The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$


Comments (1)

Comment #1769 by Carl on

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


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