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