Lemma 15.26.2. Let $(R, \mathfrak m)$ be a local domain with fraction field $K$. Let $S$ be a finite type $R$-algebra. Let $M$ be a finite $S$-module. For every valuation ring $A \subset K$ dominating $R$ there exists an ideal $I \subset \mathfrak m$ and a nonzero element $a \in I$ such that

1. $I$ is finitely generated,

2. $A$ has center on $R[\frac{I}{a}]$,

3. the fibre ring of $R \to R[\frac{I}{a}]$ at $\mathfrak m$ is not zero, and

4. the strict transform $S_{I, a}$ of $S$ along $R \to R[\frac{I}{a}]$ is flat and of finite presentation over $R$, and the strict transform $M_{I, a}$ of $M$ along $R \to R[\frac{I}{a}]$ is flat over $R$ and finitely presented over $S_{I, a}$.

Proof. Write $S = R[x_1, \ldots , x_ n]/J$ and denote $N = S \oplus M$ viewed as a module over $P = R[x_1, \ldots , x_ n]$. If we can prove the lemma in case $S$ is a polynomial algebra over $R$, then we can find $I, a$ satisfying (1), (2), (3) such that the strict transform $N_{I, a}$ of $N$ along $R \to R[\frac{I}{a}]$ is flat over $R$ and finitely presented as a module over the strict transform $P_{I, a}]$ of $P$. Since $P_{I, a} = R[\frac{I}{a}][x_1, \ldots , x_ n]$ (small detail omitted) we find that the summand $S_{I, a} \subset N_{I, a}$ is flat over $R$ and finitely presented as a module over $R[\frac{I}{a}][x_1, \ldots , x_ n]$. Hence $S_{I, a}$ is finitely presented as an $R[\frac{I}{a}]$-algebra. Moreover, the summand $M_{I, a} \subset N_{I, a}$ is flat over $R$ and finitely presented as a module over $P_{I, a}$ hence also finitely presented as a module over $S_{I, a}$, see Algebra, Lemma 10.6.4. This reduces us to the case discussed in the next paragraph.

Assume $S = R[x_1, \ldots , x_ n]$. Choose a presentation

$0 \to K \to S^{\oplus r} \to M \to 0.$

Let $M_ A$ be the quotient of $M \otimes _ R A$ by its torsion submodule, see Lemma 15.22.2. Then $M_ A$ is a finite module over $S_ A = A[x_1, \ldots , x_ n]$. By Lemma 15.22.10 we see that $M_ A$ is flat over $A$. By Lemma 15.25.6 we see that $M_ A$ is finitely presented. Hence there exist finitely many elements $k_1, \ldots , k_ t \in S_ A^{\oplus r}$ which generate the kernel of the presentation $S_ A^{\oplus r} \to M_ A$ as an $S_ A$-module. For any choice of $a \in I \subset \mathfrak m$ satisfying (1), (2), and (3) we denote $M_{I, a}$ the strict transform of $M$ along $R \to R[\frac{I}{a}]$. It is a finite module over $S_{I, a} = R[\frac{I}{a}][x_1, \ldots , x_ n]$. By Algebra, Lemma 10.70.12 we have $A = \mathop{\mathrm{colim}}\nolimits _{I, a} R[\frac{I}{a}]$. This implies that $S_ A = \mathop{\mathrm{colim}}\nolimits S_{I, a}$ and

$\mathop{\mathrm{colim}}\nolimits M \otimes _ R R[\textstyle {\frac{I}{a}}] = M \otimes _ R A$

Choose $I, a$ and lifts $k_1, \ldots , k_ t \in S_{I, a}^{\oplus r}$. Since $M_ A$ is the quotient of $M \otimes _ R A$ by torsion, we see that the images of $k_1, \ldots , k_ t$ in $M \otimes _ R A$ are annihilated by a nonzero element $\alpha \in A$. After replacing $I, a$ by a different pair (recall that the colimit is filtered), we may assume $\alpha = x/a^ n$ for some $x \in I^ n$ nonzero. Then we find that $x k_1, \ldots , x k_ t$ map to zero in $M \otimes _ R A$. Hence after replacing $I, a$ by a different pair we may assume $x k_1, \ldots , x k_ t$ map to zero in $M \otimes _ R R[\frac{I}{a}]$ for some nonzero $x \in R$. Then finally replacing $I, a$ by $xI, xa$ we find that we may assume $k_1, \ldots , k_ t$ map to $a$-power torsion elements of $M \otimes _ R R[\frac{I}{a}]$. For any such pair $(I, a)$ we set

$M'_{I, a} = S_{I, a}^{\oplus r}/ \sum S_{I, a}k_ j.$

Since $M_ A = S_ A^{\oplus r}/ \sum S_ Ak_ j$ we see that $M_ A = \mathop{\mathrm{colim}}\nolimits _{I, a} M'_{I, a}$. At this point we finally apply Algebra, Lemma 10.168.1 (3) to conclude that $M'_{I, a}$ is flat for some pair $(I, a)$ as above. This lemma does not apply a priori to the system of strict transforms

$M_{I, a} = (M \otimes _ R R[\textstyle {\frac{I}{a}}])/a\text{-power torsion}$

as the transition maps may not satisfy the assumptions of the lemma. But now, since flatness implies torsion free (Lemma 15.22.9) and since $M_{I, a}$ is the quotient of $M'_{I, a}$ (because we arranged it so the elements $k_1, \ldots , k_ t$ map to zero in $M_{I, a}$) by the $a$-power torsion submodule we also conclude that $M'_{I, a} = M_{I, a}$ for such a pair and we win. $\square$

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