I am struggling with Proposition 8.1.12e of Liu's Algebraic geometry and arithmetic curves. The setup is as follows: let $X=\text{Spec}A$, $I=(f_1,\dots,f_n)\subseteq A$ an ideal, and let $$\tilde A=\bigoplus_{d\ge0}I^d.$$ We distinguish $I\subseteq\tilde A_0=A$ and $I=\tilde A_1$ by denoting with $t_1,\dots,t_n\in\tilde A_1$ the elements corresponding to $f_1,\dots,f_n\in A$ respectively. If $\mathscr{I}=\tilde{I}$ in $X$ and $\tilde X=\text{Proj}\tilde A$, the proof of Proposition 8.1.12e says that $\mathscr{I}\mathcal{O}|_{D_+(t_i)}$ is generated by $f_i$ for every $1\le i\le n$ because $f_j=f_i(t_jt_i^{-1})$, i.e. $$\mathscr{I}\mathcal{O}|_{D_+(t_i)}=f_i\mathcal{O}_{D_+(t_i)}$$ Then, the claim is that $$\mathscr{I}\mathcal{O}_{D_+(t_i)}=\mathcal{O}_{D_+(t_i)}(1)$$ I don't understand this for two reasons:
- Since $D_+(t_i)$ is an affine scheme, how can you take the twist $\mathcal{O}_{D_+(t_i)}(1)$? Isn't it more correct to write it as $\mathcal{O}(1)|_{D_+(t_i)}$?
- Assuming $\mathcal{O}_{D_+(t_i)}(1)=\mathcal{O}(1)|_{D_+(t_i)}$, if I try to calculate it I get $$\mathcal{O}(1)_{D_+(t_i)}=(\tilde{A}(1)_{(t_i)})\tilde{}=(t_i\tilde{A}_{(t_i)})\tilde{}=t_i\mathcal{O}_{D_+(t_i)}.$$
But $t_i\mathcal{O}_{D_+(t_i)}\neq f_i\mathcal{O}_{D_+(t_i)}$, so how is $\mathscr{I}\mathcal{O}_{D_+(t_i)}=\mathcal{O}_{D_+(t_i)}(1)$ true?