Pictures for normalization of curves and surfaces

Question

Is there a place where to find pictures of normalizations of curve and surface singularities, over $\mathbb R$ and $\mathbb C$? If possible with some details on the equations used, on the algebra calculations. If you can picture in some way higher dimensional singularities I'll be interested too, but low dimension is enough. Thank you.

Answer 1

Collecting my comments and adding a bit more.

I made some pictures in this answer, but they are admittedly the two easiest cases for curves. You can find interactive plots for them here and here.

For curves, resolution of singularities, which can be achieved by blowing up, or really by taking the strict transform, is the geometric incarnation of normalization. For example, take the nodal cubic curve $y^2 = x^3$ in $\mathbb{A}^2$. Its coordinate ring $R := \frac{k[x,y]}{(y^2 - x^3)}$ is not integrally closed since $y/x$ satisfies the equation $T^2 - x = 0$. Adjoining $y/x$, we find $R[y/x]$ is integrally closed, hence is the normalization. We can realize the normalization abstractly by noting that $z := y/x$ satisfies $xz - y = 0$ and $z^2 - x = 0$, so $R[y/x] \cong \frac{k[x,y,z]}{(y^2 - x^3, xz - y, z^2 - x)}$. In geometric terms, this amounts to adding another dimension and considering the curve in $3$-space that is cut out by the original equation $y^2 = x^3$, along with the "slope equation" $xz = y$, and the integral dependence $z^2 = x$. Blowing up is basically the same thing, except we don't know the integral dependence beforehand. So we form $R_1 := \frac{k[x,y,z]}{(y^2 - x^3, xz - y)}$ and find that $\operatorname{Spec}(R_1)$ has $2$ irreducible components, one corresponding to the exceptional divisor, and the other to the strict transform, which is cut out by the integral dependence we found before.

(I do want to emphasize that these really are the easiest cases, which may belie how complicated resolving singularities by blowing up can be. Sometimes multiple blowups are required. For instance, consider the plane curve $y^2 = x^4 + x^5$. Then $y/x^2$ is the missing integral element, but we would only find it after two blowups. However, this complication is about finding the missing integral elements. Once we have found them, we can proceed as before.)

In general, we can picture the normalization of an affine variety as follows. Let $R$ be an affine domain over a field $k$ (e.g., the coordinate ring of an affine variety), let $K = \operatorname{Frac}(R)$ and let $\DeclareMathOperator{\wtR}{\widetilde{R}} \wtR$ be the integral closure of $R$ in $K$. By a theorem of Emmy Noether, then $\wtR$ is finitely generated as an $R$-module. Thus we can obtain $\wtR$ from $R$ by adjoining finitely many integral elements $f_1, \ldots, f_t \in K$. For each $i$ write $f_i = g_i/h_i$ with $g_i, h_i \in R$. For each $i$, $f_i$ satisfies a monic polynomial \begin{align*} F_i(T_i) := T_i^{m_i} + a_{m_i-1,i} T_i^{m_i-1} + \cdots + a_{0, i} \in R[T_i] \, . \end{align*} Then \begin{align*} \wtR = \frac{R[T_1, \ldots T_t]}{(h_1 T_1 - g_1, \ldots, h_t T_t - g_t, F_1(T_1), \ldots, F_t(T_t))} \end{align*} Geometrically, this means we can visualize the normalization by adding $t$ more dimensions and considering the variety cut out by the original equations defining $R$, along with $h_i T_i - g_i$ and $F_i(T_i)$ for each $i$.