195-
On the existence of Minkowski units in totally
real cyclic fields
par FRANTI0160EK MARKO
RÉSUMÉ. Soit K un corps de nombres
cyclique
réel dedegré
n qui est leproduit
de deux nombres premiers distincts et tel que le nombre de classes du n-ième corpscyclotomique
soitégal
à 1.Nous établissons certaines conditions nécessaires et suffisantes pour l’existence d’une unité de Minkowski pour K.
ABSTRACT. Let K be a
totally
realcyclic
number field ofdegree
n that is the
product
of two distinct primes and such that the class number of the n-thcyclotomic
fieldequals
1. We derivecertain necessary and sufficient conditions for the existence of a
Minkowski unit for K.
1. Introduction
The main result
(Theorem 11.1)
of[1]
states a necessary and sufficientcondition for the existence of the Minkowski unit in the
totally
realcyclic
fields of
degree pn.
These conditions are of an inductive nature asthey
relate the existence of the Minkowski unit in such a field
Kn
to the ex-istence of the Minkowski unit in the subfield Additional necessary conditions are the
isomorphism
of thefactor-group
of unitsand the group of
integers
of thep~‘-th cyclotomic
fieldtogether
withthe
surjectivity
of the norm map fromKn
to on units modulo~~1}.
For the sufficient conditions it is necessary to know the Minkowski unit of and the
generator
of overZ[(pn].
Therelationship
be-tween these
generators
is then describedusing
a coordinate element fromZ[(p.].
If the coordinate element can berepresented by
a unit of a certainring
and thepreviously
mentioned norm map on units issurjective,
theexistence of the Minkowski unit follows. This method works best when all the invertible classes can be covered
by
units as is the case for thedegrees pn
=4, 8
and 9. For thesedegrees
the existence of the Minkowski unit isequivalent
to thesurjectivity
of the norm map onunits,
see~1~,
Theorem11.2.
The
general
idea of localgenerators
and coordinates similar to[1]
extendsto
totally
realcyclic
number fields ofdegree n
for which thering 7G~~n~
ofthe
integers
of the n-thcyclotomic
field is aunique
factorisation domain.We treat in detail the case when n is a
product
of two distinctprimes.
Among
other results we prove thefollowing
statement.Theorem 1.1. Let K be a
totally
realcyclic
numberfield of degree
n =6,10
or 14. Then K contains a Minkowski unit
if
andonly if
the norm mapsfrom
K to itssubfields
aresurjective
on units.For
precise
formulation seeCorollary
5.3. This theorem is ageneraliza-
tion of Theorem 3 of
[3].
2. Notation
Throughout
the paper thesymbols
p, q will denote distinctprime
num-bers,
n apositive composite integer,
K= Kn
atotally
realcyclic
numberfield of
degree
n over the fieldQ
of rational numbers. Denoteby
a agenerator
of the Galois group G of K overQ, by
a subfield of K ofdegree
m >1 over Q
that is fixedby
thesubgroup
of Ggenerated by
0".The letters m,
l,
d will be reserved for divisors of n that arelarger
than1,
hencethey correspond
to subfields of K. LetEK~
be the group of allunits of the field and
UKm
= Further denoteby
the m-th
cyclotomic polynomial,
let(m
aprimitive
m-th root ofunity
and
put
and
- ~ ~x~ / (cpm (x) ) .
Note thatUK
is notonly
a G-module but an0-module as well.
A unit u E is called a Minkowski unit of
Km
if theconjugates
of ugenerate
this isequivalent
to the Galois moduleU Km being cyclic.
For an 0-module W denote
Wm
= andW’n
= andWm
=Wm/Wm.
An O-module W will be called structured ifWm
isa nontrivial
cyclic Om-module
for eachmin. Actually Om
asO-modules in that case.
The number of
isomorphism
classes of 0-structured modules is denotedby l~l (n) .
Weconjecture
that eachisomorphism
class of structured 0- modules containsUK
for some number field K. In any case the indexM(n)
is an upper bound for the number of isoclasses of structured Galois modulesUK
for fields K ofdegree
n.The function
SZ(m)
is defined as the sum of the exponents in theprime
factorization of m, i.e. then For a
structured module W we say that its element w has
length
k and write.1"’1/ I ~ ... 1---r ~ t ~ "I . , ~ ~ i ~
3. Coordinates and Galois structure
We start with an observation. Let m = and
put
Lemma 3.1. The ideal
of Z [x] generated by polynomials Ak for
k =1,
... , sequals (W. (x)) -
Proof. Clearly ’Pm(x)
dividesAk,
so writeAk
= Thepolyno-
mials pk are all
irreducible,
and thisimplies
that thepolynomials gk
arepairwise relatively prime.
Therefore the ideal ofgenerated by
allgk ’s equals
for a certain non-zero c E7G~~~.
This shows that the ideal Igenerated by
theAk’s
isprincipal,
and isgenerated by
Henceall coefficients of
polynomials lying
in I are divisibleby
c, however thepolynomals Ak
aremonic,
and this forces c to be 1. DExistence of a Minkowski unit of K
implies
certain structuralproperties
of the groups of units of subfields of K.
Proposition
3.2.If K
has a Minkowskiunit,
thenNorm K/ Km UK
=U Krn for
eachsubfield Km of
K.Moreover,
the O-moduleisomorphic
toOm.
’
Proof.
If K has a Minkowskiunit E,
thenby [4], Prop.1.3,
theZ[G]-module
UK
isisomorphic
to0,
where thegenerator
Ecorresponds
to 1 and theconjugation by
Qcorresponds
tomultiplication by
~.Moreover,
in thiscase
according
to[4], Prop.I.4,
and itscorollary,
each subfieldK",,
has aMinkowski unit and =
UK"~
for each subfield of K. Thegroup
UKm
isisomorphic
to and thefactor-group UK/ flpl,, UK,,
pis
isomorphic
to 0 modulo the idealgenerated by
allOa (x) - Using
Lemmap
3.1,
weget
that thisfactor-group
isisomorphic
toG(cpn()).
It is animmediate consequence of
R(m)
that the ideal of 0generated by
all(~)
p
for all
plm equals (cp.",(x)~",,(x)).
Therefore for eachmin
thefactor-group UKm/ flpl m UKm
isisomorphic
toOm = G((cp",,(x)).
Dp
Corollary
3.3. The 0-modules 0 is structured.Moreover, if
K has aMinkowski
unit,
then U =UK
is structured.Proof.
SinceUKm
=NormK/KmU
=U’Øm(X)
we haveUKm
=Um
for each0 For
special
values of n the Galois module U is structured if the normmaps on units are
surjective (without
theassumption
about the existence of the Minkowskiunit).
Proposition
3.4.If Z[(n]
is aprincipal
ideal domain andNormK/KmUK
=
UKm for
eachsubfield Km of K,
thenUK
is structured.Proofs. According
to[3], UK
isisomorphic
as a G-module to an ideal M ofthe
ring
O.Assuming
that the norm map on units aresurjective (that
isUKm for
each subfieldKm
ofK)
we obtain thatUm
=UKm corresponds
to module that is annihilatedby I1zln;zfm
and
Üm
isisomorphic
to Lemma 3.1implies
thatp
0. (x) MI (om
p is a module overZ[(m].
is aprincipal
ideal_
domain,
then so isZ[(m]
and in this caseUm
is acyclic Z[(,,,]-module.
Thefact that it is nontrivial follows from the Dirichlet unit theorem. D From now on assume that an O-module W is structured.
Keep
in mindthe
prominent example
of the structuredO-module, namely
the groupUK
of the field K that is
given by Proposition
3.4.Fix and denote
by
wm an element ofWm
whose classrepresents
a gen- erator ofWm
as anOm-module.
Note that each wm can be written aswm = where vm E W . Then each w E W can be written
uniquely
as a
product
w= flmln
wheream (w)
is apolynomial
from 0 suchthat its
degree
does not exceed thedegree
of Theexponents
am(u)
are called coordinates of w.
The elements
(as
well as aregenerators
of W asan 0- module. In order to work with the above factorizations and to determine the O-structure of
W,
we will derive conditionsdescribing
whenarbritrary polynomials am (x)
from O.For each wm and each
dim
we have = Denote am,d = Then am,m = 1 and the collection of all such am,d fordim; d ~
mis called the coordinate
system
of W.Proposition
3.5. Letmln
bepolynomials
in O. Then= 1
if
andonly if for
alldin
the conditionCd :
(mod satisfied.
Proof.
Weproceed
in stepsdepending
on the upper bound for thelength
of w =
First, O(n)
if andonly
if w E wn whichis
equivalent
toan(x)
= 0(mod ’Pn(x))
and that is conditionCn.
Nextassume that
£( w)
k. We arelooking
for conditions that areequivalent
to
£(w)
k. Theassumption ~(w) 1~
means that ¡ saywith z, E
Wz.
Chooseloin
such that£(lo)
= k and considerI I I , " ,
. It is a
product
of 10 and other terms 1 forlo.
Sinceeach zl
=ytl(X)
1 for some yi E W and E for 1=1= lo,
each
ztlo
(x)belongs
toWio.
Therefore zl EWio
if andonly
if (x) EWlo
which is the same as = 0. Since the last
equality
isequivalent
toClo .
Therefore.~(w)
if andonly
if all conditionsC,
for = 1~ are satisfied. Hence u = 1 if andonly
if all conditionsCd
for
d~n
are satisfied. 0Although
the above conditionsCd
do not describeexplicitly
the Galoisaction on wm
they provide
a means ofchecking
whether thecomputations involving
Galois action ongenerators
are correct.It is
interesting
thatonly
theleading exponent
am,d in the factorization’l/;d(X)
of with respect to
generators
wl is needed in the above theorem.The
polynomials
am,d are relatedby
various congruences.They
varydepending
on thetype
ofdecomposition
of n into theproduct
ofprimes
and therefore we will
investigate
therelationship
between am,dseparately
for each such
type.
In the next sections we will be
looking
for conditions on a coordinatesystem
am,d of U that willguarantee
the existence of a Minkowski unit inl~. For that purpose we will
investigate
therelationship
between coordinatesystems
for various fields of the samedegree
n.4. The case n =
p2
The main
priority
in the case n =p2
is not to prove new results but toexplain
our method in thesimplest
nontrivial case, show a connection with the work of~1~,
and view and reprove some of their results from a differentperspective.
First note that(mod
A structured O-module T is
given by
agenerating
settpl
and thecoordinate
a 2 . If t = then t’P x - Since coordinate ap2 p. If t = then
t’°P(")
= . Sincecoordinate 2 ? then
tp
. Sincethe congruence - 1
(mod
must havea solution which is
equivalent
toapa,p being
invertible modulo(p, ~pp (x) ) . Conversely,
if apolynomial
ap2,p is invertible modulo(p,
thenwe can define an O-module T
generated by
that is structuredand ap2,p is its coordinate. Set T =
Op2 ® Op
as an abelian group andtp2
=( l, 0), tp
=(o,1 ) .
A structure on T is defined asc(x)(a(x), b(x)) _ ( f (x), czp2,pe(x)
+b(x)c(x)),
wheree(x)
andf (x)
are thequotient
and the remainderrespectively
after the division of thepolynomial a(x) c(x)
Assume now that we have two structured O-modules T and W gene- rated
by
andrespectively,
with coordinates ap2,p andbp2 ~p respectively.
We would like to determine when T and W are
isomorphic
as O-modules.As a
special
case, for number fields K and.K~
ofdegree
n =4, 9
or 25satisfying
andUK
this wouldallow us to compare Galois structures of
UK
and!7~
as well as to determinewhether K has an Minkowski unit
(if UK
isisomorphic
to the structuredmodule
0).
O-morphism,
thenL(Tp)
CW~,
and therefore = andL(tp)
=wP~~~
for somea(x)
E 0 and EOp.
Amap L defined on
generators
in the above way is amorphism
of O-modules= 1 for E
Op implies
= 1.Proposition
3.5applied
to T = 1 if andonly
ifk(x)
- 0(mod
and+pl(x) -
0(mod
That means
Use
Proposition
3.5again
for W to see thatif and
only
ifa (x) k (x)
= 0(mod
andThe first congruence is
clearly
satisfied and the second one isequivalent
toSince there are no restrictions on the values
of kI(x)
we can choosel~~ (x)
= 1and obtain
This congruence
gives
a necessary and sufficient condition for L to be anO-morphism
from T to W.Therefore the existence of an
O-morphism
L from T to W as above isgiven by
thefollowing
congruence:If this congruence is satisfied for some
a(x), c(x),
we cancompute b(x)
asrequired
to make L anO-morphism.
Assume now that L as above is an
isomorphism.
ThenL(Tp)
=Wp
andsince = the class of
c(x)
modulopp (z)
must be invertible.Since
L(T)
= W the class of modulo’Pp2(X)
must be invertible aswell.
Conversely,
we will show that if is invertible modulo andc(x)
is invertible modulocpP(x),
then L is anisomorphism.
Assumet E
Ker(L)
andk(x), 1(x)
are coordinates of t with respect to thegenerators
Then
L(t) =
1implies a(z)k(z) m
0Then
L() p2
’ ’’ ’ - 1implies
0(mod
and since the class ofa(x)
isinvertible,
we have = 0.Therefore
L(t) muyx)~(~) -
1 and since the class of is invertible wealso have
l (x)
= 0 and thus t = 1. We have established that such map L isinjective.
Ifc(x)
is invertible modulo thenclearly L(Tp)
=Wp.
If also the class of
a(x)
modulo isinvertible,
then L also induces amap onto W =
W /Wp
and thus L issurjective.
Hence we determined the conditions for an
isomorphism
between struc-tured modules.
Proposition
4.1. Let T and W be structured 0-modulesfor
n =p2
withcoordinates ap2,p and
&p2 p respectively.
Then W as O-modulesif
andonly if
there isa(x)
E 0 that is invertible modulo(x),
EOp
that isinvertible modulo and
When
comparing
this result with[1],
Theorem11.1,
it is worthnoting
that the congruence classes there appear as a consequence of a choice of
generators
whereas congruence classes here appear because of therequire-
ment for
morphism.
The above criteria further
simplifies
for n =p2
=4, 9
and 25. This isthe case when the class number of
Z[(n]
andZ[(p] equals
1 and hence allunits in these
rings
are ~cyclotomic
units.If a
polynomial
isrepresented by
acyclotomic
unit for(a, p)
= 1modulo
’Pp2(X),
then its class modulo is alsorepresented by
a unit.Thus if
a (x)
isrepresented by
a unit modulo’Pp2(X)
thena(x)
is also repre-sented
by
a unit modulocpP(x).
Therefore the condition ofProposition
4.1is satisfied if and
only
if the invertible class ofbp2 Pa-2l
, P , modulo(p,
is
represented by
a unit modulocpp(x).
This also means that if an isomor-phism
of T and Wexists,
then there is one with = 1. We have thusproved
thefollowing
statement.Proposition
4.2.If n
=p2
=4, 9
or25,
then structured modules T and Ware
isomorphic
as O-modulesif
andonly if
the invertible classof bp2 Pa-l
modulo
represented by
a unit moduloCorollary
4.3.If K
is atotally
realcyclic
nurraberfield of degree
n =p2
=4 or
9,
then K contains a Minkowski unitif
andonly if Norm K IK p UK
=UK,
Corollary
4.4.If
K is atotally
realcyclic
numbersfield of degree n
= 25such that
NormK/KsUK
=UKs,
then K contains a Minkowski unitif
andonly if the
classof its
coordinateb25,5
modulo(5,
isrepresented by
aunit modulo
cp5(~).
Proof. Assumption
=UKP implies
that T =UK
is struc-tured. Put W = 0 in the above
proposition.
Since the coordinatebp2,p
= 1for
0,
the modulesUK
and 0 areisomorphic (meaning
a Minkowski unitexists for
K)
if andonly
if modulo(p,
has arepresentative
thatis a unit modulo It can be checked
easily
that every for n =4, 9,
every invertible class modulo
(p,
satisfies this condition. 0Corollary
4.5.M(25)
= 5.Proof.
It can beeasily
verified that the condition inCorollary
4.4 is trueonly
for 20 out of 100 invertible classes. 0 Therefore there are at most 5nonisomorphic
Galois structures ofUK
forfields K of
degree n
= 25 with =5. The case n = pq
We will
proceed
in the same way as outlined in section 4. First note that~~PP~x)~ ~Pa~x)) =
= q(mod
andOq(X)
(mod
A structured O-module T is
given by
agenerating
setf tP9, tp, tql
andthe coordinate
system
If t = then _the coordinate
system fapq,p, apq,ql.
If t = q p q , thentOp (x)
and Since
TWp(x) = Z’p
andp - q . - p
the congruences
apq,PaPq(t)
+qaP(t) -
1(mod pp(z))
andapq,qapq(t)
+pa9(t) -
1(mod
must have solutions which isequiv-
alent to apq,p
being
invertible modulo(q,
and apq,qbeing
invertiblemodulo
(p, cpp(x)).
Conversely, given
apair
such that thepolynomial
apq,p is invertible modulo(q, cpP(x))
and thepolynomial
apq,q is invertible modulo(p, cpP(x)),
we can define an O-moduleT generated by {tPq, ip,
thatis structured and
{aPq,P, aPq,q}
is its coordinatesystem.
SetT
=Op s3 Oq
as an abelian group andipq
=(1, 0, 0), ip
=(0, 1, 0), tq
=(0, 0, 1).
Let
7p(x)
and7q(x)
bepolynomials
such that +1. A
Z[x]-module
structure on T is defined asd(x)(a(x), b(x), c(x)) _ (f (x),
+b(x)d(x),
+c(x)d(x)),
wheree(x)
andare the
quotient
and the remainderrespectively
after the division of thepolynomial by
Assume now that we have two structured O-modules T and W
generated by tp, tq}
andfwpq,
wp,Wql respectively
with the coordinatesystems
{apq,P, apq,ql
and{bP9,P, bpq,ql respectively.
If L : T ~ W is an
O-morphism,
thenL(Tp)
CWp
andL(Tq)
CTXT
and thereforea(x) b(x) c(x) L(tP)
= andL(tq) _ Wq
and thereforeL(tpq)
== Wpq Wp qtp
== WP an Lq)
==for some
a(x)
E0, b(x), d(x)
EOp
andc(x), e(x)
EOq.
A mapL defined on generators in the above way is a
morphism
of O-modules if = 1 fork (x)
E0,
EOp
andm (x)
EOq implies
1.
Proposition
3.5applied
to T showsk(x)t1(x)tm (X)
= 1 if andonly
ifIc(x) -
0(mod apq,pk(x)
+~(~c) = 0 (mod
and
That means
and
. ...’....
Use
Proposition
3.5again
for W to see thatand
The first congruence is
clearly
satisfied and the later ones areequivalent
toand
Since there are no restrictions on the values
of k~ (x)
we can choosel~~ (x)
= 1 toverify
that bothand
must have solutions. These congruences
provide
a necessary and sufficient condition for L to be anO-morphism
from T to W.If the
following system
of congruencesL- 3 7 1 ‘ , ,,, L , , , , - - , , , ,
has a
solution,
then we cancompute b(x), c(x)
asrequired
to make L an0-morphism.
Assume now that L as above is an
isomorphism.
ThenL(Tp)
=Wp
andL(Tq)
=Wq
and sinceiuPO)L(x)
and = theclasses of
d(x)
modulo ande(x)
modulocp9(x)
must be invertible.Since
L(T)
= W the class of modulocpPq(x)
must be invertible as well.Conversely,
we will show that ifa(x)
is invertible modulod(x)
isinvertible modulo
Wp(x)
ande(x)
is invertible modulocpQ(x),
then L is anisomorphism.
Assume t EKer(L)
andk(x), l(x), m (x)
are coordinates of t withrespect
to thegenerators (tpq, tp, tq I.
Thenimplies a(x)k(x) -
0(mod cppQ(x))
and since the class of is inver-tible,
we have = 0. ThereforeL(t)
= Then= 1
implies
= 1 and = 1. Since theclasses of
d(x)
and are invertible we must have = 0 andm(.r)
= 0hence t = 1. We have established that such map L is
injective.
Ifd( x)
isinvertible modulo
’Pp(x)
ande(x)
is invertible modulo thenclearly
L(Tp)
=Wp
andL(Tq)
=Wq.
If also the class of modulo isinvertible,
then L also induces a map onto W = and hence L issurjective.
Hence we determined the conditions for an
isomorphism
between struc-tured modules.
Proposition
5.1. Let T and W be structured O-modulesfor n
= pq with coordinatesystems
andrespectively.
Then T =W as O-modules
if
andonly if
there there isa(x)
e 0 that is invertible modulo eOp
that is invertible modulo EOq
thatis invertible modulo and
The above criteria further
simplifies
when the class number ofZ[(pq]
equals
1 and hence all units inZ[(pq], Z[(p]
andZ[(q] cyclotomic
units. This is the case if and
only
if n =6,10,14,15, 21, 22, 26, 33, 34, 35
or38.
We will look first at the case when n
== 2p,
that is n =6, 10, 14, 22, 26, 34,
38.
If a(x)
is invertible modulop2p(z)
then there ispolynomial a*(x)
suchthat
a(x)a*(x) ==
1(mod ’P2p(X)).
Since =cpP(-x) it
isequivalent
to
a( -x)a*( -x) ==
1(mod cpP(x))
that isa(-x)
is invertible modulo’Pp(x).
But
a(-~) (mod 2)
shows that classes modulo(2, cpp(x))
repre-sented
by polynomials
that are invertible modulo coincide with those classes modulo(2, Wp (x)) represented by polynomials
invertible modulo’Pp(x).
We can thereforereplace
the first congruence ofProposition
5.1by
congruence= d~ (x) (mod (2,
for apolynomial d (x)
invertible modulo
’Pp(x).
How about congruence classes of
a(x)
modulo(p, x+1)
fora(x)
invertiblemodulo
cp2~,(x)?
For p =3, 5,11, 13
and19,
thepolynomial
1-x is invertiblemodulo and 1 - x == 2
(mod x
+1),
and since 2 is aprimitive
root modulo p, every nonzero class modulo
(p, x
+1)
isrepresented by
apolynomial
invertible modulo~2p(~)’
For theremaining
cases p = 7 and17,
thepolynomial 1- ~ + x2
is invertible moduloW2p (x)
and1- ~ -f- ~2 = 3 (mod x + 1 )
and we use the fact that 3 is now aprimitive
root modulo p tosee that each nonzero class modulo
(p, x-f-1)
isrepresented by
apolynomial
invertible modulo We have thus established that we can
always
choose such that the second congruence is satisfied.
We have
proved
thefollowing proposition.
Proposition
5.2. Let n =6,10,14, 22, 26, 34
or 38. Then the structured modules T and W areisomorphic
asO-modules if
andonly if
the inver-tible class
of
modulo(2,
can berepresented by
apolynomial
invertible modulo
’Pp (x) .
Corollary
5.3.If K
is atotally
realcyclic number field of degree
n =2p
=6, 10
or14,
then K contains a Minkowski unitif
andonly
=
UKP
and =UK2.
Corollary
5.4.If K is
atotally
realcyclic
nurraberfield of degree
n =2p
=22, 26, 34
or 38 such thatNormK/KpUK
=UKP
andNormK/K2UK
=UK2,
then K contains a Minkowski unit
if
andonly if
the classof
its coordinateb2p,p
modulo(2,
isrepresented by
a unit modulo’Pp(x).
Proof. Assumption
of the corollariesimply
that T =UK
is structured. For W = 0 the coordinatesystem
a2p,p =1,
a2p,2 = 1 and the modulesUK
and 0 areisomorphic (meaning
Minkowski unit exists forK)
if andonly
if the invertible class
b2p,p
modulo(p, ’Pp( x))
has arepresentative
that is aunit modulo
’Pp(x).
For n =2p
=6,10
powers of acyclotomic
unit 1 + xmodulo
cpP(x)
cover all nonzero classes modulo(2, ’Pp(x)).
For n =2p
= 14the
prime
2 has index 2 modulo 7 and hence modulo factors intoa
product
of two irreduciblepolynomials
ofdegree
3. Therefore there are49 invertible classes modulo
(2,
all of which can berepresented by
classes that are
products
of powers ofcyclotomic
units 1 + ~ and 1 + ~+ ~2
modulo
’P7(X).
0In relation to
Corollary
5.4 it is natural to ask how manypossible
classesb2p,p
modulo(2, cpp(x))
arerepresented by
units modulo or betteryet
what is the index of classes covered
by
units modulo in thesubroup
of invertible classes modulo
(2, ’Pp(x))?
This index coincides withM(n)
forthese
particular
values of n. Moregenerally, according
toProposition 5.1,
the index
M(n) equals
the index of all coordinatesystems
for whichbpq,p
isrepresented by
a unit modulo andbpq,q
isrepresented by
a unit modulo in the set of all coordinatesystems
Proof.
The claim can be established for n =22, 26, 34
and 38by using Corollary
5.4 andcomputing
the number of classes modulo(2, ’Pp(x))
thatcan be covered
by
units modulo For theremaining
classes n =15,21,33
and 35 we have to useProposition
5.1.Also,
in this case we usedthe
generators
for the n-thcyclotomic
units from[2].
The
computation
of these values was doneusing
the softwarepackage
PARI with the
help
of myundergraduate
studentAnthony
Disabella. DAcknowledgement
The author would like to thank the referee for a valuable
suggestion.
References
[1] L. BOUVIER, J. PAYAN Modules sur certains anneaux de Dedekind. J. Reine Angew. Math.
274/275 (1975), 278-286.
[2] R. On bases of the Stickelberger ideal and of the group of circular units of a cyclotomic field. J. Number Theory 40 (1992), 284-316.
[3] F. MARKO On the existence of p-units and Minkowski units in totally real cyclic fields. Abh.
Math. Sem. Univ. Hamburg 66 (1996), 89-111.
[4] N. MOSER Unités et nombre de classes d ’une extension Galoisienne diédrale de Q. Abh.
Math. Sem. Univ. Hamburg 48 (1979), 54-75.
Frantisek MARKO
Pennsylvania State University
76 University Drive Hazleton, PA 18202, USA and
Mathematical Institute Slovak Academy of Sciences
Stefanikova 49
814 38 Bratislava, Slovakia E-mail : fxml3psu. edu