2
$\begingroup$

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 $K$?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    May 20, 2013 at 2:11

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.