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.

`$K^{\times}/N_{L/K}L^{\times}$`

and its relation to $G(L\cap K^{ab}/K)$, where $L\cap K^{ab}$ is the maximal abelian extension of $K$ contained in $L$. (Probably you already know this, but I thought I'd mention it in case you didn't.) – Joe Silverman Mar 15 '12 at 20:21