The Stacks project

Proof. Let $Y \to Z$ be a finite étale morphism. It suffices to construct a finite étale morphism $f : U \to X$ such that $Y$ is isomorphic to an open and closed subscheme of $f^{-1}(Z)$, see Lemma 58.4.4. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(R)$ so the closed immersion $Y \to X$ is given by a surjection $R \to A$. We may write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as the filtered colimit of its $\mathbf{F}_ p$-subalgebras of finite type. By Lemma 58.14.1 we can find an $i$ and a finite étale morphism $Y_ i \to Z_ i = \mathop{\mathrm{Spec}}(A_ i)$ such that $Y = Z \times _{Z_ i} Y_ i$.

Choose a surjection $\mathbf{F}_ p[x_1, \ldots , x_ n] \to A_ i$. This determines a closed immersion

By the universal property of polynomial algebras and since $R \to A$ is surjective, we can find a commutative diagram

of $\mathbf{F}_ p$-algebras. Thus we have a commutative diagram

whose right square is cartesian. Clearly, if we can find $f_ i : U_ i \to X_ i$ finite étale such that $Y_ i$ is isomorphic to an open and closed subscheme of $f_ i^{-1}(Z_ i)$, then the base change $f : U \to X$ of $f_ i$ by $X \to X_ i$ is a solution to our problem. Thus we conclude by applying Lemma 58.32.3 to $Y_ i \to Z_ i \to X_ i = \mathbf{A}^ n_{\mathbf{F}_ p}$. $\square$