Lemma 61.5.7. Let $A$ be a ring. Let $V(I) \subset \mathop{\mathrm{Spec}}(A)$ be a closed subset which is a profinite topological space. Then there exists an ind-Zariski ring map $A \to B$ such that $\mathop{\mathrm{Spec}}(B)$ is w-local, the set of closed points is $V(IB)$, and $A/I \cong B/IB$.
Proof. Let $A \to A_ w$ and $Z \subset Y = \mathop{\mathrm{Spec}}(A_ w)$ as in Lemma 61.5.3. Let $T \subset Z$ be the inverse image of $V(I)$. Then $T \to V(I)$ is a homeomorphism by Topology, Lemma 5.17.8. Let $B = (A_ w)_ T^\sim $, see Lemma 61.5.1. It is clear that $B$ is w-local with closed points $V(IB)$. The ring map $A/I \to B/IB$ is ind-Zariski and induces a homeomorphism on underlying topological spaces. Hence it is an isomorphism by Lemma 61.3.8. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: