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Mar 16 at 6:47 history edited Taras Banakh
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Dec 15, 2023 at 7:11 answer added Taras Banakh timeline score: 1
Dec 14, 2023 at 19:41 vote accept Taras Banakh
Dec 14, 2023 at 14:37 answer added Ihromant timeline score: 2
Oct 25, 2023 at 18:28 history edited Taras Banakh CC BY-SA 4.0
corrected the definition of the projective Desargues Axiom
Oct 25, 2023 at 17:18 history edited Taras Banakh
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Oct 25, 2023 at 16:55 comment added Taras Banakh @TomDeMedts If indeed everything can be done with the projectivization trick, then it also should be done without that trich just in the affine plane. And it indeed can be done, but such a modified proof requires some different form of the Pappus Axiom (not that one, equivalent to the commutativity of the division ring at the presence of the Desargues Axiom).
Oct 25, 2023 at 16:50 comment added Taras Banakh @TomDeMedts The problem with this proof is that when you look at the Affine Pappus Axiom as a property of a projective plane, then it means that you can only apply it to some selected line (at infinity). Without the Desargues Axiom, we have no homogeneity of the projective completion, so cannot move that selected line to other places. But Hessenberg's Theorem applies the Pappus Axiom for three different lines. So, I do not see how it can be applied. Or I do not see something obvious?
Oct 25, 2023 at 14:28 comment added Tom De Medts What exactly is it that you find unconvincing about his proof? It looks completely accurate to me: just pass to the projectivization of the affine plane. There is indeed the tricky aspect that for non-Desarguesian projective planes, the removal of different lines could result in non-isomorphic affine planes, but this is not relevant for this proof.
Oct 25, 2023 at 12:15 history edited Taras Banakh CC BY-SA 4.0
added 44 characters in body
Oct 25, 2023 at 12:07 history asked Taras Banakh CC BY-SA 4.0