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.

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 $p_3=-3(a+b)(b+c)(c+a)$.

For a group representation on $V$, a polynomial invariant is a polynomial $p$ satisfying $p(gv)=p(v)$ for every $g\in G$. Homogeneous degree-$d$ invariants form $({\mathrm{Sym}}^d{V}^{\ast }{)}^G$. Using the symmetric power matters: an alternating invariant tensor contracted with repeated copies of the same commuting vector can vanish.

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.

The action $(x,y)\mapsto (\zeta x,{\zeta }^{-1}y)$ has invariant monomials exactly when their two exponents differ by a multiple of $n$. Every such monomial is a product of $x^n,y^n,xy$. Reducing by $uv=w^n$ leaves the monomials $w^j$, $u^i{w}^j$ and $v^i{w}^j$ with $i\ge 1$; their images are distinct monomials in $x,y$, proving that this is the only relation. The map $(x,y)\mapsto (x^n,y^n,xy)$ is finite and its fibres are the finite-group orbits, so the hypersurface is the geometric quotient.

// Destination: ringed-space.bigb

Choose finitely many homogeneous positive-degree invariants generating the ideal generated by all positive-degree invariants in the ambient polynomial ring. Such a finite set exists because a polynomial ring is Noetherian. Apply the Reynolds operator on a polynomial invariant ring to an expression for any homogeneous invariant in that ideal. Its coefficients become invariants of smaller degree, so induction proves that the chosen set generates the entire invariant ring as an algebra.

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.

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.