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Math Help - Galois Group, FTG

  1. #1
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    Galois Group, FTG

    (a) Find a splitting field K for the polynomial f(x)=(x^2-3)(x^2-5) \in \mathbb{Q}[x].
    (b) Determine the Galois Group of K over \mathbb{Q}.
    (c) Find all intermediate fields between \mathbb{Q} and K, by refering to the Fundamental Theorem of Galois Theory [FTG] to show you have them all.

    I'm pretty sure that (a) is \mathbb{Q}(\sqrt(3), \sqrt(5)), (b) is \mathbb{Z}_2 \times \mathbb{Z}_2 (c) \mathbb{Q}, \mathbb{Q}(\sqrt(3), \sqrt(5)), \mathbb{Q}(\sqrt(3)), \mathbb{Q}(\sqrt(5)), \mathbb{Q}(\sqrt(15)). I don't know how to justify (c) by using the Fundamental Theorem of Galois Theory. Also, I don't know how to determine the Galois Group for part (b). I know it is either \mathbb{Z}_2 \times \mathbb{Z}_2 or \mathbb{Z}_4. But, I don't know how to show this.
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  2. #2
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    Quote Originally Posted by diondex223 View Post
    (a) Find a splitting field K for the polynomial f(x)=(x^2-3)(x^2-5) \in \mathbb{Q}[x].
    We can write (x -\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5}) therefore the splitting field is \mathbb{Q}(\sqrt{3},\sqrt{5}).
    (b) Determine the Galois Group of K over \mathbb{Q}.
    Since [K:\mathbb{Q}]=4 it tells us that |G|=4 where G = \text{Gal}(K/\mathbb{Q}). If \theta is an automorphism of K then it is determined by its value on \theta (\sqrt{3}) \text{ and }\theta (\sqrt{5}). Remember that automorphisms permute zeros of polynomials therefore \theta (\sqrt{3}) is a zero of x^2 - 3 and \theta (\sqrt{5}) is a zero of x^2-5. Thus, \theta(\sqrt{3}) = \pm \sqrt{3} \text{ and }\theta (\sqrt{5}) = \pm \sqrt{5}. We see that there are at most 4 automorphisms but since |G| = 4 it means we must have an automorphism for each one of these possibilities. Define \sigma: K\to K by \sigma (\sqrt{3}) = -\sqrt{3} \text{ and }\sigma (\sqrt{5}) = \sqrt{5}, define \tau: K\to K by \tau (\sqrt{3}) = \sqrt{3}\text{ and }\tau (\sqrt{5}) = -\sqrt{5}. We see that \sigma \tau is another automorphism of K. Therefore, G = \left< \sigma,\tau \right> with \sigma^2 = \tau^2 = \text{id}. The non-trivial proper subgroups of G are: \left< \sigma\right> , \left<\tau\right>, \left< \sigma\tau \right> (just like \mathbb{Z}_2\times \mathbb{Z}_2).


    (c) Find all intermediate fields between \mathbb{Q} and K, by refering to the Fundamental Theorem of Galois Theory [FTG] to show you have them all.
    The proper non-trivial intermediate are: K^{\left< \sigma\right>}, K^{\left<\tau\right>}, K^{\left< \sigma\tau \right>}. Notice that \sqrt{5} is fixed by \sigma therefore (fill in the minor details in this argument, do you see how or you want me to?) that K^{\left< \sigma\right>} = \mathbb{Q}(\sqrt{5}) likewise K^{\left<\tau\right>} = \mathbb{Q}(\sqrt{3}). For \sigma\tau notice that \sqrt{3} \sqrt{5} is fixed because the two minus signs go away, therefore (again you need to fill in a small detail) K^{ \left< \sigma\tau \right>} = \mathbb{Q}(\sqrt{15}).
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