Lemma 60.6.8. Let $A$ be a ring. There exists a faithfully flat, ind-Zariski ring map $A \to B$ such that $B$ satisfies the equivalent conditions of Lemma 60.6.7.

Proof. We first apply Lemma 60.5.3 to see that we may assume that $\mathop{\mathrm{Spec}}(A)$ is w-local. Choose an extremally disconnected space $T$ and a surjective continuous map $T \to \pi _0(\mathop{\mathrm{Spec}}(A))$, see Topology, Lemma 5.26.9. Note that $T$ is profinite. Apply Lemma 60.6.2 to find an ind-Zariski ring map $A \to B$ such that $\pi _0(\mathop{\mathrm{Spec}}(B)) \to \pi _0(\mathop{\mathrm{Spec}}(A))$ realizes $T \to \pi _0(\mathop{\mathrm{Spec}}(A))$ and such that

$\xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[r] \ar[d] & \pi _0(\mathop{\mathrm{Spec}}(B)) \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \pi _0(\mathop{\mathrm{Spec}}(A)) }$

is cartesian in the category of topological spaces. Note that $\mathop{\mathrm{Spec}}(B)$ is w-local, that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is w-local, and that the set of closed points of $\mathop{\mathrm{Spec}}(B)$ is the inverse image of the set of closed points of $\mathop{\mathrm{Spec}}(A)$, see Lemma 60.2.5. Thus condition (3) of Lemma 60.6.7 holds for $B$. $\square$

## 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Your email address will not be published.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 09B0. Beware of the difference between the letter 'O' and the digit '0'.