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proetale.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Pro-\'etale Cohomology}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
The material in this chapter and more can be found in the preprint \cite{BS}.
\medskip\noindent
The goal of this chapter is to introduce the pro-\'etale topology and
show how it simplifies the introduction of $\ell$-adic cohomology in
algebraic geometry.
\medskip\noindent
A brief overview of the history of this material as we have understood it.
In \cite[Expos\'es V and VI]{SGA5} Grothendieck et al developed a theory for
dealing with $\ell$-adic sheaves as inverse systems of sheaves of
$\mathbf{Z}/\ell^n\mathbf{Z}$-modules.
In his second paper on the Weil conjectures (\cite{WeilII}) Deligne introduced
a derived category of $\ell$-adic sheaves as a certain 2-limit of categories
of complexes of sheaves of $\mathbf{Z}/\ell^n\mathbf{Z}$-modules on the
\'etale site of a scheme $X$. This approach is used in the paper by
Beilinson, Bernstein, and Deligne (\cite{BBD}) as the basis for their
beautiful theory of perverse sheaves. In a paper entitled ``Continuous
\'Etale Cohomology'' (\cite{Jannsen}) Uwe Jannsen discusses an important
variant of the cohomology of a $\ell$-adic sheaf on a variety over a field.
His paper is followed up by a paper of Torsten Ekedahl (\cite{Ekedahl})
who discusses the adic formalism needed to work comfortably with derived
categories defined as limits.
\medskip\noindent
The goal of this chapter is to show that, if we work with the pro-\'etale
site of a scheme, then one can avoid some of the technicalities these
authors encountered. This comes at the expense of having to work with
non-Noetherian schemes, even when one is only interested in working with
$\ell$-adic sheaves and cohomology of such on varieties over an
algebraically closed field.
\section{Some topology}
\label{section-topology}
\noindent
Some preliminaries. We have defined {\it spectral spaces} and
{\it spectral maps} of spectral spaces in
Topology, Section \ref{topology-section-spectral}.
The spectrum of a ring is a spectral space, see
Algebra, Lemma \ref{algebra-lemma-spec-spectral}.
\begin{lemma}
\label{lemma-spectral-split}
Let $X$ be a spectral space. Let $X_0 \subset X$ be the set of closed points.
The following are equivalent
\begin{enumerate}
\item Every open covering of $X$ can be refined by a finite
disjoint union decomposition $X = \coprod U_i$ with $U_i$
open and closed in $X$.
\item The composition $X_0 \to X \to \pi_0(X)$ is bijective.
\end{enumerate}
Moreover, if $X_0$ is closed in $X$ and every point of $X$ specializes
to a unique point of $X_0$, then these conditions are satisfied.
\end{lemma}
\begin{proof}
We will use without further mention that
$X_0$ is quasi-compact
(Topology, Lemma \ref{topology-lemma-closed-points-quasi-compact})
and $\pi_0(X)$ is profinite
(Topology, Lemma \ref{topology-lemma-spectral-pi0}).
Picture
$$
\xymatrix{
X_0 \ar[rd]_f \ar[r] & X \ar[d]^\pi \\
& \pi_0(X)
}
$$
If (2) holds, the continuous bijective map $f : X_0 \to \pi_0(X)$ is
a homeomorphism by
Topology, Lemma \ref{topology-lemma-bijective-map}.
Given an open covering $X = \bigcup U_i$, we get an open covering
$\pi_0(X) = \bigcup f(X_0 \cap U_i)$. By
Topology, Lemma \ref{topology-lemma-profinite-refine-open-covering}
we can find a finite open covering of the form $\pi_0(X) = \coprod V_j$
which refines this covering.
Since $X_0 \to \pi_0(X)$ is bijective each connected component of
$X$ has a unique closed point, whence is equal to the set of points
specializing to this closed point. Hence $\pi^{-1}(V_j)$ is the
set of points specializing to the points of $f^{-1}(V_j)$.
Now, if $f^{-1}(V_j) \subset X_0 \cap U_i \subset U_i$, then
it follows that $\pi^{-1}(V_j) \subset U_i$ (because the open set
$U_i$ is closed under generalizations). In this way we see
that the open covering $X = \coprod \pi^{-1}(V_j)$ refines
the covering we started out with. In this way we see that
(2) implies (1).
\medskip\noindent
Assume (1). Let $x, y \in X$ be closed points. Then we have the open covering
$X = (X \setminus \{x\}) \cup (X \setminus \{y\})$.
It follows from (1) that there exists a disjoint union decomposition
$X = U \amalg V$ with $U$ and $V$ open (and closed) and $x \in U$ and
$y \in V$. In particular we see that every connected component of $X$
has at most one closed point. By
Topology, Lemma \ref{topology-lemma-quasi-compact-closed-point}
every connected component (being closed) also does have a closed point.
Thus $X_0 \to \pi_0(X)$ is bijective. In this way we see that (1) implies (2).
\medskip\noindent
Assume $X_0$ is closed in $X$ and every point specializes to a unique
point of $X_0$. Then $X_0$ is a spectral space
(Topology, Lemma \ref{topology-lemma-spectral-sub})
consisting of closed points, hence profinite
(Topology, Lemma \ref{topology-lemma-characterize-profinite-spectral}).
Let $x, y \in X_0$ be distinct. By
Topology, Lemma \ref{topology-lemma-profinite-refine-open-covering}
we can find a disjoint union decomposition
$X_0 = U_0 \amalg V_0$ with $U_0$ and $V_0$ open and closed.
Let $\{U_i\}$ be the set of quasi-compact open subsets of $X$
such that $U_0 = X_0 \cap U_i$. Similarly, let $\{V_j\}$ be the
set of quasi-compact open subsets of $X$ such that
and $V_0 = X_0 \cap V_j$. If $U_i \cap V_j$ is nonempty for all $i, j$,
then there exists a point $\xi$ contained in all of them
(use the $U_i \cap V_j$ is constructible, hence closed in the
constructible topology, and use
Topology, Lemmas \ref{topology-lemma-constructible-hausdorff-quasi-compact} and
\ref{topology-lemma-intersection-closed-in-quasi-compact}).
However, since $X$ is sober and $V_0$ is closed in $X$,
the intersection $\bigcap U_i$ is the set of points specializing to $U_0$.
Similarly, $\bigcap V_j$ is the set of points specializing to $V_0$.
Since $U_0 \cap V_0$ is empty this is a contradiction.
Thus we find disjoint quasi-compact opens $U, V \subset X$
such that $U_0 = X_0 \cap U$ and $V_0 = X_0 \cap V$.
Observe that $X = U \cup V = U \amalg V$ as
$X_0 \subset U \cup V$ (use
Topology, Lemma \ref{topology-lemma-quasi-compact-closed-point}).
This proves that $x, y$ are not in the same connected component of $X$.
In other words, $X_0 \to \pi_0(X)$ is injective. The map is also
surjective by
Topology, Lemma \ref{topology-lemma-quasi-compact-closed-point}
and the fact that connected components are closed.
In this way we see that the final condition implies (1).
\end{proof}
\begin{example}
\label{example-not-w-local}
Let $T$ be a profinite space. Let $t \in T$ be a point and assume
that $T \setminus \{t\}$ is not quasi-compact.
Let $X = T \times \{0, 1\}$. Consider the topology on $X$
with a subbase given by the sets
$U \times \{0, 1\}$ for $U \subset T$ open, $X \setminus \{(t, 0)\}$,
and $U \times \{1\}$ for $U \subset T$ open with $t \not \in U$.
The set of closed points of $X$ is $X_0 = T \times \{0\}$ and
$(t, 1)$ is in the closure of $X_0$.
Moreover, $X_0 \to \pi_0(X)$ is a bijection.
This example shows that conditions (1) and (2) of
Lemma \ref{lemma-spectral-split} do no imply the set of closed points
is closed.
\end{example}
\noindent
It turns out it is more convenient to work with spectral
spaces which have the slightly stronger property mentioned in
the final statement of Lemma \ref{lemma-spectral-split}.
We give this property a name.
\begin{definition}
\label{definition-w-local}
A spectral space $X$ is {\it w-local} if the set of closed points $X_0$
is closed and every point of $X$ specializes to a unique closed point.
A continuous map $f : X \to Y$ of w-local spaces is {\it w-local}
if it is spectral and maps any closed point of $X$ to a closed point of $Y$.
\end{definition}
\noindent
We have seen in the proof of Lemma \ref{lemma-spectral-split}
that in this case $X_0 \to \pi_0(X)$ is a homeomorphism and that
$X_0 \cong \pi_0(X)$ is a profinite space. Moreover, a connected
component of $X$ is exactly the set of points specializing to
a given $x \in X_0$.
\begin{lemma}
\label{lemma-closed-subspace-w-local}
Let $X$ be a w-local spectral space. If $Y \subset X$ is closed,
then $Y$ is w-local.
\end{lemma}
\begin{proof}
The subset $Y_0 \subset Y$ of closed points is closed because
$Y_0 = X_0 \cap Y$. Since $X$ is $w$-local, every $y \in Y$ specializes
to a unique point of $X_0$. This specialization is in $Y$, and hence
also in $Y_0$, because $\overline{\{y\}}\subset Y$. In conclusion, $Y$
is $w$-local.
\end{proof}
\begin{lemma}
\label{lemma-silly}
Let $X$ be a spectral space. Let
$$
\xymatrix{
Y \ar[r] \ar[d] & T \ar[d] \\
X \ar[r] & \pi_0(X)
}
$$
be a cartesian diagram in the category of topological spaces
with $T$ profinite. Then $Y$ is spectral and $T = \pi_0(Y)$.
If moreover $X$ is w-local, then $Y$ is w-local, $Y \to X$ is w-local,
and the set of closed points of $Y$ is the inverse image of the
set of closed points of $X$.
\end{lemma}
\begin{proof}
Note that $Y$ is a closed subspace of $X \times T$ as $\pi_0(X)$
is a profinite space hence Hausdorff
(use Topology, Lemmas \ref{topology-lemma-spectral-pi0} and
\ref{topology-lemma-fibre-product-closed}).
Since $X \times T$ is spectral
(Topology, Lemma \ref{topology-lemma-product-spectral-spaces})
it follows that $Y$ is spectral
(Topology, Lemma \ref{topology-lemma-spectral-sub}).
Let $Y \to \pi_0(Y) \to T$ be the canonical factorization
(Topology, Lemma \ref{topology-lemma-space-connected-components}).
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. We conclude that $\pi_0(Y) \to T$
is a homeomorphism by
Topology, Lemma \ref{topology-lemma-bijective-map}.
\medskip\noindent
Next, assume that $X$ is w-local and let $X_0 \subset X$ be the
set of closed points. The inverse image $Y_0 \subset Y$ of $X_0$ in
$Y$ maps bijectively onto $T$ as $X_0 \to \pi_0(X)$ is a bijection
by Lemma \ref{lemma-spectral-split}. Moreover, $Y_0$ is quasi-compact
as a closed subset of the spectral space $Y$. Hence
$Y_0 \to \pi_0(Y) = T$ is a homeomorphism by
Topology, Lemma \ref{topology-lemma-bijective-map}.
It follows that all points of $Y_0$ are closed in $Y$.
Conversely, if $y \in Y$ is a closed point, then
it is closed in the fibre of $Y \to \pi_0(Y) = T$
and hence its image $x$ in $X$ is closed in the (homeomorphic) fibre of
$X \to \pi_0(X)$. This implies $x \in X_0$ and hence $y \in Y_0$.
Thus $Y_0$ is the collection of closed points of $Y$
and for each $y \in Y_0$ the set of generalizations of $y$ is
the fibre of $Y \to \pi_0(Y)$. The lemma follows.
\end{proof}
\section{Local isomorphisms}
\label{section-local-isomorphism}
\noindent
We start with a definition.
\begin{definition}
\label{definition-local-isomorphism}
Let $\varphi : A \to B$ be a ring map.
\begin{enumerate}
\item We say $A \to B$ is a {\it local isomorphism} if for every prime
$\mathfrak q \subset B$ there exists a $g \in B$, $g \not \in \mathfrak q$
such that $A \to B_g$ induces an open immersion $\Spec(B_g) \to \Spec(A)$.
\item We say $A \to B$ {\it identifies local rings} if for every prime
$\mathfrak q \subset B$ the canonical map
$A_{\varphi^{-1}(\mathfrak q)} \to B_\mathfrak q$ is an isomorphism.
\end{enumerate}
\end{definition}
\noindent
We list some elementary properties.
\begin{lemma}
\label{lemma-base-change-local-isomorphism}
Let $A \to B$ and $A \to A'$ be ring maps. Let $B' = B \otimes_A A'$
be the base change of $B$.
\begin{enumerate}
\item If $A \to B$ is a local isomorphism, then $A' \to B'$ is a
local isomorphism.
\item If $A \to B$ identifies local rings, then $A' \to B'$
identifies local rings.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-composition-local-isomorphism}
Let $A \to B$ and $B \to C$ be ring maps.
\begin{enumerate}
\item If $A \to B$ and $B \to C$ are local isomorphisms, then $A \to C$
is a local isomorphism.
\item If $A \to B$ and $B \to C$ identify local rings, then $A \to C$
identifies local rings.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-local-isomorphism-permanence}
Let $A$ be a ring. Let $B \to C$ be an $A$-algebra homomorphism.
\begin{enumerate}
\item If $A \to B$ and $A \to C$ are local isomorphisms, then $B \to C$
is a local isomorphism.
\item If $A \to B$ and $A \to C$ identify local rings, then $B \to C$
identifies local rings.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-local-isomorphism-implies}
Let $A \to B$ be a local isomorphism. Then
\begin{enumerate}
\item $A \to B$ is \'etale,
\item $A \to B$ identifies local rings,
\item $A \to B$ is quasi-finite.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-structure-local-isomorphism}
Let $A \to B$ be a local isomorphism. Then there exist $n \geq 0$,
$g_1, \ldots, g_n \in B$, $f_1, \ldots, f_n \in A$ such that
$(g_1, \ldots, g_n) = B$ and $A_{f_i} \cong B_{g_i}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-fully-faithful-spaces-over-X}
Let $p : (Y, \mathcal{O}_Y) \to (X, \mathcal{O}_X)$ and
$q : (Z, \mathcal{O}_Z) \to (X, \mathcal{O}_X)$
be morphisms of locally ringed spaces.
If $\mathcal{O}_Y = p^{-1}\mathcal{O}_X$, then
$$
\Mor_{\text{LRS}/(X, \mathcal{O}_X)}((Z, \mathcal{O}_Z), (Y, \mathcal{O}_Y))
\longrightarrow
\Mor_{\textit{Top}/X}(Z, Y),\quad
(f, f^\sharp) \longmapsto f
$$
is bijective. Here $\text{LRS}/(X, \mathcal{O}_X)$ is the category of
locally ringed spaces over $X$ and $\textit{Top}/X$ is the category
of topological spaces over $X$.
\end{lemma}
\begin{proof}
This is immediate from the definitions.
\end{proof}
\begin{lemma}
\label{lemma-local-isomorphism-fully-faithful}
Let $A$ be a ring. Set $X = \Spec(A)$. The functor
$$
B \longmapsto \Spec(B)
$$
from the category of $A$-algebras $B$ such that $A \to B$ identifies
local rings to the category of
topological spaces over $X$ is fully faithful.
\end{lemma}
\begin{proof}
This follows from Lemma \ref{lemma-fully-faithful-spaces-over-X}
and the fact that if $A \to B$ identifies local rings, then the pullback
of the structure sheaf of $\Spec(A)$ via $p : \Spec(B) \to \Spec(A)$
is equal to the structure sheaf of $\Spec(B)$.
\end{proof}
\section{Ind-Zariski algebra}
\label{section-ind-zariski}
\noindent
We start with a definition; please see Remark \ref{remark-slightly-stronger}
for a comparison with the corresponding definition of the article \cite{BS}.
\begin{definition}
\label{definition-ind-zariski}
A ring map $A \to B$ is said to be {\it ind-Zariski} if $B$ can be written
as a filtered colimit $B = \colim B_i$ with each $A \to B_i$ a local
isomorphism.
\end{definition}
\noindent
An example of an Ind-Zariski map is a localization $A \to S^{-1}A$, see
Algebra, Lemma \ref{algebra-lemma-localization-colimit}.
The category of ind-Zariski algebras is closed under several natural
operations.
\begin{lemma}
\label{lemma-base-change-ind-zariski}
Let $A \to B$ and $A \to A'$ be ring maps. Let $B' = B \otimes_A A'$
be the base change of $B$.
If $A \to B$ is ind-Zariski, then $A' \to B'$ is ind-Zariski.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-composition-ind-zariski}
Let $A \to B$ and $B \to C$ be ring maps. If $A \to B$ and $B \to C$
are ind-Zariski, then $A \to C$ is ind-Zariski.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-ind-zariski-permanence}
Let $A$ be a ring. Let $B \to C$ be an $A$-algebra homomorphism.
If $A \to B$ and $A \to C$ are ind-Zariski, then $B \to C$
is ind-Zariski.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-ind-ind-zariski}
A filtered colimit of ind-Zariski $A$-algebras is ind-Zariski over $A$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-ind-zariski-implies}
Let $A \to B$ be ind-Zariski. Then $A \to B$ identifies local rings,
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Constructing w-local affine schemes}
\label{section-construction}
\noindent
An affine scheme $X$ is called {\it w-local} if its underlying
topological space is w-local (Definition \ref{definition-w-local}).
It turns out given any ring $A$ there is a canonical faithfully
flat ind-Zariski ring map $A \to A_w$ such that $\Spec(A_w)$ is
w-local. The key to constructing $A_w$ is the following simple lemma.
\begin{lemma}
\label{lemma-localization}
Let $A$ be a ring. Set $X = \Spec(A)$. Let $Z \subset X$ be a locally closed
subscheme which of the form $D(f) \cap V(I)$ for some $f \in A$ and
ideal $I \subset A$. Then
\begin{enumerate}
\item there exists a multiplicative subset $S \subset A$ such that
$\Spec(S^{-1}A)$ maps by a homeomorphism to the set of points of $X$
specializing to $Z$,
\item the $A$-algebra $A_Z^\sim = S^{-1}A$ depends only on
the underlying locally closed subset $Z \subset X$,
\item $Z$ is a closed subscheme of $\Spec(A_Z^\sim)$,
\end{enumerate}
If $A \to A'$ is a ring map and $Z' \subset X' = \Spec(A')$ is a
locally closed subscheme of the same form which maps into $Z$,
then there is a unique $A$-algebra map
$A_Z^\sim \to (A')_{Z'}^\sim$.
\end{lemma}
\begin{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).
\medskip\noindent
The isomorphism class of the localization $S^{-1}A$ depends only
on the corresponding subset $\Spec(S^{-1}A) \subset \Spec(A)$, 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 $S$.
\end{proof}
\noindent
Let $A$ be a ring. Let $E \subset A$ be a finite subset. We get a
stratification of $X = \Spec(A)$ into locally closed subschemes by
looking at the vanishing behaviour of the elements of $E$. More precisely,
given a disjoint union decomposition $E = E' \amalg E''$ we set
\begin{equation}
\label{equation-stratum}
Z(E', E'') =
\bigcap\nolimits_{f \in E'} D(f) \cap \bigcap\nolimits_{f \in E''} V(f) =
D(\prod\nolimits_{f \in E'} f) \cap V( \sum\nolimits_{f \in E''} fA)
\end{equation}
The points of $Z(E', E'')$ are exactly those $x \in X$ such that
$f \in E'$ maps to a nonzero element in $\kappa(x)$ and $f \in E''$
maps to zero in $\kappa(x)$. Thus it is clear that
\begin{equation}
\label{equation-stratify}
X = \coprod\nolimits_{E = E' \amalg E''} Z(E', E'')
\end{equation}
set theoretically. Observe that each stratum is constructible.
\begin{lemma}
\label{lemma-refine}
Let $X = \Spec(A)$ as above. Given any finite stratification
$X = \coprod T_i$ by constructible subsets, there exists a finite
subset $E \subset A$ such that the stratification (\ref{equation-stratify})
refines $X = \coprod T_i$.
\end{lemma}
\begin{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 (\ref{equation-stratify}) is the one whose strata are
labeled by the vanishing pattern of the elements of $E$ which
clearly refines the given stratification.
\end{proof}
\noindent
We continue the discussion.
Given a finite subset $E \subset A$ we set
\begin{equation}
\label{equation-ring}
A_E = \prod\nolimits_{E = E' \amalg E''} A_{Z(E', E'')}^\sim
\end{equation}
with notation as in Lemma \ref{lemma-localization}. This makes sense because
(\ref{equation-stratum}) shows that each $Z(E', E'')$ has the correct shape.
We take the spectrum of this ring and denote it
\begin{equation}
\label{equation-spectrum}
X_E = \Spec(A_E) = \coprod\nolimits_{E = E' \amalg E''} X_{E', E''}
\end{equation}
with $X_{E', E''} = \Spec(A_{Z(E', E'')}^\sim)$. Note that
\begin{equation}
\label{equation-closed}
Z_E = \coprod\nolimits_{E = E' \amalg E''} Z(E', E'')
\longrightarrow
X_E
\end{equation}
is a closed subscheme. By construction the closed subscheme $Z_E$
contains all the closed points of the affine scheme $X_E$ as every point
of $X_{E', E''}$ specializes to a point of $Z(E', E'')$.
\medskip\noindent
Let $I(A)$ be the partially ordered set of all finite subsets of $A$.
This is a directed partially ordered set. For $E_1 \subset E_2$ there
is a canonical transition map $A_{E_1} \to A_{E_2}$ of $A$-algebras.
Namely, given a decomposition $E_2 = E'_2 \amalg E''_2$ we set
$E'_1 = E_1 \cap E'_2$ and $E''_1 = E_1 \cap E''_2$. Then observe that
$Z(E'_1, E''_1) \subset Z(E'_2, E''_2)$ hence a unique $A$-algebra map
$A_{Z(E'_1, E''_1)}^\sim \to A_{Z(E'_2, E''_2)}^\sim$ by
Lemma \ref{lemma-localization}. Using these maps collectively we obtain
the desired ring map $A_{E_1} \to A_{E_2}$. Observe that the corresponding
map of affine schemes
\begin{equation}
\label{equation-transition}
X_{E_2} \longrightarrow X_{E_1}
\end{equation}
maps $Z_{E_2}$ into $Z_{E_1}$. By uniqueness we obtain a system of
$A$-algebras over $I(A)$ and we set
\begin{equation}
\label{equation-colimit-ring}
A_w = \colim_{E \in I(A)} A_E
\end{equation}
This $A$-algebra is ind-Zariski and faithfully flat over $A$.
Finally, we set $X_w = \Spec(A_w)$ and endow it with the closed subscheme
$Z = \lim_{E \in I(A)} Z_E$. In a formula
\begin{equation}
\label{equation-final}
X_w = \lim_{E \in I(A)} X_E \supset Z = \lim_{E \in I(A)} Z_E
\end{equation}
\begin{lemma}
\label{lemma-make-w-local}
Let $X = \Spec(A)$ be an affine scheme. With $A \to A_w$, $X_w = \Spec(A_w)$,
and $Z \subset X_w$ as above.
\begin{enumerate}
\item $A \to A_w$ is ind-Zariski and faithfully flat,
\item $X_w \to X$ induces a bijection $Z \to X$,
\item $Z$ is the set of closed points of $X_w$,
\item $Z$ is a reduced scheme, and
\item every point of $X_w$ specializes to a unique point of $Z$.
\end{enumerate}
In particular, $X_w$ is w-local (Definition \ref{definition-w-local}).
\end{lemma}
\begin{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 \ref{algebra-lemma-ff-rings}. This proves (1).
\medskip\noindent
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 $T = \lim_{E \subset E'} T_{E'}$ 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 $\lim T_{E'} \cap Z_{E'}$ 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 $Z$.
\medskip\noindent
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
$$
X_E = \Spec(A_f) \amalg \Spec(A_{V(f)}^\sim)
$$
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.
\medskip\noindent
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).
\end{proof}
\begin{remark}
\label{remark-size-w}
Let $A$ be a ring. Let $\kappa$ be an infinite cardinal bigger or
equal than the cardinality of $A$. Then the cardinality of $A_w$
(Lemma \ref{lemma-make-w-local})
is at most $\kappa$. Namely, each $A_E$ has cardinality at most
$\kappa$ and the set of finite subsets of $A$ has cardinality at most $\kappa$
as well. Thus the result follows as $\kappa \otimes \kappa = \kappa$, see
Sets, Section \ref{sets-section-cardinals}.
\end{remark}
\begin{lemma}[Universal property of the construction]
\label{lemma-universal}
Let $A$ be a ring. Let $A \to A_w$ be the ring map constructed in
Lemma \ref{lemma-make-w-local}. For any ring map $A \to B$ such that
$\Spec(B)$ is w-local, there is a unique factorization $A \to A_w \to B$
such that $\Spec(B) \to \Spec(A_w)$ is w-local.
\end{lemma}
\begin{proof}
Denote $Y = \Spec(B)$ 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 $A_w = \colim A_E$ and that the set of closed points of
$\Spec(A_w)$ is the limit of the closed subsets $Z_E \subset X_E = \Spec(A_E)$.
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 $Z_E \to X = \Spec(A)$ is bijective, and since the strata
$Z(E', E'')$ are constructible we see that
$$
Y_0 = \coprod f^{-1}(Z(E', E'')) \cap Y_0
$$
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 \ref{lemma-fully-faithful-spaces-over-X}
this corresponds to a unique morphism of schemes
$Y \to X_{E', E''}$ over $X$. This finishes the proof.
\end{proof}
\noindent
Recall that the spectrum of a ring is profinite if and only if
every point is closed. There are in fact a whole slew of equivalent
conditions that imply this. See
Algebra, Lemma \ref{algebra-lemma-ring-with-only-minimal-primes} or
Topology, Lemma \ref{topology-lemma-characterize-profinite-spectral}.
\begin{lemma}
\label{lemma-profinite-goes-up}
Let $A$ be a ring such that $\Spec(A)$ is profinite. Let $A \to B$ be a
ring map. Then $\Spec(B)$ is profinite in each of the following cases:
\begin{enumerate}
\item if $\mathfrak q,\mathfrak q' \subset B$ lie over the same
prime of $A$, then neither $\mathfrak q \subset \mathfrak q'$, nor
$\mathfrak q' \subset \mathfrak q$,
\item $A \to B$ induces algebraic extensions of residue fields,
\item $A \to B$ is a local isomorphism,
\item $A \to B$ identifies local rings,
\item $A \to B$ is weakly \'etale,
\item $A \to B$ is quasi-finite,
\item $A \to B$ is unramified,
\item $A \to B$ is \'etale,
\item $B$ is a filtered colimit of $A$-algebras as in (1) -- (8),
\item etc.
\end{enumerate}
\end{lemma}
\begin{proof}
By the references mentioned above
(Algebra, Lemma \ref{algebra-lemma-ring-with-only-minimal-primes} or
Topology, Lemma \ref{topology-lemma-characterize-profinite-spectral})
there are no specializations between distinct points of $\Spec(A)$ and
$\Spec(B)$ is profinite if and only if there are no specializations
between distinct points of $\Spec(B)$. These specializations can only
happen in the fibres of $\Spec(B) \to \Spec(A)$. In this way we see
that (1) is true.
\medskip\noindent
The assumption in (2) implies all primes of $B$ are maximal by
Algebra, Lemma \ref{algebra-lemma-finite-residue-extension-closed}.
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 \'etale, then More on Algebra, Lemma
\ref{more-algebra-lemma-weakly-etale-residue-field-extensions}
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 \'etale ring map are quasi-finite
(Algebra, Lemmas
\ref{algebra-lemma-unramified-quasi-finite} and
\ref{algebra-lemma-etale-quasi-finite}).
If $B = \colim B_i$ is a filtered colimit of $A$-algebras, then
$\Spec(B) = \colim \Spec(B_i)$, hence if each $\Spec(B_i)$
is profinite, so is $\Spec(B)$. This proves (9).
\end{proof}
\begin{lemma}
\label{lemma-localize-along-closed-profinite}
Let $A$ be a ring. Let $V(I) \subset \Spec(A)$ be a closed subset
which is a profinite topological space. Then there exists an
ind-Zariski ring map $A \to B$ such that $\Spec(B)$ is w-local,
the set of closed points is $V(IB)$, and $A/I \cong B/IB$.
\end{lemma}
\begin{proof}
Let $A \to A_w$ and $Z \subset Y = \Spec(A_w)$ as in
Lemma \ref{lemma-make-w-local}.
Let $T \subset Z$ be the inverse image of $V(I)$.
Then $T \to V(I)$ is a homeomorphism by
Topology, Lemma \ref{topology-lemma-bijective-map}.
Let $B = (A_w)_T^\sim$, see Lemma \ref{lemma-localization}.
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 \ref{lemma-local-isomorphism-fully-faithful}.
\end{proof}
\begin{lemma}
\label{lemma-w-local-algebraic-residue-field-extensions}
Let $A$ be a ring such that $X = \Spec(A)$ is w-local. Let $I \subset A$
be the radical ideal cutting out the set $X_0$ of closed points in $X$.
Let $A \to B$ be a ring map inducing algebraic extensions on residue
fields at primes. Then
\begin{enumerate}
\item every point of $Z = V(IB)$ is a closed point of $\Spec(B)$,
\item there exists an ind-Zariski ring map $B \to C$ such that
\begin{enumerate}
\item $B/IB \to C/IC$ is an isomorphism,
\item the space $Y = \Spec(C)$ is w-local,
\item the induced map $p : Y \to X$ is w-local, and
\item $p^{-1}(X_0)$ is the set of closed points of $Y$.
\end{enumerate}
\end{enumerate}
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-profinite-goes-up} applied to $A/I \to B/IB$
all points of $Z = V(IB) = \Spec(B/IB)$ are closed, in fact $\Spec(B/IB)$
is a profinite space.
To finish the proof we apply Lemma \ref{lemma-localize-along-closed-profinite}
to $IB \subset B$.
\end{proof}
\section{Identifying local rings versus ind-Zariski}
\label{section-connected-components}
\noindent
An ind-Zariski ring map $A \to B$ identifies local rings
(Lemma \ref{lemma-ind-zariski-implies}). The converse does not hold
(Examples, Section \ref{examples-section-not-ind-etale}).
However, it turns out that there is a kind of structure theorem for
ring maps which identify local rings in terms of ind-Zariski
ring maps, see Proposition \ref{proposition-maps-wich-identify-local-rings}.
\medskip\noindent
Let $A$ be a ring. Let $X = \Spec(A)$. The space of connected
components $\pi_0(X)$ is a profinite space by
Topology, Lemma \ref{topology-lemma-spectral-pi0}
(and Algebra, Lemma \ref{algebra-lemma-spec-spectral}).
\begin{lemma}
\label{lemma-construct}
Let $A$ be a ring. Let $X = \Spec(A)$. Let $T \subset \pi_0(X)$ be a
closed subset. There exists a surjective ind-Zariski ring map $A \to B$
such that $\Spec(B) \to \Spec(A)$ induces a homeomorphism of $\Spec(B)$
with the inverse image of $T$ in $X$.
\end{lemma}
\begin{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 \ref{topology-lemma-closed-union-connected-components}.
For each $\alpha$ we have $Z_\alpha = \Spec(A_\alpha)$ where
$A \to A_\alpha$ is a local isomorphism (a localization at an idempotent).
Setting $B = \colim A_\alpha$ proves the lemma.
\end{proof}
\begin{lemma}
\label{lemma-construct-profinite}
Let $A$ be a ring and let $X = \Spec(A)$. Let $T$ be a profinite space and
let $T \to \pi_0(X)$ be a continuous map. There exists an
ind-Zariski ring map $A \to B$ such that with $Y = \Spec(B)$ the diagram
$$
\xymatrix{
Y \ar[r] \ar[d] & \pi_0(Y) \ar[d] \\
X \ar[r] & \pi_0(X)
}
$$
is cartesian in the category of topological spaces and such that
$\pi_0(Y) = T$ as spaces over $\pi_0(X)$.
\end{lemma}
\begin{proof}
Namely, write $T = \lim T_i$ as the limit of an inverse system finite
discrete spaces over a directed partially ordered set (see
Topology, Lemma \ref{topology-lemma-profinite}). For each $i$ let
$Z_i = \Im(T \to \pi_0(X) \times T_i)$. 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 \ref{lemma-construct}
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 $\Spec(B_i) = X \times_{\pi_0(X)} Z_i$ 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 \ref{lemma-local-isomorphism-fully-faithful} and
\ref{lemma-ind-zariski-implies}).
Set $B = \colim B_i$. Because $T = \lim Z_i$ we have
$X \times_{\pi_0(X)} T = \lim X \times_{\pi_0(X)} Z_i$
and hence $Y = \Spec(B) = \lim \Spec(B_i)$
fits into the cartesian diagram
$$
\xymatrix{
Y \ar[r] \ar[d] & T \ar[d] \\
X \ar[r] & \pi_0(X)
}
$$
of topological spaces. By Lemma \ref{lemma-silly}
we conclude that $T = \pi_0(Y)$.
\end{proof}
\begin{example}
\label{example-construct-space}
Let $k$ be a field. Let $T$ be a profinite topological space.
There exists an ind-Zariski ring map $k \to A$ such that
$\Spec(A)$ is homeomorphic to $T$. Namely, just apply
Lemma \ref{lemma-construct-profinite} to $T \to \pi_0(\Spec(k)) = \{*\}$.
In fact, in this case we have
$$
A = \colim \text{Map}(T_i, k)
$$
whenever we write $T = \lim T_i$ as a filtered limit with each $T_i$ finite.
\end{example}
\begin{lemma}
\label{lemma-w-local-morphism-equal-points-stalks-is-iso}
Let $A \to B$ be ring map such that
\begin{enumerate}
\item $A \to B$ identifies local rings,
\item the topological spaces $\Spec(B)$, $\Spec(A)$ are w-local,
\item $\Spec(B) \to \Spec(A)$ is w-local, and
\item $\pi_0(\Spec(B)) \to \pi_0(\Spec(A))$ is bijective.
\end{enumerate}
Then $A \to B$ is an isomorphism
\end{lemma}
\begin{proof}
Let $X_0 \subset X = \Spec(A)$ and $Y_0 \subset Y = \Spec(B)$ 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 $\Spec(A)$ 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 \ref{algebra-lemma-flat-going-down}).
Thus Algebra, Lemma \ref{algebra-lemma-unique-prime-over-localize-below}
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 \ref{algebra-lemma-characterize-zero-local}.
\end{proof}
\begin{lemma}
\label{lemma-w-local-morphism-equal-stalks-is-ind-zariski}
Let $A \to B$ be ring map such that
\begin{enumerate}
\item $A \to B$ identifies local rings,
\item the topological spaces $\Spec(B)$, $\Spec(A)$ are w-local, and
\item $\Spec(B) \to \Spec(A)$ is w-local.
\end{enumerate}
Then $A \to B$ is ind-Zariski.
\end{lemma}
\begin{proof}
Set $X = \Spec(A)$ and $Y = \Spec(B)$. 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 $X' = \Spec(A')$ the diagram
$$
\xymatrix{
X' \ar[r] \ar[d] & \pi_0(X') \ar[d] \\
X \ar[r] & \pi_0(X)
}
$$
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 \ref{lemma-construct-profinite}. By
Lemma \ref{lemma-silly} we see that $X'$ is w-local and
the set of closed points $X'_0 \subset X'$ is the inverse image of $X_0$.
\medskip\noindent
We obtain a continuous map $Y \to X'$ of underlying topological spaces
over $X$ identifying $\pi_0(Y)$ with $\pi_0(X')$. By
Lemma \ref{lemma-local-isomorphism-fully-faithful}
(and Lemma \ref{lemma-ind-zariski-implies})
this is 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 \ref{lemma-w-local-morphism-equal-points-stalks-is-iso}
we see that $Y \cong X'$ and we win.
\end{proof}
\noindent
The following proposition is a warm up for the type of result
we will prove later.