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Proving each nonzero element of a subfield of C has an inverse

  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Let S={p+qα+rα2 : p, q, r [itex]\in \mathbb{Q}[/itex]}, where α=[itex]\sqrt[3]{2}[/itex]. Then S is a subfield of [itex]\mathbb{C}[/itex]. Prove that each nonzero element of S has a multiplicative inverse in S.

    3. The attempt at a solution

    Let p, q, r[itex]\in\mathbb{Q}[/itex] such that not all of p, q, r are zero. If each nonzero element of S has a multiplicative inverse in S, then there are rational numbers a, b, c so that

    (p+qα+rα2)(a+bα+cα2)=1

    Expanding and using the fact that α3=2 and α4=2α, we get:

    (pa+2qc+2rb)+α(pb+qa+2rc)+α2(pc+qb+ra)=1

    Since 1, α, and α2 are linearly independent we may equate coefficients to give a system of equations:

    pa+2qc+2rb=1
    pb+qa+2rc=0
    pc+qb+ra=0

    Now to finish the problem I want to show that this system always has a solution in [itex]\mathbb{Q}[/itex]. I tried using Cramer's rule (i.e. showing the determinant of the matrix whose entries are the coefficients of p, q, r is equal to zero only if p=q=r=0), but I had no luck and it just turned into a mess. Likewise, back substitution was a mess. Can anyone give me a hand on how to complete this problem?

    Thanks!
     
  2. jcsd
  3. Sep 12, 2014 #2

    Ray Vickson

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    If ##x, y \in S## have multiplicative inverses in ##S## and if ##x + y \neq 0##, can you prove that ##x+y## has a multiplicative inverse in ##S##? Can you prove that ##1, \alpha, \alpha^2## all have multiplicative inverses in ##S##?
     
  4. Sep 12, 2014 #3
    Thanks for the quick response! I think I see what you're getting at. If I can prove what you wrote, then since each element of S is a linear combination of 1, α, and α2 whose coefficients are in [itex]\mathbb{Q}[/itex], it follows that each element of S has an inverse in S. Unfortunately I really am not sure how to show that if [itex]x, y, x^{-1}, y^{-1} \in S[/itex] with [itex] x \ne y [/itex], then [itex] (x+y)^{-1} \in S[/itex]. I tried to solve the equation [itex](x+y)(ax^{-1}+bx^{-1})=1[/itex] for a and b, but that didn't get me anywhere, and I'm not sure what else to try. Is it possible for you to give me a small hint to get me going in the right direction?

    Thanks again.
     
  5. Sep 13, 2014 #4

    HallsofIvy

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    Was that really the original statement of the problem? Saying "S is a subfield of [itex]mathbb{C}[/itex]. Prove that each nonzero element of S has a multiplicative inverse in S" seems very strange! A subfield is, by definition, a field so each non-zero element must have a multiplicative inverse. It would make more sense to first define S then ask to show that every nonzero element has a multiplicative inverse, as part of proving that S is a subfield.
     
  6. Sep 13, 2014 #5
    You're correct, what I wrote is a strange problem statement. It was my fault. The full problem was to show S is a subfield of C. However, I knew how to prove all of the other field axioms hold for S, so I changed the problem statement. As you correctly pointed out, I didn't change it very well. Would it be better if I wrote "confirm that each nonzero element of S has an identity"? Or should I just have written up the actual problem statement but mentioned I only need help proving the existence of multiplicative inverses?
     
  7. Sep 13, 2014 #6

    Ray Vickson

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    Many of these issues are dealt with in
    http://math.stackexchange.com/quest...proof-that-the-algebraic-numbers-form-a-field

    If you Google 'algebraic numbers' or 'algebraic number fields' you will encounter loads of material on your topic.
     
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