Theorem 7.3.17 in (Qingliu) Algebraic Geometry and Arithmetic curve

Question

I don't understand the logic [As $F$ is a finite scheme, $\mathcal{O}_X(E)|_F\cong \mathcal{O}_F$]. Let $i:F\to X$ be the closed immersion, then the isomorphism really is $\mathcal{O}_X(E)\otimes_{\mathcal{O}_X}i_* \mathcal{O}_F\cong i_* \mathcal{O}_F$. I can see that this is true if $E\cap F=\emptyset$. But lemma 3.6 that was used to construct $E$ and $F$ didn't demand that they have no intersection.

Answer 1

By lemma 7.1.29, we have $\mathcal{O}_X(E)|_F\cong \mathcal{O}_F(E|_F)$. Since $F$ is a finite scheme, every Cartier divisor over $F$ is principal (jsut unwrap the definitions of Cartier divisor) so $\mathcal{O}_F(E|_F)\cong\mathcal{O}_F$.