How to prove irreducibility of this polynomial?

Question

Let $p,q$ be primes. Prove that $y^{n }-p$ is irreducible over $\mathbb{Q}(\sqrt[n ]q) $ .

I have tried for some time, but still feel confused about how to prove it. Can anyone help me?

Answer 1

Any root of a prime will be irrational. You can imagine trying to find roots of the polynomial in R and find there are no solutions in terms of radicals of other primes.

Answer 2

Set $F=\mathbb{Q}(\sqrt[n]{q},\varepsilon)$ where $\varepsilon$ be the $n^{\mathrm{th}}$ root of unity and $E$ be the splitting field of $x^n-p$ over $F$. As the roots of $x^n-p$ in $E$ are $\beta,\varepsilon\beta,\dots,\varepsilon^{n-1}\beta$, $x^n-p=\displaystyle\prod_{k=0}^{n-1}{(x-\varepsilon^k\beta)}$. Suppose $\psi$ is an irreducible factor of $x^n-p$ in $F[x]$, then $\psi=\displaystyle\prod_{\text{for some }k}{(x-\varepsilon^k\beta)}$, so the constant of $\psi$ is $\pm\varepsilon^c\beta^s$, so $\beta^s=r\in F$. However, $\deg\psi=s$, which means $\psi=x^s-r$. As $F(\beta)=F(\varepsilon\beta)=\cdots=F(\varepsilon^{n-1}\beta)$, the minimal polynomials of $\varepsilon^k\beta$ have the same degree $s$, so $x^n-q=\prod{(x^s-r_i)}$. So we have $s\mid n$ and $p=r_i^{n/s}$. To prove $s=n$, we only have to prove that $x^m=p$ has no root in $F$ for $m\ge2,m\mid n$, i.e., $\sqrt[m]{p}\not\in F$, which is obvious.