Lemma 10.70.12. Let (R, \mathfrak m) be a local domain with fraction field K. Let R \subset A \subset K be a valuation ring which dominates R. Then
A = \mathop{\mathrm{colim}}\nolimits R[\textstyle {\frac{I}{a}}]
is a directed colimit of affine blowups R \to R[\frac{I}{a}] with the following properties
a \in I \subset \mathfrak m,
I is finitely generated, and
the fibre ring of R \to R[\frac{I}{a}] at \mathfrak m is not zero.
Proof.
Any blowup algebra R[\frac{I}{a}] is a domain contained in K see Lemma 10.70.10. The lemma simply says that A is the directed union of the ones where a \in I have properties (1), (2), (3). If R[\frac{I}{a}] \subset A and R[\frac{J}{b}] \subset A, then we have
R[\textstyle {\frac{I}{a}}] \cup R[\textstyle {\frac{J}{b}}] \subset R[\textstyle {\frac{IJ}{ab}}] \subset A
The first inclusion because x/a^ n = b^ nx/(ab)^ n and the second one because if z \in (IJ)^ n, then z = \sum x_ iy_ i with x_ i \in I^ n and y_ i \in J^ n and hence z/(ab)^ n = \sum (x_ i/a^ n)(y_ i/b^ n) is contained in A.
Consider a finite subset E \subset A. Say E = \{ e_1, \ldots , e_ n\} . Choose a nonzero a \in R such that we can write e_ i = f_ i/a for all i = 1, \ldots , n. Set I = (f_1, \ldots , f_ n, a). We claim that R[\frac{I}{a}] \subset A. This is clear as an element of R[\frac{I}{a}] can be represented as a polynomial in the elements e_ i. The lemma follows immediately from this observation.
\square
Comments (0)
There are also: