# Radical Field Extension Example

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.

## 1 Answer

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 $E'=\Bbb Q(\sqrtn\,)$, 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.

• 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 – Bradley Hill Nov 28 '17 at 23:57
• My definition of radical is in a slightly different form but yes they are the same, thank you very much for your help – Bradley Hill Nov 29 '17 at 13:22