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Thread: Galois Group, FTG

  1. #1
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    Galois Group, FTG

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

    I'm pretty sure that (a) is $\displaystyle \mathbb{Q}(\sqrt(3), \sqrt(5))$, (b) is $\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2$ (c) $\displaystyle \mathbb{Q}, \mathbb{Q}(\sqrt(3), \sqrt(5)), \mathbb{Q}(\sqrt(3)), \mathbb{Q}(\sqrt(5)), \mathbb{Q}(\sqrt(15))$. 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 $\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2$ or $\displaystyle \mathbb{Z}_4$. 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 $\displaystyle K$ for the polynomial $\displaystyle f(x)=(x^2-3)(x^2-5) \in \mathbb{Q}[x]$.
    We can write $\displaystyle (x -\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5})$ therefore the splitting field is $\displaystyle \mathbb{Q}(\sqrt{3},\sqrt{5})$.
    (b) Determine the Galois Group of $\displaystyle K$ over $\displaystyle \mathbb{Q}$.
    Since $\displaystyle [K:\mathbb{Q}]=4$ it tells us that $\displaystyle |G|=4$ where $\displaystyle G = \text{Gal}(K/\mathbb{Q})$. If $\displaystyle \theta$ is an automorphism of $\displaystyle K$ then it is determined by its value on $\displaystyle \theta (\sqrt{3}) \text{ and }\theta (\sqrt{5})$. Remember that automorphisms permute zeros of polynomials therefore $\displaystyle \theta (\sqrt{3})$ is a zero of $\displaystyle x^2 - 3$ and $\displaystyle \theta (\sqrt{5})$ is a zero of $\displaystyle x^2-5$. Thus, $\displaystyle \theta(\sqrt{3}) = \pm \sqrt{3} \text{ and }\theta (\sqrt{5}) = \pm \sqrt{5}$. We see that there are at most $\displaystyle 4$ automorphisms but since $\displaystyle |G| = 4$ it means we must have an automorphism for each one of these possibilities. Define $\displaystyle \sigma: K\to K$ by $\displaystyle \sigma (\sqrt{3}) = -\sqrt{3} \text{ and }\sigma (\sqrt{5}) = \sqrt{5}$, define $\displaystyle \tau: K\to K$ by $\displaystyle \tau (\sqrt{3}) = \sqrt{3}\text{ and }\tau (\sqrt{5}) = -\sqrt{5}$. We see that $\displaystyle \sigma \tau$ is another automorphism of $\displaystyle K$. Therefore, $\displaystyle G = \left< \sigma,\tau \right>$ with $\displaystyle \sigma^2 = \tau^2 = \text{id}$. The non-trivial proper subgroups of $\displaystyle G$ are: $\displaystyle \left< \sigma\right> , \left<\tau\right>, \left< \sigma\tau \right>$ (just like $\displaystyle \mathbb{Z}_2\times \mathbb{Z}_2$).


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