Reducibility of rational functions in several variables

0.18 MB·English

by  Arnaud Bodin

REDUCIBILITY OF RATIONAL FUNCTIONS IN
SEVERAL VARIABLES
7
0
0 ARNAUD BODIN
2

n Abstract. We prove a analogous of Stein theorem for rational
a
functions in several variables: we bound the number of reducible
J
fibers by a formula depending on the degree of the fraction.
8

]
T
N 1. Introduction
.
h Let K be an algebraically closed field. Let f = p ∈ K(x), with
t q
a x = (x ,...,x ), n > 2 and gcd(p,q) = 1, the degree of f is degf =
m 1 n
max{degp,degq}. We associate to a fraction f = p the pencil p−λq,
[ q
λ ∈ Kˆ (where we denote Kˆ = K ∪{∞} and by convention if λ = ∞
3
v then p−λq = q).
4 For each λ ∈ Kˆ write the decomposition into irreducible factors:
3
4 nλ
0 p−λq = Fri.
1 i
5 Yi=1
0 ˆ
The spectrum of f is σ(f) = {λ ∈ K | n > 1}, and the order of
/ λ
h
reducibility is ρ(f) = (n −1).
t λ∈Kˆ λ
a A fraction f is comPposite if it is the composition of a univariate
m
rational fraction of degree more than 1 with another rational function.
:
v
Theorem 1.1. Let K be an algebraically closed field of characteristic
i
X
0. Let f ∈ K(x) be non-composite then
r
a ρ(f) < (degf)2 +degf.
A theorem of Bertini and Krull implies that if f is non-composite
then σ(f) is finite and we should notice that #σ(f) 6 ρ(f). Later
on, for an algebraically closed field of characteristic zero and for a
polynomial f ∈ K[x,y], Stein [St] proved the formula ρ(f) < degf.
This formula has been generalized in several directions, see [Na1] for
references. For a rational function f ∈ C(x,y) a consequence of the
workofRuppert [Ru]onpencilofcurves, isthat#σ(f) < (degf)2. For
K algebraicallyclosed(ofanycharacteristic) andf ∈ K(x,y)Lorenzini
Date: February 2, 2008.
1
2 ARNAUDBODIN
[Lo] proved under geometric hypotheses on the pencil (p − λq) that
ρ(f) < (degf)2. This has been generalized by Vistoli [Vi] for a pencil
in several variables for an algebraically closed field of characteristic 0.
Letusgiveanexampleextractedfrom[Lo]. Letf(x,y) = x3+y3+(1+x+y)3,
xy(1+x+y)
then deg(f) = 3 and σ(f) = {1,j,j2,∞} (where {1,j,j2} are the third
roots of unity). For λ ∈ σ(f), (f = λ) is composed of three lines hence
ρ(f) = 8 = (degf)2 − 1. Then Lorenzini’s bound is optimal in two
variables.
The motivation of this work is that we develop the analogous theory
of Stein for rational function: composite fractions, kernels of Jacobian
derivatives, groups of divisors,... The method for the two variables case
isinspired fromthework of Stein [St]and thepresentation of thatwork
by Najib [Na1]. For completeness even the proofs similar to the ones of
Stein have been included. Another motivation is that with a bit more
effort we get the case of several variables by following the ideas of [Na1]
(see the articles [Na2], [Na3]).
In §2 we prove that a fraction is non-composite if and only its spec-
trum is finite. Then in §3 we introduce a theory of Jacobian derivation
and compute the kernel. Next in §4 we prove that for a non-composite
fraction in two variables ρ(f) < (degf)2 + degf. Finally in §5 we
extend this formula to several variables and we end by stating a result
for fields of any characteristic.
Acknowledgements: I wish to thank Pierre D`ebes and Salah Najib
for discussions and encouragements.
2. Composite rational functions
Let K be an algebraically closed field. Let x = (x ,...,x ), n > 2.
1 n
Definition 2.1. A rational function f ∈ K(x) is composite if there exist
g ∈ K(x) and r ∈ K(t) with degr > 2 such that
f = r ◦g.
Theorem 2.2. Let f = p ∈ K(x). The following assertions are equiv-
q
alent:
(1) f is composite;
(2) p−λq is reducible in K[x] for all λ ∈ Kˆ such that degp−λq =
degf;
(3) p−λq is reducible in K[x] for infinitely many λ ∈ Kˆ.
Before proving this result we give two corollaries.
Corollary 2.3. f is non-composite if and only if its spectrum σ(f) is
finite.
REDUCIBILITY OF RATIONAL FUNCTIONS IN SEVERAL VARIABLES 3
One aim of this paper is to give a bound for σ(f). The hard impli-
cation of this theorem (3) ⇒ (1) is in fact a reformulation of a theorem
of Bertini and Krull.
We also give a nice application pointed out to us by P. D´ebes:
Corollary 2.4. Let p ∈ K[x] irreducible. Let q ∈ K[x] with degq <
degp and gcd(p,q) = 1. Then for all but finitely many λ ∈ K, p−λq
is irreducible in K[x].
Convention : When we define a fraction F = P we will assume that
Q
gcd(P,Q) = 1.
We start with the easy part of Theorem 2.2:
Proof. (2) ⇒ (3) is trivial. Let us prove (1) ⇒ (2). Let f = p be a
q
composite rational function. There exist g = u ∈ K(x) and r ∈ K(t)
v
with k = degr > 2 such that f = r ◦ g . Let us write r = a. Let
b
λ ∈ Kˆ such that dega − λb = degr and factorize a(t) − λb(t) =
α(t−t )(t−t )···(t−t ), α ∈ K∗,t ,...,t ∈ K. Then
1 2 k 1 k
a−λb (g−t )···(g −t )
1 k
p−λq = q ·(f −λ) = q · (g) = αq .
(cid:18) b (cid:19) b(g)
Then by multiplication by vk at the numerator and denominator we
get:
(p−λq)·(vkb(g)) = αq(u−t v)···(u−t v),
1 k
which is a polynomial identity. As gcd(a,b) = 1, gcd(u,v) = 1 and
gcd(p,q) = 1 then u−t v,...,u−t v divide p−λq. Hence p−λq is
1 k
reducible in K[x]. (cid:3)
Let usreformulate theBertini-Krulltheorem inourcontext from[Sc,
Theorem 37]. It will enable us to end the proof of Theorem 2.2.
Theorem 2.5 (Bertini, Krull). Let F(x,λ) = p(x)−λq(x) ∈ K[x,λ]
an irreducible polynomial. Then the following conditions are equivalent:
(1) F(x,λ ) ∈ K[x]isreducible forallλ ∈ K such thatdeg F(x,λ ) =
0 0 x 0
deg F.
x
(2) (a) eitherthereexistφ,ψ ∈ K[x]with deg F > max{degφ,degψ},
x
and a ∈ K[λ], such that
i
n
F(x,λ) = a (λ)φ(x)n−iψ(x)i;
i
Xi=0
(b) or char(K) = π > 0 and F(x,λ) ∈ K[xπ,λ], where xπ =
(xπ,...,xπ).
1 n
4 ARNAUDBODIN
We now end the proof of Theorem 2.2:
Proof. (3) ⇒ (1) Suppose that p − λ q is reducible in K[x] for in-
0
finitely many λ ∈ Kˆ; then it is reducible for all λ ∈ K such that
0 0
deg F(x,λ ) = deg F (see Corollary 3 of Theorem 32 of [Sc]). We
x 0 x
apply Bertini-Krull theorem:
Case (a): F(x,λ) = p(x)−λq(x) can be written:
n
p(x)−λq(x) = a (λ)φ(x)n−iψ(x)i.
i
Xi=0
So we may suppose that for i = 1,...,n, deg a = 1, let us write
λ i
a (λ) = α −λβ , α ,β ∈ K. Then
i i i i i
n n
ψ i
p(x) = α φ(x)n−iψ(x)i = φn α (x),
i i
φ
Xi=0 Xi=0 (cid:16) (cid:17)
and
n n
ψ i
q(x) = β φ(x)n−iψ(x)i = φn β (x).
i i
φ
Xi=0 Xi=0 (cid:16) (cid:17)
If we set g(x) = ψ(x) ∈ K[x], and r(t) = Pni=0αiti then p(x) = r ◦g.
φ(x) Pni=0βiti q
Moreover as deg F > max{degφ,degψ} this implies n > 2 so that
x
degr > 2. Then p = f = r ◦g is a composite rational function
q
Case (b): Let π = char(K) > 0 and F(x,λ) = p(x) − λq(x) ∈
K[xπ,λ], For λ = 0 it implies that p(x) = P(xπ), then there exists
p′ ∈ K[x] such that p(x) = (p′(x))π. For λ = −1 we obtain s′ ∈ K[x]
such that p(x)+q(x) = (s′(x))π. Then q(x) = (p(x)+q(x))−p(x) =
(s′(x))π − (p′(x))π = (s′(x)−p′(x))π. Then if we set q′ = s′ − p′
we obtain q(x) = (q′(x))π. Now set r(t) = tπ and g = p′ we get
q′
f = p = p′ π = r ◦g. (cid:3)
q q′
(cid:16) (cid:17)
3. Kernel of the Jacobian derivation
We now consider the two variables case and K is an uncountable
algebraically closed field of characteristic zero.
3.1. Jacobian derivation. Let f,g ∈ K(x,y), the following formula:
∂f ∂g ∂f ∂g
D (g) = − ,
f
∂x∂y ∂y ∂x
defines a derivation D : K(x,y) → K(x,y). Notice the D (g) is the
f f
determinant of the Jacobian matrix of (f,g). We denote by C the
f
kernel of D :
f
C = {g ∈ K(x,y) | D (g) = 0}.
f f
REDUCIBILITY OF RATIONAL FUNCTIONS IN SEVERAL VARIABLES 5
Then C is a subfield of K(x,y). We have the inclusion K(f) ⊂ C .
f f
Moreover if gk ∈ C , k ∈ Z\{0} then g ∈ C .
f f
Lemma 3.1. Let f = p, g ∈ K(x,y). The following conditions are
q
equivalent:
(1) g ∈ C ;
f
(2) f and g are algebraically dependent;
(3) g is constant on irreducible components of the curves (p−λq =
0) for all but finitely many λ ∈ Kˆ;
(4) g is constant on infinitely many irreducible components of the
ˆ
curves (p−λq = 0), λ ∈ K.
Corollary 3.2. If g ∈ C is not a constant then C = C .
f f g
Proof.
• (1) ⇔ (2). We follow the idea of [Na1] instead of [St]. f and
g are algebraically dependent if and only transc K(f,g) = 1.
K
And transc K(f,g) = 1 if and only the rank of the Jacobian
K
matrix of (f,g) is less or equal to 1, which is equivalent to
g ∈ C .
f
• (2) ⇒ (3). Let f and g be algebraically dependent. Then there
exists a two variables polynomial in f and g that vanishes. Let
us write
n
R (f)gi = 0
i
Xi=0
where R (t) ∈ K[t]. Let us write f = p, g = u and R (t) =
i q v n
α(t−λ )···(t−λ ). Then
1 m
n n
p u i p
R = 0, hence R uivn−i = 0.
i i
q v q
Xi=0 (cid:16) (cid:17)(cid:16) (cid:17) Xi=0 (cid:16) (cid:17)
By multiplication by qd for d = max{degR } (in order that
i
qdR (p) are polynomials) we obtain
i q
p p
qdR un = v −qdR un−1 −··· .
n n−1
q (cid:18) q (cid:19)
(cid:16) (cid:17) (cid:16) (cid:17)
As gcd(u,v) = 1 then v divides the polynomial qdR (p), then v
n q
divides qd−m(p−λ q)···(p−λ q). Then all irreducible factors
1 m
of v divide q or p−λ q, i = 1,...,m.
i
Letλ ∈/ {∞,λ ,...,λ }. LetV beanirreduciblecomponent
1 m λ
of p−λq, then V ∩Z(v) is zero dimensional (or empty). Hence
λ
6 ARNAUDBODIN
v is not identically equal to 0 on V . Then for all but finitely
λ
many (x,y) ∈ V we get:
λ
n
R (λ)g(x,y)i = 0.
i
Xi=0
Therefore g can only reach a finite number of values c ,...,c
1 n
(therootsof n R (λ)ti). SinceV isirreducible, g isconstant
i=0 i λ
on Vλ. P
• (3) ⇒ (4). Clear.
• (4) ⇒ (1). We first give a proof that if g is constant along an
irreducible component V of (p−λq = 0) then D (g) = 0 on V
λ f λ
(we suppose that V is not in the poles of g). Let (x ,y ) ∈ V
λ 0 0 λ
and t 7→ p(t) be a local parametrization of V around (x ,y ).
λ 0 0
By definition of p(t) we have f(p(t)) = λ, this implies that:
dp d(f(p(t))
| gradf = = 0
(cid:28)dt (cid:29) dt
and by hypotheses g is constant on V this implies g(p(t)) is
λ
constant and again:
dp d(g(p(t))
| gradg = = 0.
(cid:28)dt (cid:29) dt
Then gradf and gradg are orthogonal around (x ,y ) on V to
0 0 λ
the same vector, as we are in dimension 2 this implies that the
determinant of Jacobian matrix of (f,g) is zero around (x ,y )
0 0
on V . By extension D (g) = 0 on V .
λ f λ
We now end the proof: If g is constant on infinitely many
irreducible components V of (p − λq = 0) this implies that
λ
D (g) = 0 on infinitely many V . Then D (g) = 0 in K(x,y).
f λ f
(cid:3)
3.2. Group of the divisors. Let f = p, let λ ,...,λ ∈ Kˆ, we
q 1 n
denote by G(f;λ ,...,λ ) the multiplicative group generated by all
1 n
the divisors of the polynomials p−λ q, i = 1,...,n.
i
Let
d(f) = (degf)2 +degf.
Lemma 3.3. Let F ,...,F ∈ G(f;λ ,...,λ ). If r > d(f) then there
1 r 1 n
exists a collection of integers m ,...,m (not all equal to zero) such
1 r
that
r
g = Fmi ∈ C .
i f
Yi=1
REDUCIBILITY OF RATIONAL FUNCTIONS IN SEVERAL VARIABLES 7
Proof. Let µ ∈/ {λ ,...,λ }, and let S be an irreducible component of
1 n
(p−µq = 0). Let S¯ be the projective closure of S. The functions F
i
¯
restricted to S have their poles and zeroes on the points at infinity of
S or on the intersection S ∩Z(F ) ⊂ Z(p)∩Z(q).
i
Let n : S˜ → S¯ be a normalization of S¯. The inverse image under
normalisationofthepointsatinfinity aredenotedby{γ ,...,γ }, their
1 k
number verifies k 6 degS 6 degf.
At a point δ ∈ Z(p) ∩ Z(q), the number of points of n−1(δ) is the
local number of branches of S at δ then it is less or equal than ord (S),
δ
where ord (S) denotes the order (or multiplicity) of S at δ (see e.g.
δ
[Sh], paragraph II.5.3). Then
#n−1(δ) 6 ord (S) 6 ord Z(p−µq) 6 ord Z(p−µq)·ord Z(p)
δ δ δ δ
6 mult (p−µq,p) = mult (p,q)
δ δ
where mult (p,q) is the intersection multiplicity (see e.g. [Fu]). Then
δ
by B´ezout theorem:
#n−1(δ) 6 mult (p,q) 6 degp·degq 6 (degf)2.
δ
δ∈Z(Xp)∩Z(q) δ∈Z(Xp)∩Z(q)
Thentheinverse imageunder normalisationof∪r S∩Z(F )denoted
i=1 i
by {γ ,...,γ } have less or equal than(degf)2 elements. Notice that
k+1 ℓ
ℓ 6 degf +(degf)2 = d(f).
Now let ν be the order of F at γ (i = 1,...,r; j = 1,...,ℓ).
ij i j
Consider the matrix M = (ν ). Because the degree of the divisor
ij
(F ) (seen over S˜) is zero we get ℓ ν = 0, for i = 1,...,r, that
i j=1 ij
means that columns of M are lineParly dependent. Then rkM < ℓ 6
d(f), by hypothesis r > d(f), then the rows of M are also linearly
r
dependent. Let m (µ,S),...,m (µ,S) such that m (µ,S)ν = 0,
1 r i=1 i ij
j = 1,...,ℓ. P
Consider the function g = r Fmi(λ,S). Then this function is
µ,S i=1 i
regular and does not have zeroeQs or poles at the points γj, because
r
m (µ,S)ν = 0. Then g is constant on S.
i=1 i ij µ,S
PThis construction gives a map (µ,S) 7→ (m1(µ,S),...,mr(µ,S))
from K to Zr. Since K is uncountable, there exists infinitely many
(µ,S) with the same (m ,...,m ). Then the function g = r Fmi is
1 r i=1 i
constant on infinitely many components of curves of (p−µQq = 0) and
by Lemma 3.1 this implies g ∈ C . (cid:3)
f
3.3. Non-composite rational function. Let f = p. Let G(f) be
q
the multiplicative group generated by all divisors of the polynomials
8 ARNAUDBODIN
p−λq for all λ ∈ Kˆ. In fact we have
G(f) = G(f;λ ,...,λ ).
1 n
(λ1,...[,λn)∈Kn
Definition 3.4. A family F ,...,F ∈ G(f) is f-free if (m ,...,m ) ∈
1 r 1 r
Zr is such that r Fmi ∈ C then (m ,...,m ) = (0,...,0).
i=1 i f 1 r
A f-free famiQly F1,...,Fr ∈ G(f) is f-maximal if for all F ∈ G(f),
{F ,...,F ,F} is not f-free.
1 r
Theorem 3.5. Let f ∈ K(x,y), degf > 0. Then the following condi-
tions are equivalent:
(1) degf = min{degg | g ∈ C \K};
f
(2) σ(f) is finite;
(3) C = K(f);
f
(4) f is non-composite.
Remark 3.6. This does not give a new proof of “σ(f) is finite ⇔ f is
non-composite” because we use Bertini-Krull theorem.
Remark 3.7. The proof (1) ⇒ (2) is somewhat easier than in [St],
whereas (2) ⇒ (3) is more difficult.
Proof.
• (1) ⇒ (2). Let us suppose that σ(f) is infinite. Set f = p,
q
with gcd(p,q) = 1. For all α ∈ σ(f), let F be an irreducible
α
divisor of p − αq, such that degF < degf. By Lemma 3.3
α
there exists a f-maximal family {F ,...,F } with r 6 d(f).
1 r
Moreover r > 1 because {F } is f-free: if not there exists k 6= 0
α
such that Fk ∈ C then F ∈ C , but degF < degf that
α f α f α
contradicts the hypothesis of minimality.
Nowthecollection{F ,...,F ,F }isnotf-free,sothatthere
1 r α
exist integers {m (α),...,m (α),m(α)}, with m(α) 6= 0, such
1 r
that
Fm1(α)···Fmr(α) ·Fm(α) ∈ C .
1 r α f
Since σ(f)isinfinite thenisequalto Kˆ minus a finitenumber of
values (see Theorem 2.2) then σ(f) is uncountable and the map
α 7→ (m (α),...,m (α),m(α)) is not injective. Let α 6= β such
1 r
that m (α) = m (β) = m , i = 1,...,r and m(α) = m(β) = m.
i i i
Then Fm1 ···Fmr · Fm ∈ C and Fm1 ···Fmr · Fm ∈ C , it
1 r α f 1 r β f
implies that (F /F )m ∈ C , therefore F /F ∈ C .
α β f α β f
Now deg Fα < degf, then by the hypothesis of minimality it
Fβ
proves Fα is a constant. Let a ∈ K∗ such that F = aF , by
Fβ α β
REDUCIBILITY OF RATIONAL FUNCTIONS IN SEVERAL VARIABLES 9
definition F divides p−αq, but moreover F divides p−βq (as
α α
F do). Then as F divides both p−αq and p−βq, F divides
β α α
p and q, that contradicts gcd(p,q) = 1.
• (2) ⇒ (3). Let f = p, σ(f) finite and g ∈ C , we aim at proving
q f
that g ∈ K(f). The proof will be done in several steps:
(a) Reduction to the case g = u. Let g = u ∈ C , thenf andg
qℓ v f
are algebraically dependent, then there exists a polynomial
in f and g that vanishes. As before let us write
n
R (f)gi = 0
i
Xi=0
where R (t) ∈ K[t]. As f = p, g = u then
i q v
n n
p u i p
R = 0, hence R uivn−i = 0.
i i
q v q
Xi=0 (cid:16) (cid:17)(cid:16) (cid:17) Xi=0 (cid:16) (cid:17)
By multiplication by qd for d = max{degR } (in order that
i
all qdR (p) are polynomials) we get:
i q
p p
qdR un = v −qdR un−1 −··· .
n n−1
q (cid:18) q (cid:19)
(cid:16) (cid:17) (cid:16) (cid:17)
As gcd(u,v) = 1 then v divides the polynomial qdR (p);
n q
we write vu′ = qdR (p) then
n q
u uu′
g = = .
v qdR (p)
n q
But R (p) ∈ K(p) then uu′ ∈ C , but also we have that
n q q qd f
g ∈ K(f) if and only if uu′ ∈ K(f). This proves the
qd
reduction.
(b) Reduction to the case g = qu. Let g = u ∈ C , ℓ > 0. As
qℓ f
σ(f) is finite by Lemma 3.1 we choose λ ∈ K such that
p−λq is irreducible and g ∈ C is constant (equal to c) on
f
p−λq. As g = u,we have p−λq divides u−cqℓ. We can
qℓ
write:
u−cqℓ = u′(p−λq).
Then
u u′ p
= −λ +c.
qℓ qℓ−1 q
(cid:16) (cid:17)
10 ARNAUDBODIN
As u and f = p are in C we get u′ ∈ C ; moreover
qℓ q f qℓ−1 f
u ∈ K(f) if and only if u′ ∈ K(f). By induction on
qℓ qℓ−1
ℓ > 0 this prove the reduction.
(c) Reduction to the case g = q. Let g = qu ∈ C . g is
f
constant along the irreducible curve (p − λq = 0). Then
qu = u (p−λq)+c .
1 1
Let degp = degq. Then qhuh = uh(ph − λqh) (where Ph
1
denotes the homogeneous part of higher degree of the poly-
nomial P). Then ph−λqh divides qhuh for infinitely many
λ ∈ K. As gcd(p,q) = 1 this gives a contradiction.
Hence degp 6= degq. We may assume degp > degq (oth-
erwise qu ∈ C and p ∈ C implies pu ∈ C ). Then we
f q f f
write:
p
qu = qu −λ +c ,
1 1
q
(cid:16) (cid:17)
that proves that qu ∈ C and that qu ∈ K(f) if and only
1 f
if qu ∈ K(f). The inequality degp > degq implies that
1
degu < degu. We continue by induction, qu = qu (p −
1 1 2 q
λ) + c , with degu < degu ,..., until we get degu = 0
2 2 1 n
that is u ∈ K∗. Thus we have prove firstly that qu ∈ C ,
n n f
that is to say q ∈ C , and secondly that qu ∈ K(f) if and
f
only if q ∈ K(f).
(d) Case g = q. If q ∈ C then q is constant along the irre-
f
ducible curve (p−λq = 0) then q = a(p−λq)+c, a ∈ K∗.
Then
c p
q = ∈ K = K(f).
1−a(p −λ) q
(cid:16) (cid:17)
q
• (3) ⇒ (4). Let us assume that C = K(f) and that f is com-
f
posite, then there exist r ∈ K(t), degr > 2 and g ∈ K(x,y)
such that f = r ◦ g. By the formula degf = degr · degg
we get degf > degg. Now if r = a then we have a relation
b
b(g)f = a(g), then f and g are algebraically dependent, hence
by Lemma 3.1, g ∈ C . As C = K(f), there exists s ∈ K(t)
f f
such that g = s ◦ f. Then degg > degf. That yields to a
contradiction.
• (4) ⇒ (1). Assume that f is non-composite and let g ∈ C
f
of minimal degree. By Corollary 3.2 we get C = C , then
f g
degg = min{degh | h ∈ C \K}. Then by the already proved
g
implication (1) ⇒ (3) for g, we get C = K(g). Then f ∈ C =
g f
C = K(g), then there exists r ∈ K(t) such that f = r◦g, but
g