Definition 15.26.1. Let $R$ be a ring. Let $I \subset R$ be an ideal and $a \in I$. Let $R[\frac{I}{a}]$ be the affine blowup algebra, see Algebra, Definition 10.70.1. Let $M$ be an $R$-module. The *strict transform of $M$ along $R \to R[\frac{I}{a}]$* is the $R[\frac{I}{a}]$-module

## 15.26 Blowing up and flatness

In this section we begin our discussion of results of the form: “After a blowup the strict transform becomes flat”. More results of this type may be found in Divisors, Section 31.35 and More on Flatness, Section 38.30.

The following is a very weak version of flattening by blowing up, but it is already sometimes a useful result.

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

$I$ is finitely generated,

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

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

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

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

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

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

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. $\square$

Lemma 15.26.3. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_ k(M)$. For every $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform

has $\text{Fit}_ k(M') = R'$.

**Proof.**
First observe that $\text{Fit}_ k(M \otimes _ R R') = IR' = aR'$. The first equality by Lemma 15.8.4 part (3) and the second equality by Algebra, Lemma 10.70.2. From Lemma 15.8.9 and exactness of localization we see that $M'_{\mathfrak p'}$ can be generated by $\leq k$ elements for every prime $\mathfrak p'$ of $R'$. Then $\text{Fit}_ k(M') = R'$ for example by Lemma 15.8.7.
$\square$

Lemma 15.26.4. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_ k(M)$. Assume that $M_\mathfrak p$ is free of rank $k$ for every $\mathfrak p \not\in V(I)$. Then for every $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform

is locally free of rank $k$.

**Proof.**
By Lemma 15.26.3 we have $\text{Fit}_ k(M') = R'$. By Lemma 15.8.8 it suffices to show that $\text{Fit}_{k - 1}(M') = 0$. Recall that $R' \subset R'_ a = R_ a$, see Algebra, Lemma 10.70.2. Hence it suffices to prove that $\text{Fit}_{k - 1}(M')$ maps to zero in $R'_ a = R_ a$. Since clearly $(M')_ a = M_ a$ this reduces us to showing that $\text{Fit}_{k - 1}(M_ a) = 0$ because formation of Fitting ideals commutes with base change according to Lemma 15.8.4 part (3). This is true by our assumption that $M_ a$ is finite locally free of rank $k$ (see Algebra, Lemma 10.78.2) and the already cited Lemma 15.8.8.
$\square$

Lemma 15.26.5. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $f \in R$ be an element such that $M_ f$ is finite locally free of rank $r$. Then there exists a finitely generated ideal $I \subset R$ with $V(f) = V(I)$ such that for all $a \in I$ with $R' = R[\frac{I}{a}]$ the strict transform

is locally free of rank $r$.

**Proof.**
Choose a surjection $R^{\oplus n} \to M$. Choose a finite submodule $K \subset \mathop{\mathrm{Ker}}(R^{\oplus n} \to M)$ such that $R^{\oplus n}/K \to M$ becomes an isomorphism after inverting $f$. This is possible because $M_ f$ is of finite presentation for example by Algebra, Lemma 10.78.2. Set $M_1 = R^{\oplus n}/K$ and suppose we can prove the lemma for $M_1$. Say $I \subset R$ is the corresponding ideal. Then for $a \in I$ the map

is surjective. It is also an isomorphism after inverting $a$ in $R'$ as $R'_ a = R_ f$, see Algebra, Lemma 10.70.7. But $a$ is a nonzerodivisor on $M'_1$, whence the displayed map is an isomorphism. Thus it suffices to prove the lemma in case $M$ is a finitely presented $R$-module.

Assume $M$ is a finitely presented $R$-module. Then $J = \text{Fit}_ r(M) \subset S$ is a finitely generated ideal. We claim that $I = fJ$ works.

We first check that $V(f) = V(I)$. The inclusion $V(f) \subset V(I)$ is clear. Conversely, if $f \not\in \mathfrak p$, then $\mathfrak p$ is not an element of $V(J)$ by Lemma 15.8.7. Thus $\mathfrak p \not\in V(fJ) = V(I)$.

Let $a \in I$ and set $R' = R[\frac{I}{a}]$. We may write $a = fb$ for some $b \in J$. By Algebra, Lemmas 10.70.2 and 10.70.8 we see that $J R' = b R'$ and $b$ is a nonzerodivisor in $R'$. Let $\mathfrak p' \subset R' = R[\frac{I}{a}]$ be a prime ideal. Then $JR'_{\mathfrak p'}$ is generated by $b$. It follows from Lemma 15.8.9 that $M'_{\mathfrak p'}$ can be generated by $r$ elements. Since $M'$ is finite, there exist $m_1, \ldots , m_ r \in M'$ and $g \in R'$, $g \not\in \mathfrak p'$ such that the corresponding map $(R')^{\oplus r} \to M'$ becomes surjective after inverting $g$.

Finally, consider the ideal $J' = \text{Fit}_{k - 1}(M')$. Note that $J' R'_ g$ is generated by the coefficients of relations between $m_1, \ldots , m_ r$ (compatibility of Fitting ideal with base change). Thus it suffices to show that $J' = 0$, see Lemma 15.8.8. Since $R'_ a = R_ f$ (Algebra, Lemma 10.70.7) and $M'_ a = M_ f$ is free of rank $r$ we see that $J'_ a = 0$. Since $a$ is a nonzerodivisor in $R'$ we conclude that $J' = 0$ and we win. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (4)

Comment #2330 by Guignard on

Comment #2401 by Johan on

Comment #6767 by jok on

Comment #6768 by Johan on