<s>[INST] Degree of splitting field of $X^4+2X^2+2$ over $\mathbf{Q}$ 
Find the degree of splitting field of $f=X^4+2X^2+2$ over $\mathbf{Q}$.

By Eisenstein, $f$ is irreducible. By setting $Y=X^2$, we can solve for the roots: $Y=-1\pm i \iff X=\sqrt[4]{2}e^{a\pi i/8}$, $a\in\{3,5,11,13\}$. Clearly $f$ splits in $\mathbf{Q}(\sqrt[4]{2},\zeta_{16})$. $\zeta_{16}$ is a zero of $X^8+1$, which is irreducible over $\mathbf{Q}$ by Eisenstein applied to $(X+1)^8+1$. I was not able to go further. 
Could someone help me to proceed?
 [/INST] Let $\Omega$ be the OP's splitting field.
We have a chain of 3 field extensions,
$$\tag 1 \mathbb Q \subset \mathbb Q(\sqrt 2) \subset Q(\sqrt 2)\,(\sqrt{1 + \sqrt 2}) \subset  Q(\sqrt 2)\,(\sqrt{1 + \sqrt 2})\,(i) = \mathbb F$$
It is easy to explain why the first and last extensions have degree two. For the $2^\text{nd}$ extension we have to show that
$\tag 2 x^2 = {1 + \sqrt 2}$
has no solutions over $\mathbb Q(\sqrt 2)$. If there was a solution then there exists $a,b \in \mathbb Q$ such that
$\tag 3 (a + b\sqrt2)^2 = a^2 +2ab\sqrt2 + 2b^2 = a^2+ 2b^2 + 2ab\sqrt2 = {1 + \sqrt 2}$
Then both $a$ and $b$ are nonzero and $a = \frac{1}{2b}$ and since $a^2+ 2b^2 =1$
$\tag 4 8b^4 - 4b^2 + 1 = 0$
But no rational number $b \in \mathbb Q$ satisfies the equation $\text{(4)}$.
We conclude that the field $\mathbb F$ has degree $8$ over the rationals.
We can write out the four roots $x_1, x_2, x_3 x_4 \in \Omega$ of the quartic $x^4+2x^2+2$ using the $a + bi$ rectangular format, 
\begin{align}x_1&=\sqrt{\frac{-1+\sqrt2}{2}}+i\sqrt{\frac{1+\sqrt2}2}\\ x_2&=-\sqrt{\frac{-1+\sqrt2}{2}}-i\sqrt{\frac{1+\sqrt2}2}\\ x_3&=\sqrt{\frac{-1+\sqrt2}{2}}-i\sqrt{\frac{1+\sqrt2}2}\\ x_4&=-\sqrt{\frac{-1+\sqrt2}{2}}+i\sqrt{\frac{1+\sqrt2}2}\end{align}
Since $\sqrt{1 + \sqrt 2} \; \sqrt{\sqrt 2 -1} = 1$, it is now easy to see that all the roots belong to $\mathbb F$, so $\Omega \subset \mathbb F$.
In a chain of logic,
$$  x_1 x_3 = \sqrt 2 = \in \Omega$$
$$ \sqrt 2 \, \frac{x_1 + x_3}{2}  =  \sqrt{-1+\sqrt2} \in \Omega$$
$$  (\sqrt{-1+\sqrt2})^{-1}  =  \sqrt{1+\sqrt2}  \in \Omega$$
$$ \sqrt 2 \, \frac{x_1 + x_4}{2} \sqrt{-1+\sqrt2} = i  \in \Omega$$
Since $\sqrt 2, ,(\sqrt{1 + \sqrt 2}), i \in \Omega$, $\mathbb F \subset \Omega$.
We conclude that $\Omega = \mathbb F$
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