Let $K$ be a field and $L$ an extension of $K$. I wonder how much larger the multiplicative group $L^\times$ of $L$ is than the multiplicative group $K^\times$ of $K$.


I know that if $L=K(t)$ and $t$ is transcendental over $K$, then $L^\times/K^\times$ is isomorphic to the direct product of an infinite number of copies of the integers: Indeed, any (monic) rational function can be uniquely written as the product of a finite number of powers of irreducible polynomials. 
In particular, $L^\times/K^\times$ is not finitely generated.


What happens when $L=K(t)$ and $t$ is algebraic over $K$? I can handle the case when $K$ is finite, but what happens in the infinite case? 
Even for $L=Q(\sqrt{2})$ it's not immediate to me if $L^\times/K^\times$ is finitely generated or not.



If L is a field extension of K, how big is L*/K*?
You're looking for 


* A. Brandis,
 *Über die multiplikative Struktur von Körpererweiterungen*,
Math. Z. 87 (1965), 71-73


Brandis proved that $L^\times/K^\times$ is not finitely generated whenever $K$ is infinite and $L \ne K$ (thanks Pete).
The claim is reduced to finite algebraic extensions of global fields, for which there are infinitely many prime ideals in $K$ that do not remain inert in $L$, and this does it.


