Lemma 31.33.3. In the situation of Definition 31.33.1.

1. If $X$ is flat over $S$ at all points lying over $Z$, then the strict transform of $X$ is equal to the base change $X \times _ S S'$.

2. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $\mathcal{F}$ is flat over $S$ at all points lying over $Z$, then the strict transform $\mathcal{F}'$ of $\mathcal{F}$ is equal to the pullback $\text{pr}_ X^*\mathcal{F}$.

Proof. We will prove part (2) as it implies part (1) by the definition of the strict transform of a scheme over $S$. The question is local on $X$. Thus we may assume that $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$, and that $\mathcal{F}$ corresponds to the $B$-module $N$. Then $\mathcal{F}'$ over the open $\mathop{\mathrm{Spec}}(B \otimes _ A A[\frac{I}{a}])$ of $X \times _ S S'$ corresponds to the module

$N' = (N \otimes _ A A[\textstyle {\frac{I}{a}}])/a\text{-power-torsion}$

see Properties, Lemma 28.24.5. Thus we have to show that the $a$-power-torsion of $N \otimes _ A A[\frac{I}{a}]$ is zero. Let $y \in N \otimes _ A A[\frac{I}{a}]$ with $a^ n y = 0$. If $\mathfrak q \subset B$ is a prime and $a \not\in \mathfrak q$, then $y$ maps to zero in $(N \otimes _ A A[\frac{I}{a}])_\mathfrak q$. on the other hand, if $a \in \mathfrak q$, then $N_\mathfrak q$ is a flat $A$-module and we see that $N_\mathfrak q \otimes _ A A[\frac{I}{a}] =(N \otimes _ A A[\frac{I}{a}])_\mathfrak q$ has no $a$-power torsion (as $A[\frac{I}{a}]$ doesn't). Hence $y$ maps to zero in this localization as well. We conclude that $y$ is zero by Algebra, Lemma 10.23.1. $\square$

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