An embedding of a projective variety.

I came across with a statement

Let $X\rightarrow \mathbb{P}^n$ be a map defined by a linear system $|L|$ for some line bundle $L$ on $X$. It is embeding if $H^0(\mathbb{P}^n,O(1))\rightarrow H^0(X,O(1))=H^0(X,L)$ is injective.

Here $O(1)$ stands for the line bundle defined by hyperplane on $\mathbb{P}^n$, and the induced one on $X$ (hence it coincides with $L$).

The statement above means that if global sections of $O(1)$ can be distinguished after restricting to the image of $X$, then $X\rightarrow \mathbb{P}^n$ is embedding. It seems true but I am not so convinced. How should I understand the statement above?

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The logical quantifiers of the statement are ambiguous. You seem to interpret it as "Suppose some complete linear system determines a map $X\to \mathbb{P}^n$. If the pullback on global sections in injective, then the map is an embedding." I interpret the statement as "Suppose there is an embedding $X\to \mathbb{P}^n$. If the pullback on global sections in injective, then it is given by a complete linear system." I find both of these statements dubious, though, so I'm not sure what is meant. – Matt Oct 29 '12 at 2:45
You are right. I will make my statement clearer. – M. K. Oct 29 '12 at 3:49
I don't quite see the relation between the injectivity of $\alpha:H^0(\mathbb P^n,O(1))\to H^0(X,L)$ and the fact that $L\cong O_X(1)$ defines an embedding. Indeed, $\alpha$ injective means that $X$ is not contained in a hyperplane, while the second condition means that the morphism $X\to \mathbb P(H^0(X,L)^\vee)$ corresponding to $|L|$ is a closed immersion (that is: $L$ is very ample and generated by $n+1$ global sections). By completeness, shouldn't we have $h^0(X,L)=n+1$? But this is also $h^0(\mathbb P^n,O(1))$... Where did you read that statement? – Brenin Oct 29 '12 at 10:28
@atricolf This was one of my concerns. Namely, injectivity seems to say there is a dimension $n$ linear system $\mathfrak{d} \subset |L|$ defining the map (possibly properly contained if $H^0(X, L)$ is larger than $n+1$ dimensional. The thing that concerned me much more is that in order for $X\to \mathbb{P}^n$ to be a closed immersion, $|L|$ must separate points and tangent vectors and I don't see how the global sections map being injective will imply that. – Matt Oct 29 '12 at 16:47

The statement is false. A correct statement is:

Suppose $X$ is reduced. Then $H^0(\mathbb P^n,O(1))\to H^0(X,L)$ is injective if and only if the image of the morphism $f: X\to \mathbb P^n$ is not contained in a hyperplane.

Proof. Let $s\in H^0(\mathbb P^n,O(1))$, let $t\in H^0(X, L)$ be its image. Then $t=0$ if and only if $t(x)=0$ for all $x\in X$ (we use the fact that $X$ is reduced). Now $t(x)=s(f(x))$. So $t=0$ if and only if $f(X)\subseteq (s)_0$ (the zero locus of $s$). As $(s)_0$ is a hyperplane if $s\ne 0$, our statement is proved.

In particular, for any finite morphism from a curve $X$ to $\mathbb P^1$, the map on the $H^0$'s is always injective, but $f$ is not an embedding in general.

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