My professor gave me this problem to do and I just wanted to check whether I'm on the right track or not. I want to find the splitting field of $x^{n} - a$ over $\mathbb{Q}$ where $a$ is a positive rational number. I also want to find the degree of the splitting field. Here is what I've tried
Lets first try finding the roots of this polynomial
$$x^n -a= 0$$ $$x^n = a$$
This implies that $x = a^{\frac{1}{n}}, a^{\frac{1}{n}}\omega,...,a^{\frac{1}{n}}\omega^{n-1}$ where $\omega = e^\frac{2\pi i}{n}$.
A field where the polynomial splits would be $\mathbb{Q}[a^{\frac{1}{n}}, a^{\frac{1}{n}}\omega,...,a^{\frac{1}{n}}\omega^{n-1}]$, but the splitting field needs to be the smallest field where the polynomial splits. So I think that we would only need $\mathbb{Q}[a^\frac{1}{n}, \omega]$ to get all of the roots. I think this is the splitting field for the polynomial but I'm unsure how to prove it.
Now in order to calculate the degree of this extension I tried the following
$[\mathbb{Q}[a^\frac{1}{n}, \omega]:\mathbb{Q}] = [\mathbb{Q}[a^\frac{1}{n}, \omega]:\mathbb{Q}[\omega]]\cdot[\mathbb{Q}[\omega]:\mathbb{Q}]$
I think that
$[\mathbb{Q}[a^\frac{1}{n}, \omega]:\mathbb{Q}[\omega]] = n$
because you would need all of the powers of $a^\frac{1}{n}$
$[\mathbb{Q}[\omega]:\mathbb{Q}] = \phi(n)$
Here $\phi(n)$ is the euler totient function. So I think that the degree would of the splitting field would end up being $n*\phi(n)$.
I had a few questions. First thing I wanted to know whether the splitting field that I have is correct and if it is then how you would prove it. I also wanted to know whether the degree of the splitting field is correct.
================================================================== Edit: I don't think that $\mathbb{Q}[a^\frac{1}{n}, \omega]$ will be a splitting field in the case that $a^\frac{1}{n} \in \mathbb{Q}$. For example if we have $n = 2$ and $a = 4$ then the splitting field of $x^2 -4$ should just be $\mathbb{Q}$, but I'm wondering whether what I did above works if $a^\frac{1}{n} \not\in \mathbb{Q}$