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?


2 Answers 2


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).

  • $\begingroup$ Thanks. My construction of the extension $G$ is $G(U)=\lim F_B(V)$, where $V$ ranges over all elements of $B$ which are contained in $U$, which the restrictions as transition morphisms. I have now more or less convinced myself that there is a morphism of presheaves $F\to G$ (given by $s\mapsto (s|_V)$), which induces isomorphisms on stalks. I think that finishes the proof, because the canonical morphism from the sheafification of $F$ to $G$ is then an isomorphism. $\endgroup$
    – user363520
    Jul 4, 2017 at 9:52
  • $\begingroup$ A question about your last remark: I see why it's a sheaf if the underlying topological space is noetherian (e.g. if $X$ is noetherian), because then every open subset is quasicompact and locally nilpotent sections are nilpotent. Why is quasicompactness enough? $\endgroup$
    – user363520
    Jul 4, 2017 at 10:55
  • $\begingroup$ Yes, I actually meant that every open set should be quasi compact. $\endgroup$
    – MooS
    Jul 4, 2017 at 12:41

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 $\mathscr{F}\to \mathscr{F}^+\to\mathscr {G}$.


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.

  • $\begingroup$ I'm not sure what you mean, could you expand on this? $\endgroup$
    – user363520
    Jul 3, 2017 at 20:26
  • $\begingroup$ @JürgJenatsch If you sheafify a sheaf, you get the same sheaf back (up to unique isomorphism). This follows from the universal property cited in the answer. $\endgroup$ Jul 4, 2017 at 2:07
  • $\begingroup$ It does not follow solely from the properties of sheafification. Of course you also need the properties of extending a sheaf from a base... $\endgroup$
    – MooS
    Jul 4, 2017 at 8:50
  • $\begingroup$ @JürgJenatsch I've expanded the answer. $\endgroup$
    – Caligula
    Jul 4, 2017 at 10:29
  • $\begingroup$ I don't think the first bullet point holds. In any case that's not exactly the extension I'm working with, see my comment on the other answer. $\endgroup$
    – user363520
    Jul 4, 2017 at 10:48

You must log in to answer this question.