Is Multiplication Defined in This Field Sufficient for ℂ to be a Field?

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Discussion Overview

The discussion revolves around the definition of multiplication in the context of the field of complex numbers, ℂ, and whether a specific multiplication operation defined on the Cartesian plane, ℝ x ℝ, can establish ℂ as a field. The scope includes theoretical considerations of field properties and the implications of zero divisors.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant proposes a definition of addition and multiplication on the plane ℝ x ℝ and questions if this makes ℂ a field.
  • Another participant asserts that the proposed multiplication leads to zero divisors, which would disqualify it from being a field.
  • Further clarification is provided regarding specific examples of zero divisors, such as (0,7) and (8,0), which do not have multiplicative inverses.
  • A participant concludes that under the proposed operation, ℂ fails to be an integral domain.

Areas of Agreement / Disagreement

Participants generally agree that the proposed multiplication leads to zero divisors, which prevents ℂ from being a field. However, the discussion does not reach a consensus on alternative definitions or operations that could resolve the issue.

Contextual Notes

The discussion highlights limitations related to the proposed multiplication operation, specifically the presence of zero divisors and the implications for field properties. There are unresolved questions regarding how to define multiplication in a way that would allow ℂ to maintain its field structure.

Bachelier
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if I define The plane: ##F = ℝ## x ##ℝ = \{ (a, b) | a, b ∈ ℝ \} ##

and define addition and multiplication as:
(a, b) + (c, d) := (a + c, b + d)
(a, b) · (c, d) := (ac, bd)

Then ##F## is a field. right?

would the multiplication as described here make ℂ a field?
 
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Bachelier said:
if I define The plane: ##F = ℝ## x ##ℝ = \{ (a, b) | a, b ∈ ℝ \} ##

and define addition and multiplication as:
(a, b) + (c, d) := (a + c, b + d)
(a, b) · (c, d) := (ac, bd)

Then ##F## is a field. right?

No, you have zero divisors.

would the multiplication as described here make ℂ a field?

How would you do i2 = -1?
 


pwsnafu said:
No, you have zero divisors.

You mean like (0,7).(8,0)
 
Yes, neither (7, 0) nor (0, 8) is the additive identity but neither has a mulitplicative inverse.
 
thanks. Basically ℂ will fail to be an integral domain in the first place under this operation.
 
Last edited:

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