Let $X$ be a smooth degree $d$ $(d \ge 5)$ surface in $\mathbb{P}^3$. Let $D$ be an effective Cartier divisor (hence locally of complete intersection) on $X$ and $D_{red}$ the associated reduced scheme which is also a Cartier divisor on $X$. Then, we have a natural inclusion of ideal sheaves, $0$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X(-D_{red})$. mathcal{O}_X(-D_{red}).$$ Taking the dual we have \mathcal{O}_X(D_{red}) \mathcal{O}_X(D_{red}) \to \mathcal{O}_X(D) \to 0. 0.$$ Since, $\mathcal{O}_X(D_{red})$ and$\mathcal{O}_X(D)$are locally free sheaves of rank$1$, this would imply that the kernel of the latter map is zero. This would imply that $\mathcal{O}_X(D_{red}) \cong \mathcal{O}_X(D)$. This result is very surprising. Is there a mistake in the proof or is there an explaination for this behaviour? 3 added 4 characters in body Let$X$be a smooth degree$d(d \ge 5)$surface in$\mathbb{P}^3$. Let$D$be an effective Cartier divisor (hence locally of complete intersection) on$X$and$D_{red}$the associated reduced scheme which is also a Cartier divisor on$X$. Then, we have a natural inclusion of ideal sheaves, $0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X(-D_{red})$.  Taking the dual we have $\mathcal{O}_X(D_{red}) \to \mathcal{O}_X(D) \to 0$.  Since, $\mathcal{O}X(D{red})$\mathcal{O}_X(D_{red})$ and $\mathcal{O}_X(D)$ are locally free sheaves of rank $1$, this would imply that the kernel of the latter map is zero. This would imply that $\mathcal{O}X(D{red}) \mathcal{O}_X(D_{red}) \cong \mathcal{O}_X(D)$. mathcal{O}_X(D)$. This result is very surprising. Is there a mistake in the proof or is there an explaination for this behaviour? 2 added 3 characters in body Let$X$be a smooth degree$d(d \ge 5)$surface in$\mathbb{P}^3$. Let$D$be an effective Cartier divisor (hence locally of complete intersection) on$X$and$D_{red}$the associated reduced scheme which is also a Cartier divisor on$X$. Then, we have a natural inclusion of ideal sheaves, $0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X(-D_{red})$.  Taking the dual we have $\mathcal{O}_X(D_{red}) \to \mathcal{O}_X(D) \to 0$.  Since,$\mathcal{O}X(D{red})$and$\mathcal{O}_X(D)$are locally free sheaves of rank$1$, this would imply that the kernel of the latter map is zero. This would imply that$\mathcal{O}X(D{red}) \cong \mathcal{O}_X(D)\$. This result is very surprising. Is there a mistake in the proof or is there an explaination for this behaviour?