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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 $n>1$.

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    Discriminant? 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

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This task really depends upon how much you know. It is easy to calculate the discriminants of these fields and see that they are different and so the fields must different, and even more obviously but equivalent, $p$ is the only prime that ramifies in the first and $q$ in the second. I suspect you haven't gotten there yet.

Here are some suggestions, as I don't have time to work it out.

  1. Show that the minimal polynomial of $p^{1/n} + q^{1/n}$ has degree larger than $n$.

  2. Write $q^{1/n}$ = rational linear combination of the powers of $p^{1/n}$. Multiply both sides by the lcd of the coefficients and play around to show that $p$ must divide $q$ or vice versa.

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    $p$ and $q$ are not the only primes that ramify in $\mathbb{Q}(\sqrt[n]{p})$ and $\mathbb{Q}(\sqrt[n]{q})$, respectively.2012-04-16