# Primitive radical extension with $n$-th root with Galois group smaller than $n$

It is a well-known theorem in Field Theory that if $$F$$ is a field:

1. Which contains the $$n$$-th roots of unity for some $$n\ge 1$$.

2. Is of characteristic not dividing $$n$$.

And if $$0\ne a\in F$$ is some element, then the primitive radical extension $$F(\sqrt[n]{a})/F$$ is a Galois extension with $$\textbf{Gal}(F(\sqrt[n]{a})/F)\cong \mathbb{Z}_d$$ such that $$d|n$$.

I am looking for examples (preferably over characteristic 0, is such examples exist) for such extensions where $$d.

• Sure: $a=4$, $n=2$, where $\Bbb Q(\sqrt4\,)=\Bbb Q$. Jul 4, 2018 at 21:36
• I actually thought about this case and now I'm not sure why I decided it's not a satisfying counterexample. Jul 4, 2018 at 21:42
• Probably because $\Bbb Q$ doesn’t contain the fourth roots of unity. I should have taken the base field to be the Gaussian numbers, $\Bbb Q(i)$. Otherwise, same example. Jul 5, 2018 at 1:59

Let $F$ be the extension of the rationals got by joining a primitive 15th root of unity. Take $a=8$. Then the extension $F[ 8^{1/15}]$ of $F$ will have Galois group cyclic of order 5. (Because 8 already has cubic roots in $F$).
• So basically I can only expect $d < n$ when $a$ is already a $n/d$-power of an element of F? Jul 4, 2018 at 21:43