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Jun 3, 2016 at 17:18 vote accept A.G
Jun 1, 2016 at 11:16 answer added Jason Starr timeline score: 5
Jun 1, 2016 at 10:53 comment added Jason Starr I can see how to make a counterexample with $\mathcal{F}$ a pure, torsion-free sheaf of rank $1$, and $X$ is an integral, local complete intersection scheme (that is not normal). Namely, $X$ is a $\mathbb{P}^1$-bundle over a nodal plane cubic, and $\mathcal{F}$ is the pushforward of the structure sheaf of the normalization. However, I do not see an example with $\mathcal{F}$ an invertible sheaf.
Jun 1, 2016 at 10:31 comment added Lazzaro Campeotti It must be contagious. Note that the version without "invertible" was asked and answered here: math.stackexchange.com/questions/1805890/… with the same counterexample.
Jun 1, 2016 at 10:25 comment added Jason Starr I missed that hypothesis!
Jun 1, 2016 at 10:17 comment added Lazzaro Campeotti @JasonStarr: the question asks for $F$ invertible.
Jun 1, 2016 at 10:14 comment added Jason Starr 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.
Jun 1, 2016 at 6:21 comment added A.G @Ben Lim: Isn't it even cyclic in that case?
Jun 1, 2016 at 6:03 review First posts
Jun 1, 2016 at 6:57
Jun 1, 2016 at 6:01 comment added A.G Sorry if too elementary, but I had no answer in math.stackexchange.com/questions/1805972/…
Jun 1, 2016 at 6:00 history asked A.G CC BY-SA 3.0