Lemma 10.43.1. Let $k$ be a field of characteristic $p > 0$. Let $k \subset K$ be a field extension. The following are equivalent:

$K$ is separable over $k$,

the ring $K \otimes _ k k^{1/p}$ is reduced, and

$K$ is geometrically reduced over $k$.

Lemma 10.43.1. Let $k$ be a field of characteristic $p > 0$. Let $k \subset K$ be a field extension. The following are equivalent:

$K$ is separable over $k$,

the ring $K \otimes _ k k^{1/p}$ is reduced, and

$K$ is geometrically reduced over $k$.

**Proof.**
The implication (1) $\Rightarrow $ (3) follows from Lemma 10.42.6. The implication (3) $\Rightarrow $ (2) is immediate.

Assume (2). Let $k \subset L \subset K$ be a subextension such that $L$ is a finitely generated field extension of $k$. We have to show that we can find a separating transcendence basis of $L$. The assumption implies that $L \otimes _ k k^{1/p}$ is reduced. Let $x_1, \ldots , x_ r$ be a transcendence basis of $L$ over $k$ such that the degree of inseparability of the finite extension $k(x_1, \ldots , x_ r) \subset L$ is minimal. If $L$ 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 $\alpha \in L$ which is not separable over $k(x_1, \ldots , x_ r)$. Let $P(T) \in k(x_1, \ldots , x_ r)[T]$ be the minimal polynomial of $\alpha $ over $k(x_1, \ldots , x_ r)$. After replacing $\alpha $ by $f \alpha $ for some nonzero $f \in k[x_1, \ldots , x_ r]$ we may and do assume that $P$ lies in $k[x_1, \ldots , x_ r, T]$. Because $\alpha $ is not separable $P$ is a polynomial in $T^ p$, see Fields, Lemma 9.12.1. Let $dp$ be the degree of $P$ as a polynomial in $T$. Since $P$ is the minimal polynomial of $\alpha $ the monomials

\[ x_1^{e_1} \ldots x_ r^{e_ r} \alpha ^ e \]

for $e < dp$ are linearly independent over $k$ in $L$. We claim that the element $\partial P/\partial x_ i \in k[x_1, \ldots , x_ r, T]$ is not zero for at least one $i$. Namely, if this was not the case, then $P$ is actually a polynomial in $x_1^ p, \ldots , x_ r^ p, T^ p$. In that case we can consider $P^{1/p} \in k^{1/p}[x_1, \ldots , x_ r, T]$. This would map to $P^{1/p}(x_1, \ldots , x_ r, \alpha )$ which is a nilpotent element of $k^{1/p} \otimes _ k L$ and hence zero. On the other hand, $P^{1/p}(x_1, \ldots , x_ r, \alpha )$ is a $k^{1/p}$-linear combination the monomials listed above, hence nonzero in $k^{1/p} \otimes _ k L$. This is a contradiction which proves our claim.

Thus, after renumbering, we may assume that $\partial P/\partial x_1$ is not zero. As $P$ is an irreducible polynomial in $T$ over $k(x_1, \ldots , x_ r)$ it is irreducible as a polynomial in $x_1, \ldots , x_ r, T$, hence by Gauss's lemma it is irreducible as a polynomial in $x_1$ over $k(x_2, \ldots , x_ r, T)$. Since the transcendence degree of $L$ is $r$ we see that $x_2, \ldots , x_ r, \alpha $ are algebraically independent. Hence $P(X, x_2, \ldots , x_ r, \alpha ) \in k(x_2, \ldots , x_ r, \alpha )[X]$ is irreducible. It follows that $x_1$ is separably algebraic over $k(x_2, \ldots , x_ r, \alpha )$. 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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