Tài liệu Báo cáo " GENERALIED MASON’S THEOREM " ppt
Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (154.39 KB, 6 trang )
VNU. JOURNAL OF SCIENCE, Mathematics - Physics. T.XXI, N
0
4 - 2005
GENERALIED MASON’S THEOREM
Nguyen Thanh Quang, Phan Duc Tuan
Department of Mathematics, Vinh University
Abstract. The purpose of this pap er is to give a generalization of Mason’s theorem
by the Wronskian technique over fields of characteristic 0.
Keywords: The Wronskian technicque, Marson’s theorem.
1. Introduction
Let F be a fixed algebraically closed field of characteristic 0. L et f(z)beapoly-
nomial non - constants which coefficients in F and let
n(1/f) be the number of distinct
zeros of f. Then we have the following.
Marson’s theorem. ([2]). Let a(z),b(z),c(z) be relatively prime polynomials in F and
not all constants such that a + b = c. Then
max {deg(a), deg(b), deg(c)} a
n
1
abc
− 1.
It is now well known that Mason’s Theorem implies the following corollary.
Corollary. (Fermat’s Theorem over polynomials). The equation x
n
+ y
n
= z
n
has no
solutions in non - constants and relatively prime polynomials in F if n a 3.
The main theorem in this paper is as following:
Theorem 1.1. Les f
0
,f
1
, ,f
n
be relatively primer polynomials and f
0
,f
1
, ,f
n
be lin-
early independent over F. If
f
0
+ f
1
+ + f
n
= f
n+1
,
then
max
0aian+1
deg f
i
a n
n+1
i=0
n
1
f
i
−
n(n +1)
2
.
Remark. Theorem 1.1 is a generalization of Mason’s theorem which was obtained for
case n =1.
Typeset by A
M
S-T
E
X
34
Generalied Mason’s Theorem 35
2. Proof of the main theorem
Let ϕ(x)=
f(x)
g(x)
≡ 0 be a rational function, where f(x),g(x) are non - zero and
relatively prime polynomials on F . The degree of ϕ(x), denoted by deg ϕ(x), is defined to
be deg f(x) − deg g(x). Here the n otation deg f(x)meansthedegreeofpolynomialf(x).
From the properties of polynomial, we ha ve.
Proposition 2.1. If ϕ
1
and ϕ
2
are the rational functions on F, then
1) deg(ϕ
1
ϕ
2
)=degϕ
1
+degϕ
2
2) deg
1
ϕ
1
= − deg ϕ
2
3) deg(ϕ
1
+ ϕ
2
) a max(deg ϕ
1
, deg ϕ
2
).
Definition 2.2. Let ϕ(x) ≡ 0 be a rational function on F.Foreverya ∈ F, we write
ϕ(x)=(x − α)
m
f
1
(x)
g
1
(x)
, (m ∈ Z),
where f
1
(x),f
2
(x) are relatively prime polynomials and f
1
(α) =0,g
1
(α) =0. We call m
order of ϕ at α.
Proposition 2.3. If ϕ
1
, ϕ
2
are rational functions on F and a ∈ F, then
1) ord
α
(ϕ
1
ϕ
2
)=ord
α
ϕ
1
+ ord
α
ϕ
2
2) ord
α
(
1
ϕ
1
)=−ord
α
ϕ
1
3) ord
α
(
ϕ
1
ϕ
2
)=ord
α
ϕ
1
− ord
α
ϕ
2
.
Proposition 2.4. Let ϕ(x) be a the rational function on F and let the derivatives order
k, ϕ
(k)
≡ 0. Then
ord
α
ϕ
(k)
ϕ
−k.
Proof. Let ϕ(x)=(x − α)
m
f(x)
g(x)
, whe re f(x),g(x) are relatively prime and f(α)g(α) =0.
Then, we ha ve
ϕ
(x)=(x − α)
m−1
(mf(x)+(x − α)f
(x)) + (x − α)f(x)g
(x)
g
2
(x)
.
Since ord
α
(g(x)) = 0, we have
ord
α
(ϕ
(x)) m − 1.
Therefore
ord
α
ϕ
ϕ
= ord
α
(ϕ
) − ord
α
(ϕ) −1.
36 Nguyen Thanh Quang, Phan Duc Tuan
Thus, we obtain
ord
α
ϕ
(k)
ϕ
= ord
α
ϕ
ϕ
.
ϕ
ϕ
ϕ
(k)
ϕ
(k−1)
= ord
α
ϕ
ϕ
+ ord
α
ϕ
ϕ
+ + ord
α
ϕ
(k)
ϕ
(k−1)
−k (1)
Proposition 2.5. Let ϕ
1
, ϕ
2
be rational functions on F and a ∈ F. Then
ord
α
(ϕ
1
, ϕ
2
) min{ord
α
ϕ
1
,ord
α
ϕ
2
} .
Proof. Let ord
α
ϕ
1
= m
1
and ord
α
ϕ
2
= m
2
. Then
ϕ
1
(x)=(x − α)
m
f
1
(x)
g
1
(x),
(2)
ϕ
2
(x)=(x − α)
m
f
2
(x)
g
2
(x),
(3)
where f
1
,f
2
,g
1
,g
2
are the polynomials o ver F and f
1
(α),f
2
(α),g
1
(α),g
2
(α) =0. We set
m = min(m
1
,m
2
). Then
ϕ
1
(x)+ϕ
2
(x)=(x − α)
m
(x − α)
m
1
−m
f
1
(x)g
2
(x)+(x − α)
m
2
−m
f
2
(x)g
1
(x)
f
2
(x)g
2
(x)
.
Since f
2
(α)g
2
(α) =0, we have
ord
α
(ϕ
1
+ ϕ
2
) m = min(ord
α
ϕ
1
,ord
α
ϕ
2
).
Definition 2.6. Let f
1
,f
2
, , f
n
be polynomials on F (but to a large extent what we
do depends only on formal properties of devivations). We recall that their W ronskian is
W (f
1
,f
2
, ,f
n
)=
f
1
f
2
··· f
n
f
1
f
2
··· f
n
.
.
.
.
.
.
.
.
.
.
.
.
f
(n−1)
1
f
(n−1)
2
··· f
(n−1)
n
Remark. If f
1
,f
2
, , f
n
are linearly independent on F, then W(f
1
,f
2
, ,f
n
) =0.
Proof of Theorem 1.1. Let {α
0
, α
1
, ,α
n
} be a subset of I = {0, 1, ,n+1}. Then the
equation f
0
+ f
1
+ + f
n
= f
n+1
implies W (f
α
0
, , f
α
n
)=δW (f
0
,f
1
, , f
n
), where
δ =1or−1. Because f
0
,f
1
, f
n
are linearly i ndependent, we obtain
W (f
0
,f
1
, ,f
n
) =0.
Generalied Mason’s Theorem 37
Then, we set
P (t)=
W (f
0
,f
1
, ,f
n
)
f
0
f
1
f
n
,
Q(t)=
f
0
f
1
f
n+1
W (f
0
,f
1
, ,f
n
)
.
Hence, we ha ve
f
n+1
= P(t)Q(t).
We first prove that
degQ(t) a n
n+1
i=0
n
1
f
i
.
Let α be a zero of the function Q(z). Then α is a zero of some polynomial f
i
(0 a i a
n +1). By the hypothesis that the polynomials are relatively prime, there exists a number
v(0 a v a n + 1) s uch that f
v
(α) =0.
Let {i
0
,i
1
, ,i
n
} be a subset I|{v},thenwehave
Q(t)=δ
f
i
0
f
i
1
f
i
n
W (f
0
,f
1
, ,f
n
)
f
v
.
Denote
R(t)=
W (f
i
0
,f
i
1
, , f
i
n
)
f
i
0
f
i
1
f
i
n
as the logarithmic Wronskian corresponding to {i
0
,i
1
, , i
n
}, which is
11··· 1
f
i
0
f
i
0
f
i
1
f
i
1
···
f
i
n
f
i
n
.
.
.
.
.
.
.
.
.
.
.
.
f
(n−1)
i
0
f
i
0
f
(n−1)
i
1
f
i
1
···
f
(n−1)
i
n
f
i
n
Then f
v
= R(t)Q(t)andsoord
α
R(t)=−ord
α
Q(t). Then the determinant R(t)is
asumoffollowingterms
δ
f
α
0
f
α
1
f
(n−1)
α
n
f
α
0
f
α
1
f
α
n
,
where 0 a α
0
, α
1
, ,α
n
a n +1and δ =1or−1.
By applyi ng the propositions 2.3 and 2.4, we get
ord
α
f
α
0
f
α
1
f
(n−1)
α
n
f
α
0
f
α
1
f
α
n
= ord
α
f
α
0
f
α
0
+ ord
α
f
α
1
f
α
1
+ + ord
α
f
(n−1)
α
n
f
α
n
−n
0aian+1
f
i
(a)=0
1
. (4)
38 Nguyen Thanh Quang, Phan Duc Tuan
Therefore from Proposition 2.5, we have
ord
α
R(t) a −n
0aian+1
f
i
(a)=0
1
and so
ord
α
Q(t)=−ord
α
R(t) a −n
0aian+1
f
i
(a)=0
1
.
Since this inequality holds for any zero α of Q(t), we get
deg Q(t) a n
n+1
i=0
n
1
f
i
.
Next,wewillprovethat
deg P(t) a −
n(n +1)
2
.
Here, we have P(t) as the logarithmic Wronskian corresponding to I = {0, 1, ,n} which
is
11··· 1
f
0
f
0
f
1
f
1
···
f
n
f
n
.
.
.
.
.
.
.
.
.
.
.
.
f
(n)
0
f
0
f
(n)
1
f
1
···
f
(n)
n
f
n
The determinant P (t)isasumoffollowingterms
δ
f
β
0
f
β
1
f
(n−1)
β
n
f
β
0
f
β
1
f
β
n
.
For every term, by Proposition 2.4 we have
deg
f
β
0
f
β
1
f
(n−1)
β
n
f
β
0
f
β
1
f
β
n
=deg
f
β
0
f
β
0
deg
f
β
1
f
β
1
+ +deg
f
(n)
β
n
f
β
n
= −(1 + 2 + + n)=−
n(n +1)
2
. (5)
Therefore
deg P(t) a −
n(n +1)
2
so
deg f
n+1
=degP(t)+degQ(t) a n
n+1
i=0
n
1
f
i
−
n(n +1)
2
.
Generalied Mason’s Theorem 39
By the similar arguments applying to the polynomial f
0
,f
1
, , f
n
, we have
max
0a⊂an+1
(degf
i
) a n
n+1
i=0
n
1
f
i
−
n(n +1)
2
.
Theorem 1.1 is proved.
References
1. S. Lang, Introduction to Complex Hyperbolic Spaces, Springer - Verlag, (1987).
2. S. Lang, Old and new conjecture Diophantine inequalitis, Bull.Amer.Math.Soc.,
23 (1990), 37 - 75.
3. R. C. Mason, Diophan tine Equations over Function Fields, London Math. Soc.,
Lecture Notes, Cambridge Univ. Press, Vol. 96 (1984).
4. M. Ru and J. T. - Y. Wang, A second main type inequality for holomorphic curves
intersecting hyperplanes, Preprint.
