f be a generically non-propermap. To prove |I∅ ∩ K| ≤ deg f1 + deg f2, we introduce two technical results. Write I∅ ∩ K as

(7.1)⋃

Γ≺AΓ relevant

K(Γ),

where K(Γ) consists of all y ∈ I∅ such that U⋆(f−y) = 0 has a solution in C∗×{0} of multiplicity

at least two for some matrix U ∈ TrsfΓ(Z2).

Lemma 7.1. Let Γ be a right (left) long origin relevant face of A. Then, there are no right (left)relevant faces of A other than Γ.

Proof. Let α := (α1, α2) ∈ Q− × Q+ be a supporting vector of Γ assuming it is right. Then, forany p ∈ Q∗, the vector αp := (α1, α2 + p) supports a face of A different form Γ. All right facesare determined this way.

If p > 0, then αp determines the face( {(

00

)},{(

00

)} ). Otherwise, the face determined by

αp is not relevant. �

The following result will be proven at the end of this section. Recall that for any finite setσ ⊂ Z2, we define ℓ(σ) := | conv(σ) ∩ Z2| − 1.

Proposition 7.2. Let Γ := (Γ1,Γ2) be a pertinent face of A. Then, we have

(A) |K(Γ)| ≤ (deg f1 + deg f2)/2,(B) |K(Γ)| ≤ ℓ(Γi) + deg fj/2 if Γj is a vertex of Aj for any distinct i, j ∈ {1, 2} and(C) |K(Γ)| ≤ ℓ(Γi) if Γi does not contain (0, 0) for some i ∈ {1, 2}.

Proof of Proposition 2.3. Let us split the union (7.1) into two disjoint subsets

M l ⊔M r,

formed by the contribution of left faces of A, and the other by its right ones. The result willfollow by showing that each of |M r| and |M l| is bounded by (deg f1 + deg f2)/2. By symmetry,

it suffices to give an upper bound for |M l|.Assume first that A has a long right relevant origin face (recall Definition 4.2). From Lemma 7.1,

it is the unique right relevant face. Item (A) of Proposition 7.2 shows that

|M l| ≤ (deg f1 + deg f2)/2.

Assume now that A does not have long origin right relevant faces (In Figure 6, for example,the only long relevant face is not origin).

Denote by Γ(0) the origin right short face of A, and suppose that Γ1(0) ⊂ A1 is the one-dimensional member of Γ(0) (we have Γ(0) = (a,�) in Figure 6). Then, Item (B) of Proposi-tion 7.2 shows that

|K(Γ(0))| ≤ ℓ(Γ1(0)) + deg f2/2.

We have the following easy fact:Each face Γ(i) ≺ A in the collection of all other bad left faces {Γ(1), . . . ,Γ(k)} is half-origin, and

its first member Γ1(i) ⊂ A1 does not contain (0, 0).

22

a

b

c

d

�

e

FIGURE 6. A pair having several short left bad faces

In Figure 6, for example, we have k = 3, Γ(1) = (b,�), Γ(2) = (c,�) and Γ(3) = (d, e).Therefore, Item (C) of Proposition 7.2 shows that

k∑

i=1

|K(Γ(i))| ≤k∑

i=1

ℓ(Γ1(i)).

Finally, note that the collection Γ1(0),Γ1(1), . . . ,Γ1(k) forms a chain of left faces of A1. There-

fore, the result follows from∑k

i=0 ℓ(Γ1(i)) ≤ deg f1/2. �

7.1. Proof of Proposition 2.3. Pick any matrix U ∈ TrsfΓ(Z2). The following Lemma shows

that it suffices to prove the result only for U .

Lemma 7.3. The set K(Γ) does not depend on the choice of the matrix in TrsfΓ(Z2).

Proof. The unimodularity of matrices in TrsfΓ(Z2) implies that the number of solutions to sys-

tems, as well as their multiplicities, is preserved. Moreover, all transformed systems under Uare equal when restricted to Γ. Therefore, any transformation of f − y under U gives a uniquenumber of solutions in C∗ × {0}, together with multiplicities. �

Replace (z1, z2) by (s, t) and U⋆(f − y) by g := (g1, g2), that is g1, g2 ∈ C[s, t, y1, y2].

Assume first that Γ is an origin face of A. Then, for i = 1, 2, the polynomial gi is expressed as

mi∑

j=0

tjgi,j(s)− yi.

A point y ∈ K satisfies

(7.2) g1,0 − y1 = g2,0 − y2 = D = 0,

where D ∈ C[s] is the polynomial determining | Jac g| at t = 0.On the one hand, (7.2) has at most degD distinct solutions in C3. This implies that

|K(Γ)| ≤ degD.

On the other hand, we have

D = g1,1 · g′2,0 − g2,1 · g

′1,0,

23

where the derivation is w.r.t. z1. Therefore, Item (A) of Proposition 7.2 follows form Lemma 5.2by observing that gi,j = gi|σ, for some horizontal set of points σ ⊂ N2. Item (B) of Proposition 7.2

follows similarly by additionally noting that deg gi,0 = ℓ(Γi).Assume that one of the members, say, Γ2, of Γ does not contain (0, 0). Then, the polynomial

g2 is written asmi∑

j=0

tjg2,j(s)− svtwy2,

where (v,w) ∈ N2. The polynomial D determining | Jac g| at t = 0 is a polynomial in s and y2

(g2,1 − sy2δk1) · g′1,0 − g1,1 · g

′2,0,

where δk1 is the Kronecker delta. The system (7.2) is written as g1,0 − y1 = g2,0 = D = 0. Ifk = 1, we get

|K(Γ)| ≤ deg g2,0 ≤ ℓ(Γ2),

from using similar arguments as in the previous cases (Γ origin). This proves (C) (⇒ (B)) fork = 1, and thus Lemma 5.2 yields (A).

We finish the proof by showing that f is generically non-proper ⇒ k = 1.

Assume that k 6= 1. Then, D depends only on s, and the system g2 = 0, | Jac g| = 0 has asolution in C∗ × {0}. One can check, using the chain rule for partial derivations that the sets

{| Jac g| = 0} and {U⋆| Jac f |) = 0} are equal in (C∗)2. Then, Proposition 3.6 shows that

n0 := #V◦(g2, U⋆| Jac f |) < V (supp g2, suppU

⋆Jf).

Since elements in TrsfΓ(Z2) preserve multiplicities for solutions to the transformed systems, we

deduce from Lemma 7.3 that

#V◦(f2, | Jac f |) = n0.

Finally, since V (supp g2, suppU⋆| Jac f |) = V (supp f2, supp | Jac f |) [1], we get

n0 < V (supp f2, supp Jac f),

which implies that f is not generically non-proper, a contradiction.

7.2. Cusps of the Jelonek set. In contrast to Section 5.3, singularities of the Jelonek set Jf

that are not nodes or self-intersection, can be also good candidates for isolated missing points off .

Consider the map (1.3) defined in Section 1, whose Jelonek set has a cusp coinciding with theunique missing point (0, 1). This is not a coincidence, and the reason for this goes as follows.

On the one hand, solutions in C∗ × C2 to the system (7.2) account for all cusps that aparametrized curve C (from Section 5.3) may have. On the other hand, Lemma 4.7, appliedto the unique relevant face Γ of A, where

Γ1 ={(

00

),(11

),(22

)}, Γ2 =

{(00

),(11

),(22

),(33

)},

shows that Jf is expressed as the set of all y ∈ C2 at which y = U⋆fΓ(s) for some matrix

U ∈ TrsfΓ(Z2). This makes Jf a curve whose i-th coordinate has the above parametrization.

Contact. Boulos El Hilany, Johann Radon Institute for Computational and Applied Mathematics,Altenberger Straße 69, 4040 Linz, Austria; boulos.hilani@gmail.com, boulos-elhilany.com.

24

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