Proving that a subspace of C is a field

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SUMMARY

The only value of m for which the subspace of complex numbers defined by {z ∈ C: Im(z) = m Re(z)} forms a field is m=0. This conclusion arises from analyzing the field axioms, particularly focusing on the multiplication axiom. When m≠0, the subspace does not satisfy the necessary conditions for closure under multiplication, as demonstrated by the example of squaring an element z in the subspace. Therefore, the subspace fails to meet the criteria for being a field except when m=0.

PREREQUISITES
  • Understanding of field axioms in abstract algebra
  • Familiarity with complex numbers and their representation
  • Knowledge of the Argand plane and geometric interpretation of complex numbers
  • Basic algebraic manipulation of complex numbers
NEXT STEPS
  • Study the properties of fields in abstract algebra
  • Learn about the geometric interpretation of complex numbers in the Argand plane
  • Explore the implications of closure properties in algebraic structures
  • Investigate other subspaces of complex numbers and their field properties
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Mathematics students, particularly those studying abstract algebra and complex analysis, as well as educators seeking to understand field properties in relation to complex number subspaces.

lolittaFarhat
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Homework Statement


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.

Homework Equations


Field axioms

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.
 
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lolittaFarhat said:

Homework Statement


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.

Homework Equations


Field axioms

The Attempt at a Solution


I tried to prove one direction :
- If z is in the subspace, Re z>0
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?
lolittaFarhat said:
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.
 
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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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