Lemma 60.5.3. 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 60.2.3).

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). $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0975. Beware of the difference between the letter 'O' and the digit '0'.