Is $R[X,Y]/(XY-1)$ a finitely generated $R[X]$-module?

Question

Let $R$ be a commutative Noetherian ring. Consider the localisation of $R[X]$ at the multiplicative closed set $\{1,X,...\}$ i.e. $R[X]_X \cong R[X,Y]/(XY - 1)$. Notice that the localisation map $R[X] \to R[X]_X$ is injective (as $X$ is not a zero divisor in $R[X]$) .

My question is : Is $R[X]_X\cong R[X,Y]/(XY - 1)$ necessarily a finitely generated $R[X]$-module ?

Definitely $R[X]_X$ is a finitely generated $R[X]$-algebra, so it is a finitely generated $R[X]$-module iff it is integral over $R[X]$. I can't seem to proceed further. Please help. (If need be, I'm willing to assume $R$ is an integral domain) .

Answer 1

If I'm not mistaken, this map is never finite, unless $R$ is the zero ring. Indeed, since $X$ is not a unit of $R[X]$, there is always some maximal ideal $M$ of $R[X]$ containing $X$. The prime ideals of $R[X]_{X}$ are in bijection with ideals of $R[X]$ not containing $X$, and the map on spectra $\varphi^{\ast} \colon \mathrm{Spec}(R[X]_{X}) \to \mathrm{Spec}(R[X])$ induced by the canonical localization morphism $\varphi \colon R[X] \to R[X]_{X}$ realizes this inclusion of prime ideals. Since the set of prime ideals of $R[X]$ which contain $X$ is nonempty (it contains $M$), $\varphi^{\ast}$ cannot be surjective. But every integral ring extension induces a surjective map on spectra by Going Up, so we're done.

Answer 2

More generally, a localization $A_S$ is never a finitely generated $A$-module unless the universal map $A \to A_S$ is surjective (the question is just the case where $A=R[x]$ and $S$ is the set of all monic monomials, i.e. powers of $x$).

Indeed, suppose that $\{\frac{a_1}{s_1}, \frac{a_2}{s_2}, ..., \frac{a_n}{s_n}\}$ is a finite generating set for the $A$-module $A_S$. Then, all elements of $A_S$ would be able to be written with a common denominator of $s:=s_1s_2...s_n$, so the map $a \mapsto \frac{a}{s}$ is surjective. Moreover, all elements of $A_S$ can also be written with denominator $1$, for an element being able to be written with denominator $1$ is equivalent to the same element multiplied by $\frac{1}{s}$ being able to be written with denominator $s$. Hence, the universal map $A \to A_S$, $a \mapsto \frac{a}{1}$, is also surjective.