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 32.25.6. Let $k$ be a field. If $X$ is smooth over $\mathop{\mathrm{Spec}}(k)$ then the set

\[ \{ x \in X\text{ closed such that }k \subset \kappa (x) \text{ is finite separable}\} \]

is dense in $X$.

Proof. It suffices to show that given a nonempty smooth $X$ over $k$ there exists at least one closed point whose residue field is finite separable over $k$. To see this, choose a diagram

\[ \xymatrix{ X & U \ar[l] \ar[r]^-\pi & \mathbf{A}^ d_ k } \]

with $\pi $ ├ętale, see Morphisms, Lemma 28.34.20. The morphism $\pi : U \to \mathbf{A}^ d_ k$ is open, see Morphisms, Lemma 28.34.13. By Lemma 32.25.5 we may choose a closed point $w \in \pi (U)$ whose residue field is finite separable over $k$. Pick any $x \in U$ with $\pi (x) = w$. By Morphisms, Lemma 28.34.7 the field extension $\kappa (w) \subset \kappa (x)$ is finite separable. Hence $k \subset \kappa (x)$ is finite separable. The point $x$ is a closed point of $X$ by Morphisms, Lemma 28.19.2. $\square$

Comments (2)

Comment #1393 by Stulemeijer Thierry on

A very small typo : replace with , and with .

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