Is $\sqrt{2}\in\mathbb{Q}(\sqrt[8]{3})$ or not? My hunch is that $\sqrt{2}\not\in\mathbb{Q}(\sqrt[8]{3})$. For practice, I want to compute the splitting field and its degree of $x^8-3$ over $\mathbb{Q}$. I know the roots are $\sqrt[8]{3},\sqrt[8]{3}\omega,\dots,\sqrt[8]{3}\omega^7$ where $\omega=e^{2\pi i/8}=\cos(\pi/4)+i\sin(\pi/4)$. 
So I believe the splitting field is $\mathbb{Q}(\sqrt[8]{3},\omega)=\mathbb{Q}(\sqrt[8]{3},\sqrt{2},i)=:K$. 
To find the degree, I write
$$
[K:\mathbb{Q}]=[K:\mathbb{Q}(\sqrt[8]{3},\sqrt{2})][\mathbb{Q}(\sqrt[8]{3},\sqrt{2}):\mathbb{Q}(\sqrt[8]{3})][\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}].
$$
The first term on the RHS is $2$, since $i\not\in\mathbb{Q}(\sqrt[8]{3},\sqrt{2})$. The third term on the RHS is $8$. The middle term is at most $2$, and I'm fairly sure it is two, so the degree is $32$, if I've reasoned this correctly. 
However, I can't prove what the middle term actually is. I want to show $\sqrt{2}\not\in\mathbb{Q}(\sqrt[8]{3})$, to see that it is $2$.  I figured I could also switch the order of the extensions, and instead show $[\mathbb{Q}(\sqrt{2},\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})]=8$ by showing $x^8-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$, but this still seems worse, considering the ways the polynomial could factor.
How can I do this right? Thank you.
 A: Your hunch can be justified and generalized by using some algebraic number theory, specifically the discriminant $D_K$ of a number field $K$. The discriminant has the following two important properties:


*

*If $K \subset L$, then $D_K$ divides $D_L$.

*The primes that divide $D_K$ are precisely the primes $p$ which ramify in $K$.


It is not hard to see on the one hand that $3$ ramifies in $\mathbb{Q}(\sqrt[8]{3})$ and on the other hand that $2$ is the only prime that ramifies in $\mathbb{Q}(\sqrt{2})$, so the discriminant of the former cannot divide the discriminant of the latter. 
A: Here's one approach to the problem:
If $\sqrt{2} \in \mathbb{Q}(\sqrt[8]{3})$, we have 
$$8 = [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})]\cdot [\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2 \cdot [\mathbb{Q}(\sqrt[8]{3}):\mathbb{Q}(\sqrt{2})],$$
and so $\sqrt[8]{3}$ must satisfy an irreducible polynomial of degree 4 over $\mathbb{Q}(\sqrt{2})$, call it $g(x)$.
Then necessarily, $g(x) \mid x^8-3$ over $\mathbb{Q}(\sqrt{2})$.
Thus the four roots of $g(x)$ are of the form $\sqrt[8]{3} \zeta_8^i$ for some $i$.
In particular, the constant term of $g(x)$ is of the form $\sqrt[2]{3} \zeta_8^j$.
But this constant term is in $\mathbb{Q}(\sqrt{2})$, so it must be real, hence $\zeta_8^j = \pm 1$.
But then $\sqrt{3} \in \mathbb{Q}(\sqrt{2})$, but I think you can show this is not true pretty easily.
