Exercise 12.2.11

Proof.

(a)

We use the notations of Theorem 12.2.3: $V=V^{(0)},V^{\prime }=V^{(1)},\dots ,V^{(m-1)}$ are the roots of the polynomial $h$, irreducible over $F$. Moreover ${\alpha }_1,\dots ,{\alpha }_n$ are the roots of $f$, and $L=F({\alpha }_1,\dots ,{\alpha }_n)=F(V)$.

Write

$$\mathrm{Gal}(L∕F)=\{{\sigma }_0=e,{\sigma }_1,\dots ,{\sigma }_{m-1}\},$$

where ${\sigma }_i$ is the unique $F$-automorphism of $L$ such that

$${\sigma }_i(V)=V^{(i)},0\le i\le n-1.$$

Let ${\tau }_i\in S_n$ the permutation associate to ${\sigma }_i$, defined by

$${\sigma }_i({\alpha }_j)={\alpha }_{{\tau }_i(j)},0\le i\le m-1,1\le j\le n.$$

Since $F({\alpha }_1,\dots ,{\alpha }_n)=F(V)$, there are ${\phi }_j\in F(x)$ such that

$${\alpha }_j={\phi }_{j-1}(V),1\le j\le n.$$

Then

$$\begin{array}{llll}\hfill {\alpha }_{{\tau }_i(j)}& ={\sigma }_i({\alpha }_j)& \hfill & \\ \hfill & ={\sigma }_i({\phi }_{j-1}(V))& \hfill & \\ \hfill & ={\phi }_{j-1}({\sigma }_i(V))& \hfill & \\ \hfill & ={\phi }_{j-1}(V^{(i)})& \hfill & \end{array}$$

Thus

$${\alpha }_{{\tau }_i(j)}={\phi }_{j-1}(V^{(i)}),0\le i\le m-1,1\le j\le n.$$

Therefore the orbit of the arrangement $a=({\alpha }_1,\dots ,{\alpha }_n)$ under the action of $G=\{{\tau }_0,\dots ,{\tau }_{m-1}\},G\subset S_n,G\simeq \mathrm{Gal}(L∕F)$ is given by

$$\begin{array}{ccccc}\hfill a=& ({\alpha }_1=\phi (V),\hfill & {\alpha }_2={\phi }_1(V),\hfill & \text{…,}\hfill & {\alpha }_n={\phi }_{n-1}(V))\hfill \\ \hfill {\tau }_1\cdot a=& ({\alpha }_{{\tau }_1(1)}=\phi (V^{\prime }),\hfill & {\alpha }_{{\tau }_1(2)}={\phi }_1(V^{\prime }),\hfill & \text{…,}\hfill & {\alpha }_{{\tau }_1(n)}={\phi }_{n-1}(V^{\prime }))\hfill \\ \hfill \dots & \hfill & \hfill & \hfill & \hfill \\ \hfill {\tau }_{m-1}\cdot a=& ({\alpha }_{{\tau }_{m-1}(1)}=\phi (V^{(m-1)}),\hfill & {\alpha }_{{\tau }_{m-1}(2)}={\phi }_1(V^{(m-1)}),\hfill & \text{…,}\hfill & {\alpha }_{{\tau }_{m-1}(n)}={\phi }_{n-1}(V^{(m-1)}))\hfill \\ \hfill \end{array}$$

The set of arrangements $A$ described by Galois is the orbit of the arrangement $({\alpha }_1,\dots ,{\alpha }_n)$ under the $G$-action, where $G$ is the subgroup $G$ of $S_n$ isomorphic to $\mathrm{Gal}(L∕F)$.

(b)

Let $\psi :G\to A={\mathcal{}O}_a$ defined by $\sigma \mapsto \sigma \cdot a$.

$\text{∙}$

If ${\tau }_k\cdot a={\tau }_l\cdot a$, where $a=({\alpha }_1,\dots ,{\alpha }_n)$ and ${\tau }_k,{\tau }_l\in G$, then $${\alpha }_{{\tau }_k(j)}={\alpha }_{{\tau }_l(j)},1\le j\le n.$$

If ${\sigma }_k,{\sigma }_l\in \mathrm{Gal}(L∕F)$ are the automorphisms associate to ${\tau }_k,{\tau }_l\in S_n$, then

$${\sigma }_k({\alpha }_j)={\sigma }_l({\alpha }_j),1\le j\le n.$$

Since $L=F({\alpha }_1,\dots ,{\alpha }_n)$, this implies that ${\sigma }_k={\sigma }_l$, thus ${\tau }_k={\tau }_l$, and $\psi$ is injective.

$\text{∙}$

Moreover $|G|=|A|=m$, therefore the injective map $\psi$ is also surjective.

$\psi :G\to A$ is a bijection.