Another multiplication for complex numers

[Anonymous No.4989142]

Is is possible to define another multiplication for complex number such that it obeys the laws of a field?

[Anonymous No.4989151]

obviously OP read about quaternions.

[Anonymous No.4989184]

Yes.
For any pair of non-zero complex numbers a and b,
define a*b = 1 and if either a or b equal 0 then a*b=0.

[Anonymous No.4989201]

>>4989184

Presumably they're not complex numbers if i^2 is not -1.

[Anonymous No.4989215]

>>4989201
What makes you think i*i will be the same if we change how multiplication works?

[Anonymous No.4989230]

>>4989215

Can you really call it the complex numbers if it's different? It will just be some uncountable field.

Besides, if I am not mistaken >>4989184 will only get you <span class="math">\mathbb{Z}_2[/spoiler].

[Anonymous No.4989242]

Do we have to keep addition the same?

>>4989184
That would make
1*1 = 1
1*2 = 1
1 = 2
so no.

[Anonymous No.4989252]

No. The complex numbers are not defined like that.

You can with the hypercomplex numbers.

[Anonymous No.4989282]

>>4989184
i^2 ≠ -1, hence this form of multiplication no longer defines the complex numbers.

If you set a*b = -1 for all non-zero complex numbers a,b, and a*b = 0 for a or b equal to zero you keep the definition of complex numbers but run into other issues, i.e. distributivity fails for non-zero a,b and c: a*(b+c) = -1 by definition and ab+ac = -2 by definition, hence a*(b+c) ≠ a*b + a*c.

[Anonymous No.4989301]

>>4989282

I think OP needs to be more precise about what the hell he is asking.

Say, you could keep all the algebraic complex numbers as normal and collapse all the rest to 1. This preserves i^2=-1, but loses uncountability.

[Anonymous No.4989325]

>>4989301
A field on the complex numbers would imply that if two numbers are different complex numbers, then they are different elements in the field.

So the only non-stupid way for his question to be interpreted is "are there other uncountable fields" in which case the answer is obviously yes.