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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.

( 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
up vote 3 down vote accepted

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.

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