galois-theory
Let $n \in \mathbb{N}$, then the order of the Galois Group
of $x^n-2$ coincide with $n \phi(n)$ for $n\in \{ 1 , \dots , 36 \}$
except for $n=\{ 8, 16, 24, 32 \}$ where this order is $\frac{ n \phi (n)}{2}$
and is easy to prove that for $p$ prime we have that the order
of the Galois Group of $x^p-2$ is $p(p-1)$.
What is the order of the Galois Group of $x^n-2 : n\in \mathbb{N}$ is general?
Thanks in advance.
Here is a broader setup for your question. Let $A$ be a Dedekind domain with fraction field $F$, $E/F$ be a finite Galois extension, and $B$ be the integral closure of $A$ in $E$. Pick a prime $\mathfrak p$ in $A$ and a prime $\mathfrak P$ in $B$ lying over $\mathfrak p$. The decomposition group $D(\mathfrak P|\mathfrak p)$ naturally maps by reduction mod $\mathfrak P$ to the automorphism group $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$ and Frobenius showed this is surjective. The kernel is the inertia group, so if $\mathfrak p$ is unramified in $B$ then we get an isomorphism from $D(\mathfrak P|\mathfrak p)$ to $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$, whose inverse is an embedding of the automorphism group of the residue field extension into $\text{Gal}(E/F)$.
If we take $A = {\mathbf Z}$ then we're in the number field situation and this is where Frobenius elements in Galois groups come from.
In your case you want to take $A = {\mathbf Q}[t]$, so $F = {\mathbf Q}(t)$. You did not give any assumptions about $f(x,t)$ as a polynomial in ${\mathbf Q}[x,t]$. (Stylistic quibble: I think it is better to write the polynomial as $f(t,x)$, specializing the first variable, but I'll use your notation.) Let's assume $f(x,t)$ is absolutely irreducible, so the ring $A' = {\mathbf Q}[x,t]/(f)$ is integrally closed. [EDIT: I should have included the assumption that $f$ is smooth, as otherwise $A'$ will not be integrally closed, but this "global" int. closed business is actually not so important. See comments below.] Write $F'$ for the fraction field of $A'$. After a linear change of variables we can assume $f(x,t)$ has a constant coefficient for the highest power of $x$, so $A'$ is the integral closure of $A$ in $F'$.
Saying for some rational $t_0$ that the specialization $g(x) = f(x,t_0)$ is separable in ${\mathbf Q}[x] = (A/(t-t_0))[x]$ implies the prime $(t-t_0)$ is unramified in $A'$. Let $E$ be the Galois closure of $F'/F$ and $B$ be the integral closure of $A$ in $E$. A prime ideal that is unramified in a finite extension is unramified in the Galois closure, so $(t-t_0)$ is unramified in $B$. For any prime $\mathfrak P$ in $B$ that lies over $(t-t_0)$, the residue field $B/\mathfrak P$ is a finite extension of $A/(t-t_0) = \mathbf Q$ and since $E/F$ is Galois the field $B/\mathfrak P$ is normal over $A/(t-t_0)$. These residue fields have characteristic 0, so they're separable: $B/\mathfrak P$ is a finite Galois extension of $\mathbf Q$. I leave it to you to check that $B/\mathfrak P$ is the Galois closure of $g(x) = f(t_0,x)$ over $\mathbf Q$. Then the isomorphism of $D(\mathfrak P|(t-t_0))$ with $\text{Aut}((B/\mathfrak P)/\mathbf Q) = \text{Gal}((B/\mathfrak P)/\mathbf Q)$ provides (by looking at the inverse map) an embedding of the Galois group of $g$ over $\mathbf Q$ into the Galois group of $f(x,t)$ over $F = {\mathbf Q}(t)$.
I agree with Damiano that there are problems when the specialization is not separable. In that case what happens is that the Galois group of the residue field extension is identified not with the decomposition group (a subgroup of the Galois group of $E/F$) but with the quotient group $D/I$ where $I = I(\mathfrak P|\mathfrak p)$ is the inertia group, and you don't generally expect a proper quotient group of a subgroup to naturally embed into the original group.
It is always a good idea to look at the local question first. What is the maximal pro-$p$ extension of $\mathbf{Q}_p$ ? There is a vast literature on the subject, starting with Demushkin whose results are exposed in an old Bourbaki talk by Serre on the Structure de certains pro-p-groupes (d'après Demuškin), (http://www.numdam.org/item?id=SB_1962-1964__8__145_0). They turn out to be extremely interesting groups which satisfy a kind "Poincaré duality" and have a striking presentation mysteriously similar to the presentation of the fundamental group of a compact orientable surface. Now I have to go for my yoga class; I hope you can find more on the web by yourself.
Addendum. Another good idea which I record here although I'm sure you don't need to be reminded of it is that in the local situation one should first ask what happens at primes $l\neq p$. What is the maximal pro-$l$ extension of $\mathbf{Q}_p$ ? This is much easier to answer because the only possible ramification is tame. You have the maximal unramified $l$-extension, which is a $\mathbf{Z}_l$-extension, and then a totally ramified extension of group $\mathbf{Z}_l(1)$, the projective limit of the roots of $1$ of $l$-power order. As a pro-$l$ group it admits the presentation
$\langle\tau,\sigma\mid\sigma\tau\sigma^{-1}=\tau^p\rangle$.
All this can be found in Chapter 16 of Hasse's Zahlentherie and in a paper by Albert in the Annals in the late 30s.
Addendum 2. A third good idea is to look at the function field analogues, wherein you replace $\mathbf{Q}$ by a function field $k(X)$ over a finite field $k$ of characterisitc $p$, where $X$ is a smooth projective absolutely irreducible curve over $k$. There would again be two cases, according as you are looking at the maximal pro-$l$ extension for a prime $l\neq p$ or for the prime $l=p$; the former should be much easier. You can simplify the problem by replacing $k$ by an algebraic closure. We are then in the setting of Abhyankar's conjecture (1957) which was very useful to Grothendieck as a first test for his theory of schemes. In any case, the conjecture was settled by Raynaud and Harbatter in the early 90s.
Best Answer
The splitting field of $x^n-2$ over $\mathbb{Q}$ is $K.L$ where $K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{2})$, so the order of the Galois group is $$ [K.L:\mathbb{Q}] = \frac{[K:\mathbb{Q}]\cdot[L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]} = \frac{n \phi(n)}{[K\cap L:\mathbb{Q}]}. $$ It remains to compute $m:=[K\cap L:\mathbb{Q}]$.
First show that $K\cap L = \mathbb{Q}(\sqrt[m]{2})$. For this, note that the norm $N_{L/(K\cap L)}(\sqrt[n]{2})$ is in $K\cap L$. This norm is the product of the $n/m$ conjugates of $\sqrt[n]{2}$ over $K$, so it is the product of $n/m$ of the conjugates of $\sqrt[n]{2}$ over $\mathbb{Q}$, and each of these conjugates has the form $\zeta_n^i \sqrt[n]{2}$. Hence the norm has the form $\zeta_n^i \sqrt[n]{2}^{n/m} = \zeta_n^i \sqrt[m]{2}$. Since this is in $K\cap L$, and $\zeta_n^i\in K$, it follows that $\sqrt[m]{2}\in K$, so $\sqrt[m]{2}\in K\cap L$. But $[\mathbb{Q}(\sqrt[m]{2}):\mathbb{Q}]=m=[K\cap L:\mathbb{Q}]$, so indeed $K\cap L=\mathbb{Q}(\sqrt[m]{2})$.
Next, since $\sqrt[m]{2}\in K$, and $K/\mathbb{Q}$ is abelian, it follows that $\mathbb{Q}(\sqrt[m]{2})/\mathbb{Q}$ is abelian and hence is Galois. Since $\zeta_m\sqrt[m]{2}$ is a conjugate of $\sqrt[m]{2}$ over $\mathbb{Q}$, it follows that $\zeta_m\in\mathbb{Q}(\sqrt[m]{2})$, so in particular $\zeta_m\in\mathbb{R}$, whence $m\le 2$.
The final step is to determine when $\sqrt{2}\in\mathbb{Q}(\zeta_n)$. For this, note that $\sqrt{2}\in\mathbb{Q}(\zeta_8)$ but $\sqrt{2}\notin\mathbb{Q}(\zeta_4)$, and that $\mathbb{Q}(\zeta_r)\cap\mathbb{Q}(\zeta_s)=\mathbb{Q}(\zeta_{(r,s)})$. Thus $\sqrt{2}\in\mathbb{Q}(\zeta_n)$ if and only if $8\mid n$. So the conclusion is that the splitting field of $x^n-2$ over $\mathbb{Q}$ has degree $n\phi(n)$ if $8\nmid n$, and has degree $n\phi(n)/2$ if $8\mid n$.
Question: is there a way to do this without showing that $K\cap L=\mathbb{Q}(\sqrt[m]{2})$, by using from the start that $K\cap L$ is a subfield of $\mathbb{Q}(\sqrt[n]{2})$ which is Galois over $\mathbb{Q}$, and then somehow going directly to the final step?