5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $h_1,...,h_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.