Normalization of the projective closure of affine plane curve over $\mathbb{C}$ I am trying to understand how to do explicit calculations for finding the normalization of a plane curve. The intuition is somewhat clear to me: "separate" the singularities or smooth them out (for curves at least). However, in practice, finding the integral closure of a ring in its field of fractions is not so clear.
I was reading through this thread and the comments, but I couldn't quite figure out how to see that you need to need to add "x" amount of points above a singularity.
Example: Take the curve defined on the affine plane $y^2 = x^6 - 1$. It is smooth, and its projective closure is the curve defined by $Y^2Z^4 = X^6 - Z^6$ in the coordinates $[X,Y,Z]$ on $\mathbb{P}^2$. When $Z = 0$, we get $X = 0$ and $Y$ is free, so we add one point at infinity, $[0,1,0]$. By calculating the Jacobian or looking at the affine patch where $Y \neq 0$, we see that the projective closure is singular at this point. We are supposed to get two points from this one point in the normalization (since we know it is a hyperelliptic curve and there should be no ramification at $\infty$ in this case), but how am I supposed to see it?
If I were to do it algebraically I would have to look at the integral closure of $\left(\frac{k[x,z]}{(x^6 - z^6 - z^4)}\right)_{(\bar{x},\bar{z})}$ in $k(\bar{x},\bar{y})$, and this doesn't look too enticing. If I were to go this route, how do I know how many points I have to add? Is it the rank of the integral closure as a module over the original ring?
 A: Directly finding the integral closure is pretty tricky in general. There are lots of other ways to do this. Probably the simplest is by blowing up.
All known algorithms for resolving curve singularities are described in detail in the first part of this book by Kollar. It's very readable. (The book gets harder in the later sections, but you don't need that material.) Take a look at the method that uses Newton polygons, for example. I can't find a good exposition online, but for an illustration of the power of Newton polygons, see this blog post.
If you're interested in computational methods for Riemann surfaces more generally, there is this book.
A: Let $K := \mathbb{C}(x,y : y^2 = x^6-1)$ (by this I mean the fraction field of $\mathbb{C}[x,y]/(y^2-(x^6-1))$) be the function field of your curve.  "Points" on the smooth (=normalized) projective curve are the same thing as discrete valuations on $K$.  You are looking for points where $x$ and $y$ are both "infinite" in the sense that they have negative valuation: but then $y^2 = x^6 - 1$ has valuation $2v(y) = 6v(x)$ so we have $v(y) = 3v(x)$.  Let us assume for the moment that $v(x) = -1$, i.e., $1/x$ is a uniformizer (=element of valuation $1$), which of course implies that $v(y) = -3$.
To actually find and describe such valuations, I think the best practical way is to work with power series (this is what Newton would have done!).
Suppose you find two nonconstant elements $x,y$ of $\mathbb{C}(\!(t)\!)$ (the field of Laurent series) such that $y^2 = x^6 - 1$: they define a point of the (affine) curve in that field, and by the valuative criterion of properness and separatedness they extend to a unique point of the projective closure in $\mathbb{C}[\![t]\!]$, whose value at $t=0$ (now it makes sense) defines a $\mathbb{C}$-point of the projective curve, but in fact, one of any projective completion of the original affine curve, in particular, the smooth one.  If you prefer to work with function fields, a different way of saying the same thing is that $x$ and $y$ define a field embedding $K \subseteq \mathbb{C}(\!(t)\!)$ for which the valuation on $\mathbb{C}(\!(t)\!)$ restricts to a valuation on $K$, i.e., gives a point on the smooth projective curve as mentioned above.
Now take a look at the series
$$y = t^{-3} - \frac{1}{2}t^3 - \frac{1}{8}t^9 + \cdots$$
(obtained by taking the square root of $1-t^6$ which starts with $1 - \frac{1}{2}t^6 - \frac{1}{8}t^{12} + \cdots$, and multiplying by $t^{-3}$), together with
$$x = t^{-1}$$
Clearly this defines a point $P_1$ as mentioned above.  We find this sort of thing by starting with $x = t^{-1}$ since we wanted $1/x$ to be a uniformizer (essentially, we are looking at the completion of $K$ with respect to this uniformizer/valuation) and then we try to solve for $y$.  But there is also a different solution, namely the other square root (the negative of the above)
$$-t^{-3} + \frac{1}{2}t^3 + \frac{1}{8}t^9 + \cdots$$
giving us another point $P_2$ on the normalizer, both above the (singular) point $P$ with coordinates $(X:Y:Z)=(0:1:0)$ of the singular projective curve (clearly $P_2$ is a different point from $P_1$ because the valuation of $y - x^3$ is going to differ between the two).
This is about as good a description of the two points as one could hope for, namely a power series "parametrization" of the two branches of the curve around the singular point $P$.  There is then the matter of making sure that we have found all possible points above the singular point $P$ we were trying to resolve: this can be done by counting multiplicities: clearly $x$ defines a map to $\mathbb{P}^1$ that is of degree $2$ since $K$ is of degree $2$ over $\mathbb{C}(x)$, and we found two points $P_1,P_2$ over $\infty \in \mathbb{P}^1$, so we found them all (and neither is ramified over $\infty$).  (I wish I had a cleaner, less ad hoc, argument to explain why we must have $v(x) = -1$, though.)
This is slightly more efficient than considering a sequence of blowups: here, if we work in the local coordinate chart given by $\tilde z := Z/Y = 1/y$ and $\tilde x := X/Y = x/y$ (satisfying $\tilde x^6 = \tilde z^4 + \tilde z^6$, as you mention), we find that $\tilde z$ looks like $t^3$ and $\tilde x$ looks like $t^2$ for one of the points on normalization, and their opposites for the other.  So around $P$ with $(\tilde z, \tilde x) = (0,0)$, both curve branches we found have a cusp with the same tangent line, and it takes at least two successive blowups to resolve them.  As for the connection with the integral closure, it is going to be the set of elements of $K$ which have nonnegative valuation everywhere on the affine chart being considered (including the two points we found on the normalization): in the case of $\tilde z,\tilde x$, for example, we can form $\tilde x^3/\tilde z^2 = x^3/y$, which is not in $\mathbb{C}[\tilde z,\tilde x]$ (because it takes a different value at $P_1$ and $P_2$) but is in its integral closure (the places where it is infinite are those with $y=0$ which are not in the affine patch covered by $\tilde z,\tilde x$).
