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Exercise I.XXV. of the book Geometry of Schemes by Eisenbud and Harris claims that the smallest non-affine scheme has three elements with a constructed topology and sheaf. But I am wondering if this is really the smallest non-affine scheme, so I tried with the two-elements example:

If we define $X=\{p,q\}$ with the discrete topology, and if $\mathscr O(X)=\mathscr O(\{p\})=\mathscr O(\{q\})=\text{ a field } K$ while $\mathscr O(\phi)=\{0\}$, and the restrictions are either identity or the obvious map, then $X$ is covered by affine schemes, so the pair $(X,\mathscr O)$ forms a scheme, but is obviously not affine.

So is the above constructed scheme the smallest non-affine one, as intended?
Any inappropriate point ought to be localised, and thanks in advance.

awllower
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1 Answers1

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The problem is that your structure sheaf is easily seen to not be a sheaf.

Let $k_1,k_2\in K$ thought of as sections over $\{p\}$ and $\{q\}$ then since $\{p\},\{q\}$ is an open cover for $X$ and clearly the sections $k_1$ and $k_2$ agree on overlaps, so there exists $k\in \mathscr O(X)$ with $\rho_{X\rightarrow\{p\}}(k)=k_1$ and $\rho_{X\rightarrow\{q\}}(k)=k_2$. But by picking another $k_3 \in \mathscr O(\{p\})$ we can show $\rho_{X\rightarrow\{q\}}$ is not injective, which is a contradiction since it is a morphism from a field.

Given any two-point discrete scheme (any not discrete scheme has only trivial open covers so is affine) , we can see $\mathscr O(\{p\})=\mathscr O_p$ for each $p$. So, $\mathscr O(\{p\})$ is local. Consider the map $\phi :\mathscr O(\{p\})\times \mathscr O(\{q\}) \rightarrow \mathscr O(X)$. Where $\phi(r,s)$ is the unique section of $\mathscr O(X)$ that restricts to $r$ and $s$ respectively on $\mathscr O(\{p\})$ and $\mathscr O(\{q\})$, (note this section exists and is unique by the sheaf property). This map is clearly a ring morphism and injective (and well defined by uniqueness of the section), it is clearly surjective as well as any global section $s$ restricts to $r_1 \in \mathscr O(\{p\})$ and $r_2\in \mathscr O(\{q\})$. By definition $\phi(r_1,r_2)=s$ so we are done. So any two-point scheme is affine. It is trivial to show any one-point scheme is affine. So the smallest non-affine scheme has three points.

This argument together with some induction shows $$\bigsqcup_{i=1}^n\operatorname{Spec} R_i\cong \operatorname{Spec} \prod_{i=1}^n R_i$$ which should not be that surprising. What is truly surprising is that this fails for any infinite cardinal as $\bigsqcup_{i=1}^\infty\operatorname{Spec} R_i$ is not quasicompact but $\operatorname{Spec}\prod_{i=1}^n R_i$ is.

PVAL-inactive
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  • I see. Then how about the sheafificaation of that presheaf? Or, is there a prroof that the three-elements example is truely the smallest non-affine scheme? Thanks very much. – awllower Aug 19 '13 at 06:13
  • The sheafification of that sheaf can be seen to be the affine scheme $\text{Spec}\ k\times k$. The way to arrive at a proof that the 3 point scheme is the smallest non affine is to take any 2 point sheaf cover it with any local rings on each point $R_1,R_2$ (if it is not affine it is clear that this is true) and show that the global sections must be isomorphic to $R_1\timesR_2$ using the sheaf properties and nothing else. – PVAL-inactive Aug 19 '13 at 06:46
  • Thanks again. But I cannot see how this proves that the three-point scheme is the smallest:after we showed that its global sections must be isomorphic to the product of $R_1$ and $R_2$, what does this tell us? – awllower Aug 22 '13 at 02:51
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    If you showed its global sections are $R_1\times R_2$, then the scheme is just the affine scheme $\operatorname{Spec} R_1\times R_2$ so it is affine and so any two point scheme is affine. Clearly any one point scheme is affine, so the smallest affine scheme must have three points. – PVAL-inactive Aug 22 '13 at 03:05
  • I see: And we can prove the assertion by means of the sheaf axiom? Thanks for this good answer thus. – awllower Aug 22 '13 at 03:12
  • I edited my answer to answer the question "why must any two point scheme be affine?". – PVAL-inactive Aug 22 '13 at 03:25
  • I appreciate that. It is short and wonderful! – awllower Aug 22 '13 at 03:28
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    I thought about this question even more and added even more to it. There are non-discrete two point schemes but they all have trivial open covers hence are affine. This also closely relates to the answer to this question http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics that begins with "Here's my list of false beliefs". – PVAL-inactive Aug 22 '13 at 06:10
  • So this shows that any finite discrete scheme is affine. In particular, any Nötherian zero-dimensional scheme is affine, although I am also wondering why this is true, and ask a new question. :P And here is a related question. – awllower Aug 24 '13 at 06:50
  • @awllower Are you confused about the induction step? Note the induction hypothesis essentially forces our sheaf to make the correct assignment (to be isomorphic to the affine scheme we want) except on global sections, and the proof to show the global sections of an n point discrete scheme (or disjoint union of open affines) is identical to the case for the two point scheme. – PVAL-inactive Aug 24 '13 at 07:13
  • I mean: I am wondering why a Nötherian zero-dimensional scheme is finite discrete. Sorry for bad wording. And the question has come out now. – awllower Aug 24 '13 at 07:51