Ask Your Question
1

Compute Galois group when base field is not $\mathbb{Q}$.

asked 2022-05-08 15:02:50 +0200

hfy880916 gravatar image

I am trying to compute the Galois group $\mathrm{Gal}(L/k)$ where $L/k$ is a Galois extension of number fields. $k$ is not necessarily $\mathbb{Q}$.

k = NumberField(x^2 + 1, 'i', embedding = I)
F = k.extension(x^4 - 17, 'a', embedding = 17^(1/4))
F.<a> = F.absolute_field()
G = F.galois_group()
k_gal = []
for g in G:
    if g(i) == i:
        k_gal.append(g)
Gal_F_k = G.subgroup(k_gal)

But Sage returns a TypeError.

TypeError: a fails to convert into the map's domain Number Field in a with defining polynomial x^8 + 4*x^6 - 28*x^4 + 208*x^2 + 256, but a `pushforward` method is not properly implemented

How could I fix my code?

edit retag flag offensive close merge delete

1 Answer

Sort by ยป oldest newest most voted
1

answered 2022-05-09 20:02:02 +0200

dan_fulea gravatar image

If i am correctly understanding what happens in the code, we use $$j=\sqrt {-1}\ ,\ a=\sqrt[4]{17}\ ,$$ and the tower of algebraic number fields $$ \begin{array}{c} F=\Bbb Q(a,j) \\ \uparrow \\ k = \Bbb Q(j) \\ \uparrow \\ \Bbb Q \end{array} $$ and ask for the group $G=\text{Gal}(F:\Bbb Q)$, then check for each $g$ in $G$ if it stabilizes $j$.

This can be done by no longer using $k$ as follows:

R.<x> = PolynomialRing(QQ)
F.<j, a> = NumberField([x^2 + 1, x^4 - 17])
L.<w> = F.galois_closure()
G = L.galois_group()
J = (x^2 + 1).roots(ring=L, multiplicities=False)[0]
A = (x^4 - 17).roots(ring=L, multiplicities=False)[0]
Stab_J = [ g for g in G if g(J) == J ]

Note that working like this we obtain a list, not a subgroup of $G$.

sage: Stab_J
[(), (1,2,4,3)(5,7,8,6), (1,3,4,2)(5,6,8,7), (1,4)(2,3)(5,8)(6,7)]
sage: list(G)
[(),
 (1,2,4,3)(5,7,8,6),
 (1,3,4,2)(5,6,8,7),
 (1,4)(2,3)(5,8)(6,7),
 (1,5)(2,6)(3,7)(4,8),
 (1,6)(2,8)(3,5)(4,7),
 (1,7)(2,5)(3,8)(4,6),
 (1,8)(2,7)(3,6)(4,5)]
sage:

Here, we use $J,A$ - which are elements in the "abstract field" $L$, so that there are maps $j\to J$ and $a\to A$ from $F$ to $L$.

edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2022-05-08 15:02:22 +0200

Seen: 16 times

Last updated: May 09