Let \(n>1\) and let \[ \mu_n = \{\zeta\in\CC^\times:\zeta^n=1\} \] act on \(\AA^2_\CC\) by \[ \zeta\cdot(x,y)=(\zeta x,\zeta y). \] Define \[ X = \operatorname{Specm}\CC[x,y]^{\mu_n} = \AA^2_\CC/\mu_n. \] Show that \(X\) is a normal affine variety with a unique singular point.
Solution
Put \[ S=\CC[x,y], \qquad R=S^{\mu_n}, \qquad X=\operatorname{Specm}R. \] For \(0\leq i\leq n\), set \[ z_i=x^{n-i}y^i. \]
1. The invariant ring is\[ \boxed{ R = \CC[z_0,z_1,\ldots,z_n] \subseteq \CC[x,y]. } \]Moreover, \(S\) is integral over \(R\) and\[ \dim X=2. \]
For a monomial \(x^ay^b\), \[ \zeta\cdot x^ay^b = \zeta^{a+b}x^ay^b. \] Hence the monomial is \(\mu_n\)-invariant exactly when \[ a+b\equiv0\pmod n. \] Because the monomials form a simultaneous eigenbasis for the scalar action, an invariant polynomial is a linear combination of invariant monomials.
Every invariant monomial has total degree \(qn\) for some \(q\geq0\), and every pair of nonnegative integers \((a,b)\) with \(a+b=qn\) can be decomposed as a sum of \(q\) pairs whose coordinates sum to \(n\). Thus every invariant monomial is a product of degree-\(n\) invariant monomials \[ z_i=x^{n-i}y^i. \] This proves the displayed generating set.
Since \[ x^n=z_0\in R, \qquad y^n=z_n\in R, \] the elements \(x\) and \(y\) satisfy the monic equations \[ T^n-z_0=0, \qquad T^n-z_n=0 \] over \(R\). Hence \(S\) is integral over \(R\). Integral extensions preserve Krull dimension, so \[ \dim R=\dim S=2. \]
2. The ring \(R\) is normal.
Let \[ K=\Frac R \] and suppose \[ u\in K \] is integral over \(R\). Since \(R\subseteq S\), the same monic equation shows that \(u\) is integral over \(S\). The polynomial ring \(S=\CC[x,y]\) is a UFD, hence integrally closed in its fraction field, so \[ u\in S. \]
But \(u\in\Frac R\), and every element of \(\Frac R\) is fixed by \(\mu_n\) because both numerator and denominator are invariant. Therefore \[ u\in S^{\mu_n}=R. \] Thus \(R\) is integrally closed in its fraction field and is normal.
3. Let\[ \mfm=(z_0,z_1,\ldots,z_n)\subseteq R. \]Then \(\mfm\) is the point \(\bar0\in X\) which is the image of the origin, and\[ X\setminus\{\bar0\} = D(z_0)\cup D(z_n). \]
The quotient map \[ \AA^2_\CC\longrightarrow X \] is induced by the inclusion \(R\hookrightarrow S\). All positive-degree invariants vanish at \((0,0)\), so the origin maps to \(\mfm\).
Now let \(\mathfrak p\) be a maximal ideal of \(R\) containing both \[ z_0=x^n \qquad\text{and}\qquad z_n=y^n. \] For every \(i\), \[ z_i^n = (x^n)^{n-i}(y^n)^i = z_0^{\,n-i}z_n^{\,i} \in\mathfrak p. \] Since \(\mathfrak p\) is radical, \[ z_i\in\mathfrak p \] for every \(i\). Hence \(\mathfrak p=\mfm\). Therefore the only point outside \(D(z_0)\cup D(z_n)\) is \(\bar0\).
4. The open set \(D(z_0)\) is smooth and has coordinate ring\[ \boxed{ R_{z_0} \cong \CC[a,a^{-1},t], \qquad a=x^n, \quad t=\frac yx. } \]
In \(R_{z_0}\) one has \[ t = \frac{z_1}{z_0} = \frac{x^{n-1}y}{x^n} = \frac yx. \] Every generator satisfies \[ z_i = x^{n-i}y^i = x^n\left(\frac yx\right)^i = a t^i. \] Thus \(R_{z_0}\) is generated by \[ a,\ a^{-1},\ t. \]
These elements are algebraically independent except for the invertibility of \(a\): distinct Laurent monomials \[ a^jt^i \] map to distinct monomials \[ x^{nj-i}y^i \] in \(\CC[x,x^{-1},y]\). Hence there is no further relation, giving \[ R_{z_0}\cong\CC[a,a^{-1},t]. \] This Laurent polynomial ring is regular, so \(D(z_0)\) is smooth.
5. The open set \(D(z_n)\) is smooth.
Interchanging \(x\) and \(y\) in step 4 gives \[ R_{z_n} \cong \CC[b,b^{-1},s], \qquad b=y^n, \quad s=\frac xy. \] This is again a regular Laurent polynomial ring. Hence \(D(z_n)\) is smooth.
6. The point \(\bar0\) is singular, with\[ \boxed{ \dim_\CC T_{\bar0}X=n+1. } \]
The invariant ring inherits the usual grading from \(\CC[x,y]\). Its positive degrees are multiples of \(n\), and the degree-\(n\) piece has basis \[ z_0,z_1,\ldots,z_n. \] The maximal ideal \(\mfm\) consists of the positive-degree elements. Therefore every element of \(\mfm^2\) has degree at least \(2n\). It follows that the classes of \[ z_0,\ldots,z_n \] are linearly independent in \[ \mfm/\mfm^2, \] and step 1 shows that every invariant of degree \(qn\) with \(q\geq2\) is a product of degree-\(n\) invariants, hence lies in \(\mfm^2\). Thus these classes also generate \(\mfm/\mfm^2\). Hence \[ \dim_\CC\mfm/\mfm^2=n+1. \] Thus \[ \dim_\CC T_{\bar0}X = \dim_\CC(\mfm/\mfm^2)^\vee = n+1. \]
By step 1, \[ \dim X=2, \] while \(n+1>2\) because \(n>1\). Therefore \(\bar0\) is singular.
7. The quotient has exactly one singular point:\[ \boxed{ \operatorname{Sing}(X)=\{\bar0\}. } \]
By step 3, every point other than \(\bar0\) lies in \(D(z_0)\) or \(D(z_n)\). Steps <1>4–<1>5 show that both opens are smooth. Step 6 shows that \(\bar0\) is singular. Hence it is the unique singular point.
8. Q.E.D.
Step 2 proves normality, and step 7 proves uniqueness of the singular point.