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Using field axioms to prove a set is not a field

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Let F = {a + b[itex]\sqrt[3]{2}[/itex]:a,b[itex]\in[/itex]Q}.
    Using the fact that [itex]\sqrt[3]{2}[/itex] is irrational, show that F is not a field.

    [Hint: What is the inverse of [itex]\sqrt[3]{2}[/itex] under multiplication?]


    2. Relevant equations

    For a field,
    For all c [itex]\in[/itex] F, there exists c-1 [itex]\in[/itex] F s.t. c*c-1 =1

    3. The attempt at a solution

    I am unsure how to relate the axioms to the set. Is c = (a +b[itex]\sqrt[3]{2}[/itex])?
    Or show there is no b[itex]\sqrt[3]{2}[/itex]*b-1[itex]\sqrt[3]{2}[/itex]=1?
    Or b[itex]\sqrt[3]{2}[/itex]*(b[itex]\sqrt[3]{2}[/itex])-1=1?
     
  2. jcsd
  3. Oct 12, 2012 #2

    Hurkyl

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    There are lots of ways to go about it, but if you're going to follow the hint, it specifically suggests showing that [itex]\sqrt[3]{2}[/itex] is an element that doesn't have an inverse.
     
  4. Oct 13, 2012 #3
    [itex]\sqrt[3]{2}[/itex][itex]\in[/itex]F with a=0 and b=1
    So there must be c such that c*[itex]\sqrt[3]{2}[/itex]=1
    c=1/[itex]\sqrt[3]{2}[/itex][itex]\notin[/itex]Q
    Since [itex]\sqrt[3]{2}[/itex][itex]\in[/itex]F, F must not be a field because it has no inverse under multiplication.
    Is this sound?
     
    Last edited: Oct 13, 2012
  5. Oct 13, 2012 #4
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