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

Question

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?

Answer 1

$\mathbb Q$ 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.

Answer 2

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.