Proposition

Let $A$ be an integral domain with field of fractions $K$, and suppose that $f \in A$ is nonzero and not a unit. Prove that $A[1/f]$ is not a finite $A$-module.

Solution

Suppose $A[1/f]$ is a finite $A$-module.

Let ${ \frac{a_1}{f^{b_1}}, \cdots, \frac{a_n}{f^{b_n}} }$ be a generating set where $a_i \in A$ and $b_i \geq 0$ for each $i$. Then ${ \frac{1}{f^{b_1}}, \cdots, \frac{1}{f^{b_n}} }$ generates $A[1/f]$. Let $b = \max{ b_1, \cdots, b_n }$. Then ${ 1, \frac{1}{f^{1}}, \cdots, \frac{1}{f^{b}} }$ generates $A[1/f]$.

Therefore, there exist $c_i$ such that

Then $f(c_0f^b + \cdots + c_b) = 1$. Therefore, $f$ is a unit. However, we chose $f$ to be a non-unit element. This is a contradiction, so $A[1/f]$ is not a finite $A$-module.