Some localization is not finitely-generated as an R-module Let $R$ be an integral domain with field of fractions $K$, and let $f \in R$ be a non-zero non-unit.
Prove that the subring $S=R[1/f]$ of $K$ is not finitely-generated as an $R$-module, using the fact that every element of $S$ may be written in the form $r/f^n$ with $r\in R$ and $n\geq0$, and any finite set of elements of $S$ can be written like this with a common denominator.
 A: Let $K$ be the quotient field of $A$. Let be a subring of $K$ containing $A$ and $f^{-1}$. 
We have to show that $A[f^{-1}]$ is not finitely generated.
On contrary, suppose $A[f^{-1}]$ is finitely generated. Then $f^{-1}$ is integral over $A$ by proposition of the book Commutative Algebra by Atiyah. $i.e.,$ there are $a_{1}, a_{2},...,a_{n}\in A$ such that \begin{equation}
f^{-n}+ a_{n}f^{-(n-1)}+...+ a_{1}=0
\end{equation}
By multiplying with $f^{n}$, we have 
\begin{equation}
1+ a_{n}f+...+ a_{1}f^{n}=0
\end{equation}
$i.e.,$ \begin{equation}
-f(a_{n}+...+a_{1}f^{n-1})=1
\end{equation}
Thus $f$ is a unit that yield a contradiction of the hypothesis.
A: Easy Answer: $fx-1$ is not monic. Edit: This is wrong unless $R$ is integrally closed.
Hard Answer: Suppose $S=R[1/f]$ is finitely generated and let $a_1,\dots,a_n$ be a generating set of $S$. By the fact we're allowed to use we may write the generators as $r_1/f^{e_1},\dots, r_n/f^{e_n}$. Let $e=e_1\dots e_n $ and write $1/f^{e+1}$ with respect to this generating set,
$$\frac{1}{f^{e+1}}=\sum_{i=1}^n \frac{b_i r_i}{f^{e_i}}$$
for some $b_i \in R$. Multiplying by $f^e$ we deduce that $1/f \in R$ and so $f$ is a unit which is a contradiction. 
A: Hint $\ $  One conceptual way to view this is as a generalization of the familiar case when $\rm\,R = \Bbb Z,\,$ where the Rational Root Test implies that proper fractions cannot be roots of polynomials $\rm\in \Bbb Z[x]\,$ that are monic (i.e. leading coefficient $= 1).$
Let $\rm\,e = 1/f.\:$ If $\rm\:R[e] = R\left<1,e,\ldots,e^n\right>\,$ then $\rm\,e^{n+1}$ is an $\rm\,R$-linear combination of lower powers of $\rm\,e,\,$ i.e. the proper fraction $\rm\,e\,$ is a root of a monic polynomial over $\rm\,R.\,$ Therefore the  proof of the Rational Root Test (RRT) shows that the denominator $\rm\,f\,$ divides the leading coefficient $ = 1,\, $ i.e. $\rm\,1/f\in R,\,$  contra hypothesis.
Remark $\ $ The RRT requires that the fraction be in lowest terms, i.e. that the denominator and numerator are coprime, which is true here since the numerator $= 1$.
A domain $\rm\,D\,$ is called integrally closed (in its fraction field) if none of its proper fractions are integral over $\rm\,D,\,$  i.e. they are not roots of monic polynomials over $\rm\,D.\:$ The usual proof of the Rational Root Test works in any domain where gcds exist (gcd-domain), e.g. in any UFD. Therefore gcd-domains and UFDs are integrally-closed.
