# Is a field uniquely determined by its multiplicative group/how much knows K_1 about fields?

As the title says I would like to know if $K_1(k)=k^*$ uniquely determines a field $k$. For finite fields this is clearly the case, but I suspect it is not ture in general. However I guess cooking up a counterexample is not so easy.

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to clarify the tag "algebraic k-theory", perhaps some background would be useful ... ;-) –  Martin Brandenburg Apr 23 '10 at 11:13

Let $K$ and $L$ be two algebraically closed fields of characteristic $0$. Then $K^{\times} \cong L^{\times}$ iff $K$ and $L$ have the same cardinality.

The forward direction is clear. Conversely, if $K$ is algebraically closed, then consider the short exact sequence

$1 \rightarrow K^{\times}[\operatorname{tors}] \rightarrow K^{\times} \rightarrow Q \rightarrow 1$.

The first term is isomorphic to $\mathbb{Q}/\mathbb{Z}$. In particular it is divisible, hence injective, hence the sequence splits. The group $Q$ is a uniquely divisible abelian group of cardinality equal to that of $K$, hence isomorphic to a $\mathbb{Q}$ vector space of dimension $\# K$. Thus we recover the structure of $K^{\times}$ from $\# K$.

If $K$ is uncountable, then $\# K$ also determines the isomorphism class of $K$, but if $K$ is countable there is another invariant: the transcendence degree. This gives $\aleph_0$ pairwise nonisomorphic fields with isomorphic multiplicative groups.

I believe that this construction can be modified to construct, for any cardinal $\kappa$, $\kappa$ pairwise nonisomorphic fields with isomorphic multiplicative groups: for instance, instead of algebraically closed, the argument goes through with solvably closed fields containing all the roots of unity.

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@George: this is true if the cardinality is uncountable. It is not true if the cardinality is countable, as I explained above. For instance the algebraic closure of $\mathbb{Q}$ is certainly not isomorphic to the algebraic closure of $\mathbb{Q}(t)$. –  Pete L. Clark Apr 23 '10 at 15:03

$\mathbb Q^\ast$ is isomorphic to $\{\pm1\}$ times a free abelian group of countable rank. The same is true for an imaginary quadratic field of class number $1$ and different from $\mathbb Q(\sqrt{-3})$ and $\mathbb Q(\sqrt{-1})$ (of which are some though not many). Other examples comes from a real quadratic field of class number $1$ (though this time one of the free factors come from units). Presumably there is an infinite number of those.

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I suppose you meant a free abelian group with a countably infinite basis... –  ulrich Apr 23 '10 at 10:42
You are right, what I wrote is true but invalidates the following arguments. Corrected. –  Torsten Ekedahl Apr 23 '10 at 11:04
The class number is irrelevant. If $K$ is a number field, then $K^*$ is isomorphic to the direct product of a finite cyclic group (whose order is the number of roots of unity in $K$) with a free abelian group of infinite countale rank. –  Robin Chapman Apr 23 '10 at 12:29
@Robin: Right I saw potential problems only because for some strange reason I had confused what was subgroup of what. –  Torsten Ekedahl Apr 23 '10 at 13:26

Hey, maybe I know what you are looking for. Bogomolov and Tschinkel have shown that (Milnor, doesnt really matter...) K1 (= multiplicative group) AND K2 determine the field - well, assuming its a field of a certain type (see the abstract),

http://www.math.nyu.edu/~tschinke/papers/yuri/09milnor/milnor12.pdf

It's a pretty recent result.

They also touch the question whether maps between these groups come from actual geometric maps between the fields.

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Here is a fleshing out / generalization of Robin Chapman's comment, which could be useful in the construction of further examples of nonisomorphic fields with isomorphic multiplicative groups.

Proposition: Let $R$ be a Dedekind domain with fraction field $K$. Let $\Sigma$ be the set of maximal ideals of $R$, and let $S = \# \Sigma$. Then $K^{\times}$ is isomorphic to the product of the unit group $R^{\times}$ of $R$ and a free abelian group of rank $S$.

Proof: There is a standard short exact sequence

$1 \rightarrow R^{\times} \rightarrow K^{\times} \rightarrow \operatorname{Prin}(R) \rightarrow 1$,

where $\operatorname{Prin}(R)$ is the group of principal fractional $R$-ideals. In turn $\operatorname{Prin}(R)$ is a subgroup of $\operatorname{Frac}(R)$, the group of all fractional $R$-ideals, which for a Dedekind domain is well known to be the free abelian group with basis $\Sigma$. So $\operatorname{Prin}(R)$ is a subgroup of a free abelian group of rank $S$ hence itself a free abelian group of rank at most $S$. In particular $\operatorname{Prin}(R)$ is projective, so the above sequence splits:

$K^{\times} \cong R^{\times} \times \operatorname{Prin}(R)$.

It remains to be seen that the rank of $\operatorname{Prin}(R)$ is equal to the rank of $\operatorname{Frac}(R)$.

Case 1: $S$ is finite. Then $R$ is a PID, so $\operatorname{Prin}(R) = \operatorname{Frac}(R)$.

Case 2: $S$ is infinite. Then, if the rank of $\operatorname{Prin}(R)$ were smaller than $S$, then there would exist a subset $\Sigma'$ of $\Sigma$ of cardinality less than $S$ such that every principal fractional ideal is a $\mathbb{Z}$-linear combination of maximal ideals lying in $\Sigma'$. But this is contradicted by the Chinese Remainder theorem: there exists $f \in R$ with prescribed natural number valuation $n_{\mathfrak{p}} = \operatorname{ord}_{\mathfrak{p}}(f)$ at each maximal ideal $\mathfrak{p}$ in any finite subset of $\Sigma$.

Since the number of maximal ideals in the ring of integers of a number field is $\aleph_0$, applying the Dirichlet unit theorem, we recover Robin's comment that for number fields $K$ and $L$, $K^{\times} \cong L^{\times}$ iff $\mu(K) \cong \mu(L)$. In particular, any formally real number field has multiplicative group isomorphic to the product of a cyclic group of order $2$ and a free abelian group of countable rank. (So this is another example of a countably infinite family of pairwise nonisomorphic fields with isomorphic multiplicative groups.)

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