# Abstract algebra, finite A-module

## Homework Statement

Let A be an integral domain with field of fractions K, and suppose that $$f\in A$$ is non zero and not a unit. Prove that $$A[\frac{1}{f}]$$ is not a finite A-module.
[Hint: if it has a finite set of generators then prove that $$1,f^{-1},f^{-2},...,f^{-k}$$ is a set of generators for some $$k>0$$, so that $$f^{-(k+1)}$$ can be expressed as a linear combination of this. Use this to prove that f is a unit.

## Homework Equations

We assume that [itex]f\neq 0[/tex] and $$f\in A\setminus A^{*}$$ where I denote $$A^{*}$$ as the set of units
$$A[\frac{1}{f}]=\{p(\frac{1}{f}); p(x)\in A[x]\}$$

## The Attempt at a Solution

suppose that $$A[\frac{1}{f}]$$ is a finite module, that is $$A[\frac{1}{f}]=\displaystyle\sum_{i=1}^{n}Ap_{i}$$ for $$p_{i}(x)\in A[x]$$
Let $$k=max deg p_{i}(x)$$
$$A[\frac{1}{f}]=A\cdot 1+A\cdot f^{-1}+......+A\cdot f^{-k}$$
(how did we get this equations???)
hence $$\exists a_{0},......,a_{k}\in A$$ (based on what there exists such elements??)
s.t $$f^{-(k+1)}=a_{0}+a_{1}f^{-1}+...+a_{k}f^{-k}$$ which gives
$$f^{-1}=a_{0}f^{k}+a_{1}f^{k-1}+....+a_{k}\in A$$ and we get $$f\in A^{*}$$ hence $$A[\frac{1}{f}]$$ is infinite.
Could someone explain me the second part of the proof and its conslusions?
Thank you