A POLYNOMIAL WITH 2k CRITICAL VALUES AT INFINITY

Janusz Gwoździewicz^{∗}, Maciej S_{ι}ekalski

IMUJ PREPRINT 2003/11 KRAKÓW

Abstract

We construct a polynomial f : ℂ^{2} → ℂ of degree 4k+2 with no critical points in ℂ^{2} and with 2k critical values at infinity.

Let f : ℂ^{2} → ℂ be a polynomial. We call λ ∈ ℂ a critical value at infinity of f if there exists a sequence z_{n} ∈ ℂ^{2} such that z_{n} → ∞, gradf(z_{n}) → (0, 0) and f(z_{n}) → λ as n → ∞. Several equivalent definitions of critical values at infinity are given in [Durfee]. The definition we use is due to Ha (see [Durfee]).

When studying polynomial automorphisms of ℂ^{2} we encounter polynomials with no critical points in ℂ^{2} (see, for example [Ephraim], [Gwoździewicz–Płoski]). It is natural to ask if a polynomial with no critical points in ℂ^{2} can have many critical values at infinity.

For given integer k > 0 we construct a polynomial f : ℂ^{2} → ℂ of degree 4k + 2 such that

(i) f has no critical points in ℂ^{2},

(ii) f has exactly 2k critical values at infinity.

Let A(x) = x^{k} − 1, B(x) = \frac{x^{k+1}}{k+1} − x and

f(x, y) = (yA^{2} − x)^{2} + B.

∗First author was partially supported by KBN Grant No 2P03A01522

To prove (i) let us recall that critical points of f are solutions of the system of equations \frac{∂f}{∂x} = \frac{∂f}{∂y} = 0, so we have

\{ \begin{matrix}2(yA^{2} − x)(2yAA′ − 1) + A = 0, \\ 2(yA^{2} − x)A^{2} = 0. \end{matrix}

The second equality gives A = 0 or yA^{2} − x = 0. If A = 0 then from the first equality we get x = 0 which is impossible because A(0) ≠ 0. If we assume that yA^{2} − x = 0 then A = 0 and we get a contradiction again.

Now we will show that all critical values at infinity of f are of the form −\frac{k}{k+1}a and a^{2} − \frac{k}{k+1}a for a to be any root of the polynomial A. Assume that z_{n} = (x_{n}, y_{n}) is a sequence which satisfies the four conditions

z_{n} → ∞,  (1)
\frac{∂f}{∂x} (z_{n}) = 2(y_{n}A(x_{n})^{2} − x_{n})(2y_{n}A(x_{n})A′(x_{n}) − 1) + A(x_{n}) → 0, (2)
\frac{∂f}{∂y} (z_{n}) = 2(y_{n}A(x_{n})^{2} − x_{n})A(x_{n})^{2} → 0,  (3)
f(z_{n}) = (y_{n}A(x_{n})^{2} − x_{n})^{2} + B(x_{n}) → λ ∈ ℂ.  (4)

Replacing z_{n} by a subsequence we may additionally assume that x_{n} → ∞ or x_{n} → b ∈ ℂ.

If x_{n} tends to infinity then by (3) y_{n}A(x_{n})^{2} − x_{n} → 0. Hence f(z_{n}) → ∞ (because B(x_{n}) → ∞) which contradicts (4).

Now let us suppose that x_{n} converges to b ∈ ℂ and b is not a root of the polynomial A. From (3) we have y_{n}A(x_{n})^{2} − x_{n} → 0 which gives y_{n} → b/A(b)^{2}. Hence the sequence z_{n} is bounded contrary to (1). So we have proven that x_{n} → a where a is a root of A.

¿From (4) we see that the sequence y_{n}A(x_{n})^{2} − x_{n} is convergent. There are two possibilities. The first one is y_{n}A(x_{n})^{2} − x_{n} → 0. Then f(z_{n}) → B(a) = −\frac{k}{k+1}a. The second one is y_{n}A(x_{n})^{2} − x_{n} → α ≠ 0. Then according to (2) we have 2y_{n}A(x_{n})A′(x_{n}) − 1 → 0 otherwise \frac{∂f}{∂x}(z_{n}) would not be convergent to zero. But we have

y_{n}A(x_{n})^{2} − x_{n} = \frac{A(x_{n})}{2A′(x_{n})}[(2y_{n}A(x_{n})A′(x_{n}) − 1) + 1] − x_{n} → −a.

So we see from (4) that

f(z_{n}) → a^{2} + B(a) = a^{2} − \frac{k}{k + 1}a.

Finally we obtain that the critical values at infinity of the the polynomial f are of the form −\frac{k}{k+1}a and a^{2} − \frac{k}{k+1}a for a ∈ A^{−1}(0). Taking x_{n} → a, y_{n} = x_{n}/A(x_{n})^{2} we get f(z_{n}) → −\frac{k}{k+1}a and taking x_{n} → a, y_{n} = 1/(2A(x_{n})A′(x_{n})) we get f(z_{n}) → a^{2} − \frac{k}{k+1}a.

It suffices to show that the above numbers are pairwise different. We will show that if a ≠ b and a, b ∈ A^{−1}(0) then

a^{2} − \frac{k}{k + 1}a ≠ b^{2} − \frac{k}{k + 1}b,

a^{2} − \frac{k}{k + 1}a ≠ −\frac{k}{k + 1}b.

The first inequality can be rewritten as (a−b)(a+b− \frac{k}{k+1}) ≠ 0 so it is enough to show that the equality a + b = \frac{k}{k+1} does not hold. If the above equality is true then a and b must be conjugate so we get the equality a + ā = \frac{k}{k+1}. Multiplying it by (k + 1)a we get

(k + 1)a^{2} − ka + k + 1 = 0,
a^{k} − 1 = 0.

We will show that these equalities can not hold. To do this let us consider the resultant R of the polynomials (k + 1)x^{2} − kx + k + 1 and x^{k} − 1 i.e.

R = | \begin{matrix}k + 1 & −k & k + 1 & 0 & … & 0 \\ 0 & k + 1 & −k & k + 1 & … & 0 \\ ⋮ & & ... & & & ⋮ \\ 0 & … & k + 1 & −k & k + 1 & 0 \\ 0 & … & 0 & k + 1 & −k & k + 1 \\ 1 & 0 & … & & −1 & 0 \\ 0 & 1 & 0 & … & & −1 \end{matrix} |

and the natural homomorphism ϕ : ℤ → ℤ/(k + 1)ℤ of the ring of integers to the ring of rests of modulo k + 1. Then

ϕ(R) = | \begin{matrix}0 & 1 & 0 & 0 & … & 0 \\ 0 & 0 & 1 & 0 & … & 0 \\ ⋮ & & ... & & & ⋮ \\ 0 & … & 0 & 1 & 0 & 0 \\ 0 & … & 0 & 0 & 1 & 0 \\ 1 & 0 & … & & −1 & 0 \\ 0 & 1 & 0 & … & & −1 \end{matrix} | = (−1)^{k+1}

hence we see that R ≠ 0.

To end the proof we need to show that the equality

a^{2} − \frac{k}{k + 1}a = −\frac{k}{k + 1}b  (5)

is false for any a, b ∈ A^{−1}(0). For that purpose let us notice that if k = 1 then a = b = 1 and a^{2} − \frac{1}{2}a = \frac{1}{2} ≠ −\frac{1}{2} = −\frac{1}{2}b. Let us suppose that k > 1. The equality (5) can be written as

\frac{k + 1}{k} = a^{−1} − ba^{−2} = a^{−1} + (−ba^{−2}).

We see that the numbers a^{−1} and −ba^{−2} must be conjugate hence −ba^{−2} = a. Multiplying the above equality by ka we get

ka^{2} − (k + 1)a + k = 0.

Of course a^{k} − 1 = 0. By using the same method as earlier ie. using a resultant modulo k we can show that the system of equations

kx^{2} − (k + 1)x + k = 0,
x^{k} − 1 = 0

has no solutions for k > 1. This ends the proof.

Acknowledgment: We found k critical values at infinity of f in the first version of the paper. We would like to thank to Tadeusz Krasiński who suggested that the polynomial f has 2k critical values at infinity.

References

[Durfee] A.H. Durfee, Five Definitions of Critical Point at Infinity, Singularities: The Brieskorn anniversary volume (Progr. Math. 162 (1998) 345–360

[Ephraim] R. Ephraim Special polars and curves with one place at infinity Proc. Sympos. Pure Math. 40 (1983), 353–359

[Gwoździewicz–Płoski] J. Gwoździewicz, A. Płoski, On the singularities at infinity of plane algebraic curves, Rocky Mountain Journal of Mathematics Volume 32, Number 1, Spring 2002, 139–148

Janusz Gwoździewicz, Maciej S_{ι}ekalski Department of Mathematics Technical University Al. 1000LPP 7 PL 25–314 Kielce e-mails: matjg@tu.kielce.pl

matms@tu.kielce.pl