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Math Help - Galois Thoery.

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by dead actor View Post
    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).
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  3. #3
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    If Gal(K/Q) is D_4, then you'll need a big diagram because there're 10 subgroups (thus 10 intermediate fields)
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