Galois Theoretic Proof of Fundamental Theorem of Algebra

Question

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.

Answer 1

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

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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

Answer 2

This is just my adaptation of user254665's answer from the comments, since they haven't posted their answer. Full disclosure: my selfish purpose in posting this answer is to hopefully draw more attention to my question, on the off chance that someone might figure out (or already know) a better answer along the way. I'm well aware that I can't assign the bounty to my own answer.

As set up in my question, let $x, y \in F$, so that $x+yi$ is an arbitrary element of $F(i)$ (I'm using $x$ and $y$ instead of $c$ and $d$ to avoid a naming conflict with cosine later), and we seek $a, b \in F$ so that $(a+bi)^2 = x+yi$. Let $r = \sqrt{x^2+y^2} \in F$ be the non-negative square root of the non-negative number $x^2+y^2$, and let $c = x r^{-1}$, $s = yr^{-1}$. Then it is readily verified that $c^2+s^2 = (x^2+y^2)r^{-2} = 1$, so we can re-express $x+yi$ in a sort of "polar form" $r(c+si)$ with $c^2+s^2=1$.

Inspired by how we can take square roots using the polar form in the case $F = \mathbb R$, we let $r' = \sqrt r \ge 0$, and use analogues of the half-angle formulae from trigonometry to suggest $c' := \epsilon_1 \sqrt{\frac{1+c}{2}}$ and $s' := \epsilon_2 \sqrt{\frac{1-c}{2}}$, where we choose $\epsilon_1, \epsilon_2 \in \{\pm 1\}$ appropriately - this just corresponds to a choice of a quadrant, and the exact choice can also be inspired by the case $F=\mathbb R$. Instead of making this choice explicit here, I'll defer it to the end when it will become clear from the algebra how the choice needs to be made.

Though these definitions are inspired by the half-angle formulae, we do need to justify their use using the properties of real closed fields: the well-definedness of $c'$ and $s'$ as elements of $F$ follows from the inequalities $\frac{1-c}{2}, \frac{1+c}{2} \ge 0$, which are equivalent to $-1 \le c \le 1$, which in turn follows from $c^2 \le c^2+s^2 = 1$ by cases.

Then we let $a = r'c'$ aqnd $b=r's'$. Squaring $a+bi$ gives $a^2-b^2+2abi$ $=$ $r'^2((c'^2-s'^2)+2c's'i)$ $=$ $r\left(\frac{1+c}{2}-\frac{1-c}{2} +2\epsilon_1\epsilon_2 \sqrt{\frac{1-c^2}{4}}i\right)$ $=$ $r(c+\epsilon_1\epsilon_2\sqrt{1-c^2}i)$. The fact that an appropriate choice of $\epsilon_1$ and $\epsilon_2$ exists then simply follows from the fact that $s = \pm \sqrt{1-c^2}$.