Lemma 9.12.10. Assumptions and notation as in Situation 9.12.7. If each P_ i is separable, i.e., each \alpha _ i is separable over K_{i - 1}, then
|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| = [K : F]
and the field extension K/F is separable. If one of the \alpha _ i is not separable over K_{i - 1}, then |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| < [K : F].
Proof.
If \alpha _ i is separable over K_{i - 1} then \deg _ s(P_ i) = \deg (P_ i) = [K_ i : K_{i - 1}] (last equality by Lemma 9.9.2). By multiplicativity (Lemma 9.7.7) we have
[K : F] = \prod [K_ i : K_{i - 1}] = \prod \deg (P_ i) = \prod \deg _ s(P_ i) = |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})|
where the last equality is Lemma 9.12.9. By the exact same argument we get the strict inequality |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| < [K : F] if one of the \alpha _ i is not separable over K_{i - 1}.
Finally, assume again that each \alpha _ i is separable over K_{i - 1}. We will show K/F is separable. Let \gamma = \gamma _1 \in K be arbitrary. Then we can find additional elements \gamma _2, \ldots , \gamma _ m such that K = F(\gamma _1, \ldots , \gamma _ m) (for example we could take \gamma _2 = \alpha _1, \ldots , \gamma _{n + 1} = \alpha _ n). Then we see by the last part of the lemma (already proven above) that if \gamma is not separable over F we would have the strict inequality |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| < [K : F] contradicting the very first part of the lemma (already prove above as well).
\square
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