I think that wikipedia article on Algebraic spaces contains a serious content error in the part on the definition of Algebraic spaces as quotients of schemes and I would like to discuss if it is indeed an error or I just missing something. The definition says:

An algebraic space $X$ comprises a scheme $U$ and a closed (!) subscheme $R \subset U \times U$ satisfying the following two conditions:

  1. $R$ is an equivalence relation as a subset $U \times U$;
  2. the two projections $p_i: R \to U$ onto each factor are étale maps

Some authors, such as Knutson, add an extra condition that an algebraic space has to be quasi-separated, meaning that the diagonal map is quasi-compact. One can always assume that $R$ and $U$ are affine schemes. Doing so means that the theory of algebraic spaces is not dependent on the full theory of schemes, and can indeed be used as a (more general) replacement of that theory.

I think that in this definition is a subtle problem hidden, since it is important to distinguish strictly if $U$ may be always assumed to be affine or not when we work with the condition that the $R \subset U \times U$ is assumed to be a closed subscheme as I will explain below why. So I'm not sure why it can "one can always assume that $R$ and $U$ are affine". That's doesn't matter only in the case if $U$ is moreover separated as we will see in following example:

It is well know that an arbitrary scheme $S$ is canonically an algebraic space. If we want to verify it using the characterization above we have to recognize it as a quotient of schemes, so we need a pair $(U,R)$ representing $S$ and satisfying axioms $1$ and $2$.

For $U$ we set $U:=S$ and for $R$ we need a closed subscheme of $U \times U$ and as canonical choice for $R$ might be choosen the closure $R:=\overline{\Delta(U)}$ of the image with respect the diagonal map $\Delta: U \to U \times U$. Here I'm not sure if that is a correct choice for $R$ in the case when $S$ is not separated, but it seems to be the only "canonical" choice. And if this choice for $R$ is correct, we have a problem: Why is the restriction of the projection map $p \vert _{\overline{\Delta(U)}}: \overline{\Delta(U)} \to U$ etale? (take the closure bar into account)

Clearly $p \vert _{\Delta(U)}: \Delta(U) \to U$ is etale since $U \cong \Delta(U)$ and $p \circ \Delta= id_U$, but in general if $f: X \to Y$ is a morphism of schemes, $Z \subset X$ an locally closed immersion (but not cl imm) and the restriction $f \vert _Z:Z \to Y$ is etale, then in general $f \vert _{\overline{Z}}: \overline{Z} \to Y$ is not more etale.

Thus there is no reason if we have a scheme $S (=U)$ with $\overline{\Delta(U)} \not \cong \Delta(U)$ for restriction $p \vert _{\overline{\Delta(U)}}: \overline{\Delta(U)} \to U$ to be etale. On the other hand affine schemes are separated and therefore $\overline{\Delta(U)} = \Delta(U)$, but for general scheme $S$ the requirement that $R \subset U \times U$ is closed is stronger than to say one may always assume that $R$ and $U$ are affine schemes. Or do I confuse something here? Is there maybe a better choice for $R$ involved?

In summary the concern of my question can be reduced to two question:

A: Is for arbitrary scheme $S$ the restriction of projection $p \vert _{\overline{\Delta(S)}}: \overline{\Delta(S)} \to U$ etale?

B: If A is true, then B is clear. If A is wrong, why it is sufficient to work with this definition of algebraic spaces with affine schemes in viewpoint the described problemwith closedness of $R$ in $U \times U$?

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    $\begingroup$ I don't know the answer to this question, or if it is an error, but I hope very much that MO doesn't become a place to go to report mathematical errors in Wikipedia. (Which is to say, I think that the post would be better if it foregrounded the question and perhaps gave a low-key mention of Wikipedia as the motivation.) $\endgroup$ – LSpice Aug 11 at 14:15
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    $\begingroup$ Since Wikipedia contains a lot of useful information and is consulted by many people many times, I think there is some benefit in discussing possible errors here in this forum. $\endgroup$ – MathCrawler Aug 11 at 14:27
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    $\begingroup$ In the modification I tried to point out the key questions arising from my concern on the discussed definition. I understand your concern, but I think that it highly depends on the specific case what kind of error in considered. A simple typo or a wrongly quoted basic definition from an undergrade lecture is one thing. But here the question treats a subtle problem. The question is if a whole theory can be made up from affine pieces or not. $\endgroup$ – katalaveino Aug 11 at 14:56
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    $\begingroup$ If we take $R$ to be the closure of the diagonal, I don't understand why the quotient of this equivalence relation is $S$, even before reducing to the affine case. $\endgroup$ – Will Sawin Aug 11 at 16:15
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    $\begingroup$ For a non separated scheme $S$ there is actually no closed equivalence relation whose quotient is $S$. Such equivalence relations will always produce separated algebraic spaces. (I think it's not so hard to check the valuative criterion of separatedness from the definition). So the Wikipedia definition is only a definition of separated algebraic spaces, and in particular doesn't generalize schemes. $\endgroup$ – Will Sawin Aug 11 at 17:07

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