The Stacks project

Proof. Let $\alpha \in K$. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $Q$ be the minimal polynomial of $\alpha$ over $E$. Then $Q$ divides $P$ in the polynomial ring $E[x]$, say $P = QR$. Hence, if $P$ splits completely over $K$, then so does $Q$. $\square$

Proof. Direct from the definitions. $\square$

Proof. If the characteristic is zero, then $E_{sep} = E$, and the result is clear. If the characteristic is $p > 0$, then $E_{sep}$ is the set of elements of $E$ which are separable over $F$. Then if $\alpha \in E_{sep}$ has minimal polynomial $P$ write $P = c(x - \alpha )(x - \alpha _2) \ldots (x - \alpha _ d)$ with $\alpha _2, \ldots , \alpha _ d \in E$. Since $P$ is a separable polynomial and since $\alpha _ i$ is a root of $P$, we conclude $\alpha _ i \in E_{sep}$ as desired. $\square$

Proof. Let $\mathcal{P}$ be the set of all minimal polynomials over $F$ of all elements of $E$. Set

It is clear that if $E$ is normal over $F$, then $\sigma (E) = T$ for all $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. Thus we see that (1) implies (2).

Conversely, assume (2). Pick $\beta \in T$. We can find a corresponding $\alpha \in E$ whose minimal polynomial $P \in \mathcal{P}$ annihilates $\beta$. Because $F(\alpha ) = F[x]/(P)$ we can find an element $\sigma _0 \in \mathop{\mathrm{Mor}}\nolimits _ F(F(\alpha ), \overline{F})$ mapping $\alpha$ to $\beta$. By Lemma 9.10.5 we can extend $\sigma _0$ to a $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. Whence we see that $\beta$ is in the common image of all embeddings $\sigma : E \to \overline{F}$. It follows that $\sigma (E) = T$ for any $\sigma$. Fix a $\sigma$. Now let $P \in \mathcal{P}$. Then we can write

for some $n$ and $\beta _ i \in \overline{F}$ by Lemma 9.10.2. Observe that $\beta _ i \in T$. Thus $\beta _ i = \sigma (\alpha _ i)$ for some $\alpha _ i \in E$. Thus $P = (x - \alpha _1) \ldots (x - \alpha _ n)$ splits completely over $E$. This finishes the proof. $\square$

Proof. Let $P_ i$ be the minimal polynomial of $\alpha _ i$ over $F$. Let $\alpha _ i = \alpha _{i, 1}, \alpha _{i, 2}, \ldots , \alpha _{i, d_ i}$ be the roots of $P_ i$ over $E$. Given two embeddings $\sigma , \sigma ' : E \to \overline{F}$ over $F$ we see that

because both sides are equal to the set of roots of $P_ i$ in $\overline{F}$. The elements $\alpha _{i, j}$ generate $E$ over $F$ and we find that $\sigma (E) = \sigma '(E)$. Hence $E/F$ is normal by Lemma 9.15.5. $\square$

Proof. Choose an algebraic closure $\overline{L}$ of $L$ (Theorem 9.10.4).

Let $\tau$ be as in (1). Then $\tau (M) = M$ as subfields of $\overline{L}$ by Lemma 9.15.5 and hence $\tau |_ M : M \to M$ is an automorphism.

Let $\sigma : M \to L$ be as in (2). By Lemma 9.10.5 we can extend $\sigma$ to a map $\tau : L \to \overline{L}$, i.e., such that

is commutative. By Lemma 9.15.5 we see that $\tau (L) = L$. Hence $\tau : L \to L$ is an automorphism which extends $\sigma$. $\square$

Proof. Choose an algebraic closure $\overline{F}$ of $F$. Recall that $[E : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})|$. Pick an element $\sigma _0 \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. Then the map

is injective. Thus the inequality. If equality holds, then every $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$ is gotten by precomposing $\sigma _0$ by an automorphism. Hence $\sigma (E) = \sigma _0(E)$. Thus $E$ is normal over $F$ by Lemma 9.15.5.

Conversely, assume that $E/F$ is normal. Then by Lemma 9.15.5 we have $\sigma (E) = \sigma _0(E)$ for all $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$. Thus we get an automorphism of $E$ over $F$ by setting $\tau = \sigma _0^{-1} \circ \sigma$. Whence the map displayed above is surjective. $\square$

Proof. Given $\tau$ replace $L$ by $\tau (L) \subset E$ and apply Lemma 9.15.7. $\square$