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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.

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.

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.