## 38.30 Blowing up and flatness

In this section we continue our discussion of results of the form: “After a blowup the strict transform becomes flat”, see More on Algebra, Section 15.26 and Divisors, Section 31.35. We will use the following (more or less standard) notation in this section. If $X \to S$ is a morphism of schemes, $\mathcal{F}$ is a quasi-coherent module on $X$, and $T \to S$ is a morphism of schemes, then we denote $\mathcal{F}_ T$ the pullback of $\mathcal{F}$ to the base change $X_ T = X \times _ S T$.

Lemma 38.30.2. Let $R$ be a ring and let $f \in R$. Let $r\geq 0$ be an integer. Let $R \to S$ be a ring map and let $M$ be an $S$-module. Assume

$R \to S$ is of finite presentation and flat,

every fibre ring $S \otimes _ R \kappa (\mathfrak p)$ is geometrically integral over $R$,

$M$ is a finite $S$-module,

$M_ f$ is a finitely presented $S_ f$-module,

for all $\mathfrak p \in R$, $f \not\in \mathfrak p$ with $\mathfrak q = \mathfrak pS$ the module $M_{\mathfrak q}$ is free of rank $r$ over $S_\mathfrak q$.

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 quotient

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

over $S' = S \otimes _ R R'$ satisfies the following: for every prime $\mathfrak p' \subset R'$ there exists a $g \in S'$, $g \not\in \mathfrak p'S'$ such that $M'_ g$ is a free $S'_ g$-module of rank $r$.

**Proof.**
This lemma is a generalization of More on Algebra, Lemma 15.26.5; we urge the reader to read that proof first. Choose a surjection $S^{\oplus n} \to M$, which is possible by (1). Choose a finite submodule $K \subset \mathop{\mathrm{Ker}}(S^{\oplus n} \to M)$ such that $S^{\oplus n}/K \to M$ becomes an isomorphism after inverting $f$. This is possible by (4). Set $M_1 = S^{\oplus n}/K$ 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 10.70.7. 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 $S$-module.

Assume $M$ is a finitely presented $S$-module satisfying (3). Then $J = \text{Fit}_ r(M) \subset S$ is a finitely generated ideal. By Lemma 38.9.3 we can write $S$ as a direct summand of a free $R$-module: $\bigoplus _{\alpha \in A} R = S \oplus C$. For any element $h \in S$ writing $h = \sum a_\alpha $ in the decomposition above, we say that the $a_\alpha $ are the coefficients of $h$. Let $I' \subset R$ be the ideal of coefficients of elements of $J$. Multiplication by an element of $S$ defines an $R$-linear map $S \to S$, hence $I'$ is generated by the coefficients of the generators of $J$, i.e., $I'$ is a finitely generated ideal. We claim that $I = fI'$ 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 $\mathfrak q = \mathfrak p S$ is not an element of $V(J)$ by property (5) and More on Algebra, Lemma 15.8.6. Hence there is an element of $J$ which does not map to zero in $S \otimes _ R \kappa (\mathfrak p)$. Thus there exists an element of $I'$ which is not contained in $\mathfrak p$, so $\mathfrak p \not\in V(fI') = V(I)$.

Let $a \in I$ and set $R' = R[\frac{I}{a}]$. We may write $a = fa'$ for some $a' \in I'$. By Algebra, Lemmas 10.70.2 and 10.70.8 we see that $I' R' = a'R'$ and $a'$ is a nonzerodivisor in $R'$. Set $S' = S \otimes _ S R'$. Every element $g$ of $JS' = \text{Fit}_ r(M \otimes _ S S')$ can be written as $g = \sum _\alpha c_\alpha $ for some $c_\alpha \in I'R'$. Since $I'R' = a'R'$ we can write $c_\alpha = a'c'_\alpha $ for some $c'_\alpha \in R'$ and $g = (\sum c'_\alpha )a' = g' a'$ in $S'$. Moreover, there is an $g_0 \in J$ such that $a' = c_\alpha $ for some $\alpha $. For this element we have $g_0 = g'_0 a'$ in $S'$ where $g'_0$ is a unit in $S'$. Let $\mathfrak p' \subset R'$ be a prime ideal and $\mathfrak q' = \mathfrak p'S'$. By the above we see that $JS'_{\mathfrak q'}$ is the principal ideal generated by the nonzerodivisor $a'$. It follows from More on Algebra, Lemma 15.8.8 that $M'_{\mathfrak q'}$ can be generated by $r$ elements. Since $M'$ is finite, there exist $m_1, \ldots , m_ r \in M'$ and $g \in S'$, $g \not\in \mathfrak q'$ such that the corresponding map $(S')^{\oplus r} \to M'$ becomes surjective after inverting $g$.

Finally, consider the ideal $J' = \text{Fit}_{k - 1}(M')$. Note that $J'S'_ g$ 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 More on Algebra, Lemma 15.8.7. Since $R'_ a = R_ f$ (Algebra, Lemma 10.70.7) and $M'_ a = M_ f$ we see from (5) that $J'_ a$ maps to zero in $S_{\mathfrak q''}$ for any prime $\mathfrak q'' \subset S'$ of the form $\mathfrak q'' = \mathfrak p''S'$ where $\mathfrak p'' \subset R'_ a$. Since $S'_ a \subset \prod _{\mathfrak q''\text{ as above}} S'_{\mathfrak q''}$ (as $(S'_ a)_{\mathfrak p''} \subset S'_{\mathfrak q''}$ by Lemma 38.7.4) we see that $J'R'_ a = 0$. Since $a$ is a nonzerodivisor in $R'$ we conclude that $J' = 0$ and we win.
$\square$

Lemma 38.30.3. Let $S$ be a quasi-compact and quasi-separated scheme. Let $X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent module on $X$. Let $U \subset S$ be a quasi-compact open. Assume

$X \to S$ is affine, of finite presentation, flat, geometrically integral fibres,

$\mathcal{F}$ is a module of finite type,

$\mathcal{F}_ U$ is of finite presentation,

$\mathcal{F}$ is flat over $S$ at all generic points of fibres lying over points of $U$.

Then there exists a $U$-admissible blowup $S' \to S$ and an open subscheme $V \subset X_{S'}$ such that (a) the strict transform $\mathcal{F}'$ of $\mathcal{F}$ restricts to a finitely locally free $\mathcal{O}_ V$-module and (b) $V \to S'$ is surjective.

**Proof.**
Given $\mathcal{F}/X/S$ and $U \subset S$ with hypotheses as in the lemma, denote $P$ the property “$\mathcal{F}$ is flat over $S$ at all generic points of fibres”. It is clear that $P$ is preserved under strict transform, see Divisors, Lemma 31.33.3 and Morphisms, Lemma 29.25.7. It is also clear that $P$ is local on $S$. Hence any and all observations of Remark 38.30.1 apply to the problem posed by the lemma.

Consider the function $r : U \to \mathbf{Z}_{\geq 0}$ which assigns to $u \in U$ the integer

\[ r(u) = \dim _{\kappa (\xi _ u)}(\mathcal{F}_{\xi _ u} \otimes \kappa (\xi _ u)) \]

where $\xi _ u$ is the generic point of the fibre $X_ u$. By More on Morphisms, Lemma 37.16.7 and the fact that the image of an open in $X_ S$ in $S$ is open, we see that $r(u)$ is locally constant. Accordingly $U = U_0 \amalg U_1 \amalg \ldots \amalg U_ c$ is a finite disjoint union of open and closed subschemes where $r$ is constant with value $i$ on $U_ i$. By Divisors, Lemma 31.34.5 we can find a $U$-admissible blowup to decompose $S$ into the disjoint union of two schemes, the first containing $U_0$ and the second $U_1 \cup \ldots \cup U_ c$. Repeating this $c - 1$ more times we may assume that $S$ is a disjoint union $S = S_0 \amalg S_1 \amalg \ldots \amalg S_ c$ with $U_ i \subset S_ i$. Thus we may assume the function $r$ defined above is constant, say with value $r$.

By Remark 38.30.1 we see that we may assume that we have an effective Cartier divisor $D \subset S$ whose support is $S \setminus U$. Another application of Remark 38.30.1 combined with Divisors, Lemma 31.13.2 tells us we may assume that $S = \mathop{\mathrm{Spec}}(R)$ and $D = \mathop{\mathrm{Spec}}(R/(f))$ for some nonzerodivisor $f \in R$. This case is handled by Lemma 38.30.2.
$\square$

Lemma 38.30.4. Let $A \to C$ be a finite locally free ring map of rank $d$. Let $h \in C$ be an element such that $C_ h$ is étale over $A$. Let $J \subset C$ be an ideal. Set $I = \text{Fit}_0(C/J)$ where we think of $C/J$ as a finite $A$-module. Then $IC_ h = JJ'$ for some ideal $J' \subset C_ h$. If $J$ is finitely generated so are $I$ and $J'$.

**Proof.**
We will use basic properties of Fitting ideals, see More on Algebra, Lemma 15.8.4. Then $IC$ is the Fitting ideal of $C/J \otimes _ A C$. Note that $C \to C \otimes _ A C$, $c \mapsto 1 \otimes c$ has a section (the multiplication map). By assumption $C \to C \otimes _ A C$ is étale at every prime in the image of $\mathop{\mathrm{Spec}}(C_ h)$ under this section. Hence the multiplication map $C \otimes _ A C_ h \to C_ h$ is étale in particular flat, see Algebra, Lemma 10.143.8. Hence there exists a $C_ h$-algebra such that $C \otimes _ A C_ h \cong C_ h \oplus C'$ as $C_ h$-algebras, see Algebra, Lemma 10.143.9. Thus $(C/J) \otimes _ A C_ h \cong (C_ h/J_ h) \oplus C'/I'$ as $C_ h$-modules for some ideal $I' \subset C'$. Hence $IC_ h = JJ'$ with $J' = \text{Fit}_0(C'/I')$ where we view $C'/J'$ as a $C_ h$-module.
$\square$

Lemma 38.30.5. Let $A \to B$ be an étale ring map. Let $a \in A$ be a nonzerodivisor. Let $J \subset B$ be a finite type ideal with $V(J) \subset V(aB)$. For every $\mathfrak q \subset B$ there exists a finite type ideal $I \subset A$ with $V(I) \subset V(a)$ and $g \in B$, $g \not\in \mathfrak q$ such that $IB_ g = JJ'$ for some finite type ideal $J' \subset B_ g$.

**Proof.**
We may replace $B$ by a principal localization at an element $g \in B$, $g \not\in \mathfrak q$. Thus we may assume that $B$ is standard étale, see Algebra, Proposition 10.144.4. Thus we may assume $B$ is a localization of $C = A[x]/(f)$ for some monic $f \in A[x]$ of some degree $d$. Say $B = C_ h$ for some $h \in C$. Choose elements $h_1, \ldots , h_ n \in C$ which generate $J$ over $B$. The condition $V(J) \subset V(aB)$ signifies that $a^ m = \sum b_ i h_ i$ in $B$ for some large $m$. Set $h_{n + 1} = a^ m$. As in Lemma 38.30.4 we take $I = \text{Fit}_0(C/(h_1, \ldots , h_{r + 1}))$. Since the module $C/(h_1, \ldots , h_{r + 1})$ is annihilated by $a^ m$ we see that $a^{dm} \in I$ which implies that $V(I) \subset V(a)$.
$\square$

Lemma 38.30.6. Let $S$ be a quasi-compact and quasi-separated scheme. Let $X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent module on $X$. Let $U \subset S$ be a quasi-compact open. Assume there exist finitely many commutative diagrams

\[ \xymatrix{ & X_ i \ar[r]_{j_ i} \ar[d] & X \ar[d] \\ S_ i^* \ar[r] & S_ i \ar[r]^{e_ i} & S } \]

where

$e_ i : S_ i \to S$ are quasi-compact étale morphisms and $S = \bigcup e_ i(S_ i)$,

$j_ i : X_ i \to X$ are étale morphisms and $X = \bigcup j_ i(X_ i)$,

$S^*_ i \to S_ i$ is an $e_ i^{-1}(U)$-admissible blowup such that the strict transform $\mathcal{F}_ i^*$ of $j_ i^*\mathcal{F}$ is flat over $S^*_ i$.

Then there exists a $U$-admissible blowup $S' \to S$ such that the strict transform of $\mathcal{F}$ is flat over $S'$.

**Proof.**
We claim that the hypotheses of the lemma are preserved under $U$-admissible blowups. Namely, suppose $b : S' \to S$ is a $U$-admissible blowup in the quasi-coherent sheaf of ideals $\mathcal{I}$. Moreover, let $S^*_ i \to S_ i$ be the blowup in the quasi-coherent sheaf of ideals $\mathcal{J}_ i$. Then the collection of morphisms $e'_ i : S'_ i = S_ i \times _ S S' \to S'$ and $j'_ i : X_ i' = X_ i \times _ S S' \to X \times _ S S'$ satisfy conditions (1), (2), (3) for the strict transform $\mathcal{F}'$ of $\mathcal{F}$ relative to the blowup $S' \to S$. First, observe that $S_ i'$ is the blowup of $S_ i$ in the pullback of $\mathcal{I}$, see Divisors, Lemma 31.32.3. Second, consider the blowup $S_ i^{\prime *} \to S_ i'$ of $S_ i'$ in the pullback of the ideal $\mathcal{J}_ i$. By Divisors, Lemma 31.32.12 we get a commutative diagram

\[ \xymatrix{ S_ i^{\prime *} \ar[r] \ar[rd] \ar[d] & S'_ i \ar[d] \\ S_ i^* \ar[r] & S_ i } \]

and all the morphisms in the diagram above are blowups. Hence by Divisors, Lemmas 31.33.3 and 31.33.6 we see

\begin{align*} & \text{ the strict transform of }(j'_ i)^*\mathcal{F}'\text{ under } S_ i^{\prime *} \to S_ i' \\ = & \text{ the strict transform of }j_ i^*\mathcal{F}\text{ under } S_ i^{\prime *} \to S_ i \\ = & \text{ the strict transform of }\mathcal{F}_ i'\text{ under } S_ i^{\prime *} \to S_ i' \\ = & \text{ the pullback of }\mathcal{F}_ i^*\text{ via } X_ i \times _{S_ i} S_ i^{\prime *} \to X_ i \end{align*}

which is therefore flat over $S_ i^{\prime *}$ (Morphisms, Lemma 29.25.7). Having said this, we see that all observations of Remark 38.30.1 apply to the problem of finding a $U$-admissible blowup such that the strict transform of $\mathcal{F}$ becomes flat over the base under assumptions as in the lemma. In particular, we may assume that $S \setminus U$ is the support of an effective Cartier divisor $D \subset S$. Another application of Remark 38.30.1 combined with Divisors, Lemma 31.13.2 shows we may assume that $S = \mathop{\mathrm{Spec}}(A)$ and $D = \mathop{\mathrm{Spec}}(A/(a))$ for some nonzerodivisor $a \in A$.

Pick an $i$ and $s \in S_ i$. Lemma 38.30.5 implies we can find an open neighbourhood $s \in W_ i \subset S_ i$ and a finite type quasi-coherent ideal $\mathcal{I} \subset \mathcal{O}_ S$ such that $\mathcal{I} \cdot \mathcal{O}_{W_ i} = \mathcal{J}_ i \mathcal{J}'_ i$ for some finite type quasi-coherent ideal $\mathcal{J}'_ i \subset \mathcal{O}_{W_ i}$ and such that $V(\mathcal{I}) \subset V(a) = S \setminus U$. Since $S_ i$ is quasi-compact we can replace $S_ i$ by a finite collection $W_1, \ldots , W_ n$ of these opens and assume that for each $i$ there exists a quasi-coherent sheaf of ideals $\mathcal{I}_ i \subset \mathcal{O}_ S$ such that $\mathcal{I}_ i \cdot \mathcal{O}_{S_ i} = \mathcal{J}_ i \mathcal{J}'_ i$ for some finite type quasi-coherent ideal $\mathcal{J}'_ i \subset \mathcal{O}_{S_ i}$. As in the discussion of the first paragraph of the proof, consider the blowup $S'$ of $S$ in the product $\mathcal{I}_1 \ldots \mathcal{I}_ n$ (this blowup is $U$-admissible by construction). The base change of $S' \to S$ to $S_ i$ is the blowup in

\[ \mathcal{J}_ i \cdot \mathcal{J}'_ i \mathcal{I}_1 \ldots \hat{\mathcal{I}_ i} \ldots \mathcal{I}_ n \]

which factors through the given blowup $S_ i^* \to S_ i$ (Divisors, Lemma 31.32.12). In the notation of the diagram above this means that $S_ i^{\prime *} = S_ i'$. Hence after replacing $S$ by $S'$ we arrive in the situation that $j_ i^*\mathcal{F}$ is flat over $S_ i$. Hence $j_ i^*\mathcal{F}$ is flat over $S$, see Lemma 38.2.3. By Morphisms, Lemma 29.25.13 we see that $\mathcal{F}$ is flat over $S$.
$\square$

Theorem 38.30.7. Let $S$ be a quasi-compact and quasi-separated scheme. Let $X$ be a scheme over $S$. Let $\mathcal{F}$ be a quasi-coherent module on $X$. Let $U \subset S$ be a quasi-compact open. Assume

$X$ is quasi-compact,

$X$ is locally of finite presentation over $S$,

$\mathcal{F}$ is a module of finite type,

$\mathcal{F}_ U$ is of finite presentation, and

$\mathcal{F}_ U$ is flat over $U$.

Then there exists a $U$-admissible blowup $S' \to S$ such that the strict transform $\mathcal{F}'$ of $\mathcal{F}$ is an $\mathcal{O}_{X \times _ S S'}$-module of finite presentation and flat over $S'$.

**Proof.**
We first prove that we can find a $U$-admissible blowup such that the strict transform is flat. The question is étale local on the source and the target, see Lemma 38.30.6 for a precise statement. In particular, we may assume that $S = \mathop{\mathrm{Spec}}(R)$ and $X = \mathop{\mathrm{Spec}}(A)$ are affine. For $s \in S$ write $\mathcal{F}_ s = \mathcal{F}|_{X_ s}$ (pullback of $\mathcal{F}$ to the fibre). As $X \to S$ is of finite type $d = \max _{s \in S} \dim (\text{Supp}(\mathcal{F}_ s))$ is an integer. We will do induction on $d$.

Let $x \in X$ be a point of $X$ lying over $s \in S$ with $\dim _ x(\text{Supp}(\mathcal{F}_ s)) = d$. Apply Lemma 38.3.2 to get $g : X' \to X$, $e : S' \to S$, $i : Z' \to X'$, and $\pi : Z' \to Y'$. Observe that $Y' \to S'$ is a smooth morphism of affines with geometrically irreducible fibres of dimension $d$. Because the problem is étale local it suffices to prove the theorem for $g^*\mathcal{F}/X'/S'$. Because $i : Z' \to X'$ is a closed immersion of finite presentation (and since strict transform commutes with affine pushforward, see Divisors, Lemma 31.33.4) it suffices to prove the flattening result for $\mathcal{G}$. Since $\pi $ is finite (hence also affine) it suffices to prove the flattening result for $\pi _*\mathcal{G}/Y'/S'$. Thus we may assume that $X \to S$ is a smooth morphism of affines with geometrically irreducible fibres of dimension $d$.

Next, we apply a blowup as in Lemma 38.30.3. Doing so we reach the situation where there exists an open $V \subset X$ surjecting onto $S$ such that $\mathcal{F}|_ V$ is finite locally free. Let $\xi \in X$ be the generic point of $X_ s$. Let $r = \dim _{\kappa (\xi )} \mathcal{F}_\xi \otimes \kappa (\xi )$. Choose a map $\alpha : \mathcal{O}_ X^{\oplus r} \to \mathcal{F}$ which induces an isomorphism $\kappa (\xi )^{\oplus r} \to \mathcal{F}_\xi \otimes \kappa (\xi )$. Because $\mathcal{F}$ is locally free over $V$ we find an open neighbourhood $W$ of $\xi $ where $\alpha $ is an isomorphism. Shrink $S$ to an affine open neighbourhood of $s$ such that $W \to S$ is surjective. Say $\mathcal{F}$ is the quasi-coherent module associated to the $A$-module $N$. Since $\mathcal{F}$ is flat over $S$ at all generic points of fibres (in fact at all points of $W$), we see that

\[ \alpha _\mathfrak p : A_\mathfrak p^{\oplus r} \to N_\mathfrak p \]

is universally injective for all primes $\mathfrak p$ of $R$, see Lemma 38.10.1. Hence $\alpha $ is universally injective, see Algebra, Lemma 10.82.12. Set $\mathcal{H} = \mathop{\mathrm{Coker}}(\alpha )$. By Divisors, Lemma 31.33.7 we see that, given a $U$-admissible blowup $S' \to S$ the strict transforms of $\mathcal{F}'$ and $\mathcal{H}'$ fit into an exact sequence

\[ 0 \to \mathcal{O}_{X \times _ S S'}^{\oplus r} \to \mathcal{F}' \to \mathcal{H}' \to 0 \]

Hence Lemma 38.10.1 also shows that $\mathcal{F}'$ is flat at a point $x'$ if and only if $\mathcal{H}'$ is flat at that point. In particular $\mathcal{H}_ U$ is flat over $U$ and $\mathcal{H}_ U$ is a module of finite presentation. We may apply the induction hypothesis to $\mathcal{H}$ to see that there exists a $U$-admissible blowup such that the strict transform $\mathcal{H}'$ is flat as desired.

To finish the proof of the theorem we still have to show that $\mathcal{F}'$ is a module of finite presentation (after possibly another $U$-admissible blowup). This follows from Lemma 38.11.1 as we can assume $U \subset S$ is scheme theoretically dense (see third paragraph of Remark 38.30.1). This finishes the proof of the theorem.
$\square$

## Comments (1)

Comment #8342 by Zhipu Wilson Zhao on