projective plane of a ternary ring

Given a ternary ring (R,0,1,*), one can construct a projective planeMathworldPlanetmath π such that R coordinatizes π:

  1. 1.

    points of π are of the forms (x,y) or (m), where x,yR, and mR{}. We assume that the symbol is not a member of R.

  2. 2.

    lines of π are sets of points, of the following forms

    • [m,b]:={(x,y)y=x*m*b,x,y,m,bR}{(m)};

    • [a]:={(x,y)x=a,x,y,aR}{()};

    • []:={(m)mR{}}.

  3. 3.

    incidence relationPlanetmathPlanetmath is the same as set membership .

The line [] is called the line at infinity of R, and points (m) are the points at infinity (or slope points) of R.

Proposition 1.

[m,b][a] for any m,bR and aR{}. If (m1,b1)(m2,b2), then [m1,b1][m2,b2], for any m1,m2,b1,b2R. Also, if ab, then [a][b], for any a,bR{}.


The first assertion is true, since ()[a]-[m,b], (m)[m,b]-[a] for any aR, and (0,b)[m,b]-[].

If m1m2, then (m1)[m1,b1]-[m2,b2], while (m2)[m2,b2]-[m1,b1]. If b1b2, then (0,0*m1*b1)=(0,b1)(0,b2)=(0,0*m2*b2), so that (0,b1)[m1,b1]-[m2,b2], while (0,b2)[m2,b2]-[m1,b1].

Next, if a, then (a,0)[a]-[], while (a)[]-[a].

Finally, if a,bR with ab, then (a,0)[a]-[b], while (b,0)[b]-[a].∎

This shows that no two lines have the same “coordinates”. In fact, more is true:

Proposition 2.

π, with points and lines defined above, is indeed a projective plane.


We need to verify that points and lines satisfy the axioms of projective plane.

  1. 1.

    Axiom 1: two distinct points are incidentMathworldPlanetmath with exactly one line. There are four cases:

    • Given a,bR{}, with ab, points (a),(b) lie on line [].

    • Given x,yR with xy, points (x,y),() lie on line [x].

    • Given x,y,mR with xy, there is a unique bR such that y=x*m*b. Then points (x,y),(m) lie on line [m,b].

    • Given x1,y1,x2,y2R, with (x1,y1)(x2,y2). If x1=x2=x, then points (x1,y1),(x2,y2) lie on line [x]. Otherwise, there is a unique pair (m,b) such that y1=x1*m*b and y2=x2*m*b, so that both points lie on line [m,b].

    All lines above are unique by proposition 1. From this, let us write PQ for the line where points P,Q lie on.

  2. 2.

    Axiom 2: two distinct lines are incident with exactly one point. In light of the previous axiom, all we need to show is that two distinct lines contain at least one point. There are three cases:

    • Given lines [a],[c] with a,cR{} and ac, they both contain point ().

    • Given lines [m,b],[a], if a=, then both contain point (m), and if a, then both contain point (a,a*m*b).

    • Given lines [m1,b1],[m2,b2], with (m1,b1)(m2,b2). If m1=m2=m, then both lines contain point (m). Otherwise, the equation x*m1*b1=x*m2*b2 has a unique solution for x. Say the solution is a. Then both lines contain point (a,a*m1*b1)=(a,a*m2*b2).

  3. 3.

    Axiom 3: there exists a quadrangle. The four points are (0,0),(1,1),(0),(), and all six lines (0,0)(1,1)=[ 1,0], (0,0)(0)=[ 0,0], (0,0)()=[ 0], (1,1)(0)=[ 0,1], (1,1)()=[ 1], and (0)()=[] are all distinct. Hence they form a quadrangle.

Therefore, π is a projective plane. ∎

If one removes the line [] and all the points on it, then the resulting plane is an affine plane. In this regard, R can be used to coordinatize an affine plane. It is possible to construct the affine plane from R without the use of the line at infinity.

Given a projective plane, one can also construct a ternary ring that coordinatizes the plane. See this entry ( for more detail.


  • 1 R. Artzy, Linear GeometryMathworldPlanetmath, Addison-Wesley (1965)
Title projective plane of a ternary ring
Canonical name ProjectivePlaneOfATernaryRing
Date of creation 2013-03-22 19:14:35
Last modified on 2013-03-22 19:14:35
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 11
Author CWoo (3771)
Entry type Definition
Classification msc 51A35
Classification msc 51E15
Classification msc 51A25
Related topic TernaryRingOfAProjectivePlane