Find all subfields in extension $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[4]{2})$

Question

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

Answer 1

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.

Answer 2

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.