Let $p,q$ and $r$ be primes in $\mathbb{Z}$ with $q \neq r$. Let $\sqrt[p]{q}$ denote any root of $x^p-q$ and let $\sqrt[p]{r}$ denote any root of $x^p - r$. I need to prove that $\mathbb{Q}(\sqrt[p]{q}) \neq \mathbb{Q}(\sqrt[p]{r})$.
This is exercise 14.7.8 from Dummit & Foote. By a previous problem, I know that $K = F(\sqrt[n]{a}) = F(\sqrt[n]{b})$ if and only if $a = b^i c_1^n$ and $b = a^jc_2^n$ for some $c_1,c_2 \in F$, that is, if and only if $a$ and $b$ generate the same subgroup of $F^\times$ modulo $n$-th powers. It looks to me that my problem follows from this since $q,r$ are both primes, but I am not convinced at all.
Any help or hint would be greatly appreciated.
Note. I know very little about algebraic number theory. I am also not familiar with discrimants and ramification.
You can't directly apply $7c$ here, because you don't know that the fields are Galois (which is an assumption in the previous problem). However, if they were the same field, then adjoining a primitive $p^{th}$ root of unity to both fields would preserve that fact. Thus we have that $\mathbb{Q}(\zeta_p q^\frac{1}{p})=\mathbb{Q}(\zeta_p r^\frac{1}{p})$ which by $7c$ tells you that $\exists c_1,c_2\in\mathbb{Q}$, $i,j\in\mathbb{N}$, such that $q=r^ic_1^n,r=q^jc_2^n$. Taking $n^{th}$ roots of both sides says that $$\left(\frac{q}{r^i}\right)^\frac{1}{n}\text{ and } \left(\frac{r}{q^i}\right)^\frac{1}{n}\in\mathbb{Q}$$
But $a^\frac{1}{n}$ is in $\mathbb{Q}$ iff every prime in the extended (ie including negative exponents since this is $\mathbb{Q}$) prime factorization of $a$ has exponent divisible by $n$. This is clearly not the case, so by contradiction we are done.