Let $K$ be a[n associative] division algebra (= skew field). By the “projective plane” $\mathbb{P}^2(K)$ over $K$ I mean, as usual, the set of triples $(x,y,z)$ of elements of $K$, not all zero, up to left-multiplication by a nonzero element $\lambda$ of $K$; write $(x:y:z)$ for the equivalence class (of all $(\lambda x, \lambda y, \lambda z)$). By a “line” $[u:v:w]$ in $\mathbb{P}^2(K)$, where $u,v,w$ are three elements of $K$ not all zero, I mean the set of points $(x:y:z)$ of $\mathbb{P}^2(K)$ such that $xu+yv+zw = 0$. (The line, of course, is determined by $u,v,w$ up to right-multiplication by a nonzero element.)
If $K$ is actually commutative (i.e., is a field), then the unique line $[u:v:w]$ connecting two distinct points $(x_1:y_1:z_1)$ and $(x_2:y_2:z_2)$ of $\mathbb{P}^2(K)$ is given by a simple formula, $$ \left\{ \begin{aligned} u &= y_1 z_2 - z_1 y_2 \\ v &= z_1 x_2 - x_1 z_2 \\ w &= x_1 y_2 - y_1 x_2 \\ \end{aligned} \right. \tag{*} $$ (this is the formula for the cross-product of vectors in $K^3$: the proof is a straightforward application of Cramer's rule to solve $x_1 u + y_1 v + z_1 w = 0$ and $x_2 u + y_2 v + z_2 w = 0$ in $u,v,w$).
Returning to the case where $K$ is no longer assumed commutative, the above formulas (*) no longer apply. It is nevertheless true that two distinct points in $\mathbb{P}^2(K)$ are connected by a line, but attempting to compute the coordinates of this line seems to give something messy like $$ \left\{ \begin{aligned} u &= -(y_2^{-1} x_2 - y_1^{-1} x_1)^{-1}\,(y_2^{-1} z_2 - y_1^{-1} z_1) \\ v &= -(x_2^{-1} y_2 - x_1^{-1} y_1)^{-1}\,(x_2^{-1} z_2 - x_1^{-1} z_1) \\ w &= 1 \\ \end{aligned} \right. \tag{†} $$ in the general case, along with a profusion of special cases.
Can we do better? More precisely:
Questions:
Is there a formula generalizing (*) to the case of division rings? More precisely, is there a formula for the line connecting two distinct points $(x_1:y_1:z_1)$ and $(x_2:y_2:z_2)$ of $\mathbb{P}^2(K)$ that does not involve division? Even more precisely, can we find three elements $U,V,W$ of the free algebra $\mathbb{Z}\langle x_1,y_1,z_1,x_2,y_2,z_2\rangle$ such that $[U(x_1,\ldots,z_2) : V(x_1,\ldots,z_2) : W(x_1,\ldots,z_2)]$ is defined at least somewhere, and produces the line connecting $(x_1:y_1:z_1)$ and $(x_2:y_2:z_2)$ when the formula is defined? (Or better, whenever the two points are distinct?)
If not, what is a proof that such a formula cannot exist?
If not, can we at least simplify (†) into something slightly less messy, or more symmetric? (E.g., something symmetric under cyclic permutation of $x_i,y_i,z_i$ and $u,v,w$? Again, if this cannot be done, I would welcome a proof.)
Reference request: where can I find a computational proof that the theorem of Desargues holds in $\mathbb{P}^2(K)$ based on direct coordinate calculations and that is, hopefully, not too messy? (The field case is fairly straightforward from formulas (*), so I would like to know to what extent the division algebra case can be handled analogously.)
