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I found some difficulties proving this exercise from hartshorne's book.

Let us first reduce to the affine case. Let $X = \mbox{Spec } A$ and define $X_{\mbox{red}}$ as in the exercise. For people who do not have the book nearby, it is defined as the quotient by the sheaf of nilradicals. I would like to prove that $X_{\mbox{red}} = \mbox{Spec } (A_{\mbox{red}}).$ Let $Y=X_{\mbox{red}}$ and $Z = \mbox{Spec } (A_{\mbox{red}})$.

The underlying topological spaces are homeomorphic, since the map $A \rightarrow A_{\mbox{red}}$ induces an isomorphism of posets on the prime posets of the two rings.

Let $f \in A$. On the corresponding distinguised open we have $$\mathcal{O}_{Z}(D(f)) = (A_{\mbox{red}})_{f}.$$

On the pre-sheaf $\mathcal{P}$ of $Y$, so $\mathcal{P}^+=O_Y$,

$$\mathcal{P}(D(f)) = (A_f)/\mbox{Nil}(A_f).$$

Which is isomorphic to $\mathcal{O}_{Z}(D(f))$ as localization commutes with ring quotients. Now I have to push in the sheafification and making sure I keep this isomorphism, which gives me troubles. Any hints?

A comment of Georges Elencwajg on his own answer suggest that sheafification is not needed. Can someone elaborate on this? Why not?

Community
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bbnkttp
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  • Sleeping on it made me realize the answer. The $\mathcal{P}$ is already a sheaf since it is isomorphic to a sheaf as a presheaf. Should I answer my own question? – bbnkttp Aug 10 '14 at 10:08
  • I assume that by HAG II you don't mean http://arxiv.org/abs/math/0404373? –  Aug 10 '14 at 14:34
  • No I mean Hartshorne's Algebraic Geometry book. Then exercise 3 from paragraph 2 in chapter 2. – bbnkttp Aug 10 '14 at 14:59

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Careful. You have shown that a the sections of a presheaf $\mathscr G$ over a base for the topology on some space $X$ are isomorphic to those of a sheaf $\mathscr F$ ($\mathscr G$, $\mathscr F$ are assumed to be presheaves on $X$). This is not enough to conclude abstractly that $\mathscr G$ is a sheaf isomorphic to $\mathscr F$; such a thing is clearly not true in general. But it is true that the sheaf associated to $\mathscr G$ is isomorphic to $\mathscr F$. This is exactly the definition of the structure sheaf of the reduced scheme associated to $X$ that Hartshorne uses.

Lemma Let $\mathscr G$ be a presheaf on a topological space $X$ and $\mathscr F$ be a sheaf on $X$. Suppose that for every open set $U$ in a base for the topology on $X$, $\mathscr G(U)$ is naturally isomorphic to $\mathscr F(U)$ (by this I mean, the isomorphisms commute with restriction). Then the sheaf associated to $\mathscr G$ is canonically isomorphic to $\mathscr F$.

I think the proof is clear.

Tomo
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  • Do you have a counter example? If $\mathscr F$ is a sheaf and we have a morphism of presheafs $\mathscr G \xrightarrow{f} \mathscr F$ such that $f_U$ is an isomorphisms for every open $U\subset X$, then surely $\mathscr G$ is an sheaf.. So in your lemma $\mathscr G$ is already a sheaf. – bbnkttp Nov 25 '15 at 18:44
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    @MohamedHashi The hypothesis of the Lemma is that you have isomorphisms of sections over a base for the topology, not for every open set in the topology. In your question, you have isomorphisms over a base (you call the sets 'distinguished open'), but of course there are other open sets than those of the form $D(f)$ f.s. $f\in A$. – Tomo Nov 26 '15 at 03:42
  • Ah I now see what you mean. A morphism of presheafs is not determined by what it does on a base as for sheafs. – bbnkttp Nov 26 '15 at 07:42
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    @MohamedHashi Not quite. Since the target is a sheaf, specifying a morphism from a presheaf to a sheaf over a base does specify the morphism over the full topology. But the point is that the presheaf might have sections missing, and hence it is possible that the morphism induced by the natural isomorphisms over a base is not an isomorphism over every open set. – Tomo Nov 26 '15 at 16:06
  • I get it, this was a valuable point. Thank you! – bbnkttp Nov 26 '15 at 19:35
  • Hi, could you provide a reference for the lemma you mentioned? – Christina Mar 02 '22 at 16:01
  • I don’t have a reference because I just wrote it down off the top of my head. It follows directly from the definition of sheafification. – Tomo Mar 03 '22 at 23:04