**Proof.**
The implication (1) $\Rightarrow $ (4) follows from Lemma 10.43.6. The implication (4) $\Rightarrow $ (3) is immediate.

Assume (3). Consider the ring homomorphism $m : K \otimes _ k k^{1/p} \rightarrow K$ given by

\[ \lambda \otimes \mu \rightarrow \lambda ^ p \mu ^ p \]

Note that $x^ p = m(x) \otimes 1$ for all $x \in K \otimes _ k k^{1/p}$. Since $K \otimes _ k k^{1/p}$ is reduced we see $m$ is injective. If $\{ a_1, \ldots , a_ m\} \subset K$ is $k$-linearly independent, then $\{ a_1 \otimes 1, \ldots , a_ m \otimes 1\} $ is $k^{1/p}$-linearly independent. By injectivity of $m$ we deduce that no nontrivial $k$-linear combination of $a_1^ p, \ldots , a_ m^ p$ is is zero. Hence (3) implies (2).

Assume (2). To prove (1) we may assume that $K$ is finitely generated over $k$ and we have to prove that $K$ is separably generated over $k$. Let $\{ x_1, \ldots , x_ d\} $ be a transcendence base of $K/k$. By Fields, Lemma 9.8.6 we have $[K : K'] < \infty $ where $K' = k(x_1, \ldots , x_ d)$. Choose the transcendence base such that the degree of inseparability $[K : K']_ i$ is minimal. If $K / K'$ is separable then we win. Assume this is not the case to get a contradiction. Then there exists $x_{d + 1} \in K$ which is not separable over $K'$, and in particular $[K'(x_{d+1}) : K']_ i > 1$. Then by Lemma 10.44.1 there is $1 \leq j \leq n + 1$ such that $K'' = k(x_1, \ldots , \widehat{x}_ j, \ldots , x_{d+1})$ satisfies $[K'(x_{d+1}) : K'']_ i = 1$. By multiplicativity $[K : K'']_ i < [K : K']_ i$ and we obtain the contradiction.
$\square$

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