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
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
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
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