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THE JACOBIANS OF LOWER DEGREES

Grzegorz Biernat

Institute of Mathematics and Computer Science, Czestochowa University of Technology

Abstract. In the present paper we give some relation of the number of zeros of a polyno- mial mapping in C2with a jacobian of non-maximal degree and the number of branches at infinity of one coordinate of this mapping.

1. Auxiliary facts

Let l =V

( )

T0 denote a line at infinity in the projective complex space P 2 (with homogeneous coordinates T0:T1:T2). Further it will be called infinity. If

∈ l

a then by a~∈C2 we denote the canonical image of the point a in affine part P2\V

( )

T1 ≅C2. For a polynomial h of two variables, h~ signifies a suitable dehomogenization of the homogenization of the polynomial h. So, we have

(

,

) (

1 ,

)

.

~

1 2 1 deg 1 2

1 X X h X X X

X

h = h

Let f1, f2 and g be polynomials of two variables and let C1,C2 be the closures, respectively, of the curves V

( ) ( )

f1 ,V f2 in the space .

P Assume further that 2

polynomials f1 and f2 are different from constants and write n =1 degf1, .

deg 2

2 f

n = We denote by Jf =Jac

(

f1,f2

)

(respectively, J~f =Jac

(

~f1,~f2

)

) the jacobian of the mapping f =

(

f1,f2

)

(respectively,

(

1 2

)

,~

~

~ f f

f = ).

Fact 1. If degJf <degf1+degf2−2, then

(

C ∪C

)

∩l =

(

C ∩C

)

∩l

2 1 2

1 .

Proof. Let 1 ,

f+ f2+ be the leading forms of the polynomials f1, f2, respectively. Put

(

+ +

)

+= f1 ,f2

f . Since the degree of the jacobian Jf is not maximal, then the jaco- bian Jf+ =0. It means that the homogeneous polynomials 1 ,

f+ f2+ are algebraically independent. Thus, there is a polynomial h of two complex variables of positive degree without constant term such that h o f+=0.Let

(

1, 2

)

=

i,j ij 1 2 , iY j

Y c Y

Y

h α β where cij≠0, αij ≥1

(2)

For an arbitrary point

(

a,b

)

C2\

{ (

0,0

) }

we have

( ) ( )

(

t 1f1+ a,b ,t 2 f2+ a,b

)

=0

h n n for t∈C (1)

If the point

(

0:a:b

)

∈C1∩l, then f1+

(

a,b

)

=0 and (1) reduces to identity

( )

(

2 ,

)

2 =0,

j j + n j j

t b a f

d β β where dj ≠0, βj ≥1,t∈C

It means that f2+

(

a,b

)

=0 and the point

(

0:a:b

)

∈C2∩l. Analogously, if

(

0:a:b

)

∈C2∩l, then

(

0:a:b

)

∈C1∩l. This ends the proof.

Assume further that the polynomials f1 and f2 have not common factors of positive degrees and the polynomial f1 has not irreducible multiples factors. Then the ca- nonical image a~ of a point a∈

(

C ∩C

)

∩l

2

1 is an isolated zero of the mapping

~f

and the germ

( )

f1a~

~ of the function ~1

f in the pointa~ has reduced decomposition [2]. Let

( )

f1 a~ h1Khk

~ = (2)

be suitable decomposition of the germ

( )

f1a~

~ into irreducible single factors in the ring of the germs of holomorphic functions in the point .

a~ Write

i a i =ord~h

µ and κi=mult~a

(

hi,f2

)

for 1≤i ≤k Fact 2. If 0

2

i

i n µ

κ for 1≤i ≤k, then the germ

( )

Jf a

~

~ does not vanish identi- cally on the set of zeros of all factors in the decomposition (2). In particular

.

≠0 Jf

Proof. Assume contrary that for parametrization i

( )

t

(

t i

( )

t

)

i 0 0

0 = µ

Φ of zeros of

the factor

i0

h in the decomposition (2) we have J~

(

Φ t0

( ) )

=0.

i

f Then according

to the formula (*) in [1] we have

( ( ) ) ( ( ) ) ( ( ) )

0

~

~

0 0 0

0 ~

2 1 2

2 Φ + Φ =

Φ ∂ t t J t

X t f f

n i i µi f i (3)

From another hand we have also

( ( ) )

(

~ 0

)

~

(

0

( ) ) ( )

0 ~

(

0

( ) )

0

2

2 1 Φ =

′ + Φ

′ ∂

Φ t t J t

X t f

f i i µi f i (4)

(3)

The equalities (3) and (4) have not zero solution, so

( ( ) ) ( ) (

~

( ( ) ) )

0

~

0 0 0

0 2

2

2 ′=

Φ

′−

Φ t t t f t

f

n i µi µi i

and

( ( ) )

( )

( ( ) ) ( )

0 0

0

0 2

2 2

~

~

i i

t t n t f

t f

i i

µ

µ

Φ = Φ ′

Simple integration gives κ =i n2 iµ , which contradicts assumption.

2. Basic fact

Assume that the polynomial f1 is irreducible and degJf <degf1+deg f2−2. Let g1 denotes the genus of the curve C1 and let a1,K,as be the zeros at infinity of the mapping f. According to the Fact 1 we infer that these zeros are exactly the points at infinity of the curve C1. In each point a~ we have reduced decomposition k

( )

f1 a~k h1( )k Khr( )kk

~ = for 1≤k ≤s (5)

where rk denotes the number of branches of the curve C1 in the point ak at infinity.

Write

( ) ( )k

j a k

j h

~k

=ord

µ and ( ) mult~a

(

h( )jk ,~f2

)

k

j = k

κ for 1≤ j ≤rk

Fact 3. Let p be the number of zeros of the mapping f and q the number of zeros of the mapping

(

f1,Jf

)

with respect of the multiplicity. If ( ) ( ) 0

2

jk k

j n µ

κ for

k ≤s

1≤ and 1≤ j ≤rk, then p s 1r q 2

(

1 g1

)

k k≤ + −

+

= . Moreover the number

q r

p s

k k

+

=1 is even.

Proof. For every point a~kdefine non-negative integer

(

1

)

2

1 + −

= k k

k M r

δ

where Mk is the Milnor number of the curve ~) (f1

V at the point a~k [3]. Summing we have

( )

∑ ∑

sk= k = sk= Mk + sk= rk−s

1 1

1 2

1 2

δ 1 (6)

(4)

For every function h( )jk from the decomposition (5) denote by

( )

( )

( ) ( )

( )

 

=

Φ t t k t

j k

j

k

j ϕ

µ

, the parametrization of its zeros. From the formula (*) in [1] it follows that

( ) ( )

(

( )

( ) )

( )

(

( )

( ) ) (

( )

( )

t

)

( )t ( )J

(

( )

( )

t

)

X t f f n t J

t jk

f k j k

j k

j k j k

j f k

j

k j k

j Φ − Φ

∂ Φ ∂

=

Φ ~

2 1 2

2

~

~

~ µ

σ µ

µ µ

µ (7)

whereσ =n1+n2−2−degJf ≥1.From another hand we have

( ) ( )

(

( )

( ) ) ( (

( )

( ) ) ) (

( )

( )

t

)

X t f f t J

t jk jk jk

f k

j k

j Φ

′ ∂ Φ

= Φ

2 1

~ 2

1 ~

~

µ

µ so

( ) ( )

(

( )

( ) ) ( (

( )

( ) ) ) (

( )

( )

t

)

X t f f t t J

t jk jk jk

f k j

k

j Φ

′ ∂ Φ

= Φ

2 1

~ 2

~

~

µ

µ (8)

From (7) and (8) we have

( ) ( )

(

( )

( ) ) (

( )

( ) )

( )

(

( )

( ) ) ( (

( )

( ) ) )

 

 ′

Φ + Φ

∂ Φ

= ∂

Φ t n f t t f t

X t f J

t f jk jk jk jk jk

k j

k j

2 2

2 2

1 ~ ~

~

~ µ

µ

σ µ

Since ~2

(

Φ( )

( ) )

= ( )+ higher terms,

k

ct j

t

f jk κ where c≠0 and ( )2 ( )jk ≠0,

k

j n µ

κ the

order of the second factor on the right side of the above equality is equal ( ).

k

κj

Taking into account of both sides we have

( )

(

( )

( ) ) (

( )

( ) )

( )jk

k j k

j f k

j t

X t f

J κ

σ

µ Φ +

= ∂ Φ +

2 0 1 0

~

~ ord ord

and summing

( )

∑ (

( )

( ) ) ∑ (

( )

( ) ) ∑

( )

= = = Φ + =

= ∂ Φ

+ k k k

k r

j k j r

j

k j r

j

k j f r

j k

j t

X t f

J 1 1

2 1 1 0 0

1

~

~ ord

ord κ

µ σ so

( )

~

(

1 2

)

2 1 1 1 ~

1 ~

~

,~ mult ~

~

~,

~ mult

~,

~ mult

ord f f

X f f J

f

f k k k

k a f a a

a +



= ∂ +

σ

(5)

From the Teissier lemma [3] we infer that

~ 1 ord

~

~,

mult ~ 1

2 1

~ 1 = + −



∂ M f

X f f

k

k k a

a

thus

(

1

)

ord~ ~f1+mult~

(

~f1,J~

)

=M +mult~

(

~f1,~f2

)

1

k k

k a f k a

σ a , 1≤k ≤s

Summing the above equalities over all points at infinity we have

(

σ1

)

n1+mult

(

f1,Jf

)

=

sk=1Mk+mult

(

f1,f2

)

−s By the Bezout theorem

(

f Jf

)

=n Jfq

, deg

mult 1 1 and mult

(

f1,f2

)

=n1n2−p From the above we conclude

(

n

)

n s M q p s

k k + − −

=

1

=1

1 3

and

(

n1−1

)(

n1−2

)

=

sk=1Mk+qps+2 which gives

( )( ) ( )

1

2 1 2

2 1 2 1

1

1 1

1n − =

= M + qps +

n sk k (9)

Subtracting (6) from (9) we have

( )( ) ( )

1

2 2 1

2 1 1

1 1 1

1− − −

= − −

+

=

=

s

k k

s

k k q p r

n

n δ

In the above equality the number on the left hand side is non-negative integer not less than g1 [3]. Thus

2 2 1

1 ≥ −

−p

= r g

q sk k

which proves the fact.

(6)

References

[1] Biernat G., The residue at infinity and Bezout’s theorem, Prace Naukowe IMiI, Częstochowa 2002, 25-27.

[2] Krasiński T., Poziomice wielomianów dwóch zmiennych a hipoteza jakobianowa, Acta Universi- tatis Lodziensis, Wyd. UŁ, Łódź 1991.

[3] Płoski A., O niezmiennikach osobliwości krzywych analitycznych, Materiały VIII Konferencji Szkoleniowej z Teorii Zagadnień Ekstremalnych, Wyd. UŁ, Łódź 1985, 80-93.

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