Joint results 2002

of Karim Belabas, Aïssa Derhem, and Daniel C. Mayer

On these pages, we present most recent results of our joint research, directly from the lab.
Basic bibliography:
K. Belabas, A fast algorithm to compute cubic fields, Math. Comp. 66 (1997), 1213-1237
A. Derhem, Capitulation dans les extensions quadratiques de corps de nombres cubiques cycliques, Thèse de doctorat, Université Laval, Quebec, 1988
D. C. Mayer, Multiplicities of dihedral discriminants, Math. Comp. 58 (1992), 831-847 and S55-S58
E-mail addresses:

Sextic fields associated with quadratic 3-ray class groups (2002/02/11)
Dan (02/01/28): By your intriguing email of January 23rd, 2002,
where you showed how "PARI" calculates
ray class groups for huge quadratic discriminants d and conductors f,
I was reminded of a conceptual difficulty
in the interpretation of the p-ray class rank mod f for an odd prime p.

1. The formula for the p-ray class rank mod f, rho'(f,p), of a quadratic field K with discriminant d
is very similar to the formula for the p-ring class rank mod f, rho(f,p):

rho'(f,p) = rho(p) + t + t' + w + w' - delta'_p(f)

rho(f,p) = rho(p) + t + w - delta_p(f)

where rho(p) denotes the ordinary p-class rank of K,
t and w are dependent on the decomposition behavior
of the rational prime factors q of the conductor f in K,

t = #{ q prime | q != p, v_q(f) >= 1, q congruent (d/q)(mod p) }

w = 0, if v_p(f) =0 or v_p(f) = 1, (d/p)=+-1,
w = 1, if v_p(f) =1, p|d or v_p(f) >= 2, p > 3 or p=3, v_3(f) >= 2, d incongruent -3(mod 9),
w = 2, if p = 3, v_3(f) >= 2, d congruent -3(mod 9)

t' and w' are independent from the decomposition behavior
of the rational prime factors q of the conductor f in K,

t' = #{ q prime | q != p, v_q(f) >= 1, q congruent +1(mod p) }

w' = 0, if v_p(f) <= 1,
w' = 1, if v_p(f) >= 2

and the p-defects are defined similarly as in my Vienna Congress 2001 presentation

delta'_p(f) = F_p-dim( I_p(f) / I_p(f) intersection S_fK(f)^p ) (p-ray defect mod f of K)

delta_p(f) = F_p-dim( I_p(f) / I_p(f) intersection R_fK(f)^p ), (p-ring defect mod f of K)

with the group S_f = { a in K | a congruent 1(mod f) } of generators of the ray ("Strahl") mod f of K
and the group R_f = S_fQ(f) of generators of the ring mod f of K.

Obviously we have the relation delta_p(f) <= delta'_p(f) <= t + t' + w + w'.

2. Before I try to apply this formula to your "PARI"-example,
I must mention some by far more simple cases where I can express my conceptual difficulties.

The first occurrences of simultaneous strict 3-rank increments

rho'(f,3) > rho(f,3) > rho(3)

can already be found in Ian Angell's tables of 1972, 1975,
but they have never been studied so detailed up to now.
They are in the complex case:

d=-4, f=3^2, f^2*d=-324 with t=t'=0, w=w'=1
d=-7, f=3^2, f^2*d=-567 with t=t'=0, w=w'=1
d=-8, f=3^2, f^2*d=-648 with t=t'=0, w=w'=1
d=-4, f=13, f^2*d=-676 with t=t'=1, w=w'=0
d=-19, f=7, f^2*d=-931 with t=t'=1, w=w'=0

where rho(3)=0, since |d| < 23, and thus certainly delta'_3(f)=delta_3(f)=0,

in the cyclotomic case:

d=-3, f=3^2, f^2*d=-243 with t=t'=0, w=2, w'=1 and delta'_3(f)=delta_3(f)=1
d=-3, f=19, f^2*d=-1083 with t=t'=1, w=w'=0 and delta'_3(f)=delta_3(f)=0

where again rho(3)=0, since |d| < 23, but the behavior of the 3rd roots of unity enables positive 3-defects,

and in the real case:

d=5, f=2*3^2, f^2*d=1620 with t=1, t'=0, w=w'=1 and delta'_3(f)=delta_3(f)=1
d=24, f=3^2, f^2*d=1944 with t=t'=0, w=2, w'=1 and delta'_3(f)=delta_3(f)=1
d=29, f=3^2, f^2*d=2349 with t=t'=0, w=w'=1 and delta'_3(f)=delta_3(f)=0
d=53, f=7, f^2*d=2597 with t=t'=1, w=w'=0 and delta'_3(f)=delta_3(f)=0
d=33, f=3^2, f^2*d=2673 with t=t'=0, w=2, w'=1 and delta'_3(f)=delta_3(f)=1

where also rho(3)=0, since d < 229, but the behavior of the fundamental unit enables positive 3-defects.

In all of these examples we have (using the formulas in 1.)

rho'(f,3)=2 > rho(f,3)=1 > rho(3)=0

and thus there exist (compare my scenario "Abelian network over K" in my Vienna Congress 2001 presentation)

a unique cyclic cubic extension N of K contained in the 3-ring class field F mod f of K,
which is the normal field with Gal(N|Q)=S_3 of the unique non-Galois cubic field L with discriminant f^2*d
(up to conjugates),
since m_3(d,f) = ( 3^{rho(f,3)} - 1 ) / 2 = 1,

three distinct further cyclic cubic extensions N',N'',N''' of K contained in the 3-ray class field F' mod f of K,
since m'_3(d,f) = ( 3^{rho'(f,3)} - 1 ) / 2 = 4,
one of them N'=K*L' the compositum of K
with the unique absolutely cyclic cubic field L' with discriminant f_0^2,
where f_0=f except in the case f=2*3^2 with f_0=3^2.

Obviously N',N'',N''' must either be absolutely abelian,
which is true for N' with Gal(N'|Q)=C(2)*C(3)=C(6)
or non-Galois.
But I have no idea how to explain the fields N'' and N'''!
Karim (02/01/29): Your argument proves that they don't have conductor f, but a divisor thereof.
Dan: Or does not every subgroup of index 3 in the ray class group mod f
correspond to a cyclic cubic extension of K?
Karim: They do, but the general result
[Hasse, the standard paper on cubic fields via class field theory]
is that, then the associated field is

1) Galois over Q iff s H = H [s = non-trivial automorphism of the quad. field]
2) has Galois group S_3 iff s acts non trivially on Cl_f / H.

In the latter case H has exact conductor f iff the field has the expected discriminant.

The general easy lemma is:
given K/k cyclic (Galois group generated by s),
L/K abelian associated to an admissible pair (f,H) [H a subgroup of Cl_f],
L/k is

1) Galois iff sH = H
2) abelian iff s acts trivially on Cl_f / H.
Dan: Please let me know, if you have information on systematic tables of sextic fields.
Karim: You can try
Dan (02/02/11): With the aid of Megrez's sextics,
I have found some traces concerning my conceptual problem
of interpreting the ray class rank correctly,
but I still do not have the complete insight.

First, I used Hilbert's Theorem 39
to compute the discriminant of a sextic N with cubic subfield L and quadratic subfield K:

d(N) = d(K)^3 * Norm_{K|Q}( D(N|K) ) = d(K)^3 * Norm_{K|Q}( f(N|K)^2 )
d(N) = d(L)^2 * Norm_{L|Q}( D(N|L) ) = (f^2)^2 * Norm_{L|Q}( f(N|L) )

==> d(N) = f^4 * d(K)^3

Then I considered the simplest cases with simultaneous strict rank increments
rho'(f,3)=2 > rho(f,3)=1 > rho(3)=0,
such as d(K) = -3 and f = 3^2, i. e., d(N) = 3^8 * (-3)^3.

Here, we have t=t'=0, w=2, w'=1, delta_3(9)=delta_3(3)=1 (pure cubic!)
and thus rho(f,3)=w-delta_3(9)=2-1=1 and rho'(f,3)=w+w'-delta'_3(9)=2+1-1=2.

Megrez's table of totally complex sextics with signature (0,3) includes:

(*) the normal field N with Gal(N|Q)=S(3) of the non-Galois (pure) cubic L=Q(3^{1/3})
which is a subfield of the 3-ring class field F mod 3^2 of K=Q((-3)^{1/2})
and can be represented as N=Q((-3)^{1/6}) in this special case

-177147 = 3^8*(-3)^3:
generating polynomial...x^6+3,
integral basis...[1,x,x^2,1/2*x^3+1/2,1/2*x^4+1/2*x,1/2*x^5+1/2*x^2],

(*) the compositum N'=K*L' of K with the absolutely cyclic cubic L' with discriminant 3^4=81,
which is absolutely Abelian with Gal(N'|Q)=C(6) and thus
a subfield of the 3-ray class field F' mod 3^2 of K

-177147 = 3^8*(-3)^3:
generating polynomial...x^6-3*x^3+3,
integral basis...[1,x,x^2,x^3,x^4,x^5],

(*) a further cyclic cubic extension N'' of K with a divisor of 3^2 as conductor

-19683 = 3^6*(-3)^3:
generating polynomial...x^6-x^3+1,
integral basis...[1,x,x^2,x^3,x^4,x^5],

(*) no further cyclic cubic extension N''' of K

-2187 = 3^4*(-3)^3: does not exist

I have marked the table entries, which I do not understand, with "?"
Maybe you know their meaning or where to get a documentation of the tables.
I do not see, how the field N'' arises and what its conductor and Galois group!
Further I would expect a fourth field N'''!

I have the feeling that these are important questions
for a full understanding of the ray class groups.

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