Enumerate the $\mathbb{Q}$-automorphisms of the normal extension $\mathbb{Q}(\sqrt[p]{2},\omega_p)$

Question

Calling $\omega_p$ a primitive p-th root of unity, Wikipedia states that:

1. $\mathbb{Q}(\sqrt[p]{2},\omega_p)$ is a normal extension of $\mathbb{Q}$
2. This normal extension has order $p(p-1)$

Also, I know a theorem stating that:

3. If $L:K$ is a finite normal extension inside $\mathbb{C}$, then there are precisely $[L:K]$ distinct K-automorphisms of $L$.

Hence, for the case $p=3$, there should be $6$ $\mathbb{Q}$-authomorphisms, and for the case $p=4$, there should be $12$ of them.

However, I am struggling "visualizing" these automorphisms, i.e., understanding exactly what they do.

Can someone enumerate them (at least for the cases $p=3$ and $p=4$)?

Answer 1

$\mathbb{Q}(\sqrt[p]{2},\omega_p)$ is generated (as a field) by $\sqrt[p]{2}$ and $\omega_p$. So, any field homomorphism $\mathbb{Q}(\sqrt[p]{2},\omega_p) \to \mathbb{Q}(\sqrt[p]{2},\omega_p)$ is determined by where it sends $\sqrt[p]{2}$ and $\omega_p$. Moreover, every field homomorphism $\mathbb{Q}(\sqrt[p]{2},\omega_p) \to \mathbb{Q}(\sqrt[p]{2},\omega_p)$ is an automorphism (since field homomorphisms are always injective linear maps and $\mathbb{Q}(\sqrt[p]{2},\omega_p)$ is finite-dimensional). So, we just need to classify field homomorphisms $\mathbb{Q}(\sqrt[p]{2},\omega_p) \to \mathbb{Q}(\sqrt[p]{2},\omega_p)$.

Let $\varphi$ be such a homomorphism. Since $(\sqrt[p]{2})^p = 2$, we must have $\varphi(\sqrt[p]{2})^p = \varphi(2) = 2$, and thus $\varphi(\sqrt[p]{2}) = \omega_p^i \sqrt[p]{2}$ for some $i \in \{0,\dots,p-1\}$. Likewise, since $\omega_p^p = 1$, we must have $\varphi(\omega_p) = \omega_p^j$ for some $j \in \{1,\dots,p-1\}$. There are $p$ possibilities for $i$ and $p-1$ possibilities for $j$, so there are at most $p(p-1)$ automorphisms of $\mathbb{Q}(\sqrt[p]{2},\omega_p)$.

What remains to be checked is that all of these choices actually do extend to field homomorphisms. You could do this directly by expanding the elements of $\mathbb{Q}(\sqrt[p]{2},\omega_p)$ as linear combinations of elements of the form $\sqrt[p]{2}^a \omega_p^b$, or use the presentation $$\mathbb{Q}(\sqrt[p]{2},\omega_p) = \mathbb{Q}(\sqrt[p]2)(\omega_p) \cong (\mathbb{Q}[x]/(x^p-2))[y]/(1+y+\dots+y^{p-1}) \cong \mathbb{Q}[x,y]/(x^p-2,1+y+\dots+y^{p-1})$$ to check that the required relations hold.