# A question from “Foundations of Projective Geometry” by Hartshorne.

"Foundations of Projective Geometry" by Hartshorne says the following:

The completion of the affine plane of four points is a projective plane with 7 points.

The affine plane of $4$ points is essentially a paralellogram $ABCD$. The completion will contain $A,B,C,D,[AB],[AD],[AC],[BD]$. Here $[AC]$ is the point of intersection of all lines parallel to $AC$ with the line at infinity (in other words it is an ideal point).

Hence I am getting $8$ points instead of $7$. Where am I going wrong?

• I think you only have 6 points here: $[AB]$ and $[CD]$ are the same point, because the lines $AB$ and $CD$ are parallel, and similarly $[AC]$ and $[BD]$ are the same. On the other hand, you don't have $[AD]$, so that gets you back up to 7. – user64687 Jul 2 '14 at 12:10
• @AsalBeagDubh- Sorry I meant something else. I have edited the question. ABCD taken counter-clockwise. – fierydemon Jul 2 '14 at 12:23
• OK, now you have all the points. As I mentioned in my previous comment, two of the items in your list are the same: $[AC]=[BD]$. Do you see why? – user64687 Jul 2 '14 at 12:25
• @AsalBeagDubh- Which two? $[AC]$ and $[BD]$ are not the same! They are not co-incident or parallel. – fierydemon Jul 2 '14 at 12:27
• The basic point is this: there is no reason the picture you draw (a parallelogram in the plane) should accurately reflect the true properties of the affine plane (which is an abstract system). Perhaps the following question might be suggestive: you say these two lines intersect in a point. Which of the four points of the plane is it? – user64687 Jul 2 '14 at 12:33

There are only three distinct pencils of parallel lines: $\{\overleftrightarrow{AB}, \overleftrightarrow{CD}\}$, $\{\overleftrightarrow{AC}, \overleftrightarrow{BD}\}$ and $\{\overleftrightarrow{AD}, \overleftrightarrow{BC}\}$. These give rise to three ideal points in the projective completion.
The mistake made was thinking $\overleftrightarrow{AC}$ was not parallel to $\overleftrightarrow{BD}$ apparently because of a visual illusion of intersecting diagonals while modeling the geometry as the corners of square.
We can also count everything in another way. Since each line has two points, the underlying field is $F_2$. The projective plane then corresponds to one dimensional subspaces of $\mathbb F_2^3$, of which there are $8-1$ (there are seven distinct generators of $1$ dimensional subspaces, since $0$ will not work.)