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I get stuck with what Example II.7.6.3 (Page 156 in Hartshorne's book Algebraic Geometry) claims. Let me recite the example here: Let $X$ be an elliptic curve $y^2z = x^3 - xz^2$ in $\mathbf P_k^2$ defined over an algebraically closed field $k$ with characteristic $\ne 2$. Let ${\scr L}={{\scr L}}(P_0)$ be the invertible sheaf corresponding to the divisor $(P_0)$ where $P_0=\infty$ (I think the claim is true for any closed point on the curve). Then this example claims $\scr L$ is not very ample because $\scr L$ is not generated by global sections and the reason why it is not generated by global sections is "if it were, then $P_0$ would be linearly equivalent to some other point $Q \in X$". I do not see why this is true. Could somebody give some hints, or answers?

Thank you CamSar to provide a detailed argument which gives further insights. Here I want to prove CamSar's argument "$h^0((X, \mathcal O_X(P_0))) \ge 3$ implies that there exists an effective divisor $Q$ linearly equivalent to $P_0$". Since the elliptic curve is defined over a field $k$, $H^0(X, \mathcal O_X(P_0))$ is a vector space over $k$ with finite dimension $\ge 3$ by assumption. Fix a set of generators of this vector space, then each element induces a positive divisor $D$ equivalent to $(P_0) = (\infty)$. Since for any ration function $f$, $\mathrm{deg} (\mathrm{div}(f)) = 0$, $D=(Q)$ for some closed point $Q \in X$. Since the dimension of the vector space over $k$ is $\ge 3$, at least one element of generators induces a divisor $(Q)$ with $Q \ne P_0$.

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OK, here is a hint. You can find an effective divisor linearly equivalent to $P_0$ but disjoint from it. –  Donu Arapura May 2 at 13:04

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If $\mathcal{O}_X(P_0)$ is very ample then it induce an embedding $f:X\rightarrow\mathbb{P}(H^{0}(X,\mathcal{O}_X(P_0))\cong\mathbb{P}^n$. Therefore $n\geq 2$ and $h^{0}((X,\mathcal{O}_X(P_0)))\geq 3$. Now, $h^{0}((X,\mathcal{O}_X(P_0)))\geq 3$ implies that there exists an effective divisor $Q$ linearly equivalent to $P_0$. In particular $P_0-Q$ is the divisor of a rational function $g = \frac{p}{q}$ on $X$. The rational map $$\phi:X\dashrightarrow\mathbb{P}^1,\: x\mapsto [p(x):q(x)]$$ is birational. This implies that $X$ is rational. A contradiction because $g(X) = 1$. Roughly speaking we are saying here that if a divisor of degree one on a curve $X$ is very ample then the curve $X$ is mapped to a curve of degree one ($deg(P_0) = 1$) in a projective space, that is a line. Therefore $g(X) = 0$. In this way you get the following.

Let $X$ be a curve and $P_0\in X$ be a point. Then $\mathcal{O}_X(P_0)$ is very ample if and only if $g(X) = 0$.

From the first part of the argument you see that $\mathcal{O}_X(D)$ very ample imply $h^{0}((X,\mathcal{O}_X(D)))\geq 3$. You may use Riemann-Roch theorem (pag 294) with $D = P_0$. In this case $deg(K_X-P_0) = 2g-2-1 = -1$. Then $P_0$ is non-special and $h^{1}(X,\mathcal{O}_X(P_0)) = 0$. By Riemann-Roch yu get $$\chi(\mathcal{O}_X(P_0)) = h^0(\mathcal{O}_X(P_0)) = deg(P_0)-g+1 = 1.$$ Again you find a contradiction.

In general, a divisor $D$ on a curve $X$ is ample if and only if $deg(D)>0$. If $g(X) = 0$ then $D$ is ample if and only if it is very ample if and only if $deg(D)>0$. However, if $g(X)=1$ then $D$ is very ample if and only if $deg(D)\geq 3$. Indeed, by Riemann-Roch, $deg(D)\leq 2$ implies $h^{0}(X,\mathcal{O}_X(D)) = deg(D)\leq 2$. Therefore the morphism induced by $D$ can not be an embedding. On the other hand if $deg(D)\geq 3\geq 2g+1$ then $D$ is very ample.

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Thank you CamSar, this helps to understand this subject deeper. Also I give my thanks to Donu Arapura for the hints with reminds me the linear system of a divisor. –  user41541 May 3 at 17:11

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