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Let $\pi \colon X \to \mathbb{P}^n$ be a closed embedding given via an invertible sheaf $\cal L$ with global sections $s_0, \dots, s_n$. Thus ${\cal L} \cong \pi^* {\cal O}_X$.

Why is ${\cal L} \cong {\cal O}_X(1)$?

In other words, why is saying that an invertible sheaf gives a closed embedding the same as saying that it is isomorphic to ${\cal O}_X(1)$ (= being very ample)?

(This is a part of Exercise 16.6.A of Vakil's notes on algebraic geometry.)

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  • $\begingroup$ By definition $L$ is very ample means that the corresponding rational maps $X \to \Bbb P H^0(X,L)$ is defined everywhere and an embedding. Unless you have another definiton ? $\endgroup$ – Nicolas Hemelsoet Feb 26 at 12:12
  • $\begingroup$ Vakil defines very ample as $X \cong \mathrm{Proj} S_{\cdot}$ and ${\cal L} \cong {\cal O}_{X}(1)$. $\endgroup$ – darko Feb 26 at 13:22
  • $\begingroup$ Although now I am starting to doubt that the two are really equivalent. Doesn't the closed embedding $\mathbb{P}^1 \to \mathbb{P}^2$ given as $[x:y] \mapsto [x^2:xy:y^2]$ pull ${\cal O}_{\mathbb{P}^2}(1)$ to ${\cal O}_{\mathbb{P}^1}(2)$? So maybe we are only considering standard closed embeddings coming from graded ring epimorphisms? $\endgroup$ – darko Feb 26 at 13:26

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