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 GG, if pp is a prime dividing |G||G|, then there is a pp-Sylow subgroup. We are not caring about other pp-Sylow groups here. However, one needs to also recall that a pp-group HH is always solvable. If |H|>1|H|>1, then HH admits nontrivial centre. If |H|=pn|H|=pn, then there is a sequence of subgroups {e}=H0⊂H1⊂⋯⊂Hn=H{e}=H0⊂H1⊂⋯⊂Hn=H where HiHi is normal in HH for all i=0,…,ni=0,…,n and Hi+1/HiHi+1/Hi is cyclic of order pp. This is to say |Hi|=pi|Hi|=pi.

On the other hand, we also make use of analysis (This is Gauss's idea). For every a>0a>0, there is a square root √a>0a>0. In other word, we have a positive root of the equation X2−a=0X2−a=0. On the other hand, every polynomial f(X)∈R[X]f(X)∈R[X] of odd degree has a root in RR. This is to say, such f(X)f(X) is not irreducible over RR unless degf=1deg⁡f=1.

Next we take a look at C=R(i)C=R(i), where, ii is the imaginary unit, or, algebraically speaking, a root of g(X)=X2+1g(X)=X2+1. Note, every z∈Cz∈C has a root. If we write z=a+biz=a+bi, then c=√|z|+a2,d=b|b|√|z|−a2c=|z|+a2,d=b|b||z|−a2 gives rise to (c+di)2=a+bi(c+di)2=a+bi. It follows that all polynomials f(X)∈C[X]f(X)∈C[X] of order 22 has a root (if this is not very obvious, use a change-of-variable), hence not irreducible. With this being said, CC does not have an extension of order 22. Say, if [E:C]=2[E:C]=2, then E=C[X]/(p(X))E=C[X]/(p(X)) and p(X)p(X) is irreducible. But It can only be of order 22, 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/FE/F be a finite separable extension. Then EE is contained in an extension KK such that K/FK/F is Galois. It is minimal in the sense that, in a fixed algebraic closure KaKa of KK, any other Galois extension LL of FF containing EE must contain KK as well. We have the following tower: F⊂E⊂K⊂L⊂Ka.F⊂E⊂K⊂L⊂Ka.

Proof. First of all, we can find a finite Galois extension of FF containing EE. For example, the composite of the splitting fields of the minimal polynomials for a basis for EE as a FF-vector space. The intersection of all Galois extensions is exactly what we want. □◻

The main result

The complex field CC 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 1f(z)1f(z) where ff is a polynomial. If ff has no root, then perhaps it can only be a constant.

Proof. Let's firstly make it a problem of Galois theory. Since R⊃QR⊃Q, it is of characteristic 00 (hence perfect) and every finite extension is separable. Hence, in particular, C/RC/R is finite and separable. Let L/CL/C be a finite extension. Then L/RL/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/RK/R. We need to prove that K=CK=C.

Put G=G(K/R)G=G(K/R). We want to show that |G|=2|G|=2 hence [K:R]=[K:C][C:R]=2[K:R]=[K:C][C:R]=2 and our result follows immediately. To do this, we first show that |G||G| is even. Let H⊂GH⊂G be a 22-Sylow subgroup of GG and we can say |H|=2n|H|=2n, |G|=2nm|G|=2nm and mm is even. Now we use the Galois correspondence. Put F=KHF=KH. We see K/FK/F is Galois and [K:F]=2n[K:F]=2n. It follows that [F:R]=m[F:R]=m. We claim that m=1m=1.

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

Therefore G=HG=H is a 22-group. Since Galois extension remains normal under lifting, we see K/CK/C is Galois. Let G1=G(K/C)⊂GG1=G(K/C)⊂G be the Galois group. We next claim that G1G1 is trivial. If not, then, being a 22-group, it has a subgroup G2G2 of index 22. Put F′=KG2F′=KG2, then we see [KG2:C]=G1/G2≅Z/2Z[KG2:C]=G1/G2≅Z/2Z. However, as mentioned above, CC has no extension of order 22. This contradiction implies that G1G1 is trivial and therefore K=CK=C. □◻

Why we have to prove that K=CK=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.