What fields $F$ have finitely generated multiplicative groups $F^{\times}$?

I see if the field is finite then it is cyclic and so it is finitely generated.

But what can we say if the field is infinite?

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    $\begingroup$ I’m certainly having a hard time thinking of any example other than finite fields. (Of course, all finite groups are finitely generated, so you don’t need to know the cyclic property of the group of units of a finite field to see,that they are finitely generated.) $\endgroup$ May 8, 2021 at 21:24
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    $\begingroup$ This answer shows that a subgroup of a finitely generated abelian group is finitely generated. Since $\mathbb Q^\times$ is not finitely generated, this means that $F$ must have finite characteristic. math.stackexchange.com/q/137287/7933 $\endgroup$ May 8, 2021 at 21:34
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    $\begingroup$ Since any subfield must have this property, if you can prove that$\mathbb F_p(x),$ does not, that means every element must be algebraic over $\mathbb F_p.$ $\endgroup$ May 8, 2021 at 21:42
  • $\begingroup$ @ThomasAndrews, thank you! but being $\mathbb{Q}^{\times}$ not finitely generated implies that any field of zero characteristic is not finitely generated. $\endgroup$ May 8, 2021 at 21:51

2 Answers 2


$\mathbb Q$ doesn’t have this property. This is because any finite set of rationals only use a finite set of primes in the numerator and denominator, so you can’t get numerators or denominators divisible by other primes.

In general, if $F$ has this property, then any subfield of $F$ has this property, because a subgroup of a finitely generated abelian group is finitely generated.

This means no field of characteristic zero has this property.

Now, given $F$ of characteristic $p,$ for each element $\alpha$ the subfield $\mathbb F_p(\alpha)$ is either finite, or it is isomorphic to $\mathbb F_p(x)$ where $x$ is an indeterminate.

We can prove that $F_p(x)$ does not have this property the same way we did it for $\mathbb Q,$ since $\mathbb F_p[x]$ is a unique factorization domain with infinitely many primes.

So, any such $F$ must be a subfield of the algebraic closure of some $\mathbb F_p.$

But any such infinite field has finite subfields $$F_1\subsetneq F_2\subsetneq \cdots\subsetneq F_n\subsetneq \cdots$$

and $F_{\infty}=\bigcup F_n$ is a subfield of $F.$

But any finite set of $F_{\infty}$ can only in originate in finitely many $F_i,$ and in particular, they must all be in one $F_N$ for some $N,$ so they can’t generate all of $F_{\infty}.$

Therefore, the only such $F$ are the finite fields.


As $F^\times$ is abelian, finitely generated implies countable.

Following Thomas Andrews's comment, we cannot have characteistic $0$ as subgroups of finitely generated groups are finitely generated, whereas $\Bbb Q^\times$ is not.

Considering transcendental extensions, note that If $F=\Bbb F_q(X)$, then $F^\times$ is not finitely generated. Indeed, just pick an irreducible polynomial of degree larger than all numerator and denominator polynomials of a finite generating set and note that it cannot be written as product of the generators. Hence by the same argumen as in the previous paragraph we conclude that $F$ is an algebraic extension of some $\Bbb F_p$, i.e., $F\subseteq \overline{\Bbb F_p}$. Now every generator $g_i$ lives in some $\Bbb F_{p^{m_i}}$, hence they all live in $\Bbb F_{p^{m_1\cdots m_n}}$.

We conclude that $F$ is a finite field.

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    $\begingroup$ I'm a bit surprised by your first paragraph: aren't all finitely generated groups countable, whether or not they are abelian? $\endgroup$
    – Rob Arthan
    May 8, 2021 at 22:58

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