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Field $\mathbb{F}$ is finite if and only if its multiplicative group $\mathbb{F}^{\times}$ is finitely generated.

The "$\Rightarrow$" implication is obvious, but how to prove the otherwise?

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2 Answers 2

up vote 5 down vote accepted

Let $\Bbb{F}$ be a field such that $\Bbb{F}^{\times}$ is finitely generated. Note that $\operatorname{char}\Bbb{F}\neq0$ because otherwise $\Bbb{Q}\subset\Bbb{F}$, and $\Bbb{Q}^{\times}$ is not finitely generated. Let $p=\operatorname{char}\Bbb{F}$ so that $\Bbb{F}_p\subset\Bbb{F}$. Note that $\Bbb{F}$ is an algebraic extension of $\Bbb{F}_p$, because if $T\in\Bbb{F}$ is transcendental over $\Bbb{F}_p$ then $\Bbb{F}_p(T)\subset\Bbb{F}$, and $\Bbb{F}_p(T)^{\times}$ is not finitely generated. So for every $x\in\Bbb{F}^{\times}$ there exists $n>0$ such that $x^n=1$, i.e. $\Bbb{F}^{\times}$ is torsion. As it is finitely generated and abelian, it is finite, and hence $\Bbb{F}$ is finite.

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We not only posted the same answer at the same time, we also have exactly the same notation! :D –  Martin Brandenburg Jun 21 at 7:47

Fact. Subgroups of finitely generated abelian groups are finitely generated.

Let $F$ be a field, assume that $F^\times$ is finitely generated. Since $\mathbb{Q}^\times$ is not finitely generated (in fact, $\mathbb{Q}^\times / \mathbb{Z}^\times \cong \bigoplus_p \mathbb{Z}$ by prime factor decomposition), we see that $F$ has positive characteristic $p$. Since $\mathbb{F}_p(T)^\times$ is not finitely generated (in fact, $\mathbb{F}_p(T)^\times / \mathbb{F}_p[T]^\times = \bigoplus_\mathfrak{p} \mathbb{Z}$), it follows that $F$ is algebraic over $\mathbb{F}_p$, i.e. the directed union of its finite subfields $F_i$. Then $F^\times$ is a directed union of finite subgroups $F_i^\times$. Since $F^\times$ is finitely generated, there is some $i$ with $F^\times = F_i^\times$, i.e. $F = F_i$.

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