Lemma 10.155.7. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Consider the category of pairs $(S, \mathfrak q)$ where $R \to S$ is étale and $\mathfrak q$ is a prime lying over $\mathfrak p$ such that $\kappa (\mathfrak p) = \kappa (\mathfrak q)$. This category is filtered and

\[ (R_{\mathfrak p})^ h = \mathop{\mathrm{colim}}\nolimits _{(S, \mathfrak q)} S = \mathop{\mathrm{colim}}\nolimits _{(S, \mathfrak q)} S_{\mathfrak q} \]

canonically.

**Proof.**
A morphism of pairs $(S, \mathfrak q) \to (S', \mathfrak q')$ is given by an $R$-algebra map $\varphi : S \to S'$ such that $\varphi ^{-1}(\mathfrak q') = \mathfrak q$. Let us show that the category of pairs is filtered, see Categories, Definition 4.19.1. The category contains the pair $(R, \mathfrak p)$ and hence is not empty, which proves part (1) of Categories, Definition 4.19.1. Suppose that $(S, \mathfrak q)$ and $(S', \mathfrak q')$ are two pairs. 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 $\kappa (\mathfrak p)$, hence they correspond to maximal ideals of $S \otimes \kappa (\mathfrak p)$, resp. $S' \otimes \kappa (\mathfrak p)$. Set $S'' = S \otimes _ R S'$. By the above there exists a unique prime $\mathfrak q'' \subset S''$ lying over $\mathfrak q$ and over $\mathfrak q'$ whose residue field is $\kappa (\mathfrak p)$. The ring map $R \to S''$ is étale by Lemma 10.143.3. This proves part (2) of Categories, Definition 4.19.1. Next, suppose that $\varphi , \psi : (S, \mathfrak q) \to (S', \mathfrak q')$ 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

\[ S'' = (S' \otimes _{\varphi , S, \psi } S') \otimes _{S' \otimes _ R S'} S' \]

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

\[ F'' = (F' \otimes _{\varphi , F, \psi } F') \otimes _{F' \otimes _{\kappa (\mathfrak p)} F'} F' \]

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

Recall that in the proof of Lemma 10.155.1 we constructed $(R_{\mathfrak p})^ h$ as the corresponding colimit but starting with $R_{\mathfrak p}$ and its maximal ideal $\mathfrak pR_{\mathfrak p}$. Now, given any pair $(S, \mathfrak q)$ for $(R, \mathfrak p)$ we obtain a pair $(S_{\mathfrak p}, \mathfrak qS_{\mathfrak p})$ for $(R_{\mathfrak p}, \mathfrak pR_{\mathfrak p})$. Moreover, in this situation

\[ S_{\mathfrak p} = \mathop{\mathrm{colim}}\nolimits _{f \in R, f \not\in \mathfrak p} S_ f. \]

Hence in order to show the equalities of the lemma, it suffices to show that any pair $(S_{loc}, \mathfrak q_{loc})$ for $(R_{\mathfrak p}, \mathfrak pR_{\mathfrak p})$ is of the form $(S_{\mathfrak p}, \mathfrak qS_{\mathfrak p})$ for some pair $(S, \mathfrak q)$ over $(R, \mathfrak p)$ (some details omitted). This follows from Lemma 10.143.3.
$\square$

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