Find the Galois group of the polynomial Consider the polynomial $ f(X)= X^4 + 9 \in \Bbb Q [X]$.  
Show that $f$ is irreducible and find $Gal (K/ \Bbb Q )$, where $K$ is the splitting field of $f$.  
For the first one I used Eisenstein kriterion for $f(X+1)$ and $p=2$.
$$f(X+1)=X^4+4X^3+6X^2+4X+10$$  
Now $2$ does not divide $1$, but it divides $4,6,4,10$. Moreover $2^2=4$ does not divide $10$. So f is irreducible over $\mathbb Q$.
But I have no idea how to find a Galois group of this.
 A: First of all you should find a decent representation of the splitting field, i.e. $\mathbb Q(i,\sqrt{6})$.
Then you can forget about the polynomial and just consider the extension $\mathbb Q(i,\sqrt{6})/\mathbb Q$. I think the the galois group of the extensions $\mathbb Q(\sqrt{a},\sqrt{b})/\mathbb Q$ (with $a,b$ not chosen degenerated) is well-known to be $V_4$. If it is not known to you, you should explicitly compute this.
A: Here are you first steps: 
Write $-9$ on its trigonometric form, that is, 
$$-9 = 9(\cos \pi + i \sin \pi)$$
Then the roots to your poynomial are
$$\begin{align}\xi_0 &= \sqrt [4]{9}\Big(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Big) = \sqrt{3}\Big(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\Big) = \frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2}\\\xi_1&=-\frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2}\\\xi_2&=-\frac{\sqrt{6}}{2} - i\frac{\sqrt{6}}{2}\\\xi_3 &=\frac{\sqrt{6}}{2} - i\frac{\sqrt{6}}{2} \end{align}$$
Then your splitting field is given by $L = Gal (X^4 + 9, \mathbb Q) = \mathbb Q [\sqrt{6}, i]$. Now  any $\sigma \in Aut_{\mathbb Q} L$ is complete determined by $\sigma (i)$ and $\sigma (\sqrt{6})$, and the possibilities are 
$$\sigma(\sqrt{6}) \in \{\sqrt{6}, - \sqrt{6}\} \ \ \text{and} \ \ \sigma(i) \in \{i, -i\}$$
From this you need to work on the properties of this group or order $4$. Draw the table of automorphism for guidance, and see if it it abelian or not, all in all your possibilities are $\mathbb Z_4$ and $\mathbb Z_2 \times \mathbb Z_2$. 
Hint: Look carefully to the last one.
A: The second roots of $-9$ are $\pm 3i$.
Now you can write $3i=(\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}i)^2$
Which yields two $4th$ roots $\pm\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}i$
Do this similarly for $-3i$ and you obtain:
$\pm\alpha:=\pm\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}i$ and $\beta:=\pm\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}i$.
Hence the splitting field of $f$ over $\mathbb Q$ is $\mathbb Q(\alpha,\beta)$. However, the extension is actually simple because $\alpha\cdot\beta=3$, so we have 
$\mathbb Q(\alpha,\beta)=\mathbb Q(\alpha)$. So $[K:\mathbb Q]=4$.
Each $\mathbb Q$-automorphism is determined by their image of $\alpha$.
All the possible images of $\alpha$ are:
$\tau_1(\alpha)=\alpha$
$\tau_2(\alpha)=-\alpha$
$\tau_3(\alpha)=\beta$
$\tau_4(\alpha)=-\beta$

Note:
If $\tau_3(\alpha)=\beta$, then $\tau_3(\beta)=\tau_3(\frac{3}{\alpha})=3\tau_3(\alpha^{-1})=3\tau_3(\alpha)^{-1}=\frac{3}{\beta}=\alpha$

Because each autmorphism has obviously order 2, we have that $Gal(K/\mathbb Q)\cong \mathbb Z_2\times \mathbb Z_2$
