TITLE: Local Fields with isomorphic multiplicative group
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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.