I have one question regarding an exercise that I have done:
The exercise is the following:
For the following polynomial from $\mathbb{Q}[X]$, determine a splitting field in $\mathbb{C}$ and the degree of this splitting field over $\mathbb{Q}$: The Polynomial is $X^4-2X^2-2$
We have $X^4-2X^2-2=(X+i\sqrt{\sqrt{3}-1})(X-i\sqrt{\sqrt{3}-1})(X+i\sqrt{\sqrt{3}+1})(X-i\sqrt{\sqrt{3}+1})$ so we have that $\mathbb{Q}(i\sqrt{\sqrt{3}-1},\sqrt{\sqrt{3}+1})=:L$ is a splitting field.
So now comes my question, with the Eisenstein Criterium we can see that the polynomial is irreducible, so in my mind, since $L$ is the splitting field of a fourth degree irreducible polynomial, the degree must be $4$.
But this is not the case since: $(i\sqrt{\sqrt{3}-1})^2 \in \mathbb{Q}(\sqrt{\sqrt{3}+1})$ and $i\sqrt{\sqrt{3}-1} \notin \mathbb{Q}(\sqrt{\sqrt{3}+1})$ we have $[\mathbb{Q}(i\sqrt{\sqrt{3}-1},\sqrt{\sqrt{3}+1}):\mathbb{Q}(\sqrt{\sqrt{3}+1})]=2$ and therefore $[\mathbb{Q}(i\sqrt{\sqrt{3}-1},\sqrt{\sqrt{3}+1}):\mathbb{Q}]=2 \dot 4=8$
Since the polynomial is of degree $4$, is irreducible, and contains both $i\sqrt{\sqrt{3}-1}$ and $i\sqrt{\sqrt{3}+1}$, couldn't we say that the minimalpolynomial of the extension $L$ should be of degree $4$ or lower? How then does $L$ have degree $8$ over $Q$?