Proving $f(x)=x^n-p$ is minimal of $\alpha=\sqrt[n]{p}$ over the field F (p is prime)

Question

I am having trouble with the concept of minimal polynomial,

In a homework question I have concluded the following:

$\mathbb{Q} \le \mathbb{F} \le \mathbb{C}$ - field extensions

such that $[\mathbb{F}:\mathbb{Q}]=m \in \mathbb{N}$, $[\mathbb{F}[\alpha]:\mathbb{F}]$ is finite and equals the degree of the minimal polynomial.

gcd(m,n)=1, and $f(x)$ is irreducible over $\mathbb{Q}$

Can I tell $f(x)$ is minimal of $\alpha$ over $\mathbb{F}$? How?

Answer 1

If you make a field-extension diagram showing $\Bbb Q\subset\Bbb Q(\alpha)\subset\Bbb F(\alpha)$ and $\Bbb Q\subset\Bbb F\subset\Bbb F(\alpha)$, I think you’ll see it. You have $[\Bbb Q(\alpha):\Bbb Q]=n$ and $[\Bbb Q:\Bbb F]=m$. With a couple of other facts that you surely have seen, that should do it.