A380103 Minimal conductors c of cyclic cubic number fields K with elementary bicyclic 3-class group Cl_3(K)=(3,3) and second 3-class group M=Gal(F_3^2(K)/K) of assigned coclass cc(M)=0,1,2,3,...
657, 2439, 7657, 41839, 231469
Offset: 1
Examples
We have M abelian for c=657=9*73 (two fields in a doublet), cc(M)=1 for c=2439=9*271 (two fields in a doublet), cc(M)=2 for c=7657=13*19*31 (three fields in a quartet), cc(M)=3 for c=41839=7*43*139 (two fields in a quartet), cc(M)=4 for c=231469=7*43*769 (four fields in a quartet). If the conductor c has two prime divisors, then cc(M)=1. For cc(M) > 1, exactly three prime divisors of the conductor c are required.
Links
- D. C. Mayer, The distribution of second p-class groups on coclass graphs, arXiv:1403.3833 [math.NT], 2014; J. Théor. Nombres Bordeaux 25 (2013), 401-456.
- Daniel Constantin Mayer, Magma program "CyclCoClass.m" with endless loop
Crossrefs
Analog of A379524 for real quadratic fields.
Programs
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Magma
// See Links section.
Formula
According to Theorem 3.12 on page 435 of "The distribution of second p-class groups on coclass graphs", the coclass of the non-abelian 3-group M is given by cc(M)+1=log_3(h_3(E_2)), where h_3(E_2) is the second largest 3-class number among the four unramified cyclic cubic extensions E_1,..,E_4 of the cyclic cubic field K, and log_3 denotes the logarithm with respect to the basis 3. An exception is the abelian 3-group A=(3,3) with correct cc(A)=1, where the FORMULA yields cc(A)=0.
Comments