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Suppose $X$ is a Noetherian, separated, nonsingular integral scheme of dimension $1$.

In this post, I learned that if $V$ is a vector bundle (locally free sheaf) ove $X$, then any nonzero global section $s$ gives a line subbundle $L$ (means the line bundle $L\subset V$, and $V/L$ is still a locally free sheaf). But what if $V$ has no nonzero global sections? Can we still get a line subbundle of $V$?

Any help is appreciated thanks!

Richard
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  • Yes. Hint: twist. – KReiser Mar 04 '22 at 07:53
  • @KReiser Thanks, but what do you mean by "twist"? Could you explain it more concretely? – Richard Mar 04 '22 at 07:57
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    I fully believe in your capacity to find the answer by pursuing that hint and looking through materials you have already used & referenced in our previous interactions on this site. I don't want to deprive you of the learning experience or hinder the development of your research abilities, so I'm not going to elaborate at this time. – KReiser Mar 04 '22 at 08:18
  • @KReiser Thank you, I want to state my understanding below and I wonder if you would like to check whether the disgussion is right. Suppose $X$ is projective over some noetherian ring. Since $\mathcal O(1)$ is ample, there is a $n$ s.t. $V(n)$ is generated by global sections, in particular there is a nonzero section, then I can use the argument in that post. But the reason why I ask the meaning of "twist" is that I didn't assume $X$ is projective over some noetherian ring. The condition is stated in the first line of the post. So there is no $\mathcal O(1)$, then what do you mean by twist? – Richard Mar 04 '22 at 08:56
  • There are certain ways in which curves are simpler than schemes of higher dimension; in particular, you should be able to get around your lack of assuming that $X$ is projective. – Tabes Bridges Mar 04 '22 at 11:08

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A week later, let's put this to rest.

The key observation, as Tabes Bridges notes in the comments, is that schemes of dimension one have nice properties. In fact, any noetherian separated scheme of dimension one has an ample invertible sheaf $\mathcal{L}$, see Stacks 09N7. In particular, this means that for any quasi-coherent sheaf of finite type $\mathcal{F}$ on $X$ there is an integer $n_0$ so that for all $n\geq n_0$, $\mathcal{F}\otimes\mathcal{L}^{\otimes n}$ is globally generated. Therefore we may solve the problem for $V\otimes \mathcal{L}^{\otimes n}$, obtain a sub line bundle, and then twist back to end up with a sub line bundle of $V$.

KReiser
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  • Thanks for your answering. But (in my assumption in the first line) it's not of finite type over a field, so the Hartshorne's treatment and quasi procjective properties can not be applied here. Are there other ways to "twist" a vector bundle in my general setting to get a nonzero global section? – Richard Mar 12 '22 at 03:07
  • @Richard Apologies, I read it as looking for varieties over a field. Not sure how that happened. But your claim is true, and I've fixed my answer to address what you're actually asking about. – KReiser Mar 12 '22 at 07:02