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rayman123
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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 f\neq 0[/tex] and f\in A\setminus A^{*} where I denote A^{*} as the set of units<br /> A[\frac{1}{f}]=\{p(\frac{1}{f}); p(x)\in A[x]\}<br /> <h2>The Attempt at a Solution</h2><br /> Proof by contradiction <br /> 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]<br /> Let k=max deg p_{i}(x) <br /> A[\frac{1}{f}]=A\cdot 1+A\cdot f^{-1}+...+A\cdot f^{-k}<br /> (how did we get this equations?)<br /> hence \exists a_{0},...,a_{k}\in A (based on what there exists such elements??)<br /> s.t f^{-(k+1)}=a_{0}+a_{1}f^{-1}+...+a_{k}f^{-k} which gives <br /> 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.<br /> Could someone explain me the second part of the proof and its conslusions?<br /> Thank you