# On Noetherian scheme, the nilradical ideal sheaf has $nil_X(U)^k=0$?

Let $$(X,\mathcal{O}_X)$$ be a Noetherian scheme. Define the nilradical sheaf to be the shefification of the presheaf $$U\mapsto nil\ \mathcal{O}_X(U)$$, denoted by $$nil_X$$. I want to show $$nil_X(U)^K=0$$ for some K.

Given a section $$s\in nil_X(U)$$, I can cover $$U$$ by finitely many opens $$\{U_i\}$$ such that on each $$U_i$$, there is $$s_i\in nil\ \mathcal{O}_X(U_i)$$, $$s(p)=(s_i)_p$$ for all $$p\in U_i$$. Since $$s_i$$ is nilpotent, I have that $$s|_{U_i}$$ is nilpotent. Hence $$s$$ is nilpotent since the covering is finite.

But in order for $$nil_X(U)^K=0$$ for some big $$K$$, it seems I need a finitely generated condition. I don't know how to proceed.

This is a quasi-coherent sheaf: on any affine open $$\operatorname{Spec} A \subset X$$, we have $$nil_X|_{\operatorname{Spec} A} \cong \widetilde{\sqrt{(0)}}$$. So the sections of $$nil_X$$ over $$\operatorname{Spec} A$$ are finitely generated as an $$A$$-module since $$A$$ is noetherian, and as $$X$$ is noetherian it is quasi-compact and therefore there's a finite cover by affine opens. This gives you the requisite finite generation condition.