**Proof.**
By Proposition 10.144.4 and the quasi-compactness of $\mathop{\mathrm{Spec}}(S)$ (see Lemma 10.17.8) we can find $g_1, \ldots , g_ n \in S$ generating the unit ideal of $S$ such that each $R \to S_{g_ i}$ is standard étale. If we prove the lemma for the ring map $R \to \prod _{i = 1, \ldots , n} S_{g_ i}$ then the lemma follows for the ring map $R \to S$. Hence we may assume that $S = \prod _{i = 1, \ldots , n} S_ i$ is a finite product of standard étale morphisms.

For each $i$ choose a ring map $R \to S_ i'$ as in Lemma 10.144.5 adapted to the standard étale morphism $R \to S_ i$. Set $S' = S_1' \otimes _ R \ldots \otimes _ R S_ n'$; we will use the $R$-algebra maps $S_ i' \to S'$ without further mention below. We claim this works. Properties (1) and (2) are immediate. For property (3) suppose that $\mathfrak q' \subset S'$ is a prime. Denote $\mathfrak p$ its image in $\mathop{\mathrm{Spec}}(R)$. Choose $i \in \{ 1, \ldots , n\} $ such that $\mathfrak p$ is in the image of $\mathop{\mathrm{Spec}}(S_ i) \to \mathop{\mathrm{Spec}}(R)$; this is possible by assumption. Set $\mathfrak q_ i' \subset S_ i'$ the image of $\mathfrak q'$ in the spectrum of $S_ i'$. By construction of $S'_ i$ there exists a $g'_ i \in S_ i'$ such that $R \to (S_ i')_{g_ i'}$ factors as $R \to S_ i \to (S_ i')_{g_ i'}$. Hence also $R \to S'_{g_ i'}$ factors as

\[ R \to S_ i \to (S_ i')_{g_ i'} \to S'_{g_ i'} \]

as desired.
$\square$

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