3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Sorry if too elementary, but I had no answer in math.stackexchange.com/questions/1805972/… $\endgroup$ – A.G Jun 1 '16 at 6:01
  • 4
    $\begingroup$ @Ben Lim: Isn't it even cyclic in that case? $\endgroup$ – A.G Jun 1 '16 at 6:21
  • 1
    $\begingroup$ 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. $\endgroup$ – Jason Starr Jun 1 '16 at 10:14
  • 2
    $\begingroup$ @JasonStarr: the question asks for $F$ invertible. $\endgroup$ – potentially dense Jun 1 '16 at 10:17
  • $\begingroup$ I missed that hypothesis! $\endgroup$ – Jason Starr Jun 1 '16 at 10:25
3
$\begingroup$

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 $\Gamma(U,\mathcal{O}_X)$.

$\endgroup$
  • $\begingroup$ 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 Jun 1 '16 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.