0
$\begingroup$

Does the projective closure of an affine plane of order $n$ also have order $n$?

I know that there exists a projective plane of order $n$ if and only if there exists an affine plane of order $n$, although I have not seen the proof of this statement. I am pretty sure the answer to the question above is yes, but I would like to know why.

Thank you!

$\endgroup$
1
$\begingroup$

Yes. The order of a block design is the number of blocks through a point minus the numbef of blocks containing any pair of points. In an affine plane of order $n$, each point lies on $n+1$ lines, and any pair of points lie together on one line. When you create the projective closure, every point still lies on $n+1$ lines, and there is still a single line through every pair of points.

The concept of a projective closure is half of the proof that an affine plane of order $n$ exists iff a projective plane of order $n$ exists. The other direction requires you to start with a projective plane and remove one line and the points on it. Showing the axioms for an affine plane is not hard, and I encourage you to give it a try!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.