I cranked out a small table of pairs of irreducible monic quartics in
Z[x] with different splitting fields but the same resolvent cubic, and
positive square free discriminant. Among these was
x^4 + 5*x^3 + 4*x^2 - 5*x - 4
x^4 + x^3 + 4*x^2 + 7*x + 4
Resolvent cubic x^3 - 4*x^2 - 9*x + 11
Discriminant = 10889 factors [10889]
The discriminant is prime. This looks pretty innocuous, but it gave
(assuming the bnfinit() and bnrinit() results are correct) the
following. As with a previous example, k1 is the Hilbert Class Field
of k0, k2 that of k1, and k3 that of k2:
k0 = Q(sqrt(10889)) has class number 3
k1 = splitting field of the resolvent cubic has G(k1/Q) = S3, and
class group V4, the four-group
k2 = splitting field of x^4 + 5*x^3 + 4*x^2 - 5*x - 4 has G(k2/Q) =
S4, and class number 2
The join of k2 and the splitting field of the other quartic, x^4 + x^3
+ 4*x^2 + 7*x + 4, is equal to the Extended Hilbert Class Field of k1
The Hilbert Class Field k3 of k2 has Galois group G(k3/Q) = GL(2,3).
It has non-normal, non-conjugate but arithmetically equivalent octic
subfields defined by the polynomials
x^8 - 4*x^7 - 12*x^6 + 50*x^5 + 39*x^4 - 166*x^3 - 10*x^2 + 102*x + 11
and
x^8 - 4*x^7 - 20*x^6 + 59*x^5 + 182*x^4 - 200*x^3 - 703*x^2 - 293*x +
131
These have Galois group 8T23, and the two fields have field
discriminant 10889^3 and class number 1.
As with a previous example, k3 should have class number 1.
