Let \[ X=V(x^3-y^2)\subseteq\AA^2_\CC. \] Show that \(X\) has a unique singular point, namely the origin.
Show that the germ \((X,0)\) is unibranch.
Show that the normalization of \(X\) is \(\AA^1_\CC\), using the morphism \[ \nu:\AA^1_\CC\longrightarrow X, \qquad t\longmapsto(t^2,t^3). \]
Solution
1. The coordinate ring of \(X\) is\[ \boxed{ A(X)\cong\CC[t^2,t^3]\subseteq\CC[t]. } \]
Consider the homomorphism \[ \phi:\CC[x,y]\longrightarrow\CC[t], \qquad x\longmapsto t^2, \qquad y\longmapsto t^3. \] Then \[ x^3-y^2\in\ker\phi. \] Conversely, divide any \[ p(x,y)\in\CC[x,y] \] by the monic polynomial \(y^2-x^3\) as a polynomial in \(y\) over \(\CC[x]\): \[ p=q(y^2-x^3)+a(x)+yb(x) \] for unique \[ q\in\CC[x,y], \qquad a,b\in\CC[x]. \] If \(p\in\ker\phi\), then \[ a(t^2)+t^3b(t^2)=0. \] The first summand contains only even powers of \(t\), while the second contains only odd powers of degree at least \(3\). Hence both summands vanish, so \[ a=b=0. \] Thus \[ \ker\phi=(y^2-x^3)=(x^3-y^2), \] and the induced map identifies \[ \CC[x,y]/(x^3-y^2) \cong \CC[t^2,t^3]. \]
2. The origin is the unique singular point of \(X\).
Let \[ F(x,y)=x^3-y^2. \] Then \[ F_x=3x^2, \qquad F_y=-2y. \] Since the base field has characteristic \(0\), a point of the plane curve is singular exactly when \[ F=F_x=F_y=0. \] The equations \[ 3x^2=0, \qquad -2y=0 \] force \[ x=y=0. \] The origin lies on \(X\), so it is the unique singular point.
3. The morphism\[ \nu:\AA^1_\CC\longrightarrow X, \qquad t\longmapsto(t^2,t^3), \]is finite and birational.
By step 1, the comorphism of \(\nu\) is the inclusion \[ \CC[t^2,t^3]\hookrightarrow\CC[t]. \] The element \(t\) is integral over \(\CC[t^2,t^3]\), because it satisfies \[ T^2-t^2=0 \] with coefficient \(t^2\in\CC[t^2,t^3]\). Moreover, \[ \CC[t] = \CC[t^2,t^3][t]. \] Hence \(\CC[t]\) is finite over \(\CC[t^2,t^3]\), so \(\nu\) is finite.
The two rings have the same fraction field, because \[ t=\frac{t^3}{t^2} \] in their common fraction field. Thus \[ \Frac\CC[t^2,t^3]=\CC(t)=\Frac\CC[t], \] and \(\nu\) is birational.
4. The normalization of \(X\) is \(\AA^1_\CC\).
Let \[ A=\CC[t^2,t^3] \subseteq K=\CC(t), \] and let \(\overline A\) be the integral closure of \(A\) in \(K\).
Step 3 shows that \(\CC[t]\) is integral over \(A\), so \[ \CC[t]\subseteq\overline A. \] Conversely, if \[ u\in\overline A, \] then \(u\) satisfies a monic polynomial with coefficients in \(A\), hence also a monic polynomial with coefficients in \(\CC[t]\). Since \(\CC[t]\) is a UFD, it is integrally closed in \(\CC(t)\), so \[ u\in\CC[t]. \] Therefore \[ \overline A=\CC[t]. \] Taking spectra gives \[ X^{\operatorname{norm}}\cong\Spec\CC[t]=\AA^1_\CC, \] and the normalization morphism is precisely \(\nu\).
5. The germ \((X,0)\) is unibranch.
The analytic branches of \(X\) at the origin correspond to the irreducible factors of \[ y^2-x^3 \] in the convergent power-series ring \[ \CC\{x,y\}. \] It therefore suffices to prove that this germ is irreducible there.
Regard \(y^2-x^3\) as a Weierstrass polynomial in \(y\). If it factored into two nonunits in \(\CC\{x,y\}\), the Weierstrass preparation theorem would, after absorbing units into the factors, give a factorization into monic Weierstrass polynomials in \(y\). Since the degree in \(y\) is \(2\), the only possibility is \[ y^2-x^3=(y-a(x))(y-b(x)) \] with \[ a(x),b(x)\in\CC\{x\}. \] Comparing the coefficient of \(y\) gives \[ b=-a, \] and comparing constant terms gives \[ a(x)^2=x^3. \] This is impossible: the \(x\)-adic order of a square is even, whereas \[ \operatorname{ord}_x(x^3)=3. \] Hence \(y^2-x^3\) is irreducible in \(\CC\{x,y\}\), so the germ has exactly one analytic branch. Thus \((X,0)\) is unibranch.
6. The normalization has exactly one point over the cusp:\[ \boxed{\nu^{-1}(0,0)=\{0\}.} \]
The equations \[ t^2=0, \qquad t^3=0 \] hold simultaneously exactly when \(t=0\). Thus the unique point of the normalization lying over the singularity is the origin of \(\AA^1\). This agrees with the unibranch computation in step 5.
7. Q.E.D.
Step 2 proves the singular-point claim, step 5 proves the unibranch claim, and steps <1>3–<1>4 identify the normalization and its morphism.