The Stacks project

Proof. A morphism of triples $(S, \mathfrak q, \phi ) \to (S', \mathfrak q', \phi ')$ is given by an $R$-algebra map $\varphi : S \to S'$ such that $\varphi ^{-1}(\mathfrak q') = \mathfrak q$ and such that $\phi ' \circ \varphi = \phi$. Let us show that the category of pairs is filtered, see Categories, Definition 4.19.1. The category contains the triple $(R, \mathfrak p, \kappa (\mathfrak p) \subset \kappa ^{sep})$ and hence is not empty, which proves part (1) of Categories, Definition 4.19.1. Suppose that $(S, \mathfrak q, \phi )$ and $(S', \mathfrak q', \phi ')$ are two triples. Note that $\mathfrak q$, resp. $\mathfrak q'$ correspond to primes of the fibre rings $S \otimes \kappa (\mathfrak p)$, resp. $S' \otimes \kappa (\mathfrak p)$ with residue fields finite separable over $\kappa (\mathfrak p)$ and $\phi$, resp. $\phi '$ correspond to maps into $\kappa ^{sep}$. Hence this data corresponds to $\kappa (\mathfrak p)$-algebra maps

Set $S'' = S \otimes _ R S'$. Combining the maps the above we get a unique $\kappa (\mathfrak p)$-algebra map

whose kernel corresponds to a prime $\mathfrak q'' \subset S''$ lying over $\mathfrak q$ and over $\mathfrak q'$, and whose residue field maps via $\phi ''$ to the compositum of $\phi (\kappa (\mathfrak q))$ and $\phi '(\kappa (\mathfrak q'))$ in $\kappa ^{sep}$. The ring map $R \to S''$ is étale by Lemma 10.143.3. Hence $(S'', \mathfrak q'', \phi '')$ is a triple dominating both $(S, \mathfrak q, \phi )$ and $(S', \mathfrak q', \phi ')$. This proves part (2) of Categories, Definition 4.19.1. Next, suppose that $\varphi , \psi : (S, \mathfrak q, \phi ) \to (S', \mathfrak q', \phi ')$ are two morphisms of pairs. Then $\varphi$, $\psi$, and $S' \otimes _ R S' \to S'$ are étale ring maps by Lemma 10.143.8. Consider

Arguing as above (base change of étale maps is étale, composition of étale maps is étale) we see that $S''$ is étale over $R$. The fibre ring of $S''$ over $\mathfrak p$ is

where $F', F$ are the fibre rings of $S'$ and $S$. Since $\varphi$ and $\psi$ are morphisms of triples the map $\phi ' : F' \to \kappa ^{sep}$ extends to a map $\phi '' : F'' \to \kappa ^{sep}$ which in turn corresponds to a prime ideal $\mathfrak q'' \subset S''$. The canonical map $S' \to S''$ (using the right most factor for example) is a morphism of triples $(S', \mathfrak q', \phi ') \to (S'', \mathfrak q'', \phi '')$ which equalizes $\varphi$ and $\psi$. This proves part (3) of Categories, Definition 4.19.1. Hence we conclude that the category is filtered.

We still have to show that the colimit $R_{colim}$ of the system is equal to the strict henselization of $R_{\mathfrak p}$ with respect to $\kappa ^{sep}$. To see this note that the system of triples $(S, \mathfrak q, \phi )$ contains as a subsystem the pairs $(S, \mathfrak q)$ of Lemma 10.155.7. Hence $R_{colim}$ contains $R_{\mathfrak p}^ h$ by the result of that lemma. Moreover, it is clear that $R_{\mathfrak p}^ h \subset R_{colim}$ is a directed colimit of étale ring extensions. It follows that $R_{colim}$ is henselian by Lemmas 10.153.4 and 10.154.8. Finally, by Lemma 10.144.3 we see that the residue field of $R_{colim}$ is equal to $\kappa ^{sep}$. Hence we conclude that $R_{colim}$ is strictly henselian and hence equals the strict henselization of $R_{\mathfrak p}$ as desired. Some details omitted. $\square$