Why the emphasis on Projective Space in Algebraic Geometry? 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?
 A: The most appropriate answer will depend on why you are working through a book on Riemann surfaces and algebraic curves, but I will try to give some suggestions.
Since you mention Riemann surfaces, let's start with some analogy with smooth manifolds. The Whitney embedding theorem says that any smooth manifold can be embedded in $\mathbb{R}^N$ for $N$ sufficiently large, so we can always think of a smooth manifold as a submanifold of $\mathbb{R}^N$. This occasionally helps with intuition and visualization, and can simplify some constructions. 
In the case of complex manifolds (e.g. Riemann surfaces), you might ask whether the same holds true holomorphically, i.e. whether any complex manifold can be holomorphically embedded in $\mathbb{C}^N$ for $N$ sufficiently large. It turns out that usually the answer is no. It is an easy consequence of the Liouville theorem that no compact complex manifold is a complex submanifold of $\mathbb{C}^N$. If you only care about compact complex manifolds, then $\mathbb{CP}^N$ turns out to be the best possible (see e.g. the Kodaira embedding theorem, which characterizes which compact complex manifolds are complex submanifolds of $\mathbb{CP}^N$).
If your motivation is the study of solutions to polynomial equations, then as mentioned in other answers and comments, projective spaces are the appropriate completions of affine space that allow as many solutions as possible, allowing various formulas (e.g. couting intersections) work without additional qualification.
About visualization: for curves in $\mathbb{CP}^2$, first take some affine chart $\mathbb{C}^2 \subset \mathbb{CP}^2$, and then look at the intersection with some "real slice" $\mathbb{R}^2 \subset \mathbb{C}^2$. For example if we look at the curve in $\mathbb{CP}^2$ given by the zero set of $x^2-yz$, by working on the affine chart $z\neq0$ this becomes $y = x^2$ on $\mathbb{C}^2$, and if we restrict to real $x,y$ we get a parabola.
A: 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!
A: The reason for working in projective space is so that there are "as many solutions as possible" to polynomial equations.  
There are two basic ways that a polynomial equation (or system of equations) can end up with no solutions.  One way is that you have an equation which has no solution over your field (for example, $x^2+1=0$ when working over the reals) but has a solution over a larger field; this is resolved by moving to an algebraically closed field, such as the complex numbers.  The other problematic case is the equation $0=1$.  This case can't be resolved by changing fields, but is resolved by going from affine space into projective space!
For example, consider the 2 equations $x+y=1$ and $x+y=2$.  If you try to solve these simultaneously, you get the contradiction $0=1$, so there is no solution.  But if you work in the projective plane, there is a solution.  Geometrically, these are 2 parallel lines, which intersect at the point at infinity corresponding to the direction of the parallel lines.  Algebraically, if we think of these equations in the projective plane with projective coordinates $[x:y:z]$, then the equations become $x+y=z$ and $x+y=2z$, which have the intersection $[1:-1:0]$.
A: Since there have been excellent answers on why one should work in projective space, let me add some comments on how to visualize it.
You probably know affine space $A^n$ sits in $P^n$ naturally: the complement of any hyperplane in $P^n$ is isomorphic to $A^n$. Let us denote affine coordinates by $x_1, \ldots, x_n$, and projective ones by $z_0, \ldots, z_n$, and embed $A^n$ in $P^n$ by sending $(x_1,\ldots, x_n)$ to $[1:x_1:,\ldots,x_n]$. i.e. $A^n = [x_0 \neq 0]$.
Now it is easy to see that $[x_0 = 0] \subset P^n$ is isomorphic to $P^{n-1}$ so we derive
$$
P^n = A^n \cup P^{n-1}
$$
as sets.
We can think of this $P^{n-1}$ as the "points at infinity", i.e. the different "directions" on how to travel to infinity in the affine space. For example any line in $A^n$ meets the points at infinity at exactly one point corresponding to his slope (thats why parallel lines in $A^n$ which have the same slope, meet at the same point at infinity). Another example would be a parabola $y=x^2$, when we go to infinity, the tangent direction of the graph will eventually become almost vertical, thats why its point at infinity is $[0:1:0]$ (for this homogenize to $yz = x^2$ and set $z=0$).
Lastly, i find it easy to visualize the whole $P^n$ as the quotient of the n-sphere with anitpodal points identified, more specifically picking a set of representatives as the lower hemispehere plus the equator (where of course we still need to identify on the equator). So if n=2 and our field is the reals, we can pick the 2-sphere which is easily visualizable. Now if we restrict to the lower hemisphere and the equator, we found one representative for each line through the lower hemisphere, and on the equator we still need to identify points. You can see that the lower hemisphere (excluding the equator) is $A^2$, and the points on the equator are the points at infinity: $P^1$. Its also clear why parallel lines meet, they converge to the same point on the equator.
Hopefully this was helpful.
Joachim
