What's the deal with complex numbers?

>>8467306
Yeah, and in fact we did. But a brainlet like you wouldn't understand.

>Could we find another function under which a field is closed
do you even know what you're talking about? this doesn't make any sense

>>8467323
[math]\mathbb{C}[/math] is closed under the square root function.
For all [math]z \in \mathbb{C}[/math] we have [math]\sqrt{z}\in \mathbb{C}[/math].
Generalization for a field [math]\mathbb{F}[/math]:
For all [math]x \in \mathbb{F}[/math] we have [math]f(x)\in \mathbb{F}[/math].
Are there function such as the square root that would make for useful algebraic structures?

>>8467329
lots of fields are closed under the square root function, this property is called being 'quadratically closed' https://en.wikipedia.org/wiki/Quadratically_closed_field

>>8467329
The extended real line [math]\overline{\mathbb{R}}[/math] is a trivial example.
Define [math]f(x)[/math] such that it either goes to [math]-\infty[/math] or [math]+\infty[/math].

>>8467337
Yeah, but are there particular functions like the square roots that entirely characterize something like a field?

>>8467346
how does the square root 'entirely characterize' the complex numbers?

>>8467349
Because that's how complex numbers are defined.
Define [math]i=\sqrt{-1}[/math].
Add and multiply it with reals: [math]x+iy[/math].
How about giving an answer to OP instead of making silly questions?

>>8467356
you can define the complex numbers without any reference to square root
https://en.wikipedia.org/wiki/Complex_number#Formal_construction

i'm just trying to get OP to refine the question to be more clear...

you implicitly introduced the real numbers which OP didn't mention so it seems like you're interpreting the question as 'degree two extensions of the real numbers' but that's not clear to me at all that it's what OP was interested in

>>8467366
Not an argument

>>8467366
>i'm just trying to get OP to refine the question to be more clear...
How do I refine it?
I'm a brainlet.
I expected some algebraist to give an example.

[math]\mathbb{R}[/math] is a field, which means it has multiplication and addition, which makes it really natural to consider polynomials [math]\mathbb{R}[X][/math].
The best way to define [math]\mathbb{C}[/math] is as the splitting field of [math]X^2+1[/math], which can easily be constructed as [math]\mathbb{R}[X]/(X^2+1)[/math].
It happens to also be an agebraic closure of [math]\mathbb{R}[/math], which means that any polynomial can be split into degree 1 plynomials in [math]\mathbb{C}[/math].

Now to get back to your question. It would be a natural idea to consider the quotient [math]\mathbb{R}[X]/I[/math], where I is another ideal than [math](X^2+1)[/math]. But with the fact that [math]\mathbb{R}[/math] is a principal domain, and the help of the chinese remainder theorem, any of these quotients is isomorphic to [math]\mathbb{R}^n \times \mathbb{C}^m[/math]. We don't really need other fields than [math]\mathbb{C}[/math] for this.

However, the choice of [math]\mathbb{R}[/math] as our base field is arbitrary. We could use, for instance, the finite field [math]F_p = \mathbb{Z}/p\mathbb{Z}[/math]. If P is an irreducible polynomial in this field, then [math]F_p/(P)[/math] is a new field, where P has a root (which is X). In fact, every finite field can be constructed this way.

>>8468159
i meant to write [math]F_p[X]/(P)[/math] near the end

>>8468159
oh and also, the decomposition of [math]\mathbb{R}[X]/I[/math] as [math]\mathbb{R}^n \times \mathbb{C}^m[/math] is only true if it is a field. if you use an ideal generated by a polynomial that has multiple factors, it wont be true.

well you get the idea anyway

>>8468159
Screencapped.