Let $p$ be a prime and $n$ a positive integer. I am trying to understand why the minimal polynomial of $\sqrt[n]{p}$ over the rationals must be $x^n-p$. I can show this in the case where $n=2$ but am having difficulties even for $n=3$. I must be missing something quite trivial as my notes just state this as a fact without any justification. However everything I have tried so far such as trying to show the powers of $\sqrt[n]{p}$ are linearly independent has not yielded fruit.
Can anyone tell me what am I overlooking?
You can use Eisenstein’s criterion. $p$ divides the coefficients of $x^n-p$ (except the leading coefficient $x^n$) but $p^2$ doesn’t divide $p$.