Lemma 61.5.8. Let $A$ be a ring such that $X = \mathop{\mathrm{Spec}}(A)$ 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

1. every point of $Z = V(IB)$ is a closed point of $\mathop{\mathrm{Spec}}(B)$,

2. there exists an ind-Zariski ring map $B \to C$ such that

1. $B/IB \to C/IC$ is an isomorphism,

2. the space $Y = \mathop{\mathrm{Spec}}(C)$ is w-local,

3. the induced map $p : Y \to X$ is w-local, and

4. $p^{-1}(X_0)$ is the set of closed points of $Y$.

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 $IB \subset B$. $\square$

Comment #2532 by Brian Conrad on

In (2) you should say $B \rightarrow C$ can also be taken to be flat (used in paragraph 2 of the proof of 52.6.7 in 097B).

Comment #2533 by Brian Conrad on

Oops, disregard the preceding comment (since ind-Zariski maps are always flat); it's been a long day...

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).