Let $R$ be a commutative unital ring. Let $n\geq 1$ be an integer. Suppose that $R[x_1, \dots, x_n]$ has a $R$-subalgebra $A$ such that $R[x_1, \dots, x_n]$ is a finitely generated $A$-module. Is it true that $R[x_1, \dots, x_n]$ is a free $A$-module? Is this true for $n=2$ at least?
I think so because there are no non-trivial relations satisfied by $x_1,\dots, x_n$ but I am not sure what happens when there is torsion.
Consider the ring $A:=R[x^2, x^3]$. Let $x^n\in R[x]$ and let $n=3k+r$ where $r\in\{0, 1, 2\}$. If $r=0$, then $x^n=x^{3k}.1$ and $x^{3k}\in R[x^2, x^3]$. If $r=1$, then $x^n=x^{3k}x$ and $x^{3k}\in R[x^2, x^3]$, and if $r=2$, then $x^n=x^{3k}.x^2.1$ and $x^{3k}.x^2\in R[x^2, x^3]$. Thus, $R[x]$ is a finitly generated $R[x^2, x^3]$-module with generating set $\{1, x\}$. But $R[x]$ is not a free $R[x^2, x^3]$-module. For simplicity set $R:=\mathbb{Z}_2$.