Projective Planes in the Finite Field Zp

# Projective Planes in the Finite Field Zp

On The Fano Plane page we looked at the projective plane over the field of integers modulo $2$, $\mathbb{Z}_2$ known as the Fano Plane $\mathbb{P}^2 (\mathbb{Z}_2)$ - the smallest projective plane (due to $\mathbb{Z}_2$ being the smallest field) containing $7$ points and $7$ lines.

We will now look more into the projective planes on the finite field $\mathbb{Z}_p$ where $p$ is a prime number.

Theorem 1: The projective plane $\mathbb{P}^2 (\mathbb{Z}_p)$ has $\frac{p^3 - 1}{p - 1} = p^2 + p + 1$ distinct points. |

**Proof:**Let $p$ be a prime number so that $\mathbb{Z}_p = \{ 0, 1, ..., p-1 \}$ is a field containing $p$ elements. All points in $\mathbb{P}^2(\mathbb{Z}_p)$ will be of the form:

\begin{align} \quad \mathbf{x} = [x_1, x_2, x_3] \in \mathbb{P}^2(\mathbb{Z}_p) \: , \quad x_1, x_2, x_3 \in \mathbb{Z}_p \:, \mathrm{not \: all \: zero} \end{align}

- Each of the coordinates in the point $\mathbf{x}$ can be one of $p$ elements from $\mathbb{Z}_p$ so ther are $p^3$ arrangements of the coordinates, however, $x_1, x_2, x_3$ cannot be all zero, so there are $p^3 - 1$ acceptable arrangements of the coordinates.

- $p^3 - 1$ is an over count of the number of
*distinct points*in $\mathbb{P}^2(\mathbb{Z}_p)$ since some coordinates are equivalent by scaling. Consider an arbitrary coordinate $[x_1, x_2, x_3] \in \mathbb{P}^2(\mathbb{Z})_p$. If we multiply this coordinate by any nonzero element in $\mathbb{Z}_p$, i.e, by $1, 2, ..., p-1$ then:

\begin{align} \quad \mathbf{x} = [x_1, x_2, x_3] \sim [kx_1, kx_2, kx_3] = k\mathbf{x} \end{align}

- There are $p - 1$ nonzero elements in $\mathbb{Z}_p$ which we can multiply $[x_1, x_2, x_3]$ by, and so there are $p - 1$ groups of coordinates which are equivalent to each other. Therefore, the total number of distinct points in $\mathbb{P}^2(\mathbb{Z}_p)$ is:

\begin{align} \quad \frac{p^3 - 1}{p - 1} = p^2 + p + 1 \quad \blacksquare \end{align}

Corollary 1: The projective plane $\mathbb{P}^2 (\mathbb{Z}_p)$ has $\frac{p^3 - 1}{p - 1} = p^2 + p + 1$ distinct lines. |

**Proof:**This followings immediately from the duality of points and lines of projective planes. Each line in $\mathbb{P}^2(\mathbb{Z}_p)$ is of the form $<a_1, a_2, a_3>$ where $a_1, a_2, a_3 \in \mathbb{P}^2(\mathbb{Z}_p)$ are not all zero and $<a_1, a_2, a_3> = <ka_1, ka_2, ka_3>$ for $k \in \mathbb{Z}_p \setminus \{ 0 \}$ gives us that there are $\displaystyle{\frac{p^3 - 1}{p - 1} = p^2 + p + 1}$ distinct lines in $\mathbb{P}^2(\mathbb{Z}_p)$. $\blacksquare$