Galois group of $x^4 + 2$ over $\mathbb{Q}$ [closed]

I have a problem finding Galois group of $x^4 + 2$ over $\mathbb{Q}$. Not sure whether to start. It's irreducible over $\mathbb{Q}$ and also have 4 different roots.

Note that $\varepsilon_8^i\sqrt[4]2$, for $i=1,3,5,7$, are roots of this polynomial, where $\varepsilon_8=\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}$ is the eighth primitive root of unity. Hence splitting field is $K=\mathbb Q(\varepsilon_8,\sqrt[4]2)=\mathbb Q(\sqrt[4]2,i)$. You can easily see that $|K:\mathbb Q|=8$, hence $Gal(K/\mathbb Q)$ has $8$ automorphisms. Automorphism $f$ is determined by $f(i)$ and $f(\sqrt[4]2)$, and $f(i)\in\{i,-i\}$ (roots of minimal polynomial of $i$ over $\mathbb Q$) and $f(\sqrt[4]2)\in \{\sqrt[4]2,-\sqrt[4]2,i\sqrt[4]2,-i\sqrt[4]2\}$ (roots of minimal polynomial of $\sqrt[4]2$ over $\mathbb Q$; it is $X^4-2$). So, every combination gives an automorphism.

Now you can calculate that $Gal(K/\mathbb Q)\cong\mathbb D_4$.

• Could you briefly note what calculations one should do to determine this Galois group? Apr 26 '18 at 8:49
• @GuusPalmer You can just write the Cayley table for $Gal(K/\mathbb Q)$ and compare it with Cayley table for $\mathbb D_4$.
– SMM
Apr 27 '18 at 10:55
• The only problem is that we can't use such a table at an exam. In the lectures we discussed something about the group behaving transitively. Also, when you look at compositions of automorphisms of the extension and they can be interchanged (composition switch of the two automorphisms) then you have a certain Galois group like $C_n$ or $S_n$ and otherwise $D_n$. Is this somewhat the idea or am I thinking stupid now? Apr 27 '18 at 11:26
• @GuusPalmer Galois group of a polynomial always acts on its roots, and every automorphism is determined by the values on the roots, so Galois group is a subgroup of $\mathbb S_n$, where $n$ is the (separable) degree of the polynomial. If the polynomial is irreducible, the above action is transitive. Now, above you know that $Gal(K/\mathbb Q)$ is a transitive subgroup of order 8 of $\mathbb S_4$, so you can deduce that it is $\mathbb D_4$. You can also note that it is a Sylow 2-subgroup of $\mathbb S_4$, so it must be $\mathbb D_4$. But it is really not a universal recipe.
– SMM
Apr 27 '18 at 11:57

Let $0<a\in \mathbb Q$ be an element which it not a square in $\mathbb Q$. The roots are $\pm\sqrt[4]{a}$ and $\pm i\sqrt[4]{a}$.

Now we determine $[L:\mathbb Q]$ where $L=\mathbb Q(i,\sqrt[4]{a})$ is the splitting field of $f$

We have $[L:\mathbb Q]=[L:\mathbb Q(\sqrt[4]{a})]\cdot [\mathbb Q(\sqrt[4]{a}):\mathbb Q]$

The first factor is two because the degree of the minimal polynomial of $i$ is bounded by 2. ($i$ is a root of $X^2+1$). However it can't be one because $\mathbb Q(\sqrt[4]{a})\subset \mathbb R$ and $L$ contains complex numbers, hence $[L:\mathbb Q]=8$.

The second one is $4$ because $a$ is not a square in $\mathbb Q$

There are 5 possibilities for $Gal(L/K)$. $3$ are abelian and $2$ not. But $1$ of the $2$ non-abelian groups has only normal subgroups (Q_8) in contrast to the other one (D_4).

So write down some elements of the Galois group and see if they generate a normal subgroup or not.

Note that the elements of the Galois group are determined by the images on the generators, for example:

Let $\phi_1:i\mapsto -i$, $\sqrt[4]{a}\mapsto \sqrt[4]{a}$. Then $<\phi_1>=\{\phi_1,id\}$

and $\phi_2:i\mapsto -i$, $\sqrt[4]{a}\mapsto i\sqrt[4]{a}$.

Now $<\phi_1>$ is not normal because $\phi_2\circ\phi_1\circ\phi_2^{-1}\notin<\phi_1>$, hence

$Gal(L/K)\cong D_4$

This list may also help.

Question: What is the Galois group if $a<0?$

• I think you've done $X^4-2$ by accident here. The answer of @SMM is correct. Jun 8 '16 at 8:31
• Like Nick said, this answeres a different question Jul 28 '16 at 20:09