MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
to clarify the tag "algebraic k-theory", perhaps some background would be useful ... ;-) – Martin Brandenburg Apr 23 '10 at 11:13
up vote 4 down vote accepted

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.

share|cite|improve this answer
@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.

share|cite|improve this answer
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),

It's a pretty recent result.

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

share|cite|improve this answer

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.