The Stacks project

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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 More on Flatness, Section 37.30.

Definition 15.26.1. Let $R$ be a domain. Let $M$ be an $R$-module. Let $R \subset R'$ be an extension of domains. The strict transform of $M$ along $R \to R'$1 is the torsion free $R'$-module

\[ M' = (M \otimes _ R R')/(M \otimes _ R R')_{tors}. \]

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

  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. Note that the assertion makes sense as $R[\frac{I}{a}]$ is a domain, and $R \to R[\frac{I}{a}]$ is injective, see Algebra, Lemmas 10.69.7 and 10.69.8. Before we start the proof of the Lemma, note that there is no loss in generality assuming that $S = R[x_1, \ldots , x_ n]$ is a polynomial ring over $R$. We also fix a presentation

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

Let $M_ A$ be the strict transform of $M$ along $R \to A$. It 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 10.69.9 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 $M_ A = \mathop{\mathrm{colim}}\nolimits _{I, a} M_{I, a}$. Thus we may choose $a \in I \subset R$ such that $k_1, \ldots , k_ t$ are elements of $S_{I, a}^{\oplus r}$ and map to zero in $M_{I, a}$. 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 also $M_ A = \mathop{\mathrm{colim}}\nolimits _{I, a} M'_{I, a}$. At this point we may apply Algebra, Lemma 10.162.1 (3) to conclude that $M'_{I, a}$ is flat for some pair $(I, a)$. (This lemma does not apply a priori to the system $M_{I, a}$ as the transition maps may not satisfy the assumptions of the lemma.) Since flatness implies torsion free ( Lemma 15.22.9), 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

\[ M' = (M \otimes _ R R')/a\text{-power torsion} \]

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 10.69.2. From Lemma 15.8.8 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 15.8.6. $\square$

Lemma 15.26.4. Let $R$ be a ring. Let $M$ be a finite $R$-module. Let $k \geq 0$ and $I = \text{Fit}_ k(M)$. Asssume 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

\[ M' = (M \otimes _ R R')/a\text{-power torsion} \]

is locally free of rank $k$.

Proof. By Lemma 15.26.3 we have $\text{Fit}_ k(M') = R'$. By Lemma 15.8.7 it suffices to show that $\text{Fit}_{k - 1}(M') = 0$. Recall that $R' \subset R'_ a = R_ a$, see Algebra, Lemma 10.69.2. 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 10.77.2) and the already cited Lemma 15.8.7. $\square$

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

\[ M' = (M \otimes _ R R')/a\text{-power torsion} \]

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 10.77.2. 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

\[ M_1' = (M_1 \otimes _ R R')/a\text{-power torsion} \longrightarrow M' = (M \otimes _ R R')/a\text{-power torsion} \]

is surjective. It is also an isomorphism after inverting $a$ in $R'$ as $R'_ a = R_ f$, see Algebra, Lemma 10.69.4. 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 15.8.6. 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 10.69.2 and 10.69.5 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 15.8.8 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 15.8.7. Since $R'_ a = R_ f$ (Algebra, Lemma 10.69.4) 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. $\square$

[1] This is somewhat nonstandard notation.

Comments (2)

Comment #2330 by Guignard on

Typo in the proof of Lemma : "it suffices to show that " should be "it suffices to show that ".

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