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Let $k$ be a number field. I am attempting to classify all the affine genus zero smooth geometrically integral curves up to its (smooth) compactification (or projectivization, if I'm not mistaken).

It is known that any smooth projective geometrically integral curve of genus zero with a $k$-point is isomorphic to $\mathbb{P}^1_k$.

Question 1. If $X$ is a genus zero affine curve over $k$ such that $X(k) \neq \emptyset$, is it true that the compactification of $X$ will be isomorphic to $\mathbb{P}^1_k$?

Question 2. In the case where $X$ has no $k$-points, would its compactification be a smooth projective conic in $\mathbb{P}^2_k$? Are they unique up to isomorphism?

I'm asking these questions with reference to the classification of genus zero curves in the projective case found here: Curves of genus 0.

oleout
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1 Answers1

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Question 1: Yes. Embed $X$ as an open set of a smooth projective curve $\overline{X}$. Then $X$ and $\overline{X}$ have the same genus (zero), so $\overline{X}$ is a smooth projective curve of genus zero with a rational point, or a $\Bbb P^1_k$.

Question 2: Yes. Embed $X$ as an open set of a smooth projective curve $\overline{X}$ of genus zero by (1). Now $\overline{X}$ has a closed immersion in to $\Bbb P^2_k$: the canonical bundle $\omega$ on $\overline{X}$ has degree $-2$ by Riemann-Roch, so $\omega^\vee$ is a line bundle of degree $2$ with $h^0=3$, which gives a closed immersion in to $\Bbb P^2_k$ (as any line bundle of degree $\geq 2g+1$ on a curve of genus $g$ gives a closed immersion to projective space). This gives $\overline{X}$ as a conic in $\Bbb P^2$. We also have that $\overline{X}$ still has no $k$-points: if $P\in \overline{X}$ was a $k$-point and $L\subset \Bbb P^2$ was a line not containing $P$, then the projection map from $P$ to $L$ would be degree one and induce an isomorphism between $X$ and an open subset of $\Bbb P^1_k$, giving $k$-rational points on $X$. The isomorphism classes of such projective conics are parametrized by classes in the Brauer group of $k$, which is a torsion abelian group which is far from trivial for number fields.

In general, varieties which are isomorphic after base change to a larger field are called twists of each other, and in the case where one of these varieties is $\Bbb P^n$, such varieties are called Severi-Brauer varieties. There's a great story behind them and about their connection to central simple algebras - one place to start learning about this is this other MSE post which contains some references.

KReiser
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  • Thank you for your detailed answer. So to clarify, in the case of question 2, the compactification of an affine curve $X$ over a fixed number field $k$ is a conic embedded in $\mathbb{P}^2_k$ whose class corresponds to an element of $\mathrm{Br}(k)$? Also, do you think it's possible to obtain a curve $X$ by removing a finite number of points from $\mathbb{P}^1_k$ such that $X(k) = \emptyset$? – oleout Aug 06 '21 at 02:28
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  • I don't see a difference in what you've said and what I've said. 2) No, since $k$ is infinite $\Bbb P^1_k$ has infinitely many $k$ points, so there's no way to do what you ask. But any $X$ you're considering is obtained from deleting a finite number of points from a smooth conic in $\Bbb P^2_k$.
    • – KReiser Aug 06 '21 at 02:36
  • Sorry, to me this correspondence is very unintuitive, so I rewrote it to be sure... Again, thank you so much for your answers and references. – oleout Aug 06 '21 at 02:53
  • Forgive me for probing further. Apparently, the projective line is isomorphic to the conic $xz-y^2=0$ via the map $[X:Y] \mapsto [X^2:XY:Y^2]$. I'm guessing that the trivial element of $\mathrm{Br}(k)$, which is a split quaternion algebra by definition, corresponds to this particular conic? What then would be the group operation associated to the classes of such projective conics? – oleout Aug 06 '21 at 05:32
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    Any conic isomorphic to $\Bbb P^1_k$ corresponds to the trivial element of $Br(k)$. As far as I can remember, one usually makes the correspondence between conics and quaternion algebras by sending the conic $ax^2+by^2=z^2$ to the quaternion algebra with structure constants $a,b$ - I don't quite remember how the group operation works here. – KReiser Aug 06 '21 at 05:48