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Let $X$ be a projective scheme over a noetherian ring, $\mathcal F$ an invertible sheaf on $X$, and $U$ an arbitrary open subset of $X$. Is $\Gamma(U,\mathcal F)$ a $\Gamma(U,\mathcal O_X)$-module of finite type?

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  • $\begingroup$ Sorry if too elementary, but I had no answer in math.stackexchange.com/questions/1805972/… $\endgroup$
    – A.G
    Commented Jun 1, 2016 at 6:01
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    $\begingroup$ @Ben Lim: Isn't it even cyclic in that case? $\endgroup$
    – A.G
    Commented Jun 1, 2016 at 6:21
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    $\begingroup$ Ben Lim's example is easily modified to a true counterexample. Let $X$ be $\mathbb{P}^2_k$, let $\mathcal{F}$ be the coherent sheaf that is the pushforward of the structure sheaf of a line $\mathbb{P}^1_k\subset \mathbb{P}^2_k$. Let $\infty\in \mathbb{P}^1$ be a $k$-point. Let $U$ be the open complement of that $k$-point inside $X$. Then $\Gamma(U,\mathcal{O}_X)$ is canonically $k$, but $\Gamma(U,\mathcal{F})$ is $k[t,t^{-1}]$ as a $k$-algebra, thus not a finite dimensional $k$-vector space. $\endgroup$ Commented Jun 1, 2016 at 10:14
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    $\begingroup$ @JasonStarr: the question asks for $F$ invertible. $\endgroup$ Commented Jun 1, 2016 at 10:17
  • $\begingroup$ I missed that hypothesis! $\endgroup$ Commented Jun 1, 2016 at 10:25

1 Answer 1

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Edit. There was an issue with my first example. The second example is fine (but unfortunately it does not work over an arbitrary field).

There is a more "conventional" example as well where $X$ is regular. Begin with $C$ a curve of genus $g\geq 1$. Let $\mathcal{L}$ be an invertible sheaf on $C$ that is algebraically equivalent to zero, yet not torsion (I guess that rules out finite fields). Let $\mathcal{M}$ be an invertible sheaf of degree $2g-1$. Consider the sheaf of quasi-coherent $\mathcal{O}_C$-algebras on $C$, $$\mathcal{A}=\text{Sym}^\bullet_{\mathcal{O}_C}(\mathcal{L}\lambda \oplus \mathcal{M}\mu) = \bigoplus_{(l,m)\in \mathbb{Z}_{\geq 0}^2} \left(\mathcal{L}^{\otimes l}\otimes_{\mathcal{O}_C} \mathcal{M}^{\otimes m}\right) \lambda^l\mu^m, $$ where $\lambda$ and $\mu$ are just placeholders. Let $U$ be the relative Spec construction, $U=\text{Spec}_C \mathcal{A}$. This is an $\mathbb{A}^2$-bundle over $C$ that compactifies to a $\mathbb{P}^2$-bundle $X$ over $C$. Now let $\mathcal{F}$ be the invertible ideal sheaf on $U$ whose corresponding sheaf of ideals in $\mathcal{A}$ is the (locally) principal ideal $\mathcal{M}\mu\cdot \mathcal{A}$.

The point is that $\Gamma(U,\mathcal{O}_X)$ equals $\Gamma(C,\mathcal{A})$, and this is a $\mathbb{Z}_{\geq 0}^2$-graded $k$-algebra $$\bigoplus_{(l,m)} \Gamma(C,\mathcal{L}^{\otimes l}\otimes_{\mathcal{O}_C}\mathcal{M}^{\otimes m})\lambda^\ell\mu^m,$$ whose nonzero graded pieces occur precisely for $(l,m)=(0,0)$ and for $m\geq 1$. In particular, this subsemigroup of $\mathbb{Z}_{\geq 0}^2$ is not finitely generated. Similarly, the ideal $\Gamma(U,\mathcal{F})$ in $\Gamma(U,\mathcal{O}_X)$ is a homogeneous ideal that has nonzero graded pieces precisely for $m\geq 1$. So this ideal cannot be finitely generated as an ideal in $\Gamma(U,\mathcal{O}_X)$.

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  • $\begingroup$ Thank you for your time, and sorry for not accepting the answer just now (it will take me some time to understand it well). $\endgroup$
    – A.G
    Commented Jun 1, 2016 at 22:16

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