Lemma 10.156.2. Let $k$ be a perfect field of characteristic $p > 0$. Let $K/k$ be an extension. Let $a \in K$. Then $\text{d}a = 0$ in $\Omega _{K/k}$ if and only if $a$ is a $p$th power.

**Proof.**
By Lemma 10.130.4 we see that there exists a subfield $k \subset L \subset K$ such that $k \subset L$ is a finitely generated field extension and such that $\text{d}a$ is zero in $\Omega _{L/k}$. Hence we may assume that $K$ is a finitely generated field extension of $k$.

Choose a transcendence basis $x_1, \ldots , x_ r \in K$ such that $K$ is finite separable over $k(x_1, \ldots , x_ r)$. This is possible by the definitions, see Definitions 10.44.1 and 10.41.1. We remark that the result holds for the purely transcendental subfield $k(x_1, \ldots , x_ r) \subset K$. Namely,

and any rational function all of whose partial derivatives are zero is a $p$th power. Moreover, we also have

since $k(x_1, \ldots , x_ r) \subset K$ is finite separable (computation omitted). Suppose $a \in K$ is an element such that $\text{d}a = 0$ in the module of differentials. By our choice of $x_ i$ we see that the minimal polynomial $P(T) \in k(x_1, \ldots , x_ r)[T]$ of $a$ is separable. Write

and hence

in $\Omega _{K/k}$. By the description of $\Omega _{K/k}$ above and the fact that $P$ was the minimal polynomial of $a$, we see that this implies $\text{d}a_ i = 0$. Hence $a_ i = b_ i^ p$ for each $i$. Therefore by Fields, Lemma 9.28.2 we see that $a$ is a $p$th power. $\square$

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