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

[Bachelier]

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?

 

[pwsnafu]

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?

 

[Bachelier]

No, you have zero divisors.

You mean like (0,7).(8,0)

 

[HallsofIvy]

Yes, neither (7, 0) nor (0, 8) is the additive identity but neither has a mulitplicative inverse.

 

[Bachelier]

thanks. Basically ℂ will fail to be an integral domain in the first place under this operation.