Let $X$ be a Noetherian scheme, $\mathscr{I}$ be a coherent sheaf of ideals and let $\pi:\operatorname{Bl}_Y X \rightarrow X$ be the blow up of $X$ along the closed subscheme $Y$ corresponding to $\mathscr{I}$.
In the proof of the book of Hartshorne "Algebraic Geometry", Proposition II.7.13a, it is written:
$$\mathscr{S}(U)(1) = \bigoplus_{d \geq 0} \mathscr{I}^{d+1}(U)$$
where $\mathscr{S}$ is the sheaf of graded algebras defined by $\mathscr{S} = \bigoplus_{d \geq 0} \mathscr{I}^d$, and $U$ is an open affine subset of $X$. From the above equality it follows that $\mathscr{S}(U)(1) = \mathscr{I} \cdot \mathscr{S}(U)$, the ideal generated by $\mathscr{I}$ in $\mathscr{S}(U)$, and so the inverse image ideal sheaf $\pi^{-1}\mathscr{I} \cdot \mathcal{O}_X$ is equal to $\mathcal{O}_{\operatorname{Bl}_YX}(1)$.
My question is, why it is $\mathscr{S}(U)(1) = \bigoplus_{d \geq 0} \mathscr{I}^{d+1}(U)$? Shouldn't it be $\mathscr{S}(U)(1) = \bigoplus_{d \geq -1} \mathscr{I}^{d+1}(U)$, since $\mathscr{S}(U)(1)_{-1} = \mathscr{S}(U)_0 = \mathcal{O}_X(U) \neq 0$ ? In that case, is the equality $\mathscr{S}(U)(1) = \mathscr{I} \cdot \mathscr{S}(U)$ still true?
Best Answer
First, recall that the functor which takes a graded $S$-module $M$ to the associated sheaf $\widetilde{M}$ on $\operatorname{Proj} S$ only cares about what happens in high-enough degree (see for instance exercise II.5.9(c)), so both Hartshorne's choice and your choice produce the same sheaf $\widetilde{\mathscr{I}}$ on the blowup. His choice has the advantage of making the next claim a little more obvious, though - using your choice, it's still true that the image of $\mathscr{S}(U)(1)$ in $\mathscr{S}(U)$ is the ideal generated by $\mathscr{I}$ in $\mathscr{S}(U)$, you just have the stuff in negative degree going to zero.