# Galois Thoery.

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• Apr 24th 2010, 10:14 PM
dead actor
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 25th 2010, 12:46 AM
aliceinwonderland
Quote:

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 $p(x)=x^4-2x^2-1$. The resolvent cubic of p(x) is

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

If you see the $D_4$ structure, you have a cyclic group $\mathbb{Z}_4$ is normal in $D_4$. A little bit of comutation shows that the splitting field of p(x) over Q is $K=\mathbb{Q}(\sqrt[4]{2}, i)$ 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, 11:39 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)