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.