# Compute the degree of the splitting field of $x^{12} - 2$ over $\mathbb{Q}$ and describe its Galois Group as a semidirect product

I have the polynomial $$f(x) = x^{12} - 2$$. I have to compute the degree of the splitting field over $$\mathbb{Q}$$ and describe its Galois Group as a semidirect product.

Clearly the splitting field is $$E= \mathbb{Q}(\zeta,\alpha$$), where $$\zeta$$ is a primitve $$12^{\text{th}}$$ root of unity and $$\alpha = \sqrt[12]{2}$$. The pertinent intermediate fields are $$L=\mathbb{Q}(\alpha)$$ and $$K=\mathbb{Q}(\zeta)$$. Since we have the degree of the splitting field as $$[E:\mathbb{Q}]=48$$, with $$[L:\mathbb{Q}]=12$$ and $$[K:\mathbb{Q}]=\varphi(12)=4$$

I look at $$\sigma:\zeta \mapsto \zeta,\ \alpha \mapsto\zeta \cdot \alpha$$, this automorphism fixes $$K$$ and is of order $$12$$, therefore we have $$N = \text{Gal}(E/K) = \langle \sigma \rangle \cong C_{12}$$, and we know $$N$$ is a normal subgroup of $$G = \text{Gal}(E/\mathbb{Q})$$, since $$K/\mathbb{Q}$$ is Galois.

Now, consider $$\tau_1,\ \tau_2 \in G$$, fixing $$\alpha$$ where $$\tau_1(\zeta)=\zeta^5,\ \tau_2(\zeta)=\zeta^7$$, and $$\tau_2 \circ \tau_1(\zeta)=\zeta^{11}$$. This automorphism fixes $$L$$ and is of order $$4$$, and therefore $$H = \text{Gal}(E/L) = \langle \tau_1,\tau_2 \rangle \cong K_{4}$$, the Klein four-group.

Am I correct? Is there more I can add in my reasonging? And how do I conclude that $$G= C_{12} \rtimes K_{4}$$? Any help is appreciated

• My first reaction was "just do polgalois in PARI"... and then they don't support polynomials of degree > 11 Dec 31, 2018 at 14:15

$$\newcommand{Gal}{\operatorname{Gal}}$$Preliminaries:

• $$\operatorname{irr}(\zeta, \Bbb Q) = \Phi_{12}(X) = X^4-X^2+1$$
• $$\zeta = \dfrac{\sqrt3+i}2$$
• $$\sqrt3 = \zeta + \zeta^{-1}$$
• $$i = \zeta + \zeta^5$$
• $$\Gal(\Bbb Q(\zeta)/\Bbb Q) = C_2 \times C_2$$
• Subfields are $$\Bbb Q$$, $$\Bbb Q(\sqrt3)$$, $$\Bbb Q(i)$$, $$\Bbb Q(\sqrt{-3})$$, $$\Bbb Q(\zeta)$$.

We wish to determine $$\Gal(\Bbb Q(\zeta, \alpha) / \Bbb Q(\zeta))$$, so we wish to factorize $$X^{12}-2$$ in $$\Bbb Q(\zeta)$$. We claim that it is irreducible.

Since $$\Bbb Q(\zeta)$$ has all the $$12$$th roots of unity, and $$\Bbb Z[\zeta]$$ is a Euclidean domain, it suffices to show that $$2$$ is neither a square nor a cube.

It is clear that $$2$$ is not a cube, since $$N_{\Bbb Q(\zeta)/\Bbb Q}(2) = 16$$ is not a cube.

That $$2$$ is not a square, i.e. $$\sqrt 2 \notin \Bbb Q(\zeta)$$, is clear, since none of the quadratic subfields of $$\Bbb Q(\zeta)$$ contain $$\sqrt 2$$.

Since $$X^{12} - 2 \in \Bbb Q(\zeta)[X]$$ is irreducible, we can know that $$[\Bbb Q(\zeta,\alpha) : \Bbb Q(\zeta)] = 12$$ and $$\Gal(\Bbb Q(\zeta,\alpha) : \Bbb Q(\zeta)) = C_{12}$$. Thus we have an exact sequence: $$1 \to \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q(\zeta)) \to \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q) \to \Gal(\Bbb Q(\zeta)/\Bbb Q) \to 1$$

i.e. $$1 \to C_{12} \to \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q) \to C_2 \times C_2 \to 1$$

Now $$G = \Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q) = \langle \rho, \sigma, \tau \rangle$$, where:

• $$\rho(\zeta) = \zeta$$, $$\rho(\alpha) = \zeta \alpha$$
• $$\sigma(\zeta) = \zeta^5$$, $$\sigma(\alpha) = \alpha$$
• $$\tau(\zeta) = \zeta^{11}$$, $$\tau(\alpha) = \alpha$$

It is clear that $$N = \langle \rho \rangle$$ is the normal subgroup $$\Gal(\Bbb Q(\zeta,\alpha)/\Bbb Q(\zeta))$$ with $$H = \langle \sigma, \tau \rangle$$ being its complement, thus $$G = N \rtimes H$$.

To determine the action, we need to compute $$\sigma \rho \sigma^{-1}$$ and $$\tau \rho \tau^{-1}$$:

• $$\sigma \rho \sigma^{-1} \alpha = \sigma \rho \alpha = \sigma (\zeta \alpha) = \zeta^5 \alpha$$
• $$\tau \rho \tau^{-1} \alpha = \tau \rho \alpha = \tau (\zeta \alpha) = \zeta^{11} \alpha$$

So $$\sigma \rho \sigma^{-1} = \rho^5$$ and $$\tau \rho \tau^{-1} = \rho^{11}$$, and that is the group action that determines the semidirect product. The group presentation is: $$G = \langle \rho, \sigma, \tau \mid \rho^{12} = \sigma^2 = \tau^2 = (\sigma \tau)^2 = 1, \sigma \rho \sigma^{-1} = \rho^5, \tau \rho \tau^{-1} = \rho^{11} \rangle$$

And another way of presenting the group is $$C_{12} \rtimes \operatorname{Aut}(C_{12})$$ with the identity homomorphism.

The problem with your attempt is that, while every element of $$G$$ must send $$\zeta$$ and $$\alpha$$ to one of their respective conjugates, not every such pair of conjugates must correspond to an element of $$G$$.

In other words, we have a set-theoretic injection $$G \to \{\text{conjugates of \zeta}\} \times \{\text{conjugates of \alpha}\}$$ which you withou justification assumed to be surjective.