# Galois Thoery.

• Apr 24th 2010, 09:14 PM
Galois Thoery.
I'm studying for an abstract algebra exam, and I came a cross this question that I dont understand, I feel lost. I wonder if u can help me with it . I need a detailed explanation please.
thank you.

1.
a) let K be the splitting field of the polynomial x^4-2x^2-1. let F=Q(i). Show that K is a Galois extension of F. Determine Gal(k\F).
b) determine Gal(K\Q). determine the lattice of subgroups of Gal(K\Q) and give the corresponding fixed subfields of K.
• Apr 24th 2010, 11:46 PM
aliceinwonderland
Quote:

I'm studying for an abstract algebra exam, and I came a cross this question that I dont understand, I feel lost. I wonder if u can help me with it . I need a detailed explanation please.
thank you.

1.
a) let K be the splitting field of the polynomial x^4-2x^2-1. let F=Q(i). Show that K is a Galois extension of F. Determine Gal(k\F).
b) determine Gal(K\Q). determine the lattice of subgroups of Gal(K\Q) and give the corresponding fixed subfields of K.

If you find the Gal(K/Q), you first need to find the resolvent cubic of $\displaystyle p(x)=x^4-2x^2-1$. The resolvent cubic of p(x) is

$\displaystyle f(x)=x^3 + 2x^2+4x+8$. Since $\displaystyle f(x)=(x+2)(x^2+4)$, we see that f(x) factors as linear times irreducible quartic. Check p(x) irreducible over $\displaystyle \mathbb{Q}(\sqrt{D})$. Thus $\displaystyle Gal(K/\mathbb{Q})=D_4$ (link).

If you see the $\displaystyle D_4$ structure, you have a cyclic group $\displaystyle \mathbb{Z}_4$ is normal in $\displaystyle D_4$. A little bit of comutation shows that the splitting field of p(x) over Q is $\displaystyle K=\mathbb{Q}(\sqrt[4]{2}, i)$ and K is a galois extension of F, i.e. , Gal(K/F)=[K:F] (It is because a cyclic extension is a galois extension).
• Apr 25th 2010, 10:39 AM
FancyMouse
If Gal(K/Q) is D_4, then you'll need a big diagram because there're 10 subgroups (thus 10 intermediate fields)