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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.

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1 Answer 1

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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.

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