1. Unramified Cyclic Cubic Extensions K6 of a Quadratic Field K2.A quadratic field K2 has no, resp. one, resp. more than one
unramified cyclic cubic extension K6,
iff the 3-class rank of K2 equals 0, resp. 1, resp. exceeds 1.
Since an unramified extension has relative conductor f = 1 without any prime divisors,
the only possibility for ambiguous principal ideals of K6 are
extension ideals of K2 that become principal in K6.
This fact excludes the Principal Factorization Types BETA, GAMMA, EPSILON,
discussed in the previous section
and only the Types ALPHA and DELTA remain.
Thus we have three possible cases of class number relations:
1. Either a real quadratic type ALPHA field with D2 > 0 and
h6 = (1 / 9)*h2*h32
2. Or a real quadratic type DELTA field with D2 > 0 and
h6 = (1 / 3)*h2*h32
3. Or a complex quadratic type ALPHA field with D2 < 0 and
h6 = (1 / 3)*h2*h32
2. The Galois Group of (K6)1 over K2.Arnold Scholz investigated the Galois group G = Gal((K6)1|K2) of
the 1st Hilbert 3-class field (K6)1 of K6 over K2
to get the structure of its abelian normal subgroup A = Gal((K6)1|K6) of index 3,
which is isomorphic to the 3-class group Syl3C(K6) of K6.
This is an application of class field theory,
since (K2)1 is the maximal abelian unramified 3-extension of K2
and thus the subgroup U = Gal((K6)1|(K2)1) of G = Gal((K6)1|K2)
with factor group G/U = Gal((K2)1|K2) = Syl3C(K2)
must be the minimal subgroup of G with abelian factor group, i. e.,
must coincide with the commutator subgroup G' of G.
Further, G is a 2-stage metabelian 3-group,
since G' = Gal((K6)1|(K2)1)
as a subgroup of the abelian normal subgroup A = Gal((K6)1|K6) = Syl3C(K6)
is abelian, i.e., G'' = 1.
2.1. Quadratic Fields K2 of 3-Class Rank 1.First, Scholz shortly indicates, that the case where
K2 is a quadratic field of 3-class rank 1,
i. e., with 3-class group of type (3x) (x >= 1)
and thus with 3-class number h2 = 3x
by assumption, we have G/G' = (3x)
and since in the present context the Frattini subgroup has the shape F(G) = G'*G3,
we get the 3-elementary factor group G/F(G) = G/G'*G3 = (3) of 3-rank 1.
According to the Basis Theorem of Burnside,
the minimal number of generators of G is 1, i. e.,
G is cyclic, G' = 1, (K6)1 = (K2)1,
G = (3x), Syl3C(K6) = A = (3x-1), and by the class number relation
(note that type ALPHA is impossible here for D2 > 0)
3x-1 = h6 = (1/3)*h2*h32 = (1/3)*3x*h32
and finally h3 = 1.
2.2. Quadratic Fields K2 of 3-Class Rank 2.Next, to be non-trivial but specific, Scholz assumes that K2 is a quadratic field
with 3-class group of (elementary abelian) type (3,3),
and therefore h2 = 9.
For this situation, we can determine the structure of the Galois group
G = Gal( (K6)1 | K2 )
of the 1st Hilbert 3-class field (K6)1 of K6 over K2.
All we need to know is
(1) the 3-class number h3 of the absolute cubic subfield K3 of K6
(2) the single principalization type k(j) of K2 in K6 only,
i. e., only the relevant member of the full quadruplet principalization type (k(1),...,k(4)) of K2.
In all numerical examples that occurred up to now in the
real and complex case,
only the following few configurations were discovered.
1. h3 = 3 and k(j) != 0 is not a fixed point (k(j) != j)
then C6 = (9,3) and G = G(4)(0,+-1,0) in ZEF 2a(4,4) < CF(4,4,3)
2. h3 = 3 and k(j) != 0 is a fixed point (k(j) = j)
then C6 = (9,3) and G = G(4)(1,0,0) in ZEF 2a(4,4) < CF(4,4,3)
3. h3 = 3 and k(j) = 0
then C6 = (3,3) and G = G(3)(0,0,0) in ZEF 1a(3,3) < CF(3,3,3)
4. h3 = 9 and k(j) != 0 is not a fixed point (k(j) != j)
then C6 = (27,9) and G = G(6)(0,+-1,0) in ZEF 2a(6,6) < CF(6,6,3)
5. h3 = 9 and k(j) != 0 is a fixed point (k(j) = j)
then C6 = (27,9) and G = G(6)(1,0,0) in ZEF 2a(6,6) < CF(6,6,3)
6. h3 = 9 and k(j) = 0
then C6 = (9,9) and G = G(5)(0,0,0) in ZEF 2a(5,5) < CF(5,5,3)
7. h3 = 27 and k(j) != 0 is not a fixed point (k(j) != j)
then C6 = (81,27) and G = G(8)(0,+-1,0) in ZEF 2a(8,8) < CF(8,8,3)
1) The 3-class group C3 of the absolute cubic subfield is always cyclic.
2) Group theoretically, there is a difference between the totally real principalization types
(2,0,0,0), i. e. a.1, and (1,0,0,0), i. e. a.2.
The three sextic fields with k(j) = 0 always belong to configuration 3,
but the exceptional field with k(j) != 0 to 1 in the former case, and to 2 in the latter.
3) For the totally real principalization type (0,0,0,0), i. e. a.3,
three sextic fields belong to configuration 3, and one exceptional field belongs to 6.
4) The most frequent complex principalization type is D,
where all four sextic fields belong to configuration 1, except the fixed points, which belong to 2.
5) For the complex principalization types H and G.19,
all four sextic fields belong to configuration 1.
6) Configurations 4, resp. 5, occur for certain complex principalization types:
a single exceptional field for principalization types E and G.16 and the first variant of type H,
and two exceptional fields for principalization type F, the second variant of type H,
and the variants of types G.16 and G.19.
7) Finally, configuration 7 is unique, up to now.
It occurs for a single field with discriminant D3 = -159208
and with a variant of principalization type F.13.
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