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INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES

WARSZAWA 1998

GROWTH AT INFINITY OF A POLYNOMIAL WITH A COMPACT ZERO SET

J A N U S Z G W O ´ Z D Z I E W I C Z Kielce University of Technology

Al. Tysi¸ aclecia Pa´ nstwa Polskiego 7, 25-314 Kielce, Poland E-mail: matjg@tu.kielce.pl

1. Introduction. In the article we give the explicit bound for the growth at infinity of a polynomial with a compact set of zeros. Our aim is to prove the following theorem:

Theorem 1. Let F ∈ R[X

1

, . . . , X

n

] be a polynomial of degree d > 2 such that the set F

−1

(0) is compact. Then there exist constants c, R > 0 such that

|F (x)| ≥ c|x|

d−(d−1)n

for all |x| > R.

Recall that we have a similar estimation in the complex case. Consider a polynomial map H : C

n

→ C

n

of degree d such that H

−1

(0) is finite. Then, by Koll´ ar’s theorem,

|H(z)| ≥ const.|z|

d−dn

for |z|  1 (see [Ko]). Our theorem is a real counterpart of this inequality.

2. Two lemmas. The following lemmas will be used in the proof of the main theo- rem.

Lemma 1. Let G : R

n

→ R be a polynomial of positive degree d. Then there exists a linear automorphism L : R

n

→ R

n

such that the polynomial F = G ◦ L satisfies the following conditions:

(i) All partial derivatives of F are of degree d − 1.

(ii) The sets Γ

i

= {x ∈ R

n

| ∂F/∂X

1

(x) = . . . = ∂F/∂X

i−1

(x) = ∂F/∂X

i+1

(x) = . . . = ∂F/∂X

n

(x) = 0, ∂F/∂X

i

(x) 6= 0} (1 ≤ i ≤ n) are one-dimensional submanifolds of R

n

whenever they are not empty,

(iii) For every x ∈ Γ

i

(1 ≤ i ≤ n) the differentials d

x

(∂F/∂X

1

), . . . , d

x

(∂F/∂X

i−1

), d

x

(∂F/∂X

i+1

), . . . , d

x

(∂F/∂X

n

) are linearly independent.

Research supported by KBN grant 8 P03A 07908.

1991 Mathematics Subject Classification: 14P10.

Received by the editors: December 3, 1996; in the revised form: June 2, 1998.

The paper is in final form and no version of it will be published elsewhere.

[123]

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P r o o f. Let GL(n) be the set of linear automorphisms of R

n

. We claim that {L ∈ GL(n) | deg ∂G ◦ L

∂X

1

= . . . = deg ∂G ◦ L

∂X

n

= d − 1}

is a dense subset of GL(n). Let G

d

(X

1

, . . . , X

n

) be the leading form of the polyno- mial G that is the homogeneous polynomial of degree d for which deg(G − G

d

) < d.

Consider a substitution G

d

◦ L where L(X

1

, . . . , X

n

) = ( P

n

i=1

l

i1

X

i

, . . . , P

n

i=1

l

ni

X

i

). We have (G

d

◦ L)(X

1

, . . . , X

n

) = G

d

( P

n

i=1

l

i1

X

i

, . . . , P

n

i=1

l

in

X

i

) = G

d

(l

11

, . . . , l

1n

)X

1d

+ . . . . . . + G

d

(l

n1

, . . . , l

nn

)X

nd

+ other monomials. If G

d

(l

11

, . . . , l

1n

) 6= 0, . . . , G

d

(l

n1

, . . . , l

nn

) 6= 0, then all partial derivatives of G ◦ L are of degree d − 1. Since the set {L ∈ GL(n) | G

d

(l

11

, . . . , l

1n

) 6= 0, . . . , G

d

(l

1n

, . . . , l

nn

) 6= 0} is a complement of a proper algebraic set, it is open and dense in GL(n). This proves the claim.

For any x = (x

1

, . . . , x

n

) from R

n

\ {0} we denote by [x] the corresponding point [x

1

, . . . , x

n

] of the projective space RP

n−1

. Consider the map

[grad G] : R

n

\ (grad G)

−1

(0) → RP

n−1

.

From the semialgebraic version of Sard’s lemma (see [BR], page 82) it follows that the set of regular values of this map contains an open subset U ⊂ RP

n−1

. The set V = {(v

1

, . . . , v

n

) ∈ R

n

× . . . × R

n

| det(v

ji

) 6= 0, [v

i

] ∈ U for i = 1, . . . , n} is an open subset of R

n

×. . .×R

n

. Each n-tuple v = (v

1

, . . . , v

n

) from this set yields a linear automorphism A

v

: R

n

→ R

n

, A

v

(x) = (hv

1

, xi, . . . , hv

n

, xi). Hence the set {A

v

∈ GL(n) | v ∈ V } is open in GL(n). Since GL(n) 3 A → A

−1

∈ GL(n) is an open map, {A

−1v

∈ GL(n) | v ∈ V } is also an open subset of GL(n). Thus, there exists v = (v

1

, . . . , v

n

) ∈ V such that the automorphism L = A

−1v

satisfies (i).

Let us define the polynomial F = G ◦ L. Since G = F ◦ A

v

, grad G = A

Tv

◦ grad F ◦ A

v

, where A

Tv

is the adjoint of A

v

. From this equation it follows that for any w ∈ R

n

\ {0}, [w] is a regular value of [grad F ] if and only if [A

Tv

(w)] is a regular value of [grad G]. Let e

1

= (1, . . . , 0), . . . , e

n

= (0, . . . , 1) form the standard basis of R

n

. Since A

Tv

(e

i

) = v

i

for i = 1, . . . , n, we conclude that [e

1

], . . . , [e

n

] are regular values of [grad F ]. Applying the implicit function theorem to the map [grad F ] we see that each of the sets Γ

i

= [grad F ]

−1

([e

i

]) (1 ≤ i ≤ n) is either a one-dimensional submanifold of R

n

or is empty.

This proves (ii).

We prove the third part of the lemma for Γ

n

. To simplify the notation we write

i

F for ∂F/∂X

i

. Because [e

n

] is a regular value of [grad F ], 0 ∈ R

n−1

is a regular value of the map ψ : R

n

\ (∂

n

F )

−1

(0) → R

n−1

, ψ = (∂

1

F/∂

n

F, . . . , ∂

n−1

F/∂

n

F ) that is, the map [grad F ] written in the coordinates {[x

1

, . . . , x

n

] ∈ RP

n−1

| x

n

6= 0} 3 [x

1

, . . . , x

n

] → (x

1

/x

n

, . . . , x

n−1

/x

n

) ∈ R

n−1

. Therefore, for every x ∈ Γ

n

the differen- tials d

x

(∂

1

F/∂

n

F ), . . . , d

x

(∂

n−1

F/∂

n

F ) are linearly independent. On the other hand, for x ∈ Γ

n

and i = 1, . . . , n − 1 we have d

x

(∂

i

F/∂

n

F ) = (1/∂

n

F )d

x

(∂

i

F ), therefore the differentials d

x

(∂

1

F ), . . . , d

x

(∂

n−1

F ) are also linearly independent. The proof for Γ

i

, i 6= n is similar.

Further, we denote by |x| the supremum norm |x| = max{|x

1

|, . . . , |x

n

|} for x = (x

1

, . . . , x

n

). We will also use the following convention: Using notation |x|  1 we mean that the corresponding condition is satisfied for |x| > R, where R is sufficiently large.

Lemma 2. Let F ∈ R[X

1

, . . . , X

n

] be a polynomial with a compact set of zeros and

let K = {x ∈ R

n

| ∀y ∈ R

n

|y| = |x| ⇒ |F (y)| ≥ |F (x)|}. If A ⊂ K is an unbounded

semialgebraic set , then the following conditions are equivalent :

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(i) |F (x)| ≥ c|x|

α

for |x|  1, (ii) |F (x)| ≥ c|x|

α

for |x|  1, x ∈ A.

P r o o f. The implication (i) ⇒ (ii) is obvious. Assume that (ii) is true. Since {|x|

x ∈ A} is an unbounded semialgebraic subset of R

+

, there exists a constant R > 0 such that (R, ∞) ⊂ {|x|

x ∈ A}. By (ii) we can choose R sufficiently large so that

|F (x)| ≥ c|x|

α

for |x| ≥ R, x ∈ A. Let y ∈ R

n

be an arbitrary point with |y| > R.

Then there exists x ∈ A such that |x| = |y|. By (ii) and the definition of K we get

|F (y)| ≥ |F (x)| ≥ c|x|

α

= c|y|

α

which ends the proof.

3. Proof of Theorem 1. The proof proceeds by induction on the number of vari- ables. For polynomials in one variable the theorem is obvious. Assume that the theorem holds for polynomials in n − 1 variables. We shall check that it is true for polynomials in n variables.

We shall perform some reductions:

If the theorem is true for a polynomial F , then it holds also for F ◦ L, where L : R

n

→ R

n

is a linear automorphism. Therefore, we can assume that F satisfies the conditions (i), (ii) and (iii) of Lemma 1.

The set F

−1

(0) is bounded. Hence F (x) 6= 0 for all |x| > R, where R is sufficiently large. Since for n ≥ 2 the set {x ∈ R

n

|x| > R} is connected, a sign of F restricted to {x ∈ R

n

|x| > R} does not change. Without loss of generality we can assume that F (x) > 0 for |x| > R.

Let

K = {x ∈ R

n

| ∀y ∈ R

n

|y| = |x| ⇒ |F (y)| ≥ |F (x)|}.

First, we prove the theorem under the additional assumption that K ∩ (grad F )

−1

(0) is unbounded. Let A be an unbounded connected component of this set. Since grad F (x) = 0 for x ∈ A, we conclude that F |

A

= c with some c > 0 (see [BR], Theorem 2.5.1). By Lemma 2 we get |F (x)| ≥ c|x|

0

for |x|  1 which ends the proof in this case.

Hence we may assume throughout the rest of the proof that K ∩ (grad F )

−1

(0) is bounded.

Let us define

A

i

= {x ∈ R

n

| |x

k

| < |x

i

| for k ∈ {1, . . . , n} \ {i}}, B

i,j

= {x ∈ R

n

| x

i

= x

j

, |x

k

| ≤ |x

i

| for k = 1, . . . , n}, C

i,j

= {x ∈ R

n

| x

i

= −x

j

, |x

k

| ≤ |x

i

| for k = 1, . . . , n}.

Since R

n

= S A

i

∪ S B

i,j

∪ S C

i,j

, at least one of the sets K ∩ S A

i

, K ∩ S B

i,j

, K ∩ S C

i,j

is unbounded. Let us consider three cases:

Case 1: K ∩S B

i,j

is unbounded. Then at least one of the sets K ∩B

i,j

(1 ≤ i < j ≤ n) is unbounded. Without loss of generality we can assume that this is the set K ∩ B

n−1,n

. Consider the polynomial ˜ F (X

1

, . . . , X

n−1

) = F (X

1

, . . . , X

n−1

, X

n−1

) of degree ˜ d ≤ d.

By the inductive assumption we have | ˜ F (˜ x)| ≥ c|˜ x|

d−( ˜˜ d−1)n−1

for ˜ x ∈ R

n−1

, |˜ x|  1.

If we take any x ∈ B

n−1,n

, x = (x

1

, . . . , x

n−1

, x

n−1

) and if we set ˜ x = (x

1

, . . . , x

n−1

), then |˜ x| = |x| and ˜ F (˜ x) = F (x). Hence |F (x)| ≥ c|x|

d−( ˜˜ d−1)n−1

for |x|  1, x ∈ B

n−1,n

. By Lemma 2 and by the inequality ˜ d−( ˜ d−1)

n−1

≥ d−(d−1)

n

we get |F (x)| ≥ c|x|

d−(d−1)n

for |x|  1.

Case 2: K ∩ S C

i,j

is unbounded. The proof is analogous.

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Case 3: K ∩ S A

i

is unbounded. Then at least one of the sets K ∩ A

i

(1 ≤ i ≤ n) is unbounded. Without loss of generality we can assume that this is K ∩ A

n

.

Take R > 0 large enough so that F (x) > 0 for |x| > R and let y = (y

1

, . . . , y

n

) be an arbitrary point in K ∩ A

n

with |y| > R. Consider a function f (x

1

, . . . , x

n−1

) = F (x

1

, . . . , x

n−1

, y

n

) defined for |x

i

| < |y

n

| (1 ≤ i < n). Taking into account two points, y = (y

1

, . . . , y

n−1

, y

n

) and x = (x

1

, . . . , x

n−1

, y

n

), where |x

i

| < |y

n

| (1 ≤ i < n), we see that |x| = |y|, therefore F (x) ≥ F (y). Hence the point (y

1

, . . . , y

n−1

) is a local minimum of f . Thus ∂F/∂X

1

(y) = . . . = ∂F/∂X

n−1

(y) = 0.

Summarizing, we see that for all y ∈ K ∩ A

n

, |y|  1 we have ∂F/∂X

1

(y) = . . . =

∂F/∂X

n−1

(y) = 0, ∂F/∂X

n

(y) 6= 0. Moreover, from Lemma 1 it follows that K ∩ A

n

is a one-dimensional semialgebraic manifold in a neighborhood of infinity. We want to find a parametrization of a branch at infinity of this set. To that end we employ complex algebraic geometry.

Define H

1

= ∂F/∂X

1

, . . . , H

n−1

= ∂F/∂X

n−1

and let C = {z ∈ C

n

| H

1

(z) = . . . . . . = H

n−1

(z) = 0}. Decompose C to the union of irreducible algebraic components C = C

1

∪ . . . ∪ C

s

. Treating R

n

as a subset of C

n

we see that K ∩ A

n

∩ C is unbounded.

Hence there exists a component C

i

such that K ∩ A

n

∩ C

i

is unbounded. For simplicity put Γ = C

i

.

We will check that dim

C

Γ = 1. By Lemma 1 there exists x ∈ K ∩ A

n

∩ Γ for which the differentials d

x

H

1

, . . . , d

x

H

n−1

are linearly independent. Therefore, dim

C

Γ ≤ n−rank(Γ, x) ≤ n−rank(d

x

H

1

, . . . , d

x

H

n−1

) = 1 (see [BR], pages 122–135). Furthermore, Γ is unbounded, so dim

C

Γ = 1.(

1

)

Next, we will check that deg Γ ≤ (d − 1)

n−1

. Let us recall an invariant δ of algebraic sets introduced in Lojasiewicz’s book ([ Lo] pages 419–420): Let W = W

1

∪ . . . ∪ W

s

be a decomposition of an algebraic set W to irreducible components. Then, by definition δ(W ) = P

s

i=1

deg W

i

. We will use the inequality δ(W ∩ V ) ≤ δ(W )δ(V ). Applying this property to the set C we see that deg Γ ≤ δ(C) = δ({H

1

= 0} ∩ . . . ∩ {H

n−1

= 0}) ≤ Q

n−1

i=1

δ({H

i

= 0}) ≤ (d − 1)

n−1

.

Further, we will consider C

n

as a affine part of the projective space CP

n

. We will use the natural identification between (x

1

, . . . , x

n

) ∈ C

n

and [1, x

1

, . . . , x

n

] ∈ CP

n

. With the use of this identification we can treat K, A

n

and Γ as subsets of CP

n

.

Since K ∩ A

n

∩ Γ is an unbounded set and CP

n

is compact, there exists a point a in the hyperplane at infinity {[x

0

, . . . , x

n

] ∈ CP

n

| x

0

= 0} such that a ∈ cl(K ∩ A

n

∩ Γ).

The homogeneous coordinates of a can be chosen such that a = [0, a

1

, . . . , a

n−1

, 1].

Indeed, for all x ∈ A

n

we have |x

i

| < |x

n

| for 1 ≤ i < n. Since a ∈ cl(A

n

), |a

i

| ≤ |a

n

| for 1 ≤ i < n. Therefore, the last coordinate a

n

does not vanish and by homogeneity we can assume that a

n

= 1.

Let ¯ Γ be the projective closure of the curve Γ. Since a ∈ ¯ Γ, according to [ Lo]

(pages 173–176) there exists a finite sequence of injective holomorphic parametrizations γ

i

: (D, 0) → (¯ Γ, a) (1 ≤ i ≤ l), where D = {t ∈ C | |t| < δ}, such that the curve ¯ Γ is the union γ

1

(D) ∪ . . . ∪ γ

l

(D) in some neighborhood of a. These parametrizations are of the form

γ

i

(t) = [t

di

, γ

i,1

(t), . . . , γ

i,n−1

(t), 1].

Furthermore, we can assume that the real branches of ¯ Γ are parametrized such that (t

di

, γ

i,1

(t), . . . , γ

i,n−1

(t)) ∈ R

n

if and only if t ∈ R. This can be done by substituting

(

1

) If dim

C

Γ = 0, then Γ would consist of one point.

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γ

i

i

t), where ξ

i

is an appropriate d

i

-th root of unity and by shrinking δ if necessary (see [Mi] or [Du] for the details).

Let H = H(X

0

, . . . , X

n

) be the homogenization of the polynomial ∂F/∂X

n

. Recall that it means that H is a homogeneous polynomial of degree deg H = deg ∂F/∂X

n

such that H(1, X

1

, . . . , X

n

) = ∂F/∂X

n

(X

1

, . . . , X

n

). We can calculate the intersection multiplicity of the curve ¯ Γ and the hypersurface {H = 0} at a using the formula

ι

a

(¯ Γ, {H = 0}) =

l

X

i=1

ord

0

(H ◦ γ

i

)

(see [Sh], pages 190–194). By B´ ezout’s theorem ι

a

(¯ Γ, {H = 0}) ≤ (deg ¯ Γ)(deg H) ≤ (d − 1)

n

. Hence ord

0

(H ◦ γ

i

) ≤ (d − 1)

n

for i = 1, . . . , l.

One has a ∈ cl(K ∩ A

n

∩ Γ). Hence there exists i (1 ≤ i ≤ l) such that a ∈ cl(K ∩ A

n

∩ γ

i

(D)). Since γ

i

is a proper map, 0 ∈ cl(γ

−1

(K ∩ A

n

)). Furthermore, we see by the definition of γ

i

that γ

−1

(K ∩ A

n

) is a semianalytic subset of R. Therefore there exists

 > 0 such that γ

i

((0, )) ⊂ K ∩ A

n

or γ

i

((−, 0)) ⊂ K ∩ A

n

(see [BM] for the definition and basic properties of semianalytic sets). In the rest of the proof we assume the former case (the proof for the case γ

i

((−, 0)) ⊂ K ∩ A

n

is similar). We will again treat K, A

n

and Γ as subsets of C

n

.

Set the following meromorphic map

φ : {t ∈ C | 0 < |t| < } 3 t → (γ

i,1

(t)/t

di

, . . . , γ

i,n−1

(t)/t

di

, 1/t

di

) ∈ C

n

.

Notice that φ({t ∈ C | 0 < |t| < }) ⊂ Γ and that φ((0, )) is an unbounded semialgebraic subset of K ∩ A

n

.

We estimate the order of F ◦ φ at zero. Either ord

0

(F ◦ φ) = 0 or by the equation (F ◦ φ)

0

=  ∂F

∂X

1

◦ φ



φ

01

+ . . . +  ∂F

∂X

n

◦ φ



φ

0n

=  ∂F

∂X

n

◦ φ

 φ

0n

we have ord

0

(F ◦ φ) = ord

0

(∂F/∂X

n

◦ φ) − d

i

.

On the other hand we have

∂F

∂X

n

(φ(t)) = H(1, γ

i,1

(t)/t

di

, . . . , γ

i,n−1

(t)/t

di

, 1/t

di

)

= t

−dideg H

H(t

ti

, γ

i,1

(t), . . . , γ

i,n−1

(t), 1) = t

−dideg H

H(γ

i

(t)).

Since deg H = d−1 and ord

0

(H ◦γ

i

) ≤ (d−1)

n

, we conclude that ord

0

(∂F/∂X

n

◦φ) ≤ (d − 1)

n

− d

i

(d − 1). By this inequality and the preceding equalities we have ord

0

(F ◦ φ) ≤ (d − 1)

n

− d

i

d or ord

0

(F ◦ φ) = 0.

R e m a r k. If f, g : {t ∈ C | 0 < |t| < } → C, f 6= 0, g 6= 0 are meromorphic func- tions, then there exist constants c, 

1

> 0 such that |f (t)| ≥ c|g(t)|

ord0f / ord0g

for all t ∈ C, 0 < |t| < 

1

.

The proof of this fact is simple and is left to the reader. By the remark and by the fact that φ((0, )) ⊂ A

n

implies |φ(t)| = |φ

n

(t)| for t ∈ (0, ), we obtain an inequality

|F (φ(t))| ≥ c|φ(t)|

ord0(F ◦φ)/ ord0n)

for t ∈ (0, 

1

) with some positive constants c, 

1

. By Lemma 2 we have

|F (x)| ≥ c|x|

ord0(F ◦φ)/ ord0n)

for |x|  1.

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Moreover, the exponent ord

0

(F ◦φ)/ ord

0

n

) ≥ ((d−1)

n

−d

i

d)/(−d

i

) = d−(d−1)

n

/d

i

≥ d − (d − 1)

n

or is equal zero and thus

|F (x)| ≥ c|x|

d−(d−1)n

for |x|  1.

4. Concluding remarks. In the course of the proof we have found a parametrization φ of the set K at infinity such that |F (x)| ≥ c|x|

ord0(F ◦φ)/ ord0n)

for|x|  1. By a slight modification of the proof one can show that the number ord

0

(F ◦ φ)/ ord

0

n

) is the Lojasiewicz exponent at infinity for the polynomial F , i.e. the largest exponent α for which the estimate |F (x)| ≥ const.|x|

α

is true for |x|  1.

We have checked that ord

0

(F ◦ φ) ≤ (d − 1)

n

− d ord

0

n

) or ord

0

(F ◦ φ) = 0. One can also prove the inequality (d − 1)

n−1

≤ ord

0

n

) < 0. As a result, there is only a finite number of fractions which can be the Lojasiewicz exponents for polynomials of fixed number of variables n and of fixed degree d.

So far I have not found a polynomial F for which the Lojasiewicz exponent L

(F ) = d − (d − 1)

n

. For example for the polynomial F (X

1

, . . . , X

n

) = (X

2

X

1m−1

− 1)

2

+ (X

3

−X

2m

)

2

+. . .+(X

n

−X

n−1m

)

2

+X

n2m

of degree d = 2m we have L

(F ) = d−(1/2

n−1

)d

n

. This suggests that Theorem 1 could be essentially sharpened.

I want to express my thanks to Stanis law Spodzieja for pointing out a mistake in the first version of this paper.

References

[BR] B. B e n e d e t t i, J. J. R i s l e r, Real Algebraic and Semi-algebraic Sets, Hermann, Paris, 1990.

[BM] E. B i e r s t o n e, P. D. M i l m a n, Semianalytic and subanalytic sets, Inst. Hautes ´ Etudes Sci. Publ. Math. 67 (1988), 5–42.

[Du] Z. D u s z a k, Indeks rzeczywistych odwzorowa´ n analitycznych, Doctoral Thesis, Uniwersytet L´ odzki, 1987.

[Ko] J. K o l l ´ a r, Sharp effective Nullstellensatz , J. Amer. Math. Soc. 1 (1988), 963–975.

[ Lo] S. L o j a s i e w i c z, Introduction to Complex Algebraic Geometry , Birkh¨ auser, Basel, 1991.

[Mi] J. W. M i l n o r, Singular Points of Complex Hypersurfaces, Princeton Univ. Press, Prince- ton, 1968.

[Sh] I. R. S h a f a r e v i c h, Basic Algebraic Geometry , Springer, New York, 1974.

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