Riemann surfaces are algebraic Via a very ample divisor one can embed a Riemann surface holomorphically into some $\mathbb{P}^n$. Now, we can then project the Riemann surface to $\mathbb{P}^3$, and we can even go until $\mathbb{P}^2$ if we allow nodes. 
If we have a Riemann surface $X$ and an embedding $f:X\to \mathbb{P}^2$ (where the image is smooth), then we can find the equation for $f(X)$ (you pull back the meromorphic functions $x_0/x_1$ and $x_1/x_2$ via $f$, where $x_0,x_1,x_2$ are the homogeneous coordinates of $\mathbb{P}^2$; these are algebraically dependent over $\mathbb{C}$, and thus they must satisfy some polynomial. The projective closure of the zeroes of this polynomial is exactly $f(X)$).
I've been reading Shafarevich's book Algebraic Geometry I, and on page 86 he describes how to find the equations for a mapping to $\mathbb{P}^n$. He simply states that if you have a map from $X$ to $\mathbb{P}^n$, then since we can project, we can reduce to the case that the map goes to $\mathbb{P}^2$, and we're basically done.
If the Riemann surface is in $\mathbb{P}^3$, for example, I don't see why the above method guarantees that it be defined by equations... It's image in $\mathbb{P}^2$ would possibly have nodes... Any ideas?
I'm looking to show that a compact Riemann surface in $\mathbb{P}^3$ is algebraic, that is, can be defined by polynomial equations.
 A: Compose the map with projections centered at points. To understand that, you can read the book Algebraic Curves and Riemann Surfaces by Rick Miranda, pages 98-102. That is a beautiful book. You will understand this perfectly.
A: A compact Riemann surface $X$ can always be embedded holomorphically into $\mathbb P^3( \mathbb C)$ . This is a highly non-trivial theorem.  
Claim:     It cannot in general be embedded into  $\mathbb P^2( \mathbb C)$.
 The simplest argument to support this claim  is to recall that a smooth complex curve of degree $d$ in $\mathbb P^2( \mathbb C)$ has genus $g=\frac{(d-1)(d-2)}{2}$.
Since  there exist Riemann surfaces of any genus  $g\geq 0$ on one hand, and since on the other hand most integers are not of the form $\frac{(d-1)(d-2)}{2}$ , the claim is proved.             
However any Riemann surface can be immersed, but not necessarily injectively, into $\mathbb P^2( \mathbb C)$ by skilfully (or skillfully if you prefer American English) composing an embedding into $\mathbb P^3( \mathbb C)$  with a projection onto a plane, so that the immersed curve will  have nodal singularities at worst.
Bibliography
An excellent   reference for these questions is Miranda's Algebraic Curves and Riemann Surfaces.
If you want to see  the complete proofs of the embedding theorem of compact Riemann surfaces into $\mathbb P^3( \mathbb C)$ (which implies algebraicity of said Riemann surfaces), look at Forster's Lectures on Riemann Surfaces (Springer) or at Narasimhan's Compact Riemann Surfaces (Birkhäuser).
(Caution: the analysis used in these two books  is not for the faint-hearted!)    
