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
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.
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.
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.
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.
