Omitted.

Omitted.

Let \(Y=X\) but endowed with the discrete topology. Let \(X'=\beta (Y)\), which is extremally disconnected by Theorem 1.4. The continuous map \(Y \to X\) factors as \(Y \to X' \to X\).

Let \(x, y\) be points of \(X, Y\), respectively. To show the above map is bijective, it suffices to show that the connected component of \((x,y)\) in \(X \times Y\) is the product of the connected component of \(x\) in \(X\) and the connected component of \(y\) in \(Y\). We know that the product of connected subsets is still connected (IsPreconnected.prod). The connected component of \((x, y)\) cannot be larger because its projections to \(X\) and \(Y\) are also connected.

To show that the bijective map is homeomorphic, we notice that the topology on both side are generated by the basis of form \(\text{image } U \times \text{image } V\).

Use the universal property of connected components.

Omitted.

Omitted.

The only if direction holds for any continuous map. Conversely, if \(f(a)\) specializes to \(f(b)\), then \(b \in f^{-1}(\overline{\{ f(a)\} }) = \overline{\{ a\} }\).

This is immediate from the definition.

Omitted.

Let \(T\) be the connected component containing \(x\). Let \(S = \bigcap _{\alpha \in A} Z_\alpha \) be the intersection of all open and closed subsets \(Z_\alpha \) of \(X\) containing \(x\). Note that \(S\) is closed in \(X\). Note that any finite intersection of \(Z_\alpha \)’s is a \(Z_\alpha \). Because \(T\) is connected and \(x \in T\) we have \(T \subset Z_\alpha \) for every \(Z_\alpha \), thus \(T \subset S\). It suffices to show that \(S\) is connected. If not, then there exists a disjoint union decomposition \(S = B \amalg C\) with \(B\) and \(C\) open and closed in \(S\). In particular, \(B\) and \(C\) are closed in \(X\), and so quasi-compact by assumption (1). By assumption (2) there exist quasi-compact opens \(U, V \subset X\) with \(B = S \cap U\) and \(C = S \cap V\). Then \(U \cap V \cap S = \emptyset \). Hence \(\bigcap _\alpha U \cap V \cap Z_\alpha = \emptyset \). By assumption (3) the intersection \(U \cap V\) is quasi-compact. Hence for some \(\alpha ' \in A\) we have \(U \cap V \cap Z_{\alpha '} = \emptyset \). Since \(X - (U \cup V)\) is disjoint from \(S\) and closed in \(X\) hence quasi-compact, we can use the same argument to see that \(Z_{\alpha ''} \subset U \cup V\) for some \(\alpha '' \in A\). Then \(Z_\alpha = Z_{\alpha '} \cap Z_{\alpha ''}\) is contained in \(U \cup V\) and disjoint from \(U \cap V\). Hence \(Z_\alpha = U \cap Z_\alpha \amalg V \cap Z_\alpha \) is a decomposition into two open pieces, hence \(U \cap Z_\alpha \) and \(V \cap Z_\alpha \) are open and closed in \(X\). Thus, if \(x \in B\) say, then we see that \(S \subset U \cap Z_\alpha \) and we conclude that \(C = \emptyset \).

It is clear that (a) implies (b). Assume (b). Let \(x \in X\), \(x \notin T\). Let \(x \in C \subset X\) be the connected component of \(X\) containing \(x\). By Lemma 1.15 we see that \(C = \bigcap V_\alpha \) is the intersection of all open and closed subsets \(V_\alpha \) of \(X\) which contain \(C\). In particular, any pairwise intersection \(V_\alpha \cap V_\beta \) occurs as a \(V_\alpha \) i.e. is still open and closed. As \(T\) is a union of connected components of \(X\) we see that \(C \cap T = \emptyset \). Hence \(T \cap \bigcap V_\alpha = \emptyset \). Since \(T\) is quasi-compact as a closed subset of a quasi-compact space, we deduce that \(T \cap V_\alpha = \emptyset \) for some \(\alpha \). For this \(\alpha \) we see that \(U_\alpha = X - V_\alpha \) is an open and closed subset of \(X\) which contains \(T\) and not \(x\). The lemma follows.

We will show that \(\pi _0(X)\) is Hausdorff, quasi-compact, and totally disconnected. By Lemma 1.15, every connected component of \(X\) is the intersection of the open and closed subsets containing it. Since \(\pi _0(X)\) is the image of a quasi-compact space, it is also quasi-compact. It is totally disconnected by construction. Let \(C,D \subset X\) be distinct connected components of \(X\). Write \(C = \bigcap _{\alpha } U_\alpha \) as the intersection of the open and closed subsets of \(X\) containing \(C\). Any finite intersection of \(U_\alpha \)’s is another \(U_\alpha \). Since \(\bigcap _\alpha U_\alpha \cap D = \emptyset \) we conclude that \(U_\alpha \cap D = \emptyset \) for some \(\alpha \) (connected components are closed, thus quasi-compact). Since \(U_\alpha \) is open and closed, it is the union of the connected components it contains, i.e., \(U_\alpha \) is the inverse image of some open and closed subset \(V_\alpha \subset \pi _0(X)\). This proves that the points corresponding to \(C\) and \(D\) are contained in disjoint open subsets, i.e., \(\pi _0(X)\) is Hausdorff.

The only remaining part in Lean is quasi-separatedness. Suppose that \(V\) is open compact in \(Y\), we show that there exists open compact \(U\) in \(X\) such that \(V = U \cap Y\). We can find \(U'\) in \(X\) such that \(V = U' \cap Y\). Write \(U' = \bigcup _{i \in I} U_i\) as a union of compact opens, then a finite subset \(i \in I_0\) of \(U_i\) already covers \(V\). Take \(U := \bigcup _{i \in I\_ 0} U_i\) as desired.

For any compact opens \(V_1,\, V_2 \subset Y\), there exists compact open \(U_1, U_2 \subset X\) s.t. \(V_i = U_i \cap Y_i\). Then \(V_1 \cap V_2 = (U_1 \cap U_2) \cap Y \), is still open compact in \(Y\). Here \(U_1 \cap U_2\) is quasi-compact since \(X\) is quasi-separated.

It is clear that \(\overline{\{ (x, y)\} } \subseteq \overline{\{ x\} } \times \overline{\{ y\} }\).

If \(z \notin \overline{\{ (x, y)\} }\), there exists opens \(U \subset X\), \(V \subset Y\) such that \(z \notin U \times V\), thus \((x, y) \notin U \times V\). Either \(x \notin U\) or \(y \notin V\), suppose \(x \notin U\). Then \(U \cap \overline{\{ x\} } = \emptyset \), thus \(U \times V \cap \overline{\{ x\} } \times \overline{\{ y\} } = \emptyset \). This shows that \(\overline{\{ x\} } \times \overline{\{ y\} } \subseteq \overline{\{ (x, y)\} }\).

Let \(X\), \(Y\) be spectral spaces. Denote \(p : X \times Y \to X\) and \(q : X \times Y \to Y\) the projections. We first show that \(X \times Y\) is sober. Let \(Z \subset X \times Y\) be a closed irreducible subset. Then \(p(Z) \subset X\) is irreducible and \(q(Z) \subset Y\) is irreducible. Let \(x \in X\) be the generic point of the closure of \(p(Z)\) and let \(y \in Y\) be the generic point of the closure of \(q(Z)\). If \((x, y) \notin Z\), then there exist opens \(x \in U \subset X\), \(y \in V \subset Y\) such that \(Z \cap (U \times V) = \emptyset \). Hence \(Z\) is contained in \((X - U) \times Y \cup X \times (Y - V)\). Since \(Z\) is irreducible, we see that either \(Z \subset (X - U) \times Y\) or \(Z \subset X \times (Y - V)\). In the first case \(p(Z) \subset (X - U)\) and in the second case \(q(Z) \subset (Y - V)\). Both cases are absurd as \(x\) is in the closure of \(p(Z)\) and \(y\) is in the closure of \(q(Z)\). Thus we conclude that \((x, y) \in Z\), which means that \((x, y)\) is the generic point for \(Z\) (Lemma 1.19).

A basis of the topology of \(X \times Y\) consists of opens of the form \(U \times V\) with \(U \subset X\) and \(V \subset Y\) quasi-compact open (here we use that \(X\) and \(Y\) are spectral). Then \(U \times V\) is quasi-compact as the product of quasi-compact spaces is quasi-compact. Moreover, any quasi-compact open of \(X \times Y\) is a finite union of such quasi-compact rectangles \(U \times V\). It follows that the intersection of two such is again quasi-compact (since \(X\) and \(Y\) are spectral). This concludes the proof.

Omitted.

\(X \times _S Y\) is the inverse image of the image \(\Delta (S)\) of the diagonal map \(\Delta : S \to S \times S\). It follows from Lemma 1.21 that \(\Delta (S)\) is a closed subset.

Omitted.

Note that \(X \times _S Y\) is a closed subspace of \(X \times Y\) by Lemma 1.22. (For subspace, see TopCat.pullbackIsoProdSubtype.) Since \(X \times Y\) is spectral (Lemma 1.20) it follows that \(X \times _S Y\) is spectral (Lemma 1.18). Note that we also used Lemma 1.23 to translate between categorical constructions and non-categorical constructions.

Since R1 implies Sober, Hausdorff implies quasi-separated, it only remains to show that \(X\) is prespectral. Write \(X\) as an inverse limit of finite discrete topological spaces. The preimage of singletons in every finite discrete space are open and closed subsets and forms a topological basis for the topology on \(X\). Thus \(X\) is prespectral.

By Lemma 1.25, \(T\) is spectral. By Lemma 1.17 \(\pi _0(X)\) is Hausdorff. By Lemma 1.24, \(Y\) is spectral. Let \(Y \to \pi _0(Y) \to T\) be the canonical factorization in Definition 1.13. It is clear that \(\pi _0(Y) \to T\) is surjective. The fibres of \(Y \to T\) are homeomorphic to the fibres of \(X \to \pi _0(X)\). Hence these fibres are connected. It follows that \(\pi _0(Y) \to T\) is injective by Lemma 1.14. We conclude that \(\pi _0(Y) \to T\) is a homeomorphism since it is a bijective map from a quasi-compact space to a Hausdorff space by isHomeomorph_iff_continuous_bijective.

It suffices to show that all the elements of \(B - {\mathfrak {q}}\) are all invertible in \(B_{{\mathfrak {p}}B}\). Suppose the contrary, let \(x \in B - {\mathfrak {q}}\) be an element not invertible in \(B_{{\mathfrak {p}}B}\). We can find a prime ideal \({\mathfrak {q}}' \subseteq {\mathfrak {p}}B\) of \(B\) containing \(x\). By going down there exists a prime \({\mathfrak {q}}'' \subseteq {\mathfrak {q}}\) lying over \({\mathfrak {p}}' = A \cap {\mathfrak {q}}'\). By the uniqueness of primes lying over \({\mathfrak {p}}'\) we see that \({\mathfrak {q}}' = {\mathfrak {q}}''\), which contradicts the fact that \(x \notin {\mathfrak {q}}\).

Omitted.

Omitted.

The functor \(F\) is a composition of two functors: 1. The fully faithful functor from the category of \(A\)-algebras \(B\) for which \(A \to B\) identifies local rings to the category of ringed spaces \((Y, \O _Y)\) over \(X = \operatorname{Spec}(A)\) satisfying \(\O _Y = p^{-1}\O _X\), where \(p: Y \to X\) is the structure map. 2. The functor sending a ringed space to its underlying topological space, and a morphism \((f, f^\# )\) to \(f\).

The second functor is fully faithful because \(f^\# \) is always an isomorphism, being given by the canonical identification \(f^{-1}\O _Y \cong f^{-1}p^{-1}\O _X = q^{-1}\O _X \cong \O _Z\), where \(p: Y \to X\) and \(q: Z \to X\) are the structure maps.

This follows because every étale algebra is of finite presentation and by possibly enlarging \(i\) we can ensure that \(B'\) is étale over \(A_i\).

Omitted.

An ind-Zariski ring map is a filtered colimit of local isomorphisms. A local isomorphism is flat.

This follows from general theory, because local isomorphisms are of finite presentation.

This is the consequence of the more general fact that if a property \(p\) of ring maps is stable under finite products, then Ind-\(p\) is also stable under finite products since finite product of filtered index category is again filtered.

It suffices to show that being local isomorphism is stable under finite products. Suppose \(A \to B_i\) is a local isomorphism for each \(i\) for finitely many \(i\)’s. Every prime \({\mathfrak {q}}\) in \(\prod _i B_i\) is an extension prime ideal of one of the factors, say \(B_0\). Since there exists \(f_0 \in B_0\), \(f_0 \notin {\mathfrak {q}}\) such that \(A \to (B_0)_{f_0}\) induces an open immersion, localize \(\prod _i B_i\) at the corresponding idempotent times this element \(f_0\) to obtain the desired result.

Omitted.

\(S^{-1} A = \operatorname {colim}_{f \in S} A_f\) and \(A \to A_f\) is a local isomorphism.

By [ Sta18 , Tag 04GG ] , \(R\) is Henselian if and only if for any étale ring map \(R \to S\) and prime \(\mathfrak {q}\) of \(S\) lying over \(\mathfrak {m}\) with \(\kappa = \kappa (\mathfrak {q})\), there exists a retraction \(S \to R\) of \(R \to S\).

Suppose \(R\) is Henselian and let \(R \to S \to \kappa \) be a factorisation of \(R \to \kappa \) with \(R \to S\) étale. Then \(\mathfrak {q} = \mathrm{ker}(S \to \kappa )\) is a prime ideal of \(S\) lying above \(\mathfrak {m}\). Hence we obtain a commutative diagram

where the composition of the dashed arrows is the identity of \(\kappa \). Hence \(\kappa (\mathfrak {q}) = \kappa \) and by assumption, \(R \to S\) has a retraction.

Now assume that \(R \to \kappa \) is étale-Henselian and let \(R \to S\) be étale and \(\mathfrak {q}\) be a prime ideal of \(S\) with \(\kappa (\mathfrak {q}) = \kappa \). Hence the composition \(R \to S \to \kappa (\mathfrak {q}) = \kappa \) is a factorisation of \(R \to \kappa \). Thus \(R \to S\) has a retraction.

Since \(B/{\mathfrak {m}}B\) is an étale algebra over the residue field \(k = A/{\mathfrak {m}}\), it splits as \(B/{\mathfrak {m}}B \cong k_1 \times \cdots \times k_n\) for finitely many finite separable extensions \(k_i\) of \(k\). Thus \(B/{\mathfrak {n}}\), as a quotient of \(B/{\mathfrak {m}}B\), is also a finite separable extension of \(k\). Choose an inclusion \(B/{\mathfrak {n}}\hookrightarrow k^{\textnormal{sep}}\). Then we have a commutative diagram

Since \(A \to k^{\textnormal{sep}}\) is étale-Henselian, there exists a retraction \(r : B \to A\) of \(A \to B\). But this retraction may not satisfy \({\mathfrak {n}}= r^{-1}({\mathfrak {m}})\).

To solve this, notice that there are only finitely many prime ideals \({\mathfrak {n}}_0 = {\mathfrak {n}}, {\mathfrak {n}}_1, \cdots , {\mathfrak {n}}_k\) of \(B\) lying over \({\mathfrak {m}}\) and they are all maximal ideals. (directly by the fact that \(B/{\mathfrak {m}}B \cong k_1 \times \cdots \times k_n\)). Since \(\bigcap _{i \ne 0} {\mathfrak {n}}_i \nsubseteq {\mathfrak {n}}\) (otherwise one of \({\mathfrak {n}}_i \subseteq {\mathfrak {n}}\) which is impossible), we can find element \(b \in {\mathfrak {n}}_i\) for every \(i \ne 0\) that does not fall in \({\mathfrak {n}}\). We may replace \((B, {\mathfrak {n}})\) by \((B_b, {\mathfrak {n}}B_b)\), then \(B\) is still étale over \(A\) and \({\mathfrak {n}}\) is the only prime ideal of \(B\) lying over \({\mathfrak {m}}\). Then the section \(s_b : B_b \to A\) given by the previous paragraph satisfies \({\mathfrak {n}}B_b = s_b^{-1}({\mathfrak {m}})\). Composition with the localization map \(B \to B_b\) gives the desired section \(s : B \to A\).

Let \(s: B \to A\) be the section given by Proposition 1.48 for the pair \((B, {\mathfrak {n}})\). Localizing at \({\mathfrak {m}}\), we obtain the following diagram:

The dotted arrow \(s'\) exists by the universal property of localization, since \({\mathfrak {n}}= s^{-1}({\mathfrak {m}})\). Moreover, \(s'\) is a section of \(f'\).

Since \(f_{\mathfrak {m}}\) is étale and \(s_{\mathfrak {m}}\) is a section of \(f_{\mathfrak {m}}\), it follows that \(s_{\mathfrak {m}}\) is also étale by the cancellation property of étale maps. Thus \(s'\), being a composition of a localization and an étale map, is flat. As a flat local ring homomorphism, \(s'\) is faithfully flat and hence injective. Since \(s'\) is also a section of \(f'\), it is surjective. Therefore, \(s'\) is an isomorphism, and the composition \(A \to B_{\mathfrak {n}}\) is an isomorphism.

Omitted.

Omitted.

This follows from general theory, because étale maps are of finite presentation.

An ind-Zariski ring map is a filtered colimit of local isomorphisms. A local isomorphism is étale.

Omitted.

Let \(x \rightsquigarrow y\) in \(X\) with \(x \in Z^g\). Then \(x\) specializes to some \(z \in Z\), which in turn specializes to a unique closed point \(c\) in \(X\) because \(X\) is w-local. Because \(Z\) is stable under specialization, \(c \in Z\). Also \(y\) specializes to a unique closed point \(c'\) in \(X\) and by uniqueness we have \(c' = c \in Z\). Hence \(y \in Z^g\).

Since \(Z\) is closed under specializations, we have \(Z^c = X^c \cap Z\). In particular \(Z^c\) is closed in \(Z\) and \(Z \to X\) preserves closed points. By 1.10, specializations in \(Z\) are the same as specializations in \(X\). Since \(Z\) is closed under specializations, every point in \(Z\) specializes to a unique closed point in \(X\) and therefore in \(Z\).

This follows from 1.59, because closed subsets are stable under specialization. The inclusion \(Y \to X\) is spectral as a closed embedding and hence w-local by the lemma.

Write \(X = \lim _{i \in I} X_i\) as a directed limit of finite discrete spaces. Let \(\pi _i \colon X \to X_i\) be the projection maps. A base of the open sets of \(X\) is given by the preimages \(\pi _i^{-1}(\{ z\} )\) for \(z \in X_i\) and \(i \in I\). Every open set in this base is also closed, because its complement is a finite union of such sets.

Choose an open subset in this base containing \(x\) but not \(y\), say \(U = \pi _{i_0}^{-1}(\{ z_0\} )\) for some \(i_0 \in I\) and \(z_0 \in X_{i_0}\). Then \(V = X - U\) is also open and closed, and we have \(X = U \coprod V\) with \(x \in U\) and \(y \in V\).

Since every point is closed, there are no nontrivial specializations between points. By Enumi 2 in Lemma 1.67, all constructible subsets of \(X\) are closed. In particular, every compact open subset is also closed. Since \(X\) is spectral, thus T0, for any two points there is a quasi-compact open (also closed) containing exactly one of them. Thus \(X\) is profinite by isTotallyDisconnected_of_isClopen_set.

It is a closed subspace of a spectral space, hence spectral. Therefore it is profinite by 1.62

Let \(X\) be a spectral space and let \(W\) denote \(E^g\). Since every point of \(W\) specializes to a point of \(E\) we see that an open of \(W\) which contains \(E\) is equal to \(W\). Hence since \(E\) is quasi-compact, so is \(W\). If \(x \in X\), \(x \notin W\), then \(Z = \overline{\{ x\} }\) is disjoint from \(W\). Since \(W\) is quasi-compact we can find a quasi-compact open \(U\) (since there is a open basis of quasi-compact opens of \(X\) and finitely many of them would suffice to cover \(W\)) with \(W \subset U\) and \(U \cap Z = \emptyset \). We conclude that \(W\) is an intersection of quasi-compact open subsets.

Let \(\mathfrak {B}\) be the set of subsets of \(X\) that are quasi-compact open or the complement of a quasi-compact open. By definition of the constructible topology and the Alexander subbasis theorem, it suffices to show that every covering \(X = \bigcup _{i \in I} B_i\) with \(B_i \in \mathfrak {B}\) has a finite subcover. Since for every \(B \in \mathfrak {B}\) also \(X - B \in \mathfrak {B}\), it suffices to show: If \(\{ B_i\} _{i \in I}\) is a family of elements of \(\mathfrak {B}\) such that all finite subfamilies have non-empty intersection, then \(\bigcap _{i \in I} B_i\) is non-empty.

Let \(\mathcal{S}\) be the set of families in \(\mathfrak {B}\) that are pairwise unequal, have non-empty finite intersections and empty intersection. For the sake of contradiction, suppose \(\mathcal{S}\) is non-empty. If we equip \(\mathcal{S}\) with the ordering by subset inclusion, Zorn’s Lemma yields a maximal family \(\{ B_i\} _{i \in I} \in \mathcal{S}\).

Let \(I' \subseteq I\) be the indices such that \(B_i\) is closed and set \(Z = \bigcap _{i \in I'} B_i\). This is a closed and non-empty subset of \(X\), because \(X\) is quasi-compact and all finite intersections are non-empty by assumption. Note that for any \(i_1, \ldots , i_n \in I\) the intersection \(\bigcap _{a=1}^n B_{i_a}\) is quasi-compact, hence \(Z \cap \bigcap _{a=1}^n B_{i_a}\) is non-empty by the same argument.

Suppose \(Z = Z' \cup Z'\) is reducible with \(Z', Z'' \subsetneq Z\) closed. Because \(X\) is a spectral space, there exist quasi-compact opens \(U', U'' \subseteq X\) with \(U' \cap Z'\) non-empty, \(U'' \cap Z''\) non-empty and \(U' \cap Z'' = \emptyset = U'' \cap Z'\). Then set \(B' = X - U' \in \mathfrak {B}\) and \(B'' = X - U'' \in \mathfrak {B}\). We claim that \(\{ B_i\} _{i \in I}\) with \(B'\) or \(B''\) added is still in \(\mathcal{S}\). Suppose, this is not the case. Then there exist finitely many indices \(i_a\) and \(j_b\) in \(I\) such that

By construction, \(Z \subseteq B' \cup B''\), so in particular

which is a contradiction. Hence \(\{ B_i\} _{i \in I}\) is not maximal, also a contradiction. So \(Z\) is irreducible. But for every \(i \in I - I'\), the set \(B_i\) is open and hence the non-empty open \(Z \cap B_i\) of \(Z\) contains the unique generic point of \(Z\). Hence \(\bigcap _{i \in I} B_i = Z \cap \bigcap _{i \in I - I'} B_i\) is non-empty, which is again a contradiction.

It suffices to show the first claim. Let \(x \in \overline{E}\) and let \(\{ U_i\} _{i \in I}\) be the set of quasi-compact open neighbourhoods of \(x\). Since \(X\) is spectral, these form a neighbourhood basis of \(x\), so \(\bigcap _{i \in I} U_i\) is the set of generalizations of \(x\). It hence suffices to show that \(\bigcap _{i \in I} U_i \cap E\) is non-empty. Indeed, as \(X\) is quasi-separated, any finite intersection of the \(U_i\) is another one and since \(x \in \overline{E}\), the intersection \(U_i \cap E\) is non-empty for all \(i\). Since \(U_i \cap E\) is closed in the constructible topology for all \(i\) and \(X\) is quasi-compact in the constructible topology by 1.66, the intersection \(\bigcap _{i \in I} U_i \cap E\) is non-empty.

Since \(Z\) is closed, it is quasi-compact. Hence by Lemma 1.64, we see that \(Z^g\) is closed in the constructible topology as an intersection of quasi-compact opens. Since \(Z\) is stable under specialization, the same holds for \(Z^g\) by Lemma 1.58. Thus by Lemma 1.67, the claim follows.

Since \(X^c\) is quasi-compact and \(\pi _0(X)\) is Hausdorff by Lemma 1.17, it suffices to show the map is bijective. To show injectivity let \(x, y \in X^c\) be distinct. Since \(X^c\) is profinite (Lemma 1.63), there exist open and closed subsets \(U_0, V_0 \subseteq X^c\) such that \(X^c = U_0 \coprod V_0\) by Lemma 1.61. Let \(U = U_0^g\) and \(V = V_0^g\). Because \(X\) is w-local, we get \(X = U \coprod V\) as sets. By Lemma 1.68, both \(U\) and \(V\) are closed and therefore also both open. Hence \(x\) and \(y\) lie in distinct connected components of \(X\), so the images in \(\pi _0(X)\) are distinct.

Since every connected component is closed, it is quasi-compact and hence contains a closed point. Thus the map is surjective.

We know that \(Y\) is spectral and \(T = \pi _0(Y)\) by Lemma 1.26. Assume \(X\) is w-local and let \(X^c \subset X\) be its closed points. The inverse image \(Y^c \subset Y\) of \(X^c\) maps bijectively onto \(T\), since \(X^c \to \pi _0(X)\) is a bijection by Lemma 1.69. (Note that we do not yet know whether \(Y^c\) is the set of closed points of \(Y\).) Moreover, \(Y^c\) is quasi-compact as a closed subset of the spectral space \(Y\). Hence \(Y^c \to \pi _0(Y) = T\) is a homeomorphism by isHomeomorph_iff_continuous_bijective ( [ Sta18 , Tag 08YE ] ). It follows that all points of \(Y^c\) are closed in \(Y\): if some \(y \in Y^c\) specialized to a distinct closed point \(y'\) (which also falls in \(Y\) since \(Y\) is closed), then both would map to the same point in \(T\), contradicting the bijectivity of \(Y^c \to T\).

Conversely, if \(y \in Y\) is a closed point, then it is closed in its fibre over \(T\), and its image \(x \in X\) is closed in the corresponding (homeomorphic) fibre of \(X \to \pi _0(X)\). Since a point in a w-local space is closed if and only if it is closed in its connected component (which is closed), we have \(x \in X^c\), so \(y \in Y^c\). Thus \(Y^c\) is exactly the set of closed points of \(Y\). Moreover, since \(Y^c \cong T\), each connected component contains a unique closed point, and for each \(y \in Y^c\) the set of generalizations of \(y\) is precisely the fibre of \(Y \to \pi _0(Y)\) (as any generalization of \(y\) lies in the same connected component with \(y\)). The lemma follows.

Omitted.

This is Lemma 1.60.

This follows from the fact that every point specializes to a unique closed point and the spectrum of the local ring at a point is the set of all points specialize to it.

Since \(A \to B\) is flat (can be checked on stalks of \(B\)), it has going down. Thus Lemma 1.27 shows for any prime \({\mathfrak {q}}\subset B\) lying over \({\mathfrak {p}}\subset A\) we have \(B_{{\mathfrak {q}}} = B_{{\mathfrak {p}}B}\). Since \(B_{{\mathfrak {q}}} = A_{{\mathfrak {p}}}\) by assumption, we see that \(A_{{\mathfrak {p}}} = B_{{\mathfrak {p}}B}\) for all primes \({\mathfrak {p}}\) of \(A\). Thus \(A = B\) since isomorphism can be checked locally on \(A\).

By Lemma 1.75, it suffices to show that \(Y = \operatorname{Spec}(B) \to X = \operatorname{Spec}(A)\) is bijective. Let \(X^c \subset X\) and \(Y^c \subset Y\) be the sets of closed points. By assumption \(Y^c\) maps into \(X^c\) and the induced map \(Y^c \to X^c\) is a bijection since Lemma 1.69 holds. By Lemma 1.74, we see that both \(X\) and \(Y\) , as sets, are disjoint unions of the spectra of the local rings at closed points. As \(A \to B\) identifies local rings, \(Y \to X\) is a bijection.

Let \(S = S_{f, I}\), \(Z = D(f) \cap V(I)\) and \(J = \widetilde{(I, f)}\).

Note that \(Z\) is homeomorphic to \(\mathrm{Spec}((A/I)_f)\). Since the map \(\mathrm{Spec}(S^{-1}A) \to \mathrm{Spec}(A)\) is a homeomorphism onto its image, to show (1) it suffices to show that the image is the set of points of \(\mathrm{Spec}(A)\) specializing to \(D(f) \cap V(I)\). Let \(\mathfrak {p}\) be a prime of \(A\) specializing to a \(\mathfrak {q}\) in \(Z\), so \(I \subseteq \mathfrak {q}\) and \(f \not\in \mathfrak {q}\). In particular, the image of every element of \(\mathfrak {p}\) in \((A / I)_f\) is contained in \(\mathfrak {q}\) and is therefore not invertible. Conversely, let \(\mathfrak {p}\) be a prime of \(A\) which does not specialize to a point in \(Z\), hence the image of \(\mathfrak {p}\) in \((A/I)_f\) contains the unit element. So there exists \(g \in \mathfrak {p}\) that becomes invertible in \((A / I)_f\), so \(g \in S\) and \(\mathfrak {p}\) is not in the image of \(\mathrm{Spec}(S^{-1} A)\).

By construction, the induced map \(S^{-1}A \to (A / I)_f\) is surjective and hence induces an isomorphism \(S^{-1}A / J \to (A / I)_f\). This shows (2).

Let \(g \in A\). Suppose the image of \(g\) in \((A' / I')_{f'}\) is not invertible. Then there exists a prime ideal \(\mathfrak {p}\) of \((A' / I')_{f'}\) that contains the image of \(g\). Hence \(g\) is in the preimage of \(\mathfrak {p}\) under \(A \to A'\), which lies in \(Z\) by assumption. Thus \(g\) is not invertible in \((A / I)_f\), which shows \(g \not\in S\).

\(A_{\widetilde{(I, f)}} = S^{-1} A\) for some multiplicative subset \(A\).

This is immediate from the definitions of \(D\) and \(V\).

This is immediate from the definition of \(Z(-, -)\).

Let \(x \in \mathrm{Spec}(A)\) and \(E = E' \coprod E''\) a disjoint union decomposition. Then \(x \in Z(E', E'')\) if and only if \(f(x) \neq 0\) for all \(f \in E'\) and \(f(x) = 0\) for all \(f \in E''\). Hence \(x\) is contained in

and this is the only set of this form in which \(x\) is contained.

By definition \(I_E\) is the product of the ideals defining \(Z(E', E'')\) in \(\mathrm{Spec}(A_{\widetilde{Z(E', E'')}})\). Hence the first claim. The second follows from the first, because \(\mathrm{Spec}(A) = \coprod _{E = E' \coprod E''} Z(E', E'')\) by 1.83.

Let \(x \in \mathrm{Spec}(A_E)\). Then \(x\) is in \(\mathrm{Spec}(A_{\widetilde{Z(E', E'')}})\) for some \(E = E' \coprod E''\). By 1.78, \(x\) specializes to a point in \(Z(E', E'') \subseteq V(I_E)\). If \(x\) is closed, then \(x\) specializes to itself, so \(x \in V(I_E)\).

This holds because for every decomposition \(E_2 = E_2' \coprod E_2''\) and \(E_1' = E_1 \cap E_2'\) and \(E_1'' = E_1 \cap E_2''\), the induced map \(\mathrm{Spec}(A_{\widetilde{Z(E_2', E_2'')}}) \to \mathrm{Spec}(A_{\widetilde{Z(E_1', E_1'')}})\) maps \(Z(E_2', E_2'')\) into \(Z(E_1', E_1'')\).

Omitted.

For \(E \subseteq A\) a finite subset, the \(A\)-algebra \(A_E\) is ind-Zariski as a finite product of ind-Zariski algebras by 1.41. Thus \(A_w\) is a filtered colimit of ind-Zariski algebras, hence ind-Zariski by 1.40.

Since \(V(I_w) = \lim _{E} V(I_E)\) (by Lemma 1.89) and \(V(I_E) \to \mathrm{Spec}(A_E) \to \mathrm{Spec}(A)\) is bijective for all \(E\) by Lemma 1.85, the first claim holds.

We first show that every point of \(\mathrm{Spec}(A_w)\) specializes to a point in \(V(I_w)\). For this let \(y \in \mathrm{Spec}(A_w)\). For all \(E \subseteq A\) finite, denote by \(T_E\) the closure of the image of \(y\) in \(\mathrm{Spec}(A_E)\). By Lemma 1.86, the intersection \(T_E \cap V(I_E)\) is non-empty. Since both \(T_E\) and \(V(I_E)\) are closed, the intersection is closed and hence quasi-compact. Thus \((\lim _E T_E) \cap V(I_w) = \lim _E (T_E \cap V(I_E))\) is non-empty. Apply ?? to singleton \({y}\), the subset \(\lim _E T_E\) is the closure of \(y\) in \(\mathrm{Spec}(A_w)\). (Shuold be a cofiltered version of IsCompact.nonempty_iInter_of_directed_nonempty_isCompact_isClosed.) Hence \(y\) specializes to a point in \(V(I_E)\).

For uniqueness, suppose \(y \in \mathrm{Spec}(A_w)\) specializes to \(z, z' \in V(I_w)\) with \(z \neq z'\). Denote the images of \(z, z'\) in \(\mathrm{Spec}(A)\) by \(x, x'\). Because \(V(I_w) \to \mathrm{Spec}(A_w) \to \mathrm{Spec}(A)\) is injective, \(x\) and \(x'\) are distinct. Hence we may assume there exists \(f \in A\) such that \(x \in D(f)\) and \(x' \in V(f)\). Set \(E = \{ f\} \). Then

Denote the images of \(y, z, z'\) in \(\mathrm{Spec}(A_E)\) by \(y_E, x_E, x'_E\). Since \(x_E\) maps to \(x \in D(f)\) and \(x_E'\) maps to \(x' \in V(f)\), they lie in different components of the disjoint union decomposition above. But since \(\mathrm{Spec}(A_w) \to \mathrm{Spec}(A)\) is continuous, \(y_E\) specializes to both \(x_E\) and \(x'_E\) which is not possible.

Finally, every closed point specializes to itself and is hence contained in \(V(I_{w})\). Conversely, every point in \(V(I_{w})\) specializes only to itself, so it is closed.

Immediate from ?? in Lemma 1.91.

By 1.90, \(A \to A_w\) is flat. Because \(V(I_w)\) surjects onto \(\mathrm{Spec}(A)\) by 1.91, in particular \(\mathrm{Spec}(A_w) \to \mathrm{Spec}(A)\) is surjective, hence \(A \to A_w\) is faithfully flat.

We have a factorization \(\kappa (\mathfrak {m}) = R / \mathfrak {m} \to S / \mathfrak {q} \to \kappa (\mathfrak {q})\). Hence \(S / \mathfrak {q}\) is an intermediate subring of an algebraic field extension and therefore a field.

By 1.78, \(\mathrm{Spec}(A_{\widetilde{V(I)}})\) is homeomorphic to the generalization \(V(I)^g\). Since \(\mathrm{Spec}(A)\) is w-local and \(V(I)\) is closed under specialization, also \(V(I)^g\) is closed under specialization by 1.58. Hence \(\mathrm{Spec}(A_{\widetilde{V(I)}}) \cong V(I)^g\) is w-local by 1.59.

By Enumi 2 in Lemma 1.78, we may identify \(V(IA_{\widetilde{V(I)}})\) with \(V(I)\). Since every point in \(\mathrm{Spec}(A_{\widetilde{V(I)}}) \cong V(I)^g\) specializes to a unique point in \(V(I)\), the closed subset \(V(I)\) contains all closed points of \(\mathrm{Spec}(A_{\widetilde{V(I)}})\). Hence, if \(V(I)\) only contains closed points, the closed points are exactly \(V(I)\).

By 1.94, every prime of \(B\) lying above a maximal ideal of \(A\) is maximal. In particular, the closed subset \(V(I B) = (\mathrm{Spec}(B) \to \mathrm{Spec}(A))^{-1}(V(I))\) only contains closed points.

Let \(\mathfrak {p}\) be a prime of \(A\). Then it is contained in a maximal ideal \(\mathfrak {m}\) that is the preimage of a prime \(\mathfrak {q}\) of \(B\). By going-down, there exists therefore a prime \(\mathfrak {q}' \subseteq \mathfrak {q}\) lying above \(\mathfrak {p}\).

Follows immediately from Lemma 1.95 and Corollary 1.92.

The map \(A \to A_w\) is ind-Zariski by Lemma 1.90 and the map \(A_w \to A_{w, I}\) is ind-Zariski by Lemma 1.79. Hence the composition \(A \to A_{w, I}\) is ind-Zariski by Lemma 1.42.

\(A \to A_w\) is ind-Zariski by 1.90 and therefore induces algebraic extensions of residue fields by 1.38. In particular, the closed subset \(V(I A_w)\) only contains closed points by 1.96 and the first claim follow from 1.95.

The map \(A \to A_{w, I}\) is ind-Zariski as a composition of ind-Zariski maps by 1.90 and 1.79. Since ind-Zariski is stable under base change, the same holds for \(A/I \to A_{w, I}/IA_{w, I}\). Hence it identifies local rings. By Enumi 2 in Lemma 1.78, \(V(IA_{w, I}) \to V(I A_w)\) is an isomorphism. Moreover \(V(I A_{w}) \to V(I)\) is a bijection by Enumi 1 because \(V(I A_{w}) \subseteq V(I_w)\). Hence, the composition \(\mathrm{Spec}(A_{w, I} / I A_{w, I}) = V(I A_{w, I}) \to V(I A_w) \to V(I)\) is bijective. Thus \(A / I \to A_{w, I} / I A_{w, I}\) is an isomorphism by 1.75.

Every point of \(V(IB)\) is a closed point of \(\mathrm{Spec}(B)\) by Lemma 1.96. Hence we may choose \(C = (B_w)_{\widetilde{V(IB_w)}}\) as in Definition 1.98. Enumi 1 holds by Lemma 1.101. Enumi 2 holds by Lemma 1.99. Then the set of closed points of \(C\) is \(V(IC)\), which is the preimage of \(V(I)\) under \(\mathrm{Spec}(C) \to \mathrm{Spec}(B) \to \mathrm{Spec}(A)\).

By ??, \(A \to B \to C\) is flat, so by 1.97 it suffices to show every closed point of \(A\) is in the image. But since \(\mathrm{Spec}(B) \to \mathrm{Spec}(A)\) is surjective, also \(V(IC) \cong V(IB) \to V(I)\) is surjective (the first map is an isomorphism by Enumi 2 in Lemma 1.101).

Write \(B = \operatorname {colim}_i B_i\) as a filtered colimit of étale \(A\)-algebras \(B_i\). Since localization commutes with colimits, we have \(B_{\mathfrak {n}}= \operatorname {colim}_i (B_i)_{{\mathfrak {n}}_i}\) where \({\mathfrak {n}}_i\) is the restriction of \({\mathfrak {n}}\) to \(B_i\). Then use Proposition 1.49 to conclude each \(B_i\) is isomorphic to \(A\).

Let \(\mathfrak {m}\) be a maximal ideal of \(A\), denote by \(\kappa \) the residue field \(\kappa (\mathfrak {m})\) and by \(\kappa ^{\mathrm{sep}}\) a separable closure of \(\kappa \). Let \(A_{\mathfrak {m}} \to B \to \kappa ^{\mathrm{sep}}\) be a factorisation of \(A_{\mathfrak {m}} \to \kappa ^{\mathrm{sep}}\) where \(A_{\mathfrak {m}} \to B\) is étale. Since \(A_{\mathfrak {m}} = \operatorname {colim}_{f \not\in \mathfrak {m}} A_f\), by 1.35 there exists \(f \not\in \mathfrak {m}\) and an étale \(A_f\)-algebra \(B'\) such that \(B = A_{\mathfrak {m}} \otimes _{A_f} B'\). Because \(A \to A_f \to B'\) is étale, there are finitely many prime ideals \(\mathfrak {q}_1, \ldots , \mathfrak {q}_n\) of \(B'\) lying over \(\mathfrak {m}\). Since the kernel of the composition \(B' \to B \to \kappa ^{\mathrm{sep}}\) lies over \(\mathfrak {m}\) (the kernel of \(A \to \kappa _{\mathrm{sep}}\) is \(\mathfrak {m}\)), we have \(n \ge 1\). Set \(\mathfrak {q} = \mathfrak {q}_1\). By prime avoidance, there exists \(g \in B'\) such that \(g \in \mathfrak {q}_i\) for all \(i \ge 2\) and \(g \not\in \mathfrak {q}\). Hence \(\mathfrak {q}\) is the unique prime ideal of \(B'_g\) lying over \(\mathfrak {m}\).

Let \(U\) be the image of the induced map \(\mathrm{Spec}(B’_g) \to \mathrm{Spec}(A)\). Since \(A \to B'\) is étale, \(U\) is open. Since \(\{ \mathfrak {m}\} \) is closed in \(\mathrm{Spec}(A)\), there exist \(a_i\) such that

The complement of \(U\) is closed and quasicompact, and contained in \(\mathrm{Spec}(A) - \{ \mathfrak {m}\} \), hence there exist \(a_1, \ldots , a_r\) such that

Hence the map \(A \to B'_g \times \prod _{i=1}^{r} A_{a_i}\) is étale and faithfully flat. By assumption, it therefore has a retraction \(\sigma \). Since \(\mathfrak {q}\) is the unique prime ideal of \(B'_g \times \prod _{i=1}^{r} A_{a_i}\) lying over \(\mathfrak {m}\), we have \(\sigma ^{-1}(\mathfrak {m}) = \mathfrak {q}\). In particular, we obtain a retraction

of the map \(A_{\mathfrak {m}} \to B'_{\mathfrak {q}}\). Precomposing with \(B' \to B'_{\mathfrak {q}}\) we obtain a map \(B' \to A_{\mathfrak {m}}\) and hence a map \(B = B' \otimes _{A_f} A_{\mathfrak {m}} \to A_{\mathfrak {m}}\) that is a retraction of \(A_{\mathfrak {m}} \to B\).

\(\{ B\} \) is an element of \(S(A)\), hence \(B\) itself is part of the diagram defining \(T(A)\).

Because colimits commute with tensor products, ind-flat is flat. One also checks that pro-surjective is surjective, hence the claim.

For every \(E \in S(A)\), the \(A\)-algebra \(\bigotimes _{B \in E} B\) is étale and faithfully flat. Hence the result follows from the definition of ind-étale and Lemma 1.110.

This follows from Lemma 1.111, Lemma 1.110 and Lemma 1.53.

By Lemma 1.35 there exists \(n \in \mathbb {N}\) and a faithfully flat, étale \(T^n(A)\)-algebra \(B'\) such that \(B = T^{\infty }(A) \otimes _{T^n(A)} B'\). By Lemma 1.108, there exists a map \(B' \to T^{n+1}(A)\). The identity of \(T^{\infty }(A)\) and the composition \(B' \to T^{n+1}(A) \to T^{\infty }(A)\) now induce a retraction

This follows from Proposition 1.114 and Lemma 1.106.

Let \(A = \operatorname {colim}_i A_i\) such that \(A_i\) is an étale \(k\)-algebra for all \(i\). Let \(\mathfrak {m}\) be the maximal ideal of \(A\). Since localization commutes with colimits, we obtain \(A = A_{\mathfrak {m}} = \operatorname {colim}_i (A_i)_{\mathfrak {m}_i}\) where \(\mathfrak {m}_i\) is the restriction of \(\mathfrak {m}\) to \(A_i\). Since \(A_i\) is étale over \(k\), it is a finite product of finite separable extensions of \(k\), hence \((A_i)_{\mathfrak {m}_i}\) is a finite separable extension of \(k\). Since a filtered colimit of finite separable extensions is an algebraic separable extension, we conclude.

Let \(\mathfrak {p}\) be a prime ideal of \(B\). After base change along \(A \to \kappa (\mathfrak {p}^c)\), we may assume \(A = k\) is a field. After postcomposing with the ind-étale map \(B \to B_{\mathfrak {p}}\), we may assume \(B\) is local. Hence the result follows from 1.116.

By Lemma 1.91, the ring \(A_w\) is w-local, so the set of closed points of \(\mathrm{Spec}(A_w)\) is closed. Let \(I\) be the radical ideal of \(A_w\) such that \(V(I)\) is the set of closed points of \(\mathrm{Spec}(A_w)\). Set \(B' = T^{\infty }(A_w)\). By Corollary 1.115, the local ring \(B'_{\mathfrak {m}}\) is strictly henselian for every maximal ideal \(\mathfrak {m}\) of \(B'\). The map \(A_w \to B'\) induces algebraic extensions on residue fields at primes by Lemma 1.113 and Proposition 1.117, so applying ?? to \(A_w \to B'\), we obtain an ind-Zariski ring map \(B' \to B\) such that \(B'/IB' \to B/IB\) is an isomorphism, \(V(IB)\) is the set of closed points of \(\mathrm{Spec}(B)\), \(B\) is w-local and \(A_w \to B' \to B\) is faithfully flat by Lemma 1.103. Since ind-Zariski maps identify local rings by Lemma 1.38 and every closed point of \(\mathrm{Spec}(B)\) maps to a closed point of \(B'\), the local rings at maximal ideals of \(B\) are strictly Henselian. Hence \(B\) is w-strictly local.

It remains to verify that the composition \(A \to A_w \to B' \to B\) is ind-étale and faithfully flat. This follows because both ind-étale and faithfully flat is stable under composition.

Each \(A \to A_W\) is a localization at an idempotent. Passing to filtered colimits, \(A \to A_T\) is surjective and ind-Zariski.

When \(T\) is closed, its inverse image \(Z\) is a closed union of connected components. By Lemma 1.16, we have

Thus, \(\operatorname{Spec}(A_T) = \varprojlim _W W\) is homeomorphic to \(Z\) via the natural map.

The map \(A \to A^S\) is ind-Zariski by Lemma 1.41 and \(A^S \to A^f_{q}\) is also ind-Zariski by Lemma 1.121. The composition \(A \to A^f_{q}\) is therefore ind-Zariski as well by Lemma 1.42. By Lemma 1.38, \(A \to A^f_{q}\) identifies local rings. The last sentence is a direct consequence of Lemma 1.121.

Both maps \(t_{g'}\) and \(t_{g} \circ t_{g' \circ g}\) induce the same topological map between the spectra by construction. Since all maps involved identify local rings by Lemma 1.123, by fully faithfulness in Definition 1.34 we conclude that they are equal.

By Lemma 1.123 we see that each \(A \to A^f_{q_i}\) is ind-Zariski, thus \(A^f_{\pi _0}\) is ind-Zariski by Lemma 1.40. We have the following canonical cartesian diagram

Because \(T = \lim Z_i\) and, \(Y = \operatorname{Spec}(A^f_{\pi _0}) = \lim \operatorname{Spec}(A^f_{q_i})\) fits into the cartesian diagram

of topological spaces. By Lemma 1.26 we conclude that \(T = \pi _0(Y)\).

Let \(A \to B\) be faithfully flat and identifying local rings, with \(B\) w-local and whose closed points of \(\operatorname{Spec}(B)\) are exactly \(V(IB)\). We will show that \(A \to B\) has a retraction.

Choose a continuous section to the surjective continuous map \(V(IB) \to V(I)\). This is possible because \(V(I) = \operatorname{Spec}(A)^c \cong \pi _0(\operatorname{Spec}(A))\) (by Lemma 1.69) is extremally disconnected, see Theorem 1.2. The image is a closed subspace \(T \subset \pi _0(\operatorname{Spec}(B)) \cong V(IB)\) (being the continuous image of the quasi-compact space \(V(I)\)) that maps homeomorphically onto \(\pi _0(\operatorname{Spec}(A))\).

Let \(B \to B_T\) be the ring map from Definition 1.120, which is surjective and ind-Zariski by Lemma 1.121. It also induces a homeomorphism \(\pi _0(\operatorname{Spec}(B_T)) \cong T \cong \pi _0(\operatorname{Spec}(A))\). Moreover, by Lemma 1.73, \(B_T\) is w-local and \(B \to B_T\) is w-local. Since ind-Zariski maps identify local rings (Lemma 1.38), the composition \(A \to B \to B_T\) remains w-local (Lemma 1.57) and identifies local rings (Lemma 1.31). Therefore, by Lemma 1.76, \(A \to B_T\) is an isomorphism.

Let \(A \to B\) be a faithfully flat ind-étale map with \(B\) w-local and such that the closed points of \(\operatorname{Spec}(B)\) are exactly \(V(IB)\). In this case, \(A \to B\) identifies local rings. To justify this, it suffices to check the condition at the maximal ideals of \(B\), which lie over maximal ideals of \(A\) since they map to \(V(I) = (\operatorname{Spec}(A))^c\). Since \(A\) is w-strictly local, the local rings of \(A\) at its maximal ideals are strictly Henselian.

Now fix a maximal ideal \(\mathfrak {n}\) of \(B\) and let \(\mathfrak {m} = \mathfrak {n} \cap A\) be its preimage in \(A\), which is a maximal ideal. By base change to \(A_{\mathfrak {m}}\) and applying Lemma 1.52, we reduce to the case where \(B\) is ind-étale over a strictly Henselian local ring \(A\), with a maximal ideal \(\mathfrak {n}\) lying over the maximal ideal \(\mathfrak {m}\) of \(A\). We then want to show \(B_{\mathfrak {n}} \cong A\). This follows from Lemma 1.105.

Having established that \(A \to B\) identifies local rings, we conclude by Lemma 1.128 that it admits a retraction.

With Lemma 1.130, it suffices to reduce to the case that \(B\) is furthur w-local and the closed points of \(\operatorname{Spec}(B)\) are exactly \(V(IB)\).

Let \(A \to B\) be faithfully flat and ind-étale. We may replace \(B\) by the ring \(C = B_{w, IB}\) (see Definition 1.98). To justify this, note that the map \(A \to C\) is faithfully flat by Lemma 1.103 and \(B \to C\) is ind-Zariski by Lemma 1.100, thus ind-étale by Lemma 1.54. Therefore, by Lemma 1.51, the composition \( A \to C \) is also ind-étale. Hence, we may assume \(\operatorname{Spec}(B)\) is w-local such that the set of closed points of \(\operatorname{Spec}(B)\) is \(V(IB)\) by ?? in Lemma 1.102.

Choose an extremally disconnected space \(T\) and a surjective continuous map \(T \to \pi _0(\operatorname {Spec}(C))\) by Proposition 1.5. (\(T\) can be chosen as \(\beta ((\pi _0(\operatorname {Spec}(C)))^{\textnormal{disc}})\).) Note that \(T\) is profinite. By Lemma 1.127, Definition 1.126 gives an ind-Zariski ring map \(C \to D\) such that \(\pi _0(\operatorname{Spec}(D)) \to \pi _0(\operatorname{Spec}(C))\) realizes \(T \to \pi _0(\operatorname{Spec}(C))\) and such that

is cartesian in the category of topological spaces.

Following Lemma 1.70, we note that \(\operatorname{Spec}(D)\) is w-local, that \(\operatorname{Spec}(D) \to \operatorname{Spec}(C)\) is w-local, and that the set of closed points of \(\operatorname{Spec}(D)\) is the inverse image of the set of closed points of \(\operatorname{Spec}(C)\).

Thus it is still true that the local rings of \(D\) at its maximal ideals are strictly henselian (as they are isomorphic to the local rings at the corresponding maximal ideals of \(C\) by Lemma 1.38). Hence \(D\) is w-contractible.

This is a combination of Proposition 1.118 and Lemma 1.132.

This follows from Theorem 1.133 and Proposition 1.131.

Let \(S\) be an étale \(R\)-algebra. Then \(S\) is \(R\)-flat. Also \(S \otimes _{R} S \cong S × T\) for some ring \(T\), in particular \(\mathrm{Spec}(S \otimes _{R} S) \to \mathrm{Spec}(S)\) is an open immersion, hence flat.