Galois group of $x^4 + 2$ over $\mathbb{Q}$ 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 $Gal(K/\mathbb Q)\cong\mathbb D_4$.
Let $0&lt;a\in \mathbb Q$ 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 $&lt;\phi_1&gt;=\{\phi_1,id\}$

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

Now $&lt;\phi_1&gt;$ is not normal because $\phi_2\circ\phi_1\circ\phi_2^{-1}\notin&lt;\phi_1&gt;$, hence

$Gal(L/K)\cong D_4$

This list may also help.

Question: What is the Galois group if $a&lt;0?$