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AN INEQUALITY FOR POLYNOMIAL MAPPINGS

arXiv:alg-geom/9305007v1 19 May 1993AN INEQUALITY FOR POLYNOMIAL MAPPINGSby Arkadiusz P loskiAbstract. We give an estimate of the growth of a polynonial mapping of Cn.1.

Main result.Let F= (F1, . .

. , Fn) : Cn →Cn be a polynomial mapping.We putd(F) = # F −1(w) for almost all w ∈Cn and call d(F) the geometric degree of F.Let di = deg Fifor i = 1, .

. .

, n. Then 0 ≤d(F) ≤Qni=1 di if Fi ̸≡0 for all i(cf. [6], p. 434).

Note that d(F) = 0 if and only if the polynomials F1, . .

. , Fn arealgebraically dependent.

For any w ∈Cn such that the fiber F −1(w) is finite weputδw(F) = d(F) −Xz∈F −1(w)multz(F)where multz(F) stands for the multiplicity of F at z (cf. [6], p. 256).

We follow theconvention that the sum of an empty family is zero. We have always δw(F) ≥0 forthe finite fibers F −1(w);δw(F) = 0 if and only if F is proper at w or d(F) = 0(cf.

[3]). For any z = (z1, .

. .

, zn) ∈Cn we put |z| =nmaxi=1 |zi|. We say that aninequality holds for |z| ≫1 if there is a constant R > 1 such that it holds for allz ∈Cn such that |z| ≥R.Let eFi be the homogenized polynomial Fi.

The main result of this note isTheorem 1.1. Suppose that the system of homogeneous equationseF1 = · · · = eFn = 0 has a finite number of solutions in the projective space Pn.

Thenthere is a positive constant C such that|F(z)| ≥C|z|−δ0(F )for |z| ≫1.The proof of (1.1) will be given at the end of this note. Now, we will prove thefollowingTheorem 1.2 (cf.

[1] for the case n = 2). Assume the assumptions of (1.1) andput µ =Pz∈F −1(0)multz(F).

Then there is a positive constant C such that|F(z)| ≥C|z|µ−nQi=1di+nmini=1 (di)for |z| ≫1.Proof of (1.2). Let us distinguish two cases.Typeset by AMS-TEX1

2BY ARKADIUSZ P LOSKICase 1. d(F) >nQi=1di −nmini=1 (di). Then F is proper by proposition 1.3 of [7], consequentlyµ = d(F) and the inequality follows from theorem 1.10 [7].Case 2. d(F) ≤nQi=1di−nmini=1 (di).

By definiton of δ0(F) we get δ0(F) ≤nQi=1di−nmini=1 (di)−µand the inequality follows from theorem 1.1.Remark 1.3. Let F = (F1, .

. .

, Fn) : Cn →Cn be a polynomial mapping. Then thefollowing two conditions are equivalent:(i) the system of equations F1 = · · · = Fn = 0 has a finite number of solutionsin Cn,(ii) there are constants C > 0 and q ∈R such that |F(z)| ≥C|z|qfor |z| ≫1For n > 2 the assumption of (1.1) and (1.2) is stronger than (i).

J. Koll`ar showedin [5] (Prop. 1.10) then we can take q = −nQi=1di +nmini=1 (di)in (ii).2.

Resultant of homogeneous polynomials.If H1, . .

. , Hm is a sequence of homogeneous polynomials in n + 1 variables,then we denote by V (H1, .

. .

, Hm) the set of all solutions in Pn of the systemH1 = · · · = Hm = 0. We will need some properties of the resultant of n + 1 formsin n + 1 variables.Property 2.1.

If H1, . .

. , Hn+1 are general forms of degrees d1, .

. .

, dn+1 > 0in n+ 1 variables X = (X1, . .

. , Xn+1), then their resultant ResX(H1, .

. .

, Hn+1) isa polynomial in coefficients of these forms, homogeneous of degree d1 · . .

. · dn+1/diwith respect to the coefficients of Hi.Property 2.2.

If the coefficients of H1, . .

. , Hn+1 lie in C,then ResX(H1, .

. .

, Hn+1) = 0if and only if V (H1, . .

. , Hn+1) ̸= ∅Property 2.3.

Suppose that H1, . .

. , Hn are homogeneous forms of degrees d1, .

. .

, dnwith coefficients in C such that the set V = V (H1, . .

. , Hn) ⊂Pn is finite.

Let Lbe a linear form such that V ∩V (L) = ∅. For any p ∈V we denote by µp themultiplicity of the mapping H1Ld1 , .

. .

, HnLdn: Pn \ V (L) →Cnat p. Then for any homogeneous form H of degree d > 0 we haveResX(H1, . .

. , Hn, H) = ResX(H1, .

. .

, Hn, Ld)Yp∈V HLd (p)µpThe properties (2.1) and (2.2) are well known (cf. [4]).

In order to check (2.3)let us assume that the hypersurfaces Hi = 0(1 ≤i ≤n) meet transversallyi.e.µp = 1 for all p ∈V .According to Bezout’s theorem V contains exactlyd1 · . .

. · dn points, consequently the product Qp∈VHLd (p) is a homogeneous formwithout multiple factors in the coefficients of H. The polynomials (in the coefficientsof the form H)ResX(H1, .

. .

, Hn, H)and Qp∈VHLd (p) have the same degree equal

AN INEQUALITY FOR POLYNOMIAL MAPPINGS3to d1 · . .

. · dn, so by the Nullstellensatz there is a constant R0 ∈C such thatResX(H1, .

. .

, Hn, H) = R0Qp∈VHLd (p)for every form H of degree d.Putting in the above equality H = Ld we get R0 = ResX(H1, . .

. , Hn, Ld).

Tocheck (2.3) in the general case let us note that the mapping Pn\V (L) →Cn definedin (2.3) is an analytic branched covering of degree d1 · . .

. · dn.

Let Ωbe a Zariskiopen subset of Cn contained in the set of regular values of this mapping. Hencefor any a = (a1, .

. .

, an) ∈Ωthe hypersurfaces Hi −aiLdi = 0(1 ≤i ≤n) meettransversally and we may apply the formula to the homogeneous forms H1 −a1Ld1,. .

. ,Hn −anLdn.

We obtain the property 2.3 in the general case by passing to thelimit when a →0.3. Proof of the main result.We begin withLemma 3.1 (cf.

[2], lemma 8.2). Let P(W, T ) = P0(W)T d + · · · + Pd(W) ∈C[W, T ], d = degT P(W, T ) > 0 be a polynomial of n + 1 variables (W, T ) =(W1, .

. .

, Wn, T ) such that P(0, T ) ̸≡0. Let δ = d −degT P(0, T ).

Then there is apositive constant C > 0 such that the condition P(w, t) = 0 impliesC|t|−δ ≤|w|for |t| ≫1.Proof of (3.1). If δ = 0, then the lemma follows from the theorem on continuityof roots.

Let δ > 0. Then P0(0) = · · · = Pδ−1(0) = 0 and Pδ(0) ̸= 0.

Supposet ̸= 0 and put s = t−1. The equation P(w, t) = 0 can be rewritten in the formP0(w) + · · · + Pδ(w)sδ + · · · + Pd(w)sd = 0By the Weierstrass Preparation Theorem we get sδ+Q1(w)sδ−1+· · ·+Qδ(w) = 0near the origin with holomorphic Qj,Qj(0) = 0 for j = 1, .

. .

, δ. Consequentlyfor small |s|, |w| we have: |s|δ ≤C1|w|(|s|δ−1 + · · · + 1) ≤C2|w| and the lemmafollows.Let eZ = (Z0, Z1, . .

. , Zn) = (Z0, Z), eFi = eFi( eZ) the homogenized polynomialFi = Fi(Z) and let G = c1Z1 + · · · + cnZn be a linear form such that(∗)V ( eF1, .

. .

, eFn, Z0, G) = ∅Let T be a variable. We may assume that d1, .

. .

, dn > 0. We putPG(W, T ) = Res eZ eF1( eZ) −W1Zd10 , .

. .

, eFn( eZ) −WnZdn0 , G(Z) −T Z0From the properties of resultant quoted in Section 2 we get immediatelyProperty 3.2. PG(W, T ) ∈C[W, T ],PG(w, t) = 0 if and only if the system ofequations F(Z) = w, G(Z) = t has a solution in Cn.

In particular PG(F(Z), G(Z)) = 0.Property 3.3. If the fiber F −1(w), w ∈Cn is finite, then degT PG(w, T ) =Pz∈F −1(w)multz FNow, we are in a position to prove theorem 1.1.

From property 3.3 we get(∗∗)degT PG(W, T ) −degT PG(0, T ) = δ0(F)for every form G satisfying (∗).

4BY ARKADIUSZ P LOSKILet d(F) > 0. After a change of coordinates we may assume that the formsG = Zi(1 ≤i ≤n) satisfy condition (∗).

Let Pi(W, T ) = PG(W, T ) with G = Zi.From relations Pi(F(Z), Zi) = 0(1 ≤i ≤n), property (∗∗) and lemma 3.1 weget|F(z)| ≥C|zi|−δ0(F )for |zi| ≫1for some C > 0 and (1.1) follows.Let d(F) = 0.Fix a linear form satisfying (∗).Then PG(W, T ) = P0(W)and F(Cn) = { w ∈Cn : P0(w) = 0 } is algebraic.Obviously F −1(0) = ∅(ifd(F) = 0 then every fiber of F is empty or infinite) so there is a C > 0 such that{ w ∈Cn : |w| < C } ∩F(Cn) = ∅, hence |F(z)| ≥C for all z ∈Cn which proves(1.1) because δ0(F) = 0 if d(F) = 0.References[1] J. Ch ιadzy´nski, On proper polynomial mappings, Bull. Pol.

Ac. Math.

31 (1983), 115–120. [2] J. Ch ιadzy´nski, T. Krasi´nski, On the Lojasiewicz exponent at infinity for polynomial mappingsC2 into C2 and components of polynomial automorphisms of C2, Ann.

Pol. Math.

(to appear). [3] Z. Jelonek, The sets of points at which a polynomial mapping is not proper, Preprint C.M.R.No 141 (March 1992).

[4] W. Gr¨obner, Moderne algebraische Geometrie, Wien–Insbruck, 1949. [5] J. Koll`ar, Sharp effective Nullstellensatz, Journal AMS 1 (1988), 963–975.

[6] S. Lojasiewicz, Introduction to complex analytic geometry, Birkh¨auser Verlag, 1991. [7] A. P loski, On the growth of proper polynomial mappings, Ann.

Pol. Math.

45 (1985), 297–309.Department of Mathematics,Technical University,Al. 1000 LPP 7, 25–314 Kielce,Poland

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