Let $C$ be a proper curve (so separated, of finite type, and universally closed) over a field $k$.
Why is $H^0(C, \mathcal{O}_C) = (\mathcal{O}_C(C))$ not zero?
My ideas: Because $X$ proper over $k$ (field, so Noetherian) and $\mathcal{O}_C \in Coh(C)$ , we get that $H^0(C, \mathcal{O}_C)$ is a finite dimensional $k$-vector space, but why it can't be zero space?