Galois Group (Example)

This post is about the Galois group of K over \mathbb{Q}, where K is the splitting field of f(x)=x^p-2, where p is an odd prime.

First we show that the polynomial f(x)=x^p-2 is irreducible over \mathbb{Q}. This follows immediately by Eisenstein’s Criterion, since 2\mid (-2), 2\nmid 1 and 2^2\nmid (-2).

Next, we show that the splitting field K of f(x) in \mathbb{C} is Q(\sqrt[p]{2},\omega), where \omega=e^{2\pi i/p} is a primitive p-th root of unity. The roots of f are \sqrt[p]2, \sqrt[p]2\omega, \sqrt[p]2\omega^2, \dots, \sqrt[p]2\omega^{p-1}.

The splitting field K contains \sqrt[p]2 and \omega=\frac{\sqrt[p]2\omega^2}{\sqrt[p]2\omega}. Thus \mathbb{Q}(\sqrt[p]2,\omega)\subseteq K.

On the other hand, \mathbb{Q}(\sqrt[p]2,\omega) contains all the roots of f, hence f splits in \mathbb{Q}(\sqrt[p]2, \omega). Thus K\subseteq\mathbb{Q}(\sqrt[p]2,\omega), since K is the smallest field that contains \mathbb{Q} and all the roots of f. All in all, we have that the splitting field K=\mathbb{Q}(\sqrt[p]2, \omega).

The next part involves determining the Galois group of K over \mathbb{Q}. We have |Gal(K/\mathbb{Q})|=[K:\mathbb{Q}]. Since [\mathbb{Q}(\sqrt[p]2):\mathbb{Q}]=p (minimal polynomial x^p-2), and [\mathbb{Q}(\omega):\mathbb{Q}]=p-1 (minimal polynomial the cyclotomic polynomial 1+x+x^2+\dots+x^{p-1}), thus |Gal(K/\mathbb{Q})|=p(p-1). Here we have used the lemma that suppose [F(\alpha):F]=m and [F(\beta):F]=n with \gcd(m,n)=1, then [F(\alpha,\beta):F]=mn.

What the Galois group does is it permutes the roots of f. Let \sigma be an element of the Galois group. \sigma(\sqrt[p]2) can possibly be \sqrt[p]2, \sqrt[p]\omega, \dots, \sqrt[p]2\omega^{p-1}, a total of p choices. Similarly, \sigma(\omega)=\omega, \omega^2, \dots,\omega^{p-1}, a total of p-1 choices. All these total up to p(p-1) elements, which is exactly the size of the Galois group.

The above Galois group Gal(K/\mathbb{Q}) is described by how its elements act on the generators. For a more concrete representation, we can actually prove that the Galois group above is isomorphic to the group of matrices \begin{pmatrix}a&b\\0&1\end{pmatrix}, where a,b\in\mathbb{F}_p, a\neq 0. We denote the group of matrices as M.

To show the isomorphism, we define a map \phi: Gal(K/\mathbb{Q})\to M, mapping \sigma_{a,b} to \begin{pmatrix}a&b\\0&1\end{pmatrix}.

Notation: \sigma_{a,b} is defined on the generators as follows, \sigma_{a,b}(\sqrt[p]2)=\sqrt[p]2\omega^b, \sigma_{a,b}(\omega)=\omega^a.

We can clearly see that the map \phi is bijective. To see it is a homomorphism, we compute \phi(\sigma_{a,b}\circ\sigma_{c,d})=\begin{pmatrix}ac&a+bd\\0&1\end{pmatrix}=\phi(\sigma_{a,b})\phi(\sigma_{c,d}).

About mathtuition88
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