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 $$g \cdot x$$.