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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.

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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.

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  • $\begingroup$ Is it easy to see why this minus the double point is isomorphic to $\mathbb{A}^1$ minus a point? $\endgroup$
    – user127776
    Commented Sep 16, 2020 at 15:35
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    $\begingroup$ 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. $\endgroup$ Commented Sep 16, 2020 at 15:39
  • $\begingroup$ I think so. Yes that makes sense now. $\endgroup$
    – user127776
    Commented Sep 16, 2020 at 15:52

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