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If we let $K$ be a number field, thank to the fact that we can extend it integer ring to an UFD where the group of units is finitely generated, we can show that

$K^\ast\cong K^\ast_{tor}\times \mathbb{Z}^{(B)}$

where $K^\ast_{tor}$ is finite and $B$ a numerable subset. However, this possibility of constructing an unique factorization domain is not possible in every field.

Now, suppose I have a field $K$ with the above property, i.e.

$K^\ast\cong G\times \mathbb{Z}^{(B)}$

for some finite group $G$ and denumerable set $B$, then what I can say about the multiplicative structure of $L$ where $L$ is a finite extension of $K$?

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Clearly the torsion subgroup remains finite, because adding infinitely many new roots of unity requires an infinite degree field extension. So the main question is whether the torsion-free quotient is free. –  Will Sawin May 20 '13 at 2:11

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