Lemma 33.25.6. Let $k$ be a field. If $X$ is smooth over $\mathop{\mathrm{Spec}}(k)$ then the set

is dense in $X$.

Lemma 33.25.6. 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 $ étale, see Morphisms, Lemma 29.34.20. The morphism $\pi : U \to \mathbf{A}^ d_ k$ is open, see Morphisms, Lemma 29.34.13. By Lemma 33.25.5 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 29.34.7 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 29.19.2. $\square$

Your email address will not be published.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 (6)

Comment #1393 by Stulemeijer Thierry on

Comment #1411 by Johan on

Comment #3949 by Rogozhin on

Comment #3955 by Johan on

Comment #3961 by Laurent Moret-Bailly on

Comment #4096 by Johan on