Automorphisms of a splitting field Can someone please tell me how to determine all $\mathbb{Q}$-automorphisms of  some splitting field $F$ of $X^3-3 \in \mathbb{Q}[X]$ and to determine $[F:\mathbb{Q}]$ ?
I think $[F:\mathbb{Q}]=3$, since I can construct a specific splitting field containing the three (complex) roots of the polynomial $X^3-3 $, but I don't know how to prove that it equals $3$. And for the automorphisms, I don't even know how to approach it.
 A: Let $F$ be a field.
Let $F^\times$ be the multiplicative group of $F$.
Let $(a, b) \in F^\times \times F$.
We denote by $\lambda(a, b)$ the permutation $x \rightarrow ax + b$ on $F$.
Let $(a, b), (c, d) \in F^\times \times F$.
Since $a(cx + d) + b = acx + ad + b$, $\lambda(a, b)\circ\lambda(c, d) = \lambda(ac, ad + b)$.
Hence $\{\lambda(a, b) | (a, b) \in F^\times \times F\}$ is a permutation group on $F$.
We call this group the affine general linear group of degree $1$ on $F$ and denote it by $AGL(1, F)$.
Since $\lambda(a, b)(0) = b$ and $\lambda(a, b)(1) - b = a$, $(a, b)$ is uniquely determined by $\lambda(a, b)$. Hence $|AGL(1, F)| = |F^\times \times F|$.
Proposition
Let $p, q$  be prime numbers which may or may not be equal.
Let $a$ be an integer which is divisible by $q$, but not not divisible by $q^2$.
$X^p - a$ is irreducible in $\mathbb{Q}[X]$ by the Eisenstein's criterion.
Let $K$ be the splitting field of $X^p - a$ in $\mathbb{C}$.
Then the Galois group $G$ of $K/\mathbb{Q}$ is isomorphic to $AGL(1, F)$, where $F = \mathbb{Z}/p\mathbb{Z}$.
Proof:
Let $\zeta$ be a primitive $p$-th root of unity.
Let $\alpha$ be a root of $X^p - a$ in $\mathbb{C}$.
$\alpha,\alpha\zeta,\dots,\alpha\zeta^{p-1}$ are distinct roots of $X^p - a$.
Hence $K = \mathbb{Q}(\alpha,\alpha\zeta,\dots,\alpha\zeta^{p-1})$.
Since $\zeta = \alpha\zeta/\alpha \in K$, $K = \mathbb{Q}(\alpha, \zeta)$.
Let $\sigma \in G$.
Since $\sigma(\zeta)^p = 1$, there exists an integer $b$ such that $\sigma(\zeta) = \zeta^b$.
Clearly $b$ (mod $p$) is uniquely determined by $\sigma$ and $b$ (mod $p) \in F^\times$.
On the other hand, there exists an integer $c$ such that $\sigma(\alpha) = \alpha\zeta^c$.
Clearly $c$ (mod $p$) is uniquely determined by $\sigma$.
We denote $\phi(\sigma) = \lambda(b$ mod $p、c$ mod $p) \in AGL(1, F)$.
Hence we get a map $\phi\colon G \rightarrow AGL(1, F)$.
Let $\sigma, \tau \in G$.
Suppose $\phi(\sigma) = \lambda(b$ mod $p, c$ mod $p$) and
$\phi(\tau) = \lambda(d$ mod $p, e$ mod $p$).
Then $\sigma\tau(\zeta) = \zeta^{bd}$ and
$\sigma\tau(\alpha) = \sigma(\alpha\zeta^e) = \alpha\zeta^c\zeta^{be} = \alpha\zeta^{be + c}$.
Hence $\phi(\sigma\tau) = \phi(\sigma)\phi(\tau)$.
Hence $\phi$ is a homomorphism.
It is easy to see that $\phi$ is injective.
It remains to prove that $\phi$ is surjective.
Since $[\mathbb{Q}(\zeta) \colon \mathbb{Q}] = p - 1$, $[K \colon \mathbb{Q}(\zeta)] > 1$.
Let $H$ be the Galois group of $K/\mathbb{Q}(\zeta)$.
Let $\tau \in H$.
There exists an integer $c$ such that $\tau(\alpha) = \alpha\zeta^c$.
We define a map $\psi\colon H \rightarrow F$ by $\psi(\tau) = c$ mod $p$.
It is easy to see that $\psi$ is an injective homomorphism.
Hence $|F| = p$ is divisible by $|H|$.
Since $|H| > 1$, $|H| = p$,
Hence $[K \colon \mathbb{Q}] = p(p-1)$.
Hence $\phi$ is surjective.
QED
