Showing that $\mathbb{Q}(\zeta_p, \sqrt[p]{\ell}) = \mathbb{Q}(\zeta_p + \sqrt[p]{\ell})$ for $p,\ell$ primes.

Question

We consider the polynomial $x^p - \ell$, where $p,\ell$ are both prime numbers. Let $\zeta_p$ be a $p$-th root of unity. We wish to show that $L = \mathbb{Q}(\zeta_p, \sqrt[p]{\ell})$ is the same as $K = \mathbb{Q}(\zeta_p + \sqrt[p]{\ell})$. Here are some of the facts I have already proved:

• The polymial $x^p - \ell$ is irreducible over $\mathbb{Q}$.
• $[\mathbb{Q}(\zeta_p) : \mathbb{Q}] = p-1$ and $[\mathbb{Q}(\sqrt[p]{\ell}):\mathbb{Q}] = p$.
• $L = \mathbb{Q}(\zeta_p,\sqrt[p]{\ell})$ is the splitting field of $x^p - \ell$ and has degree $[\mathbb{Q}(\zeta_p,\sqrt[p]{\ell}) : \mathbb{Q}] = p(p-1) = p^2 - p$. Furthermore, $L$ is a Galois extension of $\mathbb{Q}$.

My first attempt was to show both inclusions. One is trivial, but the other one is much, much harder, i.e. showing $\mathbb{Q}(\zeta_p,\sqrt[p]{\ell}) \subseteq \mathbb{Q}(\zeta_p + \sqrt[p]{\ell})$. I thought expanding $(\zeta_p + \sqrt[p]{\ell})^p$ with the binomial expansion would be useful, but it turns out to be uglier than I expected.

I further notice that one way to reduce the problem is to show that the only automorphism $\sigma \in \mathrm{Aut}(L/\mathbb{Q})$ that fixes $\zeta_p + \sqrt[p]{\ell}$ is the identity map $1$, so that $L$ and $K$ must coincide. Since $\sigma(\zeta_p + \sqrt[p]{\ell}) = \sigma(\zeta_p) + \sigma(\sqrt[p]{\ell})$ and $\sigma$ only permutes the roots, we see that $\zeta_p + \sqrt[p]{\ell} \mapsto \zeta_p^m + \zeta_p^n\sqrt[p]{\ell}$ under $\sigma$ for some $m,n \in \{1,\ldots,p-1\}$. Hence, it suffices to show that $$\zeta_p + \sqrt[p]{\ell} = \zeta_p^m + \zeta_p^n\sqrt[p]{\ell} \iff (m,n) = (1,0)$$ to conclude. However, I don't know where to go from there.

Any help or hint would be greatly appreciated. If there is an easier to show this, please let me know.

Answer 1

You were very near to the full solution. If $n\neq 0$, you can rewrite your equality as

$$ \sqrt[p]{\ell}=\frac{\zeta_p-\zeta_p^m}{1-\zeta_p^n} $$

So you deduce $\sqrt[p]{\ell} \in {\mathbb Q}(\zeta_p)$, and it is very easy to finish from there.

Answer 2

Take a real positive choice for $\sqrt[p]{\ell}$, and for $\zeta$ take $\exp(2i\pi /p)$. The sum $\zeta +\sqrt[p]\ell$ will be closer to real axis whereas $\zeta^m +\zeta^n\sqrt[p]\ell$ may be far away from real axis (easy to visualize for $\ell\gg p$.)

Answer 3

If $L=F(\theta)$ and $f(\theta)$ is a polynomial in $\theta$ of lesser degree than $ [L:F]$, then $f(\theta)\ne0$. Corollary: $f(\theta)=g(\theta)$ as values in $L$ implies $f(x)=g(x)$ as polynomials in $F[x]$ if $\deg f,\deg g< [L:F]$.

Consider applying this knowledge to $F=\Bbb Q(\zeta_p)$ and $\theta=\sqrt[p]{\ell}$.

While this isn't as elegant as Ewan's answer, I think the above is an important, natural and intuitive fact to have in one's arsenal when thinking about elements in an extension field given in a power basis.