
Lemma 10.152.4. Let $k$ be a field of characteristic $p > 0$. The following are equivalent:

1. the field extension $K/k$ is separable (see Definition 10.41.1), and

2. the map $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective.

Proof. Write $K$ as a directed colimit $K = \mathop{\mathrm{colim}}\nolimits _ i K_ i$ of finitely generated field extensions $k \subset K_ i$. By definition $K$ is separable if and only if each $K_ i$ is separable over $k$, and by Lemma 10.130.4 we see that $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective if and only if each $K_ i \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K_ i/\mathbf{F}_ p}$ is injective. Hence we may assume that $K/k$ is a finitely generated field extension.

Assume $k \subset K$ is a finitely generated field extension which is separable. Choose $x_1, \ldots , x_{r + 1} \in K$ as in Lemma 10.41.3. In this case there exists an irreducible polynomial $G(X_1, \ldots , X_{r + 1}) \in k[X_1, \ldots , X_{r + 1}]$ such that $G(x_1, \ldots , x_{r + 1}) = 0$ and such that $\partial G/\partial X_{r + 1}$ is not identically zero. Moreover $K$ is the field of fractions of the domain. $S = K[X_1, \ldots , X_{r + 1}]/(G)$. Write

$G = \sum a_ I X^ I, \quad X^ I = X_1^{i_1}\ldots X_{r + 1}^{i_{r + 1}}.$

Using the presentation of $S$ above we see that

$\Omega _{S/\mathbf{F}_ p} = \frac{ S \otimes _ k \Omega _ k \oplus \bigoplus \nolimits _{i = 1, \ldots , r + 1} S\text{d}X_ i }{ \langle \sum X^ I \text{d}a_ I + \sum \partial G/\partial X_ i \text{d}X_ i \rangle }$

Since $\Omega _{K/\mathbf{F}_ p}$ is the localization of the $S$-module $\Omega _{S/\mathbf{F}_ p}$ (see Lemma 10.130.8) we conclude that

$\Omega _{K/\mathbf{F}_ p} = \frac{ K \otimes _ k \Omega _ k \oplus \bigoplus \nolimits _{i = 1, \ldots , r + 1} K\text{d}X_ i }{ \langle \sum X^ I \text{d}a_ I + \sum \partial G/\partial X_ i \text{d}X_ i \rangle }$

Now, since the polynomial $\partial G/\partial X_{r + 1}$ is not identically zero we conclude that the map $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{S/\mathbf{F}_ p}$ is injective as desired.

Assume $k \subset K$ is a finitely generated field extension and that $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective. (This part of the proof is the same as the argument proving Lemma 10.43.1.) Let $x_1, \ldots , x_ r$ be a transcendence basis of $K$ over $k$ such that the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset K$ is minimal. If $K$ is separable over $k(x_1, \ldots , x_ r)$ then we win. Assume this is not the case to get a contradiction. Then there exists an element $\alpha \in K$ which is not separable over $k(x_1, \ldots , x_ r)$. Let $P(T) \in k(x_1, \ldots , x_ r)[T]$ be its minimal polynomial. Because $\alpha$ is not separable actually $P$ is a polynomial in $T^ p$. Clear denominators to get an irreducible polynomial

$G(X_1, \ldots , X_ r, T) = \sum a_{I, i} X^ I T^ i \in k[X_1, \ldots , X_ r, T]$

such that $G(x_1, \ldots , x_ r, \alpha ) = 0$ in $L$. Note that this means $k[X_1, \ldots , X_ r, T]/(G) \subset L$. We may assume that for some pair $(I_0, i_0)$ the coefficient $a_{I_0, i_0} = 1$. We claim that $\text{d}G/\text{d}X_ i$ is not identically zero for at least one $i$. Namely, if this is not the case, then $G$ is actually a polynomial in $X_1^ p, \ldots , X_ r^ p, T^ p$. Then this means that

$\sum \nolimits _{(I, i) \not= (I_0, i_0)} x^ I\alpha ^ i \text{d}a_{I, i}$

is zero in $\Omega _{K/\mathbf{F}_ p}$. Note that there is no $k$-linear relation among the elements

$\{ x^ I\alpha ^ i \mid a_{I, i} \not= 0 \text{ and } (I, i) \not= (I_0, i_0)\}$

of $K$. Hence the assumption that $K \otimes _ k \Omega _{k/\mathbf{F}_ p} \to \Omega _{K/\mathbf{F}_ p}$ is injective this implies that $\text{d}a_{I, i} = 0$ in $\Omega _{k/\mathbf{F}_ p}$ for all $(I, i)$. By Lemma 10.152.2 we see that each $a_{I, i}$ is a $p$th power, which implies that $G$ is a $p$th power contradicting the irreducibility of $G$. Thus, after renumbering, we may assume that $\text{d}G/\text{d}X_1$ is not zero. Then we see that $x_1$ is separably algebraic over $k(x_2, \ldots , x_ r, \alpha )$, and that $x_2, \ldots , x_ r, \alpha$ is a transcendence basis of $L$ over $k$. This means that the degree of inseparability of the finite extension $k(x_2, \ldots , x_ r, \alpha ) \subset L$ is less than the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset L$, which is a contradiction. $\square$

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