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Thread: Galois Group, generators

  1. #1
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    Galois Group, generators

    Suppose $\displaystyle G$ is the Galois Group over $\displaystyle \mathbb{Q}$ of one of the polynomials:
    i) $\displaystyle x^{11}-1$
    ii)$\displaystyle x^4-2$
    For your choice of polynomial, find the generators of the splitting field $\displaystyle E$ over $\displaystyle \mathbb{Q}$ 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 $\displaystyle G$ is the Galois Group over $\displaystyle \mathbb{Q}$ of one of the polynomials:
    i) $\displaystyle x^{11}-1$
    ii)$\displaystyle x^4-2$
    For your choice of polynomial, find the generators of the splitting field $\displaystyle E$ over $\displaystyle \mathbb{Q}$ 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 $\displaystyle x^{11}-1$ and you think about $\displaystyle x^4-2$. The polynomial $\displaystyle x^{11} - 1$ can be factored as $\displaystyle \prod_{j=0}^{10}(x - \zeta^j)$ where $\displaystyle \zeta = e^{2\pi i/11}$. Therefore, the splitting field is $\displaystyle \mathbb{Q}(1,\zeta,\zeta^2,...,\zeta^{10}) = \mathbb{Q}(\zeta) = K$. Now $\displaystyle [K:\mathbb{Q}]$ is equal to the degree of the minimal polynomial of $\displaystyle \zeta$ over $\displaystyle \mathbb{Q}$. Notice $\displaystyle f(x) = x^{10}+x^9+...+x+1$ is irreducible over $\displaystyle \mathbb{Q}$ (I am assuming you know this result) so $\displaystyle [K:\mathbb{Q}] = 10$. Let $\displaystyle G = \text{Gal}(K/\mathbb{Q})$ and thus $\displaystyle |G| = [K:\mathbb{Q}] = 10$. This tells us that we are looking at a Galois group that has $\displaystyle 10$ elements. If $\displaystyle \theta \in G$ is an automorphism of $\displaystyle K$ then it is completely determined by its value of $\displaystyle \theta (\zeta)$. Remember that automorphisms of $\displaystyle K$ permute the zeros of polynomials, this means that $\displaystyle \theta(\zeta)$ must be one of the zeros of $\displaystyle f(x)$, thus, $\displaystyle \theta (\zeta) = \zeta,\zeta^2,...,\zeta^{10}$. We see that there are at most $\displaystyle 10$ automorphism of $\displaystyle K$, however we know that $\displaystyle |G|=10$ which forces us to conclude that there is an automorphism $\displaystyle \sigma_k $ such that $\displaystyle \sigma_k (\zeta) = \zeta^k$ for each $\displaystyle k=0,1,2,..,10$. Therefore, $\displaystyle G = \{ \sigma_0,\sigma_1,...,\sigma_{10}\}$. It is easy to see that $\displaystyle G$ behaves like the group $\displaystyle \mathbb{Z}_{10}^{\times}$ i.e. $\displaystyle \sigma_a \sigma_b = \sigma_{ab(\bmod 11)}$. Thus, $\displaystyle G$ has a generator for example $\displaystyle \sigma_2$ i.e. $\displaystyle G = \left< \sigma_2\right>$.
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