The Stacks project

Lemma 60.5.1. Let $A$ be a ring. Set $X = \mathop{\mathrm{Spec}}(A)$. Let $Z \subset X$ be a locally closed subscheme which is of the form $D(f) \cap V(I)$ for some $f \in A$ and ideal $I \subset A$. Then

  1. there exists a multiplicative subset $S \subset A$ such that $\mathop{\mathrm{Spec}}(S^{-1}A)$ maps by a homeomorphism to the set of points of $X$ specializing to $Z$,

  2. the $A$-algebra $A_ Z^\sim = S^{-1}A$ depends only on the underlying locally closed subset $Z \subset X$,

  3. $Z$ is a closed subscheme of $\mathop{\mathrm{Spec}}(A_ Z^\sim )$,

If $A \to A'$ is a ring map and $Z' \subset X' = \mathop{\mathrm{Spec}}(A')$ is a locally closed subscheme of the same form which maps into $Z$, then there is a unique $A$-algebra map $A_ Z^\sim \to (A')_{Z'}^\sim $.

Proof. Let $S \subset A$ be the multiplicative set of elements which map to invertible elements of $\Gamma (Z, \mathcal{O}_ Z) = (A/I)_ f$. If $\mathfrak p$ is a prime of $A$ which does not specialize to $Z$, then $\mathfrak p$ generates the unit ideal in $(A/I)_ f$. Hence we can write $f^ n = g + h$ for some $n \geq 0$, $g \in \mathfrak p$, $h \in I$. Then $g \in S$ and we see that $\mathfrak p$ is not in the spectrum of $S^{-1}A$. Conversely, if $\mathfrak p$ does specialize to $Z$, say $\mathfrak p \subset \mathfrak q \supset I$ with $f \not\in \mathfrak q$, then we see that $S^{-1}A$ maps to $A_\mathfrak q$ and hence $\mathfrak p$ is in the spectrum of $S^{-1}A$. This proves (1).

The isomorphism class of the localization $S^{-1}A$ depends only on the corresponding subset $\mathop{\mathrm{Spec}}(S^{-1}A) \subset \mathop{\mathrm{Spec}}(A)$, whence (2) holds. By construction $S^{-1}A$ maps surjectively onto $(A/I)_ f$, hence (3). The final statement follows as the multiplicative subset $S' \subset A'$ corresponding to $Z'$ contains the image of the multiplicative subset $S$. $\square$


Comments (0)


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Your email address will not be published.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 096V. Beware of the difference between the letter 'O' and the digit '0'.