Lemma 10.144.6. Let $R \to S$ be a ring map. Assume that

1. $R \to S$ is étale, and

2. $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

Then there exists a ring map $R \to S'$ such that

1. $R \to S'$ is finite, finitely presented, and flat (in other words it is finite projective as an $R$-module),

2. $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(R)$ is surjective,

3. for every prime $\mathfrak q' \subset S'$ there exists a $g' \in S'$, $g' \not\in \mathfrak q'$ such that the ring map $R \to S'_{g'}$ factors as $R \to S \to S'_{g'}$.

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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