Why is shefication necessary in constructing the reduced scheme?

Question

Let $X$ be a scheme. Hartshorne defines the reduced scheme associated to $X$ as the sheafication of the presheaf $U \mapsto \mathcal{O}_X(U)_{\text{red}}$. Is there any example that shows that this presheaf need not be a sheaf?

Answer 1

Take $X=\coprod_n \mathrm{Spec}(\mathbb{Z}/p^n)$ for a prime $p$. Then $s:=(p)_n \in \prod_n \mathbb{Z}/p^n = \mathcal{O}_X(X)$ is not nilpotent, but the restriction to each $\mathrm{Spec}(\mathbb{Z}/p^n)$ is nilpotent. Hence, we have $[s] \neq [0]$ in $\mathcal{O}_X(X)_{red}$, although we have $[s]=[0]$ in each $\mathcal{O}_{X}(\mathrm{Spec}(\mathbb{Z}/p^n))_{red}$.

There are also quasi-compact counterexamples.

Answer 2

An indirect answer is that Hartshorne's sheaf constructions are often a little lacking in the elegance department. This is one example; the definition of the structure sheaf is another. If you are looking for a better way to understand the reduced scheme associated with an arbitrary scheme, think of it this way: it's a closed subscheme and thus given by a certain sheaf of ideals $\mathcal{I}$. What is $\mathcal{I}$? Simply, its sections on any open $U$ are the sections of $\mathcal{O}_U$ that are "in every prime ideal of $U$", i.e. whose image in the fiber at each point of $U$ is zero. This is visibly an ideal of $\mathcal{O}_U$ and these ideals visibly form a sheaf, since evaluation in a fiber is a local property. On any open affine set $\operatorname{Spec}(A)$, itgives the correct ideal, namely the nilradical, by virtue of the famous fact that this ideal is the intersection of all primes of $A$. In this interpretation, the nilradical sheaf of ideals is the set of "all sections of $\mathcal{O}_X$ defining the zero function on $X$".