The Stacks project

Proof. To see this, suppose that we have a factorization $x^ p - t = f g$. Taking derivatives we get $f' g + f g' = 0$. Note that neither $f' = 0$ nor $g' = 0$ as the degrees of $f$ and $g$ are smaller than $p$. Moreover, $\deg (f') < \deg (f)$ and $\deg (g') < \deg (g)$. We conclude that $f$ and $g$ have a factor in common. Thus if $x^ p - t$ is reducible, then it is of the form $x^ p - t = c f^ n$ for some irreducible $f$, $c \in F^*$, and $n > 1$. Since $p$ is a prime number this implies $n = p$ and $f$ linear, which would imply $x^ p - t$ has a root in $F$. Contradiction. $\square$

Proof. Say the characteristic of $k$ is $p$. Choose $\alpha \in F$. Then $\alpha ^ q \in E$ for some $p$-power $q$. Whereupon $(\alpha ^ q)^{q'} \in k$ for some $p$-power $q'$. Hence $\alpha ^{qq'} \in k$. $\square$

Proof. Let $p$ be the characteristic of $k$. Let $\alpha , \beta \in E$ be purely inseparable over $k$. Say $\alpha ^ q \in k$ and $\beta ^{q'} \in k$ for some $p$-powers $q, q'$. If $q''$ is a $p$-power, then $(\alpha + \beta )^{q''} = \alpha ^{q''} + \beta ^{q''}$. Hence if $q'' \geq q, q'$, then we conclude that $\alpha + \beta $ is purely inseparable over $k$. Similarly for the difference, product and quotient of $\alpha $ and $\beta $. $\square$

Proof. By induction on the degree of $E/F$. If the degree of the extension is $1$ then the result is clear (with $n = 0$). If not, then choose $\alpha \in E$, $\alpha \not\in F$. Say $\alpha ^{p^ r} \in F$ for some $r > 0$. Pick $r$ minimal and replace $\alpha $ by $\alpha ^{p^{r - 1}}$. Then $\alpha \not\in F$, but $\alpha ^ p \in F$. Then $t = \alpha ^ p$ is not a $p$th power in $F$ (because that would imply $\alpha \in F$, see Lemma 9.12.5 or its proof). Thus $F \subset F(\alpha )$ is a subextension of degree $p$ (Lemma 9.14.2). By induction we find $\alpha _1, \ldots , \alpha _ n \in E$ generating $E/F(\alpha )$ satisfying the conclusions of the lemma. The sequence $\alpha , \alpha _1, \ldots , \alpha _ n$ does the job for the extension $E/F$. $\square$

Proof. If the characteristic is zero we set $E_{sep} = E$. Assume the characteristic is $p > 0$. Let $E_{sep}$ be the set of elements of $E$ which are separable over $F$. This is a subextension by Lemma 9.12.13 and of course $E_{sep}$ is separable over $F$. Given an $\alpha $ in $E$ there exists a $p$-power $q$ such that $\alpha ^ q$ is separable over $F$. Namely, $q$ is that power of $p$ such that the minimal polynomial of $\alpha $ is of the form $P(x^ q)$ with $P$ separable algebraic, see Lemma 9.12.1. Hence $E/E_{sep}$ is purely inseparable. Uniqueness is clear. $\square$

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. $\square$

Proof. We first prove this in case $K$ is finite over $F$. Since we have multiplicativity for the usual degree (by Lemma 9.7.7) it suffices to prove one of the two formulas. By Lemma 9.14.8 we have $[K : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})|$. By the same lemma, given any $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$ the number of extensions of $\sigma $ to a map $\tau : K \to \overline{F}$ is $[K : E]_ s$. Namely, via $E \cong \sigma (E) \subset \overline{F}$ we can view $\overline{F}$ as an algebraic closure of $E$. Combined with the fact that there are $[E : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})|$ choices for $\sigma $ we obtain the result.

We omit the proof if the extensions are infinite. $\square$