Galois extension with Galois group $A_4$ Suppose $\operatorname{char} F \neq 2$ and $K/F$ is a degree three Galois extension with $\operatorname{Gal}(K/F)\cong \mathbb{Z}/(3)$. Is there a bijection between extensions $N/F$ with Galois group $A_4$ and the order four subgroups $T/(K^*)^2$ of $K^*/(K^*)^2$ stable under the action of $\operatorname{Gal}(K/F)$? 
If not, what extra assumptions are needed for such a correspondence to exist and how it will be defined?
 A: Let $G = A_4$, $H$ be the normal subgroup of order $4$ engendered with double transposition, and $H_1,H_2,H_3$ the three subgroups of $H$ of order $2$.
By the fundamental theorem of Galois theory, if $F \subset N$ is a Galois extension with galois group $G$, we have an intermediate normal extension $F \subset K \subset N$ with $Gal(N/K) = H$ and $Gal(K/F) = G/H = \Bbb Z/3 \Bbb Z$, and three extensions $K \subset L_i \subset N$ with $Gal(N/L_i) = H_i$.
$H_i$ is normal in $H$ ($H$ is abelian), so $L_i = K(\sqrt{a_i})$ for some $a_i \in F$.
$H$ is not normal in $G$, and in fact $G/H$ acts faithfully on the $H_i$, hence on the $L_i$ and the $a_i$ for some good choice of $a_i$.
Finally, the subgroup generated by the $a_i$ is a group $T$ of order $4$ in $K^*/K^{*2}$ such that $T$ is stable by $G/H$, and $T^{G/H} = 1$. 
If you pick a subgroup $T$ with $T^{G/H} = T$, you will obtain instead an extension $F \subset K \subset L_i \subset N$ where $F \subset L_i$ is Galois, and with $Gal(N/F)$ abelian. For example, if you pick $\Bbb Q \subset \Bbb Q(\cos(2\pi/7))$ and $T = \langle -1,-7\rangle$, you will end up with $\Bbb Q(\zeta_{28})$. 
Finding a subgroup $T$ with $T^{G/H} = 1$ is not obvious.
Say $Gal(K/F) = \{id,\sigma,\sigma^2\}$. 
Then candidates are of the form $\{1,a,\sigma(a),\sigma^2(a)\}$ for some $a$. This is a subgroup if and only if $\sigma^2(a)/a\sigma(a) = b^2$ for some $b \in K$. Then we have $a^2 = 1/b^2\sigma^2(b^2)$, so $a = \pm 1/b\sigma^2(b)$.
Hence the possible $T$ are of the form $\{1, \pm b\sigma(b), \pm \sigma(b)\sigma^2(b), \pm b\sigma^2(b)\}$ for some $b \in K$ and $b\sigma(b)$ not a square, which gives $N = K(\sqrt{\pm b\sigma(b)},\sqrt{\pm b\sigma^2(b)})$
And obviously if you pick $T$ that is not stable by $Gal(K/F)$ (which is the vast majority of subgroups of order $4$ of $K^*/K^{*2}$) you don't get a Galois extension $F \subset N$.
