Exercise 6.3.7

Proof.

We define a left action of the Galois group $G=\mathrm{Gal}(L∕F)$ on the set $S$ of the roots of $f$ by $\sigma \cdot \alpha =\sigma (\alpha )$, where $\sigma \in G,\alpha \in S$ (we know that $\sigma (\alpha )\in S$).

For a fixed $\alpha \in S$, define $G_{\alpha }={\mathrm{Stab}}_G(\alpha )=\{\sigma \in G|\sigma (\alpha )=\alpha \}$ the stabilizer of $\alpha$ in $G$, and ${\mathcal{}O}_{\alpha }=\{\sigma \cdot \alpha |\sigma \in G\}$ its orbit.

As $f$ is irreducible, by proposition 5.8.1, if $\alpha ,\beta$ are two roots of $f$, there exists a field isomorphism $\sigma :L\to L$, which is identity on $F$ (so $\sigma \in \mathrm{Gal}(L∕F)$, and such that $\sigma (\alpha )=\beta$.

Therefore the action of $G$ on $S$ is transitive, so there exists a unique orbit: for all $\alpha \in S$, ${\mathcal{}O}_{\alpha }=S$, thus

Indeed the separability of $f$ implies that $f$ has $n=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$ distinct roots in $L$.

As $(G:G_{\alpha })=|{\mathcal{}O}_{\alpha }|$, we obtain

So $n=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$ divides $|\mathrm{Gal}(L∕F)|$. □