I am reading Vakil's algebraic geometry. He gives a definition of very ample line bundle as followings:

Suppose $\pi: X \rightarrow \operatorname{Spec} A$ is a proper morphism, and $\mathscr{L}$ is an invertible sheaf on X. Then we say $\mathscr{L}$ is very ample over $A$ if $X \cong$ Proj $S_{\bullet}$ where $S_{\bullet}$ is a finitely generated graded ring over $A$ generated in degree 1 and $\mathscr{L} \cong \mathscr{O}_{\text {Proj} S_{\bullet}}(1)$.

Then there is one exercise

Show that $\mathscr{L}$ is very ample if and only if there exist a finite number of global sections $s_{0}, \ldots s_{n}$ of $\mathscr{L},$ with no common zeros, such that the morphism $$ \left[\mathrm{s}_{0}, \ldots, \mathrm{s}_{\mathrm{n}}\right]: \mathrm{X} \rightarrow \mathbb{P}_{\mathrm{A}}^{\mathrm{n}} $$is a closed embedding.

However, I think the exercise is wrong if we use his definition. For example, considering the closed embedding $ \mathbb{P}^{1} \rightarrow \mathbb{P}^{2}$ given by $[x: y] \mapsto\left[x^{2}: x y: y^{2}\right]$ it pulls $ \mathcal{O}_{\mathbb{P}^{2}}(1) \text { to } \mathcal{O}_{\mathbb{P}^{1}}(2)$. Obviously, $\mathcal{O}_{\mathbb{P}^{1}}(2)$ satisfies the condition in the exercise, but it is not isomorphic to $\mathcal{O}_{\mathbb{P}^{1}}(1)$.

So I think the definition of very ample line bundle is not proper. And actually we should use the exercise as the definition.

Am I right? If not, could you tell me the exact definition of very ample line bundle? Thank you very much.

  • $\begingroup$ I've edited your post to remove a \text tag inside the latex formatting that was causing a part of your question to be unreadable. Please just leave the latex environment for this in the future. $\endgroup$ – KReiser Nov 7 '19 at 21:05

The definition is fine - it's an existence statement, not a uniqueness statement. To fix your issue, we write $\Bbb P^1_k$ as $\operatorname{Proj} k[x^2,xy,y^2]$ instead of $\operatorname{Proj} k[x,y]$. As $\operatorname{Proj} k[x^2,xy,y^2]$ has $\mathcal{O}(1)=k\langle x^2,xy,y^2\rangle$, we see that the definition is satisfied.

This generalizes: for any graded ring $R$, the graded ring $R^{(d)}$ defined by $R^{(d)}=\bigoplus_{n\in\Bbb Z_{\geq 0}} R_{dn}$ has the same $\operatorname{Proj}$ as $R$. So if $\mathcal{O}_{\operatorname{Proj} R}(1)$ is very ample, then $\mathcal{O}_{\operatorname{Proj} R}(d)$ is also very ample for any $d>0$.

  • $\begingroup$ It is clear. Thank you very much! $\endgroup$ – Mike Nov 7 '19 at 21:23
  • $\begingroup$ By the way, I think you want to say that if $\mathcal{O}_{\mathrm{Proj} R^{(d)}}$$ (1)$ is very ample, then $\mathcal{O}_{\mathrm{Proj} R}$$ (d)$ is very ample? $\endgroup$ – Mike Nov 7 '19 at 21:47

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