Lemma 9.15.9. Let $E/F$ be a finite extension. We have

with equality if and only if $E$ is normal over $F$.

Lemma 9.15.9. Let $E/F$ be a finite extension. We have

\[ |\text{Aut}(E/F)| \leq [E : F]_ s \]

with equality if and only if $E$ is normal over $F$.

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

\[ \text{Aut}(E/F) \longrightarrow \mathop{Mor}\nolimits _ F(E, \overline{F}),\quad \tau \longmapsto \sigma _0 \circ \tau \]

is injective. Thus the inequality. If equality holds, then every $\sigma \in \mathop{Mor}\nolimits _ F(E, \overline{F})$ 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 $\sigma \in \mathop{Mor}\nolimits _ F(E, \overline{F})$. Thus we get an automorphism of $E$ over $F$ by setting $\tau = \sigma _0^{-1} \circ \sigma $. Whence the map displayed above is surjective. $\square$

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## Comments (1)

Comment #1769 by Carl on