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Let $F\rightarrow \mathbb{P}^{1}$ be a vector bundle, and let $\mathcal{F}$ be its sheaf of sections. I have read that the sheaf $\mathcal{F}(m)/\mathcal{F}$ is a skyscraper sheaf, but I do not know how to prove it.

I do not know if it is of any help, but I am going to show what I have thought until now. We have $$ \mathcal{F}(m)=\mathcal{F}\otimes\mathcal{O}_{\mathbb{P}^{1}}(m). $$ Since $H^{0}(\mathbb{P}^{1},\mathcal{O}_{\mathbb{P}^{1}}(m))\simeq K[X_{0},X_{1}]_{m}$, I guess that we may consider $(X_{1})$ as the distinguished point of $\mathbb{P}^{1}$ with respect to $\mathcal{F}(m)/\mathcal{F}$. Nevertheless, I do not know how to continue.

Any help would be appreciated.

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  • $\begingroup$ To speak of $F(m)/F$ you should first chose a map $F \to F(m)$. What map do you have in mind? $\endgroup$ – Sasha Oct 23 '16 at 10:11
  • $\begingroup$ @Sasha I am not sure, but it is induced in some way by the map $\mathcal{O}_{\mathbb{P}^{1}}\rightarrow \mathcal{O}_{\mathbb{P}^{1}}(m)$ induced by $K[X_{0},X_{1}]_{0}\rightarrow K[X_{0},X_{1}]_{m}, k\mapsto X_{1}^{m}\cdot k$ with our interpretation. $\endgroup$ – H. Jackson Oct 23 '16 at 12:38
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This works for any curve $C$. If $p$ a point on $C$ there is a short exact sequence $$ 0 \to \mathcal{O}(-p) \to \mathcal{O} \to k(p) \to 0. $$ Take a tensor product of this exact sequence with $\mathcal{F}(p)$. It stays exact because $F$ is a vector bundle. So, you get a short exact sequence $$ 0 \to \mathcal{F} \to \mathcal{F}(p) \to \mathcal{F}(p)\otimes k(p) \to 0. $$

Sheaf $\mathcal{F}(p)\otimes k(p) \cong \mathcal{F} \otimes k(p)$ is a skyscraper sheaf supported on $p$.

On $\mathbb{P}^1$ we have $\mathcal{O}(1) = \mathcal{O}(p)$ and $\mathcal{O}(m) = \mathcal{O}(mp)$, where $m \in \mathbb{Z}$. For positive $M$ we generalize previous argument by starting with short exact sequence $$ 0 \to \mathcal{O}(-mp) \to \mathcal{O} \to k_m(p) \to 0. $$ where $k_m(p)$ is a skyscraper sheaf supported on $x$ of length $m$. Then we tensor this short exact sequence with vector bundle $\mathcal{F}(mp)$.

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  • $\begingroup$ Maybe it is not clear in the statement of my question, but $m$ is supposed to be an integer (furthermore, I want it to be positive). Let $p$ be a point of $C$. Could we generalize your idea for the divisor $m\cdot p$? Thank you! $\endgroup$ – H. Jackson Nov 3 '16 at 19:20
  • $\begingroup$ Sorry, I confused twisting by a point (as divisor) and twisting by number (i.e. by hyperplane divisor), but on $\mathbb{P}^1$ this is the same thing! And this argument also can be generalized to the case you interested in. $\endgroup$ – Alex Nov 3 '16 at 21:04

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