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Apr 25, 2019 at 3:15 comment added Ingo Blechschmidt You circumvent this problem by defining schemes to be certain objects in $\mathrm{Sh}(\mathbf{Et})$, but let me remark for the benefit of others that the functor from schemes-as-usually-defined to $\mathrm{Sh}(\mathbf{Et})$ is not fully faithful if $\mathbf{Et}$ is defined, as in your post, using only finitely presented rings. For instance, the functor of points of $\mathrm{Spec}(\mathbb{Q})$ coincides with the functor of points of the empty scheme.
Apr 23, 2019 at 18:24 history became hot network question
Apr 23, 2019 at 16:20 vote accept Steve
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Apr 23, 2019 at 14:57 answer added Simon Henry timeline score: 9
Apr 23, 2019 at 13:28 history asked Steve CC BY-SA 4.0