Isomorphic group to the multiplicative group of a field of prime characteristic. This question is a little bit different from the one I made before.
Suppose that  I have an algebraically closed field of prime characteristic, is it possible to find an epimorphism $K^*\to K^*\times K^*$?
 A: My answer to this other question of the OP can be used to answer this question rather quickly.
$\newcommand{\tors}{\operatorname{tors}}$
Namely, first suppose that $K$ is algebraic over a finite field.  Then $K^{\times}$ is a torsion group.  It is easy to see that $K^{\times}$ does not surject onto $K^{\times} \times K^{\times}$: suppose $\varphi:  K^{\times} \rightarrow K^{\times} \times K^{\times}$ is a surjective homomorphism.  Let $x,y \in K^{\times} \times K^{\times}$ be two elements of prime order $\ell$ such that $\langle x \rangle \neq \langle y \rangle$, so $f = \langle x,y \rangle \cong (\mathbb{Z}/\ell \mathbb{Z})^2$.   Let $X \in \varphi^{-1}(x)$ and $Y \in \varphi^{-1}(y)$, and put 
$F = \langle X, Y \rangle$, so $\varphi$ restricts to a homomorphism from $F$ to $f$.  But $F$ is a finitely generated torsion subgroup of the group of units of a field, so $F$ is finite cyclic but has a noncyclic homomorphic image $f$: contradiction.
Now suppose that $K$ is not algebraic over a finite field.  Then by my previous answer, $K^{\times}$ has a direct summand $V$ which is a $\mathbb{Q}$-vector space of dimension $\# K$.  Notice that $K^{\times}[\tors]$ is a homomorphic image of $\mathbb{Q}$: $\mathbb{Q}/\mathbb{Z} = \bigoplus_{\ell} \mathbb{Q}_{\ell}/\mathbb{Z}_{\ell}$; since are in characteristic $p$, we further mod out by the factor $\mathbb{Q}_p/\mathbb{Z}_p$ to get 
a group isomorphic to $K^{\times}[\tors]$.  Since $\operatorname{dim} V$ is infinite, 
$V \cong (\mathbb{Q} \oplus V) \oplus (\mathbb{Q} \oplus V)$, which surjects onto $K^{\times} \oplus K^{\times}$.  
The result is also true for all algebraically closed fields of characteristic $0$.  
In fact, I am pretty sure that when $K$ is algebraically closed and not algebraic over a finite field, then the homomorphic images of $K$ are precisely the divisible groups of cardinality at most $\# K$.  
