Number of points on a line in a finite projective plane

I've been reading some proofs regarding finite projective planes of order n, and often they start out by assuming that each line contains n+1 points. Is this a fact that follows from the axioms for finite projective planes? Or does this relate to the order of the finite field being projected? (for example in the Fano plane, does the fact that each line contains 3 points follow from the fact that there are 3 elements {0,1,2} in Z_2?) I haven't been able to find a clear explanation of this (clearly I do not yet have a solid understanding of how such planes are constructed). If someone could enlighten me that would be excellent.

1. Suggestion: Get a copy of Hartshorne's little book on Projective Geometry. It'll answer questions like this.

2. Your question seems to me "How do we know that any two lines have the same number of points?" (since whether we call this "$n$" or "$n+1$" really doesn't matter). Here's a quick proof, starting with Hartshorne's four axioms for projective planes.

A. For any two distinct points $P$ and $Q$, there is exactly one line containing both.

B. Any two lines meet in at least one point.

(Quick corollary: any two distinct lines meet in exactly one point. Pf: Suppose they met in two distinct points, $P$ and $Q$. By axiom 1, there's exactly one line containing these two points, but our two lines were supposed to be distinct. Contradiction.)

C. There exist three noncollinear points.

D. Every line contains at least 3 points.

Small theorem: if $b$ and $c$ are distinct lines, there's a point that's on neither of them. Proof: $b$ intersects $c$ at some point $Q$ by axiom B. Let $B \ne Q$ be another point of $b$ (Axiom D), and $C \ne Q$ be another point of $c$. Consider the line $d$ containing $B$ and $C$ (Axiom A). It's distinct from both $b$ and $c$, for it contains a point $C$ that's not in $b$, and a point $B$ that's not in $C$. Let $D$ be a point of $D$ distinct from $B$ and $C$. If $D$ were in $b$, then $d$ would contain the distinct points $B$ and $D$ that are also in $b$, hence $b$ and $d$ would be identical (Axiom A). Similarly, $D$ cannot be in $c$. So $D$ is a point that's on neither $b$ nor $c$.

Now: Let $a$ and $b$ be distinct lines, meeting at a point $Q$. Let $P$ be a point not on $a$ or $b$ (chosen using the small theorem above).

Consider the map $f$ from $a$ to $b$ defined as follows:

for any point $Q \in a$, the line $PQ$ exists; it must intersect the line $b$ at some point $Q'$. Define $f(Q) = Q'$.

The map is a bijection (and proving that is a good exercise for you, working from the axioms). Indeed, it's called a "perspectivity", but that's getting further along in the story...

• Thanks for your response! I will definitely try to find a copy of this text. I see how every line must have the same number of points, and also that each point must have an equivalent number of lines passing through it. My main confusion is related to the number of points/lines in a finite projective plane, which is n^2+n+1 -IF you take the number of points on each line to be n+1 (where n is the order of the field to be projected). I understand how one arrives at the expression n^2+n+1, I just don't get why there are n+1 points on each line in the first place. Thanks! – Will Jan 26 '16 at 16:52
• Oh...that's just counting. Suppose that each line has $k$ points and each point lies on $p$ lines. Claim 1: $k = p$. Proof: for a point $P$, pick a line $b$ not containing $P$. Then for each of the $k$ points of $b$, there's a line from $P$ to that point, and every line through $P$ is of this form. So there are $k$ lines through $P$, and $k = p$. Now if you look at those $k$ lines, they contain $k*k$ points...but one of them, $P$, got counted $k-1$ extra times, so there are $k^2 - k + 1$ points total. If you write $k = n+1$, this gives your formula. – John Hughes Jan 26 '16 at 17:02
• It seems you're thinking of projective planes that result from projectivizing affine planes over finite fields. So consider the line $1 x + 0 y + 0 = 0$ for any finite field. The only solutions $(x, y)$ have $x = 0$, and $y$ takes on $n$ possible values, where $n$ is the size of your field. When you add the point at infinity, you get $n+1$ points on your (projective) line. Alternatively consider triples $(x,y,z)$ with $0x + 0y + 1z = 0$. All nonzero solutions to this have $z = 0$. For each element $a$ of $F$ there's the point $(a, 1, 0)$, and then there's the element $(1,0,0)$. That's $n+1$. – John Hughes Jan 26 '16 at 17:09