Any ring is integral over the subring of invariants under a finite group action

Question

I need to prove that if $G$ is a finite group that acts on ring $A$, and $A^G$ is the subring consisting of elements of $A$ which are invariant under all $g\in G$, then $A$ is integral over $A^G$. (Atiyah and Macdonald, Chapter 5, Exercise 12)

The hint with the problem is to state that each $x\in A$ must be a root of the polynomial $\prod_{g\in g}(t - g(x))$, but this is only guaranteed to be in $A[t]$. I can't think of any reason why this product, or some important factor of it such as $t - x$ where the $g$ involved is the identity, would only have coefficients in $A^G$, which is the only way I know of to finish the proof. Is there a better way, or am I missing some obvious fact about the groups/rings involved here?

Answer 1

The action of $G$ permutes the orbit $Gx=\{gx:g\in G\}$, hence permutes the linear factors $t-gx$.

Rearranging the linear factors does not change the resulting product, so it must be invariant. (Note $A$ must be commutative for the original product to make sense independent of order, so that's not an issue.)

Writing $G=\{g_1,\dots,g_n\}$, the coefficients are $e_i(g_1x,\cdots,g_nx)$ up to sign, where $e_i$ stands for the $i$th elementary symmetric polynomial. (See Vieta's formulas.) Applying $g$ yields $e_i(gg_1x,\dots,gg_nx)$, but the list $gg_1x,\dots,gg_nx$ is the same as $g_1x,\dots,g_nx$ just rearranged. Since $e_i$ is a symmetric polynomial, rearranging its arguments does not affect its value. Thus, the coefficients are invariant.