Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 15 / 1 / Solution

A monomial $x^a{y}^b$ is invariant precisely when $a-b$ is divisible by $n$: pulling back along the generator multiplies it by ${\zeta }^{a-b}$. Since distinct monomials are linearly independent, an invariant polynomial is a linear combination of exactly these invariant monomials. If $a\ge b$, write $a-b=kn$ and obtainIf $b\ge a$, the analogous expression uses $y^n$ and $xy$. Thus the polynomial invariant ring is generated bywhich satisfy $uv=w^n$.

To check that this is the only relation, reduce each monomial in $\mathbb{C}[U,V,W]$ using $UV=W^n$ until it has one of the formsTheir images are, respectively, $x^c{y}^c$, $x^{nk+c}{y}^c$ and $x^c{y}^{nk+c}$. These are pairwise distinct monomials in $x,y$, so no nontrivial linear combination of the reduced forms maps to zero. ThereforeThis identifies the determinant-one cyclic quotient of the affine plane with the affine hypersurface $UV=W^n$, and the quotient map is $(x,y)\mapsto (x^n,y^n,xy)$.

Apply the Jacobian criterion to $f=UV-W^n$. Its gradient isOver $\mathbb{C}$, with the printed $n\ge 3$, this vanishes only at $U=V=W=0$. Every other point is smooth. This point is exactly the image of the origin, and its inverse image is only the origin, since $x^n=y^n=0$ forces $x=y=0$. In fact the origin itself is singular: its tangent space has dimension three while the hypersurface has dimension two. Hence the quotient is smooth away from the image of the origin, with that point its unique singularity. Primality is not needed for this calculation; the same argument works for every integer $n\ge 2$.