# What is the Galois group of $x^4 +5$ over the rationals?

Consider the polynomial $x^4+5\in \mathbb{Q}[x]$ and Let $E/\mathbb{Q}$ be its splitting field. I would like to calculate $G = \operatorname{Gal}(E/ \mathbb{Q})$. It should be a standard exercise, but for some reason I get stuck at some point.

Let me say what I can tell. First, $E=\mathbb{Q}[\alpha,i]$ where $\alpha$ is any root of $x^4+5$. Thus, $E$ has the two subfields $\mathbb{Q}[\alpha]$ and $\mathbb{Q}[i]$ of degrees $4$ and $2$ over $\mathbb{Q}$ respectively. This forces the degree of $E$ to be either $4$ or $8$ depending on whether $i\in \mathbb{Q}[\alpha]$ or not (I know how to justify all the above claims), and this is the point where I'm stuck.

It seems to me that $[E:\mathbb{Q}]=8$, but I can't prove it. The four roots of the polynomial are given explicitly by $({\pm 1 \pm i \over \sqrt{2}})5^{1/4}$. The case I am trying to rule out is that either of the roots generates $E$, but I can't find the right argument.

In either case, I know what the final answer should be. Since $G$ acts faithfully on the $4$ roots, $G$ embeds into $S_4$ and thus if $|G|=8$, then it should be isomorphic to $D_4$ (from the Sylow theorems). If on the other hand $|G|=4$, having the two different quadratic subextensions $\mathbb{Q}[i]$ and $\mathbb{Q}[i\sqrt{5}]$ forces $G$ to be isomorphic to $\mathbb{Z}/2\times \mathbb{Z}/2$.

One way is: If $r=(-5)^{1/4}$, so that the splitting field is $L=\mathbb Q(r,i)$, then $Gal(L/\mathbb Q(i))$ is the cyclic group $C_4$, acting via $r\to i^k r$ ($k\in\mathbb Z$ mod 4). To see it: it is certainly a subgroup of this group; if it's a proper subgroup then $r^2$ is fixed, i.e. $r^2\in\mathbb Q(i)$. But that's not possible: $-5=(a+ib)^2$ has evidently no rational solutions. From this we get that $Gal(L/\mathbb Q)$ is $r\mapsto ri^k$, $i\mapsto\pm i$, which is the dihedral group.

[edited to become a 'pure Galois' proof]

• Great! I like this proof very much, so I will accept this answer. even though there were perfectly good answers before it (hope it's ok) – KotelKanim Sep 7 '13 at 14:55

This is a tower of quadratic extensions, and there are general tools to deal with this kind of situation. Some time ago, I explained the general machinery in a more complicated case here.

EDIT : In your example :

The number $2$ is not a square in ${\mathbb Q}[\sqrt{5}]$, because neither $2$ nor $2 \times 5$ are squares in $\mathbb Q$ (by “Extension rule 1” , see the reference).

The number $2$ is a square in ${\mathbb Q}[\sqrt[4]{5}]$ iff the equation $x^4+\frac{\sqrt{5}}{4}=0$ has a solution in ${\mathbb Q}[\sqrt{5}]$ (by Extension rule 2) and this is not the case.

So $\sqrt{2}\not\in {\mathbb Q}[\sqrt[4]{5}]$, and the extension $F={\mathbb Q}[\sqrt[4]{5},\sqrt{2}]$ has degree $8$ over $\mathbb Q$.

Since $F \subseteq {\mathbb R}$, we see that $L=F[i]={\mathbb Q}[\sqrt[4]{5},\sqrt{2},i]$ has degree $16$ over $\mathbb Q$.

The elements of ${\sf Gal}(L/{\mathbb Q})$ are easily described by their action on $\sqrt[4]{5},\sqrt{2},i$. Then, it is not hard to find the subgroup fixing $E$, and to deduce that $[E:\mathbb Q]=8$.

• Ok, thanks. I will look at the reference, but the second "fact" is not obvious to me. I mean, it is "obvious" that it is true, but it's not obvious how to prove it. Could you please elaborate on this? – KotelKanim Sep 7 '13 at 7:29
• Well, actually I think I got it. You can calculate the intersection with the real field and then it is clear. Right? – KotelKanim Sep 7 '13 at 7:31
• @KotelKanim I edited/corrected my answer, it is hopefully clearer now. – Ewan Delanoy Sep 7 '13 at 7:59
• +1 Going via the route of the degree 16 extension is a nice trick. Furthermore, it takes care of my earlier objection :-) – Jyrki Lahtonen Sep 7 '13 at 8:03

This may not be very useful to you, but here goes anyway.

We see that $-5$ is of order four in the field $K=\mathbb{F}_{13}$. Therefore its fourth roots are necessarily of order sixteen in some extension field of $K$. The smallest extension field of $K$ that contains sixteenth roots of unity is the field $\mathbb{F}_{13^4}$. Therefore $x^4+5$ is irreducible in $K[x]$. By a well known result this implies that the action of the Galois group of $x^4+5$ over $\mathbb{Q}$ as a group of permutations on its roots contains a 4-cycle. This rules out the Klein four group, and settles your question.

• Wow. This is nice. It is useful, but I did hope for a proof using only first-course-in-Galois-theory stuff. – KotelKanim Sep 7 '13 at 8:00
• Yeah. I am a bit disappointed at myself for not coming up with a more elementary way. – Jyrki Lahtonen Sep 7 '13 at 8:01

To show that $[E:\mathbb{Q}]=8$ it is enough to show that $i\notin \mathbb{Q}(\alpha)$ for $\alpha=\frac{1+i}{\sqrt{2}}5^{1/4}$ , since then $$[E:\mathbb{Q}]=[\mathbb{Q}(\alpha,i):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=2\cdot 4=8.$$

So write $$i=a+b\frac{1+i}{\sqrt{2}}5^{1/4}=a+\frac{b}{\sqrt{2}}5^{1/4}+i\frac{b}{\sqrt{2}}5^{1/4}$$ Then we must have $\frac{b}{\sqrt{2}}5^{1/4}=1$ so $5^{1/2}=\frac{2}{b^{2}}\in\mathbb{Q}$ which is not possible.