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 $Q \ne P_0$.