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We working in the following with Knutson's definition of an algebraic space (ie via equivalence relation; there is also another equivalent def via sheaves but let us work here with the following one):

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.

A couple of notes on used notations: the equivalence realtion $R \subset U \times R$ is considered as categoretical equivalence relation (also called "internal relation"), that means that 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.

Question: How one can see that an "usual" scheme $U$ is an algebraic space in the sense above? Assume wlog $U$ affine. The crucial task is to find an equivalence relation $R \subset U \times U$ corresponding to $U$ such that projections $p_i: R \to U$ are etale.

The most natural choice seems to me the image with respect the diagonal map $\Delta: U \to U \times U$, ie $R:= \Delta(U)$. $\Delta$ is always an immersion and thus $\Delta(U)$ is always a locally closed subscheme of $U \times U$.

If we take this choice for $R$, why $p_i: R \to U$ are etale? Or is it conventional to take another choice for $R$? eg the closure of the image? if yes, why?

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    $\begingroup$ The restriction of each of the projections $U \times U \rightarrow U$ to $\Delta(U)$ gives an isomrphism of $\Delta(U)$ to $U$, so it is in particular etale. $\endgroup$
    – AlexIvanov
    Commented Jul 15, 2020 at 21:57
  • $\begingroup$ @AlexI: yes I see. More precisely the restricted $p \vert _{\Delta(U)}: \Delta(U) \to U$ must be an isomorphism because it's nothing but the composition of isomorphisms $\Delta^{-1}: \Delta(U) \to U$ and $id_U= p \circ \Delta: U \xrightarrow{\Delta} U \times U \xrightarrow{\text{p}} U$. So indeed $R:=\Delta(U)$ is the right choice and not the schematic closure $\overline{\Delta(U)} \subset_c U \times U$, I think. $\endgroup$
    – user267839
    Commented Jul 15, 2020 at 23:03
  • $\begingroup$ A curious nitpick beside: Do you know if, if we assume that $\overline{\Delta(U)} \not \cong \Delta(U)$ (eg $U$ not separated). Is the restricted projection map $p \vert _{\overline{\Delta(U)}}:\overline{\Delta(U)} \subset U \times U \to U$ nevertheless etale, or is it in general wrong? $\endgroup$
    – user267839
    Commented Jul 15, 2020 at 23:04
  • $\begingroup$ Look at the example where $U$ is the affine line with doubled origin. The closure of the diagonal is then an affine line with four points "at the origin". Each of the projection maps $\overline{\Delta{U}} \rightarrow U$ is an isomorphism outside origin and over each of the origins of $U$ there lie two of the four points of $\overline{\Delta{U}}$. This map is an isomorphism locally on the source, hence etale. See also math.stackexchange.com/questions/1438886/… $\endgroup$
    – AlexIvanov
    Commented Jul 16, 2020 at 6:25
  • $\begingroup$ I understand why in your example $p: \overline{\Delta(U)} \to U$ is also etale but I'm not sure if this example covers all pathological phenomena can occure when passing from $\Delta(U)$ to $\overline{\Delta(U)}$ as subscheme of $U \times U$. Does your example prove a general argument why for arbitrary $U$ the projection $p: \overline{\Delta(U)} \to U$ should be etale? $\endgroup$
    – user267839
    Commented Jul 16, 2020 at 9:05

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