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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – awllower Aug 19 '13 at 6:13
  • $\begingroup$ 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. $\endgroup$ – PVAL-inactive Aug 19 '13 at 6:46
  • $\begingroup$ 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? $\endgroup$ – awllower Aug 22 '13 at 2:51
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    $\begingroup$ 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. $\endgroup$ – PVAL-inactive Aug 22 '13 at 3:05
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    $\begingroup$ 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 mathoverflow.net/questions/23478/… that begins with "Here's my list of false beliefs". $\endgroup$ – PVAL-inactive Aug 22 '13 at 6:10

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