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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.

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  • $\begingroup$ 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 $B$. $\endgroup$ – Youngsu Aug 22 at 15:17
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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 $R:=\mathbb{Z}_2$.

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

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