Ring of invariants

Remarks

$R^G$ is a subring of $R$, because each $g$ acts by a ring homomorphism and so fixes $0$, $1$, sums and products of fixed elements. The main case is a group $G$ acting linearly on a finite-dimensional vector space $V$ over a field $k$. Then $G$ acts on the polynomial ring $k[V]$ of polynomial functions by $(g \cdot f)(v) = f(g^{-1} v)$. Emmy Noether proved that for a finite group $G$ acting by $k$-algebra automorphisms on a finitely generated commutative $k$-algebra $R$, the ring $R^G$ is finitely generated.

Example: the symmetric group $S_n$ acts on $k[x_1, \ldots, x_n]$ by permuting the variables. Its ring of invariants is the ring of symmetric polynomials, which is a polynomial ring in the elementary symmetric polynomials by the fundamental theorem of symmetric polynomials.

Example: let $G = \set{1, g}$ act on $k[x, y]$ by $g \cdot x = -x$ and $g \cdot y = -y$, where the characteristic of $k$ is not $2$. A monomial $x^a y^b$ is fixed exactly when $a + b$ is even, so $k[x, y]^G = k[x^2, xy, y^2]$.

Non-example: in the last example, $x \notin k[x, y]^G$, because $g \cdot x = -x \ne x$.