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The Stacks project

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{\mathrm{Mor}}\nolimits _ F(E, \overline{F})|. Pick an element \sigma _0 \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F}). Then the map

\text{Aut}(E/F) \longrightarrow \mathop{\mathrm{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{\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


Comments (1)

Comment #1769 by Carl on

First line of proof should have an _s after [E:F].

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