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From every projective plane a coordinitisation can be constructed on a planar ternary ring, and conversely from every planar ternary ring a projective plane can be constructed. (For background see Weibel's survey of non-Desarguian planes).

Isomorphic planar ternary rings yield isomorphic projective planes, however there exist projective planes that can be coordinitised by non-isomorphic planar ternary rings.

Which planar ternary rings coordinitise their projective plane uniquely up to isomorphism?

(In other words there is a surjective function from isomorphism classes of planar ternary rings to isomorpism classes of projective planes; on what domain is it injective?)

As stated in Weibel's survey above two ternary rings are isomorphic if (and only if) the automorphism group of the projective plane maps any quadrilateral into any other quadrilateral. It is not clear to me, however, what this means about the ternary ring itself.

I know this class includes the alternative division rings (see Bruck and Kleinfeld - The Structure of Alternative Division Rings, Theorem B in Section 5).

P.S. I originally stated that this class contains the near-fields, but this is false.

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This is equivalent to the automorphism group of the plane being transitive on quadrangles, right? – Harry Altman Sep 14 at 23:02
Yes it is; this is stated in Wiebel's survey. – bearded_pard Sep 21 at 2:14
The spelling is Weibel. He wrote a famous book on homological algebra. – Todd Trimble Sep 22 at 23:53
Thanks Todd; I corrected the spelling in the original post. – bearded_pard Sep 24 at 3:38

3 Answers

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The answer to the problem is well known; since :

all the PTRs coordinatizing a given projective plane are isomorphic iff it is a moufang plane iff all these PTRS are isomorphic alternative division rings. http://www.math.uni-kiel.de/geometrie/klein/math/geometry/moufang.html

However the problem to give a purely algebraic proof of this result seem to be an open problem.

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Do you have a reference with a proof of this fact? – bearded_pard Sep 24 at 3:54
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A necessary and sufficient condition so that two planar ternary rings coordinatize isomorphic projective planes was obtained at 2004 by A.Grari ( http://www.springerlink.com/content/4fmhjmhy8eqxenrw/) the well know theorems on skew-fields and division ring and alternatif division ring can be deduced from this condition as particular cases.

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This isn't the question asked, which is to characterize projective planes which have only one isomorphism class of ternary ring, or equivalently whose automorphisms act transitively on quadrangles. – BS Sep 12 at 8:57
It doesn't answer my question, but it does give an alternative (perhaps more tractable) formulation of it; in Grari's terminology he shows that if one Planar Ternary Ring is isomorphic to another, it is comparitive to the first, or comparitive to one of four particular types of Planar Ternary Rings with a Zero. My question then becomes: When are two Hall Ternary Rings of that are equivalent in this way isomorphic? As an example, in this paper Grari merely shows this is the case for skewfields, when Bruck and Kleinfeld showed it is true for alternative division rings. – bearded_pard Sep 13 at 6:51
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Let pi be a projective plane coordinatized by a unique (up to isomorphism)Hall ternary ring S. If pi is finite then by the theorem of Ostrom-Wagner http://www.google.co.ma/url?sa=t&rct=j&q=flag+transitive+plane+wagner&source=web&cd=1&cad=rja&ved=0CCAQFjAA&url=http%3A%2F%2Fpages.uoregon.edu%2Fkantor%2FPAPERS%2FFlagTraPlanes.pdf&ei=AJ9TUKaRErGZ0QX1s4GYDw&usg=AFQjCNHNVNCAjd2szENPO_BVOHlucM_bEA then pi is desarguesian ; so S is a skew-field;and therefore there is no problem to solve in this case since the solution is well known.

If pi is infinite then pi is a flag-transitive plane so all the PTRs coordinatized this plane are equivalent under the operation of comparison. ( see lemma 4.1 http://www.springerlink.com/content/4fmhjmhy8eqxenrw/)
The problem posed can now be reduced to the following: When two equivalent (under comparison ) Hall ternary rings are isomorphic ?

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I think this is as good an answer as I am going to get: In the finite case it is necessary and sufficient for the ternary rings to be finite fields. It's still not clear to me in the infinite case, beyond that it includes alternative division rings. – bearded_pard Sep 21 at 2:24

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