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Let $K/\mathbb{Q}$ be a degree $4$ number field. Is it known how to determine whether the Galois closure of $K$, say $K'$, contains the 4-th roots of unity?

In the cubic case, there is a succinct answer: the Galois closure $E'$ of a cubic field $E/\mathbb{Q}$ contains the third roots of unity if and only if $E$ is a pure cubic field: that is, $E = \mathbb{Q}(\sqrt[3]{n})$ for some cube-free integer $n$.

Such a simple characterization doesn't seem to work for the quartic case, since for example the field $\mathbb{Q}(\sqrt{a}, \sqrt{-1})$ with a square-free integer $a \ne \pm 1$ contains the $4$-th roots of unity but is not of the form $\mathbb{Q}(\sqrt[4]{n})$ for a $4$-free integer $n$.

If this characterization is known classically, can someone give a reference?

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    $\begingroup$ I think you must make a case distinction according to the Galois group. If the Galois group of $K'/\mathbb{Q}$ is $C_4$, $V_4$ or $D_4$, and $i \in K'$, then $i$ must be in $K$ already. (For $D_4$, this is an elementary field theory exercise.) If the group is $A_4$, then there are no index $2$ subgroups, so $i \in K'$ can't occur. If the group is $S_4$, then $i \in K'$ is equivalent to $\mathbb{Q}(i)$ being the unique quadratic subextension of $K'/\mathbb{Q}$, so this is equivalent to the discriminant of $K$ being $-1$ times a square. (I wonder if this last case actually occurs, though.) $\endgroup$
    – R.P.
    Commented Oct 26, 2016 at 21:26
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    $\begingroup$ $x^4 + x^3 + x^2 - 2$ has Galois group $S_4$ and discriminant $-46^2$. [Ask gp: for(a=0,2,for(b=-2,2,for(c=-2,2,for(d=-2,2,P=Pol([1,a,b,c,d]);if(polisirreducible(P),if(issquare(-poldisc(P)),if(polgalois(P)[1]==4!,print(P)))))))) ] $\endgroup$
    – Noam D. Elkies
    Commented Oct 26, 2016 at 21:41
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    $\begingroup$ @René, if $K$ is $Q(\root4\of2)$, then the Galois group of the normal closure $K'$ is the dihedral group, and $i$ is in $K'$ but not in $K$. $\endgroup$
    – Gerry Myerson
    Commented Oct 26, 2016 at 21:49
  • $\begingroup$ Yes, I realized that my claim about $D_4$ was too hasty. I misremembered an algebra exercise from long ago (which is why I thought it was elementary). In the $D_4$ case, we must have a unique quadratic subextension $K/F/\mathbb{Q}$, whereas $K'/\mathbb{Q}$ acquires two more. I don't immediately see how to compute these, although there must be simple formulas in terms of generators of $K$ over $F$ and of $F$ over $\mathbb{Q}$... $\endgroup$
    – R.P.
    Commented Oct 26, 2016 at 21:57

1 Answer 1

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As explained in the comments, the only non-trivial case is where $K/\mathbb{Q}$ has Galois closure $K'/\mathbb{Q}$ with $\operatorname{Gal}(K'/\mathbb{Q})$ isomorphic to $D_4$. I will do this case here since it is too long for a comment.

In that case, since all proper subgroups of $D_4$ are contained in a subgroup of index $2$, we must have a quadratic subextension $F/\mathbb{Q}$, so write $F = \mathbb{Q}(\sqrt{d})$. Also write $K = F(\sqrt{\delta})$, with $\delta = a + b \sqrt{d}$, where $a,b$ are in $\mathbb{Q}$.

Since $D_4$ has three subgroups of index $2$, $K'$ has three quadratic subextensions, of which only one can be contained in $K$, since otherwise we would have had the Klein $4$-group as our Galois group. Now $K'$ is generated over $K$ by an element $\sqrt{\delta'}$ where $\delta' = a - b \sqrt{d}$.

Claim. We have that $\operatorname{Nm}_{F/\mathbb{Q}}(\delta)=\delta\delta'=a^2-db^2$ is not contained in $F^{\times 2} \cap \mathbb{Q}^{\times}$, which by Kummer theory is the subgroup of $\mathbb{Q}^{\times}$ generated by $\mathbb{Q}^{\times 2}$ and $d$.
Proof. This says precisely that $K/\mathbb{Q}$ is not normal: if $\operatorname{Nm}_{F/\mathbb{Q}}(\delta)=\epsilon^2$ with $\epsilon \in F$, then $(\epsilon/\sqrt{\delta})^2 = \delta'$, implying that all conjugates of $\delta$ are in $K$, contradiction.

Now, $K'$ contains $\sqrt{\delta \delta'}$, which is a square root of the rational number $a^2-db^2$, so generates a quadratic subextension $F'$, which differs from $F$ precisely because of the claim.

So $\sqrt{-1} \in K'$ is equivalent to $\overline{-1} \in \left\langle \overline{d}, \overline{a^2-db^2} \right\rangle \subset \mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$.

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  • $\begingroup$ This is a pretty satisfying answer. Do you know of a reference for everything you wrote here (and in the comments)? I would like to use this in a paper and it would be nice if there is a clean reference. Thanks! $\endgroup$
    – Stanley Yao Xiao
    Commented Oct 27, 2016 at 13:23
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    $\begingroup$ We had this (i.e. computing the Galois closure of a "repeated" quadratic extension) as an exercise back when I TA'd Galois Theory. The lecture notes were in Dutch though, so I don't know how useful they would be for you. Maybe you don't even need a reference, it's all just basic Galois theory. $\endgroup$
    – R.P.
    Commented Oct 27, 2016 at 15:25

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