What do the local sections of an nilpotent ideal sheaf look like? Let $(X, O_X)$ be a scheme. and $I$ an nilptoent ideal sheaf. i.e. $I^n=0$ for some $n$. Would this imply that each $I(U)$ is an nilpotent ideal of $O_X(U)$?

I’m submitting this as an alternative to hm2020’s answer. It’s an elaboration of what I wrote in the comments.
Let $I\subseteq \mathcal{O}_X$ be an ideal sheaf, and let $\mathcal{F}$ be the presheaf assigning to each open subset $U$ of $X$ the ideal $I(U)^n\subseteq \mathcal{O}_X(U)$. You say that $I$ is nilpotent of degree $n$ if $\mathcal{F}^\#$ is zero, where $\#$ is used to denote sheafification. But, since $\mathcal{F}$ is a separated presheaf, being a subsheaf of the sheaf $\mathcal{O}_X$, one has that $\mathcal{F}=0$ if and only if $\mathcal{F}^\#=0$ (e.g. see [1, Tag00WB]). Thus, we deduce the following:

Fact: Let $X$ be a scheme and $I$ an ideal sheaf of $\mathcal{O}_X$. Then, the following are equivalent:

*

*For all open subsets $U$ the ideal $I(U)^n$ is zero.

*The sheafification of the presheaf $U\mapsto I(U)^n$ is zero.


[1] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/
