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.