Let $X$ be a Noetherian scheme and let nil$_X$ the sheafification of the presheaf $U \mapsto \operatorname{nil} \mathcal{O}_X(U)$ with the obvious restriction maps. Show the following:
(i) For each open subset $U \subseteq X$ there exists some $k \geq 0$ such that $\operatorname{nil}_X(U)^k=0$.
(ii) The given presheaf is a sheaf, so we have nil ${ }_X(U)=\operatorname{nil} \mathcal{O}_X(U)$ for each open subset $U \subseteq X$.
I saw an answer of this question using the notion of quasi-coherent sheaf. Since I do not know about this notion, is there another way to solve it? Any help would be appreciated.
Assuming $(ii)$ (or just separatedness), you have an injection $$\operatorname{nil}_X(U) \hookrightarrow \prod_{x \in U}\operatorname{nil}_{X,x}.$$ Since $X$ is noetherian, so is $U$; in particular $U$ is quasi-compact. We can cover $U = \bigcup_{i=1}^n\operatorname{Spec}(A_i)$ with $A_i$ being noetherian rings. In a noetherian ring the nilradical is nilpotent, $\operatorname{nil}(A_i)^{k_i}=0$. You can pick up the biggest $k_i$, so that there exists $k$ with $\operatorname{nil}(A_i)^k=0$ for all $i=1,...,n$. Note that each $\operatorname{nil}_{X,x}$ is a localisation of some $A_i$ and nilradical commutes with localistion so that $\operatorname{nil}_{X,x}^k = 0$ for any $x \in U$ showing that $\operatorname{nil}_X(U)^k=0$.
To show that $\mathscr{F}=\operatorname{nil}_X^{pre}$ (the original presheaf) is a presheaf, first I hope that you can prove that it is a sheaf if $X$ is affine (hint: using the notion of sheaf of a base to reduce the work to distinguished open sets). In general, it follows from the subsequent argument: first, it suffices to prove that in the sequence $$0 \longrightarrow \mathscr{F}(U) \longrightarrow \prod_i \mathscr{F}(U_i) \longrightarrow \prod_{i,j}\mathscr{F}(U_i \cap U_j)$$ with $U = \bigcup_i U_i$ an open covering, we only need to deal with $U,U_i$ all affine. Indeed, this is again the consequence of the trick "sheaf of a base" and affine open subsets form a base of the Zariski topology). Let $U \subset X$ be an open affine and $U = \bigcup_{i} U_i$ be an (finite) open covering with $U_i$ affine. Write $U_i = \bigcup V_{ih}$ (a finite covering again) with $V_{ij}$ affine, write each $V_{ih} = \bigcup W_{ihk}$ with $W_{ihk}$ affine. Then there is a commutative diagram $$\require{AMScd} \begin{CD} \mathscr{F}(U) @>{}>> \prod_i \mathscr{F}(U_i) @>{}>> \prod_{i,j} \mathscr{F}(U_i \cap U_j) \\ @VVV @VVV @VVV \\ \prod \mathscr{F}(V_{ih}) @>{}>> \prod_{i,h,k} \mathscr{F}(W_{ihk}) @>{}>> \prod \mathscr{F}(W_{ihk} \cap W_{i'h'k'}) \\ @VVV @VVV\\ \prod F(V_{ih} \cap V_{i'h'}) @>{}>> \prod \mathscr{F}(W_{ihk} \cap W_{i'k'h'}) \end{CD} $$ Now the middle horizontal and the two verical arrows, chasing diagram should give you the exactness of the top horizontal arrow.