I want to find all intermediate subfields of extension $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[4]{2})$. I guess that $\mathbb{Q}(\sqrt[4]{2})$ is not a splitting field, since we would have polynomial $x^4-2$, what should also give us root of unity of order 4 in the field. Am I having it right?
So how can I find these subfields? Should I look at all subfields of $\mathbb{Q}(\sqrt[4]{2}, \omega)$ and then take those which are subfields of $\mathbb{Q}(\sqrt[4]{2})$?
Thanks
On
I think the following argument could work: as explained in Tom's proof, the given field K is a number field of degree 4 over Q, it is not Galois, and all possible subfields should have degree 2 over Q.
This is enough to deduce that the only field contained in K, other than Q, is the quadratic extension M generated by the square root of 2. The argument is as follows: suppose that K contains M and also another quadratic number field N. Then K must contain the compositum of M and N (the minimal field containing both M and N), call it L. L is a so-called biquadratic number field, it is an easy exercise to show that any such field (obtained as a compositum of 2 quadratic number fields) is a degree 4 Galois extension of Q (moreover, you can see that its Galois group is not cyclic, it is a product of two cyclic groups of order 2). Since K must contain L and both K and L are degree 4 number fields, we conclude that K=L, but this makes no sense, because L is Galois over Q and K is not. This contradiction shows that K can contain at most one quadratic subfield, and since we see that K contains M this answers your question.
Another way to prove the same is by using ramification arguments. Computing the discriminant of the polynomial x^4 - 2 ans knowing that your field K is contained in the splitting field of this polynomial, you deduce that K is an extension of Q ramified only at 2. Then any possible subfield M of K (which by a degree argument has to be quadratic) is a quadratic number field ramified only at 2. But it is well-known and very easy to prove that the only quadratic number fields ramifying only at 2 are the ones generated by the square root of -1, -2 or 2. We conclude by using the fact that K is real and this implies that the quadratic field it contains has to be real, so the only candidate that is left is the one generated by square root of 2.
You are correct in that it is not a splitting field of $\sqrt[4]{2}$, since in particular it is entirely real, but $X^4-2$ is irreducible with complex solutions. Thus it is not a splitting field of any polynomial, since if it were then the extension would be Galois and contain all the roots of $X^4-2$.
One perfectly legitimate strategy would be to look the subfields of some Galois extension $L/\mathbb{Q}$ containing $\mathbb{Q}(\sqrt[4]{2})$ (which can be easily determined once you identify the Galois group) and to identify those that lie inside $\mathbb{Q}(\sqrt[4]{2})$, by looking at the assosciated subgroups of $Gal(L/\mathbb{Q})$ containing $Gal(L/\mathbb{Q}(\sqrt[4]{2}))$. However, this is a lot of work (and requires a lot of theory, although this is not a bad thing in and of itself). There is a much simpler and more elementary way to approach this problem, though.
Hint:
$[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}] = 4$ since $f(X) = X^4-2$ is irreducible, so the only proper subfields are degree $2$ over $\mathbb{Q}$. If $K$ is one such, let $g(X)$ be the minimal polynomial of $\sqrt[4]{2}$ over $K$. $g$ must have degree $2$, divide $f$ and have $\sqrt[4]{2}$ as a root. This gives a small number of possible options. We can then use the fact that coefficients of $g$ lie in $K$ (which is a real field) to determine what possible values $g$ may take, and this tells us enough to be able to determine all possible subfields $K$.
It is worth noting that the actual solution is shorter than the hint, but I think it's a good exercise to work through if you haven't seen it before.