Lemma 53.10.3. Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_ X) = k$. If $X$ is Gorenstein and has genus $0$, then $X$ is isomorphic to a plane curve of degree $2$.
Proof. Consider the invertible sheaf $\mathcal{L} = \omega _ X^{\otimes -1}$ where $\omega _ X$ is as in Lemma 53.4.1. Then $\deg (\omega _ X) = -2$ by Lemma 53.8.3 and hence $\deg (\mathcal{L}) = 2$. By Lemma 53.10.2 we conclude that choosing a basis $s_0, s_1, s_2$ of the $k$-vector space of global sections of $\mathcal{L}$ we obtain a closed immersion
\[ \varphi _{(\mathcal{L}, (s_0, s_1, s_2))} : X \longrightarrow \mathbf{P}^2_ k \]
Thus $X$ is a plane curve of some degree $d$. Let $F \in k[T_0, T_1, T_2]_ d$ be its equation (Lemma 53.9.1). Because the genus of $X$ is $0$ we see that $d$ is $1$ or $2$ (Lemma 53.9.3). Observe that $F$ restricts to the zero section on $\varphi (X)$ and hence $F(s_0, s_1, s_2)$ is the zero section of $\mathcal{L}^{\otimes 2}$. Because $s_0, s_1, s_2$ are linearly independent we see that $F$ cannot be linear, i.e., $d = \deg (F) \geq 2$. Thus $d = 2$ and the proof is complete. $\square$
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