## A number of points in the set C2 \ F(C2)

Let F := (F1,F2) : C2 â C2 be a quasi-finite mapping (i. e., all fibers of F are finite). It is well known that it is only a finite number of points in the set. C2 \F(C2).
A number of points in the set C 2 \ F (C 2). Zbigniew Jelonek∗) Instytut Matematyki Polska Akademia Nauk ´ Sw. Tomasza 30, 31-027 Krak´ow and Instytut Matematyki Uniwersytet Jagiello´ nski Reymonta 4, 30-059 Krak´ow Poland e-mail: [email protected] Abstract In this paper we estimate a number of points of the set C2 \ F (C2 ) for a polynomial quasi-finite mapping F : C2 → C2 .

1

Introduction.

Let F := (F1 , F2 ) : C2 → C2 be a quasi-finite mapping (i. e., all fibers of F are finite). It is well known that it is only a finite number of points in the set C2 \ F (C2 ). It seems to be interesting to estimate the number of these points. Let deg F = max (deg F1 , deg F2 ) and let µ(f ) denote a number of points in a generic fiber of F. Our result is following: MSC Classification: 14 E 05, 14 E 07, 14 E 09. Key words: polynomial mapping, quasi-finite mapping, the set of points over which a polynomial mapping is not proper. *) This paper is partially supported by the grant of KBN number 2 PO3A 037 14.

1

Theorem. Let F := (F1 , F2 ) : C2 → C2 be a polynomial mapping. Assume that the set C2 \ F (C2 ) has k isolated points. Then k = 0 or k ≤ (degF1 degF2 ) − µ(F ) − 1. Moreover, k ≤ (degF − 1)2 . Corollary. Let F := (F1 , F2 ) : C2 → C2 be a quasi-finite mapping. Let us denote a number of points of the set C2 \ F (C2 ) by k. Then k = 0 or k ≤ (degF1 degF2 ) − µ(F ) − 1. Moreover, k ≤ (degF − 1)2 . Moreover, we show that every quasi-finite mapping F := (F1 , F2 ) : C2 → C2 of degree 2 is finite, hence surjective. We also give an example of a quasifinite mapping of degree 3 which is not surjective.

2

Proofs.

We use our methods from the paper  (see also  and ). Let us recall that a mapping f : Cn → Cn is not proper at a point y if there is no neighborhood U of y such that f −1 (cl(U )) is compact. In other words, in this case f is not proper at y if f is not a local analytic covering over y. Let Sf denote the set of points at which the mapping f is not proper. We have the following characterization of a set Sf (see , ): Theorem 1 Let f = (f1 , ..., fn ) : Cn → Cn be a dominant polynomial map. Then the set Sf is a (C-uniruled) hypersurface. Moreover, Qn

deg Sf ≤

(

deg fi ) − µ(F ) . mini=1,...,n deg fi i=1

and, deg f −1 (Sf ) ≤ (

n Y

deg fi ) − µ(F ).

i=1

Now we need some results about mappings of affine curves. Let us recall that an affine curve X has n branches at infinity if there is a projective completion X1 of X, which is smooth at infinity, such that the set X1 \ X has exactly n points. In other words n is a number of irreducible germs at infinity of X. We have 2

Lemma 2 Let X ⊂ C2 be an affine curve of degree d. Then X has at most d branches at infinity. Proof. Indeed, let X1 be a projective closure of X. Let H denote the line at infinity. By the Bezout theorem we have that (X1 , H) = d. But if a1 , ..., as are all points at infinity of a curve X1 and we have a following deSi composition of germs (X1 )ai = rk=1 V , then (H, Vik ) ≥ 1 and consequently Ps Pri Pi,k d = (X1 , H) = i=1 k=1 (H, Vik ) ≥ si=1 ri . But the latter number is exactly the number of branches of X at infinity. We also need: Lemma 3 Let X, Y be affine curves. Assume that X has d branches at infinity. Let f : X → Y be a polynomial, dominant mapping. Then the set Y \ f (X) has at most d − 1 points. Proof. Let X1 be a projective completion of X which is smooth at infinity. Let Y1 be any projective completion of Y. Then the mapping f : X → Y has a unique polynomial extension F : X1 → Y1 . Of course the mapping F is surjective. Let X1 \X = {x1 , ..., xd } and Y1 \Y = {y1 , ..., ys }, where s ≥ 1. We can assume that F (x1 ) = y1 . It means that Y \ {F (x2 ), F (x3 ), ..., F (xd )} ⊂ f (X). Hence the set Y \ f (X) has at most d − 1 points. Now we are ready to prove our main result: Theorem 4 Let F := (F1 , F2 ) : C2 → C2 be a polynomial mapping. Assume that the set C2 \ F (C2 ) has k isolated points. Then k = 0 or k ≤ (degF1 degF2 ) − µ(F ) − 1. Moreover, k ≤ (degF − 1)2 . Proof. First we show that n ≤ (degF1 degF2 ) − µ(F ) − 1. We can assume that the mapping F is dominant. Let Y denote the set of points at which the mapping F is not proper. Thus Y is a curve. Since the mapping F is proper outside Y , we have that all points from the set C2 \ F (C2 ) are on the curve Y . Let X := F −1 (Y ) and consider the mapping f := resX F : X 3 x → F (x) ∈ Y. It is an easy observation that C2 \ F (C2 ) = Y \ f (X). Thus we get by Lemma 3 that the set C2 \ F (C2 ) has at most deg X − 1 isolated points. But by Theorem 1 we have that deg X ≤ (degF1 degF2 ) − µ(F ). 3

Now we pass to the second estimation. Let us take a general coordiP m−i nates x, y in C2 and let us consider polynomials F1 (x, y) = m , i=0 ai (x)y Pm m−i F2 (x, y) = i=0 bi (x)y , where m =deg F and a0 , b0 are non-zero constants. Now form the resultant R(x, Z, W ) of polynomials F1 − Z, F2 − W (considered as polynomials of one variable y). This will have the form: R(x, Z, W ) =

s X

cj (Z, W )xs−j .

j=0

From elementary properties of a determinant we have that deg ci (Z, W ) ≤ m − 1 for i < s. A point (Z, W ) is not in the image of the mapping F if and only if the resultant R(x, Z, W ) is not zero for every x, i.e., if and only if ci (Z, W ) = 0 for i < s and cs (Z, W ) 6= 0. If polynomials ci (Z, W ) = 0 for i < s have a common divisor h then we have two possibilities: 1) the polynomial h divides cs (Z, W ) 2) the polynomial h does not divide cs (Z, W ). The first case is impossible for a dominant mapping. Indeed, in this case over every point from the curve H := {(Z, W ) : h(Z, W ) = 0} would be a curve (as a fiber of F ) and thus F (C2 ) = H. In the second case on the curve H are only non-isolated points of the set C2 \ F (C2 ) (or points from F (C2 )). Consequently, we can assume that polynomials ci (Z, W ) = 0 for i < s P have not a common divisor. It means that polynomials a := s−1 j=0 αj cj (Z, W ) Ps−1 and b := j=0 βj cj (Z, W ) (where αi , βj are sufficiently general coefficients) have not a common divisor, too. But all isolated points of the set C2 \ F (C2 ) must satisfy equations a = 0, b = 0. Since deg a ≤ m − 1, deg b ≤ m − 1, we have that the set C2 \ F (C2 ) has at most (m − 1)2 isolated points. Corollary 5 Let F := (F1 , F2 ) : C2 → C2 be a quasi-finite mapping. Let us denote a number of points of the set C2 \ F (C2 ) by k. Then k = 0 or k ≤ (degF1 degF2 ) − µ(F ) − 1. Moreover, k ≤ (degF − 1)2 . Now we show that a quasi-finite mapping of degree two is finite. For the proof we need a following result (see ): Theorem 6 A dominant polynomial mapping F := (F1 , ..., Fn ) : Cn → Cn is finite, if and only if restrictions of the mapping F to hypersurfaces V (Fi ) = {x ∈ Cn : Fi (x) = 0} are proper. 4

Now we are ready to prove: Theorem 7 A quasi-finite mapping F := (F1 , F2 ) : C2 → C2 of degree two is finite. Proof. Assume that F is not a finite mapping. Let Xi := {(x, y) ∈ C2 : Fi (x, y) = 0}, i = 1, 2. Curves X1 , X2 have degrees at most two. Since the mapping F is not finite, we have that either a mapping F restricted to the curve X1 is not proper, or a mapping F restricted to the curve X2 is not proper (see Theorem 6). We can assume that the first possibility holds. In this case the curve X1 must be a hyperbole, i.e., after linear change of coordinates we have F1 = xy − 1. Thus F (x, y) = (xy − 1, ax2 + by 2 + cxy + dx + ey + f ). Further a restriction of the mapping F to the hyperbole X1 is not proper. It means that a mapping h : X1 3 x → F2 (x) ∈ C can not be proper. But X1 has a rational parametrization x = t, y = 1/t and we have h(t, 1/t) = at2 + b/t2 + c + dt + e/t + f. If |a| + |d| 6= 0 and |b| + |e| = 6 0, then the mapping h is proper (we have then limt=0 = ∞ and limt=∞ = ∞). Hence e.g., a = 0 and d = 0, i.e., F2 = by 2 + cxy + ey + f. Now we have F (C × {0}) = (−1, f ), and we get a contradiction. Corollary 8 A quasi-finite mapping F := (F1 , F2 ) : C2 → C2 of degree two is surjective. Example 9 A mapping F : C2 3 (x, y) → (x(1 + xy) − y, xy) ∈ C2 is quasi-finite but (0, −1) 6∈ F (C2 ).

References  Jelonek, Z, The set of points at which a polynomial map is not proper, Ann. Polon. Math., 58, (1993), 259-266.  Jelonek, Z, Testing sets for properness of polynomial mappings, Math. Ann., (to appear).  Jelonek, Z, Geometry of affine polynomial mappings, Preprint PAN, 1997/570.  Shafarewich, I.R, Basic Algebraic Geometry, Springer-Verlag, (1974). 5