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There is a general notion of a geometric point in topos theory. A geometric point is a geometric morphism $Set\to T$.

There is also a notion of a geometric point in algebraic geometry. A geometric point is a point $k\to S$ where $k$ is algebraically closed.

Why do these notions agree?

The geometric points of the étale site of a scheme are maps $k\to S$ where $k$ is separably closed. Again, why are these the same as geometric morphisms from the category of sets into the étale topos?

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    $\begingroup$ Obviously by Zariski and étale toposes, I mean the appropriate slicey-guys over a scheme. $\endgroup$
    – Harry Gindi
    Commented Dec 10, 2010 at 6:44
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    $\begingroup$ The geom. pts of topos of a sober top. space (e.g., scheme, loc. Hausdorff space), are (up to equiv.) stalk functors at pts of the top. space, with distinct physical pts giving inequiv. functors. (For a scheme $S$ can use pullback along Spec($\overline{k(s)}) \rightarrow S$, but silly since Spec($\overline{k(s)}) \rightarrow {\rm{Spec}}(k(s))$ induces equiv. of Zariski topoi.) This is proved in "Sheaves in Geom. & Logic" and near end of 2nd volume of SGA4, where Grothendieck proves that geom. pts of etale topos of $S$ are (up to equiv.) pullbacks along Spec($k(s)_{\rm{sep}}) \rightarrow S$. $\endgroup$
    – BCnrd
    Commented Dec 10, 2010 at 10:04
  • $\begingroup$ Can you tell me how (or where) does Grothendieck prove that last correspondence? I know how to obtain a point of the topos (a functor from the étale topos to Sets that commutes with colimits and finite limits) given a geometric point of the scheme $Spec(k(s)_{sep})→S$. But how do I prove "the converse" (i.e. that this is an equivalence)? $\endgroup$
    – W.Rether
    Commented Jul 1, 2018 at 17:29
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    $\begingroup$ @W. Rether I think this is in Hakim's, Giraud's, or Illusie's thesis. I completely forgot which one, since it has been a while. $\endgroup$
    – Harry Gindi
    Commented Jul 2, 2018 at 7:48

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