To determine whether a field contains free abelian groups of arbitrarily large finite rank Suppose that $K$ is an algebraically closed field. There is a statement:

If $K$ is not the algebraic closure of a finite field, then $K^*$ contains free abelian groups of arbitrarily large finite rank.

Is it true? And why?
Moreover, is

$K$ is not the algebraic closure of a finite field, if and only if $K^*$ contains free abelian groups of arbitrarily large finite rank.

True?
Thanks very much.

Yes. If $K$ has characteristic zero, then it contains $\mathbb{Q}$, and $\mathbb{Q}^{\ast}$ contains a free abelian group of infinite rank (on the primes). Otherwise, $K$ contains an element $x$ transcendental over the prime subfield $\mathbb{F}_p$, so $K$ contains $\mathbb{F}_p(x)$, and $\mathbb{F}_p(x)^{\ast}$ contains a free abelian subgroup of infinite rank (on the irreducible polynomials over $\mathbb{F}_p$). 
For the second statement, it is necessary and sufficient that $K$ is not contained in the algebraic closure of a finite field. We do not need the hypothesis that $K$ is algebraically closed. 
