# presheaf of nilpotent element of the structure sheaf of a scheme

If we consider a scheme $$(X,\mathcal{O}_X)$$ and a presheaf $$\mathcal{N}$$ defined on $$X$$ by setting $$\mathcal{N}(U):=Nil(\mathcal{O}_X(U))$$ where $$U\subseteq X$$ is an open subset and $$Nil$$ means the nilpotent elements. Then how do we prove that this $$\mathcal{N}$$ is actually a sheaf? I did the identity axiom part, which was easy by using the fact that $$\mathcal{O}_X$$ is a sheaf. But now I have some problem proving the gluability axiom. I will write my approach below.

Gluability axiom. Consider an open subset $$U$$ with an open covering $$\{U_i\}_{i\in I}$$. If we have $$s_i\in \mathcal{N}(U_i)$$ such that $$s_i|_{U_i\cap U_j}=s_j|_{U_j\cap U_i}$$, then since $$\mathcal{O}_X$$ is a sheaf, we glue them to a $$s\in \mathcal{O}_X(U)$$. But now I cannot proceed, because to show that $$s\in \mathcal{N}(U)$$, is to find some $$n\in\mathbb{Z}^{>0}$$ such that $$s^n=0\in \mathcal{O}_X(U)$$, i.e. $$s^n|_{U_i}=0\in \mathcal{O}_X(U_i)$$. Unfortunately, this integer $$n$$ is not easy to find, as $$I$$ might be an infinite set (if the open covering is finite, then we can just simply take the maximum of $$n_i$$'s where $$n_i$$ is the integer such that $$s_i^{n_i}=0$$).

Any help is appreciated! Thanks!

I don't believe that $$N$$ is a sheaf in general and your work so far indicates exactly what can go wrong. Take $$X$$ to be the infinite disjoint union of the affine schemes $$X_i = \text{Spec } \mathbb{Z}[x]/x^i, i \in \mathbb{N}$$ and consider the open covering of $$X$$ given by $$U_i = X_i$$ and the local sections $$s_i = x \in \mathbb{Z}[x]/x^i$$. Each of these local sections is nilpotent but they glue to a section of $$\mathcal{O}_X(X) \cong \prod_i \mathbb{Z}[x]/x^i$$ which is not nilpotent, namely the element $$\prod_i x$$. The issue is exactly that one does not have a bound on the degree of nilpotency for an infinite cover.
• Were you assigned this as a homework problem? Maybe you need to assume that $X$ is quasicompact or Noetherian or something like that? Commented Oct 6, 2023 at 9:13