How is $Gal(\mathbb{Q}(\sqrt[4]{3}, i)$ equal to $D_8$? (dihedral group of order 8)

Question

From what I gathered, the automorphisms are as follows:

$\iota$ (the identity map) $\sigma_2: \sqrt[4]{3} \rightarrow \sqrt[4]{9}$ (squaring map) $\sigma_3: \sqrt[4]{3} \rightarrow \sqrt[4]{27}$ (cubing map) $\sigma_i: i \rightarrow -i$ (complex conjugation map)

I'm not sure what other automorphisms are possible, since in general I can't send a real number to an imaginary number (swapping them would be a significant change).

Answer 1

$\text{Gal}(\mathbb{Q}(\sqrt[4]{3}, i))$ is of order 8, because this is the order of the extension. So we can send $\sqrt[4]{3}$ to any other root of $X^4-3$ and $i$ to any other root of $X^2+1$, as doing this gives us 8 possibilities. One can see the group is generated by $i \mapsto -i$ and $\sqrt[4]{3} \mapsto i\sqrt[4]{3}$(call them $\sigma_1$ and $\sigma_2$ respectively). Now identify $\sigma_1$ with the reflection element in $D_8$ and identify $\sigma_2$ with the generator of the rotation subgroup of $D_8$. You can check this is an isomorphism. You can also see it in a different way, by noticing that $<\sigma_1>$ acts on $<\sigma_2>$ by conjugation, which in this case results exactly in inversion. So the galois group is the semidirect product of these two subgroups, the one being isomorphic to $\mathbb{Z}_2$, the other with $\mathbb{Z}_4$, so it is the semidirect product of both with the inversion action, which is exactly isomorphic to $D_8$.

Answer 2

$\mathbb{Q}( \sqrt[4]{3}, i)$ is the splitting field of $x^4 -3$ over $\mathbb{Q}$. So, $\text{Gal}(\mathbb{Q}( \sqrt[4]{3}, i)/\mathbb{Q}$ acts transitively on the roots of $x^4 -3$, since this polynomial is irreducible by Eisenstein. Thus, it embeds as a transitive subgroup of $S_4$.

Let's just do a quick general note. Let $G$ be the copy of this Galois group in $S_4$. The orbit stabilizer theorem tells us that 4 divides the order of $G$. Moreover, the order of $G$ divides $24$. So, $G$ has order either 4, 8, 12, 24. Order 4 gives either $C_4$ or $V$. Since 8 is the highest power of 2 dividing $24$, Sylow's theorem tells us all the order 8 subgroups are conjugate, and hence isomorphic. If you look at the normalizer of the subgroup $< ( 1 2), ( 3 4 )>$, you'll see that it is $<(1 2), (3 4), (1 3)(4 2)>$ and you can confirm that this is $D_8$, and so $D_8$ is the unique subgroup of order $8$ in $S_4$. If order $G$ is $12$, then $G = A_4$, and if $G$ has order 24 then $G = S_4$. So, the only options are $C_4, V, D_8, A_4, S_4$. But, since you know it has order 8, it must be $D_8$.