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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.

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    yes I blanked on the first line. But note that Qiaochu's answer holds as longs as $K$ is not algebraic over $\mathbb{F}_p$ regardless of whether or not $K$ is algebraically closed2011-07-29

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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.

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    @ShinyaSakai: The group is **multiplicative**, so the set $\{x^n\mid n\in\mathbf{Z}\}$ a free abelian group of rank **one**, i.e. cyclic. That's why Qiaochu had to use powers of irreducible polynomials to increase the rank.2011-11-02