Vol. 44, No. 4
DUKE MATHEMATICAL JOURNAL(C)
December 1977
THE RANK OF ELLIPTIC CURVES
ARMAND BRUMER AND KENNETH KRAMER
1. Introduction
Classically, methods of descent to obtain an upper bound for the rank of the
Mordell-Weil group of an elliptic curve have depended on the existence of isogenies of degree two or three and explicit equations, or on the construction of
coverings in ways suitable for computer calculation. (See [1], pp. 265-274 for a
general survey.) In [5], Mazur introduced an elegant and effective method of
descent using only mild information about the N6ron minimal model, but requiring the existence of a rational isogeny.
This paper grew out of an effort to compute the rank of some elliptic curve
factors of the Jacobians of modular curves left undone by Mazur in [6, 7] because they do not admit rational isogenies. It revolves around the general question of what can be said about an elliptic curve, given only its conductor and
perhaps some supplementary information about the N6ron model.
After some preliminaries in the next section, this paper splits into three parts.
The first part, in section 3 and 4, contains the local descent. The remaining
parts treat global questions.
The second part, in sections 5 and 6, may be read independently and performs a dual task. An analysis of two and three-division fields of a hypothetical
curve, based on 16], gives information about the N6ron model for use in the
descent. This simultaneously provides criteria for the non-existence of elliptic
curves having certain square-free conductors, extending nonexistence results
due to Ogg 14], Coates [2], Setzer [ 17], Neumann [12], and others. The novelty
here is that Diophantine considerations arising in the usual treatments are
thrown back in a natural way to the arithmetic of certain number fields. As a
result, we succeed in showing that for all primes N < 1000 where there is no
known curve of conductor N, in fact no such curve exists, except for 8 values
of N where the situation is still undecided. Since the writing of this paper R.
B611ing has eliminated N 211 and N 397 by ad hoc methods.
The third part, in section 7, gives a rather precise upper bound for the Selmer
group and hence for the rank of elliptic curves. We illustrate our results by
obtaining bounds for the rank of all known and unknown curves of prime conductor N < 1000. (See Table III). These bounds are at most 1, except for 8
conductors where the bound is 2. Our bounds suggest that, in general, curves of
prime conductor have the smallest rank compatible with the parity predictions
of Birch and Swinnerton-Dyer.
Received February 3, 1977. Revision received May 26, 1977. Research partially supported by
grants from the National Science Foundation and the Faculty Research Award Program of the City
University of New York.
715
716
ARMAND BRUMER AND KENNETH KRAMER
We exhibit in section 8 curves of high rank, including one of rank 5 with
prime conductor, and one of rank 9. In view of our bounds, the latter curve also
provides an example of a cubic field whose ideal class group has a 2-Sylow
subgroup of rank at least 5. (The previous record was 2.)
In an appendix we consider the following question, suggested by our study of
two and three-division fields. If the discriminant of a semi-stable curve E defined over the rationals is an eth power, does E admit an e-isogeny (so that e
<- 7, by recent work of Mazur)? We answer this affirmatively for e <_ 19. If it
were true in general then a curve of prime conductor N would have discriminant __.N (aside from the well-known exceptions). This would be useful both for
finding and for eliminating curves of prime conductor.
We wish to express our gratitude to Barry Mazur for a number of helpful
comments on this work. We also wish to acknowledge our use of unpublished
tables of Godwin, Smadja, and Barrucand on class numbers of cubic fields.
2. Preliminaries
Let E denote an elliptic curve defined over a field of K of characteristic different from 2. Let be an algebraic closure of K, and let G Gal(//K). Multiplication by 2 on the group
0--
E(/) yields the exact sequence of G-modules
-
-
E(R)z
E(/) E(/)
-
O.
Passing to cohomology, one obtains the Kummer exact sequence
(1)
0
E(K)/2E(K)
H’(G, E(R)2)
-
Hi(G,
E(R)).
Suppose now that E is given in the form y fix), where fix) is a monic cubic
polynomial with coefficients in K. Let An be the K-algebra
An
K[T]/(f(T))
and denote by A} the multiplicative group of invertible elements of An. Then
Ha(G, E([(),) can be identified with the kernel of the map from A/A2 to
K*/K * given by the norm. (See [1], p. 240, and other references given there).
Let x(P) denote the x-coordinate of a point P in E(K). If P1, P2, P3 are three
points in E(K) whose sum is zero, they lie on a straight line, say y rx + s,
and a simple calculation shows that in An
(x(Pa)- T)(x(Pz)- T)(x(Pz)- T)= (rT + s) 2.
Hence there is a homomorphism
A*/A*2
defined by extension from hn(P)
coset{x(P) T} whenever x(P)
vertible in An. Clearly hn induces a homomorphism
E(K)/2E(K)
A*/A*2
n/n
T is in-
717
THE RANK OF ELLIPTIC CURVES
which we also denote by hu, and which is injective. (See [10], p. 142 for an
elementary proof). This corresponds to the map of exact sequence (1). ([ 1], p.
269-271, contains a discussion of the map h and its role in the classical treatment of rational points on elliptic curves).
The algebra A and the map h do not depend on the choice of model for E, as
can be seen from the transformation rules for elliptic curves ([18], p. 180).
There are three cases describing the structure of At, depending on the number
of roots off(x) 0 in K, and hence on the number of points of order two of E
defined over K.
1. Then Ar is a cubic extension field of K and hr(P) is
Case 1. IE(K)I
represented by an element of Ar whose norm down to K lies in K *z.
Case 2. IE(K) I- 2. Then A--K K(A/2), where A is the discriminant
of E (or of the polynomial f). Let N be the norm from K(A1/2) to K. Then h(P)
has a representative of the form (Na, a) with a in K(A1/2).
Case 3. [E(K)2[ 4. Then A K @ K @ K and X(P) has representative
of the form (a, b, ab).
From now on, we view A as the appropriate direct sum of fields described
above, and speak of the components of an element of A; in this direct sum
decomposition.
3. The local picture
Let K be a complete field under a discrete rank one valuation v. Let ff be its
ring of integers, 13 its prime ideal, and k its residue field. Let E be an elliptic
curve defined over K. We shall describe the subgroup hr(E) of A./A.2 in the
semi-stable case only, though one could similarly handle the case of additive
reduction. We have not done this here, for the benefits would be outweighed by
the disadvantage of greatly increasing the number of cases.
Since we shall variously capture hr(E) from above or below, the following
well-known fact is essential.
LEMMA 3.1: We have
Thus
Ix (g)l
[E(K)2I
if char (k)#
if char (k)
2.
2.
Proof. There is a subgroup M C E(K) of finite index and isomorphic to 6 by
a result of Lutz. We then use the Euler characteristic of the kernel-cokernel
exact sequence of multiplication by 2 on
0
to obtain the desired result.
M ---> E(K)
--->
E(K)/M ---> 0
718
ARMAND BRUMER AND KENNETH KRAMER
Let/20 be the connected component of the reduction of the Neron minimal
model for E. We denote by Eo(K) the subgroup of points of E(K) whose reduction is non-singular, and by El(K) the kernel of reduction mod 1). This gives us
the exact sequence
- -
0---> El(K)
Eo(K) o(k) O.
(2)
The group E(K)/Eo(K) is the group of connected components of the N6ron
minimal model, and is finite ([ 19], p. 41).
LEMMA 3.2. Suppose that E has good or multiplicative reduction. Let F be
the maximal unramified extension of K. Then hv(Eo(F)) hv(Ea(F)). If, in addition, char(k) # 2, this group is trivial.
Proof. From the exact sequence (2) we obtain
-
-
eo(F)/2eo(F) :0(k)/2:0(k) 0
where / is the algebraic closure of k. Hence the last term vanishes, and
h(E0) v(E). When char(k) is odd, this group is trivial because Ea(F) is diel(F)/2e(F)
visible by 2.
COROLLARY 3.3. Suppose that both char(k)and [E(K) Eo(K)] are odd.
Then the image of i in A/A: consists of those elements for which
(a) each component can be represented by a unit
and
(b) the appropriate norm condition is satisfied.
Proof. The fact that [E(K) E0(K) is odd implies that h/(E) h(E0). The
previous lemma then implies that every element in the image of h/ becomes
trivial in A/A 2, in view of the commutative square
K/.t-K
$
(3)
$
But the elements of A/ which become squares in A must satisfy condition (a).
Let IE(K)2] 2 n. Then A has n + components, so that the subgroup of
A/A satisfying (a) and (b) has order 2 n. It follows from Lemma 3.1 that hu(E)
must fill out this subgroup.
IE
Remark. The index
E01 is in case of good reduction, and is odd in case
of multiplicative reduction with v(A) odd (Type Iv, with t, odd). The results of
Lemma 3.2 and Corollary 3.3 also apply, for char(k) # 2, in the cases of additive reduction of Types II, II*, IV, IV* of Kodaira’s classification. ([19], p. 46).
For the rest of this section we assume that char(k) 2. We make the standing assumption that K is unramified over q2 to facilitate closer analysis of the
719
THE RANK OF ELLIPTIC CURVES
-
formal group structure on El. We recall ([18], p. 183) that there is a formal
group law for addition on p which gives rise to an isomorphism
El(K). We
denote this map by z
P(z). Let E be given in minimal form by
-
y2 + a lxy + a3y x + a2x + a4x + a6.
Then the x-coordinate of P(z) is given by
(4)
x(P)
z
-
al z-l- a,
+ O(z)
and the law for multiplication by 2 by
zg- 2azz + (alaz- 7aa)z 4 + O(z 5)
where
O(s) means that V(ts -1) --> 0.
PROPOSITION 3.4: Suppose that E has supersingular reduction. Then AI is
a totally ramified extension field of K. Let r be a prime element in AI;. The
image of hl is the subgroup of A/A: consisting of the cosets of + uZr 4
where u runs through coset representatives for / k.
d/(z)
2z
a
Since E is supersingular, the formal group has height 2; hence a is in
2 (K is unramified) and a3 is a unit. By a suitable translation, we can
0 and a2 0 while keeping the model minimal. Thus
assume that a
p
Proof.
yZ + azy
+
x
a4x
+
a6
or
(Y + aa/2) 2= x +
a4x q-
(a6 + a/4).
Let a be a root of the right hand side; that is, the x-coordinate of a point of
order two. Then zr 2a satisfies the Eisenstein polynomial
zr
+ 4a4zr + 2(4a + a)
0
so A/; K[a] is a totally ramified cubic extension of K with prime element r.
Using the formal group parametrization (4) with a az 0 and a a unit we
find that (modulo Az)
-
x(P(z)) -= x(/’)
z
But z a
ing z
+ O(z) ,
az + O(z).
0(8) and units congruent to (mod 4zr) are squares in At. Hence tak-
2aau yields
hz(P(z)) =-
2a](2a)u
=- 1
2azr u
4-
upon using the equation for zr.
"lT41d
720
ARMAND BRUMER AND KENNETH KRAMER
Clearly the coset of + 7/’4//9. in r/ra*/a,2 depends precisely on the coset of u in
/p. The subgroup of a,/a,2
r/r obtained by letting u vary over a set of coset
representatives for 6/p therefore has order 16" p[ [" 2[. By Lemma 3.1,
this is the size of h/c(E), since E(K)2 is trivial. Hence the proposition is proved.
Note further that the subgroup of /r consisting of cosets of + 7r4u as u
varies does not depend on the choice of prime element
For some calculations it is convenient to note that
+ 2zr2u (mod A2).
LEMMA 3.5. Suppose that E has ordinary good reduction or multiplicative
reduction. Then there is a unique point of order two in Ea(K). In the decomposition of A, let the first component correspond to this point. If P is any point in
El(K), then hr(P) is represented by an element of A: with in the first component and a unit of the form / 0(2)in the other component (or components).
Proof. By assumption, the formal group El(K) has height one. Hence al is a
unit. We may therefore obtain a new minimal model of the form
y2 + xy x + a2x + a4x + a6.
1
Remark.
+ zr4u
The x-coordinates of points of order two satisfy
f(x)
x
+ (a2 + 1/4)x2 +
a4x
+
a6
O.
By Hensel’s lemma we can find one root a in K having the expansion
-1/4
a
az 4a4 + O(16).
The other two roots lie in K(A 1/) and are both easily seen to be 0(2).
We compute hr(P(z)) for z in p by looking at the component of Ar corresponding to one of these latter roots, say/3. Using (4) with a
get
z
z
x(P(z))
a2 + 0(2)
z-2(1 z a.z 2 + 0(8))
--
1,
aa
0 we
(mod squares).
Since K is unramified over (2 this is / 0(2) as desired. Moreover, since the
second coordinate does not depend on/3, the norm condition makes the first
coordinate of hi(P(z)), corresponding to a, trivial.
PROPOSITION 3.6. Suppose that E has ordinary good reduction. Then the
image of h consists of those elements of ak/ac2 for which
(a) the first component is in + 4c and the other component or components are in + 21c11), up to multiplication by a square and
(b) the appropriate norm condition is satisfied.
Proof. Let F be the maximal unramified extension of K. Then hF(E(F))
hF(E0(F)) Xv(Ea(F)) with the first equality coming from good reduction and
z
a2z
721
THE RANK OF ELLIPTIC CURVES
the second from Lemma 3.2. Hence Lemma 3.5 provides us with a description
of the entire group X(E(F)).
It follows via diagram (3) and Lemma 3.5 that if P is in E(K), then the first
coordinate of hie(P) becomes trivial in F*/F*. Hence it must be represented by
an element of + 4t7.
Let L K(A/z). It follows, as above, that any representative for the second
coordinate of Xr(P)must lie in L* (1 + 2F(al,))F(A1/2) .2 (1 + 2)L *.
We must now distinguish three cases.
Case 1. L K. Equivalently, IE(K).I 4. The image XK(E) is contained in
the group, mod squares, of elements of the form (a, b, ab) with a in + 4K
and b a unit in K. But this group has order 21U U[ 41 2c1 Ix (g)l. So
this case is done.
Case 2. L/K is an unramified quadratic extension. Hence
[E(K).[
2. Let
be defined to make the sequence
o -o
-o
u/u__.u,,/u?, -o o
exact, where is the map induced by norm. The second coordinate of X(P)
must lie in the pre-image under of the group (1 + 4) U/U of order 2. But
Hence this pre-image has order 2 26 26
[h(E)[. Since the first
coordinate of (P) is determined from the second by the norm condition, the
second coordinate must entirely fill this pre-image.
Case 3. L/K is ramified. Again IE(K)= 2. Let
(rood )}. Then the second coordinate of h(P) must
{x in Ulx
lie in the group
"
(1 + 2)L *
whose order is
z,/(l))
[1): (1,)2[ / [1, 2)[
21e : 2e.l /le : 2e l 21e : 2e l
The first coordinate is determined by the norm condition and we win again
When the curve E is defined over the real numbers N by the equation
or N
accordx), then the algebra A is isomorphic to
ing to whether the discriminant A of E is negative or positive. In the latter case,
we choose the first coordinate of A to coespond to the smallest root of
fix) 0. The following description of h(E(N)) is clear from a sketch of the real
locus of E.
yZ
.
PROPOSITION 3.7: (a) If A < O, then h(E) is trivial in A/A
(b) If A > O, then h(E) has order two, generated by
the class of (1, 1, 1) in A/ANZ.
722
ARMAND BRUMER AND KENNETH KRAMER
4. The local picture: multiplicative reduction
When E has bad reduction, careful analysis is required to determine any nontrivial contribution to the image of hK from the group E/Eo. This problem can be
handled for reduction of multiplicative type using parametrization by p-adic
theta functions. ([4], p. 197).
Suppose first that E has rational tangents at its node modulo p; that is,
E Eq is a Tate curve. To the j-invariant there corresponds a unique q in K*
with v(q) > 0 such that
+744+
j=
The discriminant of E is A
(1
q
qK,.
We have the theta function
q")Z4, so that v(A)
(1
qnz-i)
=0
qnz)
=1
which is holomorphic on K*, has simple zeros at the "lattice"
else, and satisfies the functional equation
0(g -1)
v(q) and A is in
n
(1
O(g)
.
’
cnq
--gO(Z).
O(q-lg)
By the theorem of Jacobi-Abel, the divisor D
on E if and only if it has degree zero and
q and nowhere
x
m(x) is a principal divisor
belongs to q. Moreover, if
x
0(zxS), = g(z) is a rational function on E defined over
K(x,
x). That is, g(z) has multiplicative period q.
POPOSlTON 4.1" Suppose that E is a Tare curve. Then there is a unique
point of order two in E(K) whose reduction is non-singular. In the decomposition of A, let the first coordinate correspond to this point. For each z in K* we
then
then have
_
kn(P(z))
[ (1,(1, z)z, z)
if
if IE (K) I
Proof. The three points of order two in E (m/q
2
4,
are given by the cosets
of
and of qa/Z. As the points whose reduction is non-singular are representis the only point of order two which does not
ed by units mod p the coset of
reduce to the node.
From the fact that the divisor of x e is 2(e) 2(0) we verify immediately
that
723
THE RANK OF ELLIPTIC CURVES
x(z)
x(- 1)
x(z)
x(+-q 1/2)
Cl O(-z)
z O(z) -z
()
c20(+-z q-1/2) O(+__z ql/2) O(Z)-9..
By using the functional equation for 0 we find that O(+_z q-1/2)
w-z ql/20(+_z ql/2). In order for (5) to give a homomorphism to n/-na*/a*z we must
therefore have c in K * and c2 in -W-qa/2K(AV2) .2, yielding the desired formula
for hr.
Next we assume that E has twisted multiplicative reduction. Hence, over the
unramified quadratic extension F K(rtl/2) of K, E becomes isomorphic to the
Tate curve E having the same j-invariant. For convenience we take to be a
unit. If y f(x) is a model for Eq defined over K, then we obtain a model
Y F(X) for E defined over K via the substitution
X=-0x and y=3/2y.
Let Eq(F) --. E(F) be the isomorphism defined by this substitution. Clearly
i(P) lies in E(K) precisely when r(P)= -P, where r generates Gal(F/K).
Thus, for z in F*, the point Q i(P(z)) lies in E(K) precisely when NF/rz is a
power of q.
LEMMA 4.2. Suppose that NlmZ qm, so that i(P(z)) Q is a point in
E(K). Then the first coordinate of X/(Q) is represented by TIm in K*/K .2.
Let L K(A/2), and n mv(A)/2. Then n must be an integer, and a representative for the remaining coordinate (or coordinates) of hr(Q) may be chosen
from 7rUL. If in addition K is an unramified extension of (2, the representative
may be taken from 7r(1 + 2L).
Proof. Let h(z) 0(-z)O(z) By formula (5) the first coordinate of h(Q) is
-.
.
the coset of
X(z) x(- ) nx(z) nx(- ) cnh(z)
But h(z) = h(z-lqm) -(-1) m+ h(z), using the functional equation for 0.
Hence h(z)rl tm + / is fixed by r, so belongs to K. Hence X(z) X(-1) is an
element of rlmK *z, since we have shown that c is in K *z.
Now let c be an element of L* which represents the second coordinate of
hr(Q). Let M F(A /) FL. (We allow the possibility that L K or L F).
By going up to F, where E becomes isomorphic to the Tate curve E and the
map h is given by Proposition 4.1, we see that c must also lie in zM.2.
Since M is unramified over L, the prime 7r of L also serves as a prime element for M. Hence, correcting c by an appropriate multiple of
we may as-
_
rr
sume that c is in L* z U.
Since F is an unramified extension of K and N/nz qm, we see that
mvr(q) mvl(A) is even. Hence n mvn(A)/2 is an integer and v(z) n.
Now the prime rx also serves as a prime element for F. Hence z is in 7rUe and
c is in
L*
rUU;
n
rU.
724
ARMAND BRUMER AND KENNETH KRAMER
If in addition K is an unramified extension of (, then UIU C_
Hence we can actually choose c from r(1 + 26L), as desired.
+ 26u.
PROPOSITION 4.3. Suppose that E has twisted multiplicative reduction and
that either char(k) is odd or else K is an unramified extension of q2. Then
Lemma 4.2 and the norm requirement provide necessary and sufficient conditions for membership in hc(E).
Proof. To facilitate the counting argument which we shall use, we distinguish four cases.
Case 1. v(A) is odd. Then IE(g)l 2 and IX <ESI 216r: 2 I. Since F/K is
unramified, m must be even in Lemma 4.2. Hence the first coordinate of hc(E)
is trivial and the second lies in U L *z. It follows from the norm condition that
hr(E) is contained in the subgroup S of Ac/Ac represented by the cosets of
(1, u) with u in U and Nr/u u in K *z. The kernel of the map
induced by the norm has order 21 :2c because L/K is ramified. Hence
isi 21: 2 IX(E)I and h(E) S.
Note that in this case the index
E01 is odd. Hence we get the same result
from Corollary 3.3 when char(k) is odd or from the arguments in Case 3 of
Proposition 3.6 when K is an unramified extension of
IE
Case 2. A in K .2. Then L K and IE(K)zI 4. The image of Xc must be
contained in the subgroup S of Ac/Ac consisting of the cosets of
(7] m, 7r U,
Ur and n mvlc(A)/2. Now iSl 2 iuc
41 2e7
S
Hence
hu(E).
Ih/c(E)l.
Case 3. L F. Then IE(K)[ 2. By Lemma 4.2 and the norm condition
h(E) is contained in the subgroup S of a,/a,z
/ consisting of the cosets of
(’0 m, "gt’UnU)
where n
mvr(A)/2, NL/rUo r, and Nz/ u is in U. Since L/K is now
where u is in
unramified, the kernel of the map
u,/
has order
[6r 26r1. Hence isi
u/u,
21: 2
IX(E)I and S
hr(E).
Case 4. v(A) even, but L # F and L # K. This only occurs, of course, if
char(k) 2. By Lemma 4.2, h(E) is contained in the subgroup S of
consisting of the cosets of
(m, 7.t.Knu)
725
THE RANK OF ELLIPTIC CURVES
where n
mVldA)/2, u is in
+ 2L, and NL/r u is in .qm K.,.
.,
Since we assume here that K is an unramified extension of
we have
N(1 + 2z) C_ + 4. Furthermore, since F K( 1/2) is unramified, generates the group (1 + 4/0/(1 + 2r) of order 2. The case of odd m is allowed
here, so that the map
(1 + 2z) / (1 + "rrz)z _.> (1 + 4,x) / (1 + 2) z
is surjective. Hence [Ker l I 2,1 and iSl-- 21,, 2, --IX(El Therefore S hr(E).
Finally, we note that these descriptions of hr(E) do not depend on the choice
of prime element rrr, nor on the choice of u0 in Case 3.
5. Criteria for the existence of semi-stable elliptic curves
Throughout this section we consider only semi-stable elliptic curves E defined over the rationals. Equivalently, these are the curves of square-free conductor. We denote conductor and discriminant by N and A respectively. For a
prime (, we denote by Ee the group of points of order ( defined over the algebraic closure. Let L Q(Ee) be the field obtained by adjoining the coordinates
of all points in Ee. Then L is a Galois extension of Q whose Galois group G is a
subgroup of Aut (Ee) GL2(IFe) since Ee has rank two over IFe. We quote from
Serre ([16], pp. 273-277) the following important information.
PROPOSITION 5.1. Suppose E is semi-stable. Let v be a valuation of L extending the p-adic valuation of (. Denote by Gv and Iv the decomposition and
inertial groups of v. Then
(1) L contains the group tXe of gth roots of unity, and we have the commutative diagram
Gal(L/t)
res $
--
Gal(((/ze)/)
Aut(Ee)- GL(IFe)
$ det
Aut(/ze)
Ie
.
in which the vertical maps are restriction and determinant, respectively.
(2) if p does not divide teA, then v is unramified over
(3) If p divides A, so that E has multiplicative reduction,
then
where c6 is the usual modular invariant. ,(x/ -c6 is unramified over (p and
we have the following cases:
(a) Iv is cyclic of order te if ordp(A) 0 (mod ) and p re.
(b) Iv is trivial if ord(A) 0 (mod ) and p re.
726
ARMAND BRUMER AND KENNETH KRAMER
-
(c) Iv is conjugate to the group
of order ( 1) if p ( and
0
ordp(A) 0 (mod ().
(d) I is conjugate to the group
and
of order
if p
0
ordp(A) 0 (mod ).
and E has ordinary good reduction at p then G is conjugate to a
(4) If p
and Iv is as in (3c)or (3d)above. However, the criteria for
subgroup of
0
deciding which case holds do not apply.
and E has supersingular good reduction at p, then L, is isomor(5) If p
1, and Gv is the normalizer of Iv, of
phic to IF}, hence cyclic of order
order 2(
1).
Remark.
Only the assertion about Lv in (3) above is not explicitly stated in
16]. It is however readily proved by using the Tate model to which E is isomorphic over t(x/-c6 ).
We now make some observations about the two-division field of E.
PROPOSITION 5.2. (1). If E has supersingular reduction at 2 then O(E) is a
cyclic cubic extension of ((A 1/2) unramified outside 2 and totally ramified at
2. Moreover, 2 remains prime in ((A1/2).
(2). If the formal group of Lover (zhas height one then
either E has a rational point of order two or else ((Ez) is an unramified cyclic
cubic extension of ((A1/Z) in which the primes dividing 2A split completely.
Proof. This is an immediate consequence of the ramification data recalled
in the previous proposition. It suffices to note that GLz(IF,) $3, the symmetric
group on three letters, and that (A1/z) is the subfield of (Ez) cut out by the
alternating group A3.
COROLLARY 5.3. Suppose that the curve E has no rational points of order
two.
(1) If E has ordinary good reduction or multiplicative reduction modulo 2,
then three divides the order of the ideal class group of ((A l/z) modulo the
subgroup generated by the classes of ideals lying above 2.
(2) If E is supersingular modulo 2, then: (a) A 5 (mod 8) and (b)for every
element a in (A 1/) for which the ideal (c) is the cube of an ideal prime to 2,
we have a-- (mod 2).
(3) Neither of +_A is a perfect square.
follows from Proposition 5.2(2) upon using the class field theoretic description of the extension (E2) over ((A1/z).
we
(2) Since ,(A 1/2) is an unramified quadratic extension of
A
sufficient
and
a
have
5 (mod 8). The assertion about c gives necessary
condition for the existence of a cyclic cubic extension of K
(A) having
Proof. (1)
,
THE RANK OF ELLIPTIC CURVES
727
the ramification properties in Proposition 5.2(1). In fact, let J denote the group
of ideles of K. The normic subgroup of J corresponding to the maximal elementary 3-extension of K unramified outside 2 is given by
NI=
K*JZ( v#2H Uv
x (1 +
2G,))
while the normic subgroup corresponding to the maximal elementary 3-extension of K unramified everywhere is
N2
K*J3
[]v U.
The index [N2: Na] is or 3, and the second condition in (2) is precisely that
required for N
(3) is now clear in view of (1) and (2).
.
Remark. This corollary provides information about the cubic field obtained
by adjoining a point of order two of the curve E to Given the conductor of E,
it can usually be used to determine this field uniquely. It can also be viewed as a
criterion for the existence of an elliptic curve of given square-free conductor.
As such it yields a generalization and strengthening of results of Setzer[ 17] and
Hadano [3].
To make a similar analysis for the three-division field L (E) we recall
that GL.(IF) is a group of order 48
3 24 and that G
Gal(L/) may be
identified with a subgroup of GL2(IF) as in [16], p. 305. The field L contains
(A/, p), where p is a primitive cube root of unity. In particular, G is a 2-group
if and only if A is a perfect cube. Let K (A a/). From Proposition 5.1 (2) and
consideration of the Tate model it is clear that extension L/K is unramified
outside 3 and when E is semi-stable.
LEMMA 5.4. One of the following holds:
(1) E admits a three-isogeny defined O onto an elliptic curve E’. Either E or
E’ contains a rational point of order three.
(2) Gal(L/) GL.(IF3) and in particular A is not a perfect cube.
.
Proof. Suppose A is a perfect cube. Then L is a 2-extension of (, unramifled outside 3 and
But there is only one such quadratic extension of t. The
Burnside basis theorem for p-groups then implies that Gal(L/) is cyclic, since
its maximal elementary quotient is. Hence L is a subfield of some cyclotomic
field (/z3,). In fact, n
since L is a 2-extension. Let r generate Gal(L/).
Since the determinant map in Proposition 5.1 (1) must be surjective, both eigenvalues _+ 1 must occur for or. Thus there must be a rational point of order 3 on E
in this case.
If A is not a perfect cube we may apply Proposition 21 of 16] to complete the
proof.
728
ARMAND BRUMER AND KENNETH KRAMER
For the rest of this section we assume that E is a semi-stable curve which
does not admit a rational three-isogeny. By an appropriate choice of basis, we
have
Gal(L/K)
(G,r) where
0
G
0
and r
The relations are cr
1, r
1, and r r o" ra. We have the lattice of fields in
Figure 1. Note that both cr and r have determinant -1 and thus
Gal(L/KOa)) (oT, r), the kernel of the determinant map of GaI(L/K).
L.
L
.(E)
L
(Figure l)
PROPOSITION 5.5. Suppose E is supersingular at 3. Then
(1) A +_ (mod 9) and the ideal (3)factors as ] in K ((A1/a).
(2) There exists a quadratic extension L of K which is ramified precisely at
and the real prime.
(3) For every element c in K of positive norm for which () is the square of
an ideal prime to loa we have a
(mod 0a). This applies for instance to the
fundamental unit of K.
Proof. (1) The congruence
A _+ (mod 9) may be seen from the formula
for A, since E is supersingular at 3 if and only if b 0 (mod 3). It may also be
proved by noting that by Proposition 5.1(5) the decomposition group of any
prime above 3 in L has order 16, so is a 2-Sylow subgroup of GL(IFa), conjugate to G (tr, r). Hence there exists more than one prime above 3 in K, the
fixed field of G. As there is ramification over 3 in the normal closure K(,o) of K
we must have (3) 0]p in K. Now/x _= _+ (mod 9), else 3 would be totally
ramified in K.
(2) Let P’ be the prime over 3 in L whose decomposition group is
Gp, (tr, r). Then the inertial group Ip, is (r), the unique cyclic subgroup of Gp,
of order 8. Let L be the fixed field of r. Since there is no ramification below P’
from to L., p must be the prime lying below P’ in K, and 10 remains prime in
729
THE RANK OF ELLIPTIC CURVES
Let 3/
(, 1)
0
so that T has order three and T-lz2T
(r’5. Then 3/carries P’
to a prime P of L lying over 133, and the inertial group
I(P/P3)
(y-lzy)
f’l
(o-, ’)
Hence I(P/P3) does not fix L2 and P3 ramifies in
The inertial group for a prime lying over the real prime of K acts non-trivially
on K(p). Hence it is generated by an element of order two having determinant
1. But the only element of order two which fixes L2 is z 4, of determinant + 1.
Hence the real prime of K ramifies in L..
Thus L. is a quadratic extension of K ramified precisely at P3 and the real
prime. Now the normic subgroup of the idele group J of K corresponding to the
maximal elementary 2-extension of K in which only P3 and ramify is
N=K*J(
1-I
U x (1 + :P3) x IR+ x
while the normic subgroup of the maximal elementary 2-extension of K which
is unramified everywhere is
Comparison of the two shows that the desired quadratic extension L2 can exist
if and only if we have the property indicated in (3).
PROPOSITION 5.6. Let E have either multiplicative or ordinary good reducof K (A 1/3) must be even.
tion at 3. Then the class number
Proof. From Proposition 5.1 one readily sees that the extension L of
L1 (p, A1/a) is everywhere unramified. The inertial group of the real prime
or of any prime above 3 in L/K must be contained in a subgroup or order two,
_
generated by an element of determinant -1 because of the ramification in
(p)/(. The only such elements in (or, z) are of the form 0-7-2i. Hence the fixed
field La of (0-, 72) yields a quadratic extension of K which is everywhere unramifled. Hence our claim.
Remark. If in addition A
(mod 9) then the above proposition may be
refined to assert that the order of the ideal class group of K modulo the classes
of primes over 3 remains even. To see this, note that the ideal (3) factors as pp
in K. Let P’ be a prime of L lying above p. Then I(P/p) C_ (trz) for some i.
But in fact I(P’/3) (o-z2i) because there is ramification in (p). The decomposition group G(P’/3) is contained in the normalizer (z4, o-zZ).
Let F be the fixed field of z4 -1. Then there is no residue extension between 3 and P’ f3 F. But F is a Galois extension of t. Hence there is no residue
extension between 3 and any prime over 3 in F. In particular, L3/K is split at
the primes over 3, as desired.
730
ARMAND BRUMER AND KENNETH KRAMER
The criteria we have given so far are concerned with the behavior of primes
dividing 3A. They may be viewed as requirements for the existence of an extension L over with Galois group contained in GL2(IF3) and having specified
ramification. The next criteria depend on information forced on the unramified
primes by the fact that L is the three-division field of an elliptic curve. When E
has good reduction modulo p, we denote its reduction by/. Then IFp(/3) denotes the 3-division field of/.
LEMMA 5.7: Let p be prime to 3A and assume that A is a cube modulo p.
Then
IFp(a) is a cyclic extension of IF of degree f dividing
8.
Let a be the trace of Frobenius. Then more precisely:
(1)/f p
(mod 3), then f 2 if and only if a =- 0 (mod 3). Otherwise
(2)/f p -=
(mod 3), then f <- 2 if and only if
a
0 (mod 3). Otherwise
f=4.
Proof. If the characteristic polynomial of Frobenius is given by
X
a + p (X to1) (X- o2), then the number of points Np, of/ over
the field IF, is
Nv.
where
(1
wT)(1
o)
I
(2
a + p)
runs through the n th roots of unity. In particular, we have
Nv =p+ 1-a
Np2
(p + 1) 2-
1) q- a]N,2
[(p- ) + (a 2p)]N.4.
Np4-- [(p
N;
a
2
Since A is a cube mod p and p is prime to 3A, the extension IFv(/z) of IF is
cyclic of order dividing 8. Now N,/N,, is prime to three unless p
(mod 3)
and av 0 (mod 3). Moreover Nv,/N, is prime to 3 unless p
(mod 3) and
ap 0 (mod 3). Since all nine points of are defined over IFv in any case, the
claim follows immediately.
z
COROLLARY 5.8. Suppose that E has good reduction at p--- (mod 3)
and that ,(IF) does not contain a point of order three. Let I9 be the prime of
degree one lying over p in K ((A1/z).
(1) If E has ordinary good reduction or multiplicative reduction at 3 then the
class of is not a square in the ideal class group of K.
(2) If E is supersingular at 3 then splits in the extension L2/K described in
Proposition 5.5(2).
THE RANK OF ELLIPTIC CURVES
731
+ p ap ig not divisible by three, we have av 0
Proof. Since ]/(IFp)]
(mod 3) so that [IFv(E3) IFI 8 by Lemma 5.7(1). The decomposition group
of a prime lying over p in L must therefore be cyclic of order 8, hence must be
(r). Therefore p splits in the fixed field L2 of z and remains prime in the fixed
field L3 of (o-, z2), as desired.
Remark. By explicitly listing the five elliptic curves (up to isomorphism)
defined over IF2 we find that/(IFz) contains a point of order three if and only if
/ is isomorphic to the curve yZ + y xz. Hence the conclusions above apply
for p 2 when / is not that curve, and in particular when/ is not supersingular at 2.
If 5 splits completely in the two-division field ((E), so that/(IF) contains
all the points of order two, then the inequality 2 _< I/(IFs)I _< 10 of the "Riemann hypothesis" shows that/(IFs) cannot also contain a point of order three.
The conclusions of the above corollary then apply with p 5.
COROLLARY 5.9: Suppose that E is a semi-stable curve with no rational
three-isogeny and having discriminant A
(modulo 7). Then one prime, 7,
lying over 7 in K (A1/z) splits in both the fields Lz and Lz of Figure 1, while
the other two primes over 7, and
remain prime in Lz and in L3.
,
in
Proof. Since A is not a square modulo 7, we must have a point of order two
/(IFT). The degree f= 1IF7(/3) IFT] cannot satisfy f<_ 2. Otherwise,
a
64
0 (modulo 9),
so that a7 -+
(modulo 9) and
8 ar 0 or 7 (modulo 9). But N7 must be even and satisfy the inequality 3 -< N7 <- 13 of the "Riemann hypothesis," a contradiction. Hencef 4 by
N72
Nr
Lemma 5.7(2).
Let P be a prime over 7 in L. The decomposition group Gp is a cyclic subgroup of order 4 contained in Gal(L/L) (o-z, z 2) since 7 splits completely up
to L1 ((p, A1/3). Hence Gv is one of the conjugate subgroups (r), (O’Z),
Only (zz) fixes Lz or Lz. Hence one prime 7 over 7 in K splits in both L and La,
while the others each remain prime in
L and La.
PROPOSITION 5.10. Suppose that 3 splits completely in the two-division
field of E and that E has good reduction modulo 3. Then E is supersingular at 3.
Proof. We may write E in the form
4y
f(x) x’ + bzx + 2b4x + b6
=
with minimal discriminant. Since 3 splits completely in the field obtained by
adjoining a root off(x) 0 to t, and the discriminant of the polynomialfis not
divisible by 3, we must have
f(x) =- x(x 1)(x + 1)
(modulo 3).
Hence bz -= 0 (mod 3) and this makes E supersingular at 3.
732
ARMAND BRUMER AND KENNETH KRAMER
6. Tossing out prime conductors
Throughout this section, N denotes a prime. As an application of the results
of section 5 we now show that there exists no curve of conductor N for 108 of
the 168 primes less than 1000.
In Table I we prove non-existence by three-division field criteria. Note that if
a semi-stable curve E admits a rational three-isogeny b E
E’ then E or E’
contains a rational point of order three. Miyawaki [8] has shown that for prime
conductor this occurs only if N 19 or 37. If E does not admit a rational threeisogeny, then Q(A1/3) ((N1/3) by Proposition 5.4, and we may apply the resuits of Proposition 5.5 or 5.6.
Where 19 appears in Table I, we refer only to a curve not admitting a rational three-isogeny.
-
TABLE I: Reduction Modulo 3
N
ORDINARY
SUPERSINGULAR
2, 5, 7, 13, 23, 29, 31, 41,
59, 97, 103, 137, 157, 167,
173, 193,227,239,241,257,
263,283,311,313,317,349,
367,401,419,439,457,461,
479, 491,547, 569, 607, 617,
619, 641, 661, 691, 751, 769,
787, 821,823,839, 853,907,
929, 967, 977,983
19", 107, 251,379,523,
683,971
Eliminated by
Eliminated by
Prop. 5.6
Prop. 5.5 (1)
Eliminated by
Eliminated by
Prop. 5.6
Prop. 5.5 (3)
We apply the two-division field criteria of section 5 to those prime conductors not eliminated by Table I. Neumann [12] and also Setzer [17] have
shown that there exists a curve of prime conductor N with a rational point of
order two if and only if N 17 or N is of the form u + 64. *For the starred
primes in Table Ii we refer only to curves not having a rational point of order
two.
TABLE II: Reduction Modulo 2
N
3, 17", 47, 71, 73*, 113", 127,
151, 191, 199, 271,281,337,
353", 383,409,449,463,487,
521,577,599,601,631,647,
719, 727,761,809, 857,863,
881,887,911,919, 937,953,
ORDINARY
Eliminated by
Cor. 5.3 (1)
SUPERSINGULAR
Eliminated by
Cor. 5.3 (2a)
991
149, 181,421,541,613,653,
773, 941
Eliminated by
Cor. 5.3 (1)
Eliminated by
Cor. 5.3 (2b)
1Using the same criteria, the reader will readily explain the prevalence of isogenies of degree two
or three for curves of low square-free composite conductor in Table of [9].
THE RANK OF ELLIPTIC CURVES
733
There is an interplay between the two-division field and three-division field
criteria of section 5 which we can exploit to rule out a few more conductors.
N 223,293,509" For these conductors, a curve must be ordinary at both
2 and 3 by Corollary 5.3(2) and Proposition 5.5(1). By the remark after Corollary 5.8, condition (1) of that corollary applies: the ideal class of the prime P2 of
degree one lying above 2 in (N l/a) must not be a square. But one can check
that it is, contradiction!
N 883" A curve of conductor 883 must be supersingular at 3 by Proposition 5.6, and ordinary at 2 with (V) (x/-883) by Corollary 5.3. Let
883 and K (a). Since the class number of K is odd, there is a unique
c
quadratic extension L2 of K with the ramification properties in Proposition
5.5(2). A Kummer generator for L is (a 17)(ct 12)(c 5). One checks that
all the primes over 7 in K split in L. But this contradicts Corollary 5.9.
N 587, 743: A curve of conductor 587 must be ordinary at 3 and supersingular at 2 with ((x/) (x/-587). Since the class number of ((x/-587)
is not divisible by three, the normic subgroup corresponding to the extension
(E)/@(V/-587) is N1 as given in the proof of Corollary 5.3. One checks by
class-field theory that the primes over 3 in ((/ 587) split in (E). But this
contradicts Proposition 5.10.
We find that a curve of conductor 743 must be ordinary at both 2 and 3, with
(() ((x/" 743). One checks that the ideal classes of the primes over 3 in
((x/-743) are cubes. Hence 3 splits completely in (E2) and we again have a
contradiction of Proposition 5.10.
7. Rank calculations
Let E be an elliptic curve defined over ( and let A be the corresponding
algebra defined in section 2. For each completion (p of @ we have the homomorphism
We denote by S the Selmer group, consisting of those elements of A*/A *
which are in k(E) for all primes p including oo. In view of exact sequence (1) of
section 2, the identification of the map with the map h and functoriality, this
agrees with the usual definition of S as the elements of Hi(G, E(()2) which are
trivial in HI(G, E(()) for all p. The Tate-Shafarevitch group _lZt_ is defined to
make the following sequence exact:
(6)
If V is an IF-vector space, we denote its dimension by [V]. Let r be the rank
of the Mordell-Weil group E(). Then
r
+ [_LLL2] + [E()]
[S]
and calculating [S] from a knowledge of h,(E) for all p is referred to as doing a
two-descent.
734
ARMAND BRUMER AND KENNETH KRAMER
We now consider the case in which E does not have a rational point of order
two. Then the algebra A may be identified with a cubic subfield F of the twodivision field of E. We shall need the following symbols"
r
rank of Mordell-Weil group E().
set of rational primes at which E has additive reduction.
set of rational primes at which E has multiplicative reduction with ord A
even.
F
cubic subfield of two-division field of E.
dimension of ideal class group of F modulo squares.
if A<0
2 irA>0
number of primes lying over p in F.
a
m
g
u
n
PROPOSITION 7.1. Suppose that E has no rational point of order two. Then
r+[lllz]_<g+ u+ e+
(np- 1).
P
dP
Proof. Every clement of S can bc represented by an clement s in F* with
NF/S in (2. Let H bc the subgroup of F*/F.2 consisting of those coscts represented by elements h in F of positive norm for which the ideal (h) is a square.
Let Up(E) bc the subgroup of hp(E) whose elements have coordinates in A*/A*2p
which can bc represented by units. Then the sequence
(7)
0-- H
S
-->
S
-
@
Xp(E)/Up(E)
is exact.
For primes p not in (I) m I") (I:) a the local criteria of sections 3 and 4 show that
[X(E)/U(E)] 0. For primes p in (I) m the criteria of section 4 show that
by the
[hp(E)/Up(E)] _< 1. For primes p in we have [hp(E)/Up(E)] <- np
norm condition of section 2. It is clear that [H] g + u. The desired inequality
now follows from counting dimensions in the exact sequences (6) and (7).
a
Remark. If E has a rational point of order two, similar arguments yield the
bound obtained by Mazur ([5], p. 257, Proposition 9.8(b) with p 2).
The description in sections 3 and 4 of what is permitted locally in hp(E) when
p 2 or when p is in (I9 m can often be applied to improve the bound in Proposition 7.1 without having a model for E at hand. One first narrows the possibilities
for the field F, as follows. Given the conductor of E, we consider what sort of
reduction modulo 2 and what values of the discriminant A up to multiplication
by a square are permitted by Corollary 5.3. The class-field theoretic description
of the extension ((E2)/((X/) then limits us to only finitely many possible twodivision fields for E. For example, we have the following "uniqueness" result:
735
THE RANK OF ELLIPTIC CURVES
PROPOSITION 7.2. Let E be a hypothetical semi-stable curve which does
not have a rational point of order two. Suppose its discriminant is specified up
to a square. Let I be the three-Sylow subgroup of the ideal class group of
((x/) modulo the classes of ideals over 2. Then there is at most one candidate
for the two-division field of E if
(1) E has ordinary or multiplicative reduction modulo 2 and I is cyclic
o?"
(2) E has supersingular reduction modulo 2 and I is trivial.
We illustrate below how the bound of Proposition 7.1 can be improved for
prime conductor. Examples involving square-free composite conductor may be
treated similarly, though there usually exist more cases.
_
Curves of prime conductor. Let E be a curve of prime conductor N which
does not have a rational point of order two. Then its discriminant A has the
form ___N with s odd. The sign must be chosen to satisfy the ideal class group
criterion of Corollary 5.3(1) if E is ordinary modulo 2, or the congruence A 5
(mod 8) if E is supersingular modulo 2.
Let F be a cubic subfield of the two-division field of E. The ramification data
in Proposition 5.2 imply that the discriminant of F is 2nN, with the same sign
as A. Moreover, n 0 if E is ordinary at 2 and A
(mod 4); otherwise,
n=2.
We now perform the two-descent for E. Let H be the subgroup of F*/F.2
consisting of those cosets which can be represented by elements h in F of positive norm such that the ideal (h) is a square. Now OrdN(A) and hence
[E(N) E0(IN)] is odd. It follows from Corollary 3.3 and from Proposition 3.4
or 3.6 that the Selmer group S of E is contained in H. There are no further local
requirements for membership in S except possibly at p 2 or p
Let H’ be the subgroup of H whose elements satisfy all the conditions arising
from the local descent over
and let e [H] [H’]. It is clear that if E is
supersingular at 2, one needs to know only the field F to decide whether an
element of H meets the requirements of Proposition 3.4 to be in H’. If E is
ordinary at 2 and A
(modulo 8) one again needs to know only F to decide
whether an element of H meets the requirements of Proposition 3.6. If, however, A
(modulo 8), a model for E is required to calculate ez. We illustrate
this difficulty in Example 1 after Table III. In any case, one sees that e is zero
or one.
To complete the descent at p
we note that there are no further conditions in Proposition 3.7 if A < 0, so that S H’. Since [H’]
+ g e2, it
follows from (6) that
if A <0.
r+ [2_K]
+ g(8)
.
z
,
If A > 0, then Proposition 3.7 provides an additional condition which may cut
down the dimension of H’ by at most one. Let e= [H’] IS]. Then
(9)
r+[_l_U_2]=2+g-2- if A>O.
736
ARMAND BRUMER AND KENNETH KRAMER
As Example 2 after Table III illustrates, a model for E seems necessary to determine e=.
TABLE III. In the first column of Table III we list the 60 primes N > 1000 for which a curve of
conductor N may exist, according to the results of section 6.
In the second and third columns we give all possibilities for the sign of the discriminant A and for
the type of reduction modulo 2 (ordinary ord; supersingular ss) which a curve of conductor N
is permitted to have. These are determined by using Corollary 5.3.
In the fourth column of Table III we list the type of reduction modulo 3, based on Propositions
5.5 and 5.6, although this would be needed only for a three-descent.
According to Remark 3 after the table, there corresponds to each line of Table III a unique
possible two-division field (except, of course, for the Setzer-Neumann curves!). Let F be a cubic
subfield of this field. In the fifth column we give the two-rank g of the ideal class group of F.
The sixth column of Table III contains the number 2 defined in the paragraph before formula (8)
above. In some sense, e2 measures the sharpening of the bound for the rank r of E(() accomplished
by applying the precise local information from the descent over
The seventh column contains an upper bound for r, determined by formula (8) if A < 0, or by
formula (9) and the fact that e= >- 0 if A > 0. This bound is best possible in the sense that there
exists an elliptic curve of conductor N, with the specified behavior modulo 2 and 3, whose rank
equals the bounds, except where we have noted:
(a) There exists an elliptic curve whose rank is one less than the bound, and for which an explicit
model is required to complete the two-descent at (See Example 2). We do not know of a curve
whose rank equals the bound.
(b) We do not know whether any curve of conductor N having the specified behavior modulo 2
and 3 exists. See Remark 2 after the table.
We note in the remarks those curves which have a rational point of order two, as determined by
Setzer [17] and Neumann [12], and those curves which admit a rational three-isogeny, as determined by Miyawaki [8].
.
N
Sign of A
11
17
19
37
37
43
53
61
67
73
79
83
89
89
101
109
113
131
139
163
179
197
211
229
233
*One gets r
At 2
At 3
g
ss
ord
ss
ord
ss
ord
ord
ss
ord
ord
ss
ord
ord
ord
ord
ord
ss
ord
ord
ord
0
ord
+
+
+
ss
ss
ss
ss
ord
ord
ss
ord
ord
ord
ord
ord
+
+
+
ss
ord
ord
ss
ord
ss
ss
ss
ord
ord
ord
ss
ss
ss
ord
ord
ord
0
0
0
0
0
0
0
0
0
0
e2
Remarks
0
0
0
Miyawaki
1"
Miyawaki, (a)
0
0
Setzer-Neumann
0
Setzer-Neumann
0
0
Setzer-Neumann
Setzer-Neumann
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Bound
for r
0
0
0
0
(b)
0
0 from a three-descent ([5], Corollary 9.10).
0
(b)
737
THE RANK OF ELLIPTIC CURVES
TABLE III Continued
Bound
N
233
269
277
307
307
307
307
33
347
353
359
373
389
397
431
433
443
443
467
499
503
557
557
563
571
593
593
643
659
659
673
677
677
701
709
733
739
757
797
811
827
829
Sign of A
+
+
+
+
+
+
+
+
+
+
+
+
At 2
At 3
ord
ss
ord
ss
ss
ss
ord
ord
ss
ord
ord
ss
ss
ord
ord
ord
ss
ord
ss
ord
ord
ord
ss
ord
ss
ord
ord
ord
ss
ord
ord
ord
ss
ss
ss
ord
ss
ss
ord
ord
ss
ord
ss
ss
+
859
877
ss
ss
+
947
997
+
ss
ss
ss
ord
ord
ord
ss, ord
ord
ord
ss
ss, ord
ord
ord
ord
ss
ord
ss, ord
ord
ord
ord
ord
g
0
0
0
0
0
0
ss
ss
for r
Remarks
0
Setzer-Neumann
0
0
0
0
0
0
0
0
0
0
0
0
2
0
2
Setzer-Neumann
(a)
(b)
0
0
0
0
0
0
0
0
0
ord
ord
ord
ord
ord
ord
ord
ord
ord
ord
2
0 or
0
0
0
0
0 or
0
2
0
or 0
both occur
(a)
(b)
both occur
(a)
0
0
0
0
0
0
0
0
0
0
0
0
0
0
or 0
2
2
2
0
2
0
Setzer-Neumann
0
(b)
0
(b)
(a)
0
2
(a)
0
(b)
ord
ss
ss
ss
0
0
0
ord
ord
ord
ord
0
0
0
0
0
0
0
2
(b)
(b)
(b)
Remark 1. Table III is consistent with an unpublished list of SwinnertonDyer’s containing curves of prime conductor less than 1000. However, we did
run across the curve
738
ARMAND BRUMER AND KENNETH KRAMER
y +y-x3+
34x2
+ 13x+
A
-659
which is not in that list.
Remark 2. For the 108 prime conductors N eliminated in section 6, we can
actually say a bit more: any semi-stable curve of discriminant +_Nsd6, and
where s and d are prime to 6, must have a rational point of order two or a
rational three-isogeny. On the other hand, our methods cannot eliminate a conductor N when there does exist a semi-stable curve with discriminant as above.
Remark 3. For the values of N in Table III, it follows from Proposition 7.2
that the two-division field @(E2) is uniquely determined by N and the specified
behavior modulo 2, except for N 307. The case of 307 falls into the general
situation in which A 5 (modulo 8) and, in the notation of Proposition 7.2, I is
a non-trivial cyclic group. The proof of Corollary 5.3 then shows that when the
criteria of that corollary are met there are three possible two-division fields
with supersingular reduction and one with ordinary reduction. For N 307
there is in fact a curve corresponding to each of these four fields.
Thus, Table III is arranged so that there is only one possible two-division
field for each line of the table, although there may be many curves with that
two-division field. When A < 0 and A
(mod 8) our rank calculation depends only on the two-division field. Hence, where we have forced negative
discriminant A
(mod 8), the rank of S is r for every curve with that conductor and behavior modulo 2.
Remark 4. Table III shows in particular that every curve of conductor
67, 109, 139, 179 has rank 0. This answers a question of Barry Mazur and
permits the complete determination of the rational points of the modular curve
Xo(N) for these cases. ([6], [7]).
N
It is only when A is positive or when A is congruent to modulo 8 that the
Selmer group cannot be evaluated solely from the two-division field. If A > 0
one must know the real completion corresponding to the smallest root of the
cubic giving the points of order two. If A
(mod 8) one must know the 2-adic
embedding corresponding to the point of order two in the formal group. The
following examples illustrate these observations.
Example 1. All curves of conductor 431 have the same two-division field.
The two curves
both with A
for N 503.
9x- 8
E: y + xy + y x3 x
E’: y / xy x
-431, have ranks 0 and respectively. A similar situation arises
Example 2. The following curves, for which N
groups of rank 2 and respectively.
A
443 have Selmer
THE RANK OF ELLIPTIC CURVES
E: yZ + xy + y
E’: y + xy x
739
x: 84x- 301
3x- 2
The first curve is particularly interesting in that it appears to have rank 0 while
_l__z has rank 2. This is the only non-trivial 2Ll_ we have run across for prime
conductor N < 1000.
Example 3. The cubic subfield of the two-division field of all curves of
prime conductor 18097 has class number 2. The curves
E: yZ + 15xy + y x:- 49x
E’: y + 9xy + y x:- 7x
have A
18097 and ranks 3 and respectively. This shows the combined effect
of the fact that A is positive and congruent to modulo 8.
To illustrate how our methods apply for composite conductors, we consider
the semi-stable curves in Table of [9] for which computer calculation did not
yield the rank. For example, all curves of conductor 170 with discriminant of
the form 170d must have the same two-division field, its cubic subfield F has
class number one. A calculation requiring only an equation for F and the results
of Corollary 3.3, Proposition 4.1, and Proposition 4.3 then shows that the Selmer group S of every such curve is trivial. Hence r [_z] 0. Amusingly,
since the curve 170C in [9] had been shown by computer calculation to have
S 0, the same had to be true for 170F and 170G which were left undecided.
Similar arguments show r [_z] 0 for the curves 106E, 142G, 174I,
182J, 195J also left undecided in [9], although a model must be used to complete the descent at for the curve 142G, which has positive discriminant. We
note that for the two curves above with a rational point of order three, Mazur’s
techniques ([5], Proposition 9.8(b)) give r 0, but for the curve 170F which
also has a rational point of order three, [5] yields only the bound r -< 1.
8. Curves of high rank and class groups
N6ron 11] proved that there exist elliptic curves of rank 11 over ( although
no explicit examples are known. Recently Penney and Pomerance [ 15], through
a computer search, found three curves of rank 7. While in a particularly masochistic frame of mind, one of us has attempted to find further examples. Unlike
the previously known ones, ours do not have rational points of order two and
are nearly semi-stable. It seems that there are such examples with very small
coefficients but some new techniques would be needed to locate them.
The crude bound of Proposition 7.1 can be used to get examples of cubic
fields for which the rank g of the two-Sylow subgroup of the ideal class group is
and g 2 for real
fairly large. The best results known previously were g
and complex fields respectively.
The smallest prime which is the conductor of a curve of rank three is 5077.
We have the curve
740
ARMAND BRUMER AND KENNETH KRAMER
y2 + y
7x + 6
A 5077.
The next two curves have rank at least four and prime conductor:
y 4xy + 13y =x3-x
A -812011
x3
y2 + 8xy + l ly
+ 2x
-
3x
A
y2 + 14xy + 29y =x3 + 2x2- 15x
A
x3
The curve
-953243.
177858971
has prime conductor and rank at least five, while the curve
y2
22xy + 41y
x3
+ 2x2
63x
A
44887.331171
is semi-stable of rank at least six. Their corresponding cubic fields (whose discriminants are 4A) contain (/27Z) 4 in their ideal class groups.
The curve
y2 + 525xy x3 + 228x
14972955x + (856475)
is a twist by -3 of a semi-stable curve. It has rank at least nine and discriminant
A
-36. 213. 52. 75217. 1420015714498561.
Alan Candiotti has shown that the last factor above is a prime. Hence the rank
of this curve is odd according to the Birch and Swinnerton-Dyer conjectures (if
the curve is a Weil curve!). For the convenience of the reader we list the xcoordinates of 9 independent points
x
0, 11336, 2538, -2242,
1330, -5314, -7674, -3538, -4330.
The corresponding cubic field has (7Z/2) 5 in its ideal class group by Proposition 7.1.
We wish to thank Farley Mawyer for the factorizations of discriminants and
for independent verification.
9. Appendix: The discriminant and isogenies
The following problem was suggested by the results on two and three-division fields in section 5.
Problem. Let E be a semi-stable curve defined over the rationals, with discriminant A. if A is an tn power, then e _< 7 and E admits a rational ’-isogeny.
Serre 16] has shown that if E is semi-stable and admits an e-isogeny onto E’,
then E or E’ has a rational point of order e. Mazur has recently shown that
there are points of order e defined over the rationals only if e _< 7. Hence it
suffices to verify that if A is an th power, then E admits a rational ’-isogeny.
We now do this for f < 19.
741
THE RANK OF ELLIPTIC CURVES
LEMMA 9.1. Suppose that K is an unramified extension of (e and that E is
an elliptic curve defined over K, with ordinary good reduction. Suppose, moreover, that (k), the reduction modulo re, contains a point P of order re. Then
either: (i)/5 lifts to a point of order in E(K) and K(Ee) K(tze)
or: (ii) the extension K(Ee) over K is totally and wildly ramified, and if d is the
discriminant of this extension, then orde(d) 2 2.
Proof. In the notation of section 3, we have the exact sequence
Let P be a point in E(K) whose reduction is/5. Then gP is in El(K) and we may
write gP (x(z), y(z)), where z is in the prime ideal (() of K and (x(z), y(z)) is
the formal group parametrization of El(K). To correct P by an element
Q (x(t), y(t)) in E so that P Q is a point of order g, one solves the equation
(x(z), y(z)) e(x(t), y(t))
for t. In the formal group, this amounts to solving
z die(t) tet +
+ cete+
where e gives the power series for multiplication by Since E has ordinary
.
good reduction, Oe modulo e is a power series in and Ce
the Weierstrass preparation theorem we may write
te
tOe(t)
z
(te +
ae-1 :-1
+
+ alt + ao)U(t)
0 (mod e). Thus, by
R(t)U(t)
where the coefficients ai belong to the ideal (e) and the constant term of U(t) is a
unit in K. One sees readily that orde(a0)
orde(z) and orde(al) 1.
If orde(z) >- 2 then Hensel’s lemma guarantees that there is a solution in K to
R(t) 0, and (i) holds. If orde(z) 1, then R(t) is an Eisenstein polynomial, so
that K(t) over K is totally ramified of degree g. The discriminant of this extension is (N(R’(t)))
(e) since orde(al) 1. Moreover, K(Ee) K(t, tze). But
which is totally ramified, and so
K(Ee) over K(t) is an extension of degree e
is tame. Transitivity of the discriminant now gives our claim in (ii).
PROPOSITION 9.2. If the discriminant A of a semi-stable curve E defined
over the rationals is an gth power for some prime <- 19, then E admits a
rational isogeny of degree
Proof. Suppose that E admits no rational e-isogeny. Then Gal(((Ee)/t)
GL2(IFe) according to Serre [16]. Since A is an eth power, ((Ee) is unramified
outside (, as one sees from the Tate parametrization for primes dividing A. If e
divides A, then e(Ee) is obtained by adjoining /ze to the unramified field
).
Let d be the discriminant of the extension ((Ee) over (, and let n IGL(Fe)[
(gz 1)(z g). Case by case consideration then shows that [d[ TM aa with
742
ARMAND BRUMER AND KENNETH KRAMER
if g[A or if case (i) of Lemma 9.1 holds
2
2
2_ 2
4
if case (ii) of Lemma 9.1 holds
if the reduction mod ( is supersingular.
Odlyzko 13] recently improved Minkowski’s lower bound for the discriminant to give in our case
Id[ 1/’ > (22.2) exp (-254/n)
and
[d[ 1In > (21.8) exp (-70/n)
These contradict the existence of the extension Q(Ee) over
for (<_ 19.
For instance, when
19, the largest choice for a gives
21.9946
while the first lower bound of Odlyzko is at least 22.15.
Idl
Remarks. (1) In case of prime conductor, there exists a rational isogeny of
prime degree g > 2 only for conductor 11 ( 5), 19 ( 3), or 37 (( 3),
according to [8]. There exists a rational point of order two only for the SetzerNeumann curves ([17] and [12]). In general there are infinitely many semistable curves with A an (th power if ( --< 7.
(2) One may use Odlyzko’s improved estimates under the extended Riemann
hypothesis to get a conditonal affirmative solution to the problem for -< 31.
(3) It might be noted that if A z e, the related Diophantine equation xa y2
1728z e, which looks improbable, has in fact infinitely many rational solutions, since the surface it defines is birationally isomorphic to a plane. Cf. Reviews in Number Theory, Vol. II, pp. 103ft.
Added in proof" In a letter dated July 22, 1977, R. B611ing informs us that he
has now succeeded in eliminating the lines marked (b) in Table III.
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Math. Soc. 40(1966), 193-291.
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53(1974), 199-210.
THE RANK OF ELLIPTIC CURVES
743
4. S. LANG, Elliptic Functions, Addison-Wesley, Reading, Mass., 1973.
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"Algorithm for determining the type of a singular fiber in an elliptic pencil," Modular
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BRUMER: DEPARTMENT OF MATHEMATICS, FORDHAM UNIVERSITY, BRONX, NEW YORK 10458
KRAMER: DEPARTMENT OF MATHEMATICS, CITY UNIVERSITY OF NEW YORK, QUEENS COLLEGE, FLUSHING, NEW YORK 11367