On the Galois group of generalized Laguerre polynomials
par FARSHID HAJIR
À Georges Gras,
à l’occasion de son 60ème anniversaire RÉSUMÉ. En utilisant la théorie despolygones
de Newton, onobtient un critère
simple
pour montrer que le groupe de Ga- lois d’unpolynôme
soit"grand."
Si on fixe a Fi-laseta et Lam ont montré que le
Polynôme
Generalisé deLaguerre
L~a~(x) _ ’L.7=o
est irréductiblequand
ledegré
nest assez
grand.
On utilise notre critère afin de montrer que, sousces
hypothèses,
le groupe de Galois de est soit le groupealterné, soit le groupe
symétrique,
dedegré
n,généralisant
desrésultats de Schur pour
Q == 0, 1, ::!:~, -1 -
n.ABSTRACT.
Using
thetheory
of NewtonPolygons,
we formulatea
simple
criterion for the Galois group of apolynomial
to be"large."
For a fixed aE Q - Zo,
Filaseta and Lam have shown that the nthdegree
GeneralizedLaguerre Polynomial
=irreducible for all
large enough
n. Weuse our criterion to show
that,
under theseconditions,
the Ga-lois group of is either the
alternating
or symmetric groupon n letters,
generalizing
results of Schur for a =0,1, J=-
-1- n.1. Introduction
It is a basic
problem
ofalgebra
tocompute
the Galois group of agiven
irreducible
polynomial
over a field K. If we order the monicdegree
npoly-
nomials over Z
by increasing height,
then theproportion
which consists ofirreducible
polynomials
with Galois groupSn
tends to1;
for a more pre- cise statement, see forexample Gallagher [6]. Nevertheless,
to prove that the Galois group of agiven polynomial
ofdegree n
isSn
can be difficultif n is
large.
Thealgorithmic
aspects of Galois groupcomputations
havewitnessed a number of recent
advances,
for which an excellent reference is thespecial
issue[18]
of the Journal ofSymbolic Computation, especially
Manuscrit reçu le 23 juin 2004.
Mots clefs. Galois group, Generalized Laguerre Polynomial, Newton Polygon.
This work was supported by the National Science Foundation under Grant No. 0226869.
the foreward
by Matzat, McKay,
andYokoyama. Currently,
for rationalpolynomials
ofdegree
up to15,
efficientalgorithms
areimplemented,
forinstance,
in GP-PARI and MAGMA. Animportant piece
of any suchalgo-
rithm is the collection of data
regarding
individual elements of the Galois group, for which the standard method is to factor thepolynomial
modulovarious
"good" primes (i.e.
those notdividing
itsdiscriminant),
obtain-ing
thecycle-type
of thecorresponding
Frobeniusconjugacy
classes in theGalois group.
Our first
goal
in this paper is to formulate a criterion whichexploits
the behavior of a
given polynomial
at "bad"primes
forbounding
the sizeof its Galois group from below. The criterion is
especially
efhcacious ifone
suspects
that a "medium size"prime (roughly
betweenn/2
andn)
iswildly
ramified in thesplitting
field of thepolynomial.
The criterion wegive (Theorem 2.2)
followsquite simply
from thetheory
ofp-adic
NewtonPolygons;
it is used inslightly
lessgeneral
form in Coleman[2]
and isreminiscent
of,
but distinctfrom,
a criterion of Schur[21, §1].
Our second
goal
is to illustrate theutility
of the criterionby using
itto calculate the Galois group for
specializations
of a certain one-parameterfamily
ofpolynomials,
which we now introduce. In the second volume of their influential and classic work[20], Pôlya
andSzegô
define the General- izedLaguerre Polynomial (GLP)
The
special
case a = 0 hadappeared
much earlier in the work of Abel[1,
p.
284]
andLaguerre [17],
and thegeneral
case can in fact be found in Sonin[24,
p.41]. Shortly
after thepublication
of[20],
thestudy
of thealgebraic properties
of thisfamily
oforthogonal polynomials
was initiatedby
Schur[21], (22~.
For
instance,
for the discriminant of the monicintegral polynomial
we have the
following
formula of Schur~22~:
In
particular,
if a is not in-n, -2) nz,
has norepeated
roots. For0152 =
0~1/2,1,
Schur[21], [22]
established theirreducibility
of allover
Q,
and also showed that their Galois groups are aslarge
aspossible,
namely An
if is a rational square, andSn
otherwise.A number of recent articles on the
algebraic properties
of GLP haveappeared, including
Feit[3],
Coleman[2],
Gow[9],
Filaseta-Williams[5],
Filaseta-Lam
[4],
Sell[23], Hajir [10], [11],
andHajir-Wong [12].
In par-ticular,
we have thefollowing
theorem of Filaseta and Lam[4]
on the irre-ducibility
of GLP.Theorem.
(Filaseta-Lam) If a
is afixed
rational number which is nota
negative integers,
thenfor
all butfinitely
manyintegers
n, is irreducible overQ.
In this paper, we
provide
acomplement
to the theorem of Filaseta and Lamby computing
the Galois group of when n islarge
withrespect
to a
e Q - Zo. Namely,
we prove thefollowing
result.Theorem 1.1.
Suppose a is
afixed
rational nurrcber which is not anegative integer.
Thenfor
all butfinitely
manyintegers
n, the Galois groupo f
is An if
square andSn
otherwise.Remarks. 1. The
hypothesis
that a not be anegative integer
is necessary,as in that case, is divisible
by
x for n >jal.
For astudy
of thealgebraic properties
of(x)
for a EZo, n lai,
see[10], [23]
and[11].
In the latter
work,
Theorem 2.2 is used to prove thatAn
is contained in theGalois group of
L~ n r~
for alllarge enough (with respect
ton) positive integers
r.2.
Using
a different set oftechniques,
thefollowing companion
to Theorem1.1 is
proved
in[12]:
If we 5 and a number fieldK,
then for allbut
finitely
many a eK,
is irreducible and has Galois groupAn
or
Sn
over K. For each n4, infinitely
manyQ-reducible specializations exist,
and for n =4,
there areinfinitely
manyspecializations
which areirreducible but have
D4-Galois
group overQ,
cf.~11,
Section6].
3. For
integral
a, some cases where is a square(giving
Galois groupAn )
are. a = 1 and n -1
(mod 2)
or n + 1 is an odd square([22]),
. a = rt, and n = 2
(mod 4) ([9],
it is notyet
known if all of thesepolynomials
are irreducible[5]),
. a =
-1- n,
and n > 0(mod 4) (~21), [2]),
· a =
-2 - n,
and n -1(mod 4) ([10]).
~ ~ ==
-3 - n,
and n + 1 is an odd square([23]).
See
[11, §5]
as well as the above-cited papers for more details.4. The
proofs
of the Filaseta-Lam Theorem in[4]
and of Theorem 1.1 areboth effective.
Acknowledgements.
I would like to thank theorganizers,
ChristianMaire,
Jean-Robert Belliard and Hassan
Oukhaba,
forhaving
invited me tospeak
at the Fall 2003 Conference in honor of
Georges
Gras at theUniversity
ofFranche-Comté. I am also
grateful
to the anonymous referee for a carefulreading
and comments whichimproved
the paper.2. A criterion for
having large
Galois group2.1. Newton
Polygons.
Let K be a fieldequipped
with a discrete val- uation v and acorresponding completion Kv.
We assume v isnormalized,
i.e.
v(K*) = Z,
andemploy
the same letter v to denote an extension ofthis valuation to an
algebraic
closureKv
ofKv.
For a
polynomial f (x)
= with0,
the v-adic Newton
Polygon
off (~),
denoted is defined to be thelower convex hull of the set of
points
It is the
highest polygonal
linepassing
on or below thepoints
inSv( f ).
The
points
where theslope
of the Newtonpolygon changes (including
therightmost
and leftmostpoints)
are called the corners of their x-coordinates are the breaks of
NPv( f ).
For the convenience of the
reader,
we recall the main theorem about v-adic Newton
Polygons;
for aproof see,
forinstance,
Gouvêa[8].
A very nicesurvey of the uses of the Newton
Polygon
forproving irreducibility
is Mott[19].
Forgeneralizations
to severalvariables,
see Gao[7]
and references therein. For a recent paper onusing
NewtonPolygons
tocompute
Galoisgroups, see Kolle-Schmid
[15].
Theorem 2.1
(Main
Theorem of NewtonPolygons). Suppose f (x)
EK[x]
is not divisible
by
x. Let(xo, yo), (xi, yi),
... ,(xr, Yr)
denote the successive verticesof NPv(f),
withslopes
mi =(yi - Yi-i)l(xi - xi-1).
Then thereexist
polynomials fi, ... , f r
inKv ~x~
such that2.2. Newton Index. We now suppose that K is a fixed
global field,
i.e.K is a finite extension
of Q (number
fieldcase)
or of where F is afinite field
(function
fieldcase).
Aglobal
field Kenjoys
theproperty
that for agiven
element a EK, v(a)
= 0 for all butfinitely
many valuations v of K.Definition. Given
f
EK[x]
and À a linearpolynomial
inK[x],
we definethe v-Newton Index of
f
at~1, ~ ( f , A),
to be the least commonmultiple
ofthe denominators
(in
lowestterms)
of allslopes
The NewtonIndex of
f
overK,
is defined to be the least commonmultiple
of all~ ( f , À)
as v runs over the non-archimedeanplaces
of K and A runs overthe linear
polynomials
inK[x].
Some comments on this definition are in order. It is clear
that,
for eachv and
À, ~ ( f , a)
is a divisor oflcm(1 , 2, ... ,
where n is thedegree
off ,
hence the same is true forThus,
the Newton index off
over Kis a well-defined divisor of n! . It is not clear from its
definition, however,
that it is
effectively computable;
we make some comments on this in thefollowing
Lemma. For theapplications
we have inmind,
it will sufhce to look for asingle
suitablepair v, A
such that~ ( f , A)
is divisibleby
aprime
p E
(n/2, n - 2),
but we have formalized the concept of Newton Index in order to demonstrate thepossibilites
and limitations of thisapproach
forfuture
applications.
Lemma 2.1. Let
K f
be asplitting field for f
overK,
and letR f
be the(finite)
setof
non-archimedeanplaces of
K thatramify
inKf.
ThenProo f .
Let us first note that for any linearK[x]-polynomial À, K f
is asplitting
field forf
o A.Now,
if aninteger
d > 1 divides the denominator of someslope
ofNP, (f
oÀ),
then v is ramified in(see
theproof
of Theorem
2.2).
This shows thatJV’(f)
is the least commonmultiple
ofA)
as v runs overR f
and À runs over linearK[x]-polynomials.
It remains
only
to show that for each v, we can restrict this to the leastcommon
multiple
over monicKv ~x~-polynomials
À. To seethis,
supposeA(x)
= ax + b andAI (x)
= x + b(with a
E.K*, b
EK)
so that(foÀ)(x) = ( f o ÀI)(ax).
Let h= f
oÀI
and g= f
o À. Note that ifh(x) = ¿J=ü
and
g(x)
=h(ax) = rn bjxj, then the set of slopes
of NPv(g)
is simply
a shift of that of
NPv (h) by v (a),
because for all 0j
k n,Since
v(K*) = Z,
it follows that theslopes
ofNPv (h)
andNPv (g)
havethe same denominators.
Thus, ~Y ( f , a) _ giving
the desiredconclusion. D
According
to thepreceding Lemma,
we would have an effectiveprocedure
for
calculating
if we could solve thefollowing problem.
Problem. Give an
algorithm
forcalculating
We now formulate a criterion for an irreducible
polynomial
to have"large"
Galois group. Thekey
idea appears in Coleman’scomputation
[2]
of the Galois group of the nthTaylor polynomial
of theexponential function,
whichincidentally
is the GLP 1 -n)(x).
Theorem 2.2.
Suppose
K is aglobal field
andf (x)
is an irreduciblepoly-
nomial in
K[x].
Then~f
divides the orderof
the Galois groupof f
overK.
Moreover, if ,4"f
has aprime
divisor Ê in the rangen/2
Ê n -2,
where n is the
degree of f,
then the Galois groupof f
containsAn,
in whichcase, this Galois group is
An if disc( f )
is a square in K* andSn
otherwise.Proo f . Suppose v
is a valuation ofK, ~
is a linearK[x]-polynomial, and q
is an
arbitrary
divisor of the denominator of someslope
s ofNPv (g)
whereg =
f
o À.Clearly, g
is irreducible and has the same Galois group asf .
It sufhces to show
that q
divides the order of the Galois groupof g
overK.
By
Theorem2.1,
there exists a root a EKv of g
with valuation -s.Since q divides the denominator of s, q divides the ramification index e of
Kv(a)/ Kv ([8], Proposition 5.4.2).
But e divides thedegree ~Kv (a) : Kv],
which in turn divides the order of the Galois group
of g
overKv,
hencealso over K.
If q
= Ê is aprime
in the interval(n/2, n - 2),
then theGalois group
of g
contains an£-cycle,
so it must containAn by
a theoremof Jordan
[14] (or
see, forinstance,
Hall’s book[13,
Thm 5.6.2 and5.7.2]).
It is well-known that the Galois group over K of an irreducible
polynomial
of
degree n
is contained inAn
if andonly
if its discriminant is a square inK* . 0
Remark. Schur
proved
a similar result([21, §1, II]), namely,
if the dis- criminant of a number field K ofdegree n
is divisibleby pn,
then the Galois closure L of K hasdegree [L : Q]
divisibleby
p. Ingeneral,
if p dividesthe discriminant of an irreducible
polynomial f ,
it is not easy to determine thep-valuation
of the discriminant of the stem field~ ~x~ / ( f ); thus,
each ofTheorem 2.2 and Schur’s criterion can be useful
depending
on whether wehave information about the discriminant of the field or that of the defin-
ing polynomial.
Neither criterion is useful when the discriminant off
issquare-free,
forexample,
since in that case, all the non-trivial ramification indices are 2. On the otherhand,
over base fieldQ,
irreduciblepolynomials
with
square-free
discriminant also have Galois groupSn
see e.g. Kondo[16];
theproof
of this fact uses thetriviality
of the fundamental group ofQ.
3. Proof of Theorem 1.1
We now let K =
Q.
For aprime
p, we writeNPp
inplace
ofNPv
wherev =
ordp
is thep-adic
valuation ofQ.
Lemma 3.1.
Suppose
p is aprime
in the interval(n/2,
n -2),
andf (~) _
¿j=o
E is an irreduciblepolynorrzial of degree
n withp-integral coefficients,
i.e. 0for j
=0, ... ,
n.Suppose further
that
Then,
p divides the orderof
the Galois groupof f
over~. Indeed,
thisGalois
group is An if disc( f )
E~*2
anâSn
otherwise.Proo f .
It is easy to check that( )
i is divisibleby
p if andonly
ifn - p + 1
-j p - 1.
Thegiven assumptions
thenguarantee
that(o, ordp (co ) )
and(p, o)
are the first two corners ofNPp( f ). Therefore, -ordp(co)/p
is aslope
of
NPp(f).
It follows fromi)
that and we are doneby
Theorem 2.2.D Remark. It is easy to see that the Lemma holds for p E
(1
+n/2, n - 2)
if we
replace i)
withi’)
1ordp(co) p~(2p - n -1).
We are now
ready
to prove the Main Theorem.Proof of
Theorem 1.1. We write a =A/ti
in lowestterms,
i.e.with IL 2:
1and = 1.
By assumption,
a is not anegative integer.
We willwork with the normalized
(monic, integral) polynomial
We wish to
apply
Lemma 3.1 toit,
so we letand seek an
appropriate prime
p, i.e. onesatisfying
the conditions of the Lemma.By
asuitably strong
form of Dirichlet’s theorem onprimes
in arithmeticprogressions,
there exists an effective constantD (IL)
such that if x >D(p)
and h >
x),
the interval~~ - h, x]
contains aprime
in the congru-ence class À
mod fi (see
Filaseta-Lam[4, p. 179]). Taking x
= n - 3 >we find that for some
integer Ê
E[1, n],
p= ILl!
+ À is aprime satisfying
as
long
aswhich
clearly
holds for all nlarge enough
withrespect
toÀ,
IL.We now fix a
prime
p =satisfying (3.2).
For such aprime
p, let uscheck the
hypotheses
of Lemma 3.1. We have(ny + ~c + ~) / (~c + 1 ) > n/2
if and
only
ifSince a is not a
negative integer, if M
=1,
then À ~0,
so(3.3)
holds forall r~. If ~c >
l,
wesimply
need to take n >-2(y
+À) /(p - 1)
in order toachieve
n/2
p n - 2. Our cj areintegral
so ourpolynomial
isp-integral
for every
prime
p. Before we discuss thep-adic
valuations of the coefficients cj, let us note that in the congruence class À mod ~c, the smallestmultiple of p larger
than p is(~+ 1)p, and, similarly,
thelargest multiple of p
in thiscongruence class which is less than p is
(-IL + 1)p.
Now we claimthat,
forn
large enough,
we haveas well as
Indeed,
if IL =1,
then(3.4)
holds for all n, while2,
n > -2~implies (3.4);
moreover,(3.5)
is a direct consequence of(3.2).
From(3.4), (3.5)
and(3.1),
we then read off thatordp(cj)
= 1 for0 j ~ -1,
and0
for f ::; j :::;
n. Oneeasily
checks that(3.4)
and(3.5) give exactly
p > ~ 2013 1 and n - p~,
i.e. conditionsii)
andiii)
of Lemma3.1 hold.
By
Filaseta-Lam[4],
there is aneffectively computable
constantN(a)
such thatf (x)
is irreducible for n >N(a). Thus,
all the conditions of Lemma 3.1hold,
and theproof
of the theorem iscomplete.
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Farshid HAJIR
Department of Mathematics & Statistics University of Massachusetts
Amherst, MA 01003-9318 USA haj iremath . umass . edu