ac.commutative algebra - If L is a field extension of K, how big is L*/K*?

To add to Franz's nice answer:

Let $mathcal{C}$ be the collection of groups isomorphic to the direct sum of a free abelian group of countable rank with a finite abelian group.
In the case where $L/K$ is a nontrivial finite separable extension of global fields, each of the following groups is in $mathcal{C}$ (and the first two are free):

1) The group of fractional ideals of $K$ (or divisors in the function field case)

2) The group of principal fractional ideals of $K$

3) $K^times$

4) $L^times/K^times$

Proof: The first three can be proved in succession by using infinitude of primes, finiteness of class groups, and the Dirichlet unit theorem.

4) As suggested by t3suji and Franz, Chebotarev shows that the rank is infinite. On the other hand, the following trick shows that $L^times/K^times$ is a subgroup of a group in $mathcal{C}$ (and hence in $mathcal{C}$ itself): Replace $L$ by its Galois closure. Let $sigma_1,ldots,sigma_d$ be the elements of $operatorname{Gal}(L/K)$. Then
$$x mapsto (sigma_1(x)/x,ldots,sigma_d(x)/x)$$
injects $L^times/K^times$ into $L^times times cdots times L^times$,
which is in $mathcal{C}$.