J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
P. L
LORENTE
E. N
ART
N. V
ILA
Decomposition of primes in number fields
defined by trinomials
Journal de Théorie des Nombres de Bordeaux, tome
3, n
o1 (1991),
p. 27-41
<http://www.numdam.org/item?id=JTNB_1991__3_1_27_0>
© Université Bordeaux 1, 1991, tous droits réservés.
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Decomposition
of
primes
in number fields
defined
by
trinomials.
par P.
LLORENTE,
E. NART AND N. VILAAbstract 2014 In this paper we deal with the problem of finding the prime-ideal decomposition of a prime integer in a number field K defined by an
irreducible trinomial of the type Xpm + AX + B ~
Z[X],
in terms of A and B. We also compute effectively the discriminant of K.1. Introduction
Let K be the number field defined
by
an irreducible trinomial of thetype :
-In this paper we
study
theprime-ideal decomposition
of the rationalprimes
in K. Our results extend those of Velez in[6],
where he deals with thedecomposition of p
in the case A = 0.However,
the methods aredifferent,
oursbeing
based on Newton’spolygon techniques.
The results areessentially complete
except
for a fewspecial
cases which can be handledby
anspecific
treatment(see
section2.3).
This is doneexplicitely
for p"’
= 4 or5,
so that there are noexceptions
at all forquartic
andquintic
trinomials.Let us remark that the main aim of the paper is to
give
acomplete
answer in the case/B (Theorems
3 and4).
The resultsconcerning
the other cases are
easily
obtainedapplying
the ideas of[2],
where we dealtwith the
computation
of the discriminant ofIi ,
whereas the casepiA, p ÁB
was not even considered. Wegive
also thep-valuation
of the discriminantof K in all cases
including
those not coveredby
(2~.
Mots clefs: Decomposition of primes, Discriminant, Trinomials.. Manuscrit reçu Ie 20 juillet 1990 .
2. Results
Let K =
Q(O),
where 0 is a root of an irreduciblepolynomial
of thetype :
where
n, A, B
EZ,n
> 3. For the case n = 3 see~1~ .
Let us denoteby
dand
the
respective
discriminants of 7~ and 0. Forsimplicity
we shall write inthe
sequel
N for the ideal normFor any
prime q e Z
andinteger u E
Z(or
q-adic integer
u EZq)
weshall denote
by
thegreatest exponent s
such that and we shallwrite u := .
It is well-known that we can assume that the conditions :
are not satisfied
simultaneously
for anyprime
integer
q. We shall make thisassumption
throughout
the paper.Let
F(X)
EZ[X]
be apolynomial,
q E Z aprime integer
and letbe the
decomposition
ofF(X )
as aproduct
of irreducible factors(mod q).
An
integer
ideal a of any number field L will be called"q analogous
to thepolynomial
F(X)"
if thedecomposition
of a into aproduct
ofprime
idealsof L is of the
type :
2.1.
Decomposition
of theprimes
q notdividing
n.THEOREM 1. Let q be a
prime
number such that qIn.
Let us denote a =(n -
and b =(n,
v,(B)).
Thedecomposition
ofq into a
product
ofprime
ideals of li is a ~follows :If q
JAB
and thedecomposition
of f (X )
into aproduct
of irreduciblefactors
(mod q)
is of the type :and we have
where
If q
JABD,
q isq-analogous
tof (X ).
2.2.
Decomposition
of theprimes
pdividing
nTHEOREM 2.
If p
then p isp-analogous
tof (X )
andvp(d)
= 0.From now on we assume that n =
pm,
> 3 for someprime
p e Z andTHEOREM 3.
Suppose that p
>2,
piA
and p
Let us denote :Then we h ave :
where
Moreover I = J in cases
(2.2.2)
and(2.2.4),
1 = J + 1 in case(2.2.3)
andI = J +
[(so - rn.)/2~
+ 1 in the rest of the cases.THEOREM 4.
Suppose
that 2~A,
2~’B
and let ro, ri , r, so, e, ek(1 k
where
Moreover I = J in cases
(2.2.8),
1 = J + 1 in cases(2.2.9),
1 = J + u - 1in cases
(2.2.10)
and I = J + u in cases(2.2.11).
2.3.
Quartic
andquintic
trinomialsIn this section we
complete
thegeneral
theorems above in the cases n = 4and 5. Let n =
p"‘.
Theorems2,
3 and 4give
thedecomposition
of p
in allcases
except
for thefollowing :
For the
primes
p theonly
case not coveredby
Theorem 1 is :Equations satisfying
(2.3.1)
or(2.3.2)
can be handledby
anspecific
treat-ment but the results are toodisperse
to fit them into a reasonable theorem.For
instance,
for n =4,
(2.3.2)
is notpossible
and(2.3.1)
occursonly
for p = 2 andequations :
For n =
5,
(2.3.1)
is notpossible
and(2.3.2)
occursonly
for q = 2 andequations :
THEOREM 5. The
decomposition
of 2 in the number field definedby
(2.3.3)
or
(2.3.4)
is- . I- 1
where a is an
integer
idealhaving
thefollowing decomposition :
For e > 2 and B -
1(mod 4) :
lvhereas for e > 2 and B -
3(mod 4) :
In all cases
N(p)
= = 2 andN(~2 )
= 4.Moreover,
v2 (d)
= 4 whene = 0 and in the cases
(2.3.5), (2.3.6)
andv2(d)
= 6 in the rest of the cases.3. Proofs
The
proofs
of the Theorems of Section 2 areessentially
based on an oldtechnique developed by
Oreconcerning
Newton’spolygon
of the trinomialf (X ) (cf. [3]
and[4]).
Forcommodity
of the reader we sum up the results we need of[3]
and[4]
in Theorem 6 below.We recall first some definitions about Newton’s
polygon.
LetF(X) =
X" +
a,Xn.-’
+... + an. EZ[X]
and p E ~ be aprime
number. The lower convexenvelope
r of the set ofpoints
n}(ao
=1)
inthe euclidean
2-space
determines the so-called "Newton’spolygon
ofF(X)
with respect top" .
Let~1,..., ~
be the sides of thepolygon
and.~;, hi
thelenght
of theprojections
of5’z
to the X-axis and Y-axisrespectively.
Letea = and
li
=~~..1~
for all i. IfSi
begins
at thepoint
(s,
vp(a.,»
letfor all
0 j
êi. Thepolynomial :
is called the "associated
polynomial
ofS".
We defineF(X)
to be"Si-regular"
if p does not divide the discriminant ofF(X)
will be called,r-regular"
if it isSi-regular
for all i .THEOREM 6.
(Ore [4],
Theorems 6 and8).
LetF(X)
EZ[X]
be a monicirreducible
polynomial
and let L =Q(a),
a a root ofF(X).
Let p be aprime ;
with the above notations about Newton’spolygon
r withrespecto
to p, we have thefollowing
decomposition
of p into aproduct
ofinteger
ideals of L :"
For each
i,
the ideal ax isp-analogous
toFi(Y)
ifF(X)
isSi-regular.
Moeover,
if F(X)
isr-regular
we have :where
i(a)
denotes the index of a. Thisexpression
forvp(i(a))
alsocoin-cides with the number of
points
withinteger
coordinates below thepolygon
except
for thepoints
on the X -axis and on the last ordinate.For the
proof
of theorem 1 we need a well-known lemma(cf.[5]) :
LEMMA 1. Let L be a number field of
degree
[L : Q]
= n. Let q be aprime
integer
unramified in L and let s be the number ofprime
ideals of Llying
over q.Then,
the discriminant d of L satisfiesProof of
theorerrc 1. The assertions(2.1.1)
and(2.1.2)
are astraightforward
application
of Theorem 6. For(2.1.3)
see theproof
of[2
Theorem2].
(2.1.4)
is consequence of Lemma 1 and the fact that in this case = 1if q
ramifies[2,
Theorem2]. (2.1.5)
is obvious and the assertionsconcerning
thecomputation
of are contained in[2,
Theorem1].
Theorem 2 follows from Theorem 6 and
[2,
Theorem1].
We shall deal with theproof
of Theorems 3’and 4altogether.
Theproof
of Theorem 5 is similar to those of thegeneral
theorems.Proof of
Theorem 3 and4.
SincepiA
and p/B,
we havef (X ) - (X
+B)"
(mod
p).
Let r be the Newton’spolygon
of thepolynomial :
It is easy to see that :
Let us determine first which would be the
partial
shape
of r if the twofinal
points
(n -1,r1),(n,ro)
where omitted.By
(3.2.1)
we find that inthat case r would have m - 1 sides
S’~ , ... ,
if p = 2 and one moreside
if p > 2,
each sideSk
ending
at thepoint
(ek, k) (see
figure
1).
Infact, i
= ek is thegreatest
subindex withvp(A¡) =
k and theslope
ofSk
is
1/ek
so that theseslopes
arestictly increasing.
Now,
when we considerthe two final
points
of r we find that we canalways
assure that F containsthe sides
,S~ , ... ,
if r > m, the sidesSl,...,
S,._y
if r m and p >2,
and the sides
so , ... ,
S’r-2
if r m and p = 2.Let r’
denote,
in each case, the rest of the sides of r.By
Theorem6,
the assertions(2.2.1)
and(2.2.7)
areproved.
In order to find the furtherdecomposition
of therespective
ideals a and b of Theorem 3 and 4 we shallstudy
theshape
and associatedpolynomials
of 1~’. We mustdistinguish
several cases.
Before,
note that for each 1 ~ k m, the number ofpoints
with
integer
coordinates below the sidesSi
U ... U,S’k except
for thepoints
on the X-axis and on the last ordinate is
and
Case r m, ro r, : r’ has
only
one side withlengths
of theprojections
to the axis : i = = e,
h = I if p > 2 and É = 2e , h = 2 if p = 2
(see
.
fig.
2).
Therefore - :=(~, h)
= 1 or 2according
to p > 2 or p = 2. In thelatter case the associated
polynomial
iscongruent
(mod 2)
toy2
+ Y +1,
which is irreducible.By
Theorem6,
(2.2.2)
and(2.2.8)
areproved.
SinceF(X)
isr-regular
we have :hence,
I = J in bothcases, as desired.
Case r m, ro > r, : If p >
2,
r’ has two sides,5’,
,S’’ withprojections
to the axis
.~ = e - 1, h = 1
and É’ =1, h’
= ro - rirespectively
(see
fig.
3).
If p = 2,
r’ contains the side and two more sides with the samedimensions of S and S’
above,
except
for the case ri =m, ro = m + 1
in which besides
Sm-i
there isonly
one side withprojections
to the axisi = h = 2 and associated
polynomial
congruent
(mod 2)
toY’
+ Y +1,
which is irreducible
(see
fig.
3).
By
Theorem6,
(2.2.3)
and(2.2.9)
areproved.
Since isr-regular
in any case, we have :otherwise ,
hence I = J + 1 in both cases, as desired.
Figure 3
This ends the discusion of the case r m.
Assume from now on that r = m + 1. If we
study
r’ in this case asabove,
we are led to manyp-irregular
cases. For this reason, instead of thepolynomial
f (X - B)
we seek anopportune
substituteproviding
a much moreregular
situation.Let
Q
=-nB/(n - 1)A.
Since= 0,
,C3
is ap-adic integer
and it isclear that Theorem 6 is also
applicable
to thepolynomial :
Computation
leads to :hence,
easy to check that :
hence, by
(3.2.2) :
so that so = > m.
Thus,
Newton’spolygon
r~3
ofC(X)
withrespect
to p can be alsoexpressed
as :and we need
only
tostudy
in order to find theprime-ideal
decomposition
of therespective
ideals b of Theorems 3 and 4.Again,
we have todistinguish
several cases :
Case r = m +
1, p
> 3 or p = 3 and so > m + 2 :I",
contains and one more side of dimensions~=2,~=~o2013m
(see
fig.
4).
For this latterside, - =
(é, h)
= 1 or 2according
to so - m odd or even. In the latter caseand its discriminant is
congruent
to(_l)n(n-l)/2 2Dp.
Sincevp(D) == so-m
(mod 2), (2.2.6)
isproved by
Theorem6, Moreover,
since we are in aregular
case we have : .as desired.
Figure 4
Case
r = m -f-1, p = 3
m+2 :rfi
has only
one side with é = 3 and h = 2 or 3according
to so = m + 1 or m + 2(see
fig.
4).
In the lattercase - = 3 and the associated
polynomials
isSince
(-1)~+~)/2 =
(_1)"~-~
in this case,multiplying by
B2
weget
thepolynomial
.1~(y)
=y3 -
which is irreducible(mod
if
Da m
(-1)’"~ (mod 3).
By
Theorem6,
(2.2.5)
isproved.
Since we are in aregular
case we have :Case r =
m -f-1, p =
2 :r’
hasonly
one side with t= 2 and h = so.- m + I(see
fig.5),
hence - = 1 or 2according
to so - m + 1 odd or even, orequivalently according
tov2(D) - m
even or odd. In the latter case, theassociated
polynomial
iscongruent
(mod 2)
toy2 +
1,
hence,
it is anirregular
case. In the former case Theorem 6 proves(2.2.10)
and :as desired.
Figure 5
Finally,
in order to deal with the casev2(D) - m
odd it is necessary tochange again
Newton’spolygon.
Let 2u = so - m + 1 andwhere
Do
=D/n"
=B~’-~ -
(n -
1)"-~A2.
Since = so = 2u +and there are
exactly
two summands in(3.2.3)
with v2minimum and
equal
to 2u + m -1,
hence,
v2( f(ô)) >
2u + m. From therelation :
and
being
v2((n-1)A6+nB)
= v+m, we conclude thatv2(f’(6» =
Thus Newton’s
polygon
F5
withrespect
to p of thepolynomial
f (X
+6)
isagain expressible
as :rh
=51
U ~ ~ ~ U UF§.
"’7e have now threepossibilities
(see fig.5) :
a) V2 (f (6))
= 2u + m.IP’ 6
hasonly
one side with i =2,
h = 2u + 1hence E =
(é, h)
= 1 anda=p 2, N(p)
= 2. Moreover I = +2(m-l)+u=J+u-1.
b)
v2(f(b))
= 2u+m+l. hasonly
one side with associatedpolyno-mial
congruent
(mod
2)
to y2 + Y +1,
which isirreducible,
hencea =
p, = 4. Moreover I = +
2(m - 1)
~- ~ + 1 = J + u.c) v2( f (b))
> 2u + m + 1.IF’ 6
has two sides and a =p-pl, N(p) _
=
2,
1 = J + u like in caseb).
Taking
congruence(mod
22"’+"’.+2~
of(3.2.3)
we shall be able to decide in which case falls ourpolynomial.
All summands of(3.2.3)
vanish(mod
~.+~+2~
except for thefollowing :
Dividing by
22"~+"’+~ andtaking
congruence(mod
8)
we obtain :From
(3.2.2)
weget
, hence(3.2.4)
isequal
to :which is
equal
to -1 -D2
(mod 8)
if u > 1 and to 2"’ -~ 1 -D2
if u = 1. Therefore cases
a) b)
andc)
areequivalent
to thefollowing
This ends the
proof
of(2.2.10)
and(2.2.11).
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P. Llorente - E. Nart, Effective determination of the decomposition of the rationalprimes in a cubic field, Proc. Amer. Math. Soc. 87
(1983),
579-585.[2]
P. Llorente - E. Nart - N. Vila, Discriminants of number fields defined by trinomials, Acta Arith. 43(1984),
367-373.[3]
Ö. Ore,
Zur Theorie der algebraischen Körper, Acta Math. 44(1923),
219-314.[4]
Ö. Ore, Newtonsche Polygone in der Theorie des algebraischen Körper, math. Ann. 99(1928),
84-117.[5]
R. G. Swan, Factorization of polynomials over finite fields, Pacific J. Math. 12(1962),
1099-1106.[6] W. Y. Vélez, The factorization of p in
Q(a1/pk )
and the genus field ofQ(a1/n),
Tokyo J. Math. 11
(1988),
1-19.Dept. Matematiques
Univ. Autonoma de Barcelona
08193 Bellaterra, Barcelona
Spain.
Dept.
Algebra
i Geometria Facultat de Matematiques Univ. de BarcelonaGran Via, 585 08007 Barcelona