Polynomial invariant ring ($\mathbb{C}[W{]}^G$)

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 $\zeta =e^{\pi i/n}$ and let the binary dihedral group act on ${\mathbb{C}}^2$ by $(x,y)\mapsto (\zeta x,{\zeta }^{-1}y)$ and $(x,y)\mapsto (-y,x)$. For $n\ge 2$, the invariantspresent the invariant ring as $\mathbb{C}[U,V,W]/(W^2-V{U}^2+4V^{n+1})$. First take the cyclic invariants $A=x^{2n},B=y^{2n},C=xy$ with $AB=C^{2n}$. The residual involution interchanges $A,B$ and negates $C$, so invariant normal forms are polynomials in $A+B,C^2$ plus $C(A-B)$ times such polynomials. For $n=3$ the group has order twelve and the equation is $W^2-V{U}^2+4V^4=0$.

The scalar sign action of a cyclic group of order two on ${\mathbb{C}}^2$ has invariant ring $\mathbb{C}[X^2,XY,Y^2]\cong \mathbb{C}[a,b,c]/(ac-b^2)$. It is not factorial: $a,b,c$ are pairwise nonassociate irreducibles, since every nonconstant invariant has degree at least two, while $ac=b^2$ gives two distinct factorizations.

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.

A finite group permutes the associate classes of irreducible polynomial factors of an invariant. The product of one representative from each orbit transforms by a scalar character of the group. If that character is trivial, the product is invariant. Its transitive factor orbit makes it prime in the invariant ring: an invariant polynomial divisible by one orbit factor is divisible by all of them.

For the permutation action on $n$ coordinates with $n\ge 2$, every alternating-group invariant is a symmetric polynomial plus the Vandermonde product $\Delta$ times a symmetric polynomial. Thus the invariant ring is $\mathbb{C}[e_1,\dots ,e_n,Z]/(Z^2-D(e_1,\dots ,e_n))$, where $D$ is the discriminant polynomial. The two summands are independent over the symmetric polynomial ring.