Fundamental units in
afamily of cubic fields
par VEIKKO ENNOLA
RÉSUMÉ. Soit O l’ordre maximal du corps
cubique engendré
parune racine 03B5 de
l’equation
x3 + 0,> 3. Nous prouvons que 03B5, 03B5-1 forment un
système
fondamental d’unités dansO,
si[O : Z [03B5]] ~
ABSTRACT. Let O be the maximal order of the cubic field gen- erated
by
a zero 03B5 of x3 + ~ > 3.We prove
that 03B5, 03B5 - 1
is a fundamental pair of units forO,
if[O : Z[03B5]]
1. Introduction
Many computational
methods in numbertheory depend
on the knowl-edge
of the unit group of an order in analgebraic
number field.Especially,
several
parametrized
families of cubic orders with agiven
fundamentalpair
of units are known
(see,
e.g.,[3]
and papers citedthere). However,
itseems that the results
mostly
suffer from theincompleteness
that either it is not known whether the units also form a fundamentalpair
of unitsfor the maximal order of the field
(cf.
thecorrigendum
to[4]),
or this isachieved
by imposing
a further restrictive condition.E.g.,
in the case of anon-Galois cubic field this means in
practice
that the discriminant of thedefining polynomial
is assumed to besquare-free.
In an earlier paper
[2]
wegathered together
basic arithmetic facts and further results andconjectures
about the two families of cubic fields con-taining exceptional units,
the mainemphasis laying
on the non-abelianfamily.
This is the set .~ _ whereFor
each f,
we fix 6 somehow among itsconjugates
in order toget
aunique
field
Ft.
Here .~ > 3 is a natural limitation to avoidduplication
and toexclude from the
family
three fields one of which iscyclic
and the othertwo are not
totally
real.E. Thomas
[5] proved
thaté, é - 1
is a fundamentalpair
of units for theorder
7L~~~,
and in[2, Conjecture 4.1]
weconjectured
that the same is trueManuscrit reçu le 20 juin 2003.
for the maximal order 0 =
OFt. Using
the Voronoialgorithm
in the caseswhen the discriminant D =
D(ft)
of thepolynomial flex)
is divisibleby
the square of a
prime p ~ 7,
we were able to show that theconjecture
istrue for 3 P 500.
The main
problem
in this context is whether a unit of the formis a non-trivial
pth
power inFl.
For(a, b) = (2,1)
or(1, 2)
we showed in[2]
that it is not so if p =
5,
and claimed that the same holds for anyprime
p.We have been able to prove this
by
means of a very tediouscomputation,
thedetails of which are
uninteresting.
Assuggested
in[2],
the crucialquestion
here is a successful choice of the
approximation polynomial
q. Thefollowing
construction seems to work in all cases:
Suppose
that7f
= whereyy E
Fe,
p is an oddprime,
and a and b arecoprime positive integers
lessthan p. Put 6 =
1,
if a is even and b isodd,
and 5 = -1 otherwise. Takewhere Tr denotes the trace from
Fi
toQ.
This choice is different from theone in
[2],
but so far it has worked well in each caseinvestigated.
Let j
denote the index[0 : 7G(e~~.
Our purpose is to prove thefollowing
result which shows that the
conjecture
is trueif j
is not toolarge:
Theorem.
If j
=~O : 7G(e~~ ~/3,
then e, e - 1 is afundamental pair of
units
for
the maxirrtal order 0of
thefield Fi.
Using Maple
we havecomputed
theprime
factorization of D and have verifiedthat,
for 3 $10000,
D has asquared
factork2 with k
>t/3 only
in a few cases, and that in these cases k - 0 mod7, D ~
0 mod73, so
that
(see
Lemma1) j
is a divisor ofk/7.
One can then checkthat j ~/3.
Therefore,
e, e - 1 is a fundamentalpair
of units for C7 if 3 ~ 10000.2. Basic lemmas
These lemmas are contained in
[2],
but in order to make theproof
of thetheorem
self-contained,
werepeat
theirproofs shortly
here. Notethat j2
isa divisor of
D,
infact, D/j2
is the discriminant of the fieldFt. By pm 11
cwe mean that
p~"’~ ~ c, t
c.Lemma 1.
(i) If £ =j.
2 mod7,
then 7f
D.(ii) If t
2 mod7,
23 mod49,
then72 II D, 7 ( j.
(iii)
mod49,
then73 II D, 71 II j.
Proof.
Thepolynomial
discriminant D has theexpression
Note that every
prime
divisor of D is mod 8.It is easy to see that
(i) holds,
e.g.,by
directcomputation. Suppose therefore,
that - 2 mod 7.Substituting i
= 2 + 7n in(1)
weget
Hence
73 II
Donly
for n m 3 mod7,
== 23 mod49,
and otherwise7 211 D.
.For
t 0 23
mod49,
is an Eisenstein
polynomial
modulo7,
so that 7 isfully
ramified inF~,
and(ii)
follows.For f == 23 mod
49, (iii)
is a consequence of(6 - 2)2/7
E 0. This factcan be seen in many ways, the most
straightforward
butperhaps
not thecleverest method
being
tocompute
the minimalpolynomial.
0Lemma 2. The
ring
C~ has a Z-basisof
thef orm 1,
é, a, where a =(u
+VE +
e2) / j,
and theirctegers
u, v are determinedby
0 u, vj
andProo f .
We shall use a theorem of Voronoi[1,
p.11 l,
Theorem1].
Put a =2(~
+ t+ 1),
b= ~2 - ~ -
9. Since the resultant of a and b withrespect
to .~equals
336 and b isodd,
thegcd (a, b)
is a divisor of 21.Firstly,
we must show that the simultaneous congruencesI"’to _ -11--
do not have a common solution for
any k
> 1.Suppose
thecontrary.
It follows from the identitiesthat k2 (a, b),
a contradiction.Hence C~ has an
integral
basis of therequired form,
where u and v have to be determined.Put j’ = j / (7, j ),
so that 7{ j’. By
Voronoi’s theoremwe have
where t is a solution of the congruences
Further useful identities are
It follows from
(7)
that(a, j~)
=1,
and then from theequations (3), (6)
that t =
b/a
is a solution of(5). Substituting t
=b/a
in(4)
andusing
(7)
to remove thedenominators,
we obtain after a shortcomputation
thecongruences
(2)
moduloj’.
Suppose finally
that7 j, i.e., t
- 23 mod 49. In this caseso that Voronoi’s theorem
implies u - 4, v -
3 mod 7. This is in accor-dance with
(2)
modulo 49. 03. Proof of the theorem
For the basic facts
concerning
Voronoi’salgorithm
intotally
real cubicfields,
see~1~, Chapter
IV. For any number ai EFt,
let(or ~0(’),
i =0,1, 2)
be theconjugates,
and let 19 = be thecorresponding
vectorin
Ilg3.
We choose the order of theconjugates
so thatLet A = E
C7~
be the lattice inJR3 corresponding
to 0. The theorem is an immediate consequence of thefollowing
Lemma 3.
-suppose that j t/3. Let Z
be the relative minimaof A adjacent
tof
on thepositive
x- andy-axis, respectively.
ThenProof.
Weapply
the result of Lemma 2. Since e - 1 is aunit,
it is clearthat
(e - 1)-1
is a relative minimum of A. It follows from(8)
thatso
that ~
mustsatisfy
the conditionsSince ~
EC7,
there areintegers x, y, z
such thatConsider
(10)
as asystem
of linearequations
in theunknowns x, y,
z. Thedeterminant of the
system
is-v’Ï5 / j.
Here the square root ispositive,
andto
get
the correctsign
we use(8).
It follows thatChanging
thesign of ~,
if necessary, we may assume that z > 0.From
(1)
we haveJD > P2
+ 3t -2,
so that(11), (9)
and(8) imply
Since j .~/3,
it follows that zj.
Subtract the
equations (10)
with i =1, 2. Substituting
theexpressions
of a’ and a" we obtain after a short
computation
The absolute value of the
right-hand
side is less thanWe shall show that this
expression
is less than3 j /e 1,
so that the final result will beFor that purpose it is
enough
to show thatThe
following improved
bound for e is véid: e r, whereTo see
this,
check that0, ft(r)
> 0. We then havewhich is
easily
seen to be > 2 4. For £ = 3 one cansimply compute
theapproximate
values ofE, E’, E"
and check that(14)
holds even then.We have thus
proved (13).
We contend that(j, ~ 2013 v)
= 1 whichimplies
because z j, v j ~/3,
and thepossibility z
= y = 0 is absurd.Suppose
that ~ - vand j
are both divisibleby
aprime
p. If p =7,
then Lemma 1
gives t -=
23 mod49,
but then v =- t =- 2 mod 7 leads to acontradiction with
(2).
Hencep ~
7. From(2)
we haveh(2) -
0 mod p,where
On the other hand
so that £ > 2 mod p and
h(2) _
-56 mod p, which isimpossible.
Hence
(15)
istrue, whence ~
== x-~u+(e-1)-1. Looking
at
(8),
one can see that the two last conditions(9)
areonly
satisfied forx ~- u = 0 or 1.
However,
the first condition(9)
does not hold for x + u = 1.It follows
that ~ _ (E - l)-’,
and the firstpart
of the lemma isproved.
The
proof that 77
=6-1
is very much the same. Thenumber q
satisfiesthe conditions
Again
there areintegers x, y, z
such thatWe may assume that z >
0,
and we can provethat z j
as before. Sub-tracting
theequations (17)
with i =0, 2
we obtain thefollowing analogue
of
(12):
In order to achieve the result
we have to show that the absolute value of the
right-hand
side of(18)
isless than 1. This is true if we can show that
But
(20)
followseasily
from(8).
Hence(19)
holds.Suppose
nowthat j
and t - v - 1 are both divisibleby
aprime
p.If p = 7,
then t =- 23 mod 49 and v - ~-1 - 1 mod
7,
which contradicts(2).
Thusp 0
7. Since v = ~-1 mod p, it follows from(2)
thath(~)+28
= 0 mod p.On the other
hand,
which is
impossible.
Asbefore,
we now haveso that
r~ = x + u + ~ + e-1. However,
the first and third condition(16)
areonly
satisfied for x + u-~ ~
= 0. Thiscompletes
theproof.
DReferences
[1] B. N. DELONE, D. K. FADDEEV, The Theory of Irrationalities of the Third Degree. Trudy Mat. Inst. Steklov, vol. 11 (1940); English transl., Transl. Math. Monographs, vol. 10, Amer. Math. Soc., Providence, R. I., Second printing 1978.
[2] V. ENNOLA, Cubic number fields with exceptional units. Computational Number Theory (A.
Pethö et al., eds.), de Gruyter, Berlin, 1991, pp. 103-128.
[3] H. G. GRUNDMAN, Systems of fundamental units in cubic orders. J. Number Theory 50 (1995), 119-127.
[4] M. MIGNOTTE, N. TZANAKIS, On a family of cubics. J. Number Theory 39 (1991), 41-49, Corrigendum and addendum, 41 (1992), 128.
[5] E. THOMAS, Fundamental units for orders in certain cubic number fields. J. Reine Angew.
Math. 310 (1979), 33-55.
Veikko ENNOLA
Department of Mathematics University of Turku FIN-20014, Finland E-mail : ennolaCutu.ii