# Galois Group, generators

• Mar 15th 2009, 07:30 PM
selenne431
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.
• Mar 15th 2009, 08:10 PM
ThePerfectHacker
Quote:

Originally Posted by selenne431
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. $G = \left< \sigma_2\right>$.