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?
 A: You're off to a good start. The splitting field $\Omega$ contains $\sqrt[4]{2}e^{\tfrac a8\pi i}$ for $a\in\{\pm3,\pm5\}$ and hence also
$$\frac{\sqrt[4]{2}e^{\tfrac 58\pi i}}{\sqrt[4]{2}e^{\tfrac 38\pi i}}=e^{\tfrac14\pi i}=\zeta_8,$$
which shows that $\Bbb{Q}(\zeta_8)\subset\Omega$. Over this subfield we already have
$$X^4+2X^2+2=(X^2-(-1+i))(X^2-(-1-i)),$$
where $-1-i=\zeta_8^2(-1+i)$. So if $\alpha\in\Omega$ is a root of $X^2-(-1+i)$ then
$$(\zeta_8\alpha)^2-(-1-i)=\zeta_8^2(\alpha^2-(-1+i))=0,$$
i.e. $\zeta_8\alpha$ is a root of $X^2-(-1-i)$. This shows that $\Omega$ is the splitting field of $X^2-(-1+i)$ over $\Bbb{Q}(\zeta_8)$. It now suffices to check that $-1+i$ is not a square in $\Bbb{Q}(\zeta_8)$ to conclude that $[\Omega:\Bbb{Q}]=8$.
Edit: In response to the comment below; an excellent hands-on proof of the fact that $-1+i$ is not a square in $\Bbb{Q}(\zeta_8)$ has already been given in another answer. So here's a less constructive proof:
Suppose $-1+i$ is a square in $\Bbb{Q}(\zeta_8)$. Then it is a square in its ring of integers $\Bbb{Z}[\zeta_8]$, which is a unique factorization domain, and the factorization
$$-1+i
=-(1-\zeta_8)(1+\zeta_8)
=\zeta_8^2\tfrac{1+\zeta_8}{1-\zeta_8}(1-\zeta_8)^2,$$
shows that the unit $\tfrac{1+\zeta_8}{1-\zeta_8}\in\Bbb{Z}[\zeta_8]$ is then also a square, where $\tfrac{1+\zeta_8}{1-\zeta_8}=\zeta_8(1+\zeta_8+\zeta_8^2)$.
Because $\Bbb{Z}[\zeta_8]$ is a UFD, every unit in $\Bbb{Z}[\zeta_8]$ is a cyclotomic unit, i.e. of the form
$$\zeta_8^a\big(\tfrac{1-\zeta_8^3}{1-\zeta_8}\big)^b
=\zeta_8^a(1+\zeta_8+\zeta_8^2)^b,$$
for unique $a\in\Bbb{Z}/8\Bbb{Z}$ and $b\in\Bbb{Z}$. This shows that $\zeta_8(1+\zeta_8+\zeta_8^2)$ is not a square.
A: The OP asked in a comment

is there some easy way to show that $−1+i$ is not a square in $\mathbb Q(\zeta_8)$?

Let $a,b,c,d \in \mathbb Q$ and $\zeta_8 = e^{\tfrac14\pi i}$.
Consider the number
$$\tag 1 \nu = a + b e^{\tfrac14\pi i}+ c e^{\tfrac12\pi i}  + d e^{\tfrac34\pi i}$$
Then
$\quad \nu^2 = a^2 + c^2 e^{i \pi} + 2 b d e^{i \pi} +$
$\quad\quad\quad\quad 2 a b e^{(i/4) \pi} + 2 c d e^{(5 i/4) \pi} +$
$\quad\quad\quad\quad b^2 e^{(i/2) \pi} + 2 a c e^{(i/2) \pi}  + d^2 e^{(3 i/2) \pi}+$
$\quad\quad\quad\quad 2 ad e^{(3 i/4) \pi} + 2 bc e^{(3 i/4) \pi} $
$\quad\quad\,=$
$\quad\quad\quad\quad a^2 - c^2- 2 b d +$
$\quad\quad\quad\quad (2 a b - 2 c d) e^{(1 i/4) \pi} +$
$\quad\quad\quad\quad (b^2  + 2 a c - d^2) e^{(1 i/2) \pi}+$
$\quad\quad\quad\quad (2 ad  + 2bc) e^{(3 i/4) \pi} $
Assume that $\nu^2 = i -1$. Then $a b - c d = 0$ and $ad  + bc = 0$. Multiplying the first equation by $c$ and the second equation by $a$ and then subtracting the first from the second and simplifying we get
$\quad (a^2 + c^2)d = 0$
Case 1: If both $a$ and $c$ are zero then $\nu^2 = -2bd + (b^2 -d^2)i$. So $b \ne 0$ and $d = \frac{1}{2b}$ and  
$\quad 4b^4 -4b^2 - 1 = 0$
But that would imply that $\sqrt 2$ is a rational number.
Case 2: If $d = 0$ then  $\nu^2 =
a^2  - c^2 + 2 abe^{πi/4}  + (b^2 +2ac)i + 2bc  e^{3πi/4}$. If we assume that $b \ne 0$ we must have both $a = 0$ and $c = 0$, which is impossible. But then $a^2  - c^2 = -1$ and $2ac = 1$. But then $a =\frac{1}{2c}$ and we wind up with, 
$\quad 4c^4 -4c^2 - 1 = 0$
But that would imply that $\sqrt 2$ is a rational number.
It is interesting to note that $f(x) = 4x^4 -4x^2 - 1 $ has the same splitting field as the initial polynomial the OP was examining, and it has two real roots,
$$ x = \pm\sqrt{\frac{1+\sqrt2}2}$$
A: 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 $\Omega = \mathbb F$
