Are units of rings of functions on algebraic varieties finitely generated (mod. constants)? Hello,
Consider the following question. Let $A$ be a finitely generated reduced algebra over an algebraically closed field $k$. Consider the group of units of $A$, modulo the group $k^*$. Is this group always finitely generated?
Admittedly, I did not think of that question seriously, but I will be glad to hear the answer.
Thank you,
Sasha
 A: Yes. See the beginning of section 3 of "Compactifications of subvarieties of tori" by Jenia Tevelev. He has a finitely generated integral domain $A$ (he calls it $\mathcal O(X)$) over an algebraically closed field $k$, and states that it is "well-known" that $A^\ast/k^\ast$ is a finitely generated free group. I Googled a little and I found other statements of this result, but I couldn't find a better reference.
Note that if you accept the statement for integral domains, it's not hard to show it for reduced algebras more generally.
A: I also think the answer is "yes" and I also haven't yet nailed down a precise reference.  I found though a nice analysis of the relative unit group $R^{\times}/k^{\times}$ in the case $R = k[X]$ is the coordinate ring of a regular, integral affine curve over an arbitrary field $k$ in the following paper:

Rosen, Michael
$S$-units and $S$-class group in algebraic function fields.
J. Algebra 26 (1973), 98-–108. 

It seems vaguely plausible that you could use this to prove the general (integral) case by some fibering argument, but I haven't really thought this through.  
A: As always, the algebraic geometry version is easier than the number theoretic one. For convenience (which can be tweaked) let me assume that $A$ is the coordinate ring of an affine open set of a smooth projective variety $X$ and let $K$ be the rational functions on $X$. Then we have the natural map $K^*\to \mathrm{Div} X$ given by divisor of a function. Let $D_1,\ldots, D_n$ be the divisors at infinity. The kernel of the above map is just $k^*$ and an element of $A$ is a unit if and only if its divisor is supported wholly on the $D_i$'s. Thus, we see that $A^*/k^*$ is a subgroup of the finitely generated free abelian group generated by the $D_i$'s. 
A: I translate into English Lemma 6.5 from Sansuc's paper Groupe de Brauer et arithmétique des groupes
algébriques linéaires sur un corps de nombres,  J. reine angew. Math.
327 (1981), 12-80.
Let $X$ be an algebraic variety over an arbitrary field $k$,
i.e a geometrically integral algebraic $k$-scheme.
We denote $ U(X) := k[X]^* / k^* $,
the group of units of the ring of regular functions $k[X]^*$
modulo nonzero constants.

Rosenlicht's theorem: Let $X$ and $Y$ be two algebraic $k$-varieties
and $G$ be a connected, smooth, linear algebraic $k$-group, .


(i) $U(X)$ is a finitely generated free abelian group;


(ii) $U(X\times_k Y)=U(X)\oplus U(Y)$;


(iii) $U(G)=\hat{G}(k)$ (the group of $k$-characters of $G$).

Reference: M. Rosenlicht, Toroidal algebraic groups,
Proc. AMS 12 (1961), 984–988,
scan here.
For other references see Sansuc's paper.
(i) answers the question.
Rosenlicht's proof of (i) is close to Mohan's answer.
