# Projective line with two points identified

Let's $$\mathbb{P}^1$$ be the projective line over a field. If we identify two arbitrary points of $$\mathbb{P}^1$$, is the resulting space a scheme? If so is there a explicit description of it?

I think the answer should be positive. You can consider three projective lines each of two intersecting at one point. Let's call it a triangle of projective lines. You can give a projective variety structure to this (consider three lines in general position on plane and then projectivize it). Now $$\mathbb{Z}_3$$ acts on the triangle of projective lines, by permuting them. When a finite group acts on a projective variety the quotient has the structure of a scheme. The quotient is exactly a projective line with two points identified.

This is just the nodal cubic in $$\mathbb P^2$$, for instance $$V(Y^2Z - X^2(X+Z))$$. Its normalization is $$\mathbb P^1$$, with two points lying over the node $$(0:0:1)$$. Now precompose the normalization map with an automorphism of $$\mathbb P^1$$ taking your two arbitrary points to the two points lying over the node.
• Is it easy to see why this minus the double point is isomorphic to $\mathbb{A}^1$ minus a point? Commented Sep 16, 2020 at 15:35
• Are you familiar with the "projection from a point" construction? Every line through the node meets the curve in one other point, so use can use the fact that the space of lines through the node is isomorphic to $\mathbb P^1$ to show this without too much trouble. Commented Sep 16, 2020 at 15:39