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