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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.

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closed as too localized by Alex Bartel, Felipe Voloch, Chandan Singh Dalawat, GH from MO, Igor Rivin Dec 23 '11 at 15:00

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No. For example, $\mathbb{Q}_3$ is not isomorphic to $\mathbb{Q}_5$, for the following reason: any field isomorphism would have to map $-1$ to $-1$; but $\mathbb{Q}_5$ contains a square root of $-1$, whereas $\mathbb{Q}_3$ does not. – Martin Bright Dec 23 '11 at 7:43
Never: a Galois group $G=\mathrm{Gal} (F/\mathbb{Q}_b)$ is solvable, and in fact $[[G,G],[G,G]]$ is a $p$-group (contained in the wild inertia), and for any $p$ one can choose $F$ so that it is a non-trivial $p$-group. – Homology Dec 23 '11 at 7:48
One can more simply compare the groups of roots of unity contained in the two fields. – Chandan Singh Dalawat Dec 23 '11 at 8:43
... except for ${\mathbf Q}_2$ and ${\mathbf Q}_3$, which have the same number of roots of unity. But $-2$ is a square in ${\mathbf Q}_3$ but not in ${\mathbf Q}_2$. – KConrad Dec 23 '11 at 8:48
This has also been asked on MSE, where it really belongs. I have voted to close. – Alex B. Dec 23 '11 at 9:06
up vote 10 down vote accepted

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.

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