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It is obvious that $\mathbb{Q}_r$ is topologically isomorphic to $\mathbb Q_s$ while $r$ and $s$ denote different primes. But I really don't know whether it is true in the aspect of algebra. As I failed to prove it, I think that it is false, but I can't give a counterexample.

Last I'm quite sorry that I'm new to MathJax and I don't know how to use it properly.Thanks for reading and I would appreciate it if you could solve my problem.

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Notice that if $p$ is another prime (so that $p$, $r$, $s$ are all different and say different from $2$) then $\mathbb{Q}_r$ contains $\sqrt{p}$ iff $p$ is a square mod $r$. And for any $p$ you can find $r$ and $s$ such that $(p/r)=1$ and $(p/s)=-1$. – user8268 Dec 23 '11 at 7:51
In fact, by Dirichlet's theorem (about primes in arithmetic sequences), for any $r$, $s$ you can find a prime $p$ such that $(p/r)=1$, $(p/s)=-1$ ($(./.)$ is the Legendre symbol). So it proves that all $\mathbb{Q}_r$'s are non-isomorphic as fields. – user8268 Dec 23 '11 at 7:57
same question was posted in mathoverflow:… – Paul Dec 23 '11 at 8:21
And why are they topologically isomorphic? The metric thus defined are "different" in the sense that the topology they determined are distinct, is it not true? – awllower Dec 23 '11 at 9:52
Various answers are given (in an answer stricto sensu, and in the comments) in the MO version of the question, whose link has been given by @Paul. – Pierre-Yves Gaillard Dec 23 '11 at 9:58

Never. Looking at the number of roots of unity in your field suffices to distinguish all ${\mathbb{Q}}_p$ for odd values of $p$, because the number of roots of $1$ there is precisely $p-1$. It's different for the $2$-adic numbers, since they have two roots of unity, same as the $3$-adics. But the $2$-adics have a square root of $-7$ and the $3$-adics don't, whereas the $3$-adics have a square root of $10$ and the $2$-adics don't.

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