
Lemma 10.152.2. 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,

$\Omega _{k(x_1, \ldots , x_ r)/k} = \bigoplus \nolimits _{i = 1}^ r k(x_1, \ldots , x_ r) \text{d}x_ i$

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

$\Omega _{K/k} = \bigoplus \nolimits _{i = 1}^ r K\text{d}x_ i$

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

$P(T) = T^ d + \sum \nolimits _{i = 1}^ d a_ i T^{d - i}$

and hence

$0 = \text{d}P(a) = \sum \nolimits _{i = 1}^ d a^{d - i}\text{d}a_ i$

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. $\square$

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).