Lemma 61.5.3. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme. With $A \to A_ w$, $X_ w = \mathop{\mathrm{Spec}}(A_ w)$, and $Z \subset X_ w$ as above.

1. $A \to A_ w$ is ind-Zariski and faithfully flat,

2. $X_ w \to X$ induces a bijection $Z \to X$,

3. $Z$ is the set of closed points of $X_ w$,

4. $Z$ is a reduced scheme, and

5. every point of $X_ w$ specializes to a unique point of $Z$.

In particular, $X_ w$ is w-local (Definition 61.2.3).

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

$X_ E = \mathop{\mathrm{Spec}}(A_ f) \amalg \mathop{\mathrm{Spec}}(A_{V(f)}^\sim )$

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.

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). $\square$

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