Numbers of vectors in a vector space over a finite field, with different multiplication I had a recent question in an assignment that I couldn't complete.  We are given the following:


*

*$q$ is an odd prime power.

*$(F,+,\cdot)=\text{GF}\left(q^2\right)$.

*$K$ is the $q$ element subfield of $F$.

*$\square=\text{non-zero squares of F}= \{ \alpha^{2i} \} $, where $\alpha$ is a primitive root of $\mathrm{GF}(q^2)$. 

*$\boxtimes=\text{non-squares of F}= \{ \alpha^{2i+1} \} $
(i.e. $F=\square\sqcup\boxtimes\sqcup 0 $)


*

*$o \colon F \times F \rightarrow F$ is given by
$$a o b = \left\{
  \begin{array}{lr}
    ab & a \in \square \\
    ab^q & a \in \boxtimes \\
    0 & a = 0
  \end{array}\right.$$
So basically it's regular multiplication with the Frobenius automorphism (sometimes).


For previous parts of the question, I've determined:


*

*$N$ = left near field $(F,+,o)$

*$N$ is a vector space over $K$
Now I'm asked to find the number of points in a set $P$, where each point is defined as:
$P$ = set of triples (x,y,z), not all 0, identified up to scalar multiplication of N{0}
i.e. $(x,y,z) == (n o x, n o y, n o z)$ $\forall n \in N \{ 0 \}$
So $P$ is the vector space $N^3$
I was hoping to use the follow: for a $n$ dimensional vector space over a finite field of order $q$, we have 
$\left(
\begin{array}{c}
 n \\
 k
\end{array}
\right)_q$ k-dimensional subspaces.
(http://en.wikipedia.org/wiki/Gaussian_binomial_coefficient under Applications).  But I'm unsure how to prove its allowed (or even if it is)
 A: 
Yes. The number of elements of P agrees with the usual formula for the number of elements of projective space over a field.  This follows by Orbit-Stabilizer, and the fact that the groups and stabilizers have the same size.

Let K be a finite field of size q, where q is an odd prime power, and let F be a finite field of size q2 containing K.  We define a new (unital, associative, but only partially distributive) multiplication on F, and call the resulting structure N.  N is a 2-dimensional vector space over K and is the (regular) Dickson near-field construction.
In particular, for every nonzero a in N, the map μa:N→N:x↦(a o x) is a K-automorphism of the underlying vector space of N. The set of μa for nonzero a in N forms a group, and P is the set of orbits of the nonzero elements of N3 (a 6-dimensional K-vector space) under this action of μ.
The important feature of μ is that it is sharply transitive, that is, it acts regularly on the nonzero elements of N.  If (x,y,z) is a triple of nonzero elements, then μ acts on each component regularly, and the orbit has size |N|−1.  If one of the components is 0, say z=0, then in effect μ is only acting on (x,y), and again the orbit has size |N|−1.  The only time the argument fails is if x=y=z=0, but this point is discarded for P anyways.  In other words, every orbit of μ on N3−0 has size |N|−1, and so the number of orbits is (|N|3−1)/(|N|−1) = q4 + q2 + 1, just like for the projective space over F.
