Exercise 7.1.8

Proof. Here, as in theorem 7.1.1, $F\subset L$ is a normal separable extension.

Let $\alpha \in L$, and $h$ the minimal polynomial of $\alpha$ over $F$. As $L$ is a normal extension of $F$, $h$ splits completely over $L$, so $h=\underset{i=1}{\overset{r}{\prod <!--\; FUNCTION\; APPLICATION\; -->}}(x-{\alpha }_i)$, where ${\alpha }_1,\dots ,{\alpha }_r\in L$, and the ${\alpha }_i,1\le i\le r$ are distinct since $h$ is a separable polynomial.

The Galois group $G=\mathrm{Gal}(L∕F)$ acts on the set $S=\{{\alpha }_1,\cdots ,{\alpha }_r\}$, with the action defined by $\sigma \cdot \gamma =\sigma (\gamma ),\sigma \in G,\gamma \in S$.

As $h$ is irreducible over $F$, $G$ acts transitively on $S$, so the orbit ${\mathcal{}O}_{\alpha }$ of $\alpha$ is $S$ of cardinality $r$, and $G_{\alpha }$, the stabilizer of $\alpha$ in $G$ satisfies

As $F\subset L$ is a Galois extension, the Galois group $G$ has order $n=|G|=[L:F]$. Consequently $|G_{\alpha }|=n∕r$, so $G_{\alpha }$ is a subgroup of $G$ with index $r$ and cardinality $m:=n∕r$.

Note that, or all $\sigma \in G$,

Therefore

As $h$ is the minimal polynomial of $\alpha$ over $F$, $[F(\alpha ):F]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)=r$.

Since $F\subset L$ is a Galois extension, $F(\alpha )\subset L$ also, so we find again by the Tower Theorem:

Let ${\sigma }_1,\dots ,{\sigma }_r$ a complete system of representants of the left cosets $\sigma G_{\alpha },\sigma \in G$. Then the ${\sigma }_i{G}_{\alpha }$ form a partition of $G$ :

If $\sigma \in {\sigma }_i{G}_{\alpha }$, then $\sigma ={\sigma }_i\tau ,\tau \in G_{\alpha }$, thus $\sigma (\alpha )={\sigma }_i(\tau (\alpha ))={\sigma }_i(\alpha )$. Let ${\gamma }_i={\sigma }_i(\alpha )\in S$. The image of $\alpha$ by all the elements of the left coset ${\sigma }_i{G}_{\alpha }$ is a constant equal to ${\gamma }_i={\sigma }_i(\alpha )$. As $|{\sigma }_i{G}_{\alpha }|=|G_{\alpha }|=m$,

Moreover $T:=\{{\gamma }_1,\cdots ,{\gamma }_r\}\subset \{{\alpha }_1,\cdots ,{\alpha }_r\}$, and the ${\gamma }_i,1\le i\le r,$ are distinct since

Moreover $T\subset S,|T|=|S|=r$, thus $T=S$.

Consequently $g=\underset{i=1}{\overset{r}{\prod <!--\; FUNCTION\; APPLICATION\; -->}}{(x-{\gamma }_i)}^m=\underset{i=1}{\overset{r}{\prod }}{(x-{\alpha }_i)}^m=h^m$.

Conclusion: if $F\subset L$is a Galois extension, $h$the minimal polynomial of $\alpha \in L$over $F$, and $g=\underset{\sigma \in G}{\prod <!--\; FUNCTION\; APPLICATION\; -->}(x-\sigma .\alpha )$, then $g=h^m,m\in {\mathbb{N}}^{\text{∗}}$(where $m=[L:F(\alpha )]$). □