Let $L$ be the splitting field of $x^4+1$ over ℚ. Show that $\mathrm{Aut}_\mathbb Q()$ is $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$

Question

Let $L$ be the splitting field of $x^4+1$ over $\mathbb Q$. Show that $\mathrm{Aut}_\mathbb Q()$ is $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$.

I know $[L:\mathbb Q]$ is $4$. However, why among all the groups with order $4$, $\mathrm{Aut}_\mathbb Q()$ is $\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$?

Answer 1

We have $L = \mathbb{Q}\big(\sqrt{2}(1+i),\sqrt{2}(1-i),\sqrt{2}(-1-i),\sqrt{2}(-1+i)\big) = \mathbb{Q}(\sqrt2, i)$. The extension $L \supset \mathbb{Q}$ is normal so $\left|\operatorname{Aut}_\mathbb{Q}(L)\right| = 4$.

There are two possibilities, $\operatorname{Aut}_\mathbb{Q}(L) \cong \mathbb{Z}_4$, or $\operatorname{Aut}_\mathbb{Q}(L) \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2$.

Clearly, the elements of $\operatorname{Aut}_\mathbb{Q}(L)$ generated by $\sqrt{2} \mapsto -\sqrt{2}, i \mapsto i$ and $\sqrt{2} \mapsto \sqrt{2}, i \mapsto -i$ are different and have order $2$. However, $\mathbb{Z}_4$ has only one element of order $2$, which is $\overline{2}$. Hence it must be $\operatorname{Aut}_\mathbb{Q}(L) \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2$.

Answer 2

Y.H. Chan's comment shows the standard way to solve this type of problems. Another approach would be to notice that $(x^4-1)(x^4+1)=x^8-1$ and therefore the splitting field of $x^4+1$ over $\mathbb{Q}$ is $\mathbb{Q}(\zeta_8)$. It is known that the Galois group of $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is $\mathbb{Z}_n^*$. Therefore the Galois group you are looking for is $\mathbb{Z}_8^*$, which by Gauss' theorem (or by direct check) is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.