You are here
Hometernary ring of a projective plane
Primary tabs
ternary ring of a projective plane
In the parent entry, we constructed a coordinate system in a projective plane. We used a set $\mathcal{R}$ of labels for the points on the identity line as “values” for the coordinates. In this entry, we will equip $\mathcal{R}$ with some operations based on properties of the projective plane, and turn $\mathcal{R}$ into an algebraic system. In fact, we show that $\mathcal{R}$ with the operations is a ternary ring.
Let $\pi$ be a projective plane, coordinatized by a set $\mathcal{R}$ of symbols, which includes $0$ and $1$. Let $\ell$ be the designated line of infinity, and $Y$ the intersection of $\ell$ and the $Y$axis. Below is a table showing forms of coordinates that points and lines of $\pi$ may have:
points  lines 

$(x,y)$ if they do not lie on $\ell$  $[\,m,b\,]$ if they do not pass through $Y$ 
$(m)$ if they lie $\ell$ but not $Y$  $[\,x\,]$ if they pass through $Y$ but are not $\ell$ 
$(\infty)$ if it is the point $Y$  $[\,\infty\,]$ if it is $\ell$ 
What are the relationships between the coordinates of lines and the coordinates of those points incident with the lines? There are three cases:
1. If the line is $[\,\infty\,]$, the points on it have the form $(m)$, where $m\in\mathcal{R}\cup\{\infty\}$,
2. If the line is of the form $[\,x\,]$ where $x\neq\infty$, the points on it (other than $Y$ itself) have the form $(x,y)$,
3. If the line is of the form $[\,m,b\,]$, then a point on it either has the form $(x,y)$ or is $(m)$.
The last case is the most interesting. In particular, what is the relationship among $m,b,x,y$? The following four equations show us that any three of them determine the remaining one uniquely:
 S.
$(0,b)(x,y)\cap[\,\infty\,]=(m)$,
 T.
$(m)(x,y)\cap[\,0\,]=(0,b)$,
 U.
$(m)(0,b)\cap[\,x\,]=(x,y)$,
 V.
$(m)(0,b)\cap[\,0,y\,]=(x,y)$.
Equations $S,T$ may be illustrated by the following figure:
\pspicture(4,2)(4,3) \pssetunit=25pt \psline¡¿(4,3)(1,3) \psline¡¿(4.5,2.5)(3.5,2.5) \psline¡¿(3,3)(1.5,3) \psline¡¿(4,1.35)(2.5,0) \psdots[linecolor=black,dotsize=5pt](3.75,2.5) \psdots[linecolor=black,dotsize=5pt](1.2,2.6) \psdots[linecolor=black,dotsize=5pt](2.625,2.5) \uput[r](0.05,1.155)$[\,\infty\,]$ \uput[l](2,1.155)$[\,0\,]$ \psdots[linecolor=black,dotsize=5pt](3.0794,1.1588) \psdots[linecolor=black,dotsize=5pt](0.9859,0.3145) \uput[u](3.5,1.1588)$(0,b)$ \uput[r](1.25,0.5)$(m)$ \psdots[linecolor=black,dotsize=5pt](0.75,0.675) \uput[d](0.75,0.75)$(x,y)$
and equations $U,V$ are illustrated by the next two figures below:
\pspicture(3,3)(3,3) \pssetunit=25pt \psline¡¿(4,3)(1,3) \psline¡¿(4.5,2.5)(3.5,2.5) \psline¡¿(3,3)(1.5,3) \psline¡¿(4,1.35)(2.5,0) \psdots[linecolor=black,dotsize=5pt](3.75,2.5) \psdots[linecolor=black,dotsize=5pt](1.2,2.6) \psdots[linecolor=black,dotsize=5pt](2.625,2.5) \psdots[linecolor=black,dotsize=5pt](3.0794,1.1588) \psdots[linecolor=black,dotsize=5pt](0.9859,0.3145) \uput[u](3.5,1.1588)$(0,b)$ \uput[r](1.25,0.5)$(m)$ \psdots[linecolor=black,dotsize=5pt](0.75,0.675) \uput[r](0.75,0.85)$(x,y)$ \uput[l](0.85,0.5)$[\,x\,]$ \psline¡¿(1.255,3)(0.4305,3) \pspicture(3,3)(3,3) \pssetunit=25pt \psline¡¿(4,3)(1,3) \psline¡¿(4.5,2.5)(3.5,2.5) \psline¡¿(3,3)(1.5,3) \psline¡¿(4,1.35)(2.5,0) \psdots[linecolor=black,dotsize=5pt](3.75,2.5) \psdots[linecolor=black,dotsize=5pt](1.2,2.6) \psdots[linecolor=black,dotsize=5pt](2.625,2.5) \psdots[linecolor=black,dotsize=5pt](3.0794,1.1588) \psdots[linecolor=black,dotsize=5pt](0.9859,0.3145) \uput[u](3.5,1.1588)$(0,b)$ \uput[r](1.25,0.5)$(m)$ \psdots[linecolor=black,dotsize=5pt](0.75,0.675) \uput[u](0.75,0.675)$(x,y)$ \uput[d](0.15,1.25)$[\,0,y\,]$ \psline¡¿(3,0.5417)(3.5497,3)
As a result, the four equations above define four partial functions $S,T,U,V:\mathcal{R}^{3}\to\mathcal{R}$, where
$S(b,x,y)=m,\quad T(m,x,y)=b,\quad U(x,m,b)=y,\quad\mbox{and}\quad V(y,m,b)=x.$ 
Out of the four partial functions, $T,U,V$ are total. $S$ is not, because $S(b,0,y)$ does not exist, as $(0,b)(0,y)\cap[\,\infty\,]=(\infty)$. It is easy to see that all four partial functions are onto.
In light of the discussion above, we see that $(\mathcal{R},0,1,F)$ is an algebraic system, where $F$ is any of the three total functions. In the literature, $F=U$ is the choice. For the remainder of the discussion, we write $y=x*m*b$ to denote $y=U(x,m,b)$.
Proposition 1.
$(\mathcal{R},0,1,*)$ is a ternary ring.
Proof.
There are five conditions to verify:
1. Since $(0)(0,b)\cap[\,x\,]=(x,b)$, we have $x*0*b=b$. Since $(m)(0,b)\cap[\,0\,]=(0,b)$, we have $0*m*b=b$.
2. Next, since $(m)(0,0)\cap[\,1\,]=(1,m)$, we have $1*m*0=m$, and since $(1)(0,0)\cap[\,x\,]=(x,x)$, we have $x*1*0=x$.
3. Given $m_{1}\neq m_{2}$ and $b_{1},b_{2}$, we have two distinct lines $[\,m_{1},b_{1}\,]$ and $[\,m_{2},b_{2}\,]$, which intersect at a unique point $(x,y)$, whence $x*m_{1}*b_{1}=y=x*m_{2}*b_{2}$.
4. Given $m,x,y$, have two points $(m)$ and $(x,y)$, both incident with line $(m)(x,y)$. This line is not $[\,\infty\,]$, which means that it intersects the $Y$axis $[\,0\,]$ at $(0,b)$ for some unique $b$, whence $x*m*b=y$.
5. Given $x_{1}\neq x_{2}$ and $y_{1},y_{2}$, we have two distinct points $(x_{1},y_{1})$ and $(x_{2},y_{2})$. The line $(x_{1},y_{1})(x_{2},y_{2})$ does not pass through $Y(=(\infty))$, or else it has coordinate $[\,x\,]$ for some $x$, forcing $x_{1}=x=x_{2}$. As a result $(x_{1},y_{1})(x_{2},y_{2})$ intersects $[\,\infty\,]$ at $(m)$ for some $m\neq\infty$, and intersects $[\,0\,]$ at $(0,b)$ for some $b$. In other words, $(x_{1},y_{1})(x_{2},y_{2})=[\,m,b\,]$, or, equivalently, $x_{1}*m*b=y_{1}$ and $x_{2}*m*b=y_{2}$. The pair $(m,b)$ is clearly unique.
This shows that $\mathcal{R}$, together with $0,1$ and the ternary operation $*$, is a ternary ring. ∎
It is easy to see that points $(x,y)$ on lines with coordinates of the form $[\,m,b\,]$ satisfy equations of the form $y=x*m*b$, and points $(x,y)$ (except the point $(\infty)$) on lines with coordinates of the form $[\,a\,]$, where $a\neq\infty$, satisfy equations of the form $x=a$.
Remarks.

We call $\mathcal{R}$ the ternary ring of the projective plane $\pi$ determined by the quadrangle $O,I,X,Y$. We also call $\mathcal{R}$ a coordinate ring of $\pi$. If we are given a different quadrangle, we will end up with a different coordinate ring for $\pi$, and the two ternary rings may not be isomorphic.

It can be shown that any ternary ring is a coordinate ring for some projective plane. See this entry for more detail.
References
 1 M. Hall, Jr., The Theory of Groups, Macmillan (1959)
 2 R. Artzy, Linear Geometry, AddisonWesley (1965)
Mathematics Subject Classification
51E15 no label found51N15 no label found05B25 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections