I'm working on this problem as part of a study guide and I ran into something that's tripping me up. The problem is, Let $K$ be the splitting field of $x^4+3$ over $\mathbb{Q}$. Find generators of the group $Aut(K/F)$.
If we let $w=e^{\pi i/4}$ and $\alpha=3^{1/4}$ then the roots of the above polynomial are $\alpha w,\ \alpha w^3,\ \alpha w^5,\ \alpha w^7$ and it doesn't take much work to see that the splitting field $K=\mathbb{Q}(\alpha w, w^2)=\mathbb{Q}(\alpha w, i)$.
I know that if $\beta$ is a root of some polynomial in $\mathbb{Q}[x]$, then $\sigma(\beta)$ is also a root of that polynomial for any $\sigma \in Aut(K/F)$. Thus, an automorphism may send $\alpha w$ to $\alpha w,\ \alpha w^3,\ \alpha w^5,\ \alpha w^7$ and $i$ can be sent to $\pm i$.
This much is all fine, but now I confuse myself. The set of solutions says that there are two generators for $Aut(K/F)$, which are $$\sigma: \alpha w \mapsto \alpha w^3, i \mapsto -i$$ $$\tau:\alpha w \mapsto \alpha w,\ i \mapsto -i$$ however, this doesn't make sense, because if $i \mapsto -i$, then $\alpha w=\alpha(\sqrt2/2+i\sqrt2/2) \mapsto \alpha(\sqrt2/2-i\sqrt2/2)=\alpha w^7$ and thus, $\tau$ cannot fix $\alpha w$.
I think that to fix the problem, we should just let $\tau: i \to -i$. Then $\left<\sigma, \tau\right>\cong D_4$. Am I correct in my analysis? Or have I made a misunderstanding?