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Proving that a subspace of C is a field

  1. Sep 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the only m for which the subspace of C given by {z ∈ C: Im(z) = m Re(z)} is a field is m=0.
    2. Relevant equations
    Field axioms
    3. The attempt at a solution
    I tried to prove one direction :
    - If z is in the subspace, Re z>0 and m≠0 then Arg z<Arg z^2, so z^2 is not in the subspace.
    I need a hint to prove the other direction.
     
  2. jcsd
  3. Sep 10, 2015 #2

    Mark44

    Staff: Mentor

    No. For example, z = -1 -mi is in C, but Re(z) < 0. Note that for a given value of m, C is a straight line through the origin. Which field axioms prevent this subspace from being a field?
     
  4. Sep 10, 2015 #3

    RUber

    User Avatar
    Homework Helper

    An element of the subspace z would have the form z = x + imx, with x in the reals.
    As Mark mentioned, you should check the axioms, it seems like this space satisfies most of the addition axioms. I think you were on the right track with squaring z. Using z^2 as an example, you can show that the only m that satisfies multiplication axioms (for any choice of x and a fixed m) is 0.
     
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