Lemma 61.5.1. 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 $S$. $\square$

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