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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 $K^*\to K^*\times K^*$?

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The multiplicative group of an algebraically closed field is divisible. Divisible groups are direct sums of copies of $\mathbb{Q}$ and of Prufer $p$-groups, and the cardinalities of the number of each uniquely determines the isomorphism type. So the question is equivalent to asking whether the dimension of $K^*$ modulo its torsion (as a $\mathbb{Q}$-vector space) and the dimensions of the $p$-parts of $K^*$ for each prime $p$ are each either $0$ or infinite. (In the case of $\mathbb{C}$, as your previous question, all of them are infinite). –  Arturo Magidin May 21 '12 at 20:09
    
@ArturoMagidin, i know that $t(\mathbb{\bar{F_p}}^*)\cong \bigoplus_{q\ne p}\mathbb{Z}(q^{\infty}),$ where $\mathbb{\bar{F_p}}$ is the algebraic clousure of $\mathbb{F_p}$, so in this case the dimiensions of the $p$-parts are infinite. –  Hector Pinedo May 24 '12 at 12:51

1 Answer 1

up vote 1 down vote accepted

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 $\# K$.

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Thanks again doctor Clark. I confess that just until now i read your answer. –  Hector Pinedo Sep 17 '13 at 11:46

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