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Maybe it is a silly question but i don't uderstand why the following statement is true:

"Let X be a complex space and $\pi :Y \longrightarrow X$ be a proper modification of $X$. The pull back $\pi^∗S$ of a torsion-free coherent sheaf of $\mathcal{O}_X$-module $S$ is not torsion-free in general. For a counterexample, see the example in Hans Grauert and Oswald Riemenschneider, Verschwindungssa ̈tze fu ̈r analytische Kohomologiegruppen auf komplexen Ra ̈umen (it is in german),

i. e. the pullback of the maximal ideal sheaf of the origin in $\mathbb{C}^2$ under blow-up of the origin is not torsion-free. One can say more or less that $\pi^∗S$ is torsion-free in a point $y \in Y$ if and only if $S$ is locally free in $π(y)$.''

I don't understand this example, if i take $\pi: \mathrm{BL}(0,\mathbb{C}^2) \longrightarrow X$ to be the blow up of the origin of $\mathbb{C}^2$ and i look in local coordinates $(x,y)=(x',x'y')$, the pull back of the function $f(x,y)=\alpha x+ \beta y$ is the function $f \circ \pi(x',y')=x'(\alpha+ \beta y')$ how can it be a torsion element ? by what element of $\mathcal{O}_{\mathrm{BL}(0,\mathbb{C}^2)}$ should i multiply it to get zero ?

Thanks in advance for your help

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    $\begingroup$ You are confusing $\pi ^*\mathfrak{m}$ and its image in $\mathscr{O}_{\operatorname{BL} }$. The map $\pi ^*\mathfrak{m} \rightarrow \mathscr{O}_{\operatorname{BL} }$ has a kernel, which is precisely the torsion part. $\endgroup$
    – abx
    Commented Feb 17, 2023 at 15:35
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    $\begingroup$ okay, how is defined this map $\pi ^*\mathfrak{m} \rightarrow \mathscr{O}_{\operatorname{BL} }$ if it is just the inclusion i don't see where is the kernel ? $\endgroup$
    – singularity
    Commented Feb 17, 2023 at 18:00
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    $\begingroup$ It is the pull back of the inclusion $\mathfrak{m}\hookrightarrow \mathscr{O}_X $. The point is that this pull back is not injective. $\endgroup$
    – abx
    Commented Feb 17, 2023 at 18:35
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    $\begingroup$ thanks but it is still unclear to me, the elements of $\mathfrak{M}$ are the functions of $\mathcal{O}_{\mathbb{C}^2}$ which vanishes at the origin right ?. I must misunderstand something but the pull back of the inclusion $\mathfrak{m}\hookrightarrow \mathscr{O}_X$ seems to me to be just the inclusion of $\pi ^*\mathfrak{m}$ in $\mathscr{O}_{\operatorname{BL} }$ . Is it possible to get an explicit example ? $\endgroup$
    – singularity
    Commented Feb 17, 2023 at 19:37
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    $\begingroup$ How could it be more explicit than that? Do the computation... $\endgroup$
    – abx
    Commented Feb 17, 2023 at 19:59

1 Answer 1

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To compute the pullback one can use a locally free resolution: $$ 0 \to \mathcal{O}_X \to \mathcal{O}_X \oplus \mathcal{O}_X \to \mathfrak{m} \to 0,\tag{*} $$ where the second arrow is induced by $(x,y)$ and the first by $(y,-x)$. Note that the pullbacks of these two functions to the blowup $\tilde{X}$ vanish along the exceptional divisor $E \subset \tilde{X}$, therefore they can be written as $$ x = x' \cdot e, \qquad y = y' \cdot e, $$ where $e$ is a section of the line bundle $\mathcal{O}_{\tilde{X}}(E)$ vanishing on $E$ and $x',y'$ are sections of $\mathcal{O}_{\tilde{X}}(-E)$. Therefore, the pullback of $(*)$ looks as $$ \mathcal{O}_{\tilde{X}} \to \mathcal{O}_{\tilde{X}} \oplus \mathcal{O}_{\tilde{X}} \to \pi^*\mathfrak{m} \to 0, $$ where the first arrow is given by $(y'\cdot e, -x'\cdot e)$, hence it factors as $$ \mathcal{O}_{\tilde{X}} \stackrel{e}\to \mathcal{O}_{\tilde{X}}(E) \stackrel{(y',-x')}\to \mathcal{O}_{\tilde{X}} \oplus \mathcal{O}_{\tilde{X}}. $$ It follows that there is an exact sequence $$ 0 \to \mathrm{Coker}(e) \to \pi^*\mathfrak{m} \to \mathrm{Coker}(y',-x') \to 0 $$ so it remains to note that $$ \mathrm{Coker}(e) \cong \mathcal{O}_E(E) $$ is a torsion sheaf.

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