The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$


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