540365 is the smallest discriminant of a real quadratic field K
with capitulation type c.21 (0,2,3,1),
symbolic order X3, and non-terminal group G=Gal(K2|K) in CBF2a(5,6).
Important remark:
This example shows that the strict leaf conjecture
that only terminal nodes (leaves) can occur as Gal(K2|K) is false.
However, it might still be true that a group cannot occur as Gal(K2|K),
if one of its successors is a leaf in CBFb with the same capitulation type.
We give the complete data needed to determine the capitulation type and the group G=Gal(K2|K).
Computed on January 01, 2008, at the University of Manitoba's Computer Centre, Winnipeg City [1,2].
Counter, n = 66
|
Discriminant, d = 540365
|
3-class group of type (3,3)
|
3-class number, h = 9
|
Conductor, f = 1
|
Fundamental unit, e0 = (U0 + V0*x)/T0, with x2 = d, and regulator, R
|
U0
|
V0
|
T0
|
|
R
|
15437
|
21
|
2
|
|
9.6
|
The non-Galois absolute cubic subfields (L1,L2,L3,L4)
of the four unramified cyclic cubic relative extensions N|K
|
Regulators, R
|
39.1
|
116.3
|
118.9
|
131.6
|
Class numbers, h
|
9
|
3
|
3
|
3
|
Polynomials, p(X) = X3 - C*X - D, with d(p) = i2*d
|
(C,D)
|
(122,499)
|
(68,163)
|
(768,7247)
|
(642,4961)
|
Indices, i
|
1
|
1
|
27
|
27
|
Fundamental units, e1 = (U1 + V1*x + W1*x2)/T1, with P(x) = 0
|
U1
|
16
|
-1487
|
238
|
-10294
|
V1
|
3
|
-394
|
32
|
-812
|
W1
|
0
|
60
|
1
|
41
|
T1
|
1
|
1
|
9
|
9
|
Fundamental units, e2 = (U2 + V2*x + W2*x2)/T2, with P(x) = 0
|
U2
|
83
|
-1936
|
-2870111
|
462048
|
V2
|
26
|
-531
|
-164494
|
75971
|
W2
|
2
|
80
|
7840
|
2660
|
T2
|
1
|
1
|
9
|
3
|
Splitting primes, q
|
31
|
97,181
|
19
|
67,229
|
Associated quadratic forms, F = a*X2 + b*X*Y + c*Y2
|
(a,b,c)
|
(31,715,-235)
|
(-181,651,161)
|
(19,713,-421)
|
(-229,487,331)
|
Represented primes, q
|
31
|
-181
|
19
|
-229
|
Associated ideal cubes, (x + y*d1/2)/2, with 4*q3 = x2 - d*y2
|
(x,y)
|
(649,1)
|
(1661,7)
|
(5143,7)
|
(241747,329)
|
Principalization
|
0
|
2
|
3
|
1
|
Capitulation type c.21: (0,2,3,1)
|
Group G in CBF2a(5,6)
|
Contents
|