TITLE: Why the emphasis on Projective Space in Algebraic Geometry?
QUESTION [70 upvotes]: I have no doubt this is a basic question. However, I am working through Miranda's book on Riemann surfaces and algebraic curves, and it has yet to be addressed.
Why does Miranda (and from what little I've seen, algebraic geometers in general) place so much emphasis on projective space when studying algebraic curves? Why is this the natural setting to conduct algebraic geometry in?
Also, projective spaces and curves in them are hard for me to visualize, and in general I don't have any good intuition about these objects. Do working algebraic geometers simply not visualize things as much, or are there some nice interpretations of projective spaces and algebraic curves I am missing that would make them seem more natural and give me more intuition about them?

REPLY [3 votes]: This is a less elementary answer, but I think it's worth considering.
(1) As someone else put it, the projective space allows "as many solutions as possible". This is made precise in Bézout's theorem very elegantly: if $X,Y\subset\Bbb{P}^n$ are distinct irreducible hypersurfaces, then
$$\sum_{P\in X\cap Y}i(P, X\cap Y)=\deg(X)\cdot \deg(Y),$$
where $i(P, X\cap Y)$ is the intersection index of $X,Y$ at $P$. The intersection indices can be defined in $\Bbb{A}^n$, but the nice equality above only works in $\Bbb{P}^n$.
So Bézout's theorem is only true in a projective space!
(2) One of the tools used in Riemann Surfaces is divisors, which are widely used in algebraic geometry. Although divisors can be defined in affine varieties, things work specially well in projective varieties.
For a simple example, let $C\subset \Bbb{P}^2$ be a smooth projective curve. For every function $f\in k(C)$, the divisor $\text{div}(f)$ consisting of all zeros minus all poles of $f$ has degree zero and if $f$ either has no zeros or no poles, then $f$ is constant. In particular the Riemann-Roch space $\mathcal{L}(0)=\{f\in k(C)\mid \text{div}(f)\geq 0\}$ is the space of constants. We can of course talk about Riemann-Roch theorem, which relates $\dim \mathcal{L}(D), \deg D$ and the genus $g$.
All of the above can only be said because the curve is projective!
(3) Another very important tool in algebraic geometry is sheaf cohomology. If $X$ is an affine scheme and $\mathcal{F}$ is a quasi-coherent sheaf, then Serre's vanishing theorem says that
$$H^i(X,\mathcal{F})=0\text{ for all }i>0,$$
which essentially means affine schemes are "cohomologically trivial".
But there's real action going on in projective schemes. Just to mention a simple example, one can show that
$$h^n(\Bbb{P^n},\mathcal{O}_{\Bbb{P^n}}(d))=\binom{-d-1}{n},\text { for }d\leq -(n+1)$$
$$\chi(\Bbb{P^n},\mathcal{O}_{\Bbb{P^n}}(d))=\frac{(d+1)(d+2)...(d+n)}{n!} \text{ for all }d\in\Bbb{Z}$$
where $\chi$ is the alternating sum of the $h^i(\Bbb{P^n},\mathcal{O}_{\Bbb{P}^n}(d))$, namely Euler's characteristic.
Take also the crucial Serre's Duality Theorem, which says $h^i(X,\mathcal{F})=h^{n-i}(X,\omega_X\otimes\mathcal{F}^*)$. One of the hypothesis of the theorem is that $X$ is a proper scheme, and the most natural example of a proper scheme is -- guess what -- a projective scheme!