# Thread: Galois closure

1. ## Galois closure

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

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

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

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

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

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

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

But, $\displaystyle \mathbb{Q}(\sqrt[n]{a})\cap \mathbb{Q}(\xi)$ is a subfield of both $\displaystyle \mathbb{Q}(\xi)$ and $\displaystyle \mathbb{Q}(\sqrt[n]{a})$. But subfields of $\displaystyle \mathbb{Q}(\sqrt[n]{a})$ have form $\displaystyle \mathbb{Q}(\sqrt[m]{a})$, $\displaystyle m|n$. Thus, somehow the only possibilities for this degree are $\displaystyle 1\text{ or }2$. 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 $\displaystyle \mathbb{Q}(\xi)(\sqrt[n]{a})/ \mathbb{Q}(\xi)$ is an $\displaystyle n$-Kummer extension. Therefore, the degree is equal to the order of the subgroup $\displaystyle \left< a(F^{\times})^n \right>$ in $\displaystyle F^{\times}/(F^{\times})^n$ where $\displaystyle F=\mathbb{Q}(\xi)$. But this seems to be really messy.

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

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