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According to Wikipedia there are two common ways to define algebraic spaces: they can be defined as either quotients of schemes by étale equivalence relations, or as sheaves on a big étale site that are locally isomorphic to schemes. Namely:

I) a la Knutson:

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.

Knutson adds an extra condition that the diagonal map is quasi-compact.

II) as a sheaf:

An algebraic space $\mathfrak {X}$ can be defined as a sheaf of sets $$\mathfrak {X}:(\operatorname{Sch}/S)^{\text{op}}_{\text{ét}} \to \operatorname{Sets}$$ such that

  1. There is a surjective étale morphism $h_X \to \mathfrak {X}$;
  2. the diagonal morphism $\Delta _{{\mathfrak {X}}/S}: \mathfrak {X} \to \mathfrak {X} \times \mathfrak {X}$ is representable and quasicompact (thanks to David's careful remark).

(Rmk: in II)1. we identified a scheme $X$ with its image $h_X$ wrt the Yoneda embedding $X \to \operatorname{Hom}(X,{-})$.)

Two questions:

  1. About construction I). Wikipedia moreover says that if $R$ is the trivial equivalence over each connected conponent of $U$ (i.e. for all $x,y \in U$ lying in same component then $xRy$ iff $x=y$) then the so defined algebraic space is a scheme in the usual sense. Why?

  2. Where I can find a proof/ reason that the constructions I) and II) are indeed equivalent?

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  • $\begingroup$ Regarding 1. what do you think is the algebraic space associated to a scheme? The word 'is' here is a little misleading. It's really "is the image of a scheme under the inclusion of schemes into algebraic spaces". Regarding the equivalence, I don't have a reference, but if you want the Knutson definition, to correspond exactly, you might need to add $\Delta_{\mathfrak{X}/S}$ quasicompact to II). I assume this is all in the Stacks project somewhere. It looks like most of what you want is in stacks.math.columbia.edu/tag/0264 $\endgroup$ – David Roberts Jul 14 at 0:56
  • $\begingroup$ @David Roberts: Hi. One nitpick on your first point: yes indeed if $U$ is a scheme and we want to consider it as algebraic space we endow it with trivial relation $R$. About the 2. condition in I I'm not sure. Why are in this scheme case the projections $p: R \to U$ etale? Sure, set theoretically we have in this case $\vert R \vert = \vert U \vert$ since $R$ is as a set the diagonal. But how we conclude etaleness of $p$ as a morphism schemes? $\endgroup$ – katalaveino Jul 14 at 1:20
  • $\begingroup$ You take $R = U$ as a scheme, and $R \to U\times U$ the diagonal, so that $R\to U\times U \xrightarrow{pr_i} U$ is the identity map for $i=1,2$. (Also, the connected component stuff is irrelevant.) $\endgroup$ – David Roberts Jul 14 at 5:06
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    $\begingroup$ I think "$R$ is an equivalence relation as a subset $U\times U$" is misleading. It should be an equivalence relation internal to the category of schemes. If this is equivalent to saying it is so on the underlying set of the locally ringed space(s), then fine. Otherwise, it should be changed. $\endgroup$ – David Roberts Jul 14 at 5:11
  • $\begingroup$ About the "equivalence relation" you are of course right. So essentially we consider the affine scheme $R$ as categoretical equivalence relation, ie for all $T \in (Sch)$ the set $Hom(T,R) \subset Hom(T, U \times U)= Hom(T,U) \times Hom(T,U)$ is the equivalence relation in usual sense. That's what you meant by "internal relation", right? $\endgroup$ – katalaveino Jul 15 at 0:34

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