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Abstract algebra, finite A-module

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A be an integral domain with field of fractions K, and suppose that [tex]f\in A[/tex] is non zero and not a unit. Prove that [tex]A[\frac{1}{f}][/tex] is not a finite A-module.
    [Hint: if it has a finite set of generators then prove that [tex]1,f^{-1},f^{-2},...,f^{-k}[/tex] is a set of generators for some [tex]k>0[/tex], so that [tex] f^{-(k+1)}[/tex] can be expressed as a linear combination of this. Use this to prove that f is a unit.



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



    3. The attempt at a solution
    Proof by contradiction
    suppose that [tex]A[\frac{1}{f}][/tex] is a finite module, that is [tex]A[\frac{1}{f}]=\displaystyle\sum_{i=1}^{n}Ap_{i}[/tex] for [tex]p_{i}(x)\in A[x][/tex]
    Let [tex]k=max deg p_{i}(x)[/tex]
    [tex]A[\frac{1}{f}]=A\cdot 1+A\cdot f^{-1}+......+A\cdot f^{-k}[/tex]
    (how did we get this equations???)
    hence [tex]\exists a_{0},......,a_{k}\in A[/tex] (based on what there exists such elements??)
    s.t [tex]f^{-(k+1)}=a_{0}+a_{1}f^{-1}+...+a_{k}f^{-k}[/tex] which gives
    [tex]f^{-1}=a_{0}f^{k}+a_{1}f^{k-1}+....+a_{k}\in A[/tex] and we get [tex]f\in A^{*}[/tex] hence [tex]A[\frac{1}{f}][/tex] is infinite.
    Could someone explain me the second part of the proof and its conslusions?
    Thank you
     
  2. jcsd
  3. Sep 29, 2012 #2
    I got it after all :)
     
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