/*
Base fields:
orbit 28c: Q(a)
orbits 28ab: Q(a,r3)
SIC fields:
orbit 28c (ray class field): Q(a,r1,r2,t,b1,b2,b3,i) (degree 576)
orbits 28ab: Q(a,r1,r2,r3,t,b1,b2,b3,b4,i) (degree 2304)
*/
P0:=PolynomialRing(RationalField());
Fa:=NumberField(x0^2-29);
Pa:=PolynomialRing(Fa);
Fr1:=NumberField(xa^2-2);
Pr1:=PolynomialRing(Fr1);
Fr2:=NumberField(xr1^2-7);
Pr2:=PolynomialRing(Fr2);
Fr3:=NumberField(xr2^2-5);
Pr3:=PolynomialRing(Fr3);
Ft:=NumberField(16*xr3^3-4*r1*(r2-1)*xr3^2-4*(r2+1)*xr3+r1*(r2+3));
Pt:=PolynomialRing(Ft);
Fb2:=NumberField(xt^2+(a+5));
Pb2:=PolynomialRing(Fb2);
Fb3:=NumberField(xb2^3-126*xb2-378: Check:=false); // ~5 seconds to check irreducibility
Pb3:=PolynomialRing(Fb3);
Fb4:=NumberField(xb3^2+20*a+50+(6*a+36)*r3: Check:=false); // ~200 seconds to check irreducibility
Pb4:=PolynomialRing(Fb4);
Fb1:=NumberField(xb4^2+(a+1): Check:=false); // ~5000 seconds to check irreducibility
Pb1:=PolynomialRing(Fb1);
F*:=NumberField(xb1^2+1: Check:=false); // irreducibility not checked directly. See below for indirect check.
P:=PolynomialRing(F);
tau:=((-2*r1*r2+6*r1)*t^2+(-3*r2+8)*t+(-1/2*r1*r2+1/2*r1))*i-t;
/*
CHECKING IRREDUCIBILITY OF THE DEFINING POLYNOMIAL OF F.
A direct check is likely to be somewhat time consuming. So we did an indirect check, by looking at the subfields
F = Q(a,r1,r2,t,b2,b1,i) (degree 192)
Fb4 = Q(a,r3,b4) (degree 8)
Fb3 = Q(b3) (degree 3)
Consider first G = F ∩ Fb4. The degree of G must divide GCD(192,8) = 8.
We have that a ∈ G, and r3 ∉ G (5 seconds to check). So the degree of G must be 2 or 4.
The defining polynomials of the subfields of Fb4 are
[
x0^8 + 7*x0^7 + 4*x0^6 + 23*x0^5 - 104*x0^4 + 15*x0^3 - 25*x0^2 - 80*x0 - 20,
x0^2 + x0 - 7,
x0^2 + x0 - 1,
x0^2 + x0 - 36,
x0^4 - 17*x0^2 + 36
]
from which we see that the only subfield of Fb4 of degree 4 is the one whose defining polynomial is
x0^4 - 17*x0^2 + 36. This splits over F into two quadratic factors (17 seconds):
[
x^2 - a*x + 6,
x^2 + a*x + 6
]
So G is degree 2. Since G contains a this means G = Q(a). It follows that
F ∨ Fb4 = Q(a,r1,r2,r3,t,b2,b1,b4,i) is degree 4*192 = 768.
Now consider G = Fb3 ∧ (F ∨ Fb4). The degree of G must divide GCD(768,3) = 3. If Degree(G) was 3 it
would mean that b3 ∈ F ∨ Fb4 = Q(a,r1,r2,r3,t,b2,b1,b4,i). But that would imply
Q(a,r1,r2,t,b2,b3,b1,i) was a subfield of F ∨ Fb4 = Q(a,r1,r2,r3,t,b2,b1,b4,i), and that its degree was
a divisor of 768. However, Q(a,r1,r2,t,b2,b3,b1,i) is the ray class field, with degree 9*64=576, and 576
is a not a divisor of 768. We conclude that Degree(G) is 1, implying that
Degree( Q(a,r1,r2,r3,t,b2,b3,b1,b4,i) is 2304, and that each of the defining polynomials in the tower in
d28Field.mgm is irreducible over the field below it.
*/
*