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 $\mathbb{Q}(\sqrt[4]{2})$?



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.


$[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}] = 4$ 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.

  • 1
    $\begingroup$ I have followed your hint. So, I have to split $x^4 - 2$ into linear factors and then choos any two of them and check if $\sqrt[4]{2}$ is the root of a new polynomial (deg 2) and the coefficients of a new polynomial construct field $K$ with degree 2 over $\mathbb{Q}$. Obviosly, we must take term $x-\sqrt[4]2$. Then I consequently adjoin other linear terms and I get that we only can take term $\sqrt[4]4$. Did I det the correct answer? The other question is, why our new polynomial (deg 2) must divide $x^4 - 2$? $\endgroup$
    – michael
    Jun 1 '14 at 15:01
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    $\begingroup$ @michael Yes, that's correct, so the only possible value of $g$ is $(X-\sqrt[4]{2})(X+\sqrt[4]{2})=X^2-\sqrt[4]{4}=X^2-\sqrt{2}$. (This tells us that $K$ contains $\sqrt{2}$). It must divide $x^4-2$ because if $f(X)$ is the minimal polynomial of an element $x$ over a field $K$, then $f$ divides any polynomial $g(X) \in K[X]$ such that $g(x) = 0$. You can prove this using the division algorithm and the fact that $f(X)$ is the polynomial of least degree in $K[X]$ such that $f(x) = 0$. $\endgroup$ Jun 1 '14 at 18:30

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.


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