## Tag `0535`

## 15.24. 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 15.24.1. 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 15.24.2. 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.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 15.20.10 we see that $M_A$ is flat over $A$. By Lemma 15.23.6 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 10.69.10 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 15.20.9), we also conclude that $M'_{I, a} = M_{I, a}$ for such a pair and we win. $\square$Lemma 15.24.3. 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 15.24.4. 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 15.24.3 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 15.24.5. 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. $\square$

- This is somewhat nonstandard notation. ↑

The code snippet corresponding to this tag is a part of the file `more-algebra.tex` and is located in lines 5096–5320 (see updates for more information).

```
\section{Blowing up and flatness}
\label{section-blowup-flat}
\noindent
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 \ref{flat-section-blowup-flat}.
\begin{definition}
\label{definition-strict-transform}
Let $R$ be a domain. Let $M$ be an $R$-module. Let $R \subset R'$ be an
extension of domains. The {\it strict transform of $M$ along
$R \to R'$}\footnote{This is somewhat nonstandard notation.} is
the torsion free $R'$-module
$$
M' = (M \otimes_R R')/(M \otimes_R R')_{tors}.
$$
\end{definition}
\noindent
The following is a very weak version of flattening by blowing up, but
it is already sometimes a useful result.
\begin{lemma}
\label{lemma-flatten-on-affine-blowup}
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
\begin{enumerate}
\item $I$ is finitely generated,
\item $A$ has center on $R[\frac{I}{a}]$,
\item the fibre ring of $R \to R[\frac{I}{a}]$ at $\mathfrak m$
is not zero, and
\item 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 $S_{I, a}$.
\end{enumerate}
\end{lemma}
\begin{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 \ref{algebra-lemma-blowup-domain} and
\ref{algebra-lemma-blowup-dominant}.
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 \ref{lemma-valuation-ring-torsion-free-flat}
we see that $M_A$ is flat over $A$. By
Lemma \ref{lemma-flat-finite-type-valuation-ring-finite-presentation}
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 \ref{algebra-lemma-valuation-ring-colimit-affine-blowups}
we have $A = \colim_{I, a} R[\frac{I}{a}]$.
This implies that $S_A = \colim S_{I, a}$ and
$M_A = \colim_{I, a} M_{I, a}$.
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
$M_A = \colim_{I, a} M'_{I, a}$.
At this point we may apply
Algebra, Lemma \ref{algebra-lemma-flat-finite-presentation-limit-flat} (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 \ref{lemma-flat-torsion-free}),
we also conclude that $M'_{I, a} = M_{I, a}$ for such a pair and we win.
\end{proof}
\begin{lemma}
\label{lemma-blowup-fitting-ideal}
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 $\text{Fit}_k(M') = R'$.
\end{lemma}
\begin{proof}
First observe that $\text{Fit}_k(M \otimes_R R') = IR' = aR'$.
The first equality by Lemma \ref{lemma-fitting-ideal-basics} part (3)
and the second equality by
Algebra, Lemma \ref{algebra-lemma-affine-blowup}.
From Lemma \ref{lemma-principal-fitting-ideal}
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 \ref{lemma-fitting-ideal-generate-locally}.
\end{proof}
\begin{lemma}
\label{lemma-blowup-fitting-ideal-locally-free}
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 $k$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-blowup-fitting-ideal} we have
$\text{Fit}_k(M') = R'$. By Lemma \ref{lemma-fitting-ideal-finite-locally-free}
it suffices to show that $\text{Fit}_{k - 1}(M') = 0$.
Recall that $R' \subset R'_a = R_a$, see
Algebra, Lemma \ref{algebra-lemma-affine-blowup}.
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 \ref{lemma-fitting-ideal-basics} part (3).
This is true by our assumption that
$M_a$ is finite locally free of rank $k$
(see Algebra, Lemma \ref{algebra-lemma-finite-projective})
and the already cited Lemma \ref{lemma-fitting-ideal-finite-locally-free}.
\end{proof}
\begin{lemma}
\label{lemma-blowup-module}
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 $r$.
\end{lemma}
\begin{proof}
Choose a surjection $R^{\oplus n} \to M$. Choose a finite submodule
$K \subset \Ker(R^{\oplus n} \to M)$ 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 \ref{algebra-lemma-finite-projective}.
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 \ref{algebra-lemma-blowup-in-principal}.
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.
\medskip\noindent
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.
\medskip\noindent
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 \ref{lemma-fitting-ideal-generate-locally}.
Thus $\mathfrak p \not \in V(fJ) = V(I)$.
\medskip\noindent
Let $a \in I$ and set $R' = R[\frac{I}{a}]$. We may write $a = fb$
for some $b \in J$. By Algebra, Lemmas \ref{algebra-lemma-affine-blowup} and
\ref{algebra-lemma-blowup-add-principal} 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 \ref{lemma-principal-fitting-ideal}
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 $g$.
\medskip\noindent
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 \ref{lemma-fitting-ideal-finite-locally-free}.
Since $R'_a = R_f$ (Algebra, Lemma \ref{algebra-lemma-blowup-in-principal})
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.
\end{proof}
```

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