Cyclic extension without primitive root of unity

Question

Let $F$ be a field that doesn't contain a primitive fourth root of unity. Let $L = F(\sqrt a)$ for some $a \in F - F^2$ and let $K=L(\sqrt b)$ for some $b \in L - L^2$.

If we have $N_{L/F}(b) \equiv a\bmod F^{*2}$, prove that:

1. $K/F$ is a cyclic extension.

2. $a$ is a sum of two squares in $F$.

Answer 1

We need to make the assumption that $F$ is of characteristic different from two, for otherwise both extensions $L/F$ and $K/L$ are inseparable, and there's no hope of getting a Galois extension.

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It follows right away that $L/F$ is a cyclic Galois extension of degree two. All the elements $y\in L$ can be written in the form $y=f_1+f_2\sqrt a$ with $f_1,f_2\in F$. The non-trivial $F$-automorphism $\sigma$ of $L$ is given by the recipe $\sigma(f_1+f_2\sqrt a)=f_1-f_2\sqrt a.$

Assume that there exists an element $b=f_1+f_2\sqrt a\in L$ such that $$ N_{L/F}(b)=b\sigma(b)=f_1^2-af_2^2\in a\cdot (F^*)^2.\qquad(*) $$ Because $a\neq0$ $(*)$ implies the existence of $f_3\in F^*$ such that $$ f_1^2-af_2^2=af_3^2\implies f_1^2=a(f_2^2+f_3^2). $$ We first observe that we must have $f_2\neq0$ for otherwise $a=(f_1/f_3)^2\in F^2$.

We must also have $f_2^2+f_3^2\neq0$, for otherwise $f_2^2+f_3^2=0\implies (f_3/f_2)^2=-1$ contradicting the assumption that there are no roots of unity of order four in $F$. Therefore we can write $$ a=\frac{f_1^2}{f_2^2+f_3^2}= \left(\frac{f_1f_2}{f_2^2+f_3^2}\right)^2+\left(\frac{f_1f_3}{f_2^2+f_3^2}\right)^2 $$ as a sum of two squares in $F$.

Next we prove that $K/F$ is Galois. Denote $u=\sqrt{b}$. Equation $(*)$ implies that $ \sigma(b)=af_3^2/b. $ Therefore$$v:=f_3\sqrt{a/b}=f_3\sqrt{a}u^{-1}=\sqrt{\sigma(b)}\in K.\qquad(**)$$ The polynomial $$ \begin{aligned} p(x)&=(x-u)(x+u)(x-v)(x+v)\\ &=(x^2-u^2)(x^2-v^2)\\ &=(x^2-b)(x^2-\sigma(b))\\ &=x^4-(b+\sigma(b))x^2+b\sigma(b)\\ &=x^4-2f_1x^2+a \end{aligned} $$ has its coefficients in $F$. Because $f_2\neq0$ we have $\sqrt a\in F(b)\subseteq F(u)$. The equation $(**)$ then implies that also $v\in F(u)$, so $K=F(u)$ is the splitting field of the separable polynomial $p(x)$ ($b\neq\sigma(b)$, so the roots of $p(x)$ are all simple).

Furthermore, because $K=F(u)$ is of degree four, we can conclude that $p(x)$ is irreducible. This implies that there is an $F$-automorphism $\tau$ of $K$ such that $\tau(u)=v$. Therefore $$\tau(b)=\tau(u^2)=\tau(u)^2=v^2=\sigma(b),$$ so the restriction of $\tau$ to the field $L$ coincides with $\sigma$. Consequently $(**)$ implies that $$ \tau(v)=\tau(f_3\sqrt{a}u^{-1})=f_3\sigma(\sqrt{a})\tau(u)^{-1} =-f_3\sqrt{a}v^{-1}=-f_2\sqrt{a}(f_3\sqrt{a}u^{-1})^{-1}=-u. $$ This implies easily that $\tau$ is of order four proving the missing claim.