# Free algebra implies free module

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.

• If $R$ is Noetherian, then you can take a Noether normalization over which your ring is finite and free if it is Cohen-Macaulay (and some conditions). For any ring $B$ between these two rings, the ring $R[x_1,\dots, x_n]$ is finite, but in general not free over $B$. – Youngsu Aug 22 at 15:17

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 $$R:=\mathbb{Z}_2$$.
• the non-trivial relation is $x*x^2-x^3=0$, right? Then we could take $R=\mathbb{Z}$ which seems simpler. – user693936 Aug 21 at 12:51