Given two distinct prime numbers $p$ and $q$, how can we prove that $\mathbb{Q} \left( \sqrt[n]{p}\right) \neq \mathbb{Q} \left( \sqrt[n]{q}\right)$ where $\sqrt[n]{p}$,$\sqrt[n]{q}\in \mathbb{R}$ and $n>1$.
Prove that $\mathbb{Q} \left( \sqrt[n]{p}\right) \neq \mathbb{Q} \left( \sqrt[n]{q}\right)$.
4
$\begingroup$
algebraic-number-theory
field-theory
-
1$\mathbb{Q}(p)=\mathbb{Q}(q)$. – 2012-02-29
-
0How familiar are you with algebraic number theory? You can do this using the fact that if two number fields are isomorphic then so are their rings of integers, and then using e.g. discriminants or ramification. – 2012-02-29
-
0Discriminant? Can you show me the integral basis of $ \mathbb{Q} \left( \sqrt[n]{p}\right) $ easily? Or how can you compute the discriminant? – 2012-02-29