The Stacks project

Proof. Let $Z \subset X$ be the inverse image of $T$. Then $Z$ is the intersection $Z = \bigcap Z_\alpha $ of the open and closed subsets of $X$ containing $Z$, see Topology, Lemma 5.12.12. For each $\alpha $ we have $Z_\alpha = \mathop{\mathrm{Spec}}(A_\alpha )$ where $A \to A_\alpha $ is a local isomorphism (a localization at an idempotent). Setting $B = \mathop{\mathrm{colim}}\nolimits A_\alpha $ proves the lemma. $\square$

Proof. Namely, write $T = \mathop{\mathrm{lim}}\nolimits T_ i$ as the limit of an inverse system finite discrete spaces over a directed set (see Topology, Lemma 5.22.2). For each $i$ let $Z_ i = \mathop{\mathrm{Im}}(T \to \pi _0(X) \times T_ i)$. This is a closed subset. Observe that $X \times T_ i$ is the spectrum of $A_ i = \prod _{t \in T_ i} A$ and that $A \to A_ i$ is a local isomorphism. By Lemma 61.6.1 we see that $Z_ i \subset \pi _0(X \times T_ i) = \pi _0(X) \times T_ i$ corresponds to a surjection $A_ i \to B_ i$ which is ind-Zariski such that $\mathop{\mathrm{Spec}}(B_ i) = X \times _{\pi _0(X)} Z_ i$ as subsets of $X \times T_ i$. The transition maps $T_ i \to T_{i'}$ induce maps $Z_ i \to Z_{i'}$ and $X \times _{\pi _0(X)} Z_ i \to X \times _{\pi _0(X)} Z_{i'}$. Hence ring maps $B_{i'} \to B_ i$ (Lemmas 61.3.8 and 61.4.6). Set $B = \mathop{\mathrm{colim}}\nolimits B_ i$. Because $T = \mathop{\mathrm{lim}}\nolimits Z_ i$ we have $X \times _{\pi _0(X)} T = \mathop{\mathrm{lim}}\nolimits X \times _{\pi _0(X)} Z_ i$ and hence $Y = \mathop{\mathrm{Spec}}(B) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Spec}}(B_ i)$ fits into the cartesian diagram

of topological spaces. By Lemma 61.2.5 we conclude that $T = \pi _0(Y)$. $\square$

Proof. Let $X_0 \subset X = \mathop{\mathrm{Spec}}(A)$ and $Y_0 \subset Y = \mathop{\mathrm{Spec}}(B)$ be the sets of closed points. By assumption $Y_0$ maps into $X_0$ and the induced map $Y_0 \to X_0$ is a bijection. As a space $\mathop{\mathrm{Spec}}(A)$ is the disjoint union of the spectra of the local rings of $A$ at closed points. Similarly for $B$. Hence $X \to Y$ is a bijection. Since $A \to B$ is flat we have going down (Algebra, Lemma 10.39.19). Thus Algebra, Lemma 10.41.11 shows for any prime $\mathfrak q \subset B$ lying over $\mathfrak p \subset A$ we have $B_\mathfrak q = B_\mathfrak p$. Since $B_\mathfrak q = A_\mathfrak p$ by assumption, we see that $A_\mathfrak p = B_\mathfrak p$ for all primes $\mathfrak p$ of $A$. Thus $A = B$ by Algebra, Lemma 10.23.1. $\square$

Proof. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. Let $X_0 \subset X$ and $Y_0 \subset Y$ be the set of closed points. Let $A \to A'$ be the ind-Zariski morphism of affine schemes such that with $X' = \mathop{\mathrm{Spec}}(A')$ the diagram

is cartesian in the category of topological spaces and such that $\pi _0(X') = \pi _0(Y)$ as spaces over $\pi _0(X)$, see Lemma 61.6.2. By Lemma 61.2.5 we see that $X'$ is w-local and the set of closed points $X'_0 \subset X'$ is the inverse image of $X_0$.

We obtain a continuous map $Y \to X'$ of underlying topological spaces over $X$ identifying $\pi _0(Y)$ with $\pi _0(X')$. By Lemma 61.3.8 (and Lemma 61.4.6) this corresponds to a morphism of affine schemes $Y \to X'$ over $X$. Since $Y \to X$ maps $Y_0$ into $X_0$ we see that $Y \to X'$ maps $Y_0$ into $X'_0$, i.e., $Y \to X'$ is w-local. By Lemma 61.6.4 we see that $Y \cong X'$ and we win. $\square$

Proof. Let $A \to A_ w$, resp. $B \to B_ w$ be the faithfully flat, ind-Zariski ring map constructed in Lemma 61.5.3 for $A$, resp. $B$. Since $\mathop{\mathrm{Spec}}(B_ w)$ is w-local, there exists a unique factorization $A \to A_ w \to B_ w$ such that $\mathop{\mathrm{Spec}}(B_ w) \to \mathop{\mathrm{Spec}}(A_ w)$ is w-local by Lemma 61.5.5. Note that $A_ w \to B_ w$ identifies local rings, see Lemma 61.3.4. By Lemma 61.6.5 this means $A_ w \to B_ w$ is ind-Zariski. Since $B \to B_ w$ is faithfully flat, ind-Zariski (Lemma 61.5.3) and the composition $A \to B \to B_ w$ is ind-Zariski (Lemma 61.4.3) the proposition is proved. $\square$

Proof. The equivalence of (1) and (2) follows immediately from Proposition 61.6.6.

Assume (3)(a) and (3)(b). Let $A \to B$ be faithfully flat and ind-Zariski. We will use without further mention the fact that a flat map $A \to B$ is faithfully flat if and only if every closed point of $\mathop{\mathrm{Spec}}(A)$ is in the image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. We will show that $A \to B$ has a retraction.

Let $I \subset A$ be an ideal such that $V(I) \subset \mathop{\mathrm{Spec}}(A)$ is the set of closed points of $\mathop{\mathrm{Spec}}(A)$. We may replace $B$ by the ring $C$ constructed in Lemma 61.5.8 for $A \to B$ and $I \subset A$. Thus we may assume $\mathop{\mathrm{Spec}}(B)$ is w-local such that the set of closed points of $\mathop{\mathrm{Spec}}(B)$ is $V(IB)$.

Assume $\mathop{\mathrm{Spec}}(B)$ is w-local and the set of closed points of $\mathop{\mathrm{Spec}}(B)$ is $V(IB)$. Choose a continuous section to the surjective continuous map $V(IB) \to V(I)$. This is possible as $V(I) \cong \pi _0(\mathop{\mathrm{Spec}}(A))$ is extremally disconnected, see Topology, Proposition 5.26.6. The image is a closed subspace $T \subset \pi _0(\mathop{\mathrm{Spec}}(B)) \cong V(IB)$ mapping homeomorphically onto $\pi _0(A)$. Replacing $B$ by the ind-Zariski quotient ring constructed in Lemma 61.6.1 we see that we may assume $\pi _0(\mathop{\mathrm{Spec}}(B)) \to \pi _0(\mathop{\mathrm{Spec}}(A))$ is bijective. At this point $A \to B$ is an isomorphism by Lemma 61.6.4.

Assume (1) or equivalently (2). Let $A \to A_ w$ be the ring map constructed in Lemma 61.5.3. By (1) there is a retraction $A_ w \to A$. Thus $\mathop{\mathrm{Spec}}(A)$ is homeomorphic to a closed subset of $\mathop{\mathrm{Spec}}(A_ w)$. By Lemma 61.2.4 we see (3)(a) holds. Finally, let $T \to \pi _0(A)$ be a surjective map with $T$ an extremally disconnected, quasi-compact, Hausdorff topological space (Topology, Lemma 5.26.9). Choose $A \to B$ as in Lemma 61.6.2 adapted to $T \to \pi _0(\mathop{\mathrm{Spec}}(A))$. By (1) there is a retraction $B \to A$. Thus we see that $T = \pi _0(\mathop{\mathrm{Spec}}(B)) \to \pi _0(\mathop{\mathrm{Spec}}(A))$ has a section. A formal categorical argument, using Topology, Proposition 5.26.6, implies that $\pi _0(\mathop{\mathrm{Spec}}(A))$ is extremally disconnected. $\square$

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

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 $B$. $\square$