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| On these pages, we present most recent results of our joint research, directly from the lab. |
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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 |
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E-mail addresses:
Karim.Belabas@math.u-psud.fr aderhem@yahoo.fr danielmayer@algebra.at |
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| Jordi Quer's complex quadratic field with 3-rank 5 (2002/01/29) |
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Karim (2002/01/23):
For instance, with d = -5393946914743, f=7*13*19*31,
I get a ray class group of type
[686160, 720, 360, 12, 6, 3, 3, 3, 3] (read Z/686160 x Z/720...), which has 9841 subgroups of index 3, 1944 of which correspond to the exact conductor f. [Using PARI, all the above is instantaneous.] |
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Dan (2002/01/28):
Finally, I come to your example concerning Jordi Quer's discriminant d = -5393946914743 with rho(3)=5
and f=7*13*19*31, where "PARI" computed the ray class group mod f of K as C(2^4*5*953*3^2)*C(2^4*5*3^2)*C(2^3*5*3^2)*C(2^2*3)*C(2*3)*C(3)*C(3)*C(3)*C(3) with rho'(f,3)=9 and thus with (3^9 - 1) / 2 = 9841 subgroups of index 3 (and arbitrary conductors). Since all prime factors of the conductor are congruent +1(mod 3), we have t'=4, and since 3 does not divide f, we have w=w'=0. I am not quite sure about t. Using my pocket calculator (and the quadratic reciprocity law) I had the impression that 13, 31 split in K, whereas 7, 19 remain inert (and are therefore not 3-admissible), whence t=2. |
| Karim: This is correct. |
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Dan:
By means of the formula in 1. we get
rho'(f,3) = 9 = rho(3) + t + t' + w + w' - delta'_3(f) = 5 + 2 + 4 + 0 + 0 - delta'_3(f) and I conclude delta'_3(f)=2. If also delta_3(f)=2, which I conjecture, then there are no cubic fields with discriminant f^2*d at all, since rho(f,3) = rho(3) + t + w - delta_3(f) = 5 + 2 + 0 - 2 = 5 = rho(3) (no rank increment). |
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Karim (2002/01/29):
I indeed get multiplicity 0 in 20s computing time for the huge "discriminant"
-5393946914743 * (7*13*19*31)^2 which I had chosen at random to illustrate my point that I could compute individual multiplicities for values well outside the range of my initial program. |
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Dan:
Unfortunately, I do not understand, how "PARI" found that
1944=2^3*3^5 of the 9841 subgroups of index 3 have the exact conductor f. |
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Karim:
In a stupid way:
by checking that the canonical projection pi: Cl_f --> Cl_g sends H to pi(H) which doesn't have index 3 in Cl_g, for any strict (ideal) divisor g of f. No theoretical insight to be gained here. |
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Dan:
Further, I have no interpretation of this huge number 1944,
since I only see that the 3-ray class field F' mod f of K contains the composita N' of K with the 8 cyclic cubic fields L' with conductor f (and discriminant f^2) |
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Karim:
Here, there are 8 stable subgroups (of the right conductor),
apparently all of which correspond to abelian (= C(6)) sextic fields. Could not get an equation for them... |
| Dan: I think, the ray class groups will provide us with interesting puzzles, for a while. |
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