The real number field $\mathbb{R}$, the $p$-adic number field $\mathbb{Q}_p$ and local fields Recently, I learned the following knowledge.
(1) All $\mathbb{Q}_p$ are non-archimedean local fields and $\mathbb{R}$ is an archimedean local field. They are extensions of $\mathbb{Q}$, but their topologies are different. They correspond to a place of $\mathbb{Q}$ respectively.
(2) The algebraic closure of $\mathbb{Q}_p$, denoted by $\mathbb{C}_p$, is isomorphic to $\mathbb{C}$. (Only as a field.)
I suspect that for any prime number $p$, $\mathbb{Q}_p$ is not isomorphic to $\mathbb{R}$, and for any prime numbers $p_1$ and $p_2$, $\mathbb{Q}_{p_1}$ is not isomorphic to $\mathbb{Q}_{p_2}$. But I can't prove it.
Then, my question is whether my guess is true and how to prove that $\mathbb{Q}_p$ and $\mathbb{R}$ are not isomorphic. Furthermore, whether any two different local fields are not isomorphic (only as fields)?
I have read some books on algebraic number theory and I can't find the answer, and I won't prove it myself. Is there any literature to introduce this knowledge?
 A: They can all be distinguished by the question of which square roots of integers exist. Namely, over $\mathbb{R}$ all square roots of positive integers exist. On the other hand, over $\mathbb{Q}_p, p \ge 3$ we have that by Hensel's lemma  that if $\gcd(p, n) = 1$ then $\sqrt[2]{n}$ exists iff $n$ is a quadratic residue $\bmod p$, and over $\mathbb{Q}_2$ the condition is that if $\gcd(2, n) = 1$ then $\sqrt[2]{n}$ exists iff $n \equiv 1 \bmod 8$; these conditions are not always satisfied for positive $n$ but are sometimes satisfied for negative $n$, which is enough to show that $\mathbb{R}$ is not isomorphic to $\mathbb{Q}_p$ for any $p$.
To distinguish $\mathbb{Q}_p$ from $\mathbb{Q}_q$ for distinct primes $p, q$ it suffices to observe that by the Chinese remainder theorem we can always find $n$ such that $\gcd(n, pq) = 1$ and $n$ is a quadratic residue $\bmod p$ but not a quadratic residue $\bmod q$ (with $p$ or $q$ respectively replaced by $8$ for the $\mathbb{Q}_2$ case).
A: According to your guidance, I came up with a method myself, but I'm not sure whether it is correct.
Any $n$-th root of unity in $\mathbb{Q}_p$ belongs to $\mathbb{Z}_p$. By Hensel's Lemma, the field $\mathbb{F}_p$ has a primitive $n$-th root of unity if and only if $\mathbb{Z}_p$ has a primitive $n$-th root of unity.
For any prime numbers $p_1<p_2$,  the field $\mathbb{F}_{p_1}$ does not has a primitive $(p_2-1)$-th root of unity. We deduce that $\mathbb{Q}_{p_1}$ can not be isomorphic to $\mathbb{Q}_{p_2}$.
I think the key step in this kind of problem is to use Hensel's lemma and polynomials.
