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Thread: Galois group of a polynomial

  1. #1
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    Galois group of a polynomial

    Hello.

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

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

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

    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 $\displaystyle \sigma_1: a \mapsto b=\frac{-3}{a}$ ?

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

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

    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 $\displaystyle L$ of the polynomial $\displaystyle f=x^4-4x^2+9 \in \mathbb Q[x]$ over $\displaystyle \mathbb Q$ and the galois group and all intermediate fields of the algebraic extension $\displaystyle L/\mathbb Q$.

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

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

    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 $\displaystyle \sigma_1: a \mapsto b=\frac{-3}{a}$ ?

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

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

    I hope, you can help me.

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

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

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

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

    Thank you for your very helpful answer!
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