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.
 A: 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.  
A: 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.
A: $\mathbb Q^\ast$ 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.
A: 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.)  
