# Prove that $\mathbb{Q} \left( \sqrt[n]{p}\right) \neq \mathbb{Q} \left( \sqrt[n]{q}\right)$.

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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$\mathbb{Q}(p)=\mathbb{Q}(q)$. –  Álvaro Lozano-Robledo Feb 29 '12 at 16:12
How 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. –  Qiaochu Yuan Feb 29 '12 at 16:47
Discriminant? Can you show me the integral basis of $\mathbb{Q} \left( \sqrt[n]{p}\right)$ easily? Or how can you compute the discriminant? –  Bunny Young Feb 29 '12 at 17:08

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.

I don't have time to work this all out now, so if no one has gotten it in a while, I will come back and take care of it.

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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? –  Bunny Young Feb 29 '12 at 17:06
You don't really to compute exactly the discriminants to show that they are different. What are the discriminants of the rings of integers $\Bbb Z$\sqrt[n]{p}$$ and$\Bbb Z[\sqrt[n]{q}]$? –  Andrea Mori Feb 29 '12 at 17:19
$p$ and $q$ are not the only primes that ramify in $\mathbb{Q}(\sqrt[n]{p})$ and $\mathbb{Q}(\sqrt[n]{q})$, respectively. –  Brandon Carter Apr 16 '12 at 0:59