It is obvious that $\mathbb{Q}\_r$ is topologically isomorphic to $\mathbb{Q}\_s$ where $r$ 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.



Is Q_r algebraically isomorphic to Q_s where r and s denote different primes?
For a prime $p \neq 2$, the multiplicative group $\mathbb{Q}\_p^\times$ is isomorphic to $\mathbb Z \times \mathbb Z\_p \times \mathbb Z/(p-1)$, while $\mathbb Q\_2^\times$ is isomorphic to $\mathbb Z \times \mathbb Z\_2 \times \mathbb Z/2$, cf. Serre, A course in arithmetic, Theorem II.3.2 (p. 17). Since a field isomorphism $\varphi\colon \mathbb Q\_p \to \mathbb Q\_r$ will preserve the torsion subgroup of the multiplicative subgroup, this shows that $\mathbb Q\_p \not\cong \mathbb Q\_r$ whenever $\{p,r\} \neq \{2,3\}$. The remaining case is taken care of by the fact that 
$\mathbb{Q}\_p^\times$ / $\mathbb {Q}\_p ^{\times 2}$ $ \cong \mathbb Z/2 \times \mathbb Z/2$ whenever $p \neq 2$, while $\mathbb Q\_2^\times$ / $\mathbb Q\_2^{\times 2} \cong \mathbb Z/2 \times \mathbb Z/2 \times \mathbb Z/2$, cf. loc. cit., p.18. 


