Ring of invariants
= Ring of invariants
{title2=$R^G=\{r:g(r)=r\text{ for every }g\in G\}$}
= Invariant subring
{synonym}
For a group acting on a <ring> by <ring automorphisms>, the elements fixed by every group element form a <subring> containing $1$. For finite $G$ acting on a commutative <ring>, each $r$ satisfies the monic orbit equation $\prod_{g\in G}(T-g(r))$, whose coefficients are fixed. Thus $R$ is integral over its <invariant subring>. This uses an orbit product, not division by the group order.