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my question concerns a very ample line bundle $L$ on a projective k-scheme. It gives a closed immersion

$i:X \rightarrow P^{N-1}_{k}$

to some projective space over k.

It can be viewed as induced by global sections $s_ 1,...,s_N$ of $L$.

Why do people always say that these sections form a basis of $H^{o}(X,L)$? Surely they generate $L$ in the sense of: generating in every stalk. But why is $N=dim(H^{o}(X,L)$?

Thanks a lot!

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Dear Descartes: This isn't true (take $X$ a point, for instance; then the pull-back $L$ of $\mathcal{O}(1)$ is the trivial line bundle and has only one section). – Akhil Mathew Aug 18 '11 at 18:39
The morphism $i$ is given by $x\in X \mapsto [s_1(x): \cdots : s_N(x)]$ by definition, where $(s_i)$ is a basis of $H^0(X,L)$. So by definition, $i$ goes from $X$ to $\mathbb P^{N-1}$ (with your notations). Note that this morphism does depend on a choice of a basis, but all such images of $X$ under $i$ are conjugated by an element of $PGL_N$. A more intrisic way to define $i$ is to set $i(x)=\mathrm{Ker}(s\mapsto s(x))$ which defines $i:X\to \mathbb P(H^0(X,L)^*)$. – Henri Aug 18 '11 at 20:51
Dear Descartes, your notation seems incoherent. Either call your sections $s_0,...,s_N$ or take $\mathbb P^{N-1}$ as codomain of your immersion (as correctly did Henri). – Georges Elencwajg Aug 18 '11 at 21:25
Dear Akhil, I'm not sure I understand your objection. If $X$ is a point, then it gets embedded into $\mathbb P^0_k$. Is Descartes's slight confusion betwen $N$ and $N-1$ which you are hinting at? – Georges Elencwajg Aug 18 '11 at 21:28
Sorry about that confustion, I already corrected it. – Descartes Aug 19 '11 at 5:46

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