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Math Help - Galois Group, generators

  1. #1
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    Galois Group, generators

    Suppose G is the Galois Group over \mathbb{Q} of one of the polynomials:
    i) x^{11}-1
    ii) x^4-2
    For your choice of polynomial, find the generators of the splitting field E over \mathbb{Q} and describe the elements of G by their effect on these generators. Finally, describe the group by generators and relations.

    I just need to do one of the polynomials above and I am just beginning to learn Galois Theory, so I don't understand how to do this. I guess I need a good head start on how to solve this problem. Thanks.
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  2. #2
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    Quote Originally Posted by selenne431 View Post
    Suppose G is the Galois Group over \mathbb{Q} of one of the polynomials:
    i) x^{11}-1
    ii) x^4-2
    For your choice of polynomial, find the generators of the splitting field E over \mathbb{Q} and describe the elements of G by their effect on these generators. Finally, describe the group by generators and relations.

    I just need to do one of the polynomials above and I am just beginning to learn Galois Theory, so I don't understand how to do this. I guess I need a good head start on how to solve this problem. Thanks.
    Let me do it for x^{11}-1 and you think about x^4-2. The polynomial x^{11} - 1 can be factored as \prod_{j=0}^{10}(x - \zeta^j) where \zeta = e^{2\pi i/11}. Therefore, the splitting field is \mathbb{Q}(1,\zeta,\zeta^2,...,\zeta^{10}) = \mathbb{Q}(\zeta) = K. Now [K:\mathbb{Q}] is equal to the degree of the minimal polynomial of \zeta over \mathbb{Q}. Notice f(x) = x^{10}+x^9+...+x+1 is irreducible over \mathbb{Q} (I am assuming you know this result) so [K:\mathbb{Q}] = 10. Let G = \text{Gal}(K/\mathbb{Q}) and thus |G| = [K:\mathbb{Q}] = 10. This tells us that we are looking at a Galois group that has 10 elements. If \theta \in G is an automorphism of K then it is completely determined by its value of \theta (\zeta). Remember that automorphisms of K permute the zeros of polynomials, this means that \theta(\zeta) must be one of the zeros of f(x), thus, \theta (\zeta) = \zeta,\zeta^2,...,\zeta^{10}. We see that there are at most 10 automorphism of K, however we know that |G|=10 which forces us to conclude that there is an automorphism \sigma_k such that \sigma_k (\zeta) = \zeta^k for each k=0,1,2,..,10. Therefore, G = \{ \sigma_0,\sigma_1,...,\sigma_{10}\}. It is easy to see that G behaves like the group \mathbb{Z}_{10}^{\times} i.e. \sigma_a \sigma_b = \sigma_{ab(\bmod 11)}. Thus, G has a generator for example \sigma_2 i.e. G = \left< \sigma_2\right>.
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