Is every smooth projective curve of genus 0 defined over the field of complex numbers isomorphic to a conic in the projective plane? Give an explanation for your answer.
Solution
Let \(C\) be a smooth projective curve of genus \(0\) over \(\CC\) and fix \(p\in C\).
1. There is a nonconstant meromorphic function \(f\) on \(C\) whose only pole is a simple pole at \(p\).
By the Riemann–Roch theorem, \(\ell([p]) - \ell(K_C-[p]) = 1 + 1 - 0 = 2\), so \(\ell([p])\geq 2\). The space \(L([p])\) therefore contains a function \(f\) that is not constant; its only possible pole is a simple pole at \(p\), and it has one because it is nonconstant.
2. \(f\colon C\to\PP^1\) is an isomorphism.
1. A morphism of degree \(1\) between smooth projective curves is an isomorphism.
3. \(\PP^1\) is isomorphic to the conic \(z_0 z_2=z_1^2\) in \(\PP^2\).
In projective coordinates \(t_0, t_1\) on \(\PP^1\), the map \([t_0:t_1]\mapsto[z_0:z_1:z_2]=[t_0^2:t_0 t_1:t_1^2]\) is an isomorphism onto the conic \(z_0 z_2=z_1^2\). Equivalently, it is the map given by the complete linear system of the anticanonical divisor, which has degree \(2\).
4. Q.E.D.
Steps 2 and 3 give \(C\cong\PP^1\cong V(z_0z_2-z_1^2)\), so the answer is \(\boxed{\text{yes}}\).