There are two properties of (projective, geometrically integral, and regular) curves of genus 0:
If the curve has a $k$-point, then it is isomorphic to $\mathbb{P}_k^1$
where, if $P$ is a $k$-point, the line bundle $\mathscr{L}(P)$ defines a closed embedding into $\mathbb{P}_k^1$.
All genus 0 curves can be described as conics in $\mathbb{P}_k^2$
where the degree 2 line bundle $\omega_C^{\vee}$ defines a closed embedding into $\mathbb{P}_k^2$. This is mentioned, for example, in Vakil's notes section 19.3. But $\mathbb{P}_k^1$ is not a conic in $\mathbb{P}_k^2$. What's going on here? Can you actually embed $\mathbb{P}_k^1$ as a conic in $\mathbb{P}_k^2$?
Best Answer
Yes. Combining the two facts, any conic in $\mathbb{P}^2$ with a rational point is isomorphic to $\mathbb{P}^1$ (this is a nice exercise to prove directly; you don't need to know anything about line bundles, even). For an explicit example, you can consider the Veronese embedding
$$\mathbb{P}^1 \ni (A : B) \mapsto (A^2 : AB : B^2) \in \mathbb{P}^2$$
whose image is the conic cut out by the homogeneous equation $XZ - Y^2 = 0$. This is a special case of the rational normal curve.