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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