On Hilbert's irreducibility theorem

0.2 MB·

by  Abel Castillo

ON HILBERT’S IRREDUCIBILITY THEOREM
6 ABELCASTILLOANDRAINERDIETMANN
1
0
2 Abstract. In this paper we obtain new quantitative forms of Hilbert’s Ir-
reducibility Theorem. In particular, we show that if f(X,T1,...,Ts) is an
n irreducible polynomial with integer coefficients, having Galois group G over
a the function field Q(T1,...,Ts), and K is any subgroup of G, then there are
J atmostOf,ε(Hs−1+|G/K|−1+ε)specialisationst∈Zs with|t|≤H suchthat
1 theresultingpolynomialf(X)hasGaloisgroupK overtherationals.
3

]
T
N 1. Introduction
. One ofthe fundamentalresultsinDiophantine geometryis Hilbert’s Irreducibil-
h
t ityTheorem[7],statingthatiff(X1,...,Xr,T1,...,Ts) Q[X1,...,Xr,T1,...,Ts]
a ∈
is irreducible, then there exists a specialisation t Qs such that f(X ,...,X ) =
m 1 r
∈
f(X ,...,X ,t ,...,t ) as a rational polynomial in X ,...,X still is irreducible
1 r 1 s 1 r
[
overQ[X1,...,Xr]. Infact,ifr =1thenmoreistrue: Supposethatf(X,T1,...,Ts)
∈
1 Q[X,T ,...,T ] is irreducible and of degreen in X. Consider f as a polynomialin
1 s
v X overtherationalfunctionfieldL=Q(T ,...,T ),havingrootsα ,...,α inthe
1 s 1 n
4
algebraicclosureL. Asf isirreducible,theserootsaredistinct,andwecanconsider
1
the Galois group G of f over L as a subgroup of the symmetric group S . Then
3 n
0 there exists a specialisation t Qs such that the resulting rational polynomial in
∈
0 X still is irreducible and has Galois group G over Q. In fact, if t is chosen in such
. a way that the specialisedpolynomial in X still is of degree n, and separable,then
2
0 its Galois group Gt over Q is a subgroup of G (well-defined up to conjugation, see
6 Lemma 1 for the constructionof anembedding of Gt into G) andit turns out that
1 ‘almost all’ specialisations for t preserve the Galois group, i.e. Gt = G. In this
:
v paper we are interested in getting precise quantitative forms of these statements,
i so in the setting for r =1 from above, for fixed f and any subgroup K of G let
X
r Nf(H;K)=# t Zs : t H and the splitting field of f(X,t) over Q
a { ∈ | |≤
has Galois group K ,
}
where we use to denote the maximum normof a vector. Note that without loss
|·|
of generality we can assume f to have integer coefficients, and in this arithmetic
setting we are counting integer specialisations t of bounded height H. Our first
result is the following.
Theorem 1. Let ε > 0. Suppose that f(X,T) Z[X,T ,...,T ] is irreducible.
1 s
∈
Let G be the Galois group of f(X) over Q(T ,...,T ), and let K be a subgroup of
1 s
G. Then
(1) N (H;K) Hs−1+|G/K|−1+ε,
f f,ε
≪
where G/K denotes the index of K in G.
| |
2000 Mathematics Subject Classification. 11C08,11G35, 11R32,11R45.
1
2 ABELCASTILLOANDRAINERDIETMANN
In particular,this shows that almost all specialisations for t preserve the Galois
group G of f, and specialisations leading to small subgroups of G are rare. Our
result is not the first of its kind; Cohen (see Theorem 2.1 in [2]), using the large
sieve, obtained a bound in a more general number field setting, but with exponent
s 1/2forG=K insteadofs 1+ G/K −1. Thefirsttwoboundsintheliterature
− 6 − | |
thataresensitiveto the size ofK areapparentlydue to the secondauthor[4], who
in the special case of the polynomial
Xn+T Xn−1+...+T
1 n
already obtained (1) (see [11] for very recent improvements in this special case
when in addition n 12), and due to Zywina [15]. Zywina, like Cohen, worksover
≥
general number fields rather than the rational numbers, but uses the larger sieve
instead of the large sieve and obtains the same bound (1) for the number of all
specialisations leading to a polynomial having Galois group contained in K, where
K is allowedto be any subset ofG stable under conjugation,for example a normal
subgroup. Our work makes use of recent advances on bounding the number of
points on curves instead of sieve methods, generalisingthe approachfrom [4]; note
that a somewhat similar line of attack was also used in a few previous papers (see
[12], [14], [3]) discussing the related problem of bounding the smallest admissible
specialisation in Hilbert’s irreducibility theorem. We restrict our attention to the
field of rational numbers; the method should generalise to number fields, provided
asuitableanalogueof[1]holds. ItgivessharperboundsthanCohen’sandZywina’s
results in most cases, namely as soon as K is any non-normal subgroup of G with
index exceeding 2.
To summarise our main result on Hilbert’s irreducibility theorem, let us keep the
notation from above and introduce the quantity
E (H)=# t Zs : t H and the splitting field of f(X,t) over Q
f
{ ∈ | |≤
has Galois group different from G .
}
Corollary 1. Keeping the notation and the assumptions from Theorem 1, we have
E (H) Hs−1+δG+ε,
f f,ε
≪
where
δ =max G/K −1 :K is a proper subgroup of G .
G
{| | }
Thequantityδ inCorollary1formanygroupscanbeaslargeas 1,forexample
G 2
for G = S , but for many interesting groups it can also be pretty small: For
n
example,if G=A andn 5, then δ = 1 (see [5], Theorem5.2A).Coming back
n ≥ G n
to the original question of irreducibility, still assuming r=1, let
R (H)=# t Zs : t H and f(X,t) becomes reducible in Q[X] .
f
{ ∈ | |≤ }
Corollary 2. Keeping the notation and assumptions from Theorem 1, we have
(2) R (H) Hs−1+γG+ε,
f f,ε
≪
where
(3) γ =max G/K −1 :K is an intransitive subgroup of G .
G
{| | }
Ofcoursealwaysγ δ ,butoftenγ ismuchsmallerthanδ . Asanexample,
G G G G
≤
consider f(X,t) = X5 t. The Galois group of f(X) = f(X,t) over Q(t) is the
−
dihedral group D . Clearly δ = 1, but the only proper subgroups of D have
10 G 2 10
ON HILBERT’S IRREDUCIBILITY THEOREM 3
2 or 5 elements. Those of order 5 are cyclic of order 5 and therefore transitive,
whereas those of order 2 fix one element and thus are intransitive. Consequently,
γ = 1. In this example, f(X,t) becomes reducible as soon as t is a fifth power,
G 5
so the bound (2) actually turns out to be sharp here.
Let us also remarkthat in this special case s=1 sometimes more canbe done, see
for example the papers [6] and [10].
Finally,letusreconsiderHilbert’sIrreducibilityTheoreminitsgeneralformapply-
ing to polynomials F(X ,...,X ,T ,...,T ) in r variables X ,...,X . This case
1 r 1 s 1 r
canbereducedtothespecialcaser =1byKronecker’s specialisation,seeforexam-
pleChapter9, 3in[8],ortheproofofTheorem2.5in[2],whereithasbeenshown
§
that if f(X ,...,X ,T ,...,T ) Z[X ,...,X ,T ,...,T ] is irreducible over Q,
1 r 1 s 1 r 1 s
∈
then for
J (H)=# t Zs : t H and
f
{ ∈ | |≤
f(X ,...,X ,t) becomes reducible in Q[X ,...,X ]
1 r 1 r
}
one has the upper bound
(4) J (H) Hs−1/2logH.
f f
≪
In fact, as in Corollary 2, the exponent s 1/2 in general is sharp as can be seen
−
for example by considering the polynomial
f(X ,...,X ,T ,...,T )=(X +...+X )2 (T +...+T ).
1 r 1 s 1 r 1 s
−
Like in Corollary 2, however,in special cases one can do better.
Theorem2. Letf(X ,...,X ,T ,...,T ) Z[X ,...,X ,T ,...,T ]beirreducible,
1 r 1 s 1 r 1 s
∈
and suppose that for some i 1,...,r the monomial of highest degree in X is of
i
∈{ }
the form Xnh, where n 1 and h depends at most on T ,...,T , but not on X for
i ≥ 1 s j
j =i. Moreover, let G be the Galois group of the splitting field of f(X ), considered
i
6
as a polynomial over the function field Q(X ,...,X ,X ,T ,...,T ). Then
1 i−1 i+1 1 s
J (H) Hs−γG+ε,
f f,ε
≪
where γ has been defined in (3).
G
Note that polynomials f not satisfying the assumptions of Theorem2 regarding
the form of the highest degree monomial in one of the variables can be brought
into that form after applying a suitable linear transformation on the variables
X ,...,X , that does not change the property of being reducible or irreducible
1 r
over the rationals. It seems difficult, though, to control how the relevant G and
thus γ change in this process.
G
As we have already remarked, our approach roughly follows [4], using auxiliary
varieties based on suitable Galois resolvents, and bounding the number of integral
points on these varieties. More care, however, has to be taken in constructing the
Galois resolvents in section 2 to guarantee their irreducibility. In section 3 we use
aresultfromthe literature,stemming itselffromanapplicationofthe determinant
method, to bound the number of integral points on curves which is enough to deal
withthe cases=1. Fors>1we use afibrationapproachto reduce to this special
case of curves. Theorem 1 along with Corollaries1 and 2 and Theorem 2 will then
be proved in sections 4, 5 and 6.
Acknowledgments: The authors would like to thank the anonymous referees of
a previous version of this article for several useful comments.
4 ABELCASTILLOANDRAINERDIETMANN
2. Construction of the Galois resolvents
We will now give a construction of Galois resolvents, polynomials that detect
containment of the Galois group of a polynomial in a prescribed group, as given
in Lemma 4. To this end we first need some preparations. Keeping the notation
from the introduction, we observe that the group G acts on the roots of f(X,T)
by permutations, and this gives rise to an injective homomorphism
ρ:G֒ S .
n
→
For t = (t1,...,ts) Zs, write Gt for the Galois group of the splitting field of
∈
f(X,t). To make sense out of a comparison between Gt and subgroups of G, we
constructaninjectionofGtintoGthatiscompatiblewiththechoiceofenumeration
of roots. In other words, we want the following diagram to commute.
(5) Gt(cid:31)(cid:16)❇(cid:127)ρ❇p❇t❇ι❇❇❇❇ // G(cid:15)(cid:15)(cid:127)ρ_
S
n
Lemma 1. Suppose that t Zs satisfies the conditions
∈
(6) degf(X,t)=n and ∆(t)=0,
6
where∆(T)isthediscriminantoff(X,T)viewedasapolynomialinX. Thenthere
exist injective homomorphisms ι and ρt such that the diagram in (5) commutes.
Proof. ConsidertheDedekinddomainZ(T , ,T )[T ];theconditions(6)imply
1 s−1 s
···
that the prime (T t ) is unramified in the splitting field of f(X,T). Choose a
s s
−
prime p in the splitting field of f(X,t) lying above (T t ), and the injection
s s
−
of the decomposition group of this prime into G gives rise of an injection of the
Galois group of f(X,T , ,T ,t ) over Q(T , ,T ) (see for instance [13,
1 s−1 s 1 s−1
··· ···
Section1.7,pp. 20-21]). Sincethe primepisunramifiedandthereductionf(X,T)
modulo p has degree n in X, reduction mod p sends α (T) to α (T , ,T ,t ).
i i 1 s−1 s
···
Nowsupposethatσ isanelementoftheGaloisgroupoff(X,T , ,T ,t )over
1 s−1 s
···
Q(T , ,T ), and suppose that the above injection sends σ σ with
1 s−1
··· 7→
σ(α (T))=α (T).
i j
The injection described above has the property that
σ(α (T) (mod p))=α (T) (mod p),
i j
So we caninject the Galoisgroupof f(X,T , ,T ,θ )into S by its actionon
1 s−1 s n
···
the roots off(X,T), which is precisely whatwe need forthe diagramto commute.
The proof is completed by repeating this procedure one parameter at a time. (cid:3)
Lemma 2. Let n N and z ,...,z ,w ,...,w C. Suppose that
1 n 1 n
∈ ∈
zk+...+zk =wk+...+wk
1 n 1 n
for all k 1,...,n . Then z ,...,z = w ,...,w , i.e. the z are a permu-
1 n 1 n i
∈ { } { } { }
tation of the w and vice versa.
i
Proof. This is a well known result going back to Newton. (cid:3)
ON HILBERT’S IRREDUCIBILITY THEOREM 5
Lemma 3. Let n N, let K be a subgroup of S , and let α ,...,α C be
n 1 n
∈ ∈
distinct. Further, let σ ,...,σ bea setof coset representatives for S /K, where
1 m n
{ }
m= S /K . Then there exist e ,...,e N, d ,...,d N and γ Z such that
n 1 n 1 |K|
| | ∈ ∈ ∈
all complex numbers
|K|
(7) z = d (α +γ)ke1(α +γ)ke2 (α +γ)ken
i k σi(τ(1)) σi(τ(2)) ··· σi(τ(n))
k=1 τ∈K
X X
(1 i m)
≤ ≤
are distinct.
Proof. For convenience, let us introduce the notation
e=(e ,...,e ),
1 n
wi,τ,e,k,γ =(ασi(τ(1))+γ)ke1···(ασi(τ(n))+γ)ken,
wi,τ,e,γ =wi,τ,e,1,γ.
We now show that it is possible to choose γ Z and e ,...,e N in such a way
1 n
∈ ∈
that
(8) wi,τ1,e,γ 6=wj,τ2,e,γ
for all (i,τ ) and (j,τ ) where i=j and τ ,τ K. The condition
1 2 1 2
6 ∈
wi,τ1,e,γ =wj,τ2,e,γ
is equivalent to
α +γ e1 α +γ en
σi(τ1(1)) σi(τ1(n)) =1,
α +γ ··· α +γ
(cid:18) σj(τ2(1)) (cid:19) (cid:18) σj(τ2(n)) (cid:19)
providing all denominators are different from zero. Since i = j, at least one expo-
nent e must be attached to a fraction of the form αs+γ wh6 ere α = α . Suppose
ℓ αt+γ s 6 t
that all the other exponents e where m = ℓ are fixed. Then we are left with an
m
6
equation of the form
α +γ eℓ
s
(9) =c
α +γ
(cid:18) t (cid:19)
for some c C. Now choose γ Z large enough, in terms of a sufficiently large
∈ ∈
parameter H only depending on n, such that (9) has at most one solution e N
ℓ
∈
with e H, for all possible choices of α = α , ℓ and c C. For this fixed γ, we
ℓ s t
≤ 6 ∈
have shown that for all tuples (i,τ ) and (j,τ ) where i=j, we have
1 2
6
#{e1,...,en ∈N:eℓ ≤H(1≤ℓ≤n) and wi,τ1,e,γ =wj,τ2,e,γ}≪Hn−1.
SincethereareonlyO (1)manypossibilitiestochoose(i,τ )and(j,τ )withi=j,
n 1 2
6
but there are Hn vectors e Nn where e H (1 ℓ n), by choosing H
ℓ
≫ ∈ ≤ ≤ ≤
sufficiently large we certainly can find such an exponent vector e Nn for which
∈
(8) holds true. Now fix that vector e Nn and γ, and let us write
∈
vi,k = wi,τ,e,k,γ (1 i m,1 k K ).
≤ ≤ ≤ ≤| |
τ∈K
X
If i = j, then there exists at least one k 1,..., K such that v = v : By
i,k j,k
6 ∈ { | |} 6
Lemma 2 the conditions v =v (1 k K ) would imply
i,k j,k
≤ ≤| |
wi,τ,e,γ :τ K = wj,τ,e,γ :τ K ,
{ ∈ } { ∈ }
6 ABELCASTILLOANDRAINERDIETMANN
contradicting (8). The complex numbers in (7) are now exactly of the form
|K|
z = d v (1 i m).
i k i,k
≤ ≤
k=1
X
To make them distinct, it is enough to choose d ,...,d N in such a way that
1 |K|
∈
|K|
(10) d (v v )=0
k i,k j,k
− 6
k=1
X
whenever i=j. As shownabove,for i=j there is atleast one non-zero coefficient
6 6
on the left hand side of (10), whence
|K|
# d ,...,d N|K| : d (v v )=0 and d H (1 k K )
1 |K| k i,k j,k k
{ ∈ − ≤ ≤ ≤| | }
k=1
X
H|K|−1.
≪
We can now conclude in a similar way as above: Since there are O (1) many
n
possibilities to choose i and j where i = j, but there are H|K| many vectors
6 ≫
d N|K| whered H (1 k K ),by choosingH sufficiently largewecanfind
k
av∈ectord N|K| su≤chthat≤(10)≤is|tr|uewheneveri=j. Thisfinishestheproof. (cid:3)
∈ 6
Lemma 4. Let n N, and let
∈
f(X,T)=Xn+g (T)Xn−1+...+g (T)
1 n
where g Z[T ,...,T ] (1 i n). Suppose that f(X) = f(X,T), considered
i 1 s
∈ ≤ ≤
as a polynomial in the ring Q(T)[X], has distinct roots α = α (T),...,α =
1 1 n
α (T) in the algebraic closure Q(T) of Q(T), and let G be the Galois group of
n
the corresponding splitting field operating on α (T),...,α (T). Moreover, let K
1 n
be a subgroup of G. Then there exists a Galois resolvent Φ with the following
f,K
properties:
(i) Φ is a polynomial of the form
f,K
(11) Φ (Z,T)=Zm+h (T)Zm−1+...+h (T),
f,K 1 m
where m= S /K and h Z[T ,...,T ](1 i m).
n i 1 s
| | ∈ ≤ ≤
(ii) If one specialises the parameters T ,...,T in f(X,T) to any s-tuple of
1 s
integers t=(t ,...,t ), then if the splitting field of the polynomial f(X)=
1 s
f(X,t) over Q has Galois group K, then Φ (Z) = Φ (Z,t) has an
f,K f,K
integer root z.
(iii) If one factorises Φ (Z,T) over Q[Z,T ,...,T ] into irreducible factors,
f,K 1 s
then each factor has degree at least |G| in Z.
|K|
Proof. Since the roots α (T),...,α (T) are distinct, it is possible to specialise
1 n
T ,...,T to an s-tuple t of complex numbers such that the complex roots α =
1 s 1
α (t),...,α = α (t) of f(X) = f(X,t) are all distinct. We are therefore in
1 n n
a position to invoke Lemma 3. Keeping the notation from that lemma, we find
e ,...,e N and d ,...,d N, and a γ Z, such that all the numbers z in
1 n 1 |K| i
∈ ∈ ∈
(7) are distinct. Now since replacing the variable X by X γ does not change the
−
splitting field and thus does not change the Galois group of f(X,T) over Q(T),
ON HILBERT’S IRREDUCIBILITY THEOREM 7
and also for fixed t Zs does not change the Galois group of f(X,t) over Q, we
∈
can without loss of generality assume that γ =0. We now define Φ (Z,T) to be
f,K
m m |K|
(12) Φ (Z,T)= (Z z )= Z d αke1 αken ,
f,K − i  − k σi(τ(1))··· σi(τ(n))
i=1 i=1 k=1 τ∈K
Y Y X X
 
where σ ,...,σ is a set of coset representatives for S /K.
1 m n
{ }
It is important to keep in mind that by construction the z = z (T) are distinct,
i i
sincewecanspecialiseTinsuchawaytoendupwithdistinctcomplexz . Expand-
i
ingtheexpression(12),itbecomestransparentthatΦ (Z,T)isoftheform(11),
f,K
where the h are symmetric polynomials in the z with integer coefficients. Any
i i
permutationofthe α justpermutes the z ,so the h aresymmetricpolynomialsin
i i i
the α as well, with integer coefficients. Hence, by the Fundamental Theorem on
i
symmetric functions, the h are integer polynomials in the elementary symmetric
i
polynomials in α ,...,α , which in turn by Vieta’s Theorem are of the form g .
1 n i
±
This shows that the h are integer polynomials in T ,...,T and confirms (i).
i 1 s
For the proof of (ii) and (iii) we first note that the symmetric group S operates
n
on the z via
i
|K|
̺(z )= d αke1 αken
i k ̺(σi(τ(1)))··· ̺(σi(τ(n)))
k=1 τ∈K
X X
for all ̺ S .
n
∈
To show (ii), fix any t Zs and consider
∈
|K|
z˜= d αke1 αken .
k τ(1)··· τ(n)
k=1 τ∈K
X X
Choosing the σ which is in the same coset of S /K as the identity map, one finds
i n
that z˜ is one of the z occurring on the left hand side of (12). Now suppose that
i
f(X)= f(X,t) has Galois group K over Q. Clearly, τ(z˜) =z˜for all τ K. This
∈
showsthatz˜ Q. Moreover,z˜isarootofthemonicintegerpolynomialΦ (Z,t),
f,K
∈
whence the stronger conclusion z˜ Z follows. This finishes the proof of (ii).
∈
Fortheproofof(iii),itisusefultoworkoverthefunctionfieldQ(T)ratherthanover
Q. As observed above, the z = z (T) are then distinct elements of the algebraic
i i
closure Q(T) of Q(T). As a consequence, we obtain
Stab(z )= ̺ S :̺(z )=z =σ Kσ−1 (1 i m),
i { ∈ n i i} i i ≤ ≤
which implies that
(13) ̺ S :̺(z )=z =σ σ−1Stab(z )=σ Kσ−1 (1 i,j m).
{ ∈ n i j} j i i j i ≤ ≤
Theseobservationsarecrucialforthefollowingargument: SupposethatΦ (Z,T)
f,K
factorises over Q into two factors Φ ,Φ Q[Z,T ,...,T ], i.e.
1 2 1 s
∈
(14) Φ (Z,T)=Φ (Z,T)Φ (Z,T).
f,K 1 2
We can consider Φ (Z) = Φ (Z,T) as a monic rational polynomial in Z over
f,K f,K
Q(T), andanalogouslyforΦ (Z)=Φ (Z,T). Now (12)providesa factorisationof
1 1
Φ (Z) over Q(T) into factors of the form (Z z ). Suppose that Φ has degree
f,K i 1
−
8 ABELCASTILLOANDRAINERDIETMANN
k inZ. Thenby (12), (14) anduniquenessof factorisation,Φ must be ofthe form
1
k
Φ (Z)=c (Z z )
1 · − ij
j=1
Y
for suitable c Q and distinct i 1,...,m . As shown in (13), we have
j
∈ ∈{ }
̺ S :̺(z )=z =σ Kσ−1
{ ∈ n i1 il} il i1
for all l 1,...,k . In particular, for given l 1,...,k there are exactly K
∈ { } ∈ { } | |
elements in S that map z to z , hence there are at most k K many elements in
n i1 il | |
S that map z to any root z of Φ . Therefore, if G > k K , then we can find
n i1 il 1 | | | |
an element ̺ G such that
∈
(15) ̺(z ) z ,...,z ,
i1 6∈{ i1 ik}
so ̺(z ) is no root of Φ , as all the z are distinct elements of Q(T). Now the
i1 1 i
operation of G on the z is that of field automorphisms of the splitting field of
i
Φ (Z) over Q(T). Such field automorphisms fix all elements of the ground field
f,K
Q(T)andthereforenecessarilymapanyrootofapolynomialoverQ(T)toanother
rootofthatpolynomial. AsΦ (Z)=Φ (Z,T)hascoefficientsinQ(T),weconclude
1 1
that ̺(z ) z ,...,z . This contradicts (15). Consequently, G > k K is
i1 ∈ { i1 ik} | | | |
impossible. This way we obtain the lower bound
G
k | |
≥ K
| |
for the degree in Z of any factor Φ of Φ . This finishes the proof of the lemma.
1 f,K
(cid:3)
3. Bounding the number of integer points on curves and
hypersurfaces
Lemma 5. Let f(X)=a Xn+a Xn−1+...+a C[X]. Then all roots z C
0 1 n
∈ ∈
of the equation f(z)=0 satisfy the inequality
1 a
z max k k .
| |≤ √n2−1 ·1≤k≤nvuu(cid:12)(cid:12)a0 nk (cid:12)(cid:12)
Proof. This is Theorem 3 in 27 of [9]. t(cid:12)(cid:12) (cid:0) (cid:1)(cid:12)(cid:12) (cid:3)
§ (cid:12) (cid:12)
Lemma 6. Let F Z[X ,X ] be irreducible and of degree d. Further, let P ,P
1 2 1 2
∈
be real numbers such that P ,P 1, and let
1 2
≥
N(F;P ,P )=# x Z2 :F(x)=0 and x P (1 i 2) .
1 2 i i
{ ∈ | |≤ ≤ ≤ }
Moreover, let
2
T =max Pei
i
( )
i=1
Y
with the maximum taken over all integer 2-tuples (e ,e ) for which the correspond-
1 2
ing monomial Xe1Xe2 occurs in F(X ,X ) with nonzero coefficient. Then
1 2 1 2
logP logP
(16) N(F;P ,P ) max P ,P ǫexp 1 2 .
1 2 d,ǫ 1 2
≪ { } logT
(cid:18) (cid:19)
Proof. This is Theorem 1 in [1]; see also Lemma 8 in [4] for more details. (cid:3)
ON HILBERT’S IRREDUCIBILITY THEOREM 9
It is crucial for our application of Lemma 6 in proving the following result that
the bound (16) only depends on the degree d of F, but not on its coefficients.
Lemma 7. Let F(Z;T ,...,T ) Z[Z;T ,...,T ] be irreducible, and suppose that
1 s 1 s
∈
F is monic of degree m in Z. Further, let
M (H)=# t Zs : t H such that
F
{ ∈ | |≤
F(Z;t ,...,t )=0 has an integer root z .
1 s
}
Then for every ε>0 we have
(17) M (H) Hs−1+1/m+ε.
F F,ε
≪
Proof. Without loss of generality we may assume that H 2. Then by Lemma 5,
≥
there exists a constant α 1, depending at most on F, such that whenever t Cs
≥ ∈
with t H and F(z;t ,...,t ) = 0 for some z C, then z Hα. We proceed
1 s
| | ≤ ∈ | | ≤
by induction on s. For s=1, Lemma 6 gives
M (H) # (z,t ) Z2 : z Hα, t H,F(z;t )=0
F 1 1 1
≤ { ∈ | |≤ | |≤ }
α(logH)2
Hεexp .
F,ε
≪ logT
(cid:18) (cid:19)
Now F(Z;T ) contains the monomial Zm, whence T Hαm, and we obtain
1
≥
M (H) H1/m+ε,
F F,ε
≪
as claimed. Next, let us discuss the case s > 1, assuming that the lemma has
already been proved for s 1. Let us first consider those t ,...,t Z bounded
2 s
− ∈
in modulus by H such that F(Z;T ) = F(Z;T ,t ,...,t ) still is irreducible over
1 1 2 s
the rationals, as a polynomial in Z and T . Then as above the number of per-
1
missible z and t can be bounded by O (H1/m+ε), since the term Zm is still
1 F,ε
present. Taking into account O(Hs−1) choices for t ,...,t , we end up with a
2 s
contribution of O (Hs−1+1/m+ε), which is compatible with (17). Next, let us
F,ε
discussthoset ,...,t Zboundedinmodulus byH,suchthatF(Z;T )becomes
2 s 1
∈
reducible over the rationals. As F(Z;T ,...,T ) is irreducible over the rationals,
1 s
by (4) the number of such exceptional specialisations t ,...,t can be bounded by
2 s
O (Hs−3/2+ε). Now if F(Z;T ) becomes reducible over the rationals, each irre-
F,ε 1
ducible factor must be at least linear in Z, since F(Z;T ) is monic in Z. If each
1
irreducible factor is at least quadratic in Z, then by the same argument as above
we get a contribution of O (H1/2+ε) for the number of zeros of F(Z;T ), and
F,ε 1
together with O (Hs−3/2+ε) possible choices for t ,...,t we again end up with
F,ε 2 s
a bound compatible with (17). It remains to discuss those t ,...,t Z bounded
2 s
∈
in modulus by H for which F(Z;T ) splits off a linear factor in Z. Let U denote
1
the number of such t ,...,t . We can bound U by the following ‘fibration argu-
2 s
ment’: since F is an integer polynomial, monic in Z, any linear factor of F(Z;T )
1
can be assumed to have integer coefficients and being monic in Z. Given a tu-
ple (t ,...,t ) counted by U, every choice of t Z gives rise to a monic integer
2 s 1
∈
one-variable polynomial F(Z) = F(Z;t ,t ,...,t ) of degree m having an integer
1 2 s
root z. But F(Z;T ,...,T ) is irreducible over the rationals, so by Hilbert’s Ir-
1 s
reducibility Theorem we can choose t Z such that the specialized polynomial
1
∈
G(Z;T ,...,T ) = F(Z;t ,T ,...,T ) still is irreducible over the rationals. Then
2 s 1 2 s
G still is an integer polynomial,only depending on F, and monic of degree m in z.
10 ABELCASTILLOANDRAINERDIETMANN
Therefore our inductive assumption is applicable to G, yielding
M (H) Hs−2+1/m+ε.
G F,ε
≪
On the other hand, as observed above,
M (H) U,
G
≥
since all (t ,...,t ) counted by U, for all t Z, in particular our special choice,
2 s 1
∈
give rise to a specialized F(Z) having an integer root z. Combining the latter two
bounds, we obtain
U Hs−2+1/m+ε.
F,ε
≪
Once (t ,...,t ) have been fixed, we use the trivial bound O(H) for the t ’s and
2 s 1
getatotalcontributionofO (Hs−1+1/m+ε),whichagainiscompatiblewith(17).
F,ε
This finishes the proof. (cid:3)
4. Proof of Theorem 1
Let us first briefly remark that without loss of generality we may restrict to
f(X,T) that are monic in X: For suppose that f(X,T) Z[X,T ,...,T ] of
1 s
∈
degree n in X is given. Then f(X,T) is of the form
f(X,T)=g (T)Xn+g (T)Xn−1+...+g (T)
0 1 n
for suitable g ,...,g Z[T ,...,T ]. As f is of degree n in X, the polynomial
0 n 1 s
∈
g (T)isnotidenticallyzeroandwillbezeroforatmostO (Hn−1)valuesoft Zs
0 f
∈
when t H, which is of negligible order of magnitude with respect to Theorem
| | ≤
1. Now consider the polynomial
h(X,T)=g (T)n−1f(X/g (T),T)
0 0
=Xn+g (T)Xn−1+g (T)g (T)Xn−2+...+g (T)n−1g (T)
1 0 2 0 n
in Z[X,T ,...,T ], which shares all relevant properties with f(X,T): Considered
1 s
over Q(T ,...,T ), both f and h have the same splitting field and hence the same
1 s
Galois group, and for fixed t Zs with g (t) = 0, again f and h over Q have the
0
∈ 6
same splitting field and hence the same Galois group. In particular, as f(X,T) is
irreducible over Q, the same is true for h(X,T). With respect to Theorem 1, we
maythereforewithoutlossofgeneralityassumethatf ismonicinX,i.e. g (T) 1.
0
≡
Now let Φ (Z,T) be the Galois resolvent from Lemma 4. Then for given
f,K
t Zs, if the polynomial f(X,t) has Galois group K over Q, then Φ (Z,t) has
f,K
∈
an integer root z. Factoring Φ (Z,T) over the rationals, each irreducible factor
f,K
can be assumed to have integer coefficients, being monic in Z, and having degree
at least G/K = G/K in Z. Applying Lemma 7 to each such irreducible factor
| | | | | |
of Φ (Z), we immediately obtain Theorem 1.
f,K
5. Proof of Corollary 1 and 2
Let n be the degree of X in f(X,T), so f(X,T) = g (T)Xn+O(Xn−1) for a
0
suitable g (T) Z[T ,...,T ]. The two corollaries then follow from Theorem 1 on
0 1 s
∈
noting that, by Lemma 1, if f(X,t) for some specialisationt Zs still is of degree
∈
n in X, and separable, then the Galois group K of f over Q will be a subgroup
of G. The exceptional t Zs with t H such that f(X,t) has degree less than
∈ | | ≤
n or becomes inseparable are easily seen to be of order or magnitude O (Hs−1)
f
and can therefore be neglected, as they must satisfy g (t) = 0 or ∆(t) = 0, where
0