Show that the nodal cubic \(X = V(x^2 - y^2(y-1))\) has a unique singular point.
Show that \(X\) locally has two smooth branches at zero meeting transversally.
Show that the normalization of \(X\) is \(\AA^1_\CC\).
Consider the projective closure \[ \overline X=V(x^2z-y^2(y-z))\subseteq\PP^2_\CC. \] Show that its normalization morphism \[ \nu:\PP^1_\CC\longrightarrow\overline X \] is birational, and that \(\nu^{-1}([0:0:1])\) consists of two points. Compare this to Zariski’s main theorem.
Solution
Put \[ F(x,y)=x^2-y^2(y-1)=x^2+y^2-y^3. \]
1. The origin is the unique singular point of \(X\).
The partial derivatives are \[ F_x=2x, \qquad F_y=2y-3y^2=y(2-3y). \] A singular point must satisfy \[ F=F_x=F_y=0. \] From \(F_x=0\) one gets \(x=0\), and \(F_y=0\) gives \[ y=0 \qquad\text{or}\qquad y=\frac23. \] The second possibility does not lie on \(X\), because \[ F\left(0,\frac23\right) = \frac4{27} \neq0. \] Hence the unique singular point is \[ \boxed{(0,0)}. \]
2. The germ \((X,0)\) has exactly two smooth branches, with tangent lines\[ x+iy=0 \qquad\text{and}\qquad x-iy=0, \]and the two branches meet transversally.
Let \[ s(y)=\sqrt{1-y}\in\CC\{y\} \] be the convergent square root satisfying \(s(0)=1\). In \[ \CC\{x,y\} \] one has \[ \begin{aligned} F &=x^2+y^2(1-y)\\ &=(x+iys(y))(x-iys(y)). \end{aligned} \] Thus there are two local analytic branches \[ B_+=V(x+iys(y)), \qquad B_-=V(x-iys(y)). \] Each branch is smooth at the origin because the derivative of its defining equation with respect to \(x\) is \(1\).
Since \[ s(y)=1+\text{terms of order at least }1, \] the linear terms of the two defining equations are \[ x+iy \qquad\text{and}\qquad x-iy. \] These are distinct linear forms, so the tangent lines are distinct. Hence the two smooth branches meet transversally at the origin.
3. The morphism\[ \nu_{\mathrm{aff}}:\AA^1_\CC\longrightarrow X, \qquad t\longmapsto \bigl(t(t^2+1),t^2+1\bigr), \]is finite and birational.
The image lies on \(X\), because \[ \begin{aligned} x^2-y^2(y-1) &=t^2(t^2+1)^2-(t^2+1)^2t^2\\ &=0. \end{aligned} \]
The induced inclusion of coordinate rings is \[ A = \CC[x,y]/(x^2-y^2(y-1)) \hookrightarrow \CC[t], \] with \[ x\longmapsto t(t^2+1), \qquad y\longmapsto t^2+1. \] In \(\CC[t]\), \[ t^2=y-1, \] so \(t\) is integral over \(A\). Moreover \[ \CC[t]=A[t], \] and therefore \(\CC[t]\) is finite over \(A\).
In the fraction field of \(A\), \[ t=\frac{x}{y}, \] because \(x=t(t^2+1)\) and \(y=t^2+1\). Thus \[ \Frac A=\CC(t), \] so the morphism is birational.
4. The normalization of the affine nodal cubic is\[ \boxed{X^{\operatorname{norm}}\cong\AA^1_\CC.} \]
By step 3, \(\CC[t]\) is integral over \(A\) and has the same fraction field. Let \(\overline A\) be the integral closure of \(A\) in that field. Then \[ \CC[t]\subseteq\overline A. \] Conversely, every element of \(\overline A\) is integral over \(A\), hence also integral over \(\CC[t]\) because \(A\subseteq\CC[t]\). The ring \(\CC[t]\) is a UFD and therefore integrally closed in \(\CC(t)\), so \[ \overline A\subseteq\CC[t]. \] Thus \[ \overline A=\CC[t], \] which proves the normalization statement.
5. The affine normalization from step 3 extends to the morphism\[ \nu:\PP^1_\CC\longrightarrow\overline X \]given by\[ \boxed{ \nu([u:v]) = \bigl[ u(u^2+v^2): v(u^2+v^2): v^3 \bigr]. } \]
The three homogeneous coordinate functions have degree \(3\). They do not vanish simultaneously: if \(v=0\), then \(u\neq0\) and the first coordinate is \(u^3\neq0\); if \(v\neq0\) and \(u^2+v^2=0\), then the third coordinate \(v^3\) is nonzero.
Put \[ X_0=u(u^2+v^2), \qquad Y_0=v(u^2+v^2), \qquad Z_0=v^3. \] Then \[ \begin{aligned} X_0^2Z_0 &=u^2v^3(u^2+v^2)^2,\\ Y_0^2(Y_0-Z_0) &=v^2(u^2+v^2)^2 \bigl(v(u^2+v^2)-v^3\bigr)\\ &=u^2v^3(u^2+v^2)^2. \end{aligned} \] Hence \[ X_0^2Z_0-Y_0^2(Y_0-Z_0)=0, \] so the image lies in \(\overline X\).
On the chart \(v=1\), the formula becomes \[ t=u \longmapsto [t(t^2+1):t^2+1:1], \] which is exactly the affine normalization from step 3.
6. The morphism \(\nu:\PP^1_\CC\to\overline X\) is the normalizationmorphism and is birational.
3. Thus \(\nu\) induces anisomorphism of function fields and is birational.
The source \(\PP^1_\CC\) is smooth, hence normal. The morphism \(\nu\) is nonconstant between projective curves, so it is finite. Therefore it is a finite birational morphism from a normal curve, and hence it is the normalization of \(\overline X\).
7. The fiber over the node\[ P=[0:0:1]\in\overline X \]is\[ \boxed{ \nu^{-1}(P) = \{[i:1],[-i:1]\}. } \]
If \[ \nu([u:v])=[0:0:1], \] then \(v\neq0\), so scale to \(v=1\). The first two coordinates vanish exactly when \[ u(u^2+1)=0, \qquad u^2+1=0. \] The second equation already implies the first, and it has exactly the two solutions \[ u=i, \qquad u=-i. \] Thus the fiber is the displayed two-point set.
8. The conclusion of Zariski’s Main Theorem fails for \(\nu\): thebirational morphism \(\nu\) has a disconnected fiber over the node, where thetarget \(\overline X\) is not normal.
Zariski’s Main Theorem states that a birational morphism between normal projective varieties has connected fibers. The normalization morphism \[ \nu:\PP^1_\CC\longrightarrow\overline X \] is birational and has normal source, but step 7 gives a disconnected fiber over the node. The normalization is an isomorphism over the normal locus of \(\overline X\), so \(\overline X\) is not normal at \(P\).
9. Q.E.D.
Steps <1>1–<1>2 establish the nodal singularity and its two transverse branches. Steps <1>3–<1>4 compute the affine normalization. Steps <1>5–<1>7 compute the projective normalization and the two-point fiber, and step 8 gives the requested comparison with Zariski’s Main Theorem.