I want to find an example of fields $\mathbb{Q} \lt E \lt K$ such that the extension $K:\mathbb{Q}$ is radical but the extension $E:\mathbb{Q}$ is not radical and provide justification.


Let’s see. Try adjoining a (primitive) seventh root of $1$, call it $\zeta$, and call this field $K$. Now look at the cubic extension $E=\Bbb Q(\zeta+\zeta^{-1})\supset\Bbb Q$. (The minimal polynomial for this number is $X^3+X^2-2X-1$.) Since $K$ is abelian over $\Bbb Q$, so is $E$. And $E$ is normal over $\Bbb Q$, too.

Now if $E'$ is any cubic extension of $\Bbb Q$ that is radical, it should be of form $E'=\Bbb Q(\sqrt[3]n\,)$, but these cubic extensions of $\Bbb Q$ are not normal, since you’ll need the cube roots of unity to get all cube roots of $n$. So our $E$ above is not radical.

EDIT: You have asked for another explanation, not dependent on Galois Theory. I should say that I assume that what you meant by “radical extension” was a field of the form $\Bbb Q(\sqrt[m]n\,)$, for an integer $m$ and a rational number $n$, where for convenience’s sake we assume that $n$ contains no $m$-th powers in its expansion as a product of prime powers.

There are two cases: first, the cyclotomic case where $n=\pm1$, and you’re adjoining the $m$-th roots of unity if $n=1$ and the $2m$-th roots of unity if $n=-1$. The second case covers all other values of $n$. In the first case, the degree $[\Bbb Q(\zeta_n):\Bbb Q]$ is equal to $\phi(n)$, the Euler phi-function, which is even for all values of $n$ except $1$ and $2$. This shows that the cubic extension above is not $\Bbb Q(\zeta_k)$ for any (primitive) $k$-th root of unity $\zeta_k$.

How do I show that it’s not $\Bbb Q(\sqrt[m]n\,)$ for any $n\ne\pm1$? Those radical extensions, if not real quadratic (so with degree $2$) all have embeddings into $\Bbb C$ that are not into $\Bbb R$, since the minimal polynomial for the generator has nonreal roots. In the technical language, those fields are not “totally real”. But the cubic extension I constructed for you is totally real, because the roots of $X^3+X^2-2X-1$ are all real.

  • $\begingroup$ I have not been introduced to the concepts of normal and abelian extensions, however I have looked the definitions up and I understand them, but is there any chance you could give a justification without using these concepts $\endgroup$ – Bradley Hill Nov 28 '17 at 23:57
  • $\begingroup$ My definition of radical is in a slightly different form but yes they are the same, thank you very much for your help $\endgroup$ – Bradley Hill Nov 29 '17 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.