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An axiomatic projective plane is a point-line incidence structure with the following axioms:

  1. any two distinct points are collinear (via a unique line);
  2. any two distinct lines meet in a unique point;
  3. there exists a 4-gon.

Now consider $P = \mathrm{Proj}(k[x,y,z])$, a projective plane over a commutative ring $k[x,y,z]$ with $k$ a field. Then if we consider the $k$-rational points together with the $k$-rational lines, we obtain an axiomatic projective plane.

Can we also detect an axiomatic projective plane in a general algebro-geometric plane $P = \mathrm{Proj}(A[x,y,z])$, with $A$ a commutative ring, in a natural way ? In other words: does there arise an axiomatic projective plane in some way ?

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    $\begingroup$ I probably don't understand the question, anyway: the set of $k$-rational points in $P$ is the classical projective plane over $k$, hence obviously an axiomatic projective plane. $\endgroup$
    – abx
    Commented Jun 1, 2021 at 3:22
  • $\begingroup$ @abx: Oops, I messed up mu question. I will re-phrase it. $\endgroup$
    – THC
    Commented Jun 1, 2021 at 11:19
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    $\begingroup$ What do you mean with a projective plane over an arbitrary commutative ring $A$? If you take the usual approach of considering rank 1 "subspaces" (here: submodules) of $A^3$ as points, and rank 2 "subspaces" as lines (I am being a bit sloppy for brevity), then this yields a projective plane in the axiomatic sense if and only if $A$ is a field: indeed, if $A$ isn't a field, then it admits a proper nontrivial ideal $I$, and you get "points" such as $\{ (x,0,0) : x\in I \}$ which are properly contained in other "points" (here: $\{ (x,0,0) : x\in A\}$, and they are not collinear via a unique line. $\endgroup$
    – Max Horn
    Commented Jun 1, 2021 at 13:38
  • $\begingroup$ @Max, that comment of yours sounds like an answer to me. It doesn't preclude getting a plane by some other bizarre construction. I'd say chances are low but disapproving the possibility would be hard. So I don't expect we get something significantly more complete than what you offer. $\endgroup$
    – MvG
    Commented Jun 3, 2021 at 6:23
  • $\begingroup$ Have you considered the theory of ternary rings? $\endgroup$
    – Ben McKay
    Commented Jun 10, 2021 at 16:43

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