I would like to solve the following exercise:

Let $A$ be an integral domain. Let $f$ be a non-unit element of $A$. Then show that $A[f^{-1}]$ is not finitely generated $A$-module.

Here is my attempt:

Let $K$ be the quotient field of $A$. Let $B$ be a subring of $K$ containing $A$ and $f^{-1}$.

We have to show that $A[f^{-1}]$ is not finitely generated $A$-module.

On contrary, suppose $A[f^{-1}]$ is finitely generated $A$-module. Then $f^{-1}$ is integral over $A$ by proposition 5.1 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 $f^{n}$ on both sides, 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.

Is the above argument correct?

Another solution would be highly appreciated.


Suppose $h_1=g_1/f^{k_1},h_2=g_2/f^{k_2},...,h_n=g_n/f^{k_n}\in A[f^{-1}]$, with the $g_1,...,g_n\in A$.

Then $f^{\max(k_1,...,k_n)}B\subset A$, where $B$ is the $A$-module generated by $h_1,...,h_n$.

But $f^{\max(k_1,...,k_n)}A[f^{-1}]\not\subset A$ because it contains $f^{\max(k_1,...,k_n)}(1/f^{\max(k_1,...,k_n)+1})=f^{-1}\notin A$.

Therefore, $B\neq A[f^{-1}]$ for any choice of $h_1,...,h_n$.


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