Find the splitting field of $x^4+4$ over $\mathbb{Q}$
$x^4+4$ has, as roots,
$x = \sqrt{2}e^{i\pi/4}\cdot e^{in\pi/4}$, $n=0,1,2,3$
The splitting field is, then
$$\mathbb{Q}(\sqrt{2}, \pm e^{i\pi/4}, \pm e^{i3\pi/4}) = \mathbb{Q}(\sqrt{2}, e^{i\pi/4}, e^{i3\pi/4})$$
But $e^{i3\pi/4} = (e^{i\pi/4})^3$, so we end up with
$$\mathbb{Q}(\sqrt{2}, e^{i\pi/4})$$
Am I right?
I need to calculate the degree of this splitting field. I'm thinking about taking $[\mathbb{Q}(\sqrt{2}, e^{i\pi/4}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, e^{i\pi/4}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$
$[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$
Now for the other one:
$$x = e^{i\pi/4} = \cos \pi/4 + i\sin \pi/4 = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\implies$$
$$x^8 = e^{i2\pi} = \cos 2\pi + i\sin 2\pi = -1$$
So the polynomial that has $e^{i\pi/4}$ as root is $x^8+1$. I'm certain I can find a second degree polynomial with this root, so $x^8+1$ must be irreducible, but there are so many polynomials to divide. Is there a better way?
Also, am I on the right track?
Best Answer
You are making things very complicated. Note that $$\sqrt 2 e^{\pi i/4}=\sqrt2\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right) =1+i.$$ Similarly the other zeroes of $x^4+4$ are $1-i$, $-1+i$ and $-1-i$ (draw these on the Argand diagram!). It is plain that the splitting field is $\Bbb Q(i)$. Indeed $$x^4+4=(x^2-2x+2)(x^2+2x+2).$$