Lemma 61.6.1. Let $A$ be a ring. Let $X = \mathop{\mathrm{Spec}}(A)$. Let $T \subset \pi _0(X)$ be a closed subset. There exists a surjective ind-Zariski ring map $A \to B$ such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ induces a homeomorphism of $\mathop{\mathrm{Spec}}(B)$ with the inverse image of $T$ in $X$.

Proof. Let $Z \subset X$ be the inverse image of $T$. Then $Z$ is the intersection $Z = \bigcap Z_\alpha$ of the open and closed subsets of $X$ containing $Z$, see Topology, Lemma 5.12.12. For each $\alpha$ we have $Z_\alpha = \mathop{\mathrm{Spec}}(A_\alpha )$ where $A \to A_\alpha$ is a local isomorphism (a localization at an idempotent). Setting $B = \mathop{\mathrm{colim}}\nolimits A_\alpha$ proves the lemma. $\square$

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Your email address will not be published.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