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Let $K$ be a cyclic cubic field, and suppose $a,b \in \mathcal{O}_K \setminus \{0\}$. The field $K(\sqrt{a}, \sqrt{b})$ is Galois over $K$ and generically has Galois group isomorphic to $V_4$, but it is not necessarily true that $K(\sqrt{a}, \sqrt{b})$ is Galois over $\mathbb{Q}$. My question is what are necessary and/or sufficient conditions for $K(\sqrt{a}, \sqrt{b})$ to be a Galois quartic extension over $K$ and an $A_4$-extension over $\mathbb{Q}$?

This is motivated by the phenomenon that if $L$ is an $A_4$-quartic field and $K_L$ its cubic resolvent field, then the Galois closure of $L$ is equal to $K_L(\sqrt{a}, \sqrt{b})$ for some $a,b \in \mathcal{O}_{K_L}$. I want to understand essentially the converse of this phenomenon.

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    $\begingroup$ Since your field is a compositum of $K(\sqrt{a})$ and $K(\sqrt{b})$ over $K$, it reduces to the question when $K(\sqrt{a})$ is Galois over $K$. For $K/\mathbb{Q}$ quadratic, this is a relatively elementary exercise (though still fun), but I've never considered the cyclic cubic case. The basic idea behind the answer in the quadratic case is that normality is equivalent to $\sqrt{\sigma(a)}$ being an element of $K(\sqrt{a})$, i.e. to $\sigma(a)$ being a square in $K(\sqrt{a})$, where $\sigma(a)$ is the conjugate of $a$. The rest follows by spelling out what that means. Same idea could work here. $\endgroup$
    – R.P.
    Commented Jul 17, 2022 at 16:57
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    $\begingroup$ @RP_ A priori it could be that $K(\sqrt{a},\sqrt{b})$ is Galois without $K(\sqrt{a})$ and $K(\sqrt{b})$ being Galois. Can we show this is not the case for $K$ a cyclic cubic field? $\endgroup$
    – Wojowu
    Commented Jul 17, 2022 at 17:07
  • $\begingroup$ @Wojowu Ah, of course. I was not considering that possibility, sorry. $\endgroup$
    – R.P.
    Commented Jul 17, 2022 at 17:09
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    $\begingroup$ If the Galois group is $A_4$, then indeed none of the intermediate quadratics are going to be Galois over $\mathbb{Q}$. $\endgroup$
    – Alex B.
    Commented Jul 17, 2022 at 18:21
  • $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. $\endgroup$
    – GH from MO
    Commented Jul 17, 2022 at 19:52

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A necessary and sufficient condition is that $a$, $b$ be linearly independent in $K^{\times}/K^{\times 2}$ and the Galois group of $K$ cyclically permute the classes of $a$, $b$, and $ab$ in $K^{\times}/K^{\times 2}$. As Henri is pointing out, the first of these conditions can be replaced by demanding that ($a$ not be a square in $K$ and) the norm of $a$ to $K$ be a square. This is to prevent the Galois group from being too small.

Necessity is easy to see. To see sufficiency, note that the condition guarantees that $K(\sqrt{a},\sqrt{b})$ is Galois over $\mathbb{Q}$ with the right action of the quotient $C_3$ on the normal subgroup $C_2\times C_2$, and now use the Schur-Zassenhaus theorem to conclude that the group extension is automatically split, so the Galois group is isomorphic to $A_4$.

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  • $\begingroup$ If I read that correctly, one may take $a$ to be anything square-free, and $b$ to be a conjugate of $a$? $\endgroup$
    – Stanley Yao Xiao
    Commented Jul 17, 2022 at 18:47
  • $\begingroup$ Dear Alex, your condition is necessary but not sufficient: you need in addition that the norm of $a$ (or of $b$) from $K$ to $Q$ be a square. $\endgroup$
    – Henri Cohen
    Commented Jul 17, 2022 at 18:58
  • $\begingroup$ Dear Henri, many thanks, corrected. $\endgroup$
    – Alex B.
    Commented Jul 17, 2022 at 20:43
  • $\begingroup$ @StanleyYaoXiao, no, that would not be enough. Generically, the Galois group of the Galois closure of $K(\sqrt{a})$ is a wreath product $C_2 \wr C_3$. The stated condition ensures that the Galois group is a bit smaller than that. $\endgroup$
    – Alex B.
    Commented Jul 17, 2022 at 20:47

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