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Let $E$ be an rank 1 coherent torsion free sheaf on $\mathbb{P}^3$, such that $h^0(E) = 2$, and $c_1(E) = 1$. I want to show that $E \simeq I_Y(1)$ where $I_Y$ is an ideal sheaf of a closed subscheme of $ Y \subseteq\mathbb{P}^3$ of codimension 2.

I tryed to use koszul resolution, because since $h^0(E) = 2$ a think in considering two sections $ s_1$ and $s_2 \in H^0(E)$, $Y = (s_1,s_2)_0$ and use the resolution

$0 \to \wedge^2E \to E \to^{(s_1,s_2)} \mathcal{O}_{\mathbb{P}^3} \to \mathcal{O}_{Y} \to 0$

Then we will have the exact sequence

$0 \to \wedge^2E \to E \to I_Y \to 0$

my question is how to determine $\wedge^2E$ (since it is not a vector bundle), and where the twist appear in $I_Y$, if this approach is right.

Thank you.

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    I assume that your coherent sheaf is torsion free, otherwise what you ask is wrong. If torsion free, you have a natural inclusion $E\to E^{**}$ and $E^{**}$ is a line bundle with the same first chern class and thus $E^{**}=\mathcal{O}_{\mathbb{P}^3}(1)$. The rest should be clear. – 2017-02-21
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    Mr. @Mohan you right, the sheaf should be torsion free. I will edit it. But it is still not clear how to go further, I consider a resolutoin of the sections $(s_1,s_2)$ as sections of $E^{**}$ instead of $E$? – 2017-02-21
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    As $E$ embeds in $E^{**}$ and $E^{**}$ is isomorphic to $O(1)$, $E(-1)$ is a subsheaf of $O$, so... – 2017-02-21
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    Now I get it! Since E is not locally free it is an ideal sheaf, and hence $E(-1) \simeq I$. Thank you both. – 2017-02-21

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