Two recent postings ["Arithmetically equivalent?" and "Climbing the
tower"] had a common origin, which may be of interest. A couple of
years ago, I was looking at a question about "crossed-ladders problem"
quartics sent me by Kent Holing, and a curious issue arose.
As is well known, a monic quartic polynomial
P(x) = x^4 + a*x^3 + b*x^2 + c*x + d
has the resolvent cubic
R(y) = y^3 - b*y^2 + (a*c - 4*d)*y + 4*b*d - c^2 - a^2 * d .
The issue was, when two monic quartics could have the *same* resolvent
cubic.
If r1, r2, r3, r4 are the zeroes of P(x), then r1*r2 + r3*r4 is one of
the zeroes of R(y). This expression is invariant if r1, r2, r3, r4
are all multiplied by -1. It is also invariant if (r1 + r2 + r3 + r4)/
2 is subtracted from each. Thus,
P(x), P(-x), P(x - a/2), and P(-x - a/2) all have the same resolvent
cubic.
Of course, these all define the same extension of the field of
coefficients. A few other special cases suggested themselves. Hoping
for something involving elliptic curves, I asked Prof. Allan Macleod
at (then) Paisley University, who had a web site devoted to
applications of elliptic curves to problems in recreational number
theory. He indicated that the conditions for additional tuples
yielding the same resolvent cubic involved an equation of the form
sextic = square
which is typically of genus 2, so for most given integer 4-tuples (a,
b, c, d), one can expect only finitely many "equivalent" quartics in
Q[x] with the same resolvent cubic as P(x).
However, Prof. Macleod also did a simple search, which turned up the
pair
P1(x) = x^4 + 2*x^3 + 2*x^2 + 3*x + 7 and P2(x) = x^4 + 5*x^3 +2*x^2
-10*x - 7.
Both are irreducible mod 2, so are irreducible in Z[x], with Galois
group not contained in A4. It is easily verified that both P1(x) and
P2(x) have the same resolvent cubic,
R(y) = y^3 - 2*y^2 - 22*y + 19
which is irreducible mod 3, so also irreducible in Z[y]. It follows
that both P1(x) and P2(x) have Galois group S4. Of course, all three
polynomials have the same polynomial discriminant, which is
d = 50437 = 31 * 1627.
However, it is also easily verified that P1(x) has no real zeroes and
P2(x) has four real zeroes, so they have different splitting fields.
Because the common polynomial discriminant is positive and squarefree,
and P2(x) is totally real, it follows immediately that if
k0 = Q(sqrt(d)), k1 is the splitting field of R(y), and k2 is the
splitting field of P2(x), then
k0, k1, k2 is a tower of unramified extensions, and that
k1/k0 is Abelian with Galois group C3, and k2/k1 is Abelian with
Galois group V4 = C2[x]C2, the four-group. Thus, k1 is contained in
the Hilbert Class Field of k0, and k2 is contained in the Hilbert
Class Field of k1.
Thus, k0 has class number divisible by 3, and k1 has class number
divisible by 4, and its class group has a 2-rank of at least 2. The
use of Pari-GP shows that these are in fact equalities. The class
number of k0 is exactly 3, and the class number of k1 is 4, its class
group being V4.
Pari-GP indicated that k2 had class number 2, but produced no evidence
that that its Hilbert Class Field k3 had class number greater than 1.
Prof. Franz Lemmermeyer pointed out, giving references, that, assuming
the class number of k2 is indeed 2, k3/k1 is a quaternion extension,
and consequently k3 must have an odd class number. Using Kuroda's
class number formula for 4-group extensions, then (assuming that the
class number computations for the degree-24 subfields of k3 are
correct) the class number of k3 must also be a power of 2. The only
odd power of 2 is 1, so k3 is the end of the tower.
The Galois group of k3/Q is GL(2.3). The two conjugacy classes of
octic subfields with normal closure k3 are the arithmetically
equivalent octic fields I asked about. The Galois group of the
defining octics is 8T23.
Bill Allombert allocated a huge amount of resources and ran Pari-GP's
bnfinit() on k3. The calculation took over 152 hours, and indicated
that (modulo the GRH) k3 indeed has class number 1.
I haven't forgotten about P1(x). Joining its splitting field to k2
defines at least part of the Extended Hilbert Class Field of k1.
Anyone who wants to -- have at it!
