I want to prove the following statement:
Let $X$ be a smooth hypersurface in $\mathbb P^n$, $D$ be an effective divisor on $X$. If the associated line bundle $\mathcal L(D)$ is equivalent to $\mathcal O_X(1)$ (which called the hyperplane line bundle), then $D$ is cut by some hyperplane in $\mathbb P^n$.
My approach is: If $i^* \mathcal L_1\cong i^* \mathcal L_2$, then I want to show $\mathcal L_1\cong\mathcal L_1$. In general this is not true, but I think it might be true in our case of smooth hypersurfaces. But I don't know how to prove this.
Any hint or reference would be helpful.
I will assume $n\geq 2$, otherwise everything is clear and let $d=\deg X$. Then we have an exact sequence, $0\to O_{P}(-d)\to O_P\to O_X\to 0$ (I have used $P$ for projective space and $O$ for structure sheaf), which after twisting by $O(1)$ and taking global sections gives $H^0(O_P(1))\to H^0(O_X(1))\to H^1(O_P(-d+1))$. The last term is zero and since $\mathcal{L}(D)=O_X(1)$ and $D$ is effective, $D$ is the zeroes of a section of $O_X(1)$. This section can be lifted by the above surjectivity to a section of $O_P(1)$ and then the rest is clear.