Find the degree of the splitting field of $x^4 + 1$ over $\mathbb{Q}$

Find the degree of the splitting field of $x^4 + 1$ over $\mathbb{Q}$

I think I first need to find the $4$ roots of this polynomial and then calculate $\mathbb{Q}(\mbox{root }1, \mbox{root }2, \mbox{root }3, \mbox{root }4)$, right?

I know that this polynomail has roots only in the complex field, so I need to find them:

$$x^4 + 1 = (x^2-i)(x^2+i) = (x-\sqrt{i})(x+\sqrt{i})(x-\sqrt{-i})(x+\sqrt{-i})$$

so I need to calculate

$$\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i})$$

What do I need to do in order to calculate the degree of these? I thought about doing:

$$[\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i}):\mathbb{Q}(\sqrt{i}, -\sqrt{i})][\mathbb{Q}(\sqrt{i}, -\sqrt{i}):\mathbb{Q}]$$

is this right?

Then how to calculate $[\mathbb{Q}(\sqrt{i}, -\sqrt{i}):\mathbb{Q}]$? Because it’d be the field $\mathbb{Q}$ with $\pm\sqrt{i}$, but it must contain also its multiplicative inverse $\frac{1}{\sqrt{i}}$. I’ve discovered that this field must contain at least the elements $a, b\sqrt{i}, c\frac{1}{i}$ for $a,b\in\mathbb{Q},c\in\mathbb{Q}$. But how do I know that $\frac{1}{i}$ can’t be formed with $a+b\sqrt{i}$ for example? If I find all the possible elements in the field $[\mathbb{Q}(\sqrt{i}, -\sqrt{i})]$, I can find a basis for it and then take its degree over $\mathbb{Q}$

Then, for the degree $[\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i}):\mathbb{Q}(\sqrt{i}, -\sqrt{i})]$ I should verify if $\sqrt{i}$ and $\sqrt{-i}$ are independent. If we take $w = \sqrt{i}$ then $w^2 = i$ and $w^2$ is still in $\mathbb{Q}$, so $-w^2 = -i$. Is there an element in $\mathbb{Q}(\sqrt{i}, -\sqrt{i})$ such that its square is $-i$?

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$x^4+1=\Phi_8(x)$, i.e. the roots of $x^4+1$ are the primitive eighth roots of unity and $x^4+1$ is the minimal polynomial of $\alpha=\exp\left(\frac{2\pi i}{8}\right)=\frac{1+i}{\sqrt{2}}$. The field extension $\mathbb{Q}\mapsto\mathbb{Q}(\sqrt{2})$ trivially has degree two and the same holds for the field extension $\mathbb{Q}(\sqrt{2})\mapsto\mathbb{Q}(\sqrt{2},1+i)$, since it is a complex extension and the minimal polynomial of $1+i$ over $\mathbb{Q}$ is given by $x^2-2x+2$.
$x^4+1$ has no rational root and neither $(x-\alpha)(x-\alpha^3)$, nor $(x-\alpha)(x-\alpha^5)$ nor $(x-\alpha)(x-\alpha^7)$ belong to $\mathbb{Q}[x]$, hence $x^4+1$ is irreducible over $\mathbb{Q}$.
Long story short, the degree of the splitting field of $\Phi_8(x)$ over $\mathbb{Q}$ is $\varphi(8)=4$.


$\zeta=\mathrm e^{\tfrac{i\pi}4}$ is one root of $x^4+1$. What are the other roots?

Let $\alpha$ be a root of $x^4+1$. Then $\alpha^3,\,\alpha^5,\,\alpha^7$ are also (distinct!) roots of $x^4+1$. An easy way to see this is to consider 8th roots of unity, i.e. roots of $x^8-1 = (x^4+1)(x^4-1)$. If $\alpha$ is a primitive 8th root of unity, then every odd power $\alpha^{2k+1}$ is a root of $x^4+1$ (just draw a picture). Or you could just check it directly.

This means that $\mathbb Q[\alpha]$ is the splitting field of $x^4+1$. All you have to do now is to prove that $x^4+1$ is irreducible over $\mathbb Q$ to conclude that the degree of the splitting field is $4$.

EDIT: Perhaps a better way to show that $[\mathbb Q[\alpha]:\mathbb Q] = 4$ is to first notice that since $\alpha$ is a root of $x^4+1$ that $[\mathbb Q[\alpha]:\mathbb Q] \leq 4$. Now notice that $\alpha + \alpha^7 = \sqrt 2$, so $\mathbb Q[\sqrt 2]\subseteq \mathbb Q[\alpha]$. But, $\mathbb Q[\sqrt 2]\subseteq\mathbb R$, while $\alpha$ is complex, so $\mathbb Q[\sqrt 2]\neq \mathbb Q[\alpha]$, so it must be that $[\mathbb Q[\alpha]:\mathbb Q] = 4$.

In the bellow graph are shown 8th roots of unity. Red are roots of $x^4+1$ and blue are roots of $x^4-1$.