The Galois group of $x^{4}-a$ over $\mathbb{Q}$ where $a$ is any integer $\neq 0$, $\neq \pm 1$ and is square free.

Question

This is the exercise $(g)$ of Q1 in Lang's Algebra Chapter VI

Find the Galois group of $x^{4}-a$ over $\mathbb{Q}$ where $a$ is any integer $\neq 0$, $\neq \pm 1$ and is square free.

Before this part, I have found the Galois group of $x^{4}-5$, $x^{3}-10$ and $x^{3}-x-1$ over $\mathbb{Q}$. I noticed that $5$ and $10$ are square free. However, I did not use anything related to square free during the previous proof.

I don't quite understand what is the point for $a$ is square free and $\neq 0$, $\neq \pm 1$.

Any hints are highly appreciated!!

EDIT 1:

As DonAntonio suggested, for $a>0$, the Galois group is $D_{8}$ or by different notation $D_{4}$. However, I made a mistake for $a<0$, but I don't know where I got wrong, so I posted my proof here.

For $a<0$, $-a>0$

Then, $\pm \sqrt[4]{a}=\pm \sqrt[4]{-1}\sqrt[4]{-a}=\pm (\frac{\sqrt{2}}{2}\pm \frac{\sqrt{2}\sqrt[4]{-a}}{2}i)$, and $\pm i\sqrt[4]{a}=\pm i\sqrt[4]{-1}\sqrt[4]{-a}=\pm (\frac{\sqrt{2}}{2}\pm \frac{\sqrt{2}\sqrt[4]{-a}}{2}i)$.

Thus, let $-a=b>0$, and $f(x)$ has roots $\pm (\frac{\sqrt{2}}{2}\pm \frac{\sqrt{2}\sqrt[4]{-a}}{2}i)$.

Thus, the splitting field is $\mathbb{Q}(\sqrt{2},i\sqrt[4]{b})=K$

$[\mathbb{Q}(\sqrt{2},i\sqrt[4]{b}):\mathbb{Q}(i\sqrt[4]{b})]=2$ or $1$, but it cannot be $1$ otherwise $\sqrt{2} \in \mathbb{Q}(i\sqrt[4]{b})$ but $\sqrt{2} \in \mathbb{R}$.

Thus $[\mathbb{Q}(\sqrt{2},i\sqrt[4]{b}):\mathbb{Q}(i\sqrt[4]{b})]=2$

Moreover, $[\mathbb{Q}(i\sqrt[4]{b}):Q]=4$ since $x^{4}-b=Irr(i\sqrt[4]{b},\mathbb{Q},x)$.

Thus $[\mathbb{Q}(\sqrt{2},i\sqrt[4]{b}):\mathbb{Q}]=8$

As $K$ over $\mathbb{Q}(i\sqrt[4]{b})$ is Galois of degree 2, then there exists an automorphism $\tau$ of $K$ fixing $\mathbb{Q}(i\sqrt[4]{b})$, but sending $\sqrt{2}$ to $-\sqrt{2}$.

Moreover, $x^{4}-b$ is irreducible over $\mathbb{Q}(\sqrt{2})$, also K is normal over $\mathbb{Q}(\sqrt{2})$, and thus there exists an automorphism $\sigma$ of $K$ over $\mathbb{Q}(\sqrt{2})$ mapping the root $i\sqrt[4]{b}$ to the root $-\sqrt[4]{b}$, thus id, $\sigma$, $\sigma^{2}$, $\sigma^{3}$ are distinct and $\sigma^{4}=id$, and we have the cyclic subgroup $<\sigma>$.

Since $\tau$ is not in $<\sigma>$, we have $G=<\sigma,\tau>$ as $<\sigma>$ has index 2.

Further, $\tau \sigma (\sqrt{2})=\tau (\sqrt{2})=-\sqrt{2}$, $\tau \sigma (i\sqrt[4]{b})=\tau (-\sqrt[4]{b})=-\sqrt[4]{b}$.

Meanwhile, $\sigma \tau (\sqrt{2})=\sigma (-\sqrt{2})=-\sqrt{2}$, and $\sigma \tau (i\sqrt[4]{b})=\sigma (i\sqrt[4]{b})=-\sqrt[4]{b}$

Therefore, we have $\sigma \tau=\tau \sigma$.

As $G$ is clearly not cyclic but is abelian, we have $G\cong \mathbb{Z_{4}}\times \mathbb{Z_{2}}$

Please feel free to point out where I got wrong. Thank you so much!

EDIT 2:

Okay I know where I got wrong, in the second line of my calculation, I made a huge mistake. so the root should be $\pm (\frac{\sqrt{2}\sqrt[4]{-a}}{2}\pm \frac{\sqrt{2}\sqrt[4]{-a}}{2}i)$, so the splitting field is $\mathbb{Q}(i,\sqrt{2}\sqrt[4]{-a})$. Notice that $\sqrt{2}\sqrt[4]{-a}=\sqrt[4]{2}\sqrt[4]{2}\sqrt[4]{-a}=\sqrt[4]{-4a}$. Then let $b=-4a$ and we have the splitting field is $\mathbb{Q}(i,\sqrt[4]{b})$, and we go back to the first case.

EDIT 3:

As DonAntonio pointed out during our private chat, we could also argue that by Sylow theorem --- any sylow subgroups of a certain prime are isomorphic, thus $D_{8}$ is a Sylow-2 subgroup of $S_{4}$, thus for $a<0$, the Galois group is still of order 8 which is a Sylow-2 subgroup and thus must be isomorphic to $D_{8}$

Answer 1

The conditions are there to conclude that the roots are $\;\pm \sqrt[4]a\;,\;\;\pm i\sqrt[4]a\;$ . There are two real roots, the first two in case $\;a>0\;$ , and two purely imaginary ones, otherwise all of the roots are imaginary.

For simplicity, assume $\;a>0\;$ , then $\;K:=\Bbb Q(\sqrt[4]a,\,i)\;$ is the splitting field of $\;x^4-a\;$ over $\;\Bbb Q\;$ . Any $\;\Bbb Q\,-$ automorphism of $\;K\;$ will be completely and uniquely determined by its action on $\,\sqrt[4]a,\,i\,$. Now take

$$\tau:\begin{cases}\sqrt[4]a\mapsto\sqrt[4]a\\ i\mapsto -i\end{cases}\;\;\;,\;\;\;\;\;\;\sigma: \begin{cases}\sqrt[4]a\mapsto i\sqrt[4]a\\i\mapsto i\end{cases}$$

Check the above extended to $\;K\;$ are automorphisms. By now you must already know what is $\;[K:\Bbb Q]\;$, and now the really interesting part: what are the orders of $\;\tau,\,\sigma\;$ ? And how do they relate to each other? So what is $\;\text{Gal}\left(K/\Bbb Q\right)\;$?