The Stacks project

Proof. Choose an algebraic closure $\overline{F}$. Then we can write $P = c (x - \beta _1) \ldots (x - \beta _ n)$ in $\overline{F}[x]$, see Lemma 9.10.2. Note that $c \in F^*$. Set $E = F(\beta _1, \ldots , \beta _ n)$. Then it is clear that $E$ is minimal with the requirement that $P$ splits completely over $E$.

Next, let $E'$ be another minimal field extension of $F$ such that $P$ splits completely over $E'$. Write $P = c (x - \alpha _1) \ldots (x - \alpha _ n)$ with $c \in F$ and $\alpha _ i \in E'$. Again it follows from minimality that $E' = F(\alpha _1, \ldots , \alpha _ n)$. Moreover, if we pick any $\sigma : E' \to \overline{F}$ (Lemma 9.10.5) then we immediately see that $\sigma (\alpha _ i) = \beta _{\tau (i)}$ for some permutation $\tau : \{ 1, \ldots , n\} \to \{ 1, \ldots , n\} $. Thus $\sigma (E') = E$. This implies that $E'$ is a normal extension of $F$ by Lemma 9.15.5 and that $E \cong E'$ as extensions of $F$ thereby finishing the proof. $\square$

Proof. Choose generators $\alpha _1, \ldots , \alpha _ n$ of $E$ over $F$. Let $P_1, \ldots , P_ n$ be the minimal polynomials of $\alpha _1, \ldots , \alpha _ n$ over $F$. Set $P = P_1 \ldots P_ n$. Observe that $(x - \alpha _1) \ldots (x - \alpha _ n)$ divides $P$, since each $(x - \alpha _ i)$ divides $P_ i$. Say $P = (x - \alpha _1) \ldots (x - \alpha _ n)Q$. Let $K/E$ be the splitting field of $P$ over $E$. We claim that $K$ is the splitting field of $P$ over $F$ as well (which implies that $K$ is normal over $F$). This is clear because $K/E$ is generated by the roots of $Q$ over $E$ and $E$ is generated by the roots of $(x - \alpha _1) \ldots (x - \alpha _ n)$ over $F$, hence $K$ is generated by the roots of $P$ over $F$.

Uniqueness. Suppose that $K'/E$ is a second smallest extension such that $K'/F$ is normal. Choose an algebraic closure $\overline{F}$ and an embedding $\sigma _0 : E \to \overline{F}$. By Lemma 9.10.5 we can extend $\sigma _0$ to $\sigma : K \to \overline{F}$ and $\sigma ' : K' \to \overline{F}$. By Lemma 9.15.3 we see that $\sigma (K) \cap \sigma '(K')$ is normal over $F$. By minimality we conclude that $\sigma (K) = \sigma (K')$. Thus $\sigma \circ (\sigma ')^{-1} : K' \to K$ gives an isomorphism of extensions of $E$. $\square$

Proof. Proof of (1). Let $M'$ be the smallest subextension of $L/K$ containing $M$ which is normal over $K$. By Lemma 9.16.3 this is the normal closure of $M/K$ and is finite over $K$.

Proof of (2). Let $\alpha _1, \ldots , \alpha _ n \in M'$ generate $M'$ over $M$. Let $P_1, \ldots , P_ n$ be the minimal polynomials of $\alpha _1, \ldots , \alpha _ n$ over $K$. Let $\alpha _{i, j}$ be the roots of $P_ i$ in $L$. Let $M'' = M(\alpha _{i, j})$. It follows from Lemma 9.15.6 (applied with the set of generators $M \cup \{ \alpha _{i, j}\} $) that $M''$ is normal over $K$. $\square$

Proof. Choose an algebraic closure $\overline{K}$ of $K$. Set $n = [L : K]_ s = |\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K})|$ with equality by Lemma 9.14.8. Say $\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K}) = \{ \sigma _1, \ldots , \sigma _ n\} $. Let $M' \subset \overline{K}$ be the $K$-subalgebra generated by $\sigma _ i(L)$, $i = 1, \ldots , n$. It follows from Lemma 9.15.5 that $M'$ is normal over $K$ and that it is the smallest normal subextension of $\overline{K}$ containing $\sigma _1(L)$. By uniqueness of normal closure we have $M \cong M'$. Finally, there is a surjective map

and note that $n \leq [L : K]$ by definition. $\square$