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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?

  • Do we assume they are isomorphic as topological groups? Or just as abstract groups? – Julian Rosen Aug 03 '17 at 19:22
  • @reuns Local fields have simple structure and we can describe the multiplicative group explicitly. As the question is true for finite field but false for global field, it's natural to ask about local fields. –  Aug 03 '17 at 19:38
  • @JulianRosen Abstract groups, but things may be easier when we consider they are isomorphic as topological groups. Thanks for the observation, I need to edit the question to make it clear. –  Aug 03 '17 at 19:41
  • Do you have a concrete counter-example with number fields, to check if we can take their completion at a prime ? – reuns Aug 03 '17 at 19:49
  • @reuns https://math.stackexchange.com/questions/624289/examples-of-non-isomorphic-fields-with-isomorphic-group-of-units-and-additive-gr?noredirect=1&lq=1. This isomorphism is based on some trick of infinity like how one shows the real numbers is isomorphic to the complex number as abstract groups. –  Aug 03 '17 at 19:52

1 Answers1

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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.

Lukas Heger
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