Ring is integral extension of fixed-point subring under finite automorphism group

Statement

Let $A$ be a commutative unital ring and $G$ be a finite group acting as automorphisms on $A$ (in other words, $G$ is a finite subgroup of the automorphism group of $A$). Let $B=A^{G}$ be the subring of $B$ comprising those elements fixed by every element of $G$. Then, $B$ is an integral extension of $A$. In fact, every element of $B$ satisfies a monic polynomial over $A$ of degree equal to the order of $G$.

Related facts

• Automorphism group acts transitively on fibers of spectrum over fixed-point subring

Proof

Let $x\in B$. Consider the elements $g\cdot x$ for $g\in G$. Then, all the elementary symmetric polynomials in these elements take values in $A$. Hence, we can construct a monic polynomial of degree equal to the order of $G$, with all coefficients in $A$, and whose roots are precisely the elements $g\cdot x$.