# Embedding a Riemann Surface into projective space

Let $R$ be a Riemann surface, i.e a one-dimensional complex manifold. I am getting a bit mixed up with the algebraic (schemes, varieties) and complex analytic categories (Riemann surfaces as defined above).

I know that any line bundle on a scheme $X$ with large enough degree will embed $X$ as a closed subscheme of some projective space.

I have also read the statement that a Riemann surface (here we are working in the complex analytic category) can be embedded into $\mathbb{P}_{\mathbb{C}}^n$, for some $n$, via a holomorphic line bundle. In this case, are we viewing $\mathbb{P}_{\mathbb{C}}^n$ as an object in the algebraic category (i.e viewing it as a scheme) or in the complex analytic category (i.e as a complex manifold)?

Furthermore, is this the embedding that allows us to interchange the theory of Riemann surfaces and algebraic curves? How does this embedding give a Riemann surface the structure of a closed subscheme (so now it is in the algebraic category)?

• Compact Riemann Surface=Smooth projective algebraic curve over complex numbers. – Mohan Mar 29 '18 at 0:04
• But why are they equal? Is it because of this embedding by a holomorphic line bundle? – Jadwiga Mar 29 '18 at 0:05
• Chow's theorem says that every analytic subvariety of projective space is actually algebraic: en.wikipedia.org/wiki/… and this provides the equivalence. – KReiser Mar 29 '18 at 2:16