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. $\square$

Comment #1393 by Stulemeijer Thierry on

A very small typo : replace $\pi (V)$ with $\pi (U)$, and $Pick any x\in V$ with $Pick any x\in U$.

Comment #3949 by Rogozhin on

$U$ is just a Zariski covering of $X$, but still it should probably say that you take the image of $x\in U$ in $X$ at the end.

Comment #3955 by on

Well, I really think this is fine as is. If more people complain I will change it.

Comment #3961 by Laurent Moret-Bailly on

And of course $U$ should be nonempty.

Comment #4096 by on

OK, I still think this is fine as is because the lemma that is being used will produce a nonempty U.

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