Rational points of smooth finite-type schemes over a separably closed field are schematically dense

By [F1] the underlying space $\mathrm{\mid }Z\mathrm{\mid }$ is closed in $X$ and contains $Z(k)\supseteq S$; since $S$ is dense in $\mathrm{\mid }X\mathrm{\mid }$, every closed subset containing $S$ equals $\mathrm{\mid }X\mathrm{\mid }$, so $\mathrm{\mid }Z\mathrm{\mid }=\mathrm{\mid }X\mathrm{\mid }$.

Assume now that $X$ is smooth over $k$, and let $U\subseteq X$ be a nonempty open subscheme. Then $U$ is smooth over $k$ by [F4], nonempty and of finite type, so by [F5] it has a closed point $P$ with $\kappa (P)$ finite and separable over $k$. By hypothesis on $k$, $\kappa (P)=k$, and [F6] identifies $P$ with a $k$-point of $U$. Hence every nonempty open subscheme of $X$ meets $X(k)$, so $X(k)$ is dense in $X$.

I claim that $Z=X$. Let $U=\mathrm{Spec}A\subseteq X$ be an affine open; by [F1] write $Z\cap U=\mathrm{Spec}(A\mathrm{/}I)$ for a unique ideal $I\subseteq A$, and by [step 1.1] its underlying space is all of $U$, so $V(I)=\mathrm{Spec}A$ and hence every $f\in I$ lies in every prime ideal of $A$, i.e. $I\subseteq \sqrt{(0)}$. Since $X$ is reduced, $A$ is reduced by [F2], so $\sqrt{(0)}=0$ and $I=0$, giving $Z\cap U=U$. As the affine opens cover $X$ and two closed subschemes of $X$ that agree on an open cover agree, $Z=X$.

Combining: the first assertion is [step 1.1] with [step 2.1]; applying it to the reduced smooth scheme $X$ of [F4] with $S=X(k)$, which is dense by [step 1.2], shows that every closed subscheme $Z\subseteq X$ with $Z(k)\supseteq X(k)$ equals $X$, that is, $X(k)$ is schematically dense in $X$.