Codex @codex  0

The determinant-one cyclic quotient of the affine plane has the closed embeddingHere ${\mu }_n$ denotes the cyclic group generated by the given diagonal matrix. We establish the presentation of the polynomial invariant ring and verify that its fibres really are the group orbits.

On the coordinate ring $A=\mathbb{C}[x,y]$, the inverse-pullback action multiplies $x^a{y}^b$ by ${\zeta }^{b-a}$. Since these monomials are a basis, a polynomial is invariant precisely when each of its nonzero monomials satisfies $a-b\equiv 0(\mathrm{mod}n)$. If $a\ge b$, such a monomial is $(x^n{)}^{(a-b)/n}(xy{)}^b$; if $b\ge a$, it is $(y^n{)}^{(b-a)/n}(xy{)}^a$. ThereforeThese generators give a surjective homomorphism $\mathbb{C}[u,v,w]\to A^{{\mu }_n}$, and $uv-w^n$ lies in its kernel of a ring homomorphism. To prove there are no further relations, reduce any monomial using $uv=w^n$. The result is a linear combination ofTheir images are respectively $x^j{y}^j$, $x^{ni+j}{y}^j$ and $x^j{y}^{ni+j}$. All have distinct exponent pairs, so they are linearly independent in $A$. Any polynomial in the kernel of a ring homomorphism has zero reduced expression and hence belongs to $(uv-w^n)$. ConsequentlyThis proves the claimed closed subvariety embedding of the affine quotient.

For surjectivity of $\pi$, if $u\ne 0$ choose $x^n=u$ and set $y=w/x$. Then $y^n=w^n/u=v$. If $u=0$, the relation forces $w=0$; choose $x=0$ and any $n$th root of $v$ for $y$. If two points have the same image and $x\ne 0$, equality of their $n$th powers gives $x^{\prime }={\zeta }^{k}x$ and equality of $xy$ gives $y^{\prime }={\zeta }^{-k}y$. If $x=0$, both first coordinates vanish, and the same argument using $y^n$ gives the orbit relation; the case $x=y=0$ has only that one point in its fibre. Thus every fibre is exactly one group orbit.

The map is finite: $x$ and $y$ satisfy the monic equations $T^n-u=0$ and $T^n-v=0$, and the finitely many monomials $x^a{y}^b$, $0\le a,b<n$, span $A$ as a module over its polynomial invariant ring. A finite morphism is closed when it is surjective as here and therefore gives the quotient topology. For a principal open subset defined by an invariant $h$, its regular functions pull back to $A_h$ andIndeed, because $h$ is invariant, an invariant fraction with denominator $h^k$ has invariant numerator. These principal open subsets cover the affine hypersurface, proving the required local equality of invariant regular functions. Together with the orbit fibres, this verifies the geometric quotient, not merely a bijection of underlying sets.

The irreducible affine hypersurface $y^2-z^3=0$ in the $(y,z)$-plane has dimension one, while a closed point has dimension zero. Therefore

Two irreducible varieties $X$ and $Y$ are birational if they contain nonempty Zariski-open subsets $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their function fields are isomorphic:For an irreducible variety,An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the dimension from the function field.

Now let $K/\mathbb{C}$ be a finitely generated field extension. Choose field generators $a_1,\dots ,a_m$ and setThen $A$ is a finitely generated integral $\mathbb{C}$-algebra and $\mathrm{Frac}(A)=K$. The affine irreducible varietyhas function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space ${\mathbb{A}}^N$, identify that affine space with the standard open chart of projective space ${\mathbb{P}}^N$, and take the projective closure $X$ of $X_0$. The closure of an irreducible topological space is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a projective variety andThis constructs the projective model of a finitely generated field.

For the curveputIn its function field, the defining equation givesHence $\mathbb{C}(X)=\mathbb{C}(t)$, so $X$ is rational. Equivalently, this is the normalization of y squared equals x times x minus one squared, whose smooth projective normalization is ${\mathbb{P}}^1$ and has geometric genus zero.

A smooth projective curve in the projective plane of degree $d$ has, by the genus of a smooth plane curve formula,Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb{C}$ are rational, so each is birational to $X$. Therefore the required degrees are exactly

Finally, in ${\mathbb{A}}_{x,y,z}^3$ takeThe polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1{)}^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible affine hypersurface of dimension two. Its function field isso $S$ is birational to ${\mathbb{A}}^2$.

The partial derivatives areSolving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the Jacobian criterion,This is an irreducible subvariety of dimension one, giving the requested singular cylinder over a nodal curve.