# 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:

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

2. $N$ 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)

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what are you asking? it seems that $P$ is a projective plane over $N$, not "So P is the vector space N^3" –  yoyo May 6 '11 at 12:18
You lost me at the definition of $K$. When you wrote, "$q$-dimensional subfield of $F$," did you mean $q$-element subfield of $F$"? or maybe "$q$-dimensional vector space over $F$"? I can't make sense out of "$q$-dimensional subfield of $F$." –  Gerry Myerson May 6 '11 at 12:47
@Gerry, sorry, I meant q-element subfield of F –  Zeophlite May 8 '11 at 2:25