# Is this Galois theory proof of Fundamental Theorem of Algebra correct?

I am studying Galois theory through Lang's Algebra and Dummit-Foote's Abstract Algebra. While studying the Fundamental Theorem of Algebra's proofs from both books I spent a lot of time to understand and in the process tried to simplify or rewrite the proof. I like to do this several times.

For the proof we need two facts or results as follows:

(a) There are no non-trivial finite extensions of $\Bbb R$ of odd degree.

(b) There are no quadratic extensions of $\Bbb C$.

Fundamental Theorem of Algebra : $\Bbb C$ is algebraically closed.

Proof : Since $\Bbb R$ has characteristic $0$, every finite extension is separable. Hence $\Bbb R(i)/ \Bbb R$ is separable (Because $\Bbb R(i)/ \Bbb R$ is finite extension).

$\Bbb R(i)=\Bbb C$ is contained in a finite Galois extension $K$ over $\Bbb R$. (By Corollary 23(Dummit-Foote): Let $E/F$ be any finite separable extension. Then $E$ is contained in an extension K which is Galois over $F$ and is minimal in the sense that in a fixed algebraic closure of $K$ any other Galois extension of $F$ containing $E$ contains $K$. We used $E=\Bbb R(i)$ and $F=\Bbb R$.)

Let $G$ be the Galois group of $K/ \Bbb R$.

Using fact (a), since there are no non-trivial finite extensions of $\Bbb R$ of odd degree, we have $|G|$ is even. Therefore $|G|=2^n m$ where $m$ is an odd number and $n \ge 1$.

Let $H$ be a sylow$-2-$subgroup of $G$ and $F$ be the fixed field of $H$. Hence $|G:H|=m=|F:\Bbb R|$. But again by fact (a), $|G:H|=m=1$ $\Rightarrow G=H$ is a $2-$group.

We know that p-groups have subgroups of all orders and they all are normal subgroups. Also $[K:\Bbb R]=[K: \Bbb R(i)][\Bbb R(i): \Bbb R] \Rightarrow 2^n=[K: \Bbb R(i)](2) \Rightarrow [K: \Bbb R(i)]=2^{n-1}.$

Hence Gal$(K/\Bbb R(i))$ is a $2-$group of order $2^{n-1}$ where $n \ge 1$ where $n \gt 1$ would mean that this group is non-trivial and $n=1$ would mean that it is trivial.

If $n \gt 1$, Since $2-$groups have subgroups of all orders (Being p-groups), there exists an extension of $\Bbb R(i)=\Bbb C$ of order $2$ which is contradiction to fact (b). So we can say that $n=1$ and Gal$(K/ \Bbb R(i))=1.$

Hence $K=\Bbb R(i)=\Bbb C$.

I have ommited proofs of facts (a) and (b) as they are precisely the same as in Dummit and Foote. Also I have mentioned only those things of which I want to be sure whether they are correct or not.

• Thank you very much for your proof. However, I am slightly stumped at one bit. How were you able to show that (G:H)=[F:R], or on the other hand that [F:R]=m. (by the way in my book Serge Lang Undergraduate Algebra F is the fixed field of G. Is this the same in your proof/book?) – Daniele1234 Jul 11 '17 at 10:12
• @DanielePilkington-Scimone This proof formulated by me is incomplete. Because I have proved that $K$ which is a minimal Galois extension containing $\Bbb C$ equals $\Bbb C$ itself. You can go through Ben-blum-smith's answer and our discussion where he has pointed out this flaw. Otherwise other part is correct in the proof. Also $[G:H]=[F:\Bbb R]$ comes from fundamental theorem of Galois theory. I have taken $F$ to be the fixed field of $H$ and $H$ is a sylow-2-subgroup. Hence order of $H$ is $2^n$. Which implies $[G:H]=m$. – Error 404 Jul 17 '17 at 4:36

Although there is a lot of correct material in this proof, I see two flaws. One of them is critical, the other is superficial.

Flaw #1: The critical flaw is that you are not postulating an algebraic extension of $\mathbb{C}$. Thus in some sense the proof never begins. The statement that $\mathbb{C}$ is algebraically closed is the statement that if $K$ is an algebraic extension of $\mathbb{C}$, then $K=\mathbb{C}$. Thus you should begin the proof by supposing that $K$ is any algebraic extension of $\mathbb{C}$. This is not how you defined $K$. You introduced it as a Galois extension of $\mathbb{R}$ containing $\mathbb{C}$, which is guaranteed to exist by Dummit&Foote corollary 23. Thus when you prove things about $K$, you are not proving them about any algebraic extension of $\mathbb{C}$ but only about a specific one you have constructed in the proof.

To drive the point home, you don't even need corollary 23 to construct a finite Galois extension $K$ over $\mathbb{R}$ containing $\mathbb{C}$. $K=\mathbb{C}$ is already such an extension. So you could have replaced the sentence "$\mathbb{C}$ is contained in a finite Galois extension $K$ over $\mathbb{R}$" with the sentence "$K=\mathbb{C}$ is a finite Galois extension of $\mathbb{R}$" and the logical work of the sentence wouldn't really have changed. But then if afterwards you proved that $K=\mathbb{C}$, you wouldn't have proved anything at all.

Flaw #2: It is the case that $p$-groups have subgroups of every order dividing the group order (I assume this is what you mean "all orders"), and that $p$-groups have normal subgroups of each of these orders as well, but it is not true that every subgroup of a $p$-group is normal, since nonabelian $p$-groups do exist. What is true is that for each order dividing the group order, there exist subgroups of that order, at at least one of them is normal.

• Thanks for the critical flaw. I was in an attempt to combine Lang and D&F proofs yet in a way according to my understanding so far as a newbie to Galois theory. I agree with you on the critical flaw. There is this statement in Lang's proof in the very beginning: "Every finite extension of $\Bbb R(i)$ is contained in an extension $K$ which is finite and Galois over $\Bbb R$. We must show that $K=\Bbb R(i)$." I couldn't understand his statement that how can such $K$ exist? So in order used corollary 23 and derailed from the proof as you figured it out. Can you explain me his statement? – Error 404 Sep 29 '16 at 13:45
• Regarding superficial flaw, I want to say that I meant every $p$-group has normal subgroups of orders $p,p^2,...p^{n-1}$ where $p^n$ is order of the $p$-group. I wanted 'normal' subgroups in order to be able to write Gal$(K/ \Bbb C)$. That is to make $K$ a Galois extension over $\Bbb C$. – Error 404 Sep 29 '16 at 14:00
• @VikrantDesai - regarding the superficial flaw, you're right here, it was just a matter of language. Saying "$p$-groups have subgroups of all orders and they are all normal" makes it sound like all the subgroups are normal. You could just have said "$p$-groups have normal subgroups of all possible orders" and that would have been fine. – Ben Blum-Smith Sep 29 '16 at 22:29
• @VikrantDesai - Regarding the statement by Lang. Lang is supposing an arbitrary finite extension $L$ of $\mathbb{C}$ and intends to prove that $L=\mathbb{C}$, and asserting the existence of $K\supset L$ which is finite Galois as an extension of $\mathbb{R}$. You are right to use D&F's corollary 23 for this because that is what Lang is implicitly using. $L$ is a finite extension of the characteristic zero field $\mathbb{R}$, and is therefore separable, so corollary 23 applies to imply the existence of a finite separable extension $K$ containing $L$ which is Galois over $\mathbb{R}$. cont'd... – Ben Blum-Smith Sep 30 '16 at 1:35
• Since $K\supset L\supset \mathbb{C}$, to prove that $L=\mathbb{C}$ it suffices to prove that $K=\mathbb{C}$, so the proof proceeds from there. Does that answer your question? – Ben Blum-Smith Sep 30 '16 at 1:36

Yes, I believe it is correct, although I haven't read your post in detail - there certainly is a correct proof along these lines! I usually state the two assumptions as

1. All polynomials of odd degree over the real numbers have at least one root in ${\mathbb R}$.

2. All positive real numbers have a square root.

They can both be proved the Intermediate Value Theorem, which is a moderately elementary result in calculus. Using 2, you can prove by direct calculation that all complex numbers have a square root.

So this is the most algebraic proof, in that uses the least amount of calcualus/analysis. It is also possible to avoid using Sylow's Theorem, thereby reducing the algenra required.

• Do you know Gauss' second proof of the FTA? (paultaylor.eu/misc/gauss.html in case not!) I have always had the feeling that this proof and that one are somehow very closely related. They use the same facts about $\mathbb{R}$, they are both somehow inductions on the power of 2 involved, and Gauss' proof relies on the fundamental theorem on symmetric polynomials, which can be used as the basic lemma of Galois theory (as in Galois' original development), so maybe the role of GT in this proof is somehow played by the f.t.s.p. in that one. – Ben Blum-Smith Sep 29 '16 at 13:11