Question

It is obvious that Q_r is topologically isomorphic to 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.

Answer 1

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.