Construction of the nilradical of a scheme.

Question

Let $X$ be a scheme. I see two ways to define the nilradical $\mathcal N_X$ of $X$. The first one is to take the sheafification of the presheaf $U\mapsto\operatorname{nilrad}\mathcal O_X(U)$. The second one is to again consider this presheaf, and to note that it is already a sheaf on the base for the topology of $X$ consisting of the affine open subschemes. Thus it extends to a sheaf on $X$.

Is there a quick way to show that these produce the same sheaf, without computing the sections of the sheafification?

Answer 1

This is not about the nilradical at all, this is a very general thing:

You have a presheaf $F$ on a topological space $X$, which is a sheaf (call it $F_B$ for the sake of distinction) on a base $B$ of $X$.

Then there are two ways to obtain sheaf on $X$:

1. Sheafify $F$.

2. Extend $F_B$.

Since both construction yield sheaves, that are unique up to unique isomorphism, it follows that both construction yield the same sheaf (in the sense that there is a unique isomorphism between them).

For the sheafification the uniqueness of course follows from the universal property. For the extension from the base, this is Theorem 2.7.1 on Vakil's notes 'Foundation of Algebraic Geometry'.

---

Also note that in your very special case (the nilradical) your presheaf is already a sheaf if every open subset of your scheme is quasi-compact (a very mild condition).

Answer 2

It follows at once from the properties of sheafification, which is unique up to isomorphism of sheaves.

Indeed. given any morphism of presheaves $\mathscr{F}\longrightarrow \mathscr{G}$ there exists an unique factorisation $\mathscr{F}\to \mathscr{F}^+\to\mathscr {G}$.

Edit.

Someone in the comments pointed out that you need also the extension property for presheaves defined over a basis, which sometimes are called $\mathcr{B}$-presheaves. Of course in order to apply sheafification you need to deal with an actual sheaf, and I assumed that the user was fine with the extension process.

However, let me expand this. Given a basis $\mathscr{B}$ and the $\mathscr{B}$-presheaf $U\mapsto \mathrm{nil}\,\mathscr{O}_X(U)$, the way is the following:

• (Extension to a presheaf over $X$) there is an unique presheaf $\mathscr{N}$ over $X$ such that $\mathscr{N}(U)=\mathrm{nil}\,\mathscr{O}_X(U)$;

• (Sheafification) there are a sheaf $\mathscr{N}^+$ over $X$ and a morphism $\eta :\mathscr{N}\rightarrow \mathscr{N}^+$ such that for every sheaf $\mathscr{G}$ over $X$ and for every morphism of presheaves $f:\mathscr{N}\longrightarrow \mathscr{G}$ there is an unique morphism of sheaves $g:\mathscr{N}^+\longrightarrow \mathscr{G}$ such that $f=g\circ \eta$.

In particular, the second property shows that $\mathscr{N}^+$ is unique up to isomorphism. So if you realise that the presehaf $\mathscr{N}$ is an actual sheaf you get the isomorphism $\mathscr{N}\simeq \mathscr{N}^+$ for free.