**Proof.**
The equivalence of (2) and (3) is Lemma 10.148.2. By Lemma 10.143.4 we see that (1) is equivalent to (6). Property (6) implies (5) and (4) which both in turn imply (3) (Lemmas 10.150.2, 10.151.3, and 10.151.2). Thus it suffices to show that (2) implies (1). Choose a finitely generated $k$-subalgebra $A \subset K$ such that $K$ is the fraction field of the domain $A$. Set $S = A \setminus \{ 0\} $. Since $0 = \Omega _{K/k} = S^{-1}\Omega _{A/k}$ (Lemma 10.131.8) and since $\Omega _{A/k}$ is finitely generated (Lemma 10.131.16), we can replace $A$ by a localization $A_ f$ to reduce to the case that $\Omega _{A/k} = 0$ (details omitted). Then $A$ is unramified over $k$, hence $K/k$ is finite separable for example by Lemma 10.151.5 applied with $\mathfrak q = (0)$.
$\square$

## Comments (0)

There are also: