More precisely, how does one characterize integrally closed finitely generated domains (say, over C) based on geometric properties of their varieties? Given a finitely generated domain A and its integral closure A' (in its field of fractions), what's the geometric relationship between V(A) and V(A')? 


If you can, phrase your answer in terms of complex affine varieties.



What is the geometric meaning of integral closure?
The property you are interested in is known as being *normal*. For affine varieties, the definition of normal is just that the coordinate ring is integrally closed, and the operation on varieties that corresponds to taking the integral closure of the coordinate ring is known as *normalization* (a general variety is said to be normal if it is locally isomorphic to a normal affine variety.) So far I've just restated your question; but there are a number of things known. For this we need the notion of smoothness: for varieties over $\mathbb{C}$, this should be equivalent to being a smooth manifold (the general definition is a bit technical).


1. Any smooth variety is normal.
2. The set of singular points of a normal variety has codimension $\geq 2$.


**Corollary:** For curves, normal $\iff$ smooth.


Shafarevich's *Basic Algebraic Geometry* vol. 1 is a good reference for this from the varieties point of view (and deals with smoothness more rigorously).


As regards the relationship between the varieties corresponding to $A$ and $A'$: I mostly just have intuition for curves, so I'll stick to talking about them. For curves, $A'$ is a version of $A$ with the singularities "resolved": more specifically, $V(A')$ is a smooth variety equipped with a surjective morphism of varieties from $A' \to A$ which is an isomorphism away from the preimages of the singular points of $A$. (This should be true in higher dimensions too I think: it's definitely true if one is talking about schemes, but I think it's also true that for affine varieties the map $A \to A'$ induces a map $V(A') \to V(A)$.) The two basic examples to keep in mind here are the cuspidal cubic $C\_1: y^2 = x^3$ and the nodal cubic $C\_2: y^2 = x^3 + x^2$.


In the case $C\_1$: the coordinate ring $\mathbb{C}[x, y]/(y^2 - x^3)$ has integral closure isomorphic to $\mathbb{C}[t]$, and the map of varieties here is the map from the affine line to $C\_1$ given by $t \mapsto (t^2, t^3)$. In this case the map is a bijection as sets (but not an isomorphism of affine varieties! because the inverse map cannot be expressed as a polynomial map), and the "cusp" of $C\_1$ that is visible at the point $(0,0)$ is no longer evident.


In the case $C\_2$: the coordinate ring also has integral closure isomorphic to $\mathbb{C}[t]$: this time the map is a bit more complicated, but it's $t \mapsto (t^2 -1 , t(t^2-1))$. How did I find that? In this case, looking at the curve one sees that it has a self-intersection at the origin. This means that there should be two distinct points in the normalization that have been sent to the same point in $C\_2$. Another way of stating that is that because the curve appears to have two tangent lines at the origin, there really should be two different points there, one on each tangent line. How to tell them apart? Well, as one approaches the origin from one direction, the ratio $y/x$ tends to $1$ in the limit, whereas if one approaches it from the other direction, the ratio $y/x$ tends to $-1$: so at one of our two points, $y/x=1$, and at the other one, $y/x = -1$. Since $y/x$ is well-defined everywhere else on the curve, this suggests that we want $t = y/x$ to belong to our coordinate ring at the origin. Indeed, $t^2 = x +1$, so $t$ is integral, and we can solve for $x$ and $y$ in terms of $t$ to get the original answer. So in this case we have a surjective map from the affine line to a self-intersecting curve which is injective everywhere except at the preimage of the singular point at the origin.


