Sanity check regarding Galois group of $X^8 - 3$ This is more to check if what I'm doing is correct.
Let $P(X) = X^8 - 3$. I know that the splitting field of $P(X)$ is $\Bbb Q(\zeta_8, \sqrt[8]{3}) = \Bbb Q(i, \sqrt{2}, \sqrt[8]{3})$ and so I can describe the elements of the Galois group by automorphisms 
\begin{align}
i &\longmapsto \pm i\\
\sqrt{2} &\longmapsto \pm \sqrt{2}\\
\sqrt[8]{3} &\longmapsto \zeta_8^a\sqrt[8]{3},\ \ a \in \lbrace 0, 1, \dots, 7\rbrace.
\end{align}
I'm not sure if this is a silly question, but does the action on $i$ and $\sqrt{2}$ affect the action on $\sqrt[8]{3}$? That is, for example, if I have
\begin{align}
\sigma:\ i &\longmapsto -i\\
\sqrt{2}  &\longmapsto \sqrt{2}\\
\sqrt[8]{3} & \longmapsto \zeta_8^2\sqrt[8]{3}
\end{align}
and 
\begin{align}
\tau:\ i &\longmapsto i\\
\sqrt{2}  &\longmapsto -\sqrt{2}\\
\sqrt[8]{3} & \longmapsto \zeta_8^3\sqrt[8]{3}
\end{align}
then $\sigma(\sqrt[8]{3}) = \zeta_8^2\sqrt[8]{3}$, and now, since $\zeta_8 = \frac{1}{\sqrt{2}}(1+i)$, does $\tau$ applied to this give
$$\tau(\sigma(\sqrt[8]{3})) = \tau(\zeta_8)^2\tau(\sqrt[8]{3}) = (\zeta_8^5)^2 \cdot\zeta_8^3\sqrt[8]{3} = \zeta_8^5\sqrt[8]{3}$$
so that
\begin{align}
\tau \circ \sigma :\ i &\longmapsto -i\\
\sqrt{2}  &\longmapsto -\sqrt{2}\\
\sqrt[8]{3} & \longmapsto \zeta_8^5\sqrt[8]{3}\ \ ?
\end{align}
 A: This is not a silly question, this is something to be careful about.
There's a general theorem for fields $F \subset K \subset M$ and $L \subset M$ such that $K/F$ is Galois. In this case $\text{Gal}({KL}/L)\cong\text{Gal}(K/K\cap L)$. In our situation $F=\Bbb Q$, $K= \Bbb Q(\zeta_8)$ $L=\Bbb Q(\sqrt[8]{3})$.
We have to compute the intersection $K \cap L = \Bbb Q(\zeta_8) \cap \Bbb Q(\sqrt[8]{3})$, note that since $\Bbb Q(\sqrt[8]{3}) \subset \Bbb R$, $K \cap L \subset \Bbb Q(\zeta_8) \cap \Bbb R= \Bbb Q(\sqrt{2})$, this leaves only the possibilities that $K \cap L= \Bbb Q$ or $K \cap L = \Bbb Q(\sqrt{2})$.
At this point I'll use a bit of algebraic number theory (though one can also follow with just algebra knowledge), as I know that you're familiar with that. The ring of integers of $\Bbb Q(\sqrt{2})$ is $\Bbb Z[\sqrt{2}] \cong \Bbb Z[x]/(x^2-2)$. This is known to be a PID (one can for example show that it is Euclidean (see this question) or use the Minkowski bound which is smaller than $2$ in this case). We have $\Bbb Z[\sqrt{2}]/3\Bbb Z[\sqrt{2}] \cong \Bbb Z[x]/(3,x^2-2) \cong \Bbb{F}_3[x]/(x^2-2)$. Because $2$ is a not square in $\Bbb F_3$, $x^2-2$ is irreducible over $\Bbb{F}_3$, so this ring is an integral domain, thus $3\Bbb Z[\sqrt{2}]$ is a prime ideal and $3$ is a prime element. We get by Eisenstein that $x^8-3$ is irreducible over $\Bbb Q(\sqrt{2})$. This implies that $[\Bbb Q(\sqrt{2})(\sqrt[8]{3}):\Bbb Q(\sqrt{2})]=8$, but if $K \cap L = \Bbb Q(\sqrt{2})$, then $\sqrt{2} \in \Bbb Q(\sqrt[8]{3})$, so $\Bbb Q(\sqrt{2})(\sqrt[8]{3})= \Bbb Q(\sqrt[8]{3})$ and $[\Bbb Q(\sqrt[8]{3}):\Bbb Q(\sqrt{2})]=[\Bbb Q(\sqrt[8]{3}):\Bbb Q]/[\Bbb Q(\sqrt{2}):\Bbb Q]=4$, which contradicts the previous computation. Thus we must have $\Bbb Q(\zeta_8) \cap \Bbb Q(\sqrt[8]{3}) = \Bbb Q$, so we get an isomorphism $\text{Gal}(\Bbb Q(\zeta_8, \sqrt[8]{3})/\Bbb Q(\sqrt[8]{3})) \cong \text{Gal}(\Bbb Q(\zeta_8)/\Bbb Q)$
This isomorphism tells us precisely that if we let $\sqrt[8]{3}$ fixed, then we can still do with $\zeta_8$ (and hence with $\sqrt{2}$ and $i$) everything that we can do over $\Bbb Q$, so the actions on $\zeta_8$ and $\sqrt[8]{3}$ are in this sense independent.
