**Def**: A projective space P is ** finite** if its point set is finite.

(Note that in exercise 31 you show that if dim(P) > 1 then P is finite if and only if its line set is finite.)

By virtue of the above lemma, there exists an integer q such that every line has q+1 points on it. This q is called the ** order** of the finite projective space

**Lemma 1.5.2**: Let **P** be a finite projective space of dimension d 2 and order q. Then for each point Q of **P** the quotient geometry **P**/Q has order q.

**Theorem 1.5.3**: Let **P** be a finite projective space of dimension d and order q, and let **U** be a t-dimensional subspace of **P** (1t d). Then:

- |
**U**| = q^{t}+ q^{t-1}+ ... + q + 1 = (q^{t+1}-1)/(q-1).

In particular |**P**| = q^{d}+ ... + q + 1. - The number of lines of
**U**through a fixed point of**U**equals q^{t-1}+ ... + q + 1. - The total number of lines of
**U**equals

(q^{t}+ q^{t-1}+ ... + q + 1)(q^{t-1}+ ... + q + 1)/(q+1).

**Theorem 1.5.4**: Let **P** be a finite projective space of dimension d and order q. Then

- the number of hyperplanes of
**P**is q^{d}+ ... + q + 1; - the number of hyperplanes of
**P**through a fixed point P is q^{d-1}+ ... + q + 1.

*Pf*: (a) We prove the result by induction on the dimension. If d = 1, **P** is a line and the hyperplanes are points. Since **P** has order q, there are q + 1 points in **P**. If d = 2, **P** is a plane and the hyperplanes are lines. By Theorem 1.5.3 (c), the number of lines in **P** is (q^{2} + q + 1)(q + 1)/q+1 = q^{2} + q + 1.

Now, suppose that the assertion is true for all finite projective spaces of dimension s-1. Note that we may assume that s-1 > 2 since the result has been established for dimensions 1 and 2. Let **P** be a finite projective space of dimension s, and **H** a hyperplane of **P**. By the dimension formula, every hyperplane **H**, meets **H** in a subspace of dimension s - 2 (which is > 1). Thus, any hyperplane of **P** (**H**) is spanned by an (s-2)-dimensional subspace of **H** and a point outside of **H**.

For each (s-2)-dimensional subspace **U** of **H** and each point P**P**\**H**, the subspace <**U**,P> is a hyperplane (of **P**) containing (q^{s-1} + ... + q + 1) - (q^{s-2} + ... + q + 1) = q^{s-1} points outside of **H**. Since there are q^{s} points of **P** outside of **H**, there are q = q^{s}/q^{s-1} hyperplanes **H** through **U**. By induction, there are q^{s-1} + ... + q + 1 hyperplanes of **H** (since dim(**H**) = s-1), which are (q-2)-dimensional subspaces of **P** contained in **H**. Therefore, the total number of hyperplanes of **P** is 1 + q(q^{s-1} + ... + q + 1) = q^{s} + ... + q + 1.

(b) Let P be a point of **P**, and **H** a hyperplane not through P. Any hyperplane of **P** through P meets **H** in a hyperplane of **H**. By (a) there are q^{d-1} + ... + q + 1 of these.

**Corollary 1.5.5**: A finite projective plane of order q has q^{2} + q + 1 points.

Order | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

No. | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 4 | 0 | 1 | ?? | 1 | 0 | ?? | 22 | 1 | ?? | 1 | ?? |

1. The only *known* projective planes have orders that are prime powers.

2. (*Bruck-Ryser Theorem*) If q 1 or 2 mod(4) and a projective plane of order q exists, then q is the sum of two integral squares (one of which may be 0).

3. The order 10 was eliminated by a massive computer search.