Determining all automorphisms of a splitting field

Question

Let $L$ be the splitting field of $x^3-3 \in \mathbb{Q}[X]$ over $\mathbb{Q}$. I am trying to determine all $\mathbb{Q}-Automorphisms$ of $L$

Solving $x^3-3=0$ gives the solutions $3^{\frac{1}{3}}$ , $3^{\frac{1}{3}} e^{\frac{2 \pi i}{3}}$ , $3^{\frac{1}{3}} e^{\frac{4 \pi i}{3}}$. I will call those solutions $\alpha_1, \alpha_2, \alpha_3$ This means I know that $L=\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3)$

Now I need to find $\phi: L \rightarrow L$, such that

$\phi(a+b)=\phi(a)+ \phi (b)$

and

$\phi(ab)=\phi(a) \phi (b)$

such that $\phi$ is an isomorphism and $\phi_{| K}=id_K$

I am not really sure how to find all such $\mathbb{Q}$-Automorphisms $\phi$.

Note: I did try to search Stackexchange, but I did not find anything that could help me. Keep in mind that I am a beginner and don't know Galois Groups.

Answer 1

The fundamental tool here is :

Let $\mathbb{Q}\subseteq\mathbb{Q}(\alpha)$ be a simple, algebraic extension. Let $f=f_{\alpha,\mathbb{Q}}$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and $\mathcal{R}_f(\mathbb{C})$ the set of complex roots of $f$. Let $Hom_{\ \mathbb{Q}}(\mathbb{\mathbb{Q}(\alpha),\mathbb{C}})=\{\sigma:\mathbb{Q}(\alpha)\rightarrow \mathbb{C} \ | \ \sigma_{|\mathbb{Q}}=id_\mathbb{Q} \}$ be the set of $\mathbb{Q}$-homomorphisms from $\mathbb{Q}(\alpha)$ to $\mathbb{C}$. This hold a bijection : $$ Hom_{\ \mathbb{Q}}(\mathbb{\mathbb{Q}(\alpha),\mathbb{C}})\longleftrightarrow \mathcal{R}_f(\mathbb{C}) \; \; \; \; \; \sigma\mapsto\sigma(\alpha) $$

Notice that $f(X)=X^3-3=f_{\sqrt[3]{3},\mathbb{Q}}(X)$ is the minimal polynomial of $\sqrt[3]{3}$ over $\mathbb{Q}$. The splitting field of $f$ may also be expressed as $L=\mathbb{Q}(\alpha_1,\zeta_3)$, where $\zeta_3=e^{2\pi i/3}$, because $\alpha_2=\alpha_1\cdot\zeta_3$, and $\alpha_3=\alpha_1 \cdot \zeta_3^2$. Then, you get $Hom_{\ \mathbb{Q}}(\mathbb{\mathbb{Q}(\alpha_1,\zeta_3),\mathbb{C}})=Aut_{\ \mathbb{Q}}(L)$ since $\mathbb{Q}(\alpha_1,\zeta_3)$ is a splitting field, and thus normal extension.

By applying the previous result to the simple extensions $\mathbb{Q}(\alpha_1)\subseteq\mathbb{Q}(\alpha_1,\zeta_3)$, and $\mathbb{Q}(\zeta_3)\subseteq\mathbb{Q}(\alpha_1,\zeta_3)$ you determine all the $6$ $\mathbb{Q}$-automorphims of $L$ by composition of the $2$ following $\mathbb{Q}$-automorphims :

1. $\sigma : \mathbb{Q}(\alpha_1,\zeta_3)\rightarrow \mathbb{Q}(\alpha_1,\zeta_3)$ such that $\sigma(\alpha_1)=\alpha_1 \cdot \zeta_3=\alpha_2$, and $\sigma(\zeta_3)=\zeta_3$

2. $\tau : \mathbb{Q}(\alpha_1,\zeta_3)\rightarrow \mathbb{Q}(\alpha_1,\zeta_3)$ such that $\tau(\alpha_1)=\alpha_1$, and $\sigma(\zeta_3)=\zeta_3^2$