Why are smooth projective algebraic curves over the complex numbers compact and orientable? Why smooth projective algebraic curves over complex numbers are compact and orientable? I want to use it to show algebraic smooth curves are topologically identical to toruses with different genuses.
(projective algebraic curves are zeros of a homogenous polynomial of three variables with complex coefficients in CP2. We will call a projective algebraic curve smooth in point p if its gradient is not zero at point p.)
 A: *

*Show that each smooth plane algebraic curve is closed. Potential approach: consider the restriction of the polynomial map $\mathbb{C}^3\setminus\{0\} \to \mathbb{C}$. Compose with the quotient map $\mathbb{C} \to \mathbb{C}/\mathbb{C}^\times$. Show this descends to $\mathbb{CP}^2$. Since $\mathbb{CP}^2$ is compact, this implies that each smooth plane algebraic curve is compact.


*Show that each smooth plane algebraic curve is a $1$-dimensional complex manifold. Potential approach: work in local coordinates on $\mathbb{CP}^2$ to show that each smooth plane algebraic curve is locally homeomorphic to the zero set in $\mathbb{C}^2$ of a smooth complex polynomial of two variables. Then show that these zero sets are always $1$-dimensional complex manifolds.


*We conclude that each smooth plane algebraic curve is a compact $1$-dimensional complex manifold. Every complex manifold is orientable as a real manifold (compare the Jacobians of the real and complex transition maps to see that the real Jacobians must have positive determinant -- in other words, holomorphic maps preserve orientation), so we are done.
