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
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
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. $\square$
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