2 and m = 0, "
=
2 and m = 1.
φ is a prime dividing 2 such that φ | ( ^ + 2, ω ) , ω = (1 ± Vd)/2. Finally, if the congruences of (b) are satisfied, we set: £/ = 2, Γ = 4, S = π, _ ( -π iΐπ2A-B = π4 + π4 (mod 32) " \ - (4 + π 2 ) if π 2 ^ - 5 = 5π 4 + 4π 5 + π 6 (mod 32) 2
R
where π = λ/d?, 1 + Vd if ί/ = 2, 3 (mod 4), respectively.
D
Before the proof of Theorem 1 we solve the diophantine equations of(ii). PROPOSITION 1. Let ε e ίlκ denote either the fundamental K or its conjugate. The equation:
(1.11)
u + 64v=X2,
u,veilκ,
unit of
ELLIPTIC CURVES WITH TRIVIAL CONDUCTOR
241
2
has a solution if and only if u = ±v and X Dχ = 63D# or 65Dκ, respectively, or in the following exceptional cases: = 7,
u
= 41,
u
Hi, Hi,
u = πi, = 1023, -u Hi = 65,
=
82U,
w = eu, — ε3u,
= -εu
X = εv^w. 2
Proof. We consider first the cases u = ±1 or v = ± 1 . If X = w + 64 then we have that X - 8, X + 8 e il# , and taking norms we get the solutions: rf = 7, X = ±3\/7, w = - l 2
Similarly, from X
and
d = 65, X
= 1 + 64Ϊ; we have:
} or { φ . φ 5 ? φ 5 . φ } , where 2Dχ = 9β ψ in the last case. Taking norms we get the solutions: rf = 7, Z = ± ( 2 ε 2 + 1 ) , v = β 3 ; rf = 4 1 , JΓ = ± e , v = β, rf = 65, X = ±>/65, ϋ = 1
d = 65, X = ±(2ε2 - 1), v = ε 3 .
In the case X2 + 1 = 64v setting X = (r + sVd)/2, a — s2d, and taking norms we get:
r = s (mod 2),
a 2 + 2(4 - r 2 )α + (4 + r 2 ) 2 - 167V(64t;) = 0. Let 646 2 be the discriminant of this quadratic equation. From the equality iV(64^) = b2 + r2 , we find that all the solutions of X2 + 1 = 64υ are: , , ; , X = dzε, v = ε. 2 In the case X + 64 = u, proceeding as above we find that there are no solutions. Since every solution of (1.11) with ±u or ±υ G il^ leads up to the preceding cases, we can assume that ±u, ±v ^ H^ . In this case we have that uv = ± w 2 , w e U# . Setting uw = (r + s\[d)j2, r = s (mod 2), taking norms and having in mind that N(u) = N(v) we have: N{X)2 = N(u){ί + 2 1 2 ± 2 6 tv{{uw)2)) = N(u)(l + 2 1 2 =F 2ΊN(w) ± 26r2). Taking into account all possible signs for N(u) and N(w) we find in any case a finite number of possibilities for r and N(X), from which only the following ones provide effective solutions to (1.11): r = ±2,
N(X) = ±63
or
±65,
242
SALVADOR COMALADA
and they lead to X2Dχ = 63Dχ or 65£>A: , respectively. On the other hand, it is clear that this last condition ensures the existence of a solution to (1.11). D PROPOSITION 2. Let ε e ilκ denote either the fundamental unit of K or its conjugate. The equation:
u + v=X2,
(1.12)
u,veίlκ,
XeOκ,
XφO,
has a solution if and only ifu = v and 2Dχ = X2Oκ or u = w2uo, v = w2Uo, Uo, w G ίlκ > tr(wo) a square or in the following exceptional cases: d = 2, d = 5, d = 6,
ugίljζ, uettχ> 0 0. From (1.12) we have: χ2w = uw + vw = uw + N(w)ΰw = tv(uw), since N(w) = -1 would lead to a contradiction. Now, every odd prime dividing tτ(uw) is unramified in K; hence, either tτ(uw) or j tr(uw) is a rational square. In the first case, taking w0 = uw we have u — W~1UQ, v = w~~ιtto and w~ι G il^. If \\r(uw) = r 2 , r G Z, then the prime 2 is ramified in ΛΓ and the prime ideal dividing 2 is principal. If N(u) = JV(v) = 1 we have a solution to r4 - ds2 = 1 and by a theorem of Cohn [1] we have uw = ± 1 , that is u = v , with the exceptional case d = 6, when there is the possibility wu> = ±ε2, leading to v = ε4u. If iV(w) = N(v) = - 1 , w is a square; hence, 4 2 2 G O^ and d = 2. Now, the equation r - 2s = -1 has only one solution in positive integers, namely r = s = 1 [3, D24-31]; hence we 2 have uw = ε, v = —ε w. If N(tt) = 1 and N(v) = - 1 , then w > 0; otherwise either of the sums u + v , w + £ would be negative and (1.12) would lead to a contradiction. Hence, wGil^ and (1.12) furnishes an integer Yeθκ such that y - 1, Y + 1 G U# . Taking norms we easily see that d = 5 and the solutions to (1.12) must be those given in the statement. D
ELLIPTIC CURVES WITH TRIVIAL CONDUCTOR
243
3. Let e e ίlχ denote either the fundamental unit of K = Q(Vd), d = 1 (mod 8), or its conjugate, and let φ be a prime dividing 2 such that φ 4 is principal. Then, the equation: PROPOSITION
4Y + UΫ2 =X2
(1.13)
has a solution with X, Y eDκ, ueίlκ, YOK = Φ 4 , if and only if: (i) φ is principal and there exists a generator of φ6 whose trace is a square. In this case, if φ = HOχ, the solutions 0/(1.13) are: Y = υH4,
X = rΉ,
u = υN(υ),
where r e Z, v e il# satisfy tr(vHβ) = r2. (ii) φ 2 is not principal and the following equations have an integral solution:
r4-ds2 2
= -16384,
2
\m -dn
r odd
(m,r)=l,
= 8r\
\2Sn = sm (mod r).
In this case, the solutions 0/(1.13) are: X = (m + ny/d)/2, Y, w (2/ wT = (r 2 + sy/d)/2, where m, n, r, s are a solution of (1.14). have the following exceptional cases:
= 4\,
X=t
1Q
Jy/v,
u = e5v,
γ=\—-\v9
^ \
fL±\
Vi
veii2K,
u = Vi
9
v€ίl2κ,
u = ε'v,
veίl2κ.
Proof. Suppose that there exists a solution to (1.13) with N(uY) = 16. Multiplying by UY both sides of the equation we get: X2ΰY
= 4UY2 + 167 = 4X2.
2
Hence, φ is principal generated by G = 2X/X. N{u)uG2 in (1.13) we obtain: 2
χ
2
2
4
Replacing Y = 3
= 4N(u)uG + uU G = N(u)Gtr(uG ).
Since (d, tτ(uG3)) = 1 we conclude that G is a square in DA: and 3 Xr(N(u)uG ) is a square in Z . Thus, (i) is satisfied. If we have a
244
SALVADOR COMALADA
solution of (1.13) with N(uY) = - 1 6 , taking norms directly from (1.13) we get: 2
N(X)
3
= 4tr(uΫ ).
Hence, if C = uY , we have: 12
-2 ,
2
tr(C) = r ,
N(X) = 2r,
reZ,
so that the equations of (1.14) are satisfied, and from: 3
2
C - 64N(u) = uΫ + 4YΫ = ΎX , we have that X divides C ± 64 in Dκ, which is equivalent to the congruence, l2Sn = sm (mod r). To see that the solutions given in (ii) do satisfy (1.13) we write: (C + 64)O* = φ 6 2l,
7V(2l) = r2,
since N(C + 64) = 64r2 . If r = p*1 pakk, from the first equation in (1.14) we have that all pi decompose in Q(Vd) and 21 = ψxax - φ2kak where each φ, is one of the two prime ideals dividing p, . Now, since X divides C + 64, we deduce that X2O/C = ψ 21. Hence, C + 64 = y χ 2 w , K; G il/ς: and equation (1.13) is easily obtained. We have only to check that if φ 2 is principal then we fall in the exceptional cases. We set φ2 = GOκ and Y = G2υ, v e ίlκ - If N(u) = 1, N(Y) = - 1 6 , we have N(v) = - 1 . Hence, u = ±w2, u> G il^ but u = —w2 never occurs since we would have from (1.13) the congruence X2 = — 1 (mod φ2), which is impossible. Thus, the equation becomes: 4 7 = (X-wT)(X
+ wΫ)9
and {(X - wΫ)Oκ,
(X + w7)Dκ} = {2O*, (2D*)φ 4 }.
Taking norms it can easily be deduced that d = 41 or 65 and the first and the third exceptional cases can be obtained. If N(u) = — 1, N(Y) = 16, we have that υ = ±w2 , w eίlκ, but Y < 0 implies w > 0, so that Ύ < 0, w < 0 which is impossible since 4Y + ϊiY2 =~X . Hence, 7 is a square and setting Y = H2, the equation becomes:
and we proceed as above to get the two remaining exceptional cases. D
ELLIPTIC CURVES WITH TRIVIAL CONDUCTOR
245
Proof of Theorem 1. Let's see first the equivalence between (i) and (ii): If E is a #-admissible elliptic curve defined over K, having a model (1.9), then from the equation of the discriminant:
and the conditions of Setzer's lemma, we have the following possibilities: 4 2 6 2 (1.15) B = 2 u, A -2 u = v , u,v eίlκ and A = a (mod 4 ) , 2 s 2 (1.16) B = u, A -4u = 2 υ , u,v eίlκ a n d A = -2a ( m o d 8 ) , 2 4 4 (1.17) B = 4M, A -2 u = 2 υ, u,v e ίlκ and A = 4 v ^ (mod
8 ) , 5 Ξ 4 + 8V^ 2
(mod 4
16)
if
u,veίlKy
^ ( 7 ) = ±16,
with w e i l | . Equations (2.4), (2.5) and (2.6) are completely studied in Propositions 1, 2 and 3, respectively. In our particular case, where one of the units u, υ is a square, it is easy to see that equation (2.4) implies d = - 7 , 7, 41 or 65, equation (2.5) implies d = 2 and from
248
SALVADOR COMALADA
equation (2.6) we have d = - 7 , 41 or 65. Finally, from its finite number of solutions, we can construct all the desired curves. Notice that for d = - 7 , by Stroeker's result in [6], no elliptic curves with good reduction everywhere can be found. Also, for d = 2 we have already shown in Theorem 1 that no admissible curves exist. Clearly, by Lemma 1, all the curves obtained have a global minimal model except for Eβ and E$, which cannot have it. Notice that these last curves have been obtained by setting c = 5 in (2.3). In fact, following Setzer in [4], it can be easily shown that two admissible curves over Q(\/65) attached to a solution of (2.4) or (2.6) with C\, C2 respectively, are isomorphic if and only if 2>i, Σ>2 are equivalent ideals. O 3. Counting g-admissible curves. THEOREM 3. Let K = Q(Vd), d φ 7, 41, 65, 1023 satisfy the conditions of Theorem 1. Then the number of g-admissible elliptic curves over K (up to isomorphism) is:
2 6 8 4 l^N 4 + 2fN 2f+xM
ι/(1.6) issolvablef d φ 6, ifd = 6 if either (1.4) or (1.5) is solvable and d = 3 (mod 4), if either (1.4) or (1.5) is solvable, (1.8) is unsolvable and dφl (mod 4), if (1.8) is solvable and (1.5) is unsolvable, if both (1.5) and (1.8) are solvable, ι/(1.7) is solvable,
where M is the number of solutions in positive integers (r, s) of the 4 2 equation r - ds = ±256, 2 f r. ΛΓ w ίAe number of solutions in integers (r, 5) o/7Ae equation r4 -ds2 = -16384, r = 3 (mod 4) for which the whole set of equations (1.8) has also a solution. Let / = 0, 1 according to whether the norm of the fundamental unit is —I or I, respectively. For d = 7, 41, 65, 1023, the number of g-admissible elliptic curves w 8, 6, 6, 8 respectively. Proof. Let (5 be the set of classes of isomorphic g-admissible elliptic curves. In each case, we shall consider different sets ft and we shall establish a mapping Φ: & —> S) of & onto S). We shall obtain the result about the cardinal of 0 in terms of the cardinal of 9) and the fibres of Φ . Notice that two models: y2 = x 3 + A\X2 + B\X, 2 3 2 y = χ + A2X + B2X, belong to isomorphic elliptic curves if and
ELLIPTIC CURVES WITH TRIVIAL CONDUCTOR
249
only if they are related by a dilation:
A2 = AιU\
B2 = B{U\
Ueίίk.
In fact, if a transformation (1.15) changes one model into the other we 3 2 must have S = T = 0 and R + AXR + BγR = 0, ReOκ and since d Φ 7, 4 1 , 65 (see Th. 2), R = 0. Using this argument we shall have, in any case, that Φ does not depend on the particular choice of the model. Assume that (1.6) is solvable. We define: 1/2
,
Ee Hence, if u E il^ the curves are isomorphic, and if u 0 il^ we obtain two non-isomorphic curves only when JV(w) = 1, i.e., when the norm of the fundamental unit is 1, as we must have N{A\) = N(A2) = —2 (mod 8). If both (1.5) and (1.8) are solvable we fall simultaneously in the two last cases, but, clearly, curves coming from different cases are non-isomorphic, so that the cardinal is the sum of the two last cardinals and the result follows. If, at last, we assume that (1.7) is solvable we consider: N(C) = ±256, tr(C) = r2, r G Z , (r, 2) = 1}
S) = {CeOκ;
and Φ(A, B) = -2AB/A. To show that Φ(E) e ft we put by Proposition 3: A = r/7, J5 = w//4, where u € ίlκ, N(H) = - 2 , tv(uH6) = r2. Hence, -2AB/A = -2A2B/N(A) = uHβ. Now we show that either one or two elements of (S map to every C e S) depending on whether the norm of the fundamental unit is - 1 or 1, respectively. If C e ft with tr(C) = r2 , then from: 6
C = uH ,
ueίiκ,
N(H) = 2,
He
Oκ,
we have that the elliptic curve given by: A = ±rH (XvA = - 1 (mod 4))9B = uH\ is ^-admissible and Φ(A, B) = C. If = Φ{A2, 5 2 ) since ΛΓ^j) = N{A2) we get
Bι/B2 = (Aι/A2)29 Thus,
Bι/B2eίlκ
252
SALVADOR COMALADA
and we argue as in the preceding case to get the result. For d = 7, 4 1 , 65 and 1023 we construct all the ^-admissible curves directly from the solutions of the diophantine equations in (iii) of Theorem 1. It is straightforward to check that there are 8, 6, 6, 8 curves respectively. In §5 we give explicit models for all these curves with d = 7, 41 and 65. D Assume that (1.7) is solvable. Then if the norm of the fundamental unit is 1 we must have: REMARK.
4
2
= 256,
r -ds 2
(r,2) = l. 6
Otherwise, from (r + sVd)/2 = uH , N(H) = - 2 , we should have N(u) = - 1 . Also, if the norm of the fundamental unit is - 1 and d φ 41 we must have: r4-^
2
= -256,
( r , 2 ) = 1.
Otherwise, proceeding as above we would have N(ύ) = 1 and, hence, ueίlχ so that the ^-admissible curve given by: A = rΉ, B = uH4 , should have all the 2-division points rational over K, and Theorem 2 would lead to contradiction. Let Ji, Jr be the sets of ./-invariants of all ^-admissible elliptic curves defined over imaginary or real quadratic fields, respectively. Setzer ([4], [5]) has proved that /, and JrnZ are both finite. In fact, //
=
{17 3 ,257 3 },
Λ Π Z = {17 3 , 257 3 , - 1 5 3 , 255 3 , 20 3 }. On the other hand, as a consequence of Theorems 1 and 3, we can state the following: COROLLARY.
Jr is not finite.
Proof. For any integer m = 13 (mod 32) let dm be the square-free positive integer determined by: (3.1)
7m6+l =
I6dmn2.
Taking x = 7m 3 , y = In, d = ldm we obtain a solution to (1.4) with d = 3 (mod 4). Since, (3.1) has only a finite number of solutions for a given dm , we get in this way an infinite number of real quadratic fields K = Q(y/d) admitting ^-admissible elliptic curves. By Theorem 3, there are exactly 8 such curves for each K and by [5, Th. 3] not all of them can have a rational y'-invariant. D
ELLIPTIC CURVES WITH TRIVIAL CONDUCTOR
253
4. Admissible curves. It seems very difficult to characterize all the admissible curves defined over K by means of this diophantine approach. Nevertheless, in the case 2 \ h^ it is still possible to solve it completely. In fact, under this assumption on h^, almost all the admissible curves turn out to be ^-admissible and they are already characterized by Theorem 1. More precisely, THEOREM 4. Assume 2 \ h^. All admissible elliptic curves defined over K are g-admissible except for the case: d prime, d = 1 (mod 16) and φ not principal, where
Y&κ = Φ 3 if and only if there exists a generator of φ6 whose trace is a square. The solution is then given by: X = rY, v = U, where tτ(Y2u) = r2, r e Z, except for d = 41, when there are also the solutions: l i e - 2 7 rX=—20—vu> ε-15 Γ v v
X=—j—y/ΰ,
v=eu,
u = εv,
ΛZ
ε- 17
Y= __ε-17
Y = —^—,
,
ίί2
ueίljc, 2
2
veίX κ.
254
SALVADOR COMALADA
Proof. Let us assume that (4.3) has a solution. Then, multiplying both sides of the equation by Ύ we get: (4.4)
,
V+M.
V, Λ.
I
_
If N(uv) = 1, multiplying both sides of (4.4) by ΰv we have: 2
2
2
(XY/S) = vY + aY = (XY/$)ϊwN(u). Hence, ±ΰv G \X2K. Setting ±uv = w2, again (4.4) by w , we get: tr(wti;Γ2) = wwΓ 2 + N(w)ΰwΫ2
w G ίlκ and multiplying = (X7/8)2w
,
for we must have N(w) = 1. Now, since (d, tr(wti;7 2 )) = 1, we conclude that tr(uwY2) must be a square in Z . Setting y 7 = Y\Zw~ι, u' = uw2 we have: χ =r r ' ,
v = 8/,
where tr((r) 2 w0 = ^ 2 5 ^ ^ Z .
If N(uv) = - 1 , we may assume N(u) = - 1 so that N(v) = 1 and ±ι; € i l ^ . But from (4.4) v = 1 (mod φ 2 ) . Thus, i; G il^ and we may write: a n d
Taking norms we have d = 41 and the two exceptional solutions are obtained. α Proof of Theorem 4. Let E be an admissible elliptic curve with model (2.1) and discriminant (2.2). Then, we also have: AOK = Σ)22t,
BOK = 2)4Q5,
DOK = 2> 12 ,
where 21, 55, S) are ideals of £)# and 55 = Dκ, 16O^, 40/^ or φ 4 . Since S) 12 is principal and 2 \ hk we conclude that Σ)3 and, consequently, 2)05 are principal. If 2) is principal then by a dilation we get an admissible curve isomorphic to E with a model (1.9) so that by Lemma 1 a global minimal model for it can be obtained. Otherwise,
ELLIPTIC CURVES WITH TRIVIAL CONDUCTOR
255
the non-principality of 2) is equivalent to the non-principality of φ , 4 where 9$ = φ . In this latter case we may set:
From the discriminant we have: 2
Hence, equation (4.1) is satisfied. Now it suffices to apply Proposition 4 to get the equivalence between (ii) and (iii). It is well known that for d = 1 (mod 4) condition 2 \ h^ is equivalent to ([8]): d prime or f
d = qq ,
q = q* = 3
(mod 4).
In the last case, the norm of the fundamental unit is 1 and, by the Remark at the end of §3, we have that only the plus sign is possible in (4.2). Now, it is easy to see that this equation is not solvable for these values of d. Hence, d must be a prime. In this case, the fundamental unit has norm - 1 and only the minus sign is possible in (4.2). It is clear that the solvability of (4.2) implies d = 1 (mod 16) so that the first assertion of Theorem 4 is proved. Let's assume a solution X, Y, w, v, of (4.1) such that N{Xu) = - 8 (mod 32). Let F e Dκ be a uniformizing element for φ, where YOκ = φ 3 and let us set G = F3/uY. Then, the elliptic curve given by: B = u~ιF4,
AB = 2XG2,
D = vG4,
is admissible. It is not difficult to check Setzer's conditions (a) on A, B, which are equivalent to the following ones: N(A)
ΞΞ
- 2 (mod 8),
tr(A) = 3 (mod 4),
N(B-l) = 0 (mod 8). To prove the last assertion of the theorem we count the number of admissible curves proceeding as in §3. Now, 0 will stand for the set of the isomorphic admissible elliptic curves. Notice that since d is a prime, d = 1 (mod 16) and φ is not principal, none of the conditions of Theorem 1 is satisfied. Hence, 0 will have no ^-admissible curves.
256
SALVADOR COMALADA
Assuming (4.2) solvable, we define: S) = {C e Oκ
C > 0, N(C) = -256,
We may set: = 2XG2,
(4.5)
D = vG4,
where X, Y, w, v is a solution of (4.1) and prime to 2 such that φ2t is principal. Hence, D3
ι
3
= 2l , 21 an ideal
4
~ ιYY4 = N(v)uvY — = uv~
(4.6)
By Proposition 4, = (uwY2)2, tr = r2, r e Z, so that Φ(E) G ft. To see that Φ is onto let's set C = (±r2+sVd)/2. 2 3 Then, C = uP where POK = φ , N(P) = 8, ί / G ί l | . We get an admissible elliptic curve by taking: B = u~ιF4, ^12? = ±2rPG2, (sign determined by tr(J) = - 1 (mod 4)), where i 7 , G are taken as above. Thus, 1/2
wG4
= C.
Finally, Φ ( ^ i , Bx) = Φ ( ^ 2 , # 2 ) i f f A / ^ 2 = (Bx/B2γ . Assume that £/ is attached to the solution of (4.1) Xt, 7/, wz, ι;f , / = 1, 2, in the way of (4.5). Clearly, we may assume by (4.6) that Y\ = Y2. Then we have U\/u2 = V\/v2 and since Y^w/ + 64^/ G D ^ for / = 1, 2 we conclude that W2 = U\W2, υ2 = V\W2, w G H#. Hence, £2 is attached to the solution ±2wXi, Y, U\W2 , V\W2. If w G il^, we have from (4.5) and the fact that 2ti, Qί2 are equivalent, the relation: G2
UeK*
Hence, 5i = B2U4 and from AιBι/A2B2 = (Gι/G2)2 we get ^2^ 2 > so that £Ί and E2 are isomorphic. The possibility w never occurs since we must have N{Aχ) = N(A2) = - 2 (mod 8) REMARKS.
Theorem 4.
(1)
D
= 257 is the first one to satisfy the conditions of
ELLIPTIC CURVES WITH TRIVIAL CONDUCTOR
257
(2) In case 2 \ h^ , the diophantine conditions (iii) of Theorem 1 for the existence of ^-admissible models can be reformulated as follows. d = 7, 41 or d — Iq, q = 3 (mod 4), x2 - 4dy2 = - 7 solvable, d = 2q, # ΞΞ 3 (mod 4), x2 - dy2 = -2 solvable, d prime, d = 1 (mod 16), ψ principal, r 4 — ds2 = —256 solvable. We get solutions to all these equations taking, for instance, d = 77, rf = 6, 22, 38, rf = 337, respectively. 5. A table of all admissible curves defined over K = Q(Vd), 0 < d < 100. Let E be an admissible curve defined over K = Q(Vd) ) andE is isomorphic to one of the following curves:
Ex E3
E5 Ei
E9 En E13 E\4 Eis
En E19 E21 E23
E25 E21 E29 E30 E31 E32 E33 E35
En E39 E41 E43 E45
l2 E•r.y2 = χ3 + Ax2 + Bx, A = 2: D, 1 < 1 < 46 A B D d j 4ε 6 - 2 ( ε - 1) ε3 (20) 3 4ε 6 ε 64(4ε4 + l) 3 /ε 4 - 14(β - 1) 14(ε-l)ε 6 4ε ε7 64(4ε4+l)3/ε4 2 3 6 7 - ( l + 2ε ) 16ε ε (255)3 7 2(1+2ε 2 ) 1 (256ε2 + ε) 3 ε3 2 3 1 7 -2(l+2ε ) ε (256ε2 + ε) 3 7 8ε-l 16ε2 -ε6 (-15) 3 2 6 7 - ( 8 ε - 1) 16ε -ε (-15) 3 14 - 3(ε- l)/2 16ε -ε3 (-15) 3 3 14 3(ε - 1) ε (255)3 -ε 22 -2(ε - l)/7 4ε ε3 (20) 3 3 2(ε-l)/3 38 4ε ε (20) 3 4 3 41 (3ε-l)/2 ε(ε-l)/2 ε (ε-16) /ε 41 (ε + 43)/10 (ε+ l)/2 -ε (17)% 2 2 7 3 41 -2ε ε ε (256ε + l) /ε 3 2 3 6 (257) 65 2ε -l 16ε ε 2 3 6 (257)3 65 10ε - 5 400ε (5ε) 2 6 3 65 8ε+l 16ε ε (17) 2 6 65 40ε + 5 400ε (5ε) (17)3 (ε+l) 2 /4 (ε + 3)/2 65 -ε3 (8 + ε-)2 2 6 3 2 65 5(ε + 3)/2 25(ε+l) /4 - 5 ε (8 + ε-) -2(2ε2-l) 1 65 ε3 (256ε2 - ε) 3 2 6 3 65 - 1 0 ( 2 ε - l ) 25 5 ε (256ε2 - ε) 3 3 9 3 77 3(ε 3 -l)/10 16ε -ε (-15) 3 3 9 3 77 - 3 ( ε - l ) / 5 -ε ε (255) 86 2(ε- 1)/51 4ε ε3 (20)3
258
SALVADOR COMALADA
Ei = Ef_{ for every missing /, 0 < / < 46, where σ is the nontrivial automorphism of K. In each case, ε stands for the fundamental unit of the respective field. All the curves listed above are ^-admissible except for d = 65, where E$Q , £32 , £35, £36 ? £39 and £40 clearly cannot have a global minimal model. This table has been computed using the results given in this paper. Fields having an even number of divisor classes have been treated in a straightforward way. None of them (excepting d = 65) supplies admissible elliptic curves. Acknowledgment. The author wishes to express his gratitude to Enric Nart for his help, especially in the conception of Theorem 2. REFERENCES
[1] [2] [3] [4] [5] [6] [7] [8]
J. H. E. Cohn, The diophantine equation x4 — Dy2 = 1 , Quart. J. Math. Oxford, 26 (3) (1975), 279-281. S. Comalada and E. Nart, Courbes elliptiques avec bonne reduction partout, C. R. Acad. Sci. Paris Ser. I Math., 305 (1987), 223-224. W. LeVeque, Reviews in Number Theory, Amer. Math. Soc, 2 (1974). B. Setzer, Elliptic curves over complex quadratic fields, Pacific J. Math., 74 (1) (1978), 235-250. , Elliptic curves with good reduction everywhere over quadratic fields and having rational j-invariant, Illinois J. Math., 25 (2) (1981), 233-245. R. J. Stroeker, Reduction of elliptic curves over imaginary quadratic number fields, Pacific J. Math., 108 (2) (1983), 451-463. J. T. Tate, The arithmetic of elliptic curves, Invent. Math., 23 (1974), 179-206. D. B. Zagier, Zetafuncktionen und Quadratische Kόrper, Springer-Verlag, 1981.
Received September 17, 1987 and in revised form July 17, 1989. UNIVERSITAT AUTONOMA DE BARCELONA
BELLATERRA, BARCELONA 08193 SPAIN
