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)