Let $0\to V'\to V\to M \to 0$ be an exact sequence with $V$ a vector bundle over a scheme $B$, $M \in \operatorname{Pic}(B) $. Let $H:=\mathbb{P}(V')$ be the Cartier divisor, let $p: \mathbb{P}(V) \to B$ be the projection. Let $X:=\mathbb{P}(V)$ Why is
$\mathcal O_{X}(H) = \mathcal O_{X}(1) \otimes p^{*} M $?
Any reference is much appreciated.
There is a tautological embedding $$ \mathcal{O}_X(-1) \hookrightarrow p^*V $$ and an exact sequence $0 \to p^*V' \to p^*V \to p^*M \to 0$. It is clear that $H$ is the zero locus of the composition $$ \mathcal{O}_X(-1) \to p^*V \to p^*M, $$ which is nothing but the zero locus of the corresponding section of the line bundle $$ \mathcal{O}_X(-1)^\vee \otimes p^*M \cong \mathcal{O}_X(1) \otimes p^*M. $$ This means that $\mathcal{O}_X(H) \cong \mathcal{O}_X(1) \otimes p^*M$.