Showing $K(\sqrt \alpha)/F$ is Galois if and only if $\sigma(\alpha)/\alpha$ is a unit and a square.

Question

I would like help solving is the following problem:

Assume that $K/F$ is a finite Galois extension and $\text{char} F \neq 2$. Let $G:= \text{Gal}(K/F)$ be its Galois group and let $\alpha \in K^\times$. Show that $K(\sqrt{\alpha})/F$ is a Galois extension if and only if $\frac{\sigma(\alpha)}{\alpha} \in K^{\times 2}$ for all $\sigma \in G$, where $K^{\times 2} := \{x^2 \mid x \in K^\times\}$.

I have part of a solution for the reverse implication, but I am unsure where I use the hypothesis, so I am not confident of its validity.

My argument goes as follows: if $\alpha$ is a perfect square, then $K(\sqrt{\alpha}) = K$ and the solution is trivial. Suppose $\alpha$ is not a perfect square. Then, the minimal polynomial of $\sqrt \alpha$ over $K$ is $x^2 - \alpha$. This means $[K(\sqrt\alpha) : K] = 2$. By the tower law, we have $[K(\sqrt{\alpha}), F] = [K(\sqrt{\alpha}): K] [K : F] = 2 |G|$. Given any $\sigma \in G$, we can extend it to an automorphism of $K(\sqrt \alpha)$ by choosing whether sigma will send $\sqrt \alpha$ to $+\sqrt{\sigma(\alpha)}$ or $-\sqrt{\sigma(\alpha)}$ (SEE EDIT BELOW). As $\text{char} F \neq 2$, this gives 2 choices for every $\sigma \in G$, hence we can have $2 |G|$ automorphisms, constructed in this way. As $|\text{Gal}(K(\sqrt \alpha), F)|$ is bounded above by $[K(\sqrt \alpha): F] = 2|G|$, we have constructed every possible automorphism and $|\text{Gal}(K(\sqrt \alpha), F)| = [K(\sqrt \alpha): F]$, so the extension is Galois.

As far as I can tell, this doesn't use the hypothesis on $\frac{\sigma(\alpha)}{\alpha}$, so I am sceptical.

Help with both directions of the proof would be greatly appreciated.

Edit: Following the comments from Μάρκος Καραμέρης, as $\sigma(\alpha) = \alpha . k^2$, $\sigma(\sqrt(\alpha)) = \pm k \sqrt \alpha$, for some fixed $k \in K^\times$. This gives us our extensions from $\sigma \in \text{Gal}(K/F)$ to some pair $\sigma_+, \sigma_- \in K(\sqrt \alpha)$, where $\sigma_\pm (\sqrt(\alpha)) = \pm k \sqrt \alpha$. This completes the reverse implication.

Answer 1

As I mentioned in the comments, the problem is that in general $\sqrt{\sigma(a)}$ might not be in $K(\sqrt{a})$. Notice that if $E$ is the Galois closure of $|K(\sqrt{a}):F|$ then $\sqrt{\sigma(a)}\in E$ so $|K(\sqrt{a}):F|$ is Galois if and only if $E=K(\sqrt{a}) \iff \sqrt{\sigma(a)}\in K(\sqrt{a}), \forall\sigma\in G$
Suppose $a$ is not a square.
We only need to show that $\sqrt{\sigma(a)}\in K(\sqrt{a}) \iff \sqrt{\sigma(a)}=k\sqrt{a},k\in K$. One direction is immediate: $\sigma(a)=k^2a\implies \sqrt{\sigma(a)}=k\sqrt{a}\in K(\sqrt{a})$.
For the other direction: $\sqrt{\sigma(a)}\in K(\sqrt{a}) \implies \sqrt{\sigma(a)}=k_1+k_2\sqrt{a}$ with $k_{1,2}\in K(a) \implies \sigma(a)={k_1}^2+a{k_2}^2+2k_1k_2\sqrt{a}$. Since $\sqrt{a}\not\in K(a)$ we must have $k_1=0$ or $k_2=0$, the later implies $\sqrt{\sigma(a)}\in K \implies\sqrt{a}\in K$ yielding a contradiction. Thus we must have $k_1=0$ and $\sigma(a)={k_2}^2a$ with $k_2\in K$.

Answer 2

The question actually extends to the following setting : $F$ contains the group $\mu_n$ of $n$-th roots of $1$, char$F$ does not divide $n$, $K/F$ is galois with group $G$; then for $a\in K^\times$, $L:=K(\sqrt [n]a)$ is galois over $F$ iff $s(a)/a \in $ for all $s\in G$.

Proof: For clarity, recall that the (ambiguous) notation $\sqrt [n]a$ just means an arbitrarily chosen $n$-th root of $a$ in a separable closure of $F$. Moreover, the extension $L/K$ depends only on the class $[a]$ of $a$ mod ${K^\times}^n$, so it will be convenient to write $L=K(\sqrt [n]{[a]})$. The advantage of this new notation is that $K(\sqrt [n]{[a]})=K(\sqrt [n]{[b]})$ iff $[a]=[b]$ in ${K^\times}/{K^\times}^n$.

The separability of $L/F$ being ensured by the hypothesis on char$F$, we must only show normality. Any $s\in G$ can be extended to an $F$-homomorphism $\bar s$ ("embedding") of $L$ into a separable closure. The normality of $L/F$ is then equivalent to the stability of $L$ under every $\bar s$. By definition, $(\bar s(\sqrt [n]a))^n=\bar s(a)=s(a)$, so the preliminary remarks above show that the sought for normality is equivalent to $[s(a)]=[a]$, in other words $s(a)/a\in {K^\times}^n$ for all $s\in G$. Note that all along, we only used the multiplicative structure of the fields involved, not their additive structure. This approach also allows (but cohomological tools are required) to describe explicitly the group $\bar G=Gal(L/K)$ starting from $G$. For example, if $n=2$ and $G=C_2 \times C_2$, one can derive criteria for $\bar G$ to be $C_2\times C_2 \times C_2$, or $D_4$, or $H_4$.