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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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| Multiplicities of quintic discriminants (2002/01/31) |
| Dan (2002/01/31): I guess, PARI should be able to compute quintic discriminants. |
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Karim (2002/01/31):
Definitely. It's the same program
(I assumed dihedral Galois group D_2p for p prime, out of lazyness...). |
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Dan:
For example, consider the Abelian 5-network
over the real quadratic field K with discriminant d = 5.
a) How many cyclic quintic extensions N of K, with Gal(N|Q)=D(2*5) the dihedral group of order 10, do there exist for the conductors f=5*11,5*11*31,5*11*31*41,5*11*31*41*61? The non-Galois quintic subfields L of these normal fields N have the "root"-discriminant sqrt(D)=f^2*d, i. e., sqrt(D)=15125, sqrt(D)=14535125, sqrt(D)=24433545125, sqrt(D)=90917221410125. |
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Karim:
f=5*11,
...answer: 1
f=5*11*31, ...answer: 3 f=5*11*31*41, ...answer: 13 f=5*11*31*41*61, ...answer: 51 [all this provided I did not make a mistake in my simple-minded program, of course] |
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Dan:
b) Folklore tells that a mathematician at Paris determined,
as part of the requirements for his Ph. D. thesis (about 1992), the minimal (5-admissible) prime conductor q cong (d/q)(mod 5), for which there exists a quintic field L with discriminant D=(q^2*d)^2, where d = 5. (Unfortunately, I have forgotten the name of the mathematician.) Warning: there exist smaller composed conductors f < q_{min}, such that D=(f^2*d)^2 has positive multiplicity. I wonder, if PARI, extended with your implementation of Hasse's lemma, is able to solve these problems. |
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Karim: f = 55 = 5*11 is the first composed one. Then f = 155 = 5*31, f = 205 = 5*41.
q = 211 is the first prime. |
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