The Stacks project

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 $IB \subset B$. $\square$

Comments (2)

Comment #2532 by Brian Conrad on

In (2) you should say 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...

There are also:

  • 2 comment(s) on Section 61.5: Constructing w-local affine schemes

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Your email address will not be published.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 097A. Beware of the difference between the letter 'O' and the digit '0'.