Pull back of the associated sheaf of Rees algebra

Question

Let $A$ be a Noetherian ring. Let $I\subset A$ be an ideal. Set B=$\oplus_{n\ge 0}I^n$ to be the Rees algebra associated to $I$, $\widetilde B$ to be the associated sheaf of $B$ on $\operatorname{Spec} A$. Let $f:X\to \operatorname{Spec} A$ be a proper morphism.

Definition. For a quasicoherent sheaf $\mathcal F$on $X$, define $I^n\mathcal F=\operatorname{Im}(f^*I^n\otimes_{\mathcal O_X}\mathcal F\to \mathcal F)$.

In this Stacks tag the authors write $f^*\widetilde B=\oplus_{n\ge 0}I^n\mathcal O_X$ without any explanation:

Suppose $\operatorname{Spec} C\subset X$ is an open affine subset, the section of $f^*\widetilde B$ over $\operatorname{Spec} C$ is $\oplus_{n\ge0} C\otimes_A I^n$, and with $f^*\widetilde B$ identifying to $\oplus_{n\ge 0}I^n\mathcal O_X$ we have $\oplus_{n\ge0} C\otimes_A I^n=\operatorname{Im}(\oplus _{n\ge 0}C\otimes_AI^n\to C)$, but this doesn't look correct. Anyone can help?

Answer 1

Note that $\oplus_{n\ge0} C\otimes_A I^n=\oplus _{n\ge 0}\operatorname{Im}(C\otimes_AI^n\to C)$ rather than $\operatorname{Im}(\oplus _{n\ge 0}C\otimes_AI^n\to C)$. The latter makes little sense.

In the definition of a Rees algebra, the direct sum symbol are considered as an external direct sum so as to make the resulting direct sum a graded ring.

The notation $\operatorname{Im}(C\otimes_AI^n\to C)$ is nothing but the ideal of $C$ generated by the image of $I^n$ via the obvious map $A\to C$. The latter coincides with $C\otimes_A I^n$, as predicted.