Create projective plane

Question

Please explain this https://math.stackexchange.com/a/463369/672948 in a simpler way. I am not from higher mathematics background and these terms are quite hard to understand.

I am clear upto finding $\mathbb{P}^2(F)$ ie.. $P_1=(0,0,1)$,... $P_{13}=(1,0,0)$. What does the below line mean?

equate, for example, the point $(2,1,2)$ with $2(2,1,2)=(1,2,1)=P_6$ and the point $(1,2,0)$ with the point $2(1,2,0)=(2,1,0)=P_{12}$ ?

How did they find this for a line ? $$ L_U=\{P_6,P_3,P_{13},P_9\}.$$

Answer 1

The concept you need to understand here is a finite field with $p^n$ elements where $p$ is a prime number. In this particular case the finite field has a prime number of elements where $p=3$. The three elements can be written as $\ \{0,1,2\}\ $ with the understanding that arithmetic with these elements are done using Modular arithmetic with modulus $p=3$. In particular, for multiplication you need to know that $\ 2\cdot 2=1\ $ besides the obvious $$ 1\cdot 1 = 1,\;\; 1\cdot 2 = 2\cdot 1 = 2,\;\; 0\cdot x=x\cdot 0 =0. \tag{1}$$ For addition you need to know that $\ 1+1=2\ $ and

$$ 2+2=1,\;\; 2+1 = 1+2 = 0,\;\; 0+x=x+0=x. \tag{2} $$

Given the point $\ P_6 = (1,2,1),\ $ for example, you multiply each component by $\ 2\ $ to get an equivalent point. That is, $$ 2P_6 = 2(1,2,1) = (2\cdot 1,2\cdot 2,2\cdot 1) = (2,1,2) \tag{3} $$ is an equivalent way to express the same point. Notice that this is just the same as replacing all the $2$s with $1$s and $1$s with $2$s and leaving $0$s the same.

Now, we can construct a line given one point and a direction where the direction point has last coordinate $0$. For example, given the point $\ P_3 = (0,2,1)\ $ and the direction point $\ P_{13} = (1,0,0)\ $ we add multiples of the direction point to the original point. Thus, $\ P_3+P_{13} = (0,2,1)+(1,0,0) = (1,2,1) = P_6,\ $ then $\ P_3+2P_{13} = (0,2,1)+(2,0,0) = (2,2,1) = P_9\ $ and then $\ P_2+3P_{13} = P_3\ $ which we started with since $3=0$. The direction point $\ P_{13}\ $ is also considered the point of the line "at infinity", and so now this line has four points $\ \{P_3,P_6,P_9,P_{13}\}.$ The other $12$ lines are constructed in a similar way, each with one of the four different directions.

Note: The line "at infinity" $\{P_{10},P_{11},P_{12},P_{13}\}$ contains all of the direction points "at infinity".

NOTE: The sum of equivalent points is not usually equivalent to the sum of the original points. For example, $\ P_1+P_2 = (0,1,2) = 2P_3\ $ and $\ P_1+2P_2 = 2P_{10}.\ $ Here $\ P_2\ $ is equivalent to $\ 2P_2\ $ but $\ 2P_3\ $ is not equivalent to $\ 2P_{10}.$ However, it does give us the four points $\ \{P_1,P_2,P_3,P_{10}\}\ $ of another line.

Answer 2

Well, the projective plane $P^2(K)$ over a field $K$ (say rational or real numbers in the examples below) consists of three subsets:

1. $A^2(K) = \{(a:b:1)\mid a,b\in K\}$. This set corresponds to the affine plane $K^2=\{(a,b)\mid a,b\in K\}$.

2. $L^2(K) = \{(a:1:0)\mid a\in K\}$. This is a line at infinity.

3. $\{(1:0:0)\}$. This is a point at infinity.

In the first case, the last coordinate is normalized to 1. For instance, the point $(2:1:2)$ corresponds to $(1:1/2:1)$ by multiplying the first point by $1/2$.

In the second case, the last coordinate is $0$ and the 2nd coordinate is normalized to 1. For instance, the point $(2:3:0)$ corresponds to $(2/3:1:0)$ by multiplying the first point by $1/3$.

In the third case, the 2nd and 3rd coordinates are $0$ and the first coordinate is normalized to 1. For instance, the point $(4:0:0)$ corresponds to $(1:0:0)$ by multiplying the first point by $1/4$.

Hope it helps!