1. Galois closure

Let $L$ be the Galois closure of $K=\mathbb{Q}(\sqrt[n]{a})$, where $a \in \mathbb{Q}, a>0$ and suppose $[K:\mathbb{Q}]=n$.
Prove that $[L:\mathbb{Q}]=n \varphi(n)$ or $\frac{1}{2}n\varphi(n)$
( $\varphi$ Euler's function)

2. Originally Posted by Byun
Let $L$ be the Galois closure of $K=\mathbb{Q}(\sqrt[n]{a})$, where $a \in \mathbb{Q}, a>0$ and suppose $[K:\mathbb{Q}]=n$.
Prove that $[L:\mathbb{Q}]=n \varphi(n)$ or $\frac{1}{2}n\varphi(n)$
( $\varphi$ Euler's function)
i haven't thought about the details yet but what is clear to me is that the Galois closure of $K$ would be the splitting field of the polynoimal $x^n - a,$ because $[K:\mathbb{Q}]=n,$ which means we've

assumed that $x^n - a$ is irreducible over $\mathbb{Q}.$ so if we let $\xi = \exp(2\pi i/n),$ then $L=\mathbb{Q}(\xi, \sqrt[n]{a}).$ thus $[L:\mathbb{Q}]=[\mathbb{Q}(\xi, \sqrt[n]{a}): \mathbb{Q}(\xi)] [\mathbb{Q}(\xi):\mathbb{Q}]=\varphi(n)[\mathbb{Q}(\xi, \sqrt[n]{a}): \mathbb{Q}(\xi)].$ so what you need to prove

is that the only possible values of $[\mathbb{Q}(\xi)(\sqrt[n]{a}): \mathbb{Q}(\xi)]$ are $n$ and $\frac{n}{2}.$

3. Originally Posted by NonCommAlg
is that the only possible values of $[\mathbb{Q}(\xi)(\sqrt[n]{a}): \mathbb{Q}(\xi)]$ are $n$ and $\frac{n}{2}.$
Yes, but show do you prove this?

Notice by the Natural Irrationalities theorem: Let $K = \mathbb{Q}(\sqrt[n]{a})$ and $E = \mathbb{Q}(\xi)$ and so $[KE:E] = [K:E\cap K]$.

Therefore, $[ \mathbb{Q}(\sqrt[n]{a},\xi) : \mathbb{Q}(\sqrt[n]{a})] = [\mathbb{Q}(\sqrt[n]{a}): \mathbb{Q}(\sqrt[n]{a})\cap \mathbb{Q}(\xi)]$.

But, $\mathbb{Q}(\sqrt[n]{a})\cap \mathbb{Q}(\xi)$ is a subfield of both $\mathbb{Q}(\xi)$ and $\mathbb{Q}(\sqrt[n]{a})$. But subfields of $\mathbb{Q}(\sqrt[n]{a})$ have form $\mathbb{Q}(\sqrt[m]{a})$, $m|n$. Thus, somehow the only possibilities for this degree are $1\text{ or }2$. But there does not seem to be direct way to show this. This there some kind of theorem about real subfields of a cyclotomic extension?
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The other thing I see is that $\mathbb{Q}(\xi)(\sqrt[n]{a})/ \mathbb{Q}(\xi)$ is an $n$-Kummer extension. Therefore, the degree is equal to the order of the subgroup $\left< a(F^{\times})^n \right>$ in $F^{\times}/(F^{\times})^n$ where $F=\mathbb{Q}(\xi)$. But this seems to be really messy.

4. here's a nice problem which will also prove that $[\mathbb{Q}(\xi, \sqrt[n]{a}): \mathbb{Q}(\xi)]=n \ \text{or} \ \frac{n}{2}$:

let $\xi_k=\xi^k, \ 1 \leq k \leq n, \ n > 1.$ clearly $\sum_{k=1}^n \xi_k = 0.$ now prove that if $\sum_{j=1}^m \xi_{k_j}=0,$ where $1 \leq k_1 < k_2 < \cdots < k_m \leq n,$ then $m=n \ \text{or} \ \frac{n}{2}.$