1. ## 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.

2. Originally Posted by dead actor
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).

3. If Gal(K/Q) is D_4, then you'll need a big diagram because there're 10 subgroups (thus 10 intermediate fields)