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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (19)

Comment #387 by Filip Chindea on

Comment #388 by Johan on

Comment #389 by Filip Chindea on

Comment #390 by Johan on

Comment #740 by Keenan Kidwell on

Comment #751 by Johan on

Comment #756 by Keenan Kidwell on

Comment #759 by Johan on

Comment #760 by Johan on

Comment #761 by Johan on

Comment #763 by Johan on

Comment #764 by Keenan Kidwell on

Comment #765 by Johan on

Comment #5562 by DatPham on

Comment #5745 by Johan on

Comment #5821 by DatPham on

Comment #5823 by Johan on

Comment #5824 by DatPham on

Comment #5825 by Johan on