Construction of a finite projective plane of order $p$, for any prime $p$

Question

I have this construction of a finite projective plane (FPP) of prime order $p$, but I am not sure what's going on. We have already proved that FPPs of order $q$ have $q^2+q+1$ lines and points (if that is relevant).

Take the linear space $\Bbb{Z}_p^3$, and consider the one dimensional subspaces. Notice that there are $(p^3-1)/(p-1)=p^2+p+1$ such subspaces. Let $n=p^2+p+1$, and let $P_1,\ldots,P_n$ be the subspaces, and $w_i\in P_i$ be any vector in each. Let $X_i$ be the $2$-dimensional subspace of $\Bbb{Z}_p^3$ containing $x\in\Bbb{Z}_p^3$ such that $w_i^\top x=0$. The line $L_i$ is formed by the points $P_j$ such that $P_j\subseteq X_i$. Note that $X_i$ contains $p^2-1$ nonzero vectors. Hence $L_i$ contains $(p^2-1)/(p-1)=p+1$ points.

I do not understand the structure of $\Bbb{Z}^3_p$: is this the set $$\{(a,b,c):a,b,c\in\{0,\ldots,p-1\}\}?$$ In particular, how do we know that there are $(p^3-1)/(p-1)$ one dimensional subspaces? I would have thought that this space, $\Bbb{Z}_p^3$, had just three one-dimensional subspaces generated by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. Similarly, how do we know that $X_i$ contains $p^2-1$ nonzero vectors?

Answer 1

You should read $\mathbb Z_p^3$ as $(\mathbb F_p)^3$, that is, the space of vectors consisting of three elements from the field with $p$ elements. This is a vector space over $\mathbb F_p$.

(So the elements are not just picked from the numbers $\{0,1,\ldots,p-1\}$; they come equipped with arithmetic operations modulo $p$).

There are $p^3-1$ nonzero vectors in this space; each nonzero vector is in exactly one one-dimensional subspace (namely its span), and each one-dimensional subspace contains $p$ vectors (namely: a basis vector multiplied by each of the elements of $\mathbb F_p$ in turn), of which $p-1$ are nonzero.

Therefore there must be $\frac{p^3-1}{p-1}$ linear subspaces in order to account for all of the nonzero vectors.

I would have thought that this space, $\Bbb{Z}_p^3$, had just three one-dimensional subspaces generated by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.

No, that would be akin to saying that the only one-dimensional subspaces of $\mathbb R^3$ are the axes. Every nonzero vector spans a one-dimensional subspace -- for example $\{(x,x,x)\mid x\in\mathbb F_p\}$ is such a subspace, spanned by $(1,1,1)$.

Answer 2

1. Yes, that's what's meant by $\mathbb Z_p^3$.

2. Each 1-dimensional subspace consists of all scalar multiples of one nonzero point $(a,b,c)$, so it contains exactly $p$ items. Since there are $p^3 -1 $ points to choose, and each subspace contains $p$ points, you might think you'd get $(p^3 -1 )/p$ subspaces. . That's wrong because $(0,0,0)$ is in every subspace. So to simplify, let's eliminate it: look at the $p^3 - 1$ nonzero points, and look at the nonzero items in the subspace each one spans. Now you've got $(p^3 - 1) / (p-1)$ subspaces.

3. For the third item (counting nonzero elements of $X_i$), consider what's meant by $w_i^t x = 0$: that's one linear equation in three unknowns. Take any two linearly independent solutions (over the finite field of $p$ elements), and every other solution can be written as a combination of the two. How many such combinations are there? Well, there are $p$ possibilities for the coefficients of each, so there are $p^2$ combinations. One of these is the zero combinations; all others are nonzero because the solutions were chosen linearly independent. hence there are $p^2 - 1$ such points.