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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