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

Preparation

The method is presented by Artin: we will be actively using theoriesof 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 aboutother \(p\)-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 ofsubgroups \[\{e\}=H_0 \subset H_1 \subset \cdots \subset H_n=H\] 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'sidea). For every \(a>0\), there is asquare root \(\sqrt{a}>0\). In otherword, we have a positive root of the equation \(X^2-a=0\). On the other hand, everypolynomial \(f(X) \in \mathbb{R}[X]\)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 allpolynomials \(f(X) \in \mathbb{C}[X]\)of order \(2\) has a root (if this isnot very obvious, use a change-of-variable), hence notirreducible. With this being said, \(\mathbb{C}\) does not have an extension oforder \(2\). Say, if \([E:\mathbb{C}]=2\), then \(E=\mathbb{C}[X]/(p(X))\) and \(p(X)\) is irreducible. But It can only beof order \(2\), which is absurdalready.

We also need a part of the following lemma on field extension. Inbrief, finite separable extension induces a minimal Galoisextension.

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 sensethat, 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 extensionof \(F\) containing \(E\). For example, the composite of thesplitting fields of the minimal polynomials for a basis for \(E\) as a \(F\)-vector space. The intersection of allGalois extensions is exactly what we want. \(\square\)

The main result

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

The following proof focuses on algebra and tries its best to avoidanalysis. If you are a fan of analysis, you can dive into complexanalysis 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 onlybe 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 finiteextension is separable. Hence, in particular, \(\mathbb{C}/\mathbb{R}\) is finite andseparable. Let \(L/\mathbb{C}\) be afinite extension. Then \(L/\mathbb{R}\)is still a finite and separable extension, since both the class offinite 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 wantto 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 Galoiscorrespondence. 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 mayapply the primitive element theorem to obtain \(F=\mathbb{R}(\alpha)\). \(\alpha\) is the root of an irreduciblepolynomial in \(\mathbb{R}[X]\) ofdegree \(m\). But \(m\) is odd, we must have \(m=1\).

Therefore \(G=H\) is a \(2\)-group. Since Galois extension remainsnormal under lifting, we see \(K/\mathbb{C}\) is Galois. Let \(G_1=G(K/\mathbb{C}) \subset G\) be theGalois 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, letme remind you that a Galois extension is, by definition, analgebraic extension which is normal and separable.