5
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ We not only posted the same answer at the same time, we also have exactly the same notation! :D $\endgroup$ – Martin Brandenburg Jun 21 '14 at 7:47
7
$\begingroup$

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. $F = F_i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.