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Proving that $\mathbb{Q}(\sqrt{p_1}) \ne \mathbb{Q}(\sqrt{p_2})$ where $p_1$ and $p_2$ are distinct primes.

That is,

$\mathbb{Q}(\sqrt[2]{p_1}) \ne \mathbb{Q}(\sqrt[2]{p_2})$ where $p_1$ and $p_2$ are distinct primes.

$\mathbb{Q}(\sqrt[3]{p_1}) \ne \mathbb{Q}(\sqrt[3]{p_2})$ where $p_1$ and $p_2$ are distinct primes.

$\mathbb{Q}(\sqrt[...]{p_1}) \ne \mathbb{Q}(\sqrt[...]{p_2})$ where $p_1$ and $p_2$ are distinct primes.

$\mathbb{Q}(\sqrt[p]{p_1}) \ne \mathbb{Q}(\sqrt[p]{p_2})$ where $p_1$ and $p_2$ are distinct primes.

And p is some prime.

I'm curious if there exists an elegant proof of this property and if it generalizes well for powers of $1/p_i$, where $p_i$ are also prime.

Suppose that there exists a homomorphism $\phi :\mathbb{Q}(\sqrt{p_1}) \rightarrow \mathbb{Q}(\sqrt{p_2}) $

Then $\phi (1) = 1$

And $\phi (n) = n$ , $\forall n \in \mathbb{Z}$

And $\phi (a) = a$ , $\forall a \in \mathbb{Q}$

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    $\begingroup$ Possible starting point: $\mathbb{Q}(\sqrt{p_1})$ has an element $a$ such that $a^2 = p_1$, but $\mathbb{Q}(\sqrt{p_2})$ doesn't. $\endgroup$ – Ian H Feb 8 '17 at 6:08
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    $\begingroup$ For the square roots see the few top rated questions on the list of Related questions in the right margin. Also, $\Bbb{Q}$ is an infinite set, so it seemed to me that you were not asking about finite-fields, so I replaced that tag with a more appropriate one. $\endgroup$ – Jyrki Lahtonen Feb 8 '17 at 7:12

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