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Jun 12, 2020 at 11:57 vote accept Jakob Werner
Jun 12, 2020 at 11:56 comment added Jakob Werner @Cubikova I see. So this is really only an abstract existence statement “there exists some isomorphism onto a definable functor”, and this isomorphism is in no way canonical or natural, not even after the choice of an affine covering. Thank you for your help.
Jun 8, 2020 at 9:59 comment added Cubikova Note that the choice is made up to definable bijection (or up to isomorphism of definable functors in your terminology). Indeed, I could also define the functor for $\mathbf{P}_K^1$ as sending $F$ to $\{(x,y) \in F^2 : y=1\}\cup\{(2,0)\}$. In any case, the existence of such a functor is granted by elimination of imaginaries.
Jun 8, 2020 at 9:47 comment added Cubikova In the special case of $\mathbf{P}_K^1$, you can identify such a quotient with the definable functor which sends $F$ to the (definable) set $\{(x,y)\in F^2 : y=1\} \cup \{(1,0)\}$. For $\mathbf{P}_K^n(F)$ you can cook up a similar description. For general schemes of finite type, the fact that you can find this definable (set) functor is precisely the content of elimination of imaginaries.
Jun 8, 2020 at 9:47 comment added Cubikova Indeed, the functor is isomorphic to the functor of points. In the case of $\mathbf{P}_K^n$, the functor of points is isomorphic to the functor sending $F$ to the set of $n$-dimensional subspaces of $F^{n+1}$. The usual equivalence relation on $F^{n+1}$ given by $(x_0,\ldots, x_n)\sim(y_0,\ldots,y_n)$ if and only if there is $\lambda\in F^*$ such that $\lambda x_i=y_i$ for each $i\in \{0,\ldots,n\}$ is the corresponding definable equivalence relation.
Jun 6, 2020 at 18:24 comment added Jakob Werner By the way, thank you for the hints to the literature, I will definitely have a look at it.
Jun 6, 2020 at 18:22 comment added Jakob Werner If I understand “gluing affine charts” correctly, then you are saying that also for general $\mathscr{X}$, the associated functor is the functor of points, sending $F$ to $\mathscr{X}(F) = \mathrm{Hom}_K(\mathrm{Spec}(F), \mathscr{X})$, right? Could you maybe indicate for some simple non-affine scheme like $\mathbf{P}^1_K$, how the bijection that you get to a definable functor would look like?
Jun 6, 2020 at 18:18 comment added Cubikova Thank you very much.
Jun 6, 2020 at 18:05 comment added LSpice I edited in links to the book and the individual chapters. Since you later cite Chapter 1 separately, I think you didn't mean your reference and corresponding ZBL link to refer only to the first chapter, but rather to the whole book. That's ZBL0920.03046.
Jun 6, 2020 at 18:05 history edited LSpice CC BY-SA 4.0
Links to book and chapters
Jun 6, 2020 at 17:57 history answered Cubikova CC BY-SA 4.0