Index
Perhaps at this point, you are curious as to how one could create a projective plane of order, say 31, and if such a projective plane even exists. We had enough trouble creating the projective planes of order 3 and 4!
It turns out that from ANY field, one can create a projective plane (we will call these projective coordinate planes)! Before we explore this idea further, let's recall the definition of a field.
Recall that both the real numbers and the complex numbers are fields. In addition, 
, where p is a prime number (and p is also the number of elements), is a field under modular arithmetic.
To construct a projective plane from a field, we shall use the following method. We will create a 3-dimensional vector space V= F3 over a field F by taking ordered triples (x, y, z) of our vector space where x, y, and z are all elements of our field F. It follows that if our field has k elements, then there are k3 many points in the corresponding vector space V. If our field has k elements, then it follows that each line through the origin is on k points: it is obvious that we can take any point distinct from the origin and draw it along a line with the origin. Thus, say we have (0,0,0) and (x,y,z) along a line through the origin. In order to have another point along this line, we must have some point, (rx,ry,rz) where r is some element of our field. Thus, it follows that we will only have exactly k-many points along a given line, as (rx,ry,rz) = (qx,qy,qz) if and only if r = q.
The lines that pass the through the origin are the one-dimensional vector subspaces (i.e., the set: {av | a is an element in our field F, v a nonzero vector in V}). The planes through the origin are the two-dimensional subspaces: {av + bw | a,b are in F, v and w linearly independent vectors in V. It is an easy exercise to see that our subspaces are closed.
Now, to finally create a model of a projective plane, we will see that each line of V that passes through the origin will be a "point" in our projective plane, and each plane of V will be a "line". The incidence between lines and planes is then conserved between points and lines: we will say that a one-dimensional subspace of V is incident with a 2-dimensional subspace of V precisely when the two-dimensional subspace contains the one-dimensional subspace. Notice how this mimicks the line-plane model of the real projective plane as discussed in Berger and Dorwart.
We will verify in an exercise that this is indeed a projective plane.
Now, let's look at some of the implications of all this. It follows that the projective plane created from a vector space V over a field with k elements will have order k. Thus, we know that a projective plane with prime order will always exist!
Also, note that any point of a projective plane created from a field F corresponds to a given set of points of a line through the origin in V = F3. Because we have shown that there exists a correspondance between each line through the origin and every point in our projective plane, any point can be identified as such. Thus, the collection of points that correspond to a given point are really just multiples of a given point (as each point must lie along a given line).
Many will use the notation [a,b,c] to refer to the point on a projective plane that corresponds to the line in V through the point (a,b,c) and (0,0,0). Note that for any k in our field, [ka,kb,kc] is the same point as [a,b,c]. Note that the point [0,0,0] does not exist in our projective plane.
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PROBLEM 21: (a) Show that the space created above (the projective coodinate plane) is indeed a projective plane
PROBLEM 21: (b) Construct the Fano plane from a field.
Solutions
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References. Batten, Bennet, Beutelspacher & Rosenbaum, Polster.