# Examples in Galois Theory 1 - Complex Field is Algebraically Closed

I will be writing a series of posts on important examples and applications of Galois theory, most of which are must-know. You may have know many of them already. I will try my best to make sure that the reader can learn something from these posts. In every post of this series, the background of the fundamental theorem of Galois theory (a.k.a. Galois correspondence), as well as its prerequisites, are assumed to be known.

## Preparation

The method is presented by Artin: we will be actively using theories
of the Sylow group theory. Recall that for a finite group \(G\), if \(p\) is a prime dividing \(|G|\), then there is a \(p\)-Sylow subgroup. We are not caring about
*other* \(p\)-Sylow groups here.
However, one needs to also recall that a \(p\)-group \(H\) is always solvable. If \(|H|>1\), then \(H\) admits nontrivial centre. If \(|H|=p^n\), then there is a sequence of
subgroups \[
\{e\}=H_0 \subset H_1 \subset \cdots \subset H_n=H
\] where \(H_{i}\) is normal in
\(H\) for all \(i=0,\dots,n\) and \(H_{i+1}/H_i\) is cyclic of order \(p\). This is to say \(|H_i|=p^i\).

On the other hand, we also make use of analysis (which is Gauss's
idea). For every \(a>0\), there is a
square root \(\sqrt{a}>0\). In other
word, we have a positive root of the equation \(X^2-a=0\). On the other hand, every
polynomial \(f(X) \in \mathbb{R}[X]\)
of odd degree has a root in \(\mathbb{R}\). This is to say, such \(f(X)\) is *not* irreducible over
\(\mathbb{R}\) unless \(\deg f=1\).

Next we take a look at \(\mathbb{C}=\mathbb{R}(i)\), where, \(i\) is the imaginary unit, or,
algebraically speaking, a root of \(g(X)=X^2+1\). Note, every \(z \in \mathbb{C}\) has a root. If we write
\(z=a+bi\), then \[
c=\sqrt{\frac{|z|+a}{2}}, \quad d = \frac{b}{|b|}\sqrt\frac{|z|-a}{2}
\] gives rise to \((c+di)^2=a+bi\). It follows that all
polynomials \(f(X) \in \mathbb{C}[X]\)
of order \(2\) has a root (if this is
not very obvious, use a change-of-variable), hence *not*
irreducible. With this being said, \(\mathbb{C}\) does not have an extension of
order \(2\). Say, if \([E:\mathbb{C}]=2\), then \(E=\mathbb{C}[X]/(p(X))\) and \(p(X)\) is irreducible. But It can only be
of order \(2\), which is absurd
already.

We also need a part of the following lemma on field extension. In
brief, finite separable extension induces a *minimal* Galois
extension.

Lemma.Let \(E/F\) be a finite separable extension. Then \(E\) is contained in an extension \(K\) such that \(K/F\) is Galois. It is minimal in the sense that, in a fixed algebraic closure \(K^\mathrm{a}\) of \(K\), any other Galois extension \(L\) of \(F\) containing \(E\) must contain \(K\) as well. We have the following tower: \[ F \subset E \subset K \subset L \subset K^\mathrm{a}. \]

*Proof.* First of all, we can find a finite Galois extension
of \(F\) containing \(E\). For example, the composite of the
splitting fields of the minimal polynomials for a basis for \(E\) as a \(F\)-vector space. The intersection of all
Galois extensions is exactly what we want. \(\square\)

## The main result

The complex field \(\mathbb{C}\) is algebraically closed.

The following proof focuses on algebra and tries its best to avoid analysis. If you are a fan of analysis, you can dive into complex analysis and use the maximum modulus theorem to study a polynomial. Or, you can study the behaviour of \(\frac{1}{f(z)}\) where \(f\) is a polynomial. If \(f\) has no root, then perhaps it can only be a constant.

*Proof.* Let's firstly make it a problem of Galois theory.
Since \(\mathbb{R} \supset
\mathbb{Q}\), it is of characteristic \(0\) (hence perfect) and every finite
extension is separable. Hence, in particular, \(\mathbb{C}/\mathbb{R}\) is finite and
separable. Let \(L/\mathbb{C}\) be a
finite extension. Then \(L/\mathbb{R}\)
is still a finite and separable extension, since both the class of
finite extensions and the class of separable extensions are distinguished.

Applying the lemma above, we can find a finite and Galois extension \(K/\mathbb{R}\). We need to prove that \(K=\mathbb{C}\).

Put \(G=G(K/\mathbb{R})\). We want to show that \(|G|=2\) hence \([K:\mathbb{R}]=[K:\mathbb{C}][\mathbb{C}:\mathbb{R}]=2\) and our result follows immediately. To do this, we first show that \(|G|\) is even. Let \(H \subset G\) be a \(2\)-Sylow subgroup of \(G\) and we can say \(|H|=2^n\), \(|G|=2^nm\) and \(m\) is even. Now we use the Galois correspondence. Put \(F=K^H\). We see \(K/F\) is Galois and \([K:F]=2^n\). It follows that \([F:\mathbb{R}]=m\). We claim that \(m=1\).

Indeed, applying the lemma again, we see \(F/\mathbb{R}\) is separable. Hence we may apply the primitive element theorem to obtain \(F=\mathbb{R}(\alpha)\). \(\alpha\) is the root of an irreducible polynomial in \(\mathbb{R}[X]\) of degree \(m\). But \(m\) is odd, we must have \(m=1\).

Therefore \(G=H\) is a \(2\)-group. Since Galois extension remains normal under lifting, we see \(K/\mathbb{C}\) is Galois. Let \(G_1=G(K/\mathbb{C}) \subset G\) be the Galois group. We next claim that \(G_1\) is trivial. If not, then, being a \(2\)-group, it has a subgroup \(G_2\) of index \(2\). Put \(F'=K^{G_2}\), then we see \([K^{G_2}:\mathbb{C}]=G_1/G_2 \cong\mathbb{Z}/2\mathbb{Z}\). However, as mentioned above, \(\mathbb{C}\) has no extension of order \(2\). This contradiction implies that \(G_1\) is trivial and therefore \(K=\mathbb{C}\). \(\square\)

Why we have to prove that \(K=\mathbb{C}\)? If you didn't get it, let
me remind you that a Galois extension is, by definition, an
**algebraic** extension which is normal and separable.

Examples in Galois Theory 1 - Complex Field is Algebraically Closed