The Stacks project

Proof. Let $S \subset A$ be the multiplicative set of elements which map to invertible elements of $\Gamma (Z, \mathcal{O}_ Z) = (A/I)_ f$. If $\mathfrak p$ is a prime of $A$ which does not specialize to $Z$, then $\mathfrak p$ generates the unit ideal in $(A/I)_ f$. Hence we can write $f^ n = g + h$ for some $n \geq 0$, $g \in \mathfrak p$, $h \in I$. Then $g \in S$ and we see that $\mathfrak p$ is not in the spectrum of $S^{-1}A$. Conversely, if $\mathfrak p$ does specialize to $Z$, say $\mathfrak p \subset \mathfrak q \supset I$ with $f \not\in \mathfrak q$, then we see that $S^{-1}A$ maps to $A_\mathfrak q$ and hence $\mathfrak p$ is in the spectrum of $S^{-1}A$. This proves (1).

The isomorphism class of the localization $S^{-1}A$ depends only on the corresponding subset $\mathop{\mathrm{Spec}}(S^{-1}A) \subset \mathop{\mathrm{Spec}}(A)$, whence (2) holds. By construction $S^{-1}A$ maps surjectively onto $(A/I)_ f$, hence (3). The final statement follows as the multiplicative subset $S' \subset A'$ corresponding to $Z'$ contains the image of the multiplicative subset $S$. $\square$

Proof. We may write $T_ i = \bigcup _ j U_{i, j} \cap V_{i, j}^ c$ as a finite union for some $U_{i, j}$ and $V_{i, j}$ quasi-compact open in $X$. Then we may write $U_{i, j} = \bigcup D(f_{i, j, k})$ and $V_{i, j} = \bigcup D(g_{i, j, l})$. Then we set $E = \{ f_{i, j, k}\} \cup \{ g_{i, j, l}\} $. This does the job, because the stratification (61.5.1.2) is the one whose strata are labeled by the vanishing pattern of the elements of $E$ which clearly refines the given stratification. $\square$

Proof. The map $A \to A_ w$ is ind-Zariski by construction. For every $E$ the morphism $Z_ E \to X$ is a bijection, hence (2). As $Z \subset X_ w$ we conclude $X_ w \to X$ is surjective and $A \to A_ w$ is faithfully flat by Algebra, Lemma 10.39.16. This proves (1).

Suppose that $y \in X_ w$, $y \not\in Z$. Then there exists an $E$ such that the image of $y$ in $X_ E$ is not contained in $Z_ E$. Then for all $E \subset E'$ also $y$ maps to an element of $X_{E'}$ not contained in $Z_{E'}$. Let $T_{E'} \subset X_{E'}$ be the reduced closed subscheme which is the closure of the image of $y$. It is clear that $T = \mathop{\mathrm{lim}}\nolimits _{E \subset E'} T_{E'}$ is the closure of $y$ in $X_ w$. For every $E \subset E'$ the scheme $T_{E'} \cap Z_{E'}$ is nonempty by construction of $X_{E'}$. Hence $\mathop{\mathrm{lim}}\nolimits T_{E'} \cap Z_{E'}$ is nonempty and we conclude that $T \cap Z$ is nonempty. Thus $y$ is not a closed point. It follows that every closed point of $X_ w$ is in $Z$.

Suppose that $y \in X_ w$ specializes to $z, z' \in Z$. We will show that $z = z'$ which will finish the proof of (3) and will imply (5). Let $x, x' \in X$ be the images of $z$ and $z'$. Since $Z \to X$ is bijective it suffices to show that $x = x'$. If $x \not= x'$, then there exists an $f \in A$ such that $x \in D(f)$ and $x' \in V(f)$ (or vice versa). Set $E = \{ f\} $ so that

Then we see that $z$ and $z'$ map $x_ E$ and $x'_ E$ which are in different parts of the given decomposition of $X_ E$ above. But then it impossible for $x_ E$ and $x'_ E$ to be specializations of a common point. This is the desired contradiction.

Recall that given a finite subset $E \subset A$ we have $Z_ E$ is a disjoint union of the locally closed subschemes $Z(E', E'')$ each isomorphic to the spectrum of $(A/I)_ f$ where $I$ is the ideal generated by $E''$ and $f$ the product of the elements of $E'$. Any nilpotent element $b$ of $(A/I)_ f$ is the class of $g/f^ n$ for some $g \in A$. Then setting $E' = E \cup \{ g\} $ the reader verifies that $b$ is pulls back to zero under the transition map $Z_{E'} \to Z_ E$ of the system. This proves (4). $\square$

Proof. Denote $Y = \mathop{\mathrm{Spec}}(B)$ and $Y_0 \subset Y$ the set of closed points. Denote $f : Y \to X$ the given morphism. Recall that $Y_0$ is profinite, in particular every constructible subset of $Y_0$ is open and closed. Let $E \subset A$ be a finite subset. Recall that $A_ w = \mathop{\mathrm{colim}}\nolimits A_ E$ and that the set of closed points of $\mathop{\mathrm{Spec}}(A_ w)$ is the limit of the closed subsets $Z_ E \subset X_ E = \mathop{\mathrm{Spec}}(A_ E)$. Thus it suffices to show there is a unique factorization $A \to A_ E \to B$ such that $Y \to X_ E$ maps $Y_0$ into $Z_ E$. Since $Z_ E \to X = \mathop{\mathrm{Spec}}(A)$ is bijective, and since the strata $Z(E', E'')$ are constructible we see that

is a disjoint union decomposition into open and closed subsets. As $Y_0 = \pi _0(Y)$ we obtain a corresponding decomposition of $Y$ into open and closed pieces. Thus it suffices to construct the factorization in case $f(Y_0) \subset Z(E', E'')$ for some decomposition $E = E' \amalg E''$. In this case $f(Y)$ is contained in the set of points of $X$ specializing to $Z(E', E'')$ which is homeomorphic to $X_{E', E''}$. Thus we obtain a unique continuous map $Y \to X_{E', E''}$ over $X$. By Lemma 61.3.7 this corresponds to a unique morphism of schemes $Y \to X_{E', E''}$ over $X$. This finishes the proof. $\square$

Proof. By the references mentioned above (Algebra, Lemma 10.26.5 or Topology, Lemma 5.23.8) there are no specializations between distinct points of $\mathop{\mathrm{Spec}}(A)$ and $\mathop{\mathrm{Spec}}(B)$ is profinite if and only if there are no specializations between distinct points of $\mathop{\mathrm{Spec}}(B)$. These specializations can only happen in the fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. In this way we see that (1) is true.

The assumption in (2) implies all primes of $B$ are maximal by Algebra, Lemma 10.35.9. Thus (2) holds. If $A \to B$ is a local isomorphism or identifies local rings, then the residue field extensions are trivial, so (3) and (4) follow from (2). If $A \to B$ is weakly étale, then More on Algebra, Lemma 15.104.17 tells us it induces separable algebraic residue field extensions, so (5) follows from (2). If $A \to B$ is quasi-finite, then the fibres are finite discrete topological spaces. Hence (6) follows from (1). Hence (3) follows from (1). Cases (7) and (8) follow from this as unramified and étale ring map are quasi-finite (Algebra, Lemmas 10.151.6 and 10.143.6). If $B = \mathop{\mathrm{colim}}\nolimits B_ i$ is a filtered colimit of $A$-algebras, then $\mathop{\mathrm{Spec}}(B) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Spec}}(B_ i)$ in the category of topological spaces by Limits, Lemma 32.4.2. Hence if each $\mathop{\mathrm{Spec}}(B_ i)$ is profinite, so is $\mathop{\mathrm{Spec}}(B)$ by Topology, Lemma 5.22.3. This proves (9). $\square$

Proof. Let $A \to A_ w$ and $Z \subset Y = \mathop{\mathrm{Spec}}(A_ w)$ as in Lemma 61.5.3. Let $T \subset Z$ be the inverse image of $V(I)$. Then $T \to V(I)$ is a homeomorphism by Topology, Lemma 5.17.8. Let $B = (A_ w)_ T^\sim $, see Lemma 61.5.1. It is clear that $B$ is w-local with closed points $V(IB)$. The ring map $A/I \to B/IB$ is ind-Zariski and induces a homeomorphism on underlying topological spaces. Hence it is an isomorphism by Lemma 61.3.8. $\square$

Proof. By Lemma 61.5.6 applied to $A/I \to B/IB$ all points of $Z = V(IB) = \mathop{\mathrm{Spec}}(B/IB)$ are closed, in fact $\mathop{\mathrm{Spec}}(B/IB)$ is a profinite space. To finish the proof we apply Lemma 61.5.7 to $IB \subset B$. $\square$