I think I have a simple proof that the only fields with finitely generated multiplicative groups are finite. What about if we take $K$ a number field and mod out $\mathbb{Q}^\times$ from its multiplicative group? Can $K^\times / \mathbb{Q}^\times$ be finitely generated?
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$\begingroup$ You can assume the extension is Galois. Assuming that the degree of the extension is $n>1$. You still need to use the norm map. You also need to show that infinite number of primes appear as a factor in the image of the norm map that its exponent is not divisible by $n$. For this you just need to show infinite number of primes do not split in $\mathcal{O}_K$. $\endgroup$– user127776Commented Oct 2, 2022 at 6:05
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$\begingroup$ As follows from this answer, for a field extension $K\subset L$, the group $L^*/K^*$ is finitely generated iff $K=L$ or $L$ is finite (i.e., never except in trivial cases). $\endgroup$– YCorCommented Oct 2, 2022 at 8:28
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$\begingroup$ Maybe there are refinements of this result. E.g., one could ask whether this group always has infinite $\mathbf{Q}$-rank (this time, when $K\neq L$ and $L$ is not algebraic over a finite field). This might follow from the given argument but I'm not competent enough to see this clearly. $\endgroup$– YCorCommented Oct 2, 2022 at 8:31
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