Lemma 61.8.1. Given a ring $A$ there exists a faithfully flat ind-étale $A$-algebra $C$ such that every faithfully flat étale ring map $C \to B$ has a section.

Proof. Set $T^1(A) = T(A)$ and $T^{n + 1}(A) = T(T^ n(A))$. Let

$C = \mathop{\mathrm{colim}}\nolimits T^ n(A)$

This algebra is faithfully flat over each $T^ n(A)$ and in particular over $A$, see Algebra, Lemma 10.39.20. Moreover, $C$ is ind-étale over $A$ by Lemma 61.7.4. If $C \to B$ is étale, then there exists an $n$ and an étale ring map $T^ n(A) \to B'$ such that $B = C \otimes _{T^ n(A)} B'$, see Algebra, Lemma 10.143.3. If $C \to B$ is faithfully flat, then $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(T^ n(A))$ is surjective, hence $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(T^ n(A))$ is surjective. In other words, $T^ n(A) \to B'$ is faithfully flat. By our construction, there is a $T^ n(A)$-algebra map $B' \to T^{n + 1}(A)$. This induces a $C$-algebra map $B \to C$ which finishes the proof. $\square$

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