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Math Help - Galois group of a polynomial

  1. #1
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    Galois group of a polynomial

    Hello.

    I have to find a splitting field L of the polynomial f=x^4-4x^2+9 \in \mathbb Q[x] over \mathbb Q and the galois group and all intermediate fields of the algebraic extension L/\mathbb Q.

    I found out, that f is irreducible, the roots of f are:
    a=\sqrt{2-i\sqrt{5}}, \quad b=-\sqrt{2+i\sqrt{5}}, \quad c=\sqrt{2+i\sqrt{5}}=-b, \quad d=-\sqrt{2-i\sqrt{5}}=-a and that \mathbb Q(a) is a splitting field of f.

    Furthermore I find out, that the three homomorphisms
    a \mapsto \pm b, \ a \mapsto -a have order 2, so the galois group is isomorph to \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z.

    But now I don't know how to find the fixed fields, which belongs to the homomorphisms.

    For example: Which fixed field belongs to the homomorphism \sigma_1: a \mapsto b=\frac{-3}{a} ?

    The definition says:
    \mathbb Q(a)^{\left \{ id, \sigma_1 \right \}}=\left \{ z \in \mathbb Q(a)\ |\ \sigma_1(z)=z  \right \}

    ...and this don't help me, because I need a description of the fixed field like this: \mathbb Q(\beta) (with suitable \beta).

    I hope, you can help me.

    Bye,
    Alex
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  2. #2
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    Quote Originally Posted by AlexanderW View Post
    Hello.

    I have to find a splitting field L of the polynomial f=x^4-4x^2+9 \in \mathbb Q[x] over \mathbb Q and the galois group and all intermediate fields of the algebraic extension L/\mathbb Q.

    I found out, that f is irreducible, the roots of f are:
    a=\sqrt{2-i\sqrt{5}}, \quad b=-\sqrt{2+i\sqrt{5}}, \quad c=\sqrt{2+i\sqrt{5}}=-b, \quad d=-\sqrt{2-i\sqrt{5}}=-a and that \mathbb Q(a) is a splitting field of f.

    Furthermore I find out, that the three homomorphisms
    a \mapsto \pm b, \ a \mapsto -a have order 2, so the galois group is isomorph to \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z.

    But now I don't know how to find the fixed fields, which belongs to the homomorphisms.

    For example: Which fixed field belongs to the homomorphism \sigma_1: a \mapsto b=\frac{-3}{a} ?

    The definition says:
    \mathbb Q(a)^{\left \{ id, \sigma_1 \right \}}=\left \{ z \in \mathbb Q(a)\ |\ \sigma_1(z)=z \right \}

    ...and this don't help me, because I need a description of the fixed field like this: \mathbb Q(\beta) (with suitable \beta).

    I hope, you can help me.

    Bye,
    Alex
    well, it's very straightforward: \{1,a,a^2,a^3 \} is a \mathbb{Q}-basis for \mathbb{Q}(a). so for any z \in \mathbb{Q}(a) there exist unique q_i \in \mathbb{Q} such that z=q_0 + q_1a+q_2a^2+q_3a^3.

    now \sigma_1(z)=z is equivalent to \sum_{k=0}^3q_ka^k=\sum_{k=0}^3 q_k b^k=\sum_{k=0}^3 (-3)^k q_k a^{-k}. multiplying both sides by a^3 gives us \sum_{k=0}^3q_ka^{k+3}=\sum_{k=0}^3 (-3)^kq_ka^{3-k}. we also have

    a^4=4a^2-9, \ a^5=4a^3-9a, and a^6=7a^2-36. then you'll get q_2=0, \ q_3=-q_1. therefore the fixed field of the automorphism \sigma_1 is:

    \{q_0 + q_1(a-a^3) : \ q_0, q_1 \in \mathbb{Q} \}=\mathbb{Q}(\sqrt{2}i). (note that a-a^3 is a root of the irreducible polynomial x^2 + 18).
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  3. #3
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    Smile

    Thank you for your very helpful answer!
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