Desarguesian plane

Remarks

A projective plane is Desarguesian exactly when it is isomorphic to $\mathbb{P}^2(D)$ for some division ring $D$, and then $D$ is determined up to isomorphism.

Example: $\mathbb{P}^2(\Q)$ is Desarguesian. Check: it is $\mathbb{P}^2(k)$ for the field $k = \Q$, where Desargues's theorem holds.

Non-example: the projective completion of the Moulton plane is not Desarguesian, since it contains two triangles in perspective from a point whose corresponding sides meet in three non-collinear points.

By Wedderburn's little theorem every finite division ring is a field, so the finite Desarguesian planes are the planes $\mathbb{P}^2(\F_q)$, of order $q$ a prime power. Every Desarguesian plane is a Moufang plane.