Non-Desarguesian plane

Example

A non-Desarguesian plane is a projective plane in which Desargues's theorem does not hold. A concrete one is the projective completion of the Moulton plane. Check: the Moulton plane has triangles $ABC$ and $A'B'C'$ in perspective from a point $O$. In it $AB \parallel A'B'$ and $AC \parallel A'C'$, but $BC$ is not parallel to $B'C'$. Add the line at infinity. Then $AB \cap A'B'$ and $AC \cap A'C'$ lie on the line at infinity, but $BC \cap B'C'$ does not, so the three points are not collinear.

So this plane is not a Desarguesian plane, and it is not isomorphic to $\mathbb{P}^2(D)$ for any division ring $D$.