The Stacks project

Proof. The equivalence of (1) and (2) follows immediately from Proposition 61.9.1.

Assume (3)(a), (3)(b), and (3)(c). Let $A \to B$ be faithfully flat and ind-étale. 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 $V(IB)$. In this case $A \to B$ identifies local rings by condition (3)(c) as it suffices to check this at maximal ideals of $B$ which lie over maximal ideals of $A$. Thus $A \to B$ has a retraction by Lemma 61.6.7.

Assume (1) or equivalently (2). We have (3)(c) by Lemma 61.8.5. Properties (3)(a) and (3)(b) follow from Lemma 61.6.7. $\square$

Proof. Applying Lemma 61.8.7 to $\text{id}_ A : A \to A$ we find a faithfully flat, ind-étale ring map $A \to C$ such that $C$ is w-local and such that every local ring at a maximal ideal of $C$ is strictly henselian. Choose an extremally disconnected space $T$ and a surjective continuous map $T \to \pi _0(\mathop{\mathrm{Spec}}(C))$, see Topology, Lemma 5.26.9. Note that $T$ is profinite. Apply Lemma 61.6.2 to find an ind-Zariski ring map $C \to D$ such that $\pi _0(\mathop{\mathrm{Spec}}(D)) \to \pi _0(\mathop{\mathrm{Spec}}(C))$ realizes $T \to \pi _0(\mathop{\mathrm{Spec}}(C))$ and such that

is cartesian in the category of topological spaces. Note that $\mathop{\mathrm{Spec}}(D)$ is w-local, that $\mathop{\mathrm{Spec}}(D) \to \mathop{\mathrm{Spec}}(C)$ is w-local, and that the set of closed points of $\mathop{\mathrm{Spec}}(D)$ is the inverse image of the set of closed points of $\mathop{\mathrm{Spec}}(C)$, see Lemma 61.2.5. 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 $C$). It follows from Lemma 61.11.2 that $D$ is w-contractible. $\square$

Proof. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. Choose a finite partition $X = \coprod X_ i$ and $X'_ i \to X_ i$ as in Étale Cohomology, Lemma 59.72.3. The map of topological spaces $\coprod X_ i \to X$ (where the source is the disjoint union in the category of topological spaces) has a section by Topology, Proposition 5.26.6. Hence we see that $X$ is topologically the disjoint union of the strata $X_ i$. Thus we may replace $X$ by the $X_ i$ and assume there exists a surjective finite locally free morphism $X' \to X$ such that $(X' \times _ X Y)_{red}$ is isomorphic to a finite disjoint union of copies of $X'_{red}$. Picture

The assumption on the residue fields of $A$ implies that this diagram is a fibre product diagram on underlying sets of points (details omitted). Since $X$ is extremally disconnected and $X'$ is Hausdorff (Lemma 61.5.6), the continuous map $X' \to X$ has a continuous section $\sigma $. Then $\coprod _{i = 1, \ldots , r} \sigma (X) \to Y$ is a bijective continuous map. By Topology, Lemma 5.17.8 we see that it is a homeomorphism and the proof is done. $\square$

Proof. We will use the criterion of Lemma 61.11.2. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$ and denote $f : Y \to X$ the induced morphism. As $f : Y \to X$ is a finite morphism, we see that the set of closed points $Y_0$ of $Y$ is the inverse image of the set of closed points $X_0$ of $X$. Let $y \in Y$ with image $x \in X$. Then $x$ specializes to a unique closed point $x_0 \in X$. Say $f^{-1}(\{ x_0\} ) = \{ y_1, \ldots , y_ n\} $ with $y_ i$ closed in $Y$. Since $R = \mathcal{O}_{X, x_0}$ is strictly henselian and since $f$ is finite, we see that $Y \times _{f, X} \mathop{\mathrm{Spec}}(R)$ is equal to $\coprod _{i = 1, \ldots , n} \mathop{\mathrm{Spec}}(R_ i)$ where each $R_ i$ is a local ring finite over $R$ whose maximal ideal corresponds to $y_ i$, see Algebra, Lemma 10.153.3 part (10). Then $y$ is a point of exactly one of these $\mathop{\mathrm{Spec}}(R_ i)$ and we see that $y$ specializes to exactly one of the $y_ i$. In other words, every point of $Y$ specializes to a unique point of $Y_0$. Thus $Y$ is w-local. For every $y \in Y_0$ with image $x \in X_0$ we see that $\mathcal{O}_{Y, y}$ is strictly henselian by Algebra, Lemma 10.153.4 applied to $\mathcal{O}_{X, x} \to B \otimes _ A \mathcal{O}_{X, x}$. It remains to show that $Y_0$ is extremally disconnected. To do this we look at $X_0 \times _ X Y \to X_0$ where $X_0 \subset X$ is the reduced induced scheme structure. Note that the underlying topological space of $X_0 \times _ X Y$ agrees with $Y_0$. Now the desired result follows from Lemma 61.11.5. $\square$

Proof. Let $A_ Z^\sim \to B$ be a weakly étale faithfully flat ring map. Consider the ring map

this is faithful flat and weakly étale. If $A$ is w-contractible, then there is a retraction $\sigma $. Consider the morphism

Every point of $Z \subset \mathop{\mathrm{Spec}}(A_ Z^\sim )$ maps into the component $\mathop{\mathrm{Spec}}(B)$. Since every point of $\mathop{\mathrm{Spec}}(A_ Z^\sim )$ specializes to a point of $Z$ we find a morphism $\mathop{\mathrm{Spec}}(A_ Z^\sim ) \to \mathop{\mathrm{Spec}}(B)$ as desired. $\square$