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.



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


