When is a divisible group a power of the multiplicative group of an algebraically closed field? It is known that for any algebraically closed field $\mathbb{F}$ its multiplicative group $\mathbb{F}^*$ is a divisible group, and consequently any power $\mathbb{F}^*\times\cdots\times \mathbb{F}^*.$ 
Now if we have an abelian divisible group $G,$ i would like to know under which conditions $G$ is a power of some algebraically closed field.
Thanks.
 A: Just want to mention that there is some new work done in the $k$ not being algebraically closed case. The paper is published in Journal of Algebra this year. 
A: $\newcommand{\tors}{\operatorname{tors}}$
It is possible to completely describe the structure of the multiplicative group of an algebraically closed field $F$.  (It is funny that you mention this now, since over the last few weeks I've been writing up a proof of the corresponding structure theorem for the group of $F$-rational points on an(y) abelian variety over an algebraically closed field.)
For every divisible commutative group $G$, the torsion subgroup $G[\tors]$ is again divisible, hence an injective $\mathbb{Z}$-module (by Baer's Criterion), hence is it a direct summand of $G$.  That is, $G = G[\tors] \times V$ for some subgroup $V$.   $V$ is divisible and torsionfree, and it follows that it is a $\mathbb{Q}$-vector space.  Thus when $G = F^{\times}$ for an algebraically closed field $F$ I need to tell you the structure of $G[\tors]$ and the dimension of $V$.
The torsion subgroup of $F^{\times}$ is just the subgroup of roots of unity.  So if $F$ has characteristic $0$, this is the direct limit of cyclic groups of order $n$ for all $n$, namely $\mathbb{Q}/\mathbb{Z} = \bigoplus_{\ell} \mathbb{Q}_{\ell}/\mathbb{Z}_{\ell}$.  If $F$ has characteristic $p > 0$ then there are no nontrivial $p$-power roots of unity, so the answer is $\bigoplus_{\ell \neq p} \mathbb{Q}_{\ell}/\mathbb{Z}_{\ell}$.
On to the torsionfree part $V$.  If $F$ is algebraic over a finite field then it is simply the union of its finite subfields, so there is no torsionfree part: $V = 0$.  So now assume that $F$ is not algebraic over a finite field.  I claim that in this case
$\dim V = \# F$.  First note that in all cases we have $\dim V \leq \# F$.  Now:
Case 1: Suppose $F$ has characteristic $0$ and $\# F = \aleph_0$.   Then $F \supset \mathbb{Q}$ so $F^{\times} \supset \mathbb{Q}^{\times} \cong \{ \pm 1\} \times \bigoplus_{p} \mathbb{Z}$.  Here the sum extends over all prime numbers and comes from unique factorization in $\mathbb{Z}$.  This shows that $\dim V \geq \aleph_0$.
Case 2: Suppose $F$ has characteristic $p > 0$, is not algebraic over $\mathbb{F}_p$, and $\# F = \aleph_0$.  Then there is a transcendental element $t \in F$, so $F^{\times} \supset \mathbb{F}_p[t]^{\times} \cong \mathbb{Z}/(p-1)\mathbb{Z} \times \bigoplus_{P} \mathbb{Z}$.  Here the sum extends over all monic irreducible polynomials and comes from unique factorization in $\mathbb{F}_p[t]$.  This shows that $\dim V \geq \aleph_0$.
[In these two cases I am using the fact that the quantity $\dim V$ I'm trying to compute is also $\dim_{\mathbb{Q}} F^{\times} \otimes_{\mathbb{Z}} \mathbb{Q}$.  For any commutative group $G$, let $r(G) = \dim_{\mathbb{Q}} G \otimes_{\mathbb{Z}} \mathbb{Q}$.  Then if $G_1 \subset G_2$ are commutative groups, since $\mathbb{Q}$ is a flat $\mathbb{Z}$-module, $r(G_1) \leq r(G_2)$.]
Case 3: Suppose $\# F > \aleph_0$.  Then cardinality considerations show that $\# V = \# F^{\times}$, and a $\mathbb{Q}$-vector space of uncountable cardinality has dimension equal to its cardinality.  We're done.
Finally, you ask about finite powers, but you can see that the structure there follows immediately form the complete knowledge of the structure for just one factor.  E.g. in characteristic $0$ we get
$\bigoplus_{i=1}^n F^{\times} \cong (\mathbb{Q}/\mathbb{Z})^n \oplus V$,
where $\dim V = \# F$.
I remark in passing that the result for $A(F)$ for $A$ a $g$-dimensional abelian variety is almost the same: the only thing that changes is the structure of the torsion subgroup, which is now $(\mathbb{Q}/\mathbb{Z})^{2g}$ in characteristic $0$ and $(\prod_{\ell \neq p} \mathbb{Q}_{\ell}/\mathbb{Z}_{\ell})^{2g} \times (\mathbb{Q}_p/\mathbb{Z}_p)^a$ for some $0 \leq a \leq g$ in characteristic $p > 0$.  However, the proof is harder.  I found a proof using Hilbertian subfields and group theory before* learning that two other proofs of equivalent statements appear in the literature; the first is a 1973 paper of Michael Rosen which contains a very elegant proof.
[*This cannot be completely accurate, since this 2009 paper of mine not only cites Rosen's 1973 paper but includes other results from it, enough to convince me that I must have read the paper in 2008 and thus seen Rosen's result.  But then several years passed before I took up this subject again, and I forgot that Rosen's paper gives a proof and came up with a different proof of my own.]
A: Greg Oman  in "divisible multiplicative groups of fields" obtained necessary and sufficient condition  for when a divisible group is isomorphic to  multiplicative group of a field. We can obtain two fields with isomorphic multiplicative groups such that one of them is absolutely algebraic field and another one is not.
