The radical presheaf is not a sheaf

Question

$\def\sO{\mathcal{O}} \def\sI{\mathcal{I}} $Let $(X,\sO_X)$ be a ringed space. Let $\sI\subset\sO_X$ be an ideal sheaf. We define the radical presheaf of $\sI$, denoted $\sqrt[p]{\sI}$, as the presheaf whose sections over $U\subset X$ are $\sqrt[p]{\sI}(U)=\sqrt{\sI(U)}$. I am trying to show that this presheaf is in general not a sheaf. How one would find a counterexample?

Answer 1

$\def\sO{\mathcal{O}} \def\sI{\mathcal{I}} \def\nil{\operatorname{Nil}} $I think for those topological spaces whose subspaces are all quasi-compact (i.e., Noetherian topological spaces) we can prove that the radical presheaf is a sheaf.

Here's one non Noetherian counterexample: Consider the topological space $X=\mathbb{N}=\{1,2,3,\dots\}$ whose open sets are $\varnothing$ and $U_n=\{1,\dots,n\}$, for $n\in\{1,2,\dots\}\cup\{+\infty\}$. For $n\in\{1,2,\dots\}\cup\{+\infty\}$, define $\sO_X$ to have sections over $U_n$ equal to the polynomials of degree $<n$. That is, we define $\sO_X(\varnothing)=0$, $\sO_X(X)=k[x]$ and $\sO_X(U_n)=k[x]/(x^n)$, for $n<+\infty$. On the other hand, we define the restriction morphism from some open set containing $U_n$ to $U_n$ to kill all monomials of degree $n\geq 0$. Then $\sO_X$ can be verified to be a sheaf (of rings). Denote $\nil_p\sO_X$ to the nilradical presheaf of $\sO_X$, i.e., the radical presheaf of the zero ideal sheaf of $\sO_X$. Then $x\in\Gamma(X,\sO_X)$ is a global section such that $x|_{U_n}\in\nil (k[x]/(x^n))=(\nil_p\sO_X)(U_n)$ but $x\not\in(0)=\nil k[x]=\Gamma(X,\nil_p\sO_X)$.