Very Ample sheaves Let $X$ be the non singular cubic curve $y^2z=x^3-xz^2$ in projective space of dimension 2. Let $L$ be the invertible sheaf $L(P_0)$. How does $L(P_0)$ not being generated by global sections imply that there is a point $Q$ which is equivalent to $P_0$?
I am having trouble in understanding this. It is Example(7.6.3) on Page 156 of Hartshorne.
 A: The text says

If $L(P_0)$ were generated by global sections, then $P_0$ would be linearly equivalent to some other point $Q \in X$.

This goes as follows: Suppose $L(P_0)$ is generated by global sections. Then there is a global section $s \in \Gamma(X, L(P_0))$ which generated $L(P_0)$ at the point $P_0$. The effective Cartier divisor of zeroes
$$D = (s)_0 = \{\, P \in X \,|\, s_P \in \mathfrak m_P L(P_0) \,\}$$
is then linearly equivalent to $P_0$ by Proposition 7.7. I claim that $D$ consists of one point only, which has multiplicity $1$. To see that, consider the global section $s^{\otimes 3} \in L(3P_0) = \mathcal O_X(1)$. Since it is a global section of a very ample sheaf (which is very ample with respect to the embedding $X \subset \mathbb P^2$), the associated Cartier divisor $(s^{\otimes 3})_0 = 3D$ is linearly equivalent to a Cartier divisor obtained by intersecting $X$ with a line in $\mathbb P^2$. Hence we get $$3 \deg D = \deg (s^{\otimes 3})_0 = \deg X = 3,$$
and thus $\deg D = 1$. As $D$ is effective, there is a point $Q \in X$ such that $D = 1 \cdot Q$.
