J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
L
EONARDO
C
ANGELMI
Polynomials whose Galois groups are Frobenius
groups with prime order complement
Journal de Théorie des Nombres de Bordeaux, tome
6, n
o2 (1994),
p. 391-406
<http://www.numdam.org/item?id=JTNB_1994__6_2_391_0>
© Université Bordeaux 1, 1994, tous droits réservés.
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Polynomials
whose Galois
groups
are
Frobenius
groups
with
prime
order
complement
by
LEONARDO CANGELMIRÉSUME - On donne une caractérisation effective des polynômes
irréduc-tibles de degré n à coefficients entiers dont les groupes de Galois sur Q sont des groupes de Frobenius avec noyau d’ordre n et complément d’ordre
premier.
ABSTRACT - We give an effective characterization theorem for integral
monic irreducible polynomials f of degree n whose Galois groups over Q are
Frobenius groups with kernel of order n and complement of prime order.
0. Introduction
Recently,
there has been some interest in theproblem
ofgiving
effec-tive characterizations of
polynomials
ofgiven
degree
whose Galois groupsover Q
are Frobenius groups of someparticular
kind. Recall that aFrobe-nius group can
always
be described as a semi-directproduct
N x H andthat N and H are
respectively
called the(Frobenius)
kernel and(Frobe-nius)
complement
of the group; moreover, the actionby
conjugation
of Hover N has to be
fixed-point-free,
~H)
has to divideIN 1-
1 and the groupcan
always
berepresented
as a transitivepermutation
group on~N~
ele-ments,
so that it is ofdegree
Bruen,
Jensen and Yui[BJY]
proved
a characterization theorem forpolynomials
ofprime
degree
p whoseGa-lois groups are Frobenius groups of
degree
p(in
this case, the Frobeniuscomplements necessarily
arecyclic
of orderdividing
p -1).
They posed
the
problem
offinding
analogous
characterizations forpolynomials
whose Galois groups are Frobenius ofprime
powerdegree
or Frobenius ingeneral.
In this paper, we
give
a characterization theorem forpolynomials
ofarbitrary degree n
whose Galois groups are Frobenius groups ofdegree n
and with
complements
ofprime
order p, where p is forced to divide n - 1.1991 Mathematics Subject Classification. Primary 12F10, 12Y05; Secondary 12F12, 12-04.
Key words and phrases. Effective characterization of polynomials with given Galois
groups, - Frobenius groups - - with prime order complement.
It is worth
noticing
that in thepresent
case the kernels are asgeneral
as itis
permitted by
theproperties
of a Frobenius group, thatis,
they
havejust
to be
nilpotent;
while thecomplements
are restricted to becyclic
ofprime
degree.
The method we use is derived from oneby
Williamson[Will],
whocharacterized odd
degree polynomials
whose Galois groups aregeneralized
dihedral groups.
Indeed,
these groups are Frobenius withcomplements
of order 2 and abelian
kernels;
the fundamental observation is that his method does notrequire
the kernel to beabelian,
butjust
the group to be Frobenius.Moreover,
we make the characterizationcompletely effective,
we discuss thepractical problems arising
and wejustify
theoretically
itseffectiveness
referring
to "the leastprime"
in Chebotarevdensity
theorem.In Section
I,
we recall the definition of Frobenius group and we show someinteresting
properties
of Frobenius groups withcomplement
ofprime
order: these can be described as semi-direct
products
N xZp
satisfying
somesimple
conditions.In Section
II,
we characterize monic irreduciblepolynomials,
withra-tional
integer
coefficients,
ofdegree
n, whose Galois groupsover Q
areFrobenius groups with kernel of order n and
complement
ofprime
order p; thatis,
groups of the form N xZp,
withINI
=n, which are Frobenius.
In Section
III,
wepoint
out some effective conditions which are usefulin
testing
whether apolynomial
has Galois group as above ornot;
suchconditions involve the
computation
of several resultantpolynomials
and their factorization.Then,
weapply
the characterization theorem to someexplicit
polynomials, showing
that their Galois groups are Frobenius groupswith
prime
ordercomplement.
The author is
grateful
to Prof. R. Dvornicich(University
ofPisa,
Italy)
for somesuggestions
and for several fruitful discussions.Our notations are as follows.
If S is a
set,
S)
denotes itscardinality.
If g
is an element of a group which acts on a setS, Fix g
denotes the set of the elements of S fixedby
g.For a group
G,
Aut(G)
denotes the groups ofautomorphisms
of G. If A and B are two groups, A x B denotes anypossible
semi-directproduct
of them.If n is a
positive integer, Zn, ~’n
andCn
denoterespectively
thecyclic
groupof order n, a
primitive
n-th root ofunity
and the field if n is aprime
For a
polynomial f
E7G~X~,
spl ( f ), Gal ( f ),
andD f
denote thesplitting
field of
f ,
the Galois group of thesplitting
field and the discriminant off .
For any number fieldL, DL
denotes the discriminant of Lover Q
andOL
denotes the
ring
ofintegral
elements of L.For any finite extension of number fields K
(also
written asL/K)
andany
f
EL(X~,
NL~K ( f )
denotes the normof f over
K. For any finite normalextension of number fields
L/K,
Gal(L/K)
denotes its Galois group.If k is a field or a
polynomial
ring,
andf (X)
andg(X)
are twopolynomials
in
k [X],
denotes the resultant off and g
withrespect
to the indeterminate X.
Other notations will be recalled and
adopted
in the course of the paper.I. Frobenius groups 1.0. General results
DEFINITION I.0.1. A
permutation
group G on a set S2 withIn
= n is saidto be a group of
degree
n.DEFINITION I.0.2. A transitive group G of
degree n
is said to be aregular
group
if,
andonly
if,
IFix91
=0,
forall g
EGB{1}.
DEFINITION I.0.3. A transitive group G of
degree n
is said to be aFrobe-nius group
if,
andonly if,
1,
forall g
E andIFix91
= 1,
forsome 9 E
GB{1}.
It is trivial to see that a transitive group G of
degree n
isregular if,
andonly if,
~G~
= n. On the otherhand,
a transitive group G ofdegree
n is aFrobenius group
if,
andonly if,
it has a non trivialsubgroup
H,
of index n,such that H n H9 =
{1},
forall g
EGBH
(anyone
of the stabilizers can betaken as
H).
Thisproperty
allows to deduce a verystrong
theorem aboutthe structure of Frobenius groups.
THEOREM 1.0.4
~FROBENIUS~.
Let G be a Frobenius groupof degree
n andH be a
subgroup
(of
indexn)
such that H fl H9 =111,
for
all g EGBH.
Then the subset
is a normal
subgroup
(of
ordern),
such that G = NH and N n H ={1}.
Since the
conjugates
of H areexactly
the stabilizers of thepoints
of52,
the elements of
NB{1}
areprecisely
those in G with no fixedpoints.
Nis called the
(Frobenius)
kernel ofG,
while H is said to be a(Frobenius)
complement
of G. Note that N turns out to be aregular
group ofdegree n
and that G can be written as a semi-direct
product, namely
G = N x H.Furthermore,
G satisfies some otherstrong
necessaryconditions,
as it isstated in the
following proposition.
PROPOSITION 1.0.5. Let G = N x H be a Frobenius group
of degree
n,= n and
IHI
= h. Then thefollowing
conditions hold:a) CG (H) =
CH (H)
b)
H iscore-free
(i. e., it
does not contain non-trivial normalsubgroups
of
G).
c) h I n - 1.
d) (Thompson)
N isnilpotent.
Proof. See,
e.g.,[Rob].
1.1. Frobenius groups with
prime-order
complement
Let N be a group of order n
having
anautomorphism 0
ofprime
order pand let
Zp
denote thecyclic
group of order p. We can construct the semi-directproduct
G = N xoZp:
ifZp = (T),
then we set Tx =8(x)r,
forall x E N.
Then,
we mayverify
whether G is Frobeniusby
means ofseveral different conditions.
PROPOSITION I.1.1. Let G = N
Zp,
n = andp
f
n. Then thefollowing
conditions areequivalent:
a)
G is a Frobenius groupof degree
n.b)
=Zp.
c)
d)
For any x EN,
there existsexactly
one elementy E N
such thatO
e)
e(p-l)i(x)
=1,
for
all x E N andfor
all i =1, ... , p -
1.f) Any
element inGBN
has order p.g)
G hasexactly
nSylow p-subgroups.
Proof.
a)~b):
seeProposition
1.0.5(a).
b)=>c):
by
the very definitionof G,
we have that Fix 0 =CN(T)
=hence,
F1x 0 =(1) ,
because theonly
elementscommuting
with rC)==>d):
ifx-19(x)
=y-1B(y),
with x and y inN,
thenyx-1
=0(yx-1)
and therefore = 1.
d)~e):
let y
=z0 (z) ...
then0(y)
= =y x
We may write x in the form
z-’O(z)
and hence weget
zyz-i;
by
theuniqueness
of suchexpression,
weget
zyz-1
=1,
hence y = 1. This proves the
implication
in the case i = 1. For thegeneral
case, note that condition(e)
for i = 1implies
condition(c);
on the other
hand,
by
theprimeness
of p, we have Fix 0 =Fix 0’.
So 0’
is anautomorphism
of N with the sameproperties
as 0 andtherefore the assertion holds for all i.
e)#f):
any element in GBN can be written in the formxTi,
for some x E N and some i E{1, ... , p-1}:
then(xTi)P
=1,
since(xTi)P
=...
f)~g):
G containsexactly
IGBNI
=n(p -1)
elements of orderp and
they
give
rise toexactly n subgroups
of order p.g)==>a):
Zp
hasexactly
nconjugates
and anyone of them can be writtenas
Z~,
for some x EN;
if g
EGBZp,
thenZj
=Zp
for some x E henceZ~ ,
and thereforeZp
f1ZP
= { 1 } .
Condition
(c)
of theprevious proposition
says that theautomorphism 0
is
fixed-point-free.
A Frobenius group with kernel N and
complement
Zp
will be denotedby
Fr(N, p).
It isplain
that such a group satisfies thefollowing
properties,
besides the ones claimed in
Proposition
I.1.1(cf.
Proposition
1.0.5):
- ord(B)
= p. - - N isnilpotent.
- pI IAut
In
particular,
N is the directproduct
of itsSylow subgroups: if q
n and
Nq
is theSylow q-subgroup
ofN,
then81Nq
is afixed-point-free
automor-phism
ofNQ
of order p.Conversely,
given
afixed-point-free
automorphism
of order p for each of the
Sylow subgroups
ofN,
it is determined afixed-point-free automorphism
of order p of N.Therefore,
thestudy
of groups asFr
(N, p)
is reduced to that of Frobenius groups withZp
ascomplement
and a q-group askernel;
and this isequivalent
to thestudy
offixed-point-free
automorphisms
of order p of q-groups, withq ~
p.I.2.
Cycle
structure of elements of Fr(N, p)
In
general,
given
apermutation
group G ofdegree n
andassigned
arepresentation
of G as asubgroup
S ofSn,
anyelement g
E Ggives
rise toa
partition
7r(g)
of n: infact, g
can be written as apermutation
ofdegree
n(via
thegiven
representation
ofG),
and as such it induces apartition
of ngiven
by
thelengths
of thecycles
in theexpression
of g as product
ofdisjoint
cycles.
Note that for anypermutation
of f 1.... , n~,
the new identificationof G with a
subgroup
ofSn
isgiven
by
aconjugate
of,S’;
hence thepartition
remains
unchanged.
We say that an
element g
in a group G ofdegree n
is oftype
7r, where 7ris a
given partition
of n,if,
andonly if,
= 7r, for agiven
representation
of G as a
permutation
group ofdegree
n. We will write"partitions"
inmultiplicative form;
thatis,
nil - n22 - - -
will denote thepartition
ofE
ein2given
by
el times nl, e2 times n2, ....PROPOSITION I.2.1. Let G = and
represent
G as apermutation
group
of degree
n. Then thepartition types
inducedby
G are thefollowing:
-for
theidentity of G.
-
1 -
for
each element inGBN.
-
n1 - - - nr, with some 2 such that p
~’
nj andI
n,for
eachelement in
Proof.
Eachelement g
in GBN has order p and so it isproduct
ofdisjoint
cycles
oflength
p or1;
moreover, since G is Frobenius ofdegree
n, any element can fix at most onepoint.
Hence theonly possibility
for is tobe
equal
to 1 - For the elements in note thatthey
fix nopoints
and thatp t
n .II. Characterization of
polynomials
ofdegree n
with as Galois groupsILO. Factorization
types
ofpolynomials
Let
f E
Z[X]
be a monicpolynomial
ofdegree n
and let 7r be apartition
of n. We say that
f
is oftype
7rif,
andonly if,
thepartition
of ngiven
by
the
degrees
of the irreducible factors off
is 7r. We say thatf mod q
is oftype
7r,where q
is a rationalprime, if,
andonly if,
thepartition
of ngiven
by
thedegrees
of the irreducible factorsmod q
off
is 7r.Again,
if E is anynumber
field,
we say thatf
over E is oftype
7rif,
andonly if,
thepartition
of n
given
by
thedegrees
of the irreducible factors over E off
is 7r.We have the
following deep
relations between the factorizationtypes
off
and thecycle
structure of the elements of Gal( f )
(this
can befaithfully
represented
as apermutation
group ofdegree
n, on the set of roots off ,
and any
permutation
of the roots reflects ontaking
aconjugate
of thegiven
group
representing
Gal ( f )
inSn).
LEMMA II.0.1. Let
f
EZ[X] be
a monicpolynomial of degree n
and let 7rbe a
partition
of
n.Let q
denote any rationalprime
such thatq f
D f
andLet G =
Gal ( f ) . Put L = spl( f ) .
a) (Dedekind)
If f mod q is of
type
then G contains an elementsof
type
7r.b)
(Frobenius)
If
G contains elerraentsof
types
then thedensity of
q’s
such thatf
modq is
of
type
7r ispositive.
c)
(Chebotarev)
Thedensity of q’s
such thatf
mod q isof
type
7r isequal
to theproportion
of
elements in Gof
type
7r.d) (Lagarias,
Odlyzko
etal.)
If
G contains elementsof
type
then there exists q such thatf
mod q isof
type
7r and q(DL)C,
whereC is an
effectively computable
absoluteconstant;
assuming
theGen-eralized Riemann
Hypothesis for
L,
one hasq
70(log
DL) 2.
Proof.
a)
See,
e.g.,[Jac,
pp.302-304].
b)
See[Frob].
c)
See,
e.g.,[Lang,
pp.168-170].
d)
See[LMO]
and[Oes].
11.1. Characterization theoremLet
L/Q
be a normal extension and let K be a subfield of L such thatL/K
hasdegree
n,K/Q
is acyclic
extension ofprime
degree p
andPut G = Gal
(L/Q)
and N = Gal(L/K);
then,
since(p, n)
=1,
L/Q
splits
at K and G =
ZP.
Moreover, K
is theunique
subfield of L ofdegree
pover
Q,
because N is theunique
subgroup
of G of order n. Theproduct
N x 0
Zp
is directif,
andonly if,
Fix 0 =N;
if this is not the case, thenZp
is not normal and therefore is core-free.
PROPOSITION II.1.0. Let
L/Q
be a normal extensionsof degree
pn, withGal
(L/Q)
= NZp and p t n.
Then theproduct
is notdirect if,
andonly
if,
there monic irreduciblepolynomial
f ,
withinteger
coefficients
andof degree
n, such that L =spl
( f ) .
Proof.
If theproduct
is notdirect,
Zp
is core-free and the minimalpolyno-mial of an
integer primitive
element of satisfies therequired
proper-ties. If the
product
isdirect,
L contains aunique
field ofdegree
n overQ,
In
general,
a normal extensionL/Q
whose Galois group is a non-directproduct
as abovemight
also begiven
as thesplitting
field of an irreduciblepolynomial
ofdegree
notequal
to n; anyway, theprevious proposition
saysthat we can
always
think of it as thesplitting
field of apolynomial
ofdegree
n.For the rest
of
thissubsection,
f
EZ[X]
will denote a rraonic irreduciblepolynomial of degree n
and we willput
G = Gal( f )
and L =spl
( f ).
If G = Fr
(N, p),
f
is irreducible and normal overLN,
since N isregular;
moreover, from
Proposition
1.2.1 and Lemma11.0.1,
we know that thereexists a rational
prime
q such thatf mod q
is oftype
1 - Suchconditions turn out to be also
sufficient,
as we are nowgoing
to show intwo
steps.
PROPOSITION II.1.1. Let p be a rational
prime
such that p{
n. Then thefollowing
conditions areequivalent:
-G = N ~ae
Zp, IN
= n andf is
irreducible overL~.
-p ~
I
[L : Q]
and there exists acyclic
extension kof Q of degree
p, such thatf
is irreducible and normal over k.Proof.
The first conditiontrivially implies
the other one,by taking
k =L~’.
Conversely,
sincedeg f ~
~L : Q]
and(p, n)
=1,
we havepn ~
I [L : Q].
LetE be the
splitting
field off
overk;
by
thehypothesis,
[E : k~ = n and
therefore
[E : Q]
= pn. SinceL C E,
weget
L =E; hence,
k C L andG = N xe
Zp,
where N = Gal(L/k).
PROPOSITION 11.1.2. Let
pin -
1 and let theequivalent
conditionsof
Proposition
11.1.1 hold. Put K =LN.
If
there exists a rationalprime
q,
unramified in
L,
such thatf
mod q isof
type
then G = Fr(N, p);
rraoreover, q is inert in K and
qOK splits completely
in L.Proof.
Since
[K : Q]
= p, for any rationalprime
q, unramified inL,
thereare
just
twopossibilities: either q
is inert in K or itsplits completely
in K. If the latter is the case, then
qOK
=C~1 ~ ..
C~~
and =1;
hence, reducing f
modQi
isequivalent
toreducing f
mod q. In the caseof the
hypothesis,
we would have thatf
modQi
is oftype
1 -and therefore
(see
Lemma 11.0.1(a),
whichapplies
also to number fields-cf.[vdW,
pp.190-191])
Gal(L/K)
would contain anelement,
notequal
to theidentity,
which fixes a root off ;
but this isimpossible,
since N =Gal
(L/K)
isregular
on the roots off .
This shows that q isnecessarily
inert in K.
So,
deg(qOK/q)
= p and thereforelF9p.
Since thefactor-ization of
f
overIFq
is oftype
1 . the factorization off over
Fgp,
and hence over is of
type
1’~.Being f
irreducible and normal overK,
L =K(a)
for any root a off ;
then,
splits completely
inL,
Being
= pand q
unramified inL,
thedecomposition
groupof
Qi,
GQ,,
iscyclic
of order p, i.e.GQ, cr Zp.
Putting
F =Q(a),
where a is any fixed root of
f ,
from the factorization off mod q
weget
Bi ..
B,
withdeg(Bh/q)
= p and =1,
forh =
1, ... ,
(n -
1)/p.
Comparing
the factorizations ofqOF
and ofqOL,
we see that there are
exactly
pprimes
of L aboveBh ,
while there isexactly
one
prime
above,Ci,
sayQ.
Therefore F DLGQ
andF 1J
LGQ’,
for 6’ !
This
implies
F =LGQ
andL GQ 0
LGQ’,
and soGQ , 0 G Q, -
Therefore,
Gcontains n distinct
cyclic subgroups
of order p, thehighest possible
number,
hence G has nSylow p-subgroups
and it is Frobeniusby Proposition
I.1.1.By
the sametechnique
used in theproof
of theprevious proposition,
we can show also thefollowing fact,
concerning
the kind ofprimes q
whichgive
rise to factorizations of
f
mod q oftype
1 .PROPOSITION II.1.3. Let G = Fr
(N, p),
withINI = n and p I n - 1.
Then,
for
any rationalprime
q,unramified
inL,
f mod q is of
type
1.and
only if,
q is inert in K andqOK
splits
completely
irc L.Proof.
Omitted.We are now in a
position
togive
the announcedcharacterization, just
putting together
all the above results.THEOREM II.1.4. Let p be a rational
prime
dividing
n - 1. Then G =Fr (N, p) if,
andonly if,
thefollowing
conditions hold:1.
p I [L : Q].
2. There exists an extension
k/Q,
cyclic of degree
p, such thatf
isirreducible and normal over k.
3. There exists a rational
prime
q,unramified
inL,
such thatf
mod qis
of
type
1 -p(n - 1) /p.
We will see in the next section how this characterization can be made
III. Effectiveness of the characterization
111.0. Resultants and factorization of
polynomials
over number fields
Let K =
k((3)
be a numberfield,
where k is a subfield of K andQ
is a fixed root of a
given
a monic irreduciblepolynomial g
Ek(X~,
ofdegree
m; let(31,... ,13m
denote the roots of g. Letf
EK~X~
be a monicsquare-free polynomial
ofdegree n
and al, ... , a~, be the roots off ;
wemay write
f
as apolynomial
ink[X, (3],
f
=f (X, (3).
We want to considerthe factorization
type
off
overK;
suppose that such a factorization is/i’ " ft.
For anypolynomial
EK[X]
and any rationalinteger
s, letN(s,
h,
g)
stand for the norm over k of the shiftedpolynomial
h(X -
s (3,
{3) ,
that is for
(3)).
Referring
to the factorizationalgorithm
ofpolynomials
over number fields due toTrager
(see,
also,
[vdW],
pp.
136-137),
we can affirme thefollowing:
i)
there areonly finitely
many rationalintegers s
such thatN(s,
f,
g)
is not
square-free;
ii)
for any rationalinteger s
such thatN(s,
f,g)
issquare-free,
the factorization over k ofN(s, f,
g)
isN(s,
fl,
g) ... N(s,
ft,
g).
Thus,
each irreducible factor off over
ofdegree
d maps to anirre-ducible factor of
N(s, f, g)
over k ofdegree
nd,
and vice versa.Moreover,
we are able to calculate the norms aboveby
means of thefunction
resultant,
which is available on anycomputer
algebra
system;
infact,
thefollowing equalities
hold(the
last one up to asign):
In
particular,
whenf
Ek[X]
is irreducible overk,
we maystudy
thefactorization
type
off
over the fieldk(a),
where a is a fixed root off .
In such a case, one of the factors off
is X - aand,
since we areobviously
interested in the other
factors,
we define thefollowing polynomial
which is a monic
polynomial
ink[X]
ofdegree
n(n - 1)
and it is a linearresolvent of
f when s 0
0,1. Then,
we claim thatf
is normalover k,
i.e.
spl
( f )
=k(a)
fortype
Infact,
f
is normal over kif,
andonly
if,
f
factors overk(a)
asthe
product
of n linearpolynomials,
which isequivalent
toN(s, f )
being
of
type
nn overk,
hence toR(s, f )
being
oftype
over k.III.1. Factorizations of
R(s, f )
For this subsection and
for
the next one, assumef
EZ[X]
to be a monicirreducible
polynornial
of degree
n, such that Gal( f )
= Fr(N, p),
withIN I
n, and
Put
spl ( f )
= L and K =L~’.
Fix a root a of
f
and let a’ be any other root off .
How doesf
factor overQ(a)?
Firstly,
note that[L : Q(a)]
=p and therefore any irreducible
factor of
f
overQ(a)
may be ofdegree
either 1 or p.Then,
since Gal( f )
=Fr
(N, p),
we know that there exists a rationalprime
q, unramified inL,
such that
f
mod q is oftype
1by Proposition
II.1.3,
q is inert inK and
qO K splits completely
in L.If qOL
=61 " ’
then,
proceeding
asin the
proof
ofProposition
11.1.2,
we find that the nSylow p-subgroups
ofG = Gal
( f )
are thedecomposition
groups of theprimes
of L over q,and that the
respective
decomposition
fields areprecisely
the extensionsof Q by
the roots off ,
i.e.L GQi =
Q(ai) (i
=1, ... , n).
Thisimplies
Q(a) #
Q(a’)
and soa:’ ft
Q(a).
Hence,
the minimalpolynomial
of a’ overQ(a)
hasdegree equal
to p.Thus,
we haveproved
thefollowing
fact.PROPOSITION III.1.1. Let
f
be as above and let a be a rootof f.
Thenf
over
Q(a)
isof
type
1 -and,
for
any rationalinteger
s such thatR(s, f ) is
square-free,
R(s,
f )
over Q
isof
type
Since
f
is irreducible and normal overK,
from the last remark in theprevious subsection,
we have also thefollowing
result,
involving
thefactor-ization of
R(s,
f )
over K.PROPOSITION III.1.2. Let
f
be as above.Then,
for
any rationalinteger
ssuch that
R(s,
f )
issquare-free,
R(s, f)
over K isof
type
Since we are
just
interested in the factorizationtype
ofR(s, f )
over Kand not in the actual irreducible
factors,
we canapply
again
the methoddescribed in the
previous subsection,
getting
thefollowing.
PROPOSITION III.1.3. Let
f as
aboveand g
EZ[X]
be a monic irreduciblepolynomial
such that K =Q(Q),
whereQ
is a rootof
g.Then,
for
anyrational
integer
t such thatN(t,
f,
g)
issquare-free, N(t,
f,
g)
is irreduciblefor
any rationalinteger
t such thatN(t,
R,
g)
issquare-free,
N(t,
R,
g)
overQ is
of
type
(np)n-1.
III.2. Determination of K
Recall that K is
cyclic
ofprime
degree
p overQ. Thus, being
abelian overthe
rationals,
by
Kronecker-Webertheorem,
I~’ is included in acyclotomic
field
Cl
(where
Cl
=Q((),
with C
aprimitive
1-th root ofunity),
for somepositive
integer
suchthat q
f
DK
implies q
{
DCl’
for any rationalprime
q.Assuming
l to be minimal withrespect
to thisproperty,
its factorizationis to be of the kind
p2aql ... qs,
where a E10, 1}
and qj == 1 mod p, forj
=1, ... ,
s.Moreover,
sinceDK ~
I DCl’
we haveNow observe that
f
is irreducible and normal overK; hence,
L =K(a),
for any root a of
f .
Fromthis,
it follows that andtherefore,
if q
l,
thenq
I Df.
Seeing
the factorization ofl,
we deducethat l pD f.
Thus,
if the factorization ofD f
iswith eo 2:
0,
ei 2:1,
f~ >
1,
1 mod p 1 mod p, for i =1, ... ,
rand j
=1, ... , t, putting
we
have l
m.Therefore,
K CCl
CCm
and K can be determined as one of thecyclic
subfields of
degree
pover Q
ofCm.
These arefinitely
many, sincethey
correspond
to thesubgroups
of index p of Gal(7~/m7~) *;
indeed,
their number is
precisely
1) I (p -
1).
111.3. Effective version of the characterization theorem
We are now able to
give
an effective version of Theorem 11.1.4. Forsimplicity,
we assumeGRH;
so, we can use thestrongest
boundgiven
inLemma II.0.1. This is not so
restrictive,
as it is shownby
all testedexam-ples,
where the leastprime q
is much smaller than the onegiven
theoreti-cally,
even under suchassumption.
THEOREM III.3.1. Let
f
EZ[X]
be a monic irreduciblepolynomial of
de-gree n and
Put
G = Gal( f )
and L= spl
( f ).
Let p be a rationalprime
suchthat
pin -
1. Assume GRH.If
G = Fr(N, p),
with)N)
=n,
then, putting
K =
L~’
and R =R(s, f ),
for
sorrte rationalinteger
s such thatR(s, f )
issquare-free,
we have:i. R is
of
type
(np)(n-l)/p.
ii. K is one
of
thecyclic subfields of degree
pover Q
of cm,
where mis
given
in Section111.2;
moreover,f
is irreducible over K and Rover K is
of
type nn-1.
iii. There exists a rational
prime
q, notdividing
D f,
such thatq
and
f
modq is
of
type
1.That such conditions are sufficient for G = Fr
(N, p)
wasalready proved
in Theorem
11.1.4;
thepoint
is that now we have an effective way toverify
the conditions
given
in that theorem. III.4.Examples
Although
thegiven
characterization determines if the Galois group of apolynomial
ofdegree n
is Frobenius with kernel of order n andcomplement
of
prime
order pdividing n -
1,
ageneral
method forconstructing
suchpolynomials
is notknown,
apart
from few cases.Moreover,
in some cases atheoretical method may be
outlined,
but thecomputations
involved exceedthe
capacity
of thecomputer
machines available to theauthor,
so that we are not able to exhibit manypolynomials
of therequired
degree
and withthe
required
Frobenius Galois group.Therefore,
togive
someexamples
of how our method
applies,
we must refer topolynomials
found in theliterature,
and somehowmanipulate
them.Example
1. Thesimplest example
of apolynomial
consideredby
ourmethod is a
polynomial
ofdegree
4 whose Galois group is thealternating
group of
degree
4,
xZ2, 3).
Suchpolynomials
are easy to construct.Take,
forexample,
thefollowing
one:We will prove that its Galois group is
A4,
verifying
the conditionsgiven
inTheorem 111.3.1 and
using
alsoProposition
III.1.3.-
The smallest
prime
q for whichf,
mod q factors as 1 -32 is
3. -The resolvent R =
R(-1, fl)
issquare-free
and it is irreducibleover
Q.
- The discriminant of the
polynomial
is21272132
so thepossible
cubicincluded in
C7;
aprimitive
element for such cubicfield,
which wecall
.K,
is a root of thepolynomial
We
compute
fl,g)
which turns out to besquare-free
and it isirreducible.
Then,
wecompute
N(1, R, g)
and weverify
that it issquare-free
and oftype
123.
Hence,
we claim that the Galois group off,
isA4
and that theunique
cubic field included in its
splitting
field is theunique
cubic field includedin
C7,
that is K.Example
2.Bruen,
Jensen and Yui[BJY]
constructed afamily
ofra-tional
integral polynomials
ofdegree
7 whose Galois groupsover Q
areFrobenius groups of order
21,
that is groups as Fr(Z7, 3).
From thisfamily,
we have taken thefollowing
polynomial
We
apply
our characterization theorem to thispolynomial.
- We factor
f2 mod q
for some rationalprimes
q, in order to find afactorization of
type
1 ~32 -
should we find a factorizationtype
notequal
to17,
1.3 2
or7,
we would be sure that Gal(f2) 0
Fr(Z7, 3)
(see
Proposition
I.2.1).
Theprime
3gives
therequired
factorization.- We
compute
the resolvent R =R(-1, f2),
which turns out to besquare-free
and factorsover Q
into 2 irreduciblepolynomials
ofdegree
21, Rl
andR2,
as weexpected.
-The discriminant of
f2
isequal
to224710.
Therefore,
theunique
possible
cubic field to consider isK,
the cubic field included inC7.
The norm
N(1,
f2, g)
issquare-free
and it is irreducible. The norm=
N(1,
Ri,
g)N(1,
R2,
g)
issquare-free
as well and it isof
type
216.
Hence,
we have verified that the Galois group off2
is Fr(Z7, 3)
and wehave shown that the cubic field included in its
splitting
field is K.Example
3. To find otherexplicit polynomials,
we refer to aproposition
by
Williamson. If gi and g2 are two monicpolynomials
ink[X],
where kis any number
field,
letS(gl, g2)
denote thepolynomial
whose roots are allthe sums of one root of gi and one root of g2; we can
easily
compute
suchpolynomial by
thefollowing
formula:PROPOSITION III.4.1. Let 91 and 92 be irreducible monic
polynomials
over a numberfield
k and letEl
andE2
be theirrespective
splitting fields
over k.If
E1
andE2
arelinearly disjoint
overk,
then thepolynomial
S(gl, 92)
isirreducible over k and its
splitting field
over k isElE2.
Proof.
See[Will]
-We will
apply
such a result to thepolynomials f,
andf2,
which areirreducible over K. We
compute
thepolynomial
f3(X)
=S(fl, f2)(X),
which is
equal
toSince the
splitting
fields over K off,
andf2
haverelatively
prime
degrees
over
K,
they
arelinearly disjoint
overK,
so thatf3
is irreducible over Kand the Galois group of its
splitting
field over K isZ2
xZ2
xZ7:
being
the
degree
off3
equal
to28,
we deduce thatf3
is also normal over K. It isalso obvious that the
degree
of thesplitting
field off over Q
is amultiple
of
3,
since such field includes K.So,
conditions 1 and 2 of Theorem 11.1.4are satisfied and the
only
left verification is about the factorizationtype
off3
mod q: the smallestprime q
whichyields
thetype
1’39
is 11.Hence,
Gal (/3)
=Fr(Z2
xZ2
xZ7, 3)
and the cubic field included inspl ( f3) is K.
Example
4. Weproceed
as in theprevious
case. From the mentionedfamily
ofpolynomials by
Bruen,
Jensen andYui,
we take anotherpolyno-mial,
whose
splitting
fieldagain
includes K as above. We construct thepolyno-mial
fs(X)
=S( f2,
f4)(X)
and weverify
that it is irreducible overQ.
Thisimplies
that thesplitting
field off2
and/4
aredifferent,
hence thatthey
aredisjoint
over K and that the Galois group off5
over K isZ7
xZ7.
As in theprevious
example,
conditions 1 and 2 of Theorem 11.1.4 areplainly
satisfied.Then,
we find that 11 is the smallestprime q
such that the factorizationtype
off5
mod q is 1 .316.
Therefore,
Gal(/5) = Fr(Z7
xZ7, 3)
and the cubic field included inspl ( f5)
is K. We do notreport
f5,
because itsgreatest
coefficients are48-digit
numbers.REFERENCES
[BJY]
A. A. Bruen, C. U. Jensen, N. Yui, Polynomials with Frobenius groups of primedeegre as Galois groups II, J. Number Theory 24 (1986), 305-359.
[Frob]
G. Frobenius, Über Beziehungen zwischen den Primidealen eines algebraischenKörpers und den Substitution seiner Gruppe, S. B. Akad. Wiss. Berlin (1896), 689-705.
[Jac]
N. Jacobson, Basic algebra I, 2nd ed., Freeman, New York, 1985.[Lang]
S. Lang, Algebraic number theory, GTM 110, Springer-Verlag, New York, 1986.[LMO]
J. C. Lagarias, H. L. Montgomery, A. M. Odlyzko, A bound for the least prime ideal in the Chebotarev density theorem, Invent. Math. 54(1979),
271-296.[Oes]
J. Oesterlé, Versions effectives du théorème de Chebotarev sous l’hypothèse de Riemann généralisé, Astérisque 61 (1979), 165-167.[Rob]
D. J. S. Robinson, A course in the theory of groups, GTM 80, Springer-Verlag,New York, 1982.
[Trag]
B. M. Trager, Algebraic factoring and rational function integration, ACM Sym-posium on Symbolic and Algebraic Computation 1976 (Jenks, ed.), ACM Inc.,New York, 1976, pp. 219-226.
[vdW]
B. L. van der Waerden, Modern algebra, 2nd ed., vol. I, Ungar, New York, 1953.[Will]
C. J. Williamson, Odd degree polynomials with dihedral Galois groups, J. NumberTheory 34 (1990), 153-173.
L. Cangelmi
Dipartimento di Metodi e Modelli Matematici per le Scienze Applicate
Universita di Roma "La Sapienza" Via A. Scarpa 10
00161 Roma, Italy