Galois theory: Fundamental theorem of algebra

This lecture is part of an online graduate course on Galois theory.

We use Galois theory to give a (mostly) algebraic proof that the complex numbers form an algebraically closed field.

[caspermadlener]

Here is a cool proof I came up with before knowing about Galois theory!

We can assume that the polynomial we want to find the roots for has real coefficients, because we could otherwise multiply it with its complex conjugate.
We assume that real polynomials of odd degree have a real root.
We also assume the square root of a complex number is a complex number.
We define P*, such that for every two 'different zeros' {α, β} of P≠0, P*(α+β)=0.
If deg(P)=n, deg(P*)=n(n–1)/2.
If P(x)=(x–a)(x–b)(x–c), P*(x)=(x–a–b)(x–a–c)(x–b–c).
Suppose deg(P)≡2ⁿ⁺¹ mod 2ⁿ⁺², then deg(P*)≡2ⁿ mod 2ⁿ⁺¹, which is really cool.
The goal is to first proof it for polynomials with degree 2 mod 4, next with 4 mod 8, then 8 mod 16, etc.
I proof 2 mod 4, the rest goes analog.
If deg(P)≡2 mod 4, we can find a real solution 2a of P*, because that has odd degree.
That means that a–b and a+b are two solutions of P, so P(x+a) has two solutions b and -b.
Let Q(x)=P(x+a).
Now, Q(x) and Q(-x) have two zeros in common, so we can take the greatest common diviser of Q(x) and Q(-x), which isn't constant, and Q(x) divided by this common diviser.
We get two polynomials whose degree sum up to the degree of Q, which means they either both have odd degree, or one of them has degree 2 mod 4, and the other 0 mod 4.
If one has degree 2 mod 4, and is smaller than the degree of Q, it has a complex root by principle of induction.
The only possible case without (yet) a complex root is when the greatest common diviser of Q(x) and Q(-x) is Q(x), meaning that Q(-x)=Q(x)*c.
This either means Q(-x)=Q(x) or Q(-x)=-Q(x), where the latter case implies that the polynomial Q is of odd degree, so it has a real root.
If Q(-x)=Q(x), we get Q(x)=R(x²), and R has a complex root by induction, so we can take the square root of this complex number.

By the way, it is possible to take the square root of a complex number by only taking square roots of positive real numbers, and including i, as long as there is a method to differentiate positive and negative real numbers.

[cameronspalding]

If you assume a polynomial f(z) of degree at least one has no roots then |f(z)| is bounded below so 1/f(z) is a constant, so f(z) must be a non zero constant which is a contradiction

[tim-cca]

3:56 algebraic object? what he means are complex and real are not algebraic extension of rational?

[cameronspalding]

I’m confused: how can X be a finite normal extension of R containing C if R and C are infinite

[constantijndekker]

Was having a little bit of trouble why, assuming C is NOT algebraically closed, there must exist such a normal extention X of R. I get that you can construct a normal extension over C by taking the splirring field of a polynomial with no roots in C but why must it be normal over R also?