finite projective planes have q^2+q+1 points and q^2+q+1 lines

finite projective planes have

q

2

+

q

+

1

points and

q

2

+

q

+

1

lines

Given

a

finite projective plane

that contains a

quadrangle

OXYZ (i.e. no
three of these four points are on one line).

To prove

: the plane has

q

2

+

q

+

1

points and

q

2

+

q

+

1

lines for some integer

q

, and there are

q

+

1

points on each line and

q

+

1

lines through each point.

Let

x

and

y

be the lines OX and
OY, which must exist by the axioms. By the

assumption

OXYZ is a quadrangle
these lines are distinct and Z is not on them. Let there be

p

points X

i

on

x

other than O, for each of them one line ZX

i

exists, and is distinct
(one lines cannot

pass through

two X

i

unless it is

x

but that’s not a
line through Z).

Conversely

every line through Z must

intersect

x

in a
unique point (two lines intersecting in Z cannot intersect at another point,
and Z is not a point on

x

). So there are

p

+

1

lines through Z (OZ is one
of them). By the same reasoning, using

y

, there are

q

+

1

lines through Z
so

p

=

q

. We also found

q

+

1

points (including O) on

y

and the same number
on

x

. Intersecting the

q

+

1

lines through Z with XY (on which Z does not
lie, the quadrangle again) reveals at least

q

+

1

distinct points there and
at most

q

+

1

because for each point there there is a line through it and Z.

The lines not through O intersect

x

in one of the

q

points X

i

and

y

in one of the

q

points Y

j

. There are

q

2

possibilities and each of them
is a distinct line, because there is only one line through a given X

i

and
Y

j

. The lines that do pass through O intersect XY in one of the

q

+

1

points there, again one line for each such point and vice versa. That’s

q

+

1

lines through O and

q

2

not through O,

q

2

+

q

+

1

in all.

There are

q

+

1

lines through X (to each of the points of

y

) and

q

+

1

lines through Y (to each of the points of

x

). Intersect the

q

lines
through X other than XY with the

q

lines through Y other than XY, these

q

2

intersections

are all distinct because for any P there’s only one line
PX and one line PY. Note we did not use the line XY. Conversely for any P not
on XY there must be some PX and some PY, so there are exactly

q

2

points not
on XY. Add the

q

+

1

points on XY for a total of

q

2

+

q

+

1

.

The constructions above already showed

q

+

1

lines through some points (X, Y
and Z), by the same games as before that implies for each of them

q

+

1

points
on every line not through that point. We also saw

q

+

1

points on some lines
(

x

,

y

, XY) which implies for each of them

q

+

1

lines through every point
not on that line. Such reasoning covers

q

2

items on first application and
rapidly mops up stragglers on repeated application.

Some form of this proof is standard math lore; this version was half remembered and half reconstructed.

Title

finite projective planes have

q

2

+

q

+

1

points and

q

2

+

q

+

1

lines

Canonical name

FiniteProjectivePlanesHaveQ2q1PointsAndQ2q1Lines

Date of creation

2013-03-22 15:11:18

Last modified on

2013-03-22 15:11:18

Owner

marijke (8873)

Last modified by

marijke (8873)

Numerical id

4

Author

marijke (8873)

Entry type

Proof

Classification

msc 51A35

Classification

msc 05B25

Classification

msc 51E15

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