Cfrgtkky

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 $sigma:zeta mapsto zeta, alpha mapstozeta cdot alpha$, 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

$newcommand{Gal}{operatorname{Gal}}$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.