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If a finite group has no nontrivial homomorphism to ${\mathbb{C}}^{\times }$, its polynomial invariant ring is a unique factorization domain. Factor an invariant in the ambient polynomial ring and collect its irreducible factors into orbit products. Each orbit product transforms by a linear character, hence is invariant. It is prime in the invariant ring, and the invariant factorization is a product of these primes.

Geometric invariant theory constructs algebraic quotients of group actions by a reductive algebraic group using invariant sections and a chosen linearization of a line bundle. An affine quotient is the spectrum of the polynomial invariant ring; a projective quotient uses homogeneous invariants and excludes the points where all positive-degree invariants vanish. Stability separates the orbits which give well-behaved geometric quotient points.

Invariant theory studies functions and tensors unchanged by a group action. For a linear representation, the polynomial invariant ring records invariant polynomial functions on the representation space. Tensor invariants and commuting algebra actions connect it with Schur–Weyl duality.

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 $n\ge 2$.

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.

Put $A=\mathbb{C}[V]={⨁}_{d\ge 0}{A}_d$ and $R=A^G$. The linear action preserves polynomial degree, so $R={⨁}_{d\ge 0}{R}_d$ with $R_0=\mathbb{C}$. We prove Hilbert finite generation for reductive invariants by combining finite ideal generation in $A$ with a degree-lowering argument in $R$.

Over $\mathbb{C}$, representations of a reductive algebraic group are completely reducible. On each finite-dimensional $A_d$, project its isotypic component for the trivial representation onto itself and its nontrivial summands onto zero. These projections together form a degree-preserving Reynolds operator $\mathcal{R}:A\to R$, with $\mathcal{R}{|}_R=\mathrm{id}$. The crucial additional property isHere is why it holds rather than being an arbitrary projection property. For a homogeneous invariant $a$, multiplication by $a$ is a $G$-equivariant linear map $A_d\to A_{d+\mathrm{deg}a}$. It sends invariant vectors to invariant vectors. On a nontrivial irreducible summand, its composition with projection onto the target's trivial summand is zero, since there is no nonzero equivariant map from a nontrivial irreducible representation to the trivial representation. It therefore commutes with the Reynolds operators. Summing the homogeneous components of $a$ and $h$ proves the displayed module property in general.

Let $R_{+}={⨁}_{d>0}{R}_d$ and form the ideal $I=A{R}_{+}$ in the ambient polynomial ring. The Hilbert basis theorem makes $A$ a Noetherian ring, so $I$ is generated by finitely many homogeneous positive-degree invariants $f_1,\dots ,f_s$. To justify choosing the generators from this particular set, start with any finite ideal generating set; each of its members is a finite sum of multiples of homogeneous members of $R_{+}$, and the union of those finitely many invariants then generates $I$. If $I=0$, there are no positive-degree invariants and the answer is simply $R=\mathbb{C}$.

Otherwise consider a homogeneous invariant $f$ of degree $d>0$. Since $f\in I$, writewith $h_i=0$ when that degree is negative. Taking the degree-$d$ component of any ideal expression gives exactly such an expression. Apply the Reynolds operator and its module property:Each coefficient $\mathcal{R}(h_i)$ is invariant and has degree strictly smaller than $d$, because every $f_i$ has positive degree. Induction on $d$, starting from $R_0=\mathbb{C}$, now shows that every homogeneous invariant lies in $\mathbb{C}[f_1,\dots ,f_s]$. An arbitrary invariant is a finite sum of invariant homogeneous components, soThis proves finite generation as an algebra. No Noetherian property of the polynomial invariant ring was assumed in obtaining the generators.

Let $R=\mathbb{C}[V]=\mathrm{Sym}(V^{\ast })$, with its degree grading, and let $A=R^G$ be the polynomial invariant ring. For a compact group average polynomials over $G$; for a complex reductive group average over its Zariski-dense compact form from part (i). Each degree piece is finite dimensional, so this defines a polynomial and gives the Reynolds operator on a polynomial invariant ringIt preserves degree. For the reductive case compact invariance is full group invariance by Zariski density.

Let $A_{+}$ be the positive-degree invariants and form the ideal $I=R{A}_{+}$. The Hilbert basis theorem makes $R$ a Noetherian ring, so finitely many homogeneous invariants $f_1,\dots ,f_s\in A_{+}$ generate $I$ as an $R$-ideal. Indeed start with a finite ideal-generating set and collect the finitely many invariants occurring in expressions for those generators; then replace them by homogeneous pieces.

We claim that these same $f_i$ generate $A$ as an algebra. For a homogeneous invariant $f$ of degree $d>0$, write $f={\sum }_i{a}_i{f}_i$ with homogeneous coefficients of degree $d-\mathrm{deg}{f}_i$; terms of negative degree are absent. Apply the Reynolds operator on a polynomial invariant ring to get $f={\sum }_i\mathcal{R}(a_i)f_i$. The averaged coefficients are invariant and have smaller degree than $d$. Starting from the constants, induction on degree puts them and then $f$ in $\mathbb{C}[f_1,\dots ,f_s]$. Taking homogeneous pieces covers every invariant. This proves Hilbert finite generation for reductive invariants:Noetherianity alone would not suffice for a subring; the Reynolds module property is the step that makes the induction work.

Write $x,y$ for the two coordinates. The PDF gives $\mu =e^{\pi i/3}$, a primitive sixth root of unity. The matrices satisfyThese relations reduce every word to ${\sigma }^j$ or $\tau {\sigma }^j$, $0\le j<6$. The six first matrices are diagonal and distinct, and the six second matrices are antidiagonal and distinct. Thus $G$ has exactly twelve elements; it is a binary dihedral group.

First calculate the polynomial invariant ring of $⟨\sigma ⟩$. A monomial $x^a{y}^b$ is invariant exactly when $a-b$ is divisible by $6$. Removing the smaller exponent shows that it is a product of powers ofThe sole relation is $AB=C^6$. Indeed, reduction by this relation leaves monomials $C^j{A}^a$ and $C^j{B}^b$ with $b>0$, whose images in $\mathbb{C}[x,y]$ have distinct exponent pairs. HenceOn these generators, $\tau$ interchanges $A,B$ and sends $C$ to $-C$. Put $S=A+B$ and $T=A-B$. Since $2$ is invertible, the preceding ring isEvery element has a unique form $P(S,C)+TQ(S,C)$. The induced involution fixes $S$ and negates both $C$ and $T$. Its invariants are therefore exactly the expressionsSetWe obtain the relationThe unique normal form above also proves that there are no further relations: $P_0(U,V)+W{Q}_0(U,V)=0$ forces both polynomials to vanish. Thus the polynomial invariant ring is

For completeness, the algebraic quotient by a finite group is the affine variety with this coordinate ring. Each element $h$ of $\mathbb{C}[x,y]$ satisfies the monic orbit polynomial ${\prod }_{g\in G}(Z-gh)$ with invariant coefficients, so the quotient map is a finite morphism. Invariants separate distinct finite orbits: interpolate a polynomial taking value $0$ on one orbit and $1$ on the other, and average it over $G$. Thus its fibres are precisely the orbits. The defining polynomial and resulting algebraic quotient by a finite group areUsing the converted TeX's fourth root would instead give a different group and a different binary dihedral invariant hypersurface; the sixth root from the original PDF is essential.

The coordinate ring is $\mathbb{C}[W]=\mathrm{Sym}(W^{\ast })$. The induced action is contragredient on functions:It is a group action by degree-preserving algebra automorphisms, extending the dual action on linear polynomial functions. Its polynomial invariant ring isFor a monic polynomial $f(t)={\prod }_{i=1}^d(t-r_i)$, its polynomial discriminant is $\mathrm{disc}(f)={\prod }_{i<j}(r_i-r_j{)}^2$. This is symmetric in the roots, hence polynomial in the coefficients, and is zero exactly when a root is repeated.

For the alternating group action, let $e_1,\dots ,e_n$ be the elementary symmetric polynomials and put $\Delta ={\prod }_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any transposition, decomposeNormality and index two of $A_n$ show that $p_{+}$ is symmetric and $p_{-}$ transforms by the sign character of $S_n$. Any alternating polynomial vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_{-}=\Delta q$ with $q$ symmetric. The Fundamental theorem of symmetric polynomials yieldsThe sum is direct, since a polynomial that is both symmetric and alternating is zero in characteristic zero. Also ${\Delta }^2=D(e_1,\dots ,e_n)$, where $D$ is the discriminant polynomial ofThus, more precisely than the requested quotient assertion,Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The direct sum forces $a(e)=b(e)=0$, and algebraic independence of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.

Now let $G$ be finite with no nontrivial linear characters, and write $R=\mathbb{C}[W]$, $S=R^G$. The polynomial ring $R$ is a unique factorization domain. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal{O}$, choose representatives and form its orbit product of polynomial factorsFor every $g$, $g\cdot q_{\mathcal{O}}=\theta (g)q_{\mathcal{O}}$ for a nonzero scalar $\theta (g)$. Applying two group elements proves that $\theta :G\to {\mathbb{C}}^{\times }$ is a homomorphism. The hypothesis forces $\theta =1$, so $q_{\mathcal{O}}\in S$.

This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in \mathcal{O}$ divides $a$ or $b$ in $R$. Invariance of that chosen polynomial makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the integral domain $R$. Thus $q_{\mathcal{O}}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore $\mathbb{C}[W{]}^G$ is a unique factorization domain. This gives a direct proof, without assuming a divisor-class-group theorem.

For a failure in characteristic zero, let the cyclic group of order two act on ${\mathbb{C}}^2$ by $-I$. Its coordinate invariants areEvery invariant monomial has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^i{c}^j$ and $b{a}^i{c}^j$ map to distinct monomials. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, butgives two different irreducible factorizations. Hence this invariant ring is not a unique factorization domain. The nontrivial sign character is precisely the kind of character excluded in the preceding theorem.

For a group acting linearly on $W$, act on the coordinate ring by $(g\cdot f)(w)=f(g^{-1}w)$. The polynomial invariant ring is the fixed subalgebra $\mathbb{C}[W{]}^G$. It inherits the degree grading. Even over the complex numbers it need not be a unique factorization domain, as the quadratic cone invariant ring demonstrates.

Let the symmetric group $S_3$ act by $(x,y)\mapsto (\alpha x,{\alpha }^{-1}y)$ and $(x,y)\mapsto (y,x)$, where ${\alpha }^3=1$ and $\alpha \ne 1$. Rotation invariance requires the exponents of each monomial to differ by a multiple of three. After symmetrizing, factor out a power of $xy$; the remainder is $x^{3k}+y^{3k}$. These power sums are polynomials in $x^3+y^3$ and $x^3{y}^3=(xy{)}^3$, by the recurrence $s_k=(x^3+y^3)s_{k-1}-(xy{)}^3{s}_{k-2}$. Hence the entire polynomial invariant ring is generated by $xy$ and $x^3+y^3$, which are algebraically independent. Their degrees are two and three, giving one invariant in each of degrees two, three and four, and none in degree one.

Realize the standard representation of the symmetric group on $W=\{x\in {\mathbb{C}}^4:{\sum }_i{x}_i=0\}$. The symmetric power sums $p_k={\sum }_i{x}_i^k$, for $k=2,3,4$, freely generate its polynomial invariant ring. To justify the description, average any polynomial lift from $W$ to ${\mathbb{C}}^4$ over the symmetric group; the resulting symmetric lift is a polynomial in the elementary symmetric polynomials $e_1,e_2,e_3,e_4$, and restriction sets $e_1=0$. The remaining three generators are algebraically independent. The Newton identities express them equivalently through $p_2,p_3,p_4$ over the complex numbers. In coordinates $x=(a,b,c,-a-b-c)$, one has $p_2=a^2+b^2+c^2+(a+b+c{)}^2$ and $p_3=-3(a+b)(b+c)(c+a)$.

In characteristic zero, averaging over a compact group, or over a Zariski-dense compact form of a complex reductive group, gives a degree-preserving projection onto the polynomial invariant ring. Its module property is $\mathcal{R}(af)=a\mathcal{R}(f)$ for invariant $a$. This property turns finite ideal generation in the ambient polynomial ring into finite algebra generation of the invariant subring.