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.
 A: 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.
A: 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


*

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

*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.
