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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.

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    yes, once you have in hand one of the roots of the polynomial, you need to write down explicitly what the other roots are. Then you need to notice what other irrationality needs to be adjoined to the setup to get all three in any field. You'll see that there is a unique quadratic extension of $\mathbb Q$ there, and three different cubic extensions. This is an exercise that everyone must do for her/himself. – 2012-11-07

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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 $p态c$ 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

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    @BenjaLim You may write a new answer. – 2012-12-15