The main problem is:
Is possible to see that a complete non singular abstract curve is projective using Riemann-Roch?
For abstract non singular curve i mean that $(X,\mathcal{O}_X)$ is a $k$ space of function $(X,\mathcal{O}_X)$ over a notherian irreducible space $X$ of dimension $1$, with all local rings $(\mathcal{O}_X)_x$ regular (for each point $x\in X$) that can be covered by $U_i$ ($i\in \bar{n}$) affine open sets (where $U_i$ is an affine open set iff $U_i\cong spec(A_i)$ where $A_i$ is a finitely generated $k$-algebra without nilpotent).
Complete mean that is separable and that for any $k$-variety $Y$ the projection map from $X\times Y$ in $Y$ in closed under the Zariski topology of the product.
All the machinery using to prove the R-R seems lecit under this hypotesis exept the fact that $H^i(X)$ (that is isomorphic to the Check chomology) is a $k$ vector space of finite dimesion.
So the problem is: If $X$ is an abstract complete curve over an algebrically closed field $k$, is $H^i(X)$ a finite dimensional $k$ vector space?
Note: The Grothendieck vanishing theorem reduce the question at the case of $i=0$ and $i=1$.
Someone has any suggestion please?
This might be overkill, but the following should work: a complete variety $\pi:X\to \operatorname{Spec} k$ for any field $k$ is proper, so the direct image and higher direct images of $\mathcal{O}_X$ are all coherent. But $R^i\pi_*\mathcal{O}_X$ is just $H^i(X)$, and since the coherent sheaves on $\operatorname{Spec} k$ are just the finite-dimensional vector spaces over $k$, we see that $H^i(X)$ is finite-dimensional for any complete variety over a field.