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Let $\mathbb{P}^1$ the projective line (considered as scheme) and $\mathcal{F}$ a quasi corent sheaf of finite type on on it.

Denote by $\mathcal{F}_T \subset \mathcal{F}$ it's torsion subsheaf.

My question is why the support $Supp(\mathcal{F}_T):= \{x \in \mathbb{P}^1 \vert \mathcal{F}_{T,x} \neq 0 \}$ of $\mathcal{F}_T$ is a finite set?

My considerations: Consider at arbitrary point $x$ the stalk of $\mathcal{F}$. By classification theorem for finitely generated modules over Dedekind rings we obtain

$$\mathcal{F}_x = \mathcal{O}_{P^1}^{r_x} \oplus \mathcal{F}_{T,x}$$

Locally $r_x$ is constant.

Futhermore by definition of torsion we can redefine the torsion elements

$$\mathcal{F}_{T,x}= \{s_x \in \mathcal{F}_x \vert s_{\eta} = 0 \}$$

where $s_{\eta}$ is the canonical image of $s_x$ under induced localisation map $\mathcal{F}_x \to \mathcal{F}_x \otimes \mathcal{O}_{P^1,\eta} = \mathcal{F}_x \otimes k(T)$

How does it imply that almost everywhere $\mathcal{F}_{T,x}$ is zero.

Remark: I guess the argument I'm looking for works generally for irreducible curves.

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  • $\begingroup$ If $s_{\eta}=0$, then $s$ is $0$ on some nonempty open set, which means it's supported on a proper closed subset. Do you see where to go from here? $\endgroup$ – KReiser Apr 22 at 17:39
  • $\begingroup$ @KReiser: I think the first conclusion about open sets is by Nakayama. Futhermore every closed set not containing the generic point should be finite (zero locus of polynomial(s) with one variable... guess that this is the point why it doesn't works for higher dimension). And finally it suffice to consider only finitely elements $s_x$ (=the generators) of the torsion section and build the union of their (finite) support,right? Or do you mean another argument? $\endgroup$ – KarlPeter Apr 22 at 17:49

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