Divisor on cubic curve $y^2z=x^3-xz^2$

Question

Let $X$ be the nonsingular cubic curve $y^2z=x^3-xz^2$. Let $P_0=(0,1,0)$. Then the line bundle associated to $3P_0$ is $O_X(1)$. I already know that $3P_0$ is produced by cut the curve with $z=0$. But why this is enough to get it is $O_X(1)$.

Answer 1

Your approach is correct. One nice thing about the divisor and line bundle correspondence is that the divisor is the divisor of any rational section. Sections of $\mathcal{O}_{\mathbb{P}^2}(1)$ pullback to sections of $\mathcal{O}_X(1)$, and the associated divisor of a pulled-back section gives you a divisor $D$ such that $\mathcal{O}_X(D)=\mathcal{O}_X(1)$. You did this computation and everything works!