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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!

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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.

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  • $\begingroup$ Thank you for your clear explanation! $\endgroup$
    – Mizutsuki
    Oct 6, 2023 at 9:10
  • $\begingroup$ Were you assigned this as a homework problem? Maybe you need to assume that $X$ is quasicompact or Noetherian or something like that? $\endgroup$ Oct 6, 2023 at 9:13
  • $\begingroup$ No, this is just a small problem 6.2.E I saw from Vakil's AG note. I was trying to define it without the help of the machinery of distinguished affine base he introduced. But my approach did not work and this showed me the importance of working with the distinguished affine base. Thank you again! $\endgroup$
    – Mizutsuki
    Oct 6, 2023 at 13:23

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