Lemma 9.14.8. Let K/F be a finite extension. Let \overline{F} be an algebraic closure of F. Then [K : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})|.
Proof. We first prove this when K/F is purely inseparable. Namely, we claim that in this case there is a unique map K \to \overline{F}. This can be seen by choosing a sequence of elements \alpha _1, \ldots , \alpha _ n \in K as in Lemma 9.14.5. The irreducible polynomial of \alpha _ i over F(\alpha _1, \ldots , \alpha _{i - 1}) is x^ p - \alpha _ i^ p. Applying Lemma 9.12.9 we see that |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| = 1. On the other hand, [K : F]_ s = 1 in this case hence the equality holds.
Let's return to a general finite extension K/F. In this case choose F \subset K_ s \subset K as in Lemma 9.14.6. By Lemma 9.12.11 we have |\mathop{\mathrm{Mor}}\nolimits _ F(K_ s, \overline{F})| = [K_ s : F] = [K : F]_ s. On the other hand, every field map \sigma ' : K_ s \to \overline{F} extends to a unique field map \sigma : K \to \overline{F} by the result of the previous paragraph. In other words |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| = |\mathop{\mathrm{Mor}}\nolimits _ F(K_ s, \overline{F})| and the proof is done. \square
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