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

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

1 Answer 1



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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .