Lemma 9.15.5. Let $E/F$ be an algebraic extension of fields. Let $\overline{F}$ be an algebraic closure of $F$. The following are equivalent

1. $E$ is normal over $F$, and

2. for every pair $\sigma , \sigma ' \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$ we have $\sigma (E) = \sigma '(E)$.

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

$T = \{ \beta \in \overline{F} \mid P(\beta ) = 0\text{ for some }P \in \mathcal{P}\}$

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

$P = (x - \beta _1) \ldots (x - \beta _ n)$

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. $\square$

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