TITLE: Galois Theoretic Proof of Fundamental Theorem of Algebra
QUESTION [11 upvotes]: The Galois Theory proof involves proving that $F(i)$ is algebraically closed for any real closed field $F$ (where $i$ is a square root of $-1$), where we may define a real closed field as an ordered field in which every odd degree polynomial has a root and every positive number has a square root.
The last step in the proof seems a bit ad-hoc though; it involves showing that $F(i)$ has no quadratic extension, which in turn involves showing that every element of $F(i)$ has a square root, which involves solving $(a+bi)^2=c+di$ for $a, b \in F$.
Is there a more Galois-theoretic approach to the last step that I'm not aware of? Or, for that matter, some other deeper reason why the computation works out?
Edit Here's the proof that I have that $F(i)$ is closed under square roots, to illustrate how it's ad-hoc. I'd prefer a solution that doesn't involve explicitly solving for a square root - is there a more abstract reason why $F(i)$ should be closed under taking square roots?
We can explicitly solve $(a+bi)^2=c+di$ for $a$ and $b$ in terms of $c$ and $d$. We have $(a+bi)^2 = a^2-b^2+2abi$, yielding the system of equations $$a^2-b^2=c$$ and $$2ab=d.$$ Since every element of $F$ already has a square root in $F(i)$, we may assume $d \neq 0$, and substitute $b = d/(2a)$ into the first equation, which gives $a^2-d^2/(4a^2) = c$, or $4a^4-4ca^2-d^2=0$. Observing that the discriminant of this quadratic is positive, we can then see that $$a^2 := \frac{4c + \sqrt{16c^2+16d^2}}{8} \in F$$ is positive, because $16c^2+16d^2>(4c)^2$ (and by $\sqrt{\bullet}$ I mean the positive square root), and therefore has two square roots in $F$. Letting $a$ be such a square root, we set $b = d/2a$ and call it a day.

REPLY [4 votes]: Recall :
Lemma 1 : $F$ has only one quadratic extension.
proof: Since $F$ is an ordered field with square roots, $F^\times/(F^\times)^2$ has only one non-trivial element. QED

Let $\Omega$ be an algebraic closed field containing $F(i)$ (in the following everything will be in $\Omega$).
Assume that $F(i)$ has a quadratic extension $K=F(i,x)$ with $x \in \Omega$ such that $x^2 = a+ib \in F(i)$. The $F$-Galois conjugates of $x$ are among $x$, $-x$ and the squares roots of $a-ib$. Denote $L=F(i,x,y)$ with $y \in \Omega$ such that $y^2=a-ib$.
The degree of $L/K$ is either $1$ or $2$. We have the field extensions :
$$ F \subset F(i) \subset K \subset L.$$
1st Case : $L/K$ is of degree 2. Then $L/F$ is of degree 8 and $\mathrm{Gal}(L/F(i))$ is isomorphic to $V_4$ (because $L/F(i)$ is generated by $(x,y)$ and the Galois group is generated $(x,y) \mapsto (-x,y)$ and $(x,y) \mapsto (x,-y)$).
So $\mathrm{Gal}(L/F(i))$ is a group of order 8 such that $V_4$ is a normal subgroup and does not have two quotients isomorphic to $C_2$ (by Lemma 1).

The groups $(C_2)^3$, $C_2 \times C_4$ and $H_4$ have at least two quotients isomorphic to $C_2$.
$C_8$ does not contain $V_4$.
$D_8$ does not contain $V_4$ as a normal subgroup.

So $\mathrm{Gal}(L/F(i))$ does not exist.
2nd Case : $L=K$. Then $L/F$ is Galois of order $4$. If $\mathrm{Gal}(L/F) = C_2 \times C_2$ then it contradicts Lemma 1. If $\mathrm{Gal}(L/F) = C_4$, then it contradicts the following Lemma 2 (since $-1$ is not the sum of two squares in $F$). So $\mathrm{Gal}(L/F)$ does not exist.
Lemma 2 : If $F$ is any field and $d \in F$ is a non-square element, then the following are equivalent:
(i) $F(\sqrt{d})$ is included in a Galois extension of $F$ cyclic of degree 4.
(ii) $d$ is the sum of two squares in $F$.
Proof : See 'Serre, Topic in Galois theory, Thm 1.2.4'. I prove (i) $\Rightarrow$ (ii), which is needed.
Let $L/F$ cyclic of degree $4$ containing $F(\sqrt{d})$ and $\sigma$ a generator of $\mathrm{Gal}(L/F)$. Let $\alpha \in L$, such that $\alpha^2 \in F(\sqrt{d})$ and $L=F(\sqrt{d},\alpha)$. Write $\alpha^2 = a+b\sqrt{d} \in F(\sqrt{d})$ and let $\beta = \sigma(\alpha)$.
So we have the field extensions:
$$F \subset F(\sqrt{d}) \subset L$$
Claim 1 : $\alpha\beta \in F(\sqrt{d})$. Since $L/K$ is cyclic, the automorphism $\sigma^2$ is the non trivial element of $\mathrm{Gal}(L/F(\sqrt{d}))=$ $\mathrm{Gal}\Big(F(\sqrt{d})(\alpha)/F(\sqrt{d})\Big)$, so $\sigma^2(\alpha) = -\alpha$ and $\sigma^2(\beta) = \sigma^3(\alpha)=-\beta$. So $\alpha\beta$ is fixed by $\mathrm{Gal}(L/F(\sqrt{d}))$.
Claim 2 : $(\alpha\beta)^2 = a^2 - b^2d$. Since $\sigma_{|F(\sqrt{d})}$ is a generator of $\mathrm{Gal}(F(\sqrt{d})/F)$, one has $\sigma(\sqrt{d})=-\sqrt{d}$ so $\beta^2 = \sigma(\alpha^2) = \sigma(a+b\sqrt{d}) = a-b\sqrt{d}$
Write $\alpha\beta = u + v\sqrt{d} \in F(\sqrt{d})$. Since $(u + b\sqrt{d})^2 =a^2 - b^2d$ is in $F$ and $u + b\sqrt{d} = \alpha \beta$ is not in $F$ (because it is not fixed by $\sigma$), we conclude that $u=0$.
So $(\alpha\beta)^2 = (v\sqrt{d})^2 = a^2-b^2d$ and $(v^2+b^2)d = a^2$. So
$d$ is the sum of two squares because the quotient of sums of two squares is also a sum of two squares. More precisely, one has:
$$d = \frac{a^2}{v^2+b^2} = \frac{a^2(v^2+b^2)}{(v^2+b^2)^2}=  \left( \frac{av}{v^2+b^2} \right)^2 + \left( \frac{ab}{v^2+b^2} \right)^2.$$
QED