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My question is this: Is the projective plane $P(K^3)$ (the points are the one-dimensional subspaces and the lines are the two-dimensional subspaces) for a division ring $K$ isomorphic to its dual?

I know that this is true if $K$ is a field, and I also see several ways to prove it (by using the dual vector space or a bilinear form). However, none of these proofs are in an obvious way still correct if $K$ is only a division ring.

Thus, I am not at all sure how to prove the above statement (if it is correct at all) without running into some problem with non-commutativity. Can anybody help?

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I know how to construct a dualization in $\mathbb RP^2$ using only the axioms for the projective plane. However, I have no proof that the construction will preserve incidence in all Desarguesian planes. If someone managed to proove that theorem, then we'd have a proof that planes over skew fields are self-dual as well. Should I post details on my construction, even while the proof is still missing? – MvG Nov 24 '12 at 20:57

You having trouble proving it because it is false. A projective plane over a division ring is isomorphic to its dual if and only if the division ring admits an anti-automorphism. Since there exist division rings not admitting anti-automorphisms, the projective plane over such a division ring is not isomorphic to its dual. It is isomorphic to the projective plane over the opposite division ring : the one with multiplication reversed.

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