M
ASANARI
K
IDA
Good reduction of elliptic curves over imaginary
quadratic fields
Journal de Théorie des Nombres de Bordeaux, tome
13, n
o1 (2001),
p. 201-209
<http://www.numdam.org/item?id=JTNB_2001__13_1_201_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Good
reduction of
elliptic
curves overimaginary
quadratic
fields
par MASANARI KIDA
RÉSUMÉ. Nous montrons que l’invariant
modulaire j
d’une courbeelliptique
définie sur un corpsquadratique imaginaire
ayantpar-tout bonne réduction vérifie certaines
équations diophantiennes,
sous réserve que soient vérifiées certaines
hypothèses
relatives àl’arithmétique
du corps. En résolvantexplicitement
cesequations
dans l’anneau des entiers du corps, nous montrons que de telles
courbes n’existent pas sur certains corps
quadratiques imaginaires.
Nos résultats
généralisent
des résultats antérieurs de Setzer etStroeker.
ABSTRACT. We prove that the
j-invariant
of anelliptic
curvedefined over an imaginary
quadratic
number fieldhaving good
re-duction
everywhere
satisfies certainDiophantine
equations undersome
hypothesis
on the arithmetic of thequadratic
field.By
solv-ing
theDiophantine
equationsexplicitly
in therings
ofquadratic
integers, we show the non-existence of such
elliptic
curve forcer-tain
imaginary quadratic
fields. This extends the results due toSetzer and Stroeker.
Introduction
It is well-known that there is no
elliptic
curvehaving good
reductioneverywhere
over the fieldQ
of rational numbers. This result is furthergeneralized
to certainquadratic
fields and some other fields(see
[9]
and thereferences
there).
In this paper, we are
particularly
interested in the case ofimaginary
quadratic
fields. For this case, thefollowing
nice result due to Setzer and Stroeker is known.Theorem
(Setzer
[11,
Theorem5],
Stroeker[16, (1.9) Theorem]).
Let k bean
imaginary quadratic
field. If
the class numberof k
isprime to 6,
thenthere is no
elliptic
curvedefined
over khaving
good
reductioneverywhere.
Manuscrit reçu le 26 octobre 1999.
This research was supported in part by Grant-in-Aid for Encouragement of Young Scientists (No. 11740009), Ministry of Education, Science, Sports and Culture, Japan.
Moreover,
Setzergives
a criterion for the non-existence ofelliptic
curveshaving good
reductioneverywhere
by
using
certain Galois-theoretic proper-ties of the two division fields of theelliptic
curves([11,
p.249,
Proposition]).
The aim of this paper is to prove the non-existence of such
elliptic
curvesunder somewhat weaker
assumptions
than Setzer and Stroeker’s in the above theorem. For that purpose, we reduce theproblem
offinding
such curves tosolving
certainDiophantine
equations
(cf.
Theorem2)
undersome
hypothesis. Solving
theseDiophantine
equations
explicitly,
we shallshow the
following
theorem.Theorem 1. There is no
elliptic
curvehaving good
reductioneverywhere
over thefollowing four
imaginary
quadratic
fields:
The class numbers of these four fields are all 2. Therefore the non-existence of
elliptic
curveshaving good
reductioneverywhere
for these fieldsdoes not follow from Setzer and Stroeker’s theorem above.
Moreover,
Set-zer’s criterion cannot beapplied
to these fields(cf.
[11,
Theorem 4(a)]).
This paper consists of three sections. In the first
section,
we show thatthe
j-invariant
of anelliptic
curve over animaginary
quadratic
number fieldhaving good
reductioneverywhere
satisfies a set ofDiophantine
equations.
In the second
section,
weexplain
how to solve theseDiophantine
equations.
Some numericalexamples
and theproof
of Theorem 1 aregiven
in the third section.1.
Diophantine
equations
We shall prove the
following
theorem in this section. For the definition of thej-invariant
of anelliptic
curve, we refer to[1,
p. 364(7.1)].
Theorem 2. Let k be an
imaginary
quadratic
field. Suppose
ksatisfies
thefollowing
assumptions:
(i)
The unit grouPof k
consistsof
~1;
(ii)
The class numberof k
isprime
to3;
(iii)
If the
class numbersof k
is divisibleby
2,
then ever y ideal classof
order2 contains an ideal c such that
c2
=(r)
with some rationalinteger
r.Then the
j-invariant j
of
anelliptic
curvedefined
over khaving good
re-duction
everywhere satisfies
thefollowing
setof Diophantine
equations
inthe
ring
of
integers
of
k:Proof.
It is known that thej-invariant j
of anelliptic
curvehaving
good
reduction
everywhere
satisfies thefollowing
conditions for every discretevaluation v of the
ring
ofintegers
of k(see [2, (1.3)]):
It follows from
(2a)
that j
is analgebraic integer.
Furthermore, j
generates
an ideal which is a cubeby
(2b):
(j)
=63.
By
theassumption
on the classnumber of
k,
the ideal b isprincipal,
i.e.,
b =(x).
Hence,
we haveHere we used the
assumption
on the unit group. It is easy to see that thegenerator x
can be taken in thering
ofalgebraic
integers
of k.Now
by
(2c),
there exists an ideal a such thatSuppose
first that a in(4)
isprincipal,
say a =(y).
Then weget j
-1728 =
±y 2
with somey E k. It is
readily
seen from thisequation
that y is analgebraic
integer.
Combining
this with(3),
we obtain:1::.y2
=±~ 20131728.
Changing
thesigns
of thevariables,
we haveCi : y2
= X3 ±
1728.If a is not
principal,
then the class of a is of order 2 in the ideal classgroup. The class of a contains an ideal c such that
c2
=(r)
with r E Zby
the thirdassumption.
This means that thereexists y
satisfying
a =(y)c.
Hence it follows
a2
=(y2)c2
=(ry2).
Combining
this with(4),
we havej -1728
=~ry2.
Thisimplies
thatry2
is analgebraic
integer.
Substituting
(3)
into thisequation and,
if necessary,changing
thesigns
of thevariables,
we obtain
ry2
= x3 :1::.1728.
Multiplying by
r3
thenyields
(r2y)2
=(rX)3 ±
1728r3.
Since(r2y)2
=r3 .
ry2,
we see thatr2y
isintegral. Replacing
(rx, r2y)
by
(x, y),
we haveC~.
We have thusproved
Theorem 2. DSome remarks are in order. The first
assumption
in Theorem 2 is satisfiedif k is neither
Q(B/2013l)
nor~(~).
The class number of each field is one. Hence the non-existence for these two fields follows from Setzer and Stroeker’s theorem.The
resulting
Diophantine equations
Ct
in Theorem 2 defineelliptic
curves over
Q. Thus,
toapply
the theorem to aspecific imaginary quadratic
field,
we have to determine theintegral
points
in thering
ofquadratic
integers
of anelliptic
curve defined overQ.
This isactually
possible
by
themethod we will
explain
in the next section.2.
Integral points
on anelliptic
curveLet k = be an
imaginary
quadratic
field where m is acurve
given
by
a short Weierstrassequation
with a,
b E Z. In thissection,
we discuss how to find theintegral
points
ofE in
7Gk.
First we consider the method of
computing
the Mordell-Weil groupE(k).
We denote
by
E(-m)
thequadratic
twist of Ecorresponding
to thequadratic
extensionk/Q,
which is definedby
theequation
These two curves E and
E(-m)
are both definedover Q
and areisomorphic
over k. Anexplicit isomorphism
over k isgiven
by
By
thisisomorphism,
we have cE(k).
Note that thegenerator
Q of the Galois group of thequadratic
extension acts onby
P ~ -P.Hereafter,
for afinitely generated
Abelian groupA,
letAn
be the kernelof the
multiplication-by-n
map and setA(2)
=A/A2
-We now consider the
following
map:In the map
definition,
the addition is that of the group law on E. Is is easy to see that the map is ahomomorphism. By considering
the actionof Q, it is verified that the intersection of and is a subset of
E(~)2.
Therefore,
the sum of the groups on the left hand side is direct.Similarly,
it can be seen that the map isinjective. Moreover,
the cokernel of cp is killed
by
themultiplication-by-2
map.Indeed,
for anyP E
E(k)(2),
take EE(Q)(2) C~(E(-m)(Q))(2),
thenwe have
cp(P
+pu,
P -Pa)
= 2P.Summing
up the abovearguments,
we cancompute
E(k)
by
thefollowing
procedure.
Step
1Compute
thegenerators
ofE(Q)
and~((E(-~)(Q))(2)).
For this purpose, severalpseudo algorithms
are known(see [18])
andthey
areimplemented
in some numbertheory packages
such as SIMATH([6]).
Step
2 Determine whether thegenerators
found in thepreceding
step
and allpossible
sums of them have a halfpoint
(i.e.,
apoint
Q
suchthat P =
2Q
for agiven
P)
inE(k)
or not. If there exists a halfpoint,
thencompute
it andenlarge
the group. This can be doneby
analgorithm
due toWashington
([17,
Proposition
4]).
Repeat
thisStep
3 Fill up the2-power
torsionpoints.
Anexplicit
computation
of the divisionpolynomials,
forexample,
enables us toaccomplish
it.It should be
emphasized
that we cancompute
E(k)
because the curveE is defined over
Q.
Computing
Mordell-Weil groups over analgebraic
number field
larger
than Q
is still difficult ingeneral
(cf. [3]).
Having
computed
the Mordell-Weil group, we now turn our attention tocomputing
theintegral
points.
Let
Pl,
P2, ... , P,
be thegenerators
ofE(k).
Suppose
that P =plP1
+P2 P2
+ ..-E(k)
is anintegral
point
with pl, P2, ... , ps E Z.Us-ing
Baker’stheory
forelliptic
logarithms
(cf.
[5]),
Smart andStephens
([14],
see also[15,
Chapter
XIII])
give
a smallcomputable
upper bound forjp2!? - "
Hence we are able to find all theintegral
points
inE(k).
Here we make some remarks for a
practical
computation.
To use theSmart-Stephens algorithm,
local minimal models of the curve E and thecanonical
heights
and theelliptic logarithms
of thegenerators
ofE(k)
arerequired.
Weimplemented
thealgorithms
(Tate’s
algorithm
[13,
Chapter
IV,
§9],
Silverman’salgorithm
[12]
andAlgorithm
7.4.8 in[1],
respectively)
for
computing
these invariants on KASH([4]).
We refer to[10]
for more details.Though Algorithm
7.4.8 in[1]
only
describes how tocompute
theelliptic
logarithm
of apoint
in the real locus of anelliptic
curve, it isenough
for our purpose as thefollowing
proposition
shows.Proposition.
(resp.
~~_~",~~
be theelliptic
logarithm
on E(resp.
E(-m))
and A(resp.
thecorresponding
period
lattice. Then thefollowing
diagram is
commutative: 1where the
map ai
isdefined by
(5).
Proof.
Let el, e2 and e3 be the roots of thepolynomial
where
b2, b4
andb6
are the standard invariants of E(for
thedefinitions,
see[1,
p. 364(7.1)]
forexample)
andei, e2
ande’
thecorresponding
objects
for
E(-m).
It is easy to see thatthey
are relatedby
Hence we have
Noting
that the inverse map of O(resp.
isgiven
by
where
(resp. p(z;
A-m))
is the Weierstrassp-function
relative tothe lattice A
(resp. A-,),
we have to showThe first half is obtained from the
following elementary
calculations:Differentiating
the both sides of the above obtainedequation
by
z, wehave
/ ’Y B
This
completes
theproof
of theproposition.
DIf P =
2Q
holds,
then it iseasy to see agrees with
ether 2
7(i
= 1 or2)
our
+w1,
where A = +ZW2-
We can2 2 2 2 7
choose the correct one
by
computing
the p
function at these values.Since the
generators
ofE(k)
always
arise from thepoints
in orE(-m) Q),
it isenough
tocompute $
or for theQ-rational points.
3.Examples
In this
section,
weapply
Theorem 2 to certainspecific imaginary
qua-dratic fields.
Among
61imaginary
quadratic
(1
m100,
m issquare-free),
the non-existence of anelliptic
curvehaving good
reduction isproved
for thefollowing
44 values of mby
Setzer and Stroeker’s Theorem in the introduction and Setzer’s criterion([11,
Thereom4(a)]):
m = 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 30, 33, 34, 39, 41, 42, 43, 46
47, 55, 57, 58, 62, 66, 67, 69, 70, 71, 73, 77, 78, 9, 82, 85, 86, 93, 94, 95, 97.
TABLE 1. The Mordell-Weil groups
C’
The bold-faced numbers
corresponds
to the fields whose class numbers areprime
to six.In our
forthcoming
paper[8],
we show the non-existence for m = 74by
a different method.
On the other
hand,
theonly
knownexamples
ofelliptic
curveshaving
good
reductioneverywhere
defined over animaginary quadratic
field ofsmall discriminant are the
eight
curves(up
toisomorphism)
defined over~( -65)
foundby
Setzer[11].
Among
the fifteen fields whichremain,
thefollowing
four fieldssatisfy
theassumptions
on the class number in Theorem 2 andthey
alsosatisfy
thethird
assumption
of the theorem on the existence of the rationalinteger
r.Note that r is a rational
integer
thatnecessarily
divides the discriminantof the
quadratic field,
since r satisfiesc2
= (r)
in 1~.Although
there may bemore than one r for one
field,
as for the Mordell-Weil groupcomputation,
and r2 be the r’s. In each case, m =
rlr2 holds. Then we have the
following
isomorphism
overQ:
Similarly
we can showGenerally speaking,
the smaller r wechoose,
the faster thecomputation
is.The Mordell-Weil groups and the
integral
points
of the curvesCt (k)
(t
=1 or
r)
arecompiled
in Table 1 and Table2,
respectively.
By
Theorem2,
we have the candidates of thej-invariants
ofelliptic
curveshaving good
reductioneverywhere
as follows.An
algorithm
forcomputing
all theelliptic
curveshaving good
reductioneverywhere
for agiven j-invariant
ispresented
in[7].
Alternatively,
we may use the criterion of Comalada and Nart in[2].
Their criterion tells us whichalgebraic
integers
appear as thej-invariants
ofelliptic
curveshaving good
reduction
everywhere. By
either of thesemethods,
it is shown that there is noelliptic
curvehaving good
reductioneverywhere having
thej-invariants
in the above list. We thus obtain Theorem 1.References
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1993.
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fields of class number 1. Math. Comp. 68 (1999), no. 227, 1187-1200.
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(1999), no. 228, 1679-1685.
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1999.
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235-250.
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York, 1994.
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Proc. Cambridge Philos. Soc. 122 (1997), no. 1, 9-16.
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Press, Cambridge, 1998.
[16] R.J. STROEKER, Reduction of elliptic curves over imaginary quadratic number fields. Pacific
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no. 177, 371-384.
[18] H.G. ZIMMER, Basic algorithms for elliptic curves. Number theory (Eger, 1996), de Gruyter,
Berlin, 1998, pp. 541-595. Masanari KIDA
Department of Mathematics
The University of Electro-Communications
Chofu, Tokyo 182-8585
Japan