Hole in the axioms of Hartshorne's "Foundations of Projective Geometry"?

You can deduce it, but I think it's a bit more complicated than Mariano Suárez-Álvarez's answer makes it sound. Here's one way to do it.

First, suppose the empty set is a line. Since every line is parallel to the empty set, (2) says that every $x\in\mathbb{X}$ is in a unique line. Combined with (1), this means the unique line through $x$ must also pass through every other point, and so $\mathbb{X}$ itself is a line, which violates (3).

Now suppose a singleton $\{x\}$ is a line. For any $y\neq x$, consider the line $\ell$ such that $x,y\in\ell$. By (3), there exists a point $z$ which is not on $\ell$. By (2), there exists a line $\ell'$ through $z$ which is parallel to $\ell$. But now $\{x\}$ and $\ell$ are two different lines through $x$ which are parallel to $\ell'$, which violates (2).


This "extra" axiom, that each line contains at least two points, is there to put the idea of an affine geometry in the context of a more general type of incidence geometry.

A linear space is an incidence geometry $(\mathcal{P}, \mathcal{L})$, for which

  • Each pair of points is contained in a single line
  • Each line contains at least two points

By ensuring that our lines contain two points, we can define the rank of the geometric objects in terms of how many points it takes to generate the object (lines are generated by two distinct points, planes by 3 noncollinear lines, etc.).

An affine plane is then a linear space for which there exists at least 3 noncollinear points, and for each line $\ell$ and point $x \not\in \ell$, there is a unique line $\ell^{\prime}$ containing $x$, that is disjoint from $\ell$. As pointed out in Eric's excellent answer, these conditions make it so that the requirement of at least two points per line is now redundant. But it still holds, and by stating it explicitly we make it clear that an affine space is a special kind of linear space.


Suppose a line L has 1 point x. Pick any two points different from x: the line L' through them is parallel to L. Can you see the problem?

Tags:

Plane Geometry

Affine Geometry

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