Lemma 61.6.8. Let $A$ be a ring. There exists a faithfully flat, ind-Zariski ring map $A \to B$ such that $B$ satisfies the equivalent conditions of Lemma 61.6.7.

Proof. We first apply Lemma 61.5.3 to see that we may assume that $\mathop{\mathrm{Spec}}(A)$ is w-local. Choose an extremally disconnected space $T$ and a surjective continuous map $T \to \pi _0(\mathop{\mathrm{Spec}}(A))$, see Topology, Lemma 5.26.9. Note that $T$ is profinite. Apply Lemma 61.6.2 to find an ind-Zariski ring map $A \to B$ such that $\pi _0(\mathop{\mathrm{Spec}}(B)) \to \pi _0(\mathop{\mathrm{Spec}}(A))$ realizes $T \to \pi _0(\mathop{\mathrm{Spec}}(A))$ and such that

$\xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[r] \ar[d] & \pi _0(\mathop{\mathrm{Spec}}(B)) \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \pi _0(\mathop{\mathrm{Spec}}(A)) }$

is cartesian in the category of topological spaces. Note that $\mathop{\mathrm{Spec}}(B)$ is w-local, that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is w-local, and that the set of closed points of $\mathop{\mathrm{Spec}}(B)$ is the inverse image of the set of closed points of $\mathop{\mathrm{Spec}}(A)$, see Lemma 61.2.5. Thus condition (3) of Lemma 61.6.7 holds for $B$. $\square$

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