The Stacks project

Proof. Write $K$ as a directed colimit $K = \mathop{\mathrm{colim}}\nolimits _ i K_ i$ of finitely generated field extensions $K_ i/k$. By definition $K$ is separable if and only if each $K_ i$ is separable over $k$, and by Lemma 10.131.5 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 $K/k$ is a finitely generated field extension which is separable. Choose $x_1, \ldots , x_{r + 1} \in K$ as in Lemma 10.42.3. 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

Using the presentation of $S$ above we see that

Since $\Omega _{K/\mathbf{F}_ p}$ is the localization of the $S$-module $\Omega _{S/\mathbf{F}_ p}$ (see Lemma 10.131.8) we conclude that

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 $K/k$ 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 10.44.2.) 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

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

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

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 10.158.2 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. $\square$