The Stacks project

Proof. Note that $P'$ has degree $< \deg (P)$. Hence if $P$ and $P'$ are not relatively prime, then $(P, P') = (R)$ where $R$ is a polynomial of degree $< \deg (P)$ contradicting the irreducibility of $P$. This proves we have the dichotomy between (1) and (2).

Assume we are in case (2) and $P = a_ d x^ d + \ldots + a_0$. Then $P' = da_ d x^{d - 1} + \ldots + a_1$. In characteristic $0$ we see that this forces $a_ d, \ldots , a_1 = 0$ which would mean $P$ is constant a contradiction. Thus we conclude that the characteristic $p$ is positive. In this case the condition $P' = 0$ forces $a_ i = 0$ whenever $p$ does not divide $i$. In other words, $P(x) = P_1(x^ p)$ for some nonconstant polynomial $P_1$. Clearly, $P_1$ is irreducible as well. By induction on the degree we see that $P_1(x) = Q(x^ q)$ as in the statement of the lemma, hence $P(x) = Q(x^{pq})$ and the lemma is proved. $\square$

Proof. We will use Lemma 9.12.1 without further mention. Let $P$ be the minimal polynomial of $\alpha $ over $F$. Let $Q$ be the minimal polynomial of $\alpha $ over $E$. Then $Q$ divides $P$ in the polynomial ring $E[x]$, say $P = QR$. Then $P' = Q'R + QR'$. Thus if $Q' = 0$, then $Q$ divides $P$ and $P'$ hence $P' = 0$ by the lemma. This proves (1). Part (2) follows immediately from (1) and the definitions. $\square$

Proof. Suppose that $\alpha \in \overline{F}$ is a root of both $P$ and $P'$. Then $P = (x - \alpha )Q$ for some polynomial $Q$. Taking derivatives we obtain $P' = Q + (x - \alpha )Q'$. Thus $\alpha $ is a root of $Q$. Hence we see that if $P$ and $P'$ have a common root, then $P$ does not have pairwise distinct roots. Conversely, if $P$ has a repeated root, i.e., $(x - \alpha )^2$ divides $P$, then $\alpha $ is a root of both $P$ and $P'$. Combined with Lemma 9.11.2 this proves the lemma. $\square$

Proof. Clearly, $\alpha $ is a root of $P(x^ p)$ if and only if $\alpha ^ p$ is a root of $P$. In other words, the roots of $P(x^ p)$ are the roots of $x^ p - \beta $, where $\beta $ is a root of $P$. Thus it suffices to show that the map $\overline{F} \to \overline{F}$, $\alpha \mapsto \alpha ^ p$ is bijective. It is surjective, as $\overline{F}$ is algebraically closed which means that every element has a $p$th root. It is injective because $\alpha ^ p = \beta ^ p$ implies $(\alpha - \beta )^ p = 0$ because the characteristic is $p$. And of course in a field $x^ p = 0$ implies $x = 0$. $\square$

Proof. Let $(\beta _1, \ldots , \beta _ n)$ be an element of the right hand side. We construct a map of fields corresponding to it by induction. Namely, we set $\varphi _0 : K_0 \to \overline{F}$ equal to the given map $K_0 = F \subset \overline{F}$. Having constructed $\varphi _{i - 1} : K_{i - 1} \to \overline{F}$ we observe that $K_ i = K_{i - 1}[x]/(P_ i)$. Hence we can set $\varphi _ i$ equal to the unique map $K_ i \to \overline{F}$ inducing $\varphi _{i - 1}$ on $K_{i - 1}$ and mapping $x$ to $\beta _ i$. This works precisely as $\beta _ i$ is a root of $P_ i^\varphi $. Uniqueness implies that the two constructions are mutually inverse. $\square$

Proof. This follows immediately from Lemma 9.12.8. Observe that a key ingredient we are tacitly using here is the well-definedness of the separable degree of an irreducible polynomial which was observed just prior to Definition 9.12.6. $\square$

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

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). $\square$

Proof. This is a corollary of Lemma 9.12.10. Namely, since $K/F$ is finite we can find finitely many elements $\alpha _1, \ldots , \alpha _ n \in K$ generating $K$ over $F$ (for example we can choose the $\alpha _ i$ to be a basis of $K$ over $F$). If $K/F$ is separable, then each $\alpha _ i$ is separable over $F(\alpha _1, \ldots , \alpha _{i - 1})$ by Lemma 9.12.3 and we get equality by Lemma 9.12.10. On the other hand, if we have equality, then no matter how we choose $\alpha _1, \ldots , \alpha _ n$ we get that $\alpha _1$ is separable over $F$ by Lemma 9.12.10. Since we can start the sequence with an arbitrary element of $K$ it follows that $K$ is separable over $F$. $\square$

Proof. Choose $\alpha \in F$. Then $\alpha $ is separable algebraic over $E$. Let $P = x^ d + \sum _{i < d} a_ i x^ i$ be the minimal polynomial of $\alpha $ over $E$. Each $a_ i$ is separable algebraic over $k$. Consider the tower of fields

Because $a_ i$ is separable algebraic over $k$ it is separable algebraic over $k(a_0, \ldots , a_{i - 1})$ by Lemma 9.12.3. Finally, $\alpha $ is separable algebraic over $k(a_0, \ldots , a_{d - 1})$ because it is a root of $P$ which is irreducible (as it is irreducible over the possibly bigger field $E$) and separable (as it is separable over $E$). Thus $k(a_0, \ldots , a_{d - 1}, \alpha )$ is separable over $k$ by Lemma 9.12.10 and we conclude that $\alpha $ is separable over $k$ as desired. $\square$

Proof. Let $\alpha , \beta \in E$ be separable over $k$. Then $\beta $ is separable over $k(\alpha )$ by Lemma 9.12.3. Thus we can apply Lemma 9.12.12 to $k(\alpha , \beta )$ to see that $k(\alpha , \beta )$ is separable over $k$. $\square$