Local Fields with isomorphic multiplicative group If two local fields have isomorphic multiplicative groups (as abstract groups), can we conclude that they are isomorphic as fields? What if they are isomorphic as topological groups?
 A: The key to this question is the structure of the unit group of local fields. We have the following result:
Let $K$ is a local field and let $q=p^f$ be the cardinality of the residue field.
If $K$ has characteristic $0$, then $$K^\times \cong \Bbb Z \oplus \Bbb Z/(q-1)\Bbb Z \oplus \Bbb Z/p^a\Bbb Z \oplus \Bbb Z_p^d$$
Where $d=[K:\Bbb Q_p]$ and $a \geq 0$ is an integer determined by the group of $p^n$-th roots of unity in $K$.
If $K$ has characteristic $p$, then $$K^\times \cong \Bbb Z \oplus \Bbb Z/(q-1) \Bbb Z \oplus \Bbb Z_p^{\Bbb N}$$
Both are isomorphisms of topological groups.
This is theorem 2.5.7 of Neukirch's Algebraic Number Theory
Let's treat finite characteristic first.
Every local field $K$ of characteristic $p$ is isomorphic to $\Bbb F_q((x))$ for some $q=p^n$. An argument by Hensel's lemma applied to the $q$-th cyclotomic polynomial shows that $q$ is the cardinality of the residue field. By the above theorem, the cardinality of the residue field uniquely determines and is determined by the structure of the topological group $K^\times$ and also the structure of $K^\times$ as an abstract group (we only need to know the torsion), so we get that two local fields of characteristic $p$ are isomorphic as fields iff their unit groups are isomorphic as topological groups iff their unit groups are isomorphic as abstract groups.
If $K$ is a local field of characteristic $0$ and residue characteristic $p$, then we see from the decomposition in the above theorem that the unit groups of two local fields with the same degree $d$ over $\Bbb Q_p$ that contain the same roots of unity have to be isomorphic. To construct such an example, cosider $\Bbb Q_2(\sqrt{2})$ and $\Bbb Q_2(\sqrt{-2})$, these are distinct due to Kummer theory, as $-1$ is not a square in $\Bbb Q_2$. Both are ramified extensions of degree $2$ of $\Bbb Q_2$ that contain the same roots of unity (only $\pm 1$), so their unit groups are isomorphic as topological groups by the above structure theorem for the unit groups. They can't be isomorphic as fields either, since in one field $2$ is a square and in the other one it isn't.
