## 30 Nov Fundamental Theorem of Algebra using Galois theory

In this article we shall prove that $\mathbb{C}$ is algebraically closed. Here we consider $\mathbb{C}$ as a splitting field of the polynomial $x^2+1$.

The proof uses very little analysis and most of it is Galois theory. The only facts from analysis which will be used in proof are :

• Positive real numbers have square roots.
• Every polynomial of odd degree with real coefficients has a real root.

Both of these facts are consequences of the intermediate value theorem.

We first show that every element of $\mathbb{C}$ has a square root in $\mathbb{C}$. For $a,b\in\mathbb{R}$ let

$c=\pm\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}$

$d=\pm\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}$

We choose the sign of $c$ and $d$ such that $cd$ has the same sign as b. Then we can see that $(c+\iota d)^2=a+\iota b$.

Now we show that every polynomial $f(x)\in \mathbb{R}[x]$ splits in $\mathbb{C}$. This is equivalent to showing that the splitting field of $f(x)(x^2+1)$ over $\mathbb{R}$ is $\mathbb{C}$. Let $E$ be the splitting field of $f(x)(x^2+1)$ over $\mathbb{R}$ for a fixed $f$. As $\mathbb{R}$ has characteristic zero, $f(x)(x^2+1)$ is a separable polynomial and hence $E/\mathbb{R}$ is a Galois extension. Let $G=Gal(E/\mathbb{R})$.

Let $H$ be a 2-Sylow subgroup of $G$. Let $M=E^H$, the subfield of E fixed by $H$. Then degree of the extension $M/\mathbb{R}$ is $|G/H|$ which is odd. Hence, for any $\alpha\in M$ it’s minimal polynomial over $\mathbb{R}$ has odd degree. But as any odd degree polynomial has a root in $\mathbb{R}$, $\alpha\in\mathbb{R}$. Thus, $M=\mathbb{R}$ and hence $G$ is a 2-group.

If $G\neq (1)$ it has a subgroup $N$ of index $2$. Then degree of $E^N$ over $\mathbb{C}$ is 2 and hence is generated by the square root of an element in $\mathbb{C}$. But as $\mathbb{C}$ has all its square roots we get that $G=(1)$. Hence, $E=\mathbb{R}$

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